mixed problem with integral boundary condition for a...

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Journal of Applied Mathematics and Stochastic Analysis, 16:1 (2003), 69-79. Printed in the U.S.A. ©2003 by North Atlantic Science Publishing Company 69 MIXED PROBLEM WITH INTEGRAL BOUNDARY CONDITION FOR A HIGH ORDER MIXED TYPE PARTIAL DIFFERENTIAL EQUATION M. DENCHE and A.L. MARHOUNE Université Mentouri Constantine Institut de Mathematiques 25000 Constantine, Algeria (Received April, 2000; Revised March, 2002) In this paper, we study a mixed problem with integral boundary conditions for a high order partial differential equation of mixed type. We prove the existence and uniqueness of the solution. The proof is based on energy inequality, and on the density of the range of the operator generated by the considered problem. Key words: Integral Boundary Condition , Energy Inequalities, Equation of Mixed Type, Sobolev Spaces. AMS subject classifications: 35B45, 35G10, 35M10. 1 Introduction In the rectangle , we consider the equation (1) where is bounded for , and has bounded partial derivatives such that and , , for i To equation (1) we add the initial conditions (2) the boundary conditions for , , (3) for , , (4) and integral condition

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Page 1: MIXED PROBLEM WITH INTEGRAL BOUNDARY CONDITION FOR A …downloads.hindawi.com/archive/2003/869021.pdf · differential equation of mixed type. We prove the existence and uniqueness

Journal of Applied Mathematics and Stochastic Analysis, 16:1 (2003), 69-79.Printed in the U.S.A. ©2003 by North Atlantic Science Publishing Company

69

MIXED PROBLEM WITH INTEGRAL BOUNDARYCONDITION FOR A HIGH ORDER MIXED TYPE

PARTIAL DIFFERENTIAL EQUATION

M. DENCHE and A.L. MARHOUNEUniversité Mentouri Constantine

Institut de Mathematiques25000 Constantine, Algeria

(Received April, 2000; Revised March, 2002)

In this paper, we study a mixed problem with integral boundary conditions for a high order partialdifferential equation of mixed type. We prove the existence and uniqueness of the solution. Theproof is based on energy inequality, and on the density of the range of the operator generated by theconsidered problem. Key words: Integral Boundary Condition , Energy Inequalities, Equation of Mixed Type,Sobolev Spaces. AMS subject classifications: 35B45, 35G10, 35M10.

1 Introduction

In the rectangle , we consider the equation� � ��� �� � ��� �

� � � � � �� � ��� �� � � ���� ���� � � �� �� �� ��

� ��

�� � �

� � (1)

where is bounded for , and has bounded partial derivatives such ��� �� � � � ��� �� � � �

that and , , for � � � � � � � � �� ��� �� � ��� � �� ����� � ������� ��� �i

��

To equation (1) we add the initial conditions

� � ��� �� � ���� � � ��� �� � ��� � � ��� ���� ����� � (2)

the boundary conditions

���

���� �� � �� � � � � � � � � ��� � for , , (3)�

���

���� �� � �� � � � � � � � � ��� �for , , (4)�

and integral condition

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70 M. DENCHE and A.L. MARHOUNE

��

� � ��� � �� � � ��� ��� � (5)

where and are known functions which satisfy the compatibility conditions given in� �equations (3)-(5). Integral boundary conditions for evolution problems have various applications inchemical engineering, thermoelasticity, underground water flow and population dynamics; seefor example Choi and Chan [6], Ewing and Lin [7], Shi [12], and Shi and Shillor [13].Boundary value problems for parabolic equations with integral boundary conditions areinvestigated by Batten [1], Bouziani and Benouar [2], Cannon [3, 4], Cannon, Perez Esteva andVan der Hoeck [5], Ionkin [8], Kamynin [9], Kartynnik [10], Shi [12], Yurchuk [14] and manyreferences therein. A method was developed in [2], [10], and [14] based on energyinequalities. In this paper, our objective is to extend this method to high order partialdifferential equations of mixed type.

