research article a semilinear wave equation with a boundary...
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Research ArticleA Semilinear Wave Equation with a BoundaryCondition of Many-Point Type Global Existence andStability of Weak Solutions
Giai Giang Vo
Department of Mathematics Faculty of Information Technology Huflit University 155 Su Van Hanh Street District 10Ho Chi Minh City 700000 Vietnam
Correspondence should be addressed to Giai Giang Vo ggiangvogmailcom
Received 19 July 2015 Accepted 28 September 2015
Academic Editor Khalil Ezzinbi
Copyright copy 2015 Giai Giang VoThis is an open access article distributed under the Creative CommonsAttribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper is devoted to the study of a wave equationwith a boundary condition ofmany-point typeThe existence of weak solutionsis proved by using the Galerkin method Also the uniqueness and the stability of solutions are established
1 Introduction
Recently initial-boundary value problems of wave equationshave appeared more and more in mechanics they have beendeeply studied bymany authors andwe can refer to theworks[1ndash13] In this paper we consider the following nonlinearwave equation with a boundary condition of many-pointtype
119906119905119905minus 120588119906
119909119909+ 119891 (119909 119906) = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905)
= int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
1 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(1)
where 119891 119892 ℎ 119896 120590 1199060 1199061are given real functions and 120588 120585
1
1205852 120585
119873are positive constants such that 0 = 120585
0lt 120585
1lt sdot sdot sdot lt
120585119873
lt 1
Dang and Alain [4] studied the global existence of thefollowing problem
119906119905119905minus 119906
119909119909+1003816100381610038161003816119906119905
1003816100381610038161003816
120572 sign (119906119905) = 0
119906 (1 119905) = 0
119906119909(0 119905) = 119892 (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(2)
where 0 lt 120572 lt 1 is a constant and 119892 1199060 1199061are given
functionsIn [11] Santos considered the following problem
119906119905119905minus 120588 (119905) 119906
119909119909= 0
119906 (0 119905) = 0
119906 (1 119905) + int
119905
0
119892 (119905 minus 119904) 120588 (119904) 119906119909(1 119904) 119889119904 = 0
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(3)
in which 119892 120588 1199060 1199061are given functions He studied the
asymptotic behavior of the solutions of problem (3) with
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2015 Article ID 531872 16 pageshttpdxdoiorg1011552015531872
2 Abstract and Applied Analysis
respect to the time variable In this case problem (3) is amathematical model for a linear one-dimensional motion ofan elastic bar connected with a viscoelastic element at an endof the bar
Applying the Mikusinski operational calculus D Takaciand A Takaci [12] gave the formula for finding the exactsolutions of the linear wave equation given by
119906119905119905minus 119906
119909119909+ 119870119906 + 120582119906
119905= 119891 (119909 119905)
119906119909(0 119905) = V (119905)
minus119906119909(1 119905) = 120575119906 (1 119905) + 120576119906
119905(1 119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(4)
where 119870 120582 120575 120576 ℎ 120596 are given nonnegative constants and119891 119892 119906
0 1199061are given functions Also the unknown function
119906(119909 119905) and the unknown boundary value V(119905) satisfy thefollowing integral equation
V (119905) = 119892 (119905) + ℎ119906 (0 119905)
minus 120596ℎint
119905
0
sin (120596 (119905 minus 119904)) 119906 (0 119904) 119889119904
(5)
It is worth noting that the function V(119905) is deduced froma Cauchy problem for an ordinary differential equation atthe boundary condition 119905 = 0 Indeed if V(119905) satisfies thefollowing Cauchy problem
V10158401015840 (119905) + 2119901V1015840 (119905) + 1199010V (119905)
= 119902119906119905119905(0 119905) + 119902
0119906119905(0 119905) + 119902
1119906 (0 119905)
V (0) = V0
V1015840 (0) = V1
(6)
where 119901 1199010 119902 119902
0 1199021 V0 V1are constants such that 119901
0gt 119901
2we can easily show that
V (119905) = 119892 (119905) + ℎ119906 (0 119905) + int
119905
0
119896 (119905 minus 119904) 119906 (0 119904) 119889119904 (7)
where
119892 (119905) = exp (minus119901119905) [V1minus 119901V
0+ 119901119902119906
0(0) minus 119902119906
1(0)]
sin120596119905
120596
+ exp (minus119901119905) [V0minus 119902119906
0(0)] cos120596119905 minus
1199020
120596
1199060(0)
119896 (119905) = exp (minus119901119905) [119902 (1199012minus 120596
2) minus 119901119902
0+ 119902
1]
sin120596119905
120596
+ (1199020minus 2119901119902) cos120596119905
(8)
with 120596 = radic1199010minus 119901
2 ℎ = 119902 + 119902
0120596 In the special case of 119901 =
1199020= 119902
1= 0 we obtain (5)
Besides Nguyen and Giang Vo [8] obtained the asymp-totic behavior of the weak solution of the following initial-boundary value problem as 120576 rarr 0
+
119906119905119905minus 119906
119909119909+ 119870119906 + 120582119906
119905= 119891 (119909 119905)
119906 (1 119905) = 0
119906119909(0 119905) = 119892 (119905) + ℎ119906 (0 119905) + 120576119906
119905(0 119905)
minus int
119905
0
119896 (119905 minus 119904) 119906 (0 119904) 119889119904
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(9)
in which 119870 120582 ℎ isin R 120576 gt 0 are given constants and119891 119892 119896 119906
0 1199061are given functions Problem (9) is said to be a
mathematical model describing a shock problem involving alinear viscoelastic bar
The organization of this paper is as follows First of allwe establish the global existence and uniqueness of weaksolutions of problem (1) The proof is based on the Galerkinmethod associated with a priori estimates and the weakcompact method Finally here we prove that this solution isstable in the sense of continuous dependence on the givendata (119891 119892 ℎ 119896 120590)This paper is a relative generalization of theworks [4 6 11 12]
2 Preliminaries
For convenience we denote by ⟨sdot sdot⟩ and sdot respectivelythe scalar product and the norm in 119871
2(0 1) Also we define a
closed subspace of the Sobolev space 1198671(0 1) as follows
119867 = 119906 isin 1198671(0 1) 119906 (1) = 0 (10)
with the following scalar product and norm
⟨119906 V⟩119867
= ⟨119906119909 V119909⟩
119906119867
=1003817100381710038171003817119906119909
1003817100381710038171003817
(11)
Then it is easy to show the following
Lemma 1 The embedding 119867 997893rarr 1198620([0 1]) is compact and
1199061198620([01])
le 119906119867
le 1199061198671(01)
le radic2 119906119867
forall119906 isin 119867
(12)
On the other hand we also have the following result
Lemma 2 Let 120576 gt 0 Then
V21198620([01])
le 1205761003817100381710038171003817V119909
1003817100381710038171003817
2
+ (1 +
1
120576
) V2
forall119906 isin 1198671(0 1)
(13)
The proof of the lemma is straightforward we omit thedetails
Abstract and Applied Analysis 3
Remark 3 Let 120576 = 1 we get
V1198620([01])
le (1003817100381710038171003817V119909
1003817100381710038171003817
2
+ 2 V2)12
le radic2 V1198671(01)
forall119906 isin 1198671(0 1)
(14)
Next if 119882 is a real Banach space with norm sdot 119882
119871119901(0 119879119882) consists of equivalence classes of strongly mea-
surable functions 119906 (0 119879) rarr 119882 such that 119906119871119901(0119879119882)
lt infinwith
119906119871119901(0119879119882)
=
(int
119879
0119906 (119905)
119901
119882119889119905)
1119901
if 1 le 119901 lt infin
ess sup0lt119905lt119879
119906 (119905)119882
if 119901 = infin
(15)
It is not difficult to prove that 119871119901(0 119879119882) is a Banach spaceLet 119882
0 119882
1 and 119882
2be Banach spaces satisfying 119882
0sub
1198821
sub 1198822 We further assume that 119882
0and 119882
2are reflexive
and the imbedding 1198820997893rarr 119882
1is compact Set
119882(0 119879)
= 119906 119906 isin 119871119901(0 119879119882
0) 119906
119905isin 119871
119902(0 119879119882
2)
(16)
where 1 le 119901 119902 le infinThen119882(0 119879) is a Banach space with thenorm
119906119882(0119879)
= 119906119871119901(01198791198820)
+1003817100381710038171003817119906119905
1003817100381710038171003817119871119902(01198791198822)
(17)
We also have the following lemma
Lemma 4 (see [14]) Let 1 lt 119901 119902 lt infin The embedding119882(0 119879) 997893rarr 119871
119901(0 119879119882
1) is compact
3 Global Existence and Uniqueness ofWeak Solutions
To investigate the existence of a unique weak solution ofproblem (1) the following assumptions are needed
(1198601) 119906
0isin 119867 and 119906
1isin 119871
2(0 1)
(1198602) 120590 isin 119867
1(0 119879) 119896 isin 119882
11(0 119879)
(1198603) 119891119863
2119891 isin 119862
0([0 1] times R) satisfy the following condi-
tions
(i) There exist positive functions 1205901 1205902
isin 1198711(0 1)
such that
int
119906
0
119891 (119909 119904) 119889119904 ge minus1205901(119909) 119906
2minus 120590
2(119909)
ae 119909 isin [0 1] forall119906 isin R
(18)
(ii) There exist positive functions isin 1198712(0 1) and
119891 isin 119862
0(R) such that
10038161003816100381610038161198632119891 (119909 119906)
1003816100381610038161003816le (119909)
119891 (119906) ae 119909 isin [0 1] forall119906 isin R (19)
(1198604) 119892 isin 119862
2(R) there exist constants 119886 119887 gt 0 such that
int
119906
0
119892 (119904) 119889119904 ge minus1198861199062minus 119887 forall119906 isin R (20)
(1198605) ℎ isin 119862
0(R119873+1) there exist constants 119888 119889 gt 0 such that
1003816100381610038161003816ℎ (119906
0 1199061 119906
119873)1003816100381610038161003816le 119888
119873
sum
119894=0
1003816100381610038161003816119906119894
1003816100381610038161003816+ 119889
forall119906119894isin R 119894 = 0119873
(21)
(1198606) For each 119872 gt 0 there exists a constant 119888
119872gt 0 such
that
1003816100381610038161003816ℎ (119906
0 119906
119873) minus ℎ (V
0 V
119873)1003816100381610038161003816le 119888
119872
119873
sum
119894=0
1003816100381610038161003816119906119894minus V
119894
1003816100381610038161003816
forall1003816100381610038161003816119906119894
10038161003816100381610038161003816100381610038161003816V119894
1003816100381610038161003816le 119872 119894 = 0119873
(22)
With these assumptions we have the following theorem
Theorem 5 Let assumptions (1198601)ndash(119860
6) hold Then problem
(1) has a unique weak solution 119906 such that
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (120585119894 sdot) isin 119867
1(0 119879) 119894 = 0119873
(23)
Remark 6 In the special case of119873 = 0 and 1198921015840(119905) gt minus1 for all
119905 isin R we have obtained the same results in the paper [6]
Proof ofTheorem 5Themain tool of this proof is the Galerkinmethod The procedure includes four steps as follows
(i) Galerkin approximation(ii) A priori estimates(iii) Limiting process(iv) Uniqueness of the weak solutions
Step 1 (Galerkin approximation) We use a special orthonor-mal base of119867
120593119896(119909) = radic
2
(1 + 1205832
119896)
cos (120583119896119909)
120583119896= (2119896 minus 1)
120587
2
119896 = 1 2
(24)
Now we are looking for the approximate solution of problem(1) in the form
119906119898(119909 119905) =
119898
sum
119896=1
120596119898119896
(119905) 120593119896(119909) (25)
4 Abstract and Applied Analysis
where the coefficient functions 120596119898119896
(119905) satisfy the followingsystem of nonlinear differential equations
⟨119906119898
119905119905(sdot 119905) 120593
119896⟩ + 120588 ⟨119906
119898(sdot 119905) 120593
119896⟩119867
+ V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩ = 0 119896 = 1119898
(26)
with
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
(27)
119906119898(sdot 0) equiv 119906
119898
0=
119898
sum
119896=1
119886119898119896
120593119896997888rarr 119906
0
strongly in 1198671(0 1)
(28)
119906119898
119905(sdot 0) equiv 119906
119898
1=
119898
sum
119896=1
119887119898119896
120593119896997888rarr 119906
1
strongly in 1198712(0 1)
(29)
By substituting 120588119896= radic120588120583
119896 we can rewrite the system of (26)ndash
(29) as follows
12059610158401015840
119898119896(119905) + 120588
2
119896120596119898119896
(119905) = minus
1
1003817100381710038171003817120593119896
1003817100381710038171003817
2[V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩]
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
120596119898119896
(0) = 119886119898119896
1205961015840
119898119896(0) = 119887
119898119896
119896 = 1119898
(30)
Therefore we obtain
120596119898119896
(119905) = 119886119898119896
cos (120588119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
minus
2
120593119896(0)
int
119905
0
119889119904
sdot int
119904
0
sin [120588119896(119905 minus 119904)]
120588119896
119896 (119904 minus 120591)
sdot ℎ (119906119898(1205850 120591) 119906
119898(120585119873 120591)) 119889120591 minus
2
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
[120590 (119904) + 119892 (119906119898(1205850 119904))] 119889119904
minus
2
1205932
119896(0)
int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(119904)) 120593
