existence of time periodic solutions for the nicholson's blowflies model with newtonian diffusion

Research Article

Received 31 March 2009 Published online 4 September 2009 in Wiley InterScience

(www.interscience.wiley.com) DOI: 10.1002/mma.1228MOS subject classification: 35 B 10; 35 K 55; 35 K 65

Existence of time periodic solutions for theNicholsons blowflies model with Newtoniandiffusion

Ying Yang, Ran Zhang, Chunhua Jin and Jingxue YinCommunicated by B. Straughan

This paper is concerned with the existence of periodic solutions of the Nicholsons blowflies model with Newtoniandiffusion. By constructing some suitable Lyapunov functionals and combining with LeraySchauder fixed point theorem,we establish the existence of nonnegative time periodic solutions. Copyright 2009 John Wiley & Sons, Ltd.

Keywords: Nicholsons blowflies model; Newtonian diffusion; periodic solution

1. Introduction

In this paper, we mainly study the existence of nonnegative time periodic solutions for the following one-dimensional Nicholsonsblowflies model with Newtonian diffusion

ut

= 2um

x2u+pu(x, t)eau(x,t)+g(x, t)+

tt

e(ts)u(x, s)ds, x (0,1), tR (1)

subject to the homogeneous Dirichlet boundary value condition

u(0, t)=u(1, t)=0, tR (2)where m>1, , p and a are positive constants, is a nonnegative constant, is a constant and g(x, t) is the known function whichsatisfy some structure conditions.

In 1990, Gurney et al. [1] proposed the following delay equation

du

dt=u(t)+pu(t)eau(t) (3)

which models the population of the Australian sheep-blowfly Lucilia cuprina, where p is the maximum per capita daily egg productionrate, 1 / a is the size at which the blowfly population reproduces at its maximum rate, is the per capita daily adult death rate,and is the generation time. Spatial structure may make it impossible for organisms to encounter each other in proportion totheir average density [2]. The random collision of individuals assumed in the above models may not represent interactions amongorganisms. Taking the spatial structure into account, Yang and So [3] extended (3) to linear diffusion equation model, which takesthe following form in one-dimensional case:

ut

= 2u

x2u(x, t)+pu(x, t)eau(x,t) (4)

However, it is well known that for the linear diffusion equation, the propagation velocity of perturbations is infinite, namely for anyinitial datum u0(x) 0, there holds u(x, t)>0 whenever t>0. In other words, this phenomenon implies that as long as the blowfly

Department of Mathematics, Jilin University, Changchun 130012, Peoples Republic of ChinaCorrespondence to: Chunhua Jin, Department of Mathematics, Jilin University, Changchun 130012, Peoples Republic China.E-mail: [email protected], [email protected]

Contract/grant sponsor: National Science Foundation of ChinaContract/grant sponsor: Specific Foundation for PhD Specialities of Educational Department of ChinaContract/grant sponsor: Postdoctoral Science Foundation of China

922

Copyright 2009 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2010, 33 922934

Y. YANG ET AL.

population density is not identically equal to zero at the initial time t=0, then the density will become positive as long as t>0, thatis, the blowfly would spread to the entire area immediately. In fact, it is actually that the blowfly population could not propagatein infinite speed, that is not consistent with biologic phenomena in the real world. Hence, it should be more reasonable to introducesome nonlinear diffusion with the property of finite speed of propagation, namely for any initial datum u0 with compact support,the support of u(, t) remains compact for any t>0. This means that the blowfly population would spread in the area in a finitespeed as time goes on. Furthermore, it is well known that factors influencing the variation of the population density at time tand position x not only include the migration of the population, the contribution due to deaths and births, but also include thepopulation density during the time between t and t. Inspired by this idea, it is more natural to introduce the nonlinear diffusionversion of the Equation (4), namely

ut

= 2um

x2u+pu(x, t)eau(x,t)+g(x, t)+

tt

e(ts)u(x, s)ds

where m>1. The advantage of this modified version lies in that it involves slow diffusion rather than fast diffusion, and so it is moresuitable to describe the blowfly population propagation phenomena accurately.