2 Preliminaries

In this paper, we prove existence and uniqueness of a strong solution of the problem stated inequations (1)-(5). For this, we consider the problem in equations (1)-(5) as a solution of theoperator equation

� � � , (6)

where with domain of definition consisting of functions ,� � � � � � � �� � �� � � � ���� � �� ����

such that, and, satisfies conditions in problems (3)-(5);� � � �� �� �� �� �� ���� � ��

� � � � � � ���� � � �� �

� �

the operator is considered from to , where is the Banach space consisting of all� � �functions satisfying equations (3)-(5), with the finite norm � � �����

� � � �� � � � � �� � �� � � � ������ �

� � � �� � �� ��

� �� � ��

� �

� �

� � �� � �� � � ��������

� �

�� ���� � � �� �� ��

�� � � �� ��

� �� � �� � � ��� sup (7)� � � �

��

���

� � ��� �� �� ��

� � �� �� � � � � � ��

and is the Hilbert space of vector-valued functions obtained by completion of � ��� � �� � �

the space with respect to the norm� ��� � � ��� �� � � ��� ���� �� � ��� �� �

� � � �� � � �� �� � � � �

���

��� ��

� ��� �� � �� � ���� � �� � �� � � ���� �� � �� �

� (8)

where is an arbitrary number such that .� �� � � � We then establish an energy inequality:

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Mixed Problem with Integral Boundary Condition 71

� � � � � ! � �� � (9)

and we show that the operator has the closure .� � A solution of the operator equation is called a strong solution ofDefinition 1: � � �the problem in equations (1)-(5). Inequality (9) can be extended to , i.e., � � �� �

� � � � � ! � � " � � �� �� � .

From this inequality, we obtain the uniqueness of a strong solution if it exists, and the equalityof sets and Thus, to prove the existence of a strong solution of the problem in# �� � # �� ��equations (1)-(5) for any , it remains to prove that the set is dense in .� � # �� � (2)Lemma 1: For any function satisfying the condition , for � � $ % &

��

exp� � $ ��� � �� ��� � �� � �� � �� ��� � � ���

���

� �& � �

�� �� exp (10)� � $���� � �� ������ �� ��

��

Proof: Starting from exp and integrating by parts, we� �� �

� � ��� ��

�� � $���� � �� ����

obtain equation (10). We note the above lemma holds for weaker conditions on .Remark 1: Lemma 2: For , � �

� �� �

� �� �

�� �� �� ��

� �

�� � �� �� � �� � �� ��� � �� �

� �� �

��

(11)

Proof: Starting from and using elementary inequalities yields��

��� � �� � �

�� �� �� ��

� �

� �

��

� � ��

(11). (2)Lemma 3: For satisfying the condition in equation , � �

� ���� ���

� �

� �� ��� ��

� ��

� �

� �� �exp� � $ ��� � ��� � � � ��

� �

�(x, �� � �� � �� ���

� � � �� ������ � �� �

�� �� ��

��

exp (12)� � $���� � ��� ��

��

Proof: Integrating the expression exp by parts for using��

�� � $�� � � �� �� �

�� �� ��

��� �

� � �� �

elementary inequalities and Lemma 2, we obtain expression (12).

3. An Energy Inequality and Its Consequences

Theorem 1: For any function we have the inequality � � �� �

� � � � � ! � � � , (13)

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72 M. DENCHE and A.L. MARHOUNE

where constant , with the constant satisfying! � $��$�exp max

min(��' (��� � �� � �

� �

� �� � )& �

�*' �

$ % & $ � � � �� % � �� � *�� and � � � �*� � (14)

DenoteProof:

+ � �� � �� � , , � � � ��� �� � �� �� �

�� �

� �� � � where �We consider the quadratic formula

Re exp (15)� �� �

��

� � $�� + ������

with the constant satisfying condition (14); obtained by multiplying equation (1) by$exp ; integrating over , where , with ; and by� � $��+ � � � ��� �� � ��� � � � � � � � �taking the real part. Integrating by parts times in formula (15) with the use of boundary�conditions in equations (3), (4), and (5), we obtain