119896⟩ 119889119904
119896 = 1119898
(31)
Applying the Schauder fixed-point theorem it is not difficultthat system (31) has a solution (120596
1198981 1205961198982
120596119898119898
) on aninterval [0 119879
119898] This implies that in system (26)ndash(29) there
exists a solution to 119906119898(119905) on [0 119879
119898] Moreover we can extend
the approximate solution 119906119898 to the whole interval [0 119879] (see
[15])
Step 2 (a priori estimates) In (26) we replace 120593119896(119909) by
119906119898
119905(119909 119905) Then integrating from 0 to 119905 we have after some
calculations
119864119898(119905) = 119864
119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 minus 2120590 (119905) 119906119898(0 119905)
+ 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904 minus 2int
119906119898(0119905)
0
119892 (119904) 119889119904
minus 2int
1
0
int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904 119889119909 minus 2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 +
5
sum
119896=1
119868119896(119905)
(32)
where
119864119898(119905) =
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867 (33)
We will estimate respectively the following terms on theright-hand side of (32)
Estimating 1198681(119905) Using (33) and Lemma 1 we infer that
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816le
1003817100381710038171003817119906119898(sdot 119905)
10038171003817100381710038171198620([01])
le1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817119867
le
1
radic120588
radic119864119898(119905)
(34)
Hence
1198681(119905) = minus2120590 (119905) 119906
119898(0 119905) le 120576
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+
1
120576
1205902(119905)
le 120576119864119898(119905) +
1
120576
1205902
119871infin(0119879)
forall120576 gt 0
(35)
Abstract and Applied Analysis 5
Estimating 1198682(119905) Using Lemma 1 and inequality (34) we
arrive at
1198682(119905) = 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904
le
1
120588
int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119889119904 + 120588int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816
1003816100381610038161003816119906119898(0 119904)
1003816100381610038161003816
2
119889119904
le int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119864119898(119904) 119889119904 +
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
(36)
Estimating 1198683(119905) From assumption (119860
4) we have
1198683(119905) = minus2int
119906119898(0119905)
0
119892 (119904) 119889119904 le 21198861003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+ 2119887 (37)
On the other hand we see that
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
=
10038171003817100381710038171003817100381710038171003817
119906119898
0+ int
119905
0
119906119898
119905(sdot 119904) 119889119904
10038171003817100381710038171003817100381710038171003817
2
le 21003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119905 int
119905
0
119864119898(119904) 119889119904
(38)
It follows from (37) (38) and Lemma 2 that
1198683(119905) le 2120576
119886
120588
119864119898(119905) + 4119886119905 (1 +
1
120576
)int
119905
0
119864119898(119904) 119889119904
+ 4119886 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119887
(39)
Estimating 1198684(119905) Owing to assumption (119860
3)-(i) (38) and
Lemma 2 it is not difficult to show that
1198684(119905) = minus2int
1
0
119889119909int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904
le 2int
1
0
1205901(119909)
1003816100381610038161003816119906119898(119909 119905)
1003816100381610038161003816
2
119889119909 + 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
le 120576
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
119864119898(119905)
+ 4119905 (1 +
1
120576
)10038171003817100381710038171205901
10038171003817100381710038171198711(01)
int
119905
0
119864119898(119904) 119889119904
+ 4 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2 10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
(40)
Estimating 1198685(119905)Applying integration by parts it follows that
1198685(119905) = minus2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
= minus2119906119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 1198691(119905) + 119869
2(119905) + 119869
3(119905)
(41)
We can estimate the integrals in the right-hand side of (41) asfollows
1198691(119905) = minus2119906
119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 21198881003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
119873
sum
119894=0
int
119905
0
|119896 (119905 minus 119904)|1003816100381610038161003816119906119898(120585119894 119904)
1003816100381610038161003816119889119904
+ 21198891003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816int
119905
0
|119896 (119905 minus 119904)| 119889119904 le 2 (119873 + 1)
119888
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119905 minus 119904)| radic119864119898(119904)119889119904 + 2
119889
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119904)| 119889119904 le 2120576119864119898(119905) + (119873 + 1)
2
sdot
1198882
120576120588
1198962
1198712(0119879)
int
119905
0
119864119898(119904) 119889119904 +
1198892
120576120588
1198962
1198711(0119879)
(42)
1198692(119905) = 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
|119896 (0)| int
119905
0
119864119898(119904) 119889119904 + 2
119889
radic120588
|119896 (0)|
sdot int
119905
0
radic119864119898(119904)119889119904 le [1 + 2 (119873 + 1)
119888
120588
|119896 (0)|]
sdot int
119905
0
119864119898(119904) 119889119904 +
1198892
120588
1198791198962(0)
(43)
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
respect to the time variable In this case problem (3) is amathematical model for a linear one-dimensional motion ofan elastic bar connected with a viscoelastic element at an endof the bar
Applying the Mikusinski operational calculus D Takaciand A Takaci [12] gave the formula for finding the exactsolutions of the linear wave equation given by
119906119905119905minus 119906
119909119909+ 119870119906 + 120582119906
119905= 119891 (119909 119905)
119906119909(0 119905) = V (119905)
minus119906119909(1 119905) = 120575119906 (1 119905) + 120576119906
119905(1 119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(4)
where 119870 120582 120575 120576 ℎ 120596 are given nonnegative constants and119891 119892 119906
0 1199061are given functions Also the unknown function
119906(119909 119905) and the unknown boundary value V(119905) satisfy thefollowing integral equation
V (119905) = 119892 (119905) + ℎ119906 (0 119905)
minus 120596ℎint
119905
0
sin (120596 (119905 minus 119904)) 119906 (0 119904) 119889119904
(5)
It is worth noting that the function V(119905) is deduced froma Cauchy problem for an ordinary differential equation atthe boundary condition 119905 = 0 Indeed if V(119905) satisfies thefollowing Cauchy problem
V10158401015840 (119905) + 2119901V1015840 (119905) + 1199010V (119905)
= 119902119906119905119905(0 119905) + 119902
0119906119905(0 119905) + 119902
1119906 (0 119905)
V (0) = V0
V1015840 (0) = V1
(6)
where 119901 1199010 119902 119902
0 1199021 V0 V1are constants such that 119901
0gt 119901
2we can easily show that
V (119905) = 119892 (119905) + ℎ119906 (0 119905) + int
119905
0
119896 (119905 minus 119904) 119906 (0 119904) 119889119904 (7)
where
119892 (119905) = exp (minus119901119905) [V1minus 119901V
0+ 119901119902119906
0(0) minus 119902119906
1(0)]
sin120596119905
120596
+ exp (minus119901119905) [V0minus 119902119906
0(0)] cos120596119905 minus
1199020
120596
1199060(0)
119896 (119905) = exp (minus119901119905) [119902 (1199012minus 120596
2) minus 119901119902
0+ 119902
1]
sin120596119905
120596
+ (1199020minus 2119901119902) cos120596119905
(8)
with 120596 = radic1199010minus 119901
2 ℎ = 119902 + 119902
0120596 In the special case of 119901 =
1199020= 119902
1= 0 we obtain (5)
Besides Nguyen and Giang Vo [8] obtained the asymp-totic behavior of the weak solution of the following initial-boundary value problem as 120576 rarr 0
+
119906119905119905minus 119906
119909119909+ 119870119906 + 120582119906
119905= 119891 (119909 119905)
119906 (1 119905) = 0
119906119909(0 119905) = 119892 (119905) + ℎ119906 (0 119905) + 120576119906
119905(0 119905)
minus int
119905
0
119896 (119905 minus 119904) 119906 (0 119904) 119889119904
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
(9)
in which 119870 120582 ℎ isin R 120576 gt 0 are given constants and119891 119892 119896 119906
0 1199061are given functions Problem (9) is said to be a
mathematical model describing a shock problem involving alinear viscoelastic bar
The organization of this paper is as follows First of allwe establish the global existence and uniqueness of weaksolutions of problem (1) The proof is based on the Galerkinmethod associated with a priori estimates and the weakcompact method Finally here we prove that this solution isstable in the sense of continuous dependence on the givendata (119891 119892 ℎ 119896 120590)This paper is a relative generalization of theworks [4 6 11 12]
2 Preliminaries
For convenience we denote by ⟨sdot sdot⟩ and sdot respectivelythe scalar product and the norm in 119871
2(0 1) Also we define a
closed subspace of the Sobolev space 1198671(0 1) as follows
119867 = 119906 isin 1198671(0 1) 119906 (1) = 0 (10)
with the following scalar product and norm
⟨119906 V⟩119867
= ⟨119906119909 V119909⟩
119906119867
=1003817100381710038171003817119906119909
1003817100381710038171003817
(11)
Then it is easy to show the following
Lemma 1 The embedding 119867 997893rarr 1198620([0 1]) is compact and
1199061198620([01])
le 119906119867
le 1199061198671(01)
le radic2 119906119867
forall119906 isin 119867
(12)
On the other hand we also have the following result
Lemma 2 Let 120576 gt 0 Then
V21198620([01])
le 1205761003817100381710038171003817V119909
1003817100381710038171003817
2
+ (1 +
1
120576
) V2
forall119906 isin 1198671(0 1)
(13)
The proof of the lemma is straightforward we omit thedetails
Abstract and Applied Analysis 3
Remark 3 Let 120576 = 1 we get
V1198620([01])
le (1003817100381710038171003817V119909
1003817100381710038171003817
2
+ 2 V2)12
le radic2 V1198671(01)
forall119906 isin 1198671(0 1)
(14)
Next if 119882 is a real Banach space with norm sdot 119882
119871119901(0 119879119882) consists of equivalence classes of strongly mea-
surable functions 119906 (0 119879) rarr 119882 such that 119906119871119901(0119879119882)
lt infinwith
119906119871119901(0119879119882)
=
(int
119879
0119906 (119905)
119901
119882119889119905)
1119901
if 1 le 119901 lt infin
ess sup0lt119905lt119879
119906 (119905)119882
if 119901 = infin
(15)
It is not difficult to prove that 119871119901(0 119879119882) is a Banach spaceLet 119882
0 119882
1 and 119882
2be Banach spaces satisfying 119882
0sub
1198821
sub 1198822 We further assume that 119882
0and 119882
2are reflexive
and the imbedding 1198820997893rarr 119882
1is compact Set
119882(0 119879)
= 119906 119906 isin 119871119901(0 119879119882
0) 119906
119905isin 119871
119902(0 119879119882
2)
(16)
where 1 le 119901 119902 le infinThen119882(0 119879) is a Banach space with thenorm
119906119882(0119879)
= 119906119871119901(01198791198820)
+1003817100381710038171003817119906119905
1003817100381710038171003817119871119902(01198791198822)
(17)
We also have the following lemma
Lemma 4 (see [14]) Let 1 lt 119901 119902 lt infin The embedding119882(0 119879) 997893rarr 119871
119901(0 119879119882
1) is compact
3 Global Existence and Uniqueness ofWeak Solutions
To investigate the existence of a unique weak solution ofproblem (1) the following assumptions are needed
(1198601) 119906
0isin 119867 and 119906
1isin 119871
2(0 1)
(1198602) 120590 isin 119867
1(0 119879) 119896 isin 119882
11(0 119879)
(1198603) 119891119863
2119891 isin 119862
0([0 1] times R) satisfy the following condi-
tions
(i) There exist positive functions 1205901 1205902
isin 1198711(0 1)
such that
int
119906
0
119891 (119909 119904) 119889119904 ge minus1205901(119909) 119906
2minus 120590
2(119909)
ae 119909 isin [0 1] forall119906 isin R
(18)
(ii) There exist positive functions isin 1198712(0 1) and
119891 isin 119862
0(R) such that
10038161003816100381610038161198632119891 (119909 119906)
1003816100381610038161003816le (119909)
119891 (119906) ae 119909 isin [0 1] forall119906 isin R (19)
(1198604) 119892 isin 119862
2(R) there exist constants 119886 119887 gt 0 such that
int
119906
0
119892 (119904) 119889119904 ge minus1198861199062minus 119887 forall119906 isin R (20)
(1198605) ℎ isin 119862
0(R119873+1) there exist constants 119888 119889 gt 0 such that
1003816100381610038161003816ℎ (119906
0 1199061 119906
119873)1003816100381610038161003816le 119888
119873
sum
119894=0
1003816100381610038161003816119906119894
1003816100381610038161003816+ 119889
forall119906119894isin R 119894 = 0119873
(21)
(1198606) For each 119872 gt 0 there exists a constant 119888
119872gt 0 such
that
1003816100381610038161003816ℎ (119906
0 119906
119873) minus ℎ (V
0 V
119873)1003816100381610038161003816le 119888
119872
119873
sum
119894=0
1003816100381610038161003816119906119894minus V
119894
1003816100381610038161003816
forall1003816100381610038161003816119906119894