The linear diffusive Nicholsons blowflies equation with Dirichlet boundary conditions was studied by So and Yang [4], wherethe global attractivity of the equilibrium was proved. Some numerical and Hopf bifurcation analysis of this model was also carriedout by So et al. [5]. On the other hand, the existence of periodic solutions for evolutionary equations with delays has been studiedby some researchers, see for examples [6--13] and the references therein. But as far as we know, few works are concerned withthe existence of time periodic solutions for Newtonian filtration equation with delay. In this paper, our purpose is to study theexistence of nonnegative time periodic solutions for one-dimensional Nicholsons blowflies model with Newtonian diffusion. Froma practical point of view, the population of the blowflies changes periodically over time. In the present paper, we try to explainthis phenomenon from mathematical point of view. By constructing some suitable Lyapunov functionals, the a priori estimatesfor all possible periodic solutions are obtained, then combining with LeraySchauder fixed point theorem, we finally establish theexistence of time periodic solutions. In the last part of the paper, we will give an numerical example to valid our theoretic result.

2. The main result and its proof

Because of the degeneracy for Equation (1), the problem (1)(2) might not have classical solutions, in general, and hence we firstintroduce the definition of weak solutions in the sense of the following

Definition 2.1A function u is said to be a weak solution of the problem of (1)(2), if

u{w;wL,wmL(0, T;W1,20 (0,1)),

wm

tL2(Q)

}

and for any C(Q) with (x,0)=(x,T) and (0, t)=(1, t)=0, the following integral equality holds: T0

10

{ut

um

xx

u+pu(t)eau(t)+g(x, t)+( tt

e(ts)u(x, s)ds)

}dxdt=0

The main result of this paper is the following.

Theorem 2.1Let 0gC(Q), g(x, t) 0 and g(x, t+T)=g(x, t) for some T>0, where Q= (0,1)(0, T). Then the problem (1)(2) has at least onenonnegative T-periodic solution.

To study the existence of periodic solutions, let us first consider the regularized problem

ut

= 2

x2(u+um )u+pu(t)eau(t)+g(x, t)+

tt

e(ts)uds, x (0,1), tR (5)

u(0, t) = u(1, t), tR (6)u(x, t) = u(x, t+T), x [0,1], tR (7)

The desired solution of the problem (1)(2) will be obtained by the limit of some subsequence of solutions u of the regularizedproblem. However, we need first to establish the existence of solutions u, which can be done by using the LeraySchauder fixedpoint theorem. For this purpose, we need the following lemma for the a priori estimates on solutions u.

Lemma 2.1Assume that u is a nonnegative T-periodic solution of the equation

ut= 2

x2(u+um)+

(u+pu(t)eau(t)+g(x, t)+

tt

e(ts)u(x, s)ds)

(8)


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Y. YANG ET AL.

satisfying the boundary value condition (6), where [0,1], 00 is a constant which depend on m and r.

ProofMultiplying Equation (8) by ur and integrating over Q, we obtain

T0

10

utur dxdt =

T0

10

2

x2(u+um) ur dxdt+

(

T0

10

ur+1 dxdt+ T0

10

pu(t)ureau(t) dxdt+ T0

10

gur dxdt

+ T0

10

ur tt

e(ts)u(x, s)dsdxdt)

Since u is T-periodic, T0

10

ut

ur dxdt= 1r+1

T0

10

ur+1t

dxdt=0

then we have T0

10

x

(u+um)ur

xdxdt =

T0

10

ur+1 dxdt+p T0

10

u(t)ureau(t) dxdt+ T0

10

g(x, t)ur dxdt

+ T0

10

ur tt

e(ts)u(x, s)dsdxdt

p T0

10

u(t)ureau(t) dxdt+ T0

10

g(x, t)ur dxdt+ T0

10

ur tt

e(ts)u(x, s)dsdxdt

C T0

10

ur dxdt+ T0

10

ur tt

e(ts)u(x, s)dsdxdt

On the other hand, we have

T0

10

ur tt

e(ts)u(x, s)dsdxdt e|| T0

10

ur tt

u(x, s)dsdxdt

e|| T0

10

ur T

u(x, s)dsdxdt

e||

1

T0

10

um+r dxdt+r/m1 T0

10

( T

u(x, s)ds

)(m+r)/mdxdt

= e||1

T0

10

um+r dxdt+r/m1 T 10

( T

u(x, s)ds

)(m+r)/mdx

e||[1

T0

10

um+r dxdt+r/m1 T(T+)r/m 10

T

u(x, s)(m+r)/mdsdx]

e||[1

T0

10

um+r dxdt+r/m1 T(T+)r/m([

T

]+1

) 10

T0

u(x, s)(m+r)/mdsdx]

e||[1

T0

10

um+r dxdt+r/m1 T(T+)r/m([

T

]+1

)2

10

T0

u(x, s)m+r dsdx

+r/m1 T2(T+)r/m([

T

]+1

)1/ (m1)2

]

1C T0

10

um+r dxdt+r/m1 2C 10

T0

u(x, s)m+r dsdx+r/m1 1/ (m1)2 C

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Y. YANG ET AL.