Re exp exp� � � �� � � �

� �� ��

�� �

� � $�� + ���� � � � $���� � �� ����� � ��

� � � $���� � �� �����Re exp (16)� �� �

�� � �� �� �� ��

�� �

� �

�� ��

By substituting the expression of in formula (15), using elementary inequalities and the+ inequality

� �� �

� �,

��� �� ��� � ��' �

�� �� �

���

� �

� �� � �� � �� ��� � � � �� � � where ,�

yields:

Re exp ( exp� � � �� � � �

� �'

��� ��

� ���

�� � $�� + ���� � � �� � � $���� � �� ����� ��

� � � � $���� � �� ����� �

� ��� �

� �� �� exp . (17)� ��

From equation (1) we have:

� � � � � �� �� �� ��

� ��� � � � � �

� �� � � ��� �

�� . (18)

Using elementary inequalities, we obtain

� � � ' �� �� �����

� �

� �

��� ����

��� �� � exp� � $��� �� � �� � � ����� �� ��

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Mixed Problem with Integral Boundary Condition 73

� � � $���� � �� � � ����� � � � �� �� ��� �

� �� �� exp � �� �

� �

��

� �� � *� � � $���� � �� ������ �

�� �� ��

�� �� exp , (19)� ��

��

where � - �� By integrating the last term on the right-hand side of expression (16) and combining theobtained results with Lemmas 1 and 3, and the inequalities in equations (17), (18), and (19), weget:

( exp' )��� � & ��� � � �

� � �� �

��

���

�� ��

� � � � � $���� � �� ���� � �� � �� �� � �� � �� ��� � � � � �� ��

� �� � �� �� % � � $ ��� � �� ��� ���� ���

� �

� ���� ��

� �� ��� �

� �

� �� �� � � ��

� �

exp �

+ exp� � & ��� �

� �� �� � � $���� � �� ����� ��

� � � $ ��� � �� �� � � $���� � �� � � ������ � �

� �� ��� ��� �� �� �� �� ��

� �� � � � � �exp + exp � � � � ��

� � �

� �+1 ��

��

� � � $��� �� � �� � � ����� � � *' �� �� ��

���

� �

� �

��� ����

��� �� � exp � �� ��

� � � $ ��� � �� ����

����� ���

exp � � ��

� �$ � � � � �� � � �� � � $���� � �� ������ � �� � *�* �� ��� �

��

� �

���� � exp (20)� ��

Using elementary inequalities and condition (14), we obtain:

( ' )��� � & ��� � �

� ��

� ��

���

� ��

� � �� � �� ���� � �� � �� �� � �� � �� ��� � � � �� ��

� �� � �� ������

����

��

�� � ��

% � � $ � �� � �� �� � �� � �� ����exp �����

� � �

� �� ��� ��� & ��

� �� �

��� � �� � � ��

� �

� �� � �� �� � �� � �� � � ������ �

�� ��� ��� �� �� �� �� ��

� �� � � � � �� � � ��

� � �

� ��� ����

��

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74 M. DENCHE and A.L. MARHOUNE

� �

������

� �

� � �

� ��� ���� � �

�� �� �� ��

� ����� �

� ��� � �� �� � �� � � ���� � �� � �� ��� � � �� ��

� . (21)

As the left-hand side of equation (21) is independent of , by replacing the right-hand side by�its upper bound with respect to in the interval , we obtain the desired inequality. � .�� / �

Lemma 4: The operator from to admits a closure.� � Suppose that is a sequence such thatProof: � � � � �� �0

1 � ��0 in (22)

and

� 1 �0 � in (23)