10038161003816100381610038161003816100381610038161003816V119894
1003816100381610038161003816le 119872 119894 = 0119873
(22)
With these assumptions we have the following theorem
Theorem 5 Let assumptions (1198601)ndash(119860
6) hold Then problem
(1) has a unique weak solution 119906 such that
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (120585119894 sdot) isin 119867
1(0 119879) 119894 = 0119873
(23)
Remark 6 In the special case of119873 = 0 and 1198921015840(119905) gt minus1 for all
119905 isin R we have obtained the same results in the paper [6]
Proof ofTheorem 5Themain tool of this proof is the Galerkinmethod The procedure includes four steps as follows
(i) Galerkin approximation(ii) A priori estimates(iii) Limiting process(iv) Uniqueness of the weak solutions
Step 1 (Galerkin approximation) We use a special orthonor-mal base of119867
120593119896(119909) = radic
2
(1 + 1205832
119896)
cos (120583119896119909)
120583119896= (2119896 minus 1)
120587
2
119896 = 1 2
(24)
Now we are looking for the approximate solution of problem(1) in the form
119906119898(119909 119905) =
119898
sum
119896=1
120596119898119896
(119905) 120593119896(119909) (25)
4 Abstract and Applied Analysis
where the coefficient functions 120596119898119896
(119905) satisfy the followingsystem of nonlinear differential equations
⟨119906119898
119905119905(sdot 119905) 120593
119896⟩ + 120588 ⟨119906
119898(sdot 119905) 120593
119896⟩119867
+ V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩ = 0 119896 = 1119898
(26)
with
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
(27)
119906119898(sdot 0) equiv 119906
119898
0=
119898
sum
119896=1
119886119898119896
120593119896997888rarr 119906
0
strongly in 1198671(0 1)
(28)
119906119898
119905(sdot 0) equiv 119906
119898
1=
119898
sum
119896=1
119887119898119896
120593119896997888rarr 119906
1
strongly in 1198712(0 1)
(29)
By substituting 120588119896= radic120588120583
119896 we can rewrite the system of (26)ndash
(29) as follows
12059610158401015840
119898119896(119905) + 120588
2
119896120596119898119896
(119905) = minus
1
1003817100381710038171003817120593119896
1003817100381710038171003817
2[V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩]
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
120596119898119896
(0) = 119886119898119896
1205961015840
119898119896(0) = 119887
119898119896
119896 = 1119898
(30)
Therefore we obtain
120596119898119896
(119905) = 119886119898119896
cos (120588119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
minus
2
120593119896(0)
int
119905
0
119889119904
sdot int
119904
0
sin [120588119896(119905 minus 119904)]
120588119896
119896 (119904 minus 120591)
sdot ℎ (119906119898(1205850 120591) 119906
119898(120585119873 120591)) 119889120591 minus
2
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
[120590 (119904) + 119892 (119906119898(1205850 119904))] 119889119904
minus
2
1205932
119896(0)
int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(119904)) 120593
119896⟩ 119889119904
119896 = 1119898
(31)
Applying the Schauder fixed-point theorem it is not difficultthat system (31) has a solution (120596
1198981 1205961198982
120596119898119898
) on aninterval [0 119879
119898] This implies that in system (26)ndash(29) there
exists a solution to 119906119898(119905) on [0 119879
119898] Moreover we can extend
the approximate solution 119906119898 to the whole interval [0 119879] (see
[15])
Step 2 (a priori estimates) In (26) we replace 120593119896(119909) by
119906119898
119905(119909 119905) Then integrating from 0 to 119905 we have after some
calculations
119864119898(119905) = 119864
119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 minus 2120590 (119905) 119906119898(0 119905)
+ 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904 minus 2int
119906119898(0119905)
0
119892 (119904) 119889119904
minus 2int
1
0
int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904 119889119909 minus 2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 +
5
sum
119896=1
119868119896(119905)
(32)
where
119864119898(119905) =
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867 (33)
We will estimate respectively the following terms on theright-hand side of (32)
Estimating 1198681(119905) Using (33) and Lemma 1 we infer that
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816le
1003817100381710038171003817119906119898(sdot 119905)
10038171003817100381710038171198620([01])
le1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817119867
le
1
radic120588
radic119864119898(119905)
(34)
Hence
1198681(119905) = minus2120590 (119905) 119906
119898(0 119905) le 120576
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+
1
120576
1205902(119905)
le 120576119864119898(119905) +
1
120576
1205902
119871infin(0119879)
forall120576 gt 0
(35)
Abstract and Applied Analysis 5
Estimating 1198682(119905) Using Lemma 1 and inequality (34) we
arrive at
1198682(119905) = 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904
le
1
120588
int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119889119904 + 120588int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816
1003816100381610038161003816119906119898(0 119904)
1003816100381610038161003816
2
119889119904
le int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119864119898(119904) 119889119904 +
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
(36)
Estimating 1198683(119905) From assumption (119860
4) we have
1198683(119905) = minus2int
119906119898(0119905)
0
119892 (119904) 119889119904 le 21198861003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+ 2119887 (37)
On the other hand we see that
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
=
10038171003817100381710038171003817100381710038171003817
119906119898
0+ int
119905
0
119906119898
119905(sdot 119904) 119889119904
10038171003817100381710038171003817100381710038171003817
2
le 21003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119905 int
119905
0
119864119898(119904) 119889119904
(38)
It follows from (37) (38) and Lemma 2 that
1198683(119905) le 2120576
119886
120588
119864119898(119905) + 4119886119905 (1 +
1
120576
)int
119905
0
119864119898(119904) 119889119904
+ 4119886 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119887
(39)
Estimating 1198684(119905) Owing to assumption (119860
3)-(i) (38) and
Lemma 2 it is not difficult to show that
1198684(119905) = minus2int
1
0
119889119909int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904
le 2int
1
0
1205901(119909)
1003816100381610038161003816119906119898(119909 119905)
1003816100381610038161003816
2
119889119909 + 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
le 120576
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
119864119898(119905)
+ 4119905 (1 +
1
120576
)10038171003817100381710038171205901
10038171003817100381710038171198711(01)
int
119905
0
119864119898(119904) 119889119904
+ 4 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2 10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
(40)
Estimating 1198685(119905)Applying integration by parts it follows that
1198685(119905) = minus2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
= minus2119906119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 1198691(119905) + 119869
2(119905) + 119869
3(119905)
(41)
We can estimate the integrals in the right-hand side of (41) asfollows
1198691(119905) = minus2119906
119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 21198881003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
119873
sum
119894=0
int
119905
0
|119896 (119905 minus 119904)|1003816100381610038161003816119906119898(120585119894 119904)
1003816100381610038161003816119889119904
+ 21198891003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816int
119905
0
|119896 (119905 minus 119904)| 119889119904 le 2 (119873 + 1)
119888
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119905 minus 119904)| radic119864119898(119904)119889119904 + 2
119889
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119904)| 119889119904 le 2120576119864119898(119905) + (119873 + 1)
2
sdot
1198882
120576120588
1198962
1198712(0119879)
int
119905
0
119864119898(119904) 119889119904 +
1198892
120576120588
1198962
1198711(0119879)
(42)
1198692(119905) = 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
|119896 (0)| int
119905
0
119864119898(119904) 119889119904 + 2
119889
radic120588
|119896 (0)|
sdot int
119905
0
radic119864119898(119904)119889119904 le [1 + 2 (119873 + 1)
119888
120588
|119896 (0)|]
sdot int
119905
0
119864119898(119904) 119889119904 +
1198892
120588
1198791198962(0)
(43)
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
Remark 3 Let 120576 = 1 we get
V1198620([01])
le (1003817100381710038171003817V119909
1003817100381710038171003817
2
+ 2 V2)12
le radic2 V1198671(01)
forall119906 isin 1198671(0 1)
(14)
Next if 119882 is a real Banach space with norm sdot 119882
119871119901(0 119879119882) consists of equivalence classes of strongly mea-
surable functions 119906 (0 119879) rarr 119882 such that 119906119871119901(0119879119882)
lt infinwith
119906119871119901(0119879119882)
=
(int
119879
0119906 (119905)
119901
119882119889119905)
1119901
if 1 le 119901 lt infin
ess sup0lt119905lt119879
119906 (119905)119882
if 119901 = infin
(15)
It is not difficult to prove that 119871119901(0 119879119882) is a Banach spaceLet 119882
0 119882
1 and 119882
2be Banach spaces satisfying 119882
0sub
1198821
sub 1198822 We further assume that 119882
0and 119882
2are reflexive
and the imbedding 1198820997893rarr 119882
1is compact Set
119882(0 119879)
= 119906 119906 isin 119871119901(0 119879119882
0) 119906
119905isin 119871
119902(0 119879119882
2)
(16)
where 1 le 119901 119902 le infinThen119882(0 119879) is a Banach space with thenorm
119906119882(0119879)
= 119906119871119901(01198791198820)
+1003817100381710038171003817119906119905
1003817100381710038171003817119871119902(01198791198822)
(17)
We also have the following lemma
Lemma 4 (see [14]) Let 1 lt 119901 119902 lt infin The embedding119882(0 119879) 997893rarr 119871
119901(0 119879119882
1) is compact
3 Global Existence and Uniqueness ofWeak Solutions
To investigate the existence of a unique weak solution ofproblem (1) the following assumptions are needed
(1198601) 119906
0isin 119867 and 119906
1isin 119871
2(0 1)
(1198602) 120590 isin 119867
1(0 119879) 119896 isin 119882
11(0 119879)
(1198603) 119891119863
2119891 isin 119862
0([0 1] times R) satisfy the following condi-
tions
(i) There exist positive functions 1205901 1205902
isin 1198711(0 1)
such that
int
119906
0
119891 (119909 119904) 119889119904 ge minus1205901(119909) 119906
2minus 120590
2(119909)
ae 119909 isin [0 1] forall119906 isin R
(18)
(ii) There exist positive functions isin 1198712(0 1) and
119891 isin 119862
0(R) such that
10038161003816100381610038161198632119891 (119909 119906)
1003816100381610038161003816le (119909)
119891 (119906) ae 119909 isin [0 1] forall119906 isin R (19)
(1198604) 119892 isin 119862
2(R) there exist constants 119886 119887 gt 0 such that
int
119906
0
119892 (119904) 119889119904 ge minus1198861199062minus 119887 forall119906 isin R (20)
(1198605) ℎ isin 119862
0(R119873+1) there exist constants 119888 119889 gt 0 such that
1003816100381610038161003816ℎ (119906
0 1199061 119906
119873)1003816100381610038161003816le 119888
119873
sum
119894=0
1003816100381610038161003816119906119894
1003816100381610038161003816+ 119889
forall119906119894isin R 119894 = 0119873
(21)
(1198606) For each 119872 gt 0 there exists a constant 119888
119872gt 0 such
that
1003816100381610038161003816ℎ (119906
0 119906
119873) minus ℎ (V
0 V
119873)1003816100381610038161003816le 119888
119872
119873
sum
119894=0
1003816100381610038161003816119906119894minus V
119894
1003816100381610038161003816
forall1003816100381610038161003816119906119894
10038161003816100381610038161003816100381610038161003816V119894
1003816100381610038161003816le 119872 119894 = 0119873
(22)
With these assumptions we have the following theorem
Theorem 5 Let assumptions (1198601)ndash(119860
6) hold Then problem
(1) has a unique weak solution 119906 such that
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (120585119894 sdot) isin 119867
1(0 119879) 119894 = 0119873
(23)
Remark 6 In the special case of119873 = 0 and 1198921015840(119905) gt minus1 for all
119905 isin R we have obtained the same results in the paper [6]
Proof ofTheorem 5Themain tool of this proof is the Galerkinmethod The procedure includes four steps as follows
(i) Galerkin approximation(ii) A priori estimates(iii) Limiting process(iv) Uniqueness of the weak solutions
Step 1 (Galerkin approximation) We use a special orthonor-mal base of119867
120593119896(119909) = radic
2
(1 + 1205832
119896)
cos (120583119896119909)
120583119896= (2119896 minus 1)
120587
2
119896 = 1 2
(24)
Now we are looking for the approximate solution of problem(1) in the form
119906119898(119909 119905) =
119898
sum
119896=1
120596119898119896
(119905) 120593119896(119909) (25)
4 Abstract and Applied Analysis
where the coefficient functions 120596119898119896
(119905) satisfy the followingsystem of nonlinear differential equations
⟨119906119898
119905119905(sdot 119905) 120593
119896⟩ + 120588 ⟨119906
119898(sdot 119905) 120593
119896⟩119867