Then we know that T0

10

x

(u+um)ur

xdxdt C

T0

10

ur dxdt+ T0

10

ur tt

e(ts)u(x, s)dsdxdt

0C T0

10

um+r dxdt+Cr/m0 +1C T0

10

um+r dxdt

+r/m1 2C 10

T0

u(x, s)m+r dsdx+r/m1 1/ (m1)2 C

On the other hand, we can see

T0

10

x

(u+um)ur

xdxdt =

T0

10

(+mum1)rur1u2x dxdt

T0

10

mrum+r2u2x dxdt

= 4mr(m+r)2

T0

10

x u(m+r)/22

dxdt

Therefore, we obtain

4mr

(m+r)2 T0

10

x u(m+r)/22

dxdt 0C T0

10

um+r dxdt+Cr/m0 +1C T0

10

um+r dxdt+r/m1 2C 10

T0

u(x, s)m+r dsdx

+r/m1 1/ (m1)2 C

(0+1+r/m1 2)C T0

10

x u(m+r)/22

dxdt+C

Letting 0, 1, we have T0

10

x u(m+r)/22

dxdtC(m,r) (9)

Using Poincar inequality, we get T0

10

um+r dxdtC1(m,r)

which completes the proof of Lemma 2.1.

Lemma 2.2Assume that u is a nonnegative T-periodic solution of Equation (8) satisfying the boundary value condition (6). Then we have

T0

10

umx2

dxdtC

where C>0 is a constant.

ProofIn fact, choosing r=m in (9), we obtain

T0

10

umx2

dxdtC


T0

10

x (u+um)2

dxdtC

where C>0 is a constant.


925

Y. YANG ET AL.

ProofIn fact, by direct computations, we have

T0

10

x (u+um)2

dxdt = T0

10

x

(u+um) x

(u+um)dxdt

= T0

(u+um)(ux+mum1ux)1

0

dt T0

10

2

x2(u+um)(u+um)dxdt

= T0

10

2

x2(u+um)(u+um)dxdt

= T0

10

(u+um)(utu+pu(t)eau(t)+g+

tt

e(ts)u(x, s)ds)

dxdt

= T0

10

[(u+um)utu(u+um)+pu(t)(u+um)+g(u+um)

+(u+um) tt

e(ts)u(x, s)ds]dxdt

T0

10

pu(t)(u+um)dxdt+ T0

10

g(u+um)dxdt+ T0

10

u(u+um)dxdt

+ T0

10

(u+um) tt

e(ts)u(x, s)dsdxdt

p T0

10

u2(t)dxdt+p T0

10

u2 dxdt+p T0

10

u2(t)dxdt+ T0

10

g2 dxdt

+p T0

10

|um|2 dxdt+ T0

10

u2 dxdt+ T0

10

g2 dxdt+ T0

10

|um|2 dxdt

+ T0

10

u2 dxdt+ T0

10

um+1 dxdt+ T0

10

u

tt

e(ts)u(x, s)dsdxdt

+ T0

10

um tt

e(ts)u(x, s)dsdxdt

C T0

10

u2(t)dxdt+C T0

10

u2 dxdt+ T0

10

g2 dxdt+C T0

10

|um|2 dxdt

C


uL(Q)Cwhere C>0 is a constant.

ProofDenote u(t) :=u(x, t), u(t+) :=u(x, t+).

Define

V(t)= 10

(1

2u2+ 1

2

x (u+um)2)

dx+ 0

10

u2(t+)dxd (10)

Therefore, we conclude that

V (t) = 10

[uut+ x (u+u

m)t

(x

(u+um))]

dx+ 10

[u2u2(t)]dx

= 10

[uut

2

x2(u+um)(+mum1)ut

]dx+

10

[u2u2(t)]dx

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Y. YANG ET AL.