We must show that The fact that results directly from� � � �� ��� � � � �� � �2� � �� �� � �

the continuity of the trace operators ��� Introducing the operator

��� 3 � � 3�� �� �� ��

��3 � � � � �� � ��� �� ��� ��

�� � �

� �

defined on the domain of functions verifying� � � 3 � � ������ ����

3 � �� � �� � �� � � �� � �� � �� � � �� � �� �� ��� ����3 � 3 � 3�� �� ��

� �

� �� �

we note that is dense in the Hilbert space obtained from the completion of with� � � � ����� �

respect to the norm

� � 3 � �� � �� 3 ������ �

� �

Additionally, since� � �� � �

0 0 ��� � �� �3���� � �� � �� 3���� � ��� � �� 3����� � ��� � �lim lim0 1 4 0 1 4

� �

this holds for every function , and yields .3 � � � � � � ���

Theorem 1 is valid for strong solutions, i.e., we have the inequality

� � � � � ! � " � � �� �� � , ,

hence we obtain (1)-(5) Corollary 1: A strong solution of the problem in equations is unique if it exists,and depends continuously on .� � �� ��� � � �

The range of the operator is closed in , and Corollary 2: # �� � � # �� � � # �� ��

4 Solvability of the Problem

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Mixed Problem with Integral Boundary Condition 75

To prove the solvability of the problem in equations (1)-(5), it is sufficient to show that is# �� �dense in . The proof is based on the following lemma.

Lemma 5: Suppose that is also bounded. Let � �� ������� �� � �

���

� � � � � � �� � � �� � � 5 ��

� �� � 6� � �� � � �7� � � �� � � � ���� � � � If, for and some , we have

��

� ���� � �� ���� � ��� � � � (24)

where is an arbitrary number such that , then .� � � � � � � � Proof: The equality (24) can be written as follows:

� �� � �� ���� � � � �� �� � �� ������ �� �

� �� � ��� � � �� �� �� ��

� � � � �� ��

� � �� �

� �� � (25)

If we introduce the smoothing operators with respect to [14] and ,� , � �8 � � �, �� � � � � � 9���� ��

then these operators provide the solutions of the problems

� �: ������ � : ��� � :�����

(26)g���� � ����

and

� � : ��� � :������: �����

99�

(27): ��� � ��9

��� and also have the following properties: for any , the functions and: � � ��� � : � �, �:�

� �� �

: � �, � : � ��� � : � � : � � ,9 � � 9 � 9 � �� ��� ��� � � �� are in such that and . Moreover, commutes

with , so and for .��� � �

� �9� � : � : �� 1 � : � : �� 1 �� 1 �� � �

Now, for given , we introduce the function��� ��

3��� �� � �� � �� �� � � � � � � � ���� � �� �� � � � � ��� �� � � � ��

� � �� � �

� �

� �� � ����

� �� � � � � � �

Integrating by parts, we obtain

�� � ��3 � ,3 � �� � �� � 3��� ���� � ��� � �� ��

and (28)�Then from equality (25), we have

� ; 3���� � < ��� 3������ �� �

� ��� ��

� (29)

where , and Replacing in; 3 � �� � ��3 � ,3 < ��� � � � �� ��� � �� ��� �� ��� � � � ��� �� ��

� �

� �

equation (29) by the smoothed function , and using the relation, � � ����

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76 M. DENCHE and A.L. MARHOUNE

< ���, � , < � � � , , �� � � � � � � ��< � ��� � � ���

� �

we get � � �� � �

� � �< ��� �� �� �� ��

�3 9 9; ���� � < ��� 3 ���� � 3 �����9�

� ��� � � (30)

By passing to the limit, we satisfy the equality (30) for all functions satisfying the conditions inequations (2)-(5), such that for .� �

�� �� �� �� ��

� � � � � ��� � � � ��

� �

The operator has, on , a continuous inverse defined by the relation< ��� � ��� ���

< ���: � � � �� = = :� �� � = �� �

� � � � �

��

� ������ � �� � �

� � �

�� � � � � �� �� � � � � �� � � � �

� ! � � = � �����

� � � ���� ��>

�� � (31)

where

��

�� �< ���:�� � �� (32)