+ V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩ = 0 119896 = 1119898
(26)
with
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
(27)
119906119898(sdot 0) equiv 119906
119898
0=
119898
sum
119896=1
119886119898119896
120593119896997888rarr 119906
0
strongly in 1198671(0 1)
(28)
119906119898
119905(sdot 0) equiv 119906
119898
1=
119898
sum
119896=1
119887119898119896
120593119896997888rarr 119906
1
strongly in 1198712(0 1)
(29)
By substituting 120588119896= radic120588120583
119896 we can rewrite the system of (26)ndash
(29) as follows
12059610158401015840
119898119896(119905) + 120588
2
119896120596119898119896
(119905) = minus
1
1003817100381710038171003817120593119896
1003817100381710038171003817
2[V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩]
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
120596119898119896
(0) = 119886119898119896
1205961015840
119898119896(0) = 119887
119898119896
119896 = 1119898
(30)
Therefore we obtain
120596119898119896
(119905) = 119886119898119896
cos (120588119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
minus
2
120593119896(0)
int
119905
0
119889119904
sdot int
119904
0
sin [120588119896(119905 minus 119904)]
120588119896
119896 (119904 minus 120591)
sdot ℎ (119906119898(1205850 120591) 119906
119898(120585119873 120591)) 119889120591 minus
2
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
[120590 (119904) + 119892 (119906119898(1205850 119904))] 119889119904
minus
2
1205932
119896(0)
int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(119904)) 120593
119896⟩ 119889119904
119896 = 1119898
(31)
Applying the Schauder fixed-point theorem it is not difficultthat system (31) has a solution (120596
1198981 1205961198982
120596119898119898
) on aninterval [0 119879
119898] This implies that in system (26)ndash(29) there
exists a solution to 119906119898(119905) on [0 119879
119898] Moreover we can extend
the approximate solution 119906119898 to the whole interval [0 119879] (see
[15])
Step 2 (a priori estimates) In (26) we replace 120593119896(119909) by
119906119898
119905(119909 119905) Then integrating from 0 to 119905 we have after some
calculations
119864119898(119905) = 119864
119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 minus 2120590 (119905) 119906119898(0 119905)
+ 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904 minus 2int
119906119898(0119905)
0
119892 (119904) 119889119904
minus 2int
1
0
int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904 119889119909 minus 2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 +
5
sum
119896=1
119868119896(119905)
(32)
where
119864119898(119905) =
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867 (33)
We will estimate respectively the following terms on theright-hand side of (32)
Estimating 1198681(119905) Using (33) and Lemma 1 we infer that
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816le
1003817100381710038171003817119906119898(sdot 119905)
10038171003817100381710038171198620([01])
le1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817119867
le
1
radic120588
radic119864119898(119905)
(34)
Hence
1198681(119905) = minus2120590 (119905) 119906
119898(0 119905) le 120576
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+
1
120576
1205902(119905)
le 120576119864119898(119905) +
1
120576
1205902
119871infin(0119879)
forall120576 gt 0
(35)
Abstract and Applied Analysis 5
Estimating 1198682(119905) Using Lemma 1 and inequality (34) we
arrive at
1198682(119905) = 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904
le
1
120588
int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119889119904 + 120588int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816
1003816100381610038161003816119906119898(0 119904)
1003816100381610038161003816
2
119889119904
le int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119864119898(119904) 119889119904 +
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
(36)
Estimating 1198683(119905) From assumption (119860
4) we have
1198683(119905) = minus2int
119906119898(0119905)
0
119892 (119904) 119889119904 le 21198861003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+ 2119887 (37)
On the other hand we see that
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
=
10038171003817100381710038171003817100381710038171003817
119906119898
0+ int
119905
0
119906119898
119905(sdot 119904) 119889119904
10038171003817100381710038171003817100381710038171003817
2
le 21003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119905 int
119905
0
119864119898(119904) 119889119904
(38)
It follows from (37) (38) and Lemma 2 that
1198683(119905) le 2120576
119886
120588
119864119898(119905) + 4119886119905 (1 +
1
120576
)int
119905
0
119864119898(119904) 119889119904
+ 4119886 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119887
(39)
Estimating 1198684(119905) Owing to assumption (119860
3)-(i) (38) and
Lemma 2 it is not difficult to show that
1198684(119905) = minus2int
1
0
119889119909int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904
le 2int
1
0
1205901(119909)
1003816100381610038161003816119906119898(119909 119905)
1003816100381610038161003816
2
119889119909 + 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
le 120576
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
119864119898(119905)
+ 4119905 (1 +
1
120576
)10038171003817100381710038171205901
10038171003817100381710038171198711(01)
int
119905
0
119864119898(119904) 119889119904
+ 4 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2 10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
(40)
Estimating 1198685(119905)Applying integration by parts it follows that
1198685(119905) = minus2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
= minus2119906119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 1198691(119905) + 119869
2(119905) + 119869
3(119905)
(41)
We can estimate the integrals in the right-hand side of (41) asfollows
1198691(119905) = minus2119906
119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 21198881003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
119873
sum
119894=0
int
119905
0
|119896 (119905 minus 119904)|1003816100381610038161003816119906119898(120585119894 119904)
1003816100381610038161003816119889119904
+ 21198891003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816int
119905
0
|119896 (119905 minus 119904)| 119889119904 le 2 (119873 + 1)
119888
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119905 minus 119904)| radic119864119898(119904)119889119904 + 2
119889
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119904)| 119889119904 le 2120576119864119898(119905) + (119873 + 1)
2
sdot
1198882
120576120588
1198962
1198712(0119879)
int
119905
0
119864119898(119904) 119889119904 +
1198892
120576120588
1198962
1198711(0119879)
(42)
1198692(119905) = 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
|119896 (0)| int
119905
0
119864119898(119904) 119889119904 + 2
119889
radic120588
|119896 (0)|
sdot int
119905
0
radic119864119898(119904)119889119904 le [1 + 2 (119873 + 1)
119888
120588
|119896 (0)|]
sdot int
119905
0
119864119898(119904) 119889119904 +
1198892
120588
1198791198962(0)
(43)
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
where the coefficient functions 120596119898119896
(119905) satisfy the followingsystem of nonlinear differential equations
⟨119906119898
119905119905(sdot 119905) 120593
119896⟩ + 120588 ⟨119906
119898(sdot 119905) 120593
119896⟩119867
+ V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩ = 0 119896 = 1119898
(26)
with
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
(27)
119906119898(sdot 0) equiv 119906
119898
0=
119898
sum
119896=1
119886119898119896
120593119896997888rarr 119906
0
strongly in 1198671(0 1)
(28)
119906119898
119905(sdot 0) equiv 119906
119898
1=
119898
sum
119896=1
119887119898119896
120593119896997888rarr 119906
1
strongly in 1198712(0 1)
(29)
By substituting 120588119896= radic120588120583
119896 we can rewrite the system of (26)ndash
(29) as follows
12059610158401015840
119898119896(119905) + 120588
2
119896120596119898119896
(119905) = minus
1
1003817100381710038171003817120593119896
1003817100381710038171003817
2[V119898(119905) 120593
119896(0)
+ ⟨119891 (sdot 119906119898(sdot 119905)) 120593
119896⟩]
V119898(119905) = int
119905
0
119896 (119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119904) 119906
119898(120585119873 119904)) 119889119904
+ 119892 (119906119898(1205850 119905)) + 120590 (119905)
120596119898119896
(0) = 119886119898119896
1205961015840
119898119896(0) = 119887
119898119896
119896 = 1119898
(30)
Therefore we obtain
120596119898119896
(119905) = 119886119898119896
cos (120588119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
minus
2
120593119896(0)
int
119905
0
119889119904
sdot int
119904
0
sin [120588119896(119905 minus 119904)]
120588119896
119896 (119904 minus 120591)
sdot ℎ (119906119898(1205850 120591) 119906
119898(120585119873 120591)) 119889120591 minus
2
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
[120590 (119904) + 119892 (119906119898(1205850 119904))] 119889119904
minus
2
1205932
119896(0)
int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(119904)) 120593
119896⟩ 119889119904
119896 = 1119898
(31)
Applying the Schauder fixed-point theorem it is not difficultthat system (31) has a solution (120596
1198981 1205961198982
120596119898119898
) on aninterval [0 119879
119898] This implies that in system (26)ndash(29) there
exists a solution to 119906119898(119905) on [0 119879
119898] Moreover we can extend
the approximate solution 119906119898 to the whole interval [0 119879] (see
[15])
Step 2 (a priori estimates) In (26) we replace 120593119896(119909) by
119906119898
119905(119909 119905) Then integrating from 0 to 119905 we have after some
calculations
119864119898(119905) = 119864
119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 minus 2120590 (119905) 119906119898(0 119905)
+ 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904 minus 2int
119906119898(0119905)
0
119892 (119904) 119889119904
minus 2int
1
0
int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904 119889119909 minus 2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 +
5
sum
119896=1
119868119896(119905)
(32)
where
119864119898(119905) =
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867 (33)
We will estimate respectively the following terms on theright-hand side of (32)
Estimating 1198681(119905) Using (33) and Lemma 1 we infer that
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816le
1003817100381710038171003817119906119898(sdot 119905)
10038171003817100381710038171198620([01])
le1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817119867
le
1
radic120588
radic119864119898(119905)
(34)
Hence
1198681(119905) = minus2120590 (119905) 119906
119898(0 119905) le 120576
1003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+
1
120576
1205902(119905)
le 120576119864119898(119905) +
1
120576
1205902
119871infin(0119879)
forall120576 gt 0
(35)
Abstract and Applied Analysis 5
Estimating 1198682(119905) Using Lemma 1 and inequality (34) we
arrive at
1198682(119905) = 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904
le
1
120588
int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119889119904 + 120588int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816
1003816100381610038161003816119906119898(0 119904)
1003816100381610038161003816
2
119889119904
le int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119864119898(119904) 119889119904 +
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
(36)
Estimating 1198683(119905) From assumption (119860
4) we have
1198683(119905) = minus2int
119906119898(0119905)
0
119892 (119904) 119889119904 le 21198861003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+ 2119887 (37)
On the other hand we see that
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
=
10038171003817100381710038171003817100381710038171003817
119906119898
0+ int
119905
0
119906119898
119905(sdot 119904) 119889119904
10038171003817100381710038171003817100381710038171003817
2
le 21003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119905 int
119905
0
119864119898(119904) 119889119904
(38)
It follows from (37) (38) and Lemma 2 that
1198683(119905) le 2120576
119886
120588
119864119898(119905) + 4119886119905 (1 +
1
120576
)int
119905
0
119864119898(119904) 119889119904
+ 4119886 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119887
(39)
Estimating 1198684(119905) Owing to assumption (119860
3)-(i) (38) and
Lemma 2 it is not difficult to show that
1198684(119905) = minus2int
1
0
119889119909int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904
le 2int
1
0
1205901(119909)
1003816100381610038161003816119906119898(119909 119905)
1003816100381610038161003816
2
119889119909 + 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
le 120576
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
119864119898(119905)
+ 4119905 (1 +
1
120576
)10038171003817100381710038171205901
10038171003817100381710038171198711(01)
int
119905
0
119864119898(119904) 119889119904
+ 4 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2 10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
(40)
Estimating 1198685(119905)Applying integration by parts it follows that