= 10

u

[2

x2(u+um)u+pu(t)eau(t)+g(t, x)+

tt

e(ts)u(x, s)ds]dx

+ 10

(+mum1)ut[utu+pu(t)eau(t)+g(t, x)+

tt

e(ts)u(x, s)ds]dx+

10

[u2u2(t)]dx

10

[ x

(u+um)ux

+u2+pu(t)ueau(t)+|u||g(x, t)|]dx+

10

[fu2t +f |ut||u|+pfu(t)|ut|+f |ut||g|]dx

+ 10

[u2u2(t)]dx+ 10

u

tt

e(ts)u(s)dsdx+ 10

f |ut| tt

e(ts)u(s)dsdx

10

(mum1

ux2

+(+1)u2)

dx+ 10

[pu(t)u+ugfu2t +fu|ut|+pfu(t)|ut|+f |ut|g]dxdt

+ 10

u

tt

e(ts)u(s)dsdx+ 10

f |ut| tt

e(ts)u(s)dsdx

= 10

(mum1

ux2

+(+1)u2)

dx+ 10

Fdx

since 1

0u2 dx3

10

um+1 dx+2/ (m1)3 3 10

u(m+1)/2x

2

dx+2/ (m1)3

c1 1

0u2 dx+

10

u(m+1)/2x

2

dx

+(+2+c1)2/ (m1)3 +

10

Fdx

where

0 < 30

f = +mum1

F = fu2t +fu|ut|+pfu(t)|ut|+f |ut||g|+pu(t)u+u|g|+u tt

e(ts)u(s)dsdx+fut tt

e(ts)u(s)ds

Define

c2= (+2+c1)2/ (m1)3then we have

V (t)c1 1

0u2 dx+

10

u(m+1)/2x

2

dx

+c2+

10

Fdx (11)

On the other hand, we can see

T0

10

Fdxdt T0

10

[fu2t +fu|ut|+pfu(t)|ut|+f |ut||g|+pu(t)u+u|g|+u

tt

e(ts)u(s)ds

+fut tt

e(ts)u(s)ds]dxdt

T0

10

[fu2t +

4

2fu2t +

24fu2+ 5p

2fu2t +

p

25fu2(t)+ 6

2fu2t

+ 126

f |g|2+pu2(t)+pu2+u2+g2+f |ut| tt

e(ts)u(s)ds]dxdt+C

T0

10

um+1 dxdt+C


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Y. YANG ET AL.

Since, T0

10

f |ut| tt

e(ts)u(s)dsdxdt e|| T0

10

f |ut| tt

u(s)dsdxdt

e||[72

T0

10

fu2t dxdt+1

27

T0

10

f

( tt

u(s)ds

)2dxdt

]

e||[72

T0

10

fu2t dxdt+

27

T0

10

f

tt

u2(s)dsdxdt

]

e||[72

T0

10

fu2t dxdt+

27

T0

10

T

u2(s)dsdxdt+m27

T0

10

um1 T

u2(s)dsdxdt

]

e||[72

T0

10

fu2t dxdt+T

27

([ T

]+1

) T0

10

u2 dxdt+ m27

T0

10

u2mdxdt

+m27

T0

10

( t

u2(s)ds

)2dxdt

]

e||[72

T0

10

fu2t dxdt+T

27

([ T

]+1

) T0

10

u2 dxdt+ m27

T0

10

u2mdxdt

+mT27

(T+)([

T

]+1

) T0

10

u4 dxdt

]

Choosing 4, 5, 6, 7 small appropriately, we obtain, T0

T0

Fdxdt

Q

[(1+ 4

2+ 5p

2+ 6

2+ 7e

||2

)fu2t +

24fu2+ p

25fu2(t)

+ 126

f |g2|+pu2(t)+(p+1)u2+g2+e|| 72

T0

10

fu2t dxdt

+e|| T27

([ T

]+1

) T0

10

u2 dxdt+e||m27

T0

10

u2mdxdt

+e||mT27

(T+)([

T

]+1

) T0

10

u4 dxdt

]dxdt+C

T0

10

um+1 dxdt+C

By Lemma 2.1, uLrC, for any r>0, we obtain T0

10

FdxdtC (12)

Since V(t) and u are T-periodic and from (11), (12), we get

0= T0

V (t)dtc1 T

0

10

u2 dxdt+ T0

10

u(m+1)/2x

2

dxdt

+c2T+

T0

10

Fdxdt

That is

T0

10

u2 dxdtC, T0

10

u(m+1)/2x

2

dxdtC

Then by Lemmas 2.1 and 2.3, we have T0

V(t)dt T0

10

(1

2u2+ 1

2

x (u+um)2)dxdt+

T0

0

10

u2(t+)dxddtC

Since V is continuous, there exists that a t0 [0, T] satisfies

V(t0)C

TC

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Y. YANG ET AL.