Hence the function can be represented in the form Then,� � � �� � ��� �� ��

� � � �� � �� , < < �

�< � ��� �� ��� �

� �� < ���< ��� , where

< ���: � = :� � ��� � = �� �

� �

� �

� �?��

? � ��� �� �� ��

@��

� ?�

� � �

� � �� � � � �?�� � ?� @

? @ � ?� @

@ �

� � �� �� �� �� � �

�� � �� ! �

� �� �

� �� �� �� �� � �

?�� A��

?A � � ��� �� �� �� �?� ��>

@��

� A�

� � �

��

�A�� � A� @ ?� �

A @ � A� @

@ �

� � �

� (33)

Consequently, equation (30) becomes� �� �

� ��� �� ��

�3 9 9 9; ���� � < ��� �3 � < 3 ������9�

� � �� (34)

in which is the adjoint of the operator .< ��� < ���9� �

The left-hand side of equation 34) is a continuous linear functional of Hence the� ����

function has the derivatives B � 3 � < 3 � � ���� � ��� �� � � � ����� � � �9 9 9 � B � B

�� �� ��� ���

� �

�� �

� � �� �� and the following conditions are satisfied

� B � B�� ����� ���

� �

� �� � � �� � � �� � �� � �� � � �� �� � (35)

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Mixed Problem with Integral Boundary Condition 77

The operators are bounded in For sufficiently small, we have < ��� � ���� < ��� �9 9� � ���� �� �� �

� 8 � < ��� � ���� � � �, hence the operator, has a bounded inverse in In addition, � 9�

� < ������

� 9

��

�� � ���� , are bounded operators in . So from the equality,�

� B�� �� �� ��

9 ?� 3 � < ��� � 3

?��

� � ?

� 9 ? 9 �� ? 9

�� ?� � � �� �8 � < ���� � � � � �� �� � � ��

� (36)

We conclude that the function has the derivatives , . Taking into3 � � ��� � � ��9 � 3�� ��

� 9

�� �

account equations (36) and (35), we have

� �� ��8 � < ���� � � �� � � �� � ��� � � �9 ?� 3 � < ��� � 3

�� �� ��?��

���

� 9 ? 9 �� ? 9

� ? �� ?� � �� (37)

and

. (38)� �� ��8 � < ���� � � �� � � ��� � � �9 ?� 3 � < ��� � 3

�� �� ��?��

���

� 9 ? 9 �� ? 9

� ? �� ?� � ��

Similarly, for sufficiently small, in each fixed point , the operators � �� � .�� �/ � � � ��� < �����

� 9

��

are bounded in , and the operator is continuous inversible in . And so� ��� 8 � < ��� � ���� �9� �

from equations (37) and (38), satisfies the conditions39�

� 3�� ���

� 9

�� � �� � � �� � ���

� 3�� ���

� 9

�� � �� � � ���. (39)

So, for sufficiently small, the function has the same properties as In addition, � 3 B � 39 9� ��

satisfies the integral condition in equation (28). Putting exp in equation (29), where the constant satisfies � �$ �3 � � �� � $� �

� �

� 9�� � �

$ � � % �� �

��

� and using equation (27), we obtain

� � �� � �

9 � � ��� �� �� ��

�3exp exp (40)�$��3 ; 3���� � � < ��� � � $�� ���� � < ��� ������

9

� �

Integrating each term in the left-hand side of equation (40) by parts and taking the real parts,we have

Re exp exp� �� �

� � $ � �� �� � �� ��

< ��� � � $�� ���� % ��� �� � � $���� � �� ����� ��

� � ��

� � � $���� � �� ������ � � � �� �� ���

�� exp (41)� ��

��

Re exp (42)� � � �� < ��� ���� % � � $���� � �� ������� �� �

� � �� �� � �� ��

�3 � �9 �

���

Page 10: MIXED PROBLEM WITH INTEGRAL BOUNDARY CONDITION FOR A …downloads.hindawi.com/archive/2003/869021.pdf · differential equation of mixed type. We prove the existence and uniqueness

78 M. DENCHE and A.L. MARHOUNE

Now, using inequalities (41) and (42) in equation (40) with the choice of indicated above, we$have 2Re exp ; then for , we obtain 2Re exp ,� �

� �9�$��3 ; 3���� � � 1 � �$��3; 3���� � �� �

i.e., 2Re exp 2Re exp .� �� �

��$���� � ��C3C ���� � �$��3,3���� � �

Since Re exp , we conclude that ; hence , which ends the�� �$��3,3���� � � 3 � � � �

proof of this lemma. �

Theorem 2: The range of coincides with .# �� � � Proof: Since is a Hilbert space, we have if and only if the relation # �� � �

� �� �

���

� � � � �� �� �� ��

� ���� � �� ����� � �� � �� � � � �� � ���

� �� �� �� �� ��

� �

(43)

for arbitrary and , implies that , and . � � � ��� � � � � � � � � � �� � � � �

Putting in relation (43), we obtain Taking � � �� � �� � �� ����� � ��� �� ��

� � � � � � �� ���� �� ��� ��� ��� ��, and using Lemma 5, we obtain , then .

Consequently, we have" � � �� �

��

���

� � � � �� �� �� ��

� ���� � �� � � � �� � ��� ��� � ��

� ��

� �

� (44)

The range of the trace operator is everywhere dense in Hilbert space with the norm�� � � �� �

� �� �� � � � ��

���

��� ��

� �� ��� � �� � � �� ��� � ��

��

� �

hence .� � � � ��� ��� � �

References

[1] Batten, Jr., G.W., Second-order correct boundary conditions for the numerical solution ofthe mixed boundary problem for parabolic equations, (1963), 405-413.Math. Comp. 17

[2] Bouziani, A. and Benouar, N.E., Mixed problem with integral conditions for a third orderparabolic equation, (1998), 47-58.Kobe J. Math. 15

[3] Cannon, J.R., The solution of the heat equation subject to the specification of energy,Quart. Appl. Math. 21 (1963), 155-160.

[4] Cannon, J.R., The one-dimensional heat equation, In: Encyclopedia of Math. and itsAppl. , Addison-Wesley, Mento Park, CA (1984).23

[5] Cannon, J.R., Perez Esteva, S., and Van Der Hoek, J., A Galerkin procedure for thediffusion equation subject to the specification of mass, (1987),SIAM. J. Numer. Anal. 24499-515.

[6] Choi, Y.S. and Chan, K.Y., A parabolic equation with nonlocal boundary conditionsarising from electrochemistry, (1992), 317-331.Nonl. Anal. 18

Page 11: MIXED PROBLEM WITH INTEGRAL BOUNDARY CONDITION FOR A …downloads.hindawi.com/archive/2003/869021.pdf · differential equation of mixed type. We prove the existence and uniqueness

Mixed Problem with Integral Boundary Condition 79

[7] Ewing, R.E. and Lin, T., A class of parameter estimation techniques for fluid flow inporous media, (1991), 89-97.Adv. Water Resources 14

[8] Ionkin, N.I., Solution of a boundary value problem in heat condition with a nonclassicalboundary condition, (1977), 294-304.Diff. Urav. 13

[9] Kamynin, N.I., A boundary value problem in the theory of the heat condition withnonclassical boundary condition, (1964), 33-59.USSR Comput. Math. and Math. Phys. 4

[10] Kartynnik, A.V., Three-point boundary value problem with an integral space-variablecondition for a second-order parabolic equation, (1990), 1160-1166.Differ. Eqns. 26

[11] Rivlin, T.J., , John Wiley & Sons, New York 1974.The Chebyshev Polynomials[12] Shi, P., Weak solution to evolution problem with a nonlocal constraint, SIAM J. Anal. 24

91993), 46-58.[13] Shi, P. and Shillor, M., ,Design of Contact Patterns in One-Dimensional Thermoelasticity

In: Theoretical Aspects of Industrial Design, SIAM, Philadelphia, PA 1992.[14] Yurchuk, N.I., Mixed problem with an integral condition for certain parabolic equations,

Differ. Eqns. (1986), 1457-1463.22

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