1198685(119905) = minus2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
= minus2119906119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 1198691(119905) + 119869
2(119905) + 119869
3(119905)
(41)
We can estimate the integrals in the right-hand side of (41) asfollows
1198691(119905) = minus2119906
119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 21198881003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
119873
sum
119894=0
int
119905
0
|119896 (119905 minus 119904)|1003816100381610038161003816119906119898(120585119894 119904)
1003816100381610038161003816119889119904
+ 21198891003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816int
119905
0
|119896 (119905 minus 119904)| 119889119904 le 2 (119873 + 1)
119888
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119905 minus 119904)| radic119864119898(119904)119889119904 + 2
119889
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119904)| 119889119904 le 2120576119864119898(119905) + (119873 + 1)
2
sdot
1198882
120576120588
1198962
1198712(0119879)
int
119905
0
119864119898(119904) 119889119904 +
1198892
120576120588
1198962
1198711(0119879)
(42)
1198692(119905) = 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
|119896 (0)| int
119905
0
119864119898(119904) 119889119904 + 2
119889
radic120588
|119896 (0)|
sdot int
119905
0
radic119864119898(119904)119889119904 le [1 + 2 (119873 + 1)
119888
120588
|119896 (0)|]
sdot int
119905
0
119864119898(119904) 119889119904 +
1198892
120588
1198791198962(0)
(43)
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
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Differential EquationsInternational Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Estimating 1198682(119905) Using Lemma 1 and inequality (34) we
arrive at
1198682(119905) = 2int
119905
0
1205901015840(119904) 119906
119898(0 119904) 119889119904
le
1
120588
int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119889119904 + 120588int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816
1003816100381610038161003816119906119898(0 119904)
1003816100381610038161003816
2
119889119904
le int
119905
0
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816119864119898(119904) 119889119904 +
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
(36)
Estimating 1198683(119905) From assumption (119860
4) we have
1198683(119905) = minus2int
119906119898(0119905)
0
119892 (119904) 119889119904 le 21198861003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
2
+ 2119887 (37)
On the other hand we see that
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
=
10038171003817100381710038171003817100381710038171003817
119906119898
0+ int
119905
0
119906119898
119905(sdot 119904) 119889119904
10038171003817100381710038171003817100381710038171003817
2
le 21003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119905 int
119905
0
119864119898(119904) 119889119904
(38)
It follows from (37) (38) and Lemma 2 that
1198683(119905) le 2120576
119886
120588
119864119898(119905) + 4119886119905 (1 +
1
120576
)int
119905
0
119864119898(119904) 119889119904
+ 4119886 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 2119887
(39)
Estimating 1198684(119905) Owing to assumption (119860
3)-(i) (38) and
Lemma 2 it is not difficult to show that
1198684(119905) = minus2int
1
0
119889119909int
119906119898(119909119905)
0
119891 (119909 119904) 119889119904
le 2int
1
0
1205901(119909)
1003816100381610038161003816119906119898(119909 119905)
1003816100381610038161003816
2
119889119909 + 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
le 120576
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
119864119898(119905)
+ 4119905 (1 +
1
120576
)10038171003817100381710038171205901
10038171003817100381710038171198711(01)
int
119905
0
119864119898(119904) 119889119904
+ 4 (1 +
1
120576
)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2 10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
(40)
Estimating 1198685(119905)Applying integration by parts it follows that
1198685(119905) = minus2int
119905
0
119906119898
119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
= minus2119906119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
+ 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
equiv 1198691(119905) + 119869
2(119905) + 119869
3(119905)
(41)
We can estimate the integrals in the right-hand side of (41) asfollows
1198691(119905) = minus2119906
119898(0 119905)
sdot int
119905
0
119896 (119905 minus 119904) ℎ (119906119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 21198881003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816
119873
sum
119894=0
int
119905
0
|119896 (119905 minus 119904)|1003816100381610038161003816119906119898(120585119894 119904)
1003816100381610038161003816119889119904
+ 21198891003816100381610038161003816119906119898(0 119905)
1003816100381610038161003816int
119905
0
|119896 (119905 minus 119904)| 119889119904 le 2 (119873 + 1)
119888
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119905 minus 119904)| radic119864119898(119904)119889119904 + 2
119889
radic120588
sdot radic119864119898(119905) int
119905
0
|119896 (119904)| 119889119904 le 2120576119864119898(119905) + (119873 + 1)
2
sdot
1198882
120576120588
1198962
1198712(0119879)
int
119905
0
119864119898(119904) 119889119904 +
1198892
120576120588
1198962
1198711(0119879)
(42)
1198692(119905) = 2119896 (0)
sdot int
119905
0
119906119898(0 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
|119896 (0)| int
119905
0
119864119898(119904) 119889119904 + 2
119889
radic120588
|119896 (0)|
sdot int
119905
0
radic119864119898(119904)119889119904 le [1 + 2 (119873 + 1)
119888
120588
|119896 (0)|]
sdot int
119905
0
119864119898(119904) 119889119904 +
1198892
120588
1198791198962(0)
(43)
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
1198693(119905) = 2int
119905
0
119906119898(0 119903) 119889119903
sdot int
119903
0
1198961015840(119903 minus 119904) ℎ (119906
119898(1205850 119904) 119906
119898(120585119873 119904)) 119889119904
le 2 (119873 + 1)
119888
120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904 + 2
119889
radic120588
int
119905
0
radic119864119898(119903)119889119903
sdot int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 le 2 (119873 + 1)
119888
120588
[int
119905
0
119864119898(119903) 119889119903
+ int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903]
+ int
119905
0
119864119898(119903) 119889119903 +
1198892119905
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(44)
Going in for the Cauchy-Schwartz inequality we arrive at
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
le int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119904 int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
(45)
Consequently
int
119905
0
(int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816radic119864
119898(119904)119889119904)
2
119889119903
le
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119889119903int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119864119898(119904) 119889119904
=
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
int
119905
0
119864119898(119904) 119889119904 int
119905
119904
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816119889119903
le
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
int
119905
0
119864119898(119904) 119889119904
(46)
It follows from (44) and (46) that
1198693(119905)
le [1 + 2 (119873 + 1)
119888
120588
(1 +
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
)]int
119905
0
119864119898(119903) 119889119903
+
1198892119879
120588
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
(47)
We deduce from the estimates of 1198691(119905) 119869
2(119905) 119869
3(119905) that
1198685(119905) le 2120576119864
119898(119905) + 119863
1
119879int
119905
0
119864119898(119904) 119889119904 + 119863
2
119879 (48)
with
1198631
119879= (119873 + 1)
2 1198882
120576120588
1198962
1198712(0119879)
+ 2 (119873 + 1)
119888
120588
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ |119896 (0)| + 1) + 2
1198632
119879=
1198892
120576120588
[120576119879
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1198962
1198711(0119879)
+ 1205761198791198962(0)]
(49)
On the other hand by (28) (29) and assumptions (1198601)ndash(119860
3)
119864119898(0) + 2120590 (0) 119906
119898
0(0) + 2int
119906119898
0(0)
0
119892 (119904) 119889119904
+ 2int
1
0
119889119909int
119906119898
0(119909)
0
119891 (119909 119904) 119889119904 le 119862
(50)
where 119862 is a positive constant depending only on1199060 1199061 119891 119892 120590
Combining (32) (35) (36) (39) (40) (48) and (50) weobtain after some rearrangements
119864119898(119905) le 120576119863
3
119879119864119898(119905) + int
119905
0
1198634
119879(119904) 119864
119898(119904) 119889119904 + 119863
5
119879 (51)
where
1198633
119879=
2
120588
10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 2
119886
120588
+ 3
1198634
119879(119904) =
100381610038161003816100381610038161205901015840(119904)
10038161003816100381610038161003816+ 4119879 (1 +
1
120576
) (10038171003817100381710038171205881
10038171003817100381710038171198711(01)
+ 119886) + 1198631
119879
1198635
119879=
1
120576
1205902
119871infin(0119879)
+
1
120588
100381710038171003817100381710038171205901015840100381710038171003817100381710038171198711(0119879)
+ 210038171003817100381710038171205902
10038171003817100381710038171198711(01)
+ 4 (1 +
1
120576
) (10038171003817100381710038171205901
10038171003817100381710038171198711(01)
+ 119886)1003817100381710038171003817119906119898
0
1003817100381710038171003817
2
+ 1198632
119879+ 119862 + 2119887
(52)
Choosing 1205761198633
119879= 12 by Gronwallrsquos inequality we have
119864119898(119905) le 2119863
5
119879exp(2int
119905
0
1198634
119879(119904) 119889119904) le 119863
119879 (53)
where119863119879is a positive constant depending on 119879
Next we will require the following lemma
Lemma 7 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (54)
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
Proof of Lemma 7 We put
119896119898119894
(119905) =
119898
sum
119896=1
cos (120583119896120585119894)
sin (120588119896119905)
120588119896
(55)
119892119898119894
(119905) =
119898
sum
119896=1
120593119896(120585119894) [119886
119898119896cos (120588
119896119905) + 119887
119898119896
sin (120588119896119905)
120588119896
]
minus 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨119891 (sdot 119906119898(sdot 119904)) 120593
119896⟩ 119889119904
(56)
In view of (25) and (31) 119906119898(120585119894 119905) can be rewritten as follows
119906119898(120585119894 119905) = 119892
119898119894(119905) minus 2int
119905
0
119896119898119894
(119905 minus 119904) V119898(119904) 119889119904 (57)
In connection with 119892119898119894
(119905) we have the following lemma
Lemma 8 There exists a positive constant 119862119879depending only
on 119879 such that
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 119862119879 forall119905 isin [0 119879] 119894 = 0119873 (58)
Proof of Lemma 8 We define
1198921015840
119898119894(119905) = minus119901
119898119894(119905) + 119902
119898119894(119905) minus 119903
119898119894(119905) minus 119904
119898119894(119905) (59)
with
119901119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 120588
119896119886119898119896
cos (120583119896120585119894) sin (120588
119896119905) (60)
119902119898119894
(119905) =
119898
sum
119896=1
120593119896(0) 119887
119898119896cos (120583
119896120585119894) cos (120588
119896119905) (61)
119903119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sin (120588119896119905)
120588119896
⟨119891 (sdot 119906119898
0) 120593
119896⟩ (62)
119904119898119894
(119905) = 2
119898
sum
119896=1
cos (120583119896120585119894)
120593119896(0)
sdot int
119905
0
sin [120588119896(119905 minus 119904)]
120588119896
⟨1198632119891 (sdot 119906
119898(sdot 119904))
sdot 119906119898
119905(sdot 119904) 120593
119896⟩ 119889119904
(63)
On the other hand use the inequality
(119886 + 119887 + 119888 + 119889)2le 4 (119886
2+ 119887
2+ 119888
2+ 119889
2)
forall119886 119887 119888 119889 isin R
(64)
Hence
int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904 le 4 sum
119895isin119901119902119903119904
int
119905
0
1003816100381610038161003816119895119898119894
(119904)1003816100381610038161003816
2
119889119904
equiv 1198701(119905) + 119870
2(119905) + 119870
3(119905) + 119870
4(119905)
(65)
We will estimate each term on the right-hand side of thisinequality
Estimating 1198701(119905) = 4 int
119905
0|119901119898119894
(119904)|2119889119904 Thanks to (60) and (65)
1198701(119905) = 120588int
119905
0
119898
sum
119896=1
120593119896(0)
sdot 120583119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)] + sin [120583
119896(radic120588119904 minus 120585
119894)]
2
119889119904
le 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 + 120585
119894)]
2
119889119904
+ 2120588int
119905
0
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin [120583119896(radic120588119904 minus 120585
119894)]
2
119889119904
= 2radic120588int
radic120588119905+120585119894
120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
+ 2radic120588int
radic120588119905minus120585119894