Hence, if t0tt0+T , we obtain

V(t) = V(t0)+ tt0V (s)ds

C+ tt0

c1

10

u2+

u(m+1)/2x

2 dx+c2+

10

Fdx

ds

C+ T0

10

c1

u2+

u(m+1)/2x

2+F

dxdt+c2T

C

On the other hand, we can see that

10

x (u+um)2

dx = 10

|(+mum1)ux|2 dx

10

|mum1ux|2 dx= 10

umx2

dx

By the definition of V , we see that

10

(u2+

umx2)

dx 10

(u2+

x (u+um)2)dx2V(t)C

which implies

sup0tT

10

(u2+

umx2)dxC

It follows from the definition of u that

|u(x, t)|m=|um(x, t)| =um(0, t)+

x0

um(s, t)s

ds

= x0

um(s, t)s

ds

10

umx dx

(

sup0tT

10

umx2

dx

)1/2C1/2

that is

uL(Q)C1/ (2m) :=C0which completes the proof of Lemma 2.4.

By means of the above proved lemmas and the LeraySchauder fixed point theorem, we can obtain the existence of solutionsu of the regularized problem as follows.

Lemma 2.5The regularized problem (6)(8) has a nonnegative T-periodic solution.

ProofDenote by CT (Q) the set of all continuous functions u with the T-periodicity in t. We study the following regularized equation:

ut

= 2

x2(u+|u|m1u)u+g(x, t) (13)

where 0gCT (Q). We claim that, if the problem (13), (6), (7) has a unique T-periodic solution u, then u must be nonnegative. Infact, multiplying (13) by u and integrating over Q, we obtain

Q

ut

udxdt=

Q

2

x2(u+|u|m1u)udxdt

Qu2dxdt+

Qgudxdt


929

Y. YANG ET AL.

where u =min{0, u(x, t), (x, t)Q}. Making use of integral by part, we have Q

x

(u+|u|m1u)ux dxdt+

Qu2dxdt=

Qgudxdt0

Since Q

x

(u+|u|m1u)ux dxdt=

Q(+m|u|m1)

ux2

dxdt0

then we get

Qu2dxdt0

Therefore,

u =0, a.e. in QBy the definition of u, we see that

u0, a.e. in Q

Consequently, we can rewrite the Equation (13) as

ut

= 2

x2(u+um)u+g(x, t), x (0,1), tR (14)

Hence, we know that, if the problem (14), (6), (7) has a T-periodic solution, it must be nonnegative.On the other hand, with an argument similar to [14], we claim that for any gCT (Q), the problem

ut

= 2

x2(u+um)+g(x, t), x (0,1), tR

u(0, t) = u(1, t), tRu(x, t) = u(x, t+T), x (0,1), tR

has a unique solution uC(Q). By constructing a homotopy, it is easy to obtain that the problem (14), (6), (7) admits a solutionuC(Q), and we can also see that it is nonnegative. According to Equation (14), we see that uC2,1T (Q).

Next, we consider the periodic problem of the homotopy equation for regularized problem

ut

= 2

x2(u+um)u+G(x, t), x (0,1), tR (15)

u(0, t) = u(1, t)=0, tR (16)where for any v(x, t)CT (Q),

G(x, t)=pv(x, t)eav(x,t)+g(x, t)+ tt

e(ts)u(x, s)ds

Then problem (15)(16) admits a unique solution uC2,1T (Q). Define the mapping

L :CT (Q)[0,1]CT (Q), (v,) uAs C2,1T (Q) can be compactly embedded into CT (Q), L is compact. By Lemma 2.4, we know that for any fixed point u of the mappingL there is a constant C0 independent of and , such that

uLC0Then in applying LeraySchauders fixed point theorem, we know that the problem (5)(7) admits a solution u.

In addition, we can also easily get the L2 boundedness of um / t as follows.

Lemma 2.6Assume that u is a nonnegative T-periodic solution of Equation (5) satisfying the boundary value condition (6). Then there exists aconstant C>0 such that

Q

umt2

dxdtC

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Y. YANG ET AL.