minus120585119894
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896
sin (120583119896119904)]
2
119889119904
(66)
Now we will need the following lemma
Lemma 9 Let 119886 119887 isin R 119886 lt 119887 and 119888119896isin R 119896 = 1119898 Then
int
119887
119886
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
le 2 (max |119886| |119887| + 2) int
1
0
[
119898
sum
119896=1
119888119896sin (120583
119896119904)]
2
119889119904
(67)
The proof of this lemma is simple we omit the detailsApplying Lemma 9 we deduce from (28) (66) and
assumption (1198601) that
1198701(119905) le 8radic120588 (radic120588119879 + 120585
119894+ 2)
sdot int
1
0
[
119898
sum
119896=1
120593119896(0) 120583
119896119886119898119896sin (120583
119896119904)]
2
119889119904
le 8radic120588 (radic120588119879 + 120585119894+ 2)
10038171003817100381710038171199060119898
1003817100381710038171003817
2
119867le 119862
119879
(68)
where 119862119879always indicates a constant depending on 119879
Estimating 1198702(119905) = 4 int
119905
0|119902119898119894
(119904)|2119889119904 Similarly we also obtain
1198702(119905) le 8(119879 +
120585119894+ 2
radic120588
)10038171003817100381710038171199061119898
1003817100381710038171003817
2
le 119862119879 (69)
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
Estimating 1198703(119905) = 4 int
119905
0|119903119898119894
(119904)|2119889119904We see that
119903119898119894
(119904) =
2
radic120588
119898
sum
119896=1
1
120583119896
int
1
0
sin (120583119896radic
120588119904) cos (120583119896120585119894) cos (120583
119896119909)
sdot 119891 (119909 119906119898
0(119909)) 119889119909 =
1
2radic120588
sdot int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894+ 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894minus 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
+
1
2radic120588
int
1
0
119898
sum
119896=1
1
120583119896
sin [120583119896(radic120588119904 + 120585
119894minus 119909)]
+ sin [120583119896(radic120588119904 minus 120585
119894+ 119909)] 119891 (119909 119906
119898
0(119909)) 119889119909
(70)
To estimate1198703(119905) we need the following lemma
Lemma 10 Let 119898 isin N One always has
119878119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=0
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
lt 1 +
4
120587
forall119911 isin R (71)
Proof of Lemma 10 First we assume that 0 lt 119911 le 1 Set 119872 =
[119911minus1] (which is an integer part of 119911minus1) We consider two cases
of119898
Case 1 (119898 gt 119872 + 1) Then
119878119898(119911) le
119872
sum
119896=1
10038161003816100381610038161003816sin 120583
119895119911
10038161003816100381610038161003816
120583119895
+
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
equiv 1198781
119898(119911) + 119878
2
119898(119911)
(72)
Since | sin120593| le |120593| for all 120593 isin R we get
1198781
119898(119911) =
119872
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872
sum
119896=1
120583119896119911
120583119896
= 119872119911 le 1 (73)
Moreover the function 120593 997891rarr (sin120593)120593 is decreasing on(0 1205872] hence sin(1205872)120593 ge 120593 for all 120593 isin [0 1] it followsthat
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816
sin ((119898 + 119896 minus 1) 2) 120587119911 sin ((119898 minus 119896 + 1) 2) 120587119911
sin (1205871199112)
10038161003816100381610038161003816100381610038161003816
le
1
sin (1205871199112)
le
1
119911
119896 = 1119898
(74)
On the other hand it is easy to verify the following equalityby the induction
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
=
1
120583119872+1
119898
sum
119896=119872+1
sin 120583119896119911
+
119898minus1
sum
119896=119872+1
[(
1
120583119896+1
minus
1
120583119896
)
119898
sum
119894=119896+1
sin 120583119894119911]
(75)
Using (74) and (75) we arrive at
1198782
119898(119911) =
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
1
120583119872+1
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=119872+1
sin 120583119896119911
1003816100381610038161003816100381610038161003816100381610038161003816
+
119898minus1
sum
119896=119872+1
[(
1
120583119896
minus
1
120583119896+1
)
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119894=119896+1
sin 120583119894119911
1003816100381610038161003816100381610038161003816100381610038161003816
]
le
1
120583119872+1
1
119911
+
119898minus1
sum
119896=119872+1
(
1
120583119896
minus
1
120583119896+1
)
1
119911
= (
2
120583119872+1
minus
1
120583119898
)
1
119911
lt
4
120587
(76)
Consequently it follows from (72) (73) and (76) that
119878119898(119911) le 119878
1
119898(119911) + 119878
2
119898(119911) lt 1 +
4
120587
(77)
Case 2 (1 le 119898 le 119872 + 1) We have
119878119898(119911) le
1003816100381610038161003816100381610038161003816100381610038161003816
119898
sum
119896=1
sin 120583119896119911
120583119896
1003816100381610038161003816100381610038161003816100381610038161003816
le
119872+1
sum
119896=1
1003816100381610038161003816sin 120583
1198961199111003816100381610038161003816
120583119896
le
119872+1
sum
119896=1
120583119896119911
120583119896
= (119872 + 1) 119911 lt 1 +
4
120587
(78)
Combining (77) and (78) we conclude that
119878119898(119911) lt 1 +
4
120587
forall119898 isin N 119911 isin (0 1] (79)
Since 119878119898(0) = 0 and the function 119878
119898is even periodic with
the period 2 thus inequality (79) holds for all119898 isin N 119911 isin RThe proof of Lemma 10 is complete
By (70) and Lemma 10 it leads to
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816le
1
2radic120588
int
1
0
4 (1 +
4
120587
)1003816100381610038161003816119891 (119909 119906
119898
0(119909))
1003816100381610038161003816119889119909
=
2
radic120588
(1 +
4
120587
)1003817100381710038171003817119891 (sdot 119906
119898
0)10038171003817100381710038171198711(01)
(80)
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
Hence it follows from (28) (80) and assumptions (1198601) (119860
3)
that
1198703(119905) = 4int
119905
0
1003816100381610038161003816119903119898119894
(119904)1003816100381610038161003816
2
119889119904
le
16119879
120588
(1 +
4
120587
)
21003817100381710038171003817119891 (sdot 119906
119898
0)1003817100381710038171003817
2
1198711(01)
le 119862119879
(81)
Estimating 1198704(119905) = 4 int
119905
0|119904119898119894
(119904)|2119889119904 Proving in the same way
as (81) we get
1198704(119905) = 4int
119905
0
1003816100381610038161003816119904119898119894
(119904)1003816100381610038161003816
2
119889119904 le
81198792
120588
(1 +
4
120587
)
2
sdot int
119905
0
10038171003817100381710038171198632119891 (sdot 119906
119898(sdot 119904)) 119906
119898
119905(sdot 119904)
1003817100381710038171003817
2
1198711(01)
119889119904
(82)
On the other hand using assumption (1198603)-(ii) then
10038161003816100381610038161198632119891 (119909 119906
119898(119909 119904))
1003816100381610038161003816le (119909) sup
|119911|le119879
119891 (119911) = 119862
119879 (119909) (83)
with119863119879= radic119863
119879120588
Applying the Cauchy-Schwartz inequality we clearly get
1198704(119905) le
81198792
120588
(1 +
4
120587
)
2
1198622
119879
2int
119905
0
1003817100381710038171003817119906119898
119905(sdot 119904)
1003817100381710038171003817
2
119889119904
le 119862119879
(84)
Combining (65) (68) (69) (81) and (84) we obtainLemma 8
Remark 11 Lemma 3 in [4] is a special case of Lemma 8 with119892 = ℎ = 0 and 120588 = 1
We now return to the proof of Lemma 7Note that it follows from (27) that
V1015840119898(119905) = 120590
1015840(119905) + 119892
1015840(119906119898(1205850 119905)) 119906
119898
119905(1205850 119905) + 119896 (0)
sdot ℎ (119906119898(1205850 119905) 119906
119898(1205851 119905) 119906
119898(120585119873 119905))
+ int
119905
0
1198961015840(119905 minus 119904)
sdot ℎ (119906119898(1205850 119904) 119906
119898(1205851 119905) 119906
119898(120585119873 119904)) 119889119904
(85)
On account of (53) and assumptions (1198602) (119860
4) and (119860
5) we
get10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ sup|119911|le119879
100381610038161003816100381610038161198921015840(119911)
10038161003816100381610038161003816
1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
+ (|119896 (0)| +
100381710038171003817100381710038171198961015840100381710038171003817100381710038171198711(0119879)
) [(119873 + 1) 119888119863119879+ 119889]
le
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816+ 119862
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816+ 1)
(86)
Therefore it is easy to see that
10038161003816100381610038161003816V1015840119898(119905)
10038161003816100381610038161003816
2
le 2
100381610038161003816100381610038161205901015840(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1003816100381610038161003816119906119898
119905(1205850 119905)
1003816100381610038161003816
2
+ 1) (87)
Applying Lemma 10 and the imbedding 1198671(0 1) 997893rarr
1198620([0 1]) we deduce from (27) and (57) that
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816=
10038161003816100381610038161003816100381610038161003816
1198921015840
119898119894(119905) minus 2119896
119898119894(119905) V
119898(0)
minus 2int
119905
0
119896119898119894
(119905 minus 119904) V1015840119898(119904) 119889119904
10038161003816100381610038161003816100381610038161003816
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+
2
radic120588
(1
+
4
120587
)1003816100381610038161003816V119898(0)
1003816100381610038161003816+
2
radic120588
(1 +
4
120587
)int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904
le
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816+ 119862
119879(1 + int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816119889119904)
(88)
Thus
1003816100381610038161003816119906119898
119905(120585119894 119905)
1003816100381610038161003816
2
le 2
100381610038161003816100381610038161198921015840
119898119894(119905)
10038161003816100381610038161003816
2
+ 41198622
119879(1 + 119905int
119905
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904)
(89)
Using theCauchy-Schwartz inequality and Lemma 8 thenweobtain from (87) and (89) that
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 41198622
119879119905 + 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 41198622
119879int
119905
0
120591 119889120591int
120591
0
10038161003816100381610038161003816V1015840119898(119904)
10038161003816100381610038161003816
2
119889119904
le 41198622
119879119905 +
16
3
1198624
1198791199053+ 4119862
2
119879119905
10038171003817100381710038171003817120590101584010038171003817100381710038171003817
2
1198712(0119879)
+ 2int
119905
0
100381610038161003816100381610038161198921015840
119898119894(119904)
10038161003816100381610038161003816
2
119889119904
+ 161198624
119879int
119905
0
120591 119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
le 119862119879+ 119862
119879int
119905
0
119889120591int
120591
0
1003816100381610038161003816119906119898
119905(1205850 119904)
1003816100381610038161003816
2
119889119904
(90)
Hence
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le (119873 + 1)119862119879[1 + int
119905
0
(
119873
sum
119894=0
int
120591
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904)119889120591]
(91)
By the Gronwall inequality we get Lemma 7 This completesthe proof of Lemma 7
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
Step 3 (limiting process) Due to (53) and (54) applying theBanach-Alaoglu theorem we can extract a subsequence ofsequence 119906
119898 still labeled by the same notations such that
119906119898
997888rarr 119906 weaklylowastin 119871infin
(0 1198791198671(0 1))
119906119898
119905997888rarr 119906
119905weaklylowastin 119871
infin(0 119879 119871
2(0 1))
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) weakly in 119867
1(0 119879)
(92)
Thanks to Lemma 4 and the compactness of the imbedding1198671(0 119879) 997893rarr 119862
0([0 119879]) they lead us to
119906119898
997888rarr 119906
strongly in 1198712((0 1) times (0 119879)) ae in (0 1) times (0 119879)
119906119898(120585119894 sdot) 997888rarr 119906 (120585
119894sdot) strongly in 119862
0([0 119879])
(93)
By (93)2and assumptions (119860
2) (119860
4) and (119860
6) we have
V119898(119905) 997888rarr int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
+ 119892 (119906 (1205850 119905)) + 120590 (119905) = V (119905)
(94)
strongly in 1198620([0 119879])
Also we apply the inequality1003816100381610038161003816119891 (119909 119906
119898(119909 119905)) minus 119891 (119909 119906 (119909 119905))
1003816100381610038161003816
le 119889119879
1003816100381610038161003816119906119898(119909 119905) minus 119906 (119909 119905)
1003816100381610038161003816
(95)
where 119889119879= 119863
21198911198620([01]times[minus119879119879])
By (93)1and (95) we get
119891 (sdot 119906119898) 997888rarr 119891 (sdot 119906)
strongly in 1198712((0 1) times (0 119879))
(96)
Passing to the limit in (26) by (92)12 (94) and (96) we have
119906 satisfying the equation119889
119889119905
⟨119906119905(sdot 119905) 120593⟩ + 120588 ⟨119906 (sdot 119905) 120593⟩
119867+ V (119905) 120593 (0)
+ ⟨119891 (sdot 119906 (sdot 119905)) 120593⟩ = 0
(97)
for all 120593 isin 119867 ae 119905 isin [0 119879]On the other hand it is easy to show a similar way as in
[4 p 588]119906 (119909 0) = 119906
0(119909)
119906119905(119909 0) = 119906
1(119909)
(98)
The existence of global solutions is proved
Step 4 (uniqueness of the weak solutions) Let 1199061and 119906
2be
two weak solutions of problem (1)Then 119906 = 1199061minus119906
2is a weak
solution of the following problem
119906119905119905minus 120588119906
119909119909+ 119891 (sdot 119906
1) minus 119891 (sdot 119906
2) = 0
120588119906119909(0 119905) = V (119905)
119906 (1 119905) = 0
119906 (119909 0) = 119906119905(119909 0) = 0
(99)
in which
V (119905) =
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119905)) + int
119905
0
119896 (119905 minus 119904)
sdot
2
sum
119894=1
(minus1)119894minus1
ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
(100)
To prove 1199061= 119906
2 then the following lemma is needed
Lemma 12 Let 119906 be the weak solution of the followingproblem
119906119905119905minus 120588119906
119909119909+ 119865 = 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
119906 (1 119905) = 0
120588119906119909(0 119905) = V (119905)
119906 (119909 0) = 1199060(119909)
119906119905(119909 0) = 119906
1(119909)
119906 isin 119871infin
(0 119879119867)
119906119905isin 119871
infin(0 119879 119871
2(0 1))
119906 (0 sdot) isin 1198671(0 119879)
V isin 1198712(0 119879)
119865 isin 1198711(0 119879 119871
2(0 1))
(101)
Then we have1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+ 2int
119905
0
V (119904) 119906119905(0 119904) 119889119904
+ 2int
119905
0
⟨119865 (sdot 119904) 119906119905(sdot 119904)⟩ 119889119904 ge
10038171003817100381710038171199061
1003817100381710038171003817
2
+ 12058810038171003817100381710038171199060
1003817100381710038171003817
2
119867
119886119890 119905 isin [0 119879]
(102)
Equality holds in case of 1199060= 119906
1= 0
The proof of Lemma 12 is the same as Lemma 21 in [16]
Applying Lemma 12 with 1199060
= 1199061
= 0 119865 = 119891(sdot 1199061) minus
119891(sdot 1199062) we get
119864 (119905) = minus2int
119905
0
⟨119891 (sdot 1199061) minus 119891 (sdot 119906
2) 119906
1015840(119904)⟩ 119889119904
minus 2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
minus 2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 + 2120588119896 (0)
sdot int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
+ 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 equiv 119869
1(119905) + 119869
2(119905)
+ 1198693(119905) + 119869
4(119905) + 119869
5(119905)
(103)
with 119864(119905) = 119906119905(sdot 119905)
2+ 120588119906(sdot 119905)
2
119867
Now we can estimate the integrals in the right-hand sideof (103) as follows
First term 1198691(119905) using assumption (119860
3) it follows from
Lemma 1 and (103) that
1198691(119905)
= minus2int
119905
0
⟨119891 (sdot 1199061(sdot 119904)) minus 119891 (sdot 119906
2(sdot 119904)) 119906
119905(sdot 119904)⟩ 119889119904
le
1198891198721
radic120588
int
119905
0
(1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119904)2) 119889119904
le
1198891198721
radic120588
int
119905
0
119864 (119904) 119889119904
(104)
where
1198891198721
=1003817100381710038171003817119863211989110038171003817100381710038171198620([01]times[minus119872119872])
119872 =
2
sum
119894=1
1003817100381710038171003817119906119894
1003817100381710038171003817119871infin(0119879119867)
(105)
Second term 1198692(119905) set 119889
1198722= 119892
1198622([minus119872119872])
Integrating byparts then we arrive at
1198692(119905) = minus2int
119905
0
119906119905(0 119904)
2
sum
119894=1
(minus1)119894minus1
119892 (119906119894(1205850 119904)) 119889119904
= minus2int
119905
0
119906119905(0 119904) 119889119904
sdot int
1
0
119889
119889120579
119892 (1199062(1205850 119904) + 120579119906 (120585
0 119904)) 119889120579
= minus1199062(0 119905) int
1
0
1198921015840(1199062(1205850 119905) + 120579119906 (120585
0 119905)) 119889120579
+ int
119905
0
1199062(0 119904) 119889119904 int
1
0
11989210158401015840(1199062(1205850 119904) + 120579119906 (120585
0 119904))
sdot (1199061015840
2(1205850 119904) + 120579119906
1015840(1205850 119904)) 119889120579 le 119889
1198722[1199062(0 119905)
+ int
119905
0
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) 119906
2(0 119904) 119889119904]
(106)
On the other hand we easily show that
119906 (sdot 119905)2le 119905int
119905
0
1003817100381710038171003817119906119905(sdot 119904)
1003817100381710038171003817
2
119889119904 le 119879int
119905
0
119864 (119904) 119889119904 (107)
Applying Lemma 2 we deduce from (107) that
1199062(0 119905) le 119906 (sdot 119905)
2
1198620([01])
le 120576 119906 (sdot 119905)2
119867+ (1 +
1
120576
) 119906 (sdot 119905)2
le
120576
120588
119864 (119905) + 119879(1 +
1
120576
)int
119905
0
119864 (119904) 119889119904 forall120576 gt 0
(108)
Thus it follows from (106) and (108) that
1198692(119905) le 120576
1198891198722
120588
119864 (119905) + int
119905
0
119889 (119904) 119864 (119904) 119889119904 (109)
in which
119889 (119904) =
120576
120588
1198891198722
(
100381610038161003816100381610038161199061015840
1(1205850 119904)
10038161003816100381610038161003816+
100381610038161003816100381610038161199061015840
2(1205850 119904)
10038161003816100381610038161003816) + (1 +
1
120576
)
sdot 1198791198891198722
(
100381710038171003817100381710038171199061015840
1(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+
100381710038171003817100381710038171199061015840
2(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+ 1)
(110)
Third term 1198693(119905) using the Cauchy-Schwartz inequality we
have from (103) (1198602) and (119860
6) that
1198693(119905) = minus2120588119906 (0 119905) int
119905
0
119896 (119905 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2radic119864 (119905) int
119905
0
|119896 (119905 minus 119904)|
119873
sum
119894=0
119888119872119894
radic119864 (119904)119889119904 le
1
4
119864 (119905)
+ 4 (119873 + 1)21198882
119872119879 119896
2
119871infin(0119879)
int
119905
0
119864 (119904) 119889119904
(111)
In addition we can similarly prove as in (111) for the fourthand fifth terms as follows
1198694(119905) = 2120588119896 (0) int
119905
0
119906 (0 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904 le 2120588 |119896 (0)|
sdot int
119905
0
1
radic120588
radic119864 (119904) (119873 + 1) 119888119872
1
radic120588
radic119864 (119904)119889119904 = 2 (119873
+ 1) 119888119872
|119896 (0)| int
119905
0
119864 (119904) 119889119904
1198695(119905) = 2120588int
119905
0
119906 (0 119903) 119889119903 int
119903
0
1198961015840(119903 minus 119904)
2
sum
119894=1
(minus1)119894minus1
sdot ℎ (119906119894(1205850 119904) 119906
119894(120585119873 119904)) 119889119904
le 2int
119905
0
radic119864 (119903) int
119903
0
100381610038161003816100381610038161198961015840(119903 minus 119904)
10038161003816100381610038161003816(119873 + 1)
sdot 119888119872radic119864 (119904)119889119904 119889119903 le (119873 + 1) 119888
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 1)
sdot int
119905
0
119864 (119904) 119889119904
(112)
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Abstract and Applied Analysis
Choosing 41205761198891198722
le 120588 the combination of (103) (104) (109)and (111)-(112) shows that
119864 (119905) le int
119905
0
119889 (119904) 119864 (119904) 119889119904 (113)
where
119889 (119904) = 2119889 (119904) + 8 (119873 + 1)
21198882
119872119879 119896
2
119871infin(0119879)
+ 2 (119873 + 1) 119888119872
(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+ 2 |119896 (0)| + 1)
+ 2
1198891198721
radic120588
(114)
By the Gronwall inequality we see that 119864(119905) equiv 0 that is 1199061equiv
1199062This completes the proof of Theorem 5
4 Stability of the Weak Solutions
In this section let (1199060 1199061) isin 119867
1(0 1) times 119871
2(0 1) be fixed
functions Also we assume that 1205901 1205902
isin 1198711(0 1) isin
1198712(0 1)
119891 isin 119862
0(R) and 119886 119887 119889 119888 119888
119872are fixed functions
constants satisfying assumptions (1198603)ndash(119860
6) (independent of
119891 119892 and ℎ) Applying Theorem 5 then problem (1) has aunique weak solution 119906 depending on 120590 119896 119891 119892 ℎ We denote
119906 = 119906 (120590 119896 119891 119892 ℎ) (115)
where 120590 119896 119891 119892 ℎ satisfy assumptions (1198602)ndash(119860
6)
Then the stability of the solutions of problem (1) is givenas follows
Theorem 13 Let (1198601)ndash(119860
6) hold Then the solutions of prob-
lem (1) are stable with respect to the data (120590 119896 119891 119892 ℎ) in thefollowing sense
If (120590119895 119896119895 119891119895 119892119895 ℎ119895) and (120590 119896 119891 119892 ℎ) satisfy the assump-tions (119860
2)ndash(119860
6) such that
(120590119895 119896119895) 997888rarr (120590 119896) 119904119905119903119900119899119892119897119910 119894119899 [119882
11(0 119879)]
2
(119891119895 119892119895 ℎ119895) 997888rarr (119891 119892 ℎ)
(116)
strongly in 1198620([0 1] times [minus119872119872]) times 119862
1([minus119872119872]) times
1198620([minus119872119872]
119873+1) as 119895 rarr infin for all 119872 gt 0
Then
(119906119895 119906119895
119905) 997888rarr (119906 119906
119905) 119904119905119903119900119899119892119897119910 119894119899 119871
infin(0 119879119867
1(0 1)) 119871
infin(0 119879 119871
2(0 1)) as 119895 997888rarr infin (117)
where 119906119895 = 119906(120590119895 119896119895 119891119895 119892119895 ℎ119895) 119906 = 119906(120590 119896 119891 119892 ℎ)
Proof of Theorem 13 Firstly we assume that
100381710038171003817100381710038171205901198951003817100381710038171003817100381711988211(0119879)
+ 12059011988211(0119879)
le 120590lowast
100381710038171003817100381710038171198961198951003817100381710038171003817100381711988211(0119879)
+ 11989611988211(0119879)
le 119896lowast
(118)
where 120590lowast 119896lowastare fixed positive constants
On the other hand by the proof ofTheorem 5 the a prioriestimates of the sequence 119906
119898 satisfy
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(119)
where 119862119879is a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
Due to (119) and (92) we conclude that
119862119879ge lim inf119898rarrinfin
(1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 1205881003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867
+
119873
sum
119894=0
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904) ge lim inf119898rarrinfin
1003817100381710038171003817119906119898
119905(sdot 119905)
1003817100381710038171003817
2
+ 120588
sdot lim inf119898rarrinfin
1003817100381710038171003817119906119898(sdot 119905)
1003817100381710038171003817
2
119867+
119873
sum
119894=0
lim inf119898rarrinfin
int
119905
0
1003816100381610038161003816119906119898
119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
ge1003817100381710038171003817119906119905(sdot 119905)
1003817100381710038171003817
2
+ 120588 119906 (sdot 119905)2
119867+
119873
sum
119894=1
int
119905
0
1003816100381610038161003816119906119905(120585119894 119904)
1003816100381610038161003816
2
119889119904
(120)In addition we can prove in a similar way above thatthe solution 119906
119895 of problem (1) corresponding to the data(120590119895 119896119895 119891119895 119892119895 ℎ119895) also satisfies
10038171003817100381710038171003817119906119895
119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817119906119895(sdot 119905)
10038171003817100381710038171003817
2
119867+
119873
sum
119894=0
int
119905
0
10038161003816100381610038161003816119906119895
119905(120585119894 119904)
10038161003816100381610038161003816
2
119889119904
le 119862119879 forall119905 isin [0 119879]
(121)
with 119862119879being a constant depending only on 119906
0 1199061 1205901 1205902
119891119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
We set119891119895= 119891
119895minus 119891
119892119895= 119892
119895minus 119892
ℎ119895= ℎ
119895minus ℎ
119896119895= 119896
119895minus 119896
120590119895= 120590
119895minus 120590
V119895 = 119906119895minus 119906
(122)
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 13
Then V119895 satisfies the following problem
V119895119905119905minus 120588V119895
119909119909+ 119865
119895= 0 0 lt 119909 lt 1 0 lt 119905 lt 119879
V119895 (1 119905) = 0
V119895119905(0 119905) = 119908
119895(119905)
V119895 (119909 0) = V119895119905(119909 0) = 0
(123)
where
119865119895(119909 119905) =
119891119895(119909 119906
119895) + 119891 (119909 119906
119895) minus 119891 (119909 119906)
119908119895(119905)
= 120590119895(119905) + 119892
119895(119906119895(1205850 119905)) + 119892 (119906
119895(1205850 119905))
minus 119892 (119906 (1205850 119905))
+ int
119905
0
119896119895(119905 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896119895(119905 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ int
119905
0
119896 (119905 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
119896 (119905 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
(124)
Applying Lemma 12 with 1199060= 119906
1= 0 we see that
119864119895(119905) = minus2int
119905
0
⟨119891 (sdot 119906119895) minus 119891 (sdot 119906) V119895
119905(sdot 119904)⟩ 119889119904
minus 2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904 minus 2int
119905
0
120590119895(119904)
sdot V119895119905(0 119904) 119889119904 minus 2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
minus 2int
119905
0
[119892 (119906119895(1205850 119905)) minus 119892 (119906 (120585
0 119905))] V119895
119905(0 119904) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdotℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2int
119905
0
V119895119905(0 119903) 119889119903 int
119903
0
119896119895(119903 minus 119904)
sdot ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus 2 [int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
minus int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904]
equiv 1198701(119905) + 119870
2(119905) + sdot sdot sdot + 119870
8(119905)
(125)
where
119864119895(119905) =
10038171003817100381710038171003817V119895119905(sdot 119905)
10038171003817100381710038171003817
2
+ 120588
10038171003817100381710038171003817V119895 (sdot 119905)
10038171003817100381710038171003817
2
119867 (126)
Let 119872 = (radic119862119879
+ radic119862119879)radic120588 Now we can estimate eight
integrals in the right-hand side of (125) as follows
Estimating 1198701(119905) From assumption (119860
3) it yields
1198701(119905)
= minus2int
119905
0
⟨119891 (sdot 119906119895(sdot 119904)) minus 119891 (sdot 119906 (sdot 119904)) V119895
119905(sdot 119904)⟩ 119889119904
le
119889119872
radic120588
int
119905
0
119864119895(119904) 119889119904
(127)
with 119889119872
= 11986321198911198620([01]times[minus119872119872])
Estimating 1198702(119905) It is easy to show that
1198702(119905) = minus2int
119905
0
⟨119891119895(sdot 119906
119895) V119895
119905(sdot 119904)⟩ 119889119904
le 119879
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+ int
119905
0
119864119895(119904) 119889119904
(128)
Estimating 1198703(119905) Since 120588V2
119895(0 119905) le 119864
119895(119905) by Cauchy-
Schwartz inequality then
1198703(119905) = minus2int
119905
0
120590119895(119904) V119895
119905(0 119904) 119889119904
le 120576119864119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1015840100381710038171003817100381710038171198711(0119879)
+
1
120576120588
10038161003816100381610038161003816120590119895(119905)
10038161003816100381610038161003816
2
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(129)
Moreover using the imbedding 11988211
(0 119879) 997893rarr 1198620([0 119879])
there exists a positive constant 119896119879such that
V1198620([0119879])
le 119896119879V
11988211(0119879)
forallV isin 11988211
(0 119879) (130)
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Abstract and Applied Analysis
Thus
1198703(119905) le 120576119864
119895(119905) +
1
120588
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
+
1198962
119879
120576120588
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+ int
119905
0
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816119864119895(119904) 119889119904
(131)
Estimating 1198704(119905) By help of assumption (119860
4) and (121) we
get
1198704(119905) = minus2int
119905
0
119892119895(119906119895(1205850 119904)) V119895
119905(0 119904) 119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904 +
1
120576120588
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198620([minus119872119872])
+
10038171003817100381710038171003817119892119895
101584010038171003817100381710038171003817
2
1198620([minus119872119872])
int
119905
0
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816
2
119889119904
le 120576119864 (119905) +
1
120588
int
119905
0
119864119895(119904) 119889119904
+ (119862119879+
1
120576120588
)
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
forall120576 gt 0
(132)
Estimating 1198705(119905) Proving in a similar way to (109) we also
obtain
1198705(119905)
= minus2int
119905
0
[119892 (119906119895(1205850 119904)) minus 119892 (119906 (120585
0 119904))] V119895
119905(0 119904) 119889119904
le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904
(133)
where
119889119895(119904) =
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
[
120576
120588
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ (1 +
1
120576
)
sdot 119879 (
10038171003817100381710038171003817119906119895
119905(1205850 sdot)
100381710038171003817100381710038171198711(0119879)
+1003817100381710038171003817119906119905(1205850 sdot)
10038171003817100381710038171198711(0119879)
+ 1)]
(134)
On the other hand we have
V1198711(0119879)
le radic119879 V1198712(0119879)
forallV isin 1198712(0 119879) (135)
Therefore
119889119895(119904)
le
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+ 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
=119889119895(119904)
(136)
It implies
1198705(119905) le
120576
120588
119864119895(119905)
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ int
119905
0
119889119895(119904) 119864 (119904) 119889119904 (137)
Estimating 1198706(119905) By reusing the inequalities (121) and (135)
and assumption (1198602) then
1198706(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904)
ℎ119895(119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
1003817100381710038171003817100381711989611989510038171003817100381710038171003817
2
11988211(0119879)
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)le 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + 119896
2
lowast(1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
(138)
Estimating 1198707(119905) Similarly from assumption (119860
5) we also
obtain
1198707(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896119895(119903 minus 119904) ℎ (119906
119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
le 120576119864119895(119905) +
2
120588
int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)
sdot
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
ℎ2
1198620([minus119872119872]
119873+1)equiv 120576119864
119895(119905) +
2
120588
sdot int
119905
0
119864119895(119904) 119889119904 + (1 + 119879119896
2
119879+
1
120576120588
)119889119872
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
(139)
Estimating 1198708(119905) Due to assumptions (119860
2) and (119860
6) we get
1198708(119905) = minus2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906119895(1205850 119904) 119906
119895(120585119873 119904)) 119889119904
+ 2int
119905
0
V119895119905(0 119903) 119889119903
sdot int
119903
0
119896 (119903 minus 119904) ℎ (119906 (1205850 119904) 119906 (120585
119873 119904)) 119889119904
le 120576119864119895(119905) + 119872
1
119879int
119905
0
119864119895(119904) 119889119904
(140)
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 15
where
1198721
119879=
1
120588
[(119873 + 1)21198882
119872(
10038171003817100381710038171003817119896101584010038171003817100381710038171003817
2
1198711(0119879)
+
1
120576120588
1198962
1198712(0119879)
)
+ 2 (119873 + 1) 119888119872
|119896 (0)| + 1]
(141)
Thus combining (125)ndash(128) (131) (132) and (137)ndash(140)
119864119895(119905) le 120576119872
2
119879119864119895(119905) + 119872
3
119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)
+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
+ int
119905
0
[
120576
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
(
10038161003816100381610038161003816119906119895
119905(1205850 119904)
10038161003816100381610038161003816+1003816100381610038161003816119906119905(1205850 119904)
1003816100381610038161003816)
+
10038161003816100381610038161003816120590119895
1015840(119904)
10038161003816100381610038161003816+ 119872
4
119879]119864
119895(119904) 119889119904
(142)
in which
1198722
119879=
1
120588
100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 5
1198723
119879
= (1198962
lowast+
119889119872) (1 + 119879119896
2
119879+
1
120576120588
) +
1 + 1198962
119879+ 120576
120576120588
+ 119879
+ 119862119879
1198724
119879
= 11987932
(1 +
1
120576
) (1 + radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+
119889119872
radic120588
+ 1198721
119879+
5
120588
+ 1
(143)
Choosing 1205761198722
119879= 12 and applying the Gronwall inequality
we deduce from (120) (121) and (142) that
119864119895(119905) le 2119872
3
119879exp [2119879119872
4
119879
+ 2radic119879
120576
120588
(radic119862119879+ radic119862
119879)100381710038171003817100381711989210038171003817100381710038171198622([minus119872119872])
+ 2120590lowast]
sdot (
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(144)
This shows that10038171003817100381710038171003817V119895119905
10038171003817100381710038171003817
2
119871infin(0119879119871
2(01))
+
10038171003817100381710038171003817V11989510038171003817100381710038171003817
2
119871infin(0119879119867
1(01))
le 119872119879(
10038171003817100381710038171003817
119891119895
10038171003817100381710038171003817
2
1198620([01]times[minus119872119872])
+
10038171003817100381710038171003817119892119895
10038171003817100381710038171003817
2
1198621([minus119872119872])
+
10038171003817100381710038171003817
ℎ119895
10038171003817100381710038171003817
2
1198620([minus119872119872]
119873+1)+
10038171003817100381710038171003817
119896119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
10038171003817100381710038171003817
2
11988211(0119879)
+
10038171003817100381710038171003817120590119895
1003817100381710038171003817100381711988211(0119879)
)
(145)
with 119872119879
being a constant depending only on1199060 1199061 1205901 1205902 119891
119891 120590
lowast 119896lowast 119886 119887 119889 119888 120588 119879
This completes the proof of Theorem 13
Remark 14 If we use the inequality 120588V2119895(120585119894 119905) le 119864
119895(119905) 119894 =
0119873 then with regard to (126) and (145) conclusion (117) inTheorem 13 can be extended as follows
(119906119895 119906119895
119905 119906119895(1205850 sdot) 119906
119895(1205851 sdot) 119906
119895(120585119873 sdot))
997888rarr (119906 119906119905 119906 (120585
0 sdot) 119906 (120585
1 sdot) 119906 (120585
119873 sdot))
(146)
strongly in 119871infin(0 119879119867
1(0 1)) times 119871
infin(0 119879 119871
2(0 1)) times
[1198620([0 119879])]
119873+1 as 119895 rarr infin
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
References
[1] M Aassila M M Cavalcanti and V N Domingos CavalcantildquoExistence and uniform decay of the wave equation withnonlinear boundary damping and boundary memory sourcetermrdquo Calculus of Variations and Partial Differential Equationsvol 15 no 2 pp 155ndash180 2002
[2] V Barbu I Lasiecka and M A Rammaha ldquoOn nonlinearwave equations with degenerate damping and source termsrdquoTransactions of the American Mathematical Society vol 357 no7 pp 2571ndash2611 2005
[3] L Bociu P Radu and D Toundykov ldquoRegular solutions ofwave equations with super-critical sources and exponential-to-logarithmic dampingrdquo Evolution Equations and ControlTheoryvol 2 no 2 pp 255ndash279 2013
[4] A D Dang and D P Alain ldquoMixed problem for some semi-linear wave equation with a nonhomogeneous conditionrdquoNonlinear Analysis Theory Methods amp Applications vol 12 no6 pp 581ndash592 1988
[5] T X Le and G Giang Vo ldquoThe shock of a rigid body and anonlinear viscoelastic bar associated with a nonlinearboundaryconditionrdquo J Science Natural Sciences Uni Peda HCM vol 8pp 70ndash81 2006
[6] L T Nguyen and D T Bui ldquoA nonlinear wave equation asso-ciated with a nonlinear integral equation involving boundaryvaluerdquo Electronic Journal of Differential Equations vol 2004 pp1ndash21 2004
[7] L T Nguyen U V Le and T T Nguyen ldquoOn a shock probleminvolving a linear viscoelastic barrdquo Nonlinear Analysis TheoryMethods amp Applications vol 63 no 2 pp 198ndash224 2005
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Abstract and Applied Analysis
[8] L T Nguyen and G Giang Vo ldquoA wave equation associatedwith mixed nonhomogeneous conditions global existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 7 pp 1526ndash1546 2007
[9] L T Nguyen and G Giang Vo ldquoA nonlinear wave equationassociated with nonlinear boundary conditions existence andasymptotic expansion of solutionsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 66 no 12 pp 2852ndash2880 2007
[10] P Radu ldquoWeak solutions to the cauchy problem of a semilinearwave equation with damping and source termsrdquo Advances inDifferential Equations vol 10 no 11 pp 1261ndash1300 2005
[11] M L Santos ldquoAsymptotic behavior of solutions to wave equa-tions with a memory condition at the boundaryrdquo ElectronicJournal of Differential Equations vol 2001 no 73 pp 1ndash11 2001
[12] D Takaci and A Takaci ldquoOn the approximate solution of amathematical model of a viscoelastic barrdquo Nonlinear AnalysisTheory Methods amp Applications vol 67 no 5 pp 1560ndash15692007
[13] Q Tiehu ldquoGlobal solvability of nonlinear wave equation with aviscoelastic boundary conditionrdquo Chinese Annals of Mathemat-ics Series B vol 3 1993
[14] J L Lions Quelques Methodes de Resolution des Problemes auxLimites Non-Lineaires Dunod Gauthier-Villars Paris France1969
[15] E Coddington andN LevinsonTheory of Ordinary DifferentialEquations McGraw-Hill 1955
[16] J L Lions and W A Strauss ldquoSome nonlinear evolutionequationsrdquo Bulletin de la Societe Mathematique de France vol93 p 79 1965
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Stochastic AnalysisInternational Journal of