ProofMultiplying Equation (5) by ( / t)(u+um) and integrating over Q, we obtain

Q

ut

t

(u+um)dxdt =

Q

2

x2(u+um)

t(u+um)dxdt

Qut

(u+um)dxdt

+

Q[pu(t)eau(t)+g(x, t)]

t(u+um)dxdt

+

Q tt

e(ts)u(s)ds t

(u+um)dxdt

By direct computations, we have

T0

10

2

x2(u+um)

t(u+um)dxdt=1

2

T0

10

t

(x

(u+um))2

dxdt=0

Therefore, by Lemma 2.4, we obtain

T0

10

u2t dxdt+ T0

10

mum1u2t dxdt p T0

10

u(t)eau(t) t

(u+um)dxdt+ T0

10

g(x, t)t

(u+um)dxdt

+

Q

( tt

e(ts)u(s)ds)

t

(u+um)dxdt

C T0

T0

t (u+um) dxdt

2C72

T0

10

|ut|2 dxdt+ C27

+ C82

T0

10

|mum1ut|2 dxdt+ C28

T0

10

|ut|2 dxdt+C+ C82

T0

t0mum1|ut|2 dxdt

where 7=2 / (C). Letting 8 small appropriately, we have T0

10

mum1u2t dxdtC

Then, we can see

T0

10

umt2

dxdt = T0

10

m2u2(m1)u2t dxdt

mCm1 T0

10

mum1u2t dxdtC

which completes the proof of Lemma 2.6.

We are now in a position to present the proof of the existence of periodic solutions for the problem (1)(2).

The Proof of Theorem 2.1Let =1 / n (n=1,2,. . .) and we note un for the solution of the problem (6)(8). According to Lemmas 2.2, 2.4 and 2.6, we see that

unL(Q) C0umnxL2(Q)

C1T

umnt2

L2(Q) C

Hence, there exists a subsequence {un}n=1, supposed to be {un}n=1 itself, and a function

u{u;uL; umL(0, T;W1,20 (0,1));

um

tL2(Q)

}


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Y. YANG ET AL.

such that

un(x, t) u(x, t) in L2(Q)

umn (x, t)x

um(x, t)

xin L2(Q)

pun(x, t)eaun(t) pu(x, t)eau(t) in L2(Q)Furthermore, for any C(Q) with (x,0)=(x,T) and (0, t)=(1, t)=0, one has

T0

10

1

nun

2x2

dxdt0

and T0

10

( tt

e(ts)un(x, s)ds)dxdt

T0

10

( tt

e(ts)u(x, s)ds)dxdt

Letting n in T0

10

{unt+

1

nun

2x2

umn

xx

un+pun(t)eaun(t)+g(x, t)+( tt

e(ts)un(x, s)ds)

}dxdt=0

we have T0

10

{ut

um

xx

u+pu(t)eau(t)+g(x, t)+( tt

e(ts)u(x, s)ds)

}dxdt=0

we see that u satisfies the integral identity in the definition of weak solutions. Hence, the problem of (1)(2) admits a nonnegativeT-periodic solution

u{u;uL; umL(0, T;W1,20 (0,1));

um

tL2(Q)

}

Since g(x, t) 0, we see that the T-periodic solution is nontrivial. The proof of Theorem 2.1 is complete.

3. Numerical simulation

In this section, we give an example to valid our theoretic result in Section 2.Setting m=2, =3, p=a=1 and =0 in (1), we get following test example:

ut

= 2(u2)

x23u+u(x, t)eu(x,t)+g(x, t), x (0,1) (17)

where we choose =0.1 and g(x, t)0 given by

g(x, t) =(3+ 1

2cos(2t)

)x(1x)exp

(sin (2t)

4

)+2exp

(sin (2t)

2

)

x(1x)exp

(sin (2(t))

4

)exp

[x(1x)exp

(sin (2(t))

4

)]

Next, we will implement the numerical simulation in two steps: (1) use finite difference method to get a large sparse system ofnonlinear equations and (2) use the Newton-GMRES subspace method to solve the large nonlinear system.

Step 1: Set xn=nh, n=1,. . . ,N+1 with h=0.01 and N=100 in spatial direction, and tm=mk, m=1,. . . ,M+1 with k=0.01 (herewe choose k such that =m0k for some positive integer m0) and M=100 in temporal direction. We want to obtain numericalsolution at (xn, tm), denoted by {umn }m=1,. . .,Mn=1,. . .,N . The test example (17) can be rewritten as

ut

=2u2u

x2+2

(ux

)2u+u(x, t)eu(x,t)+g(x, t), x (0,1) (18)

hence, we propose following finite difference scheme:

um+1n umnk

= 2um+1num+1n+1 2um+1n +um+1n1

h2+2

(um+1n+1 um+1n1

2h

)2um+1n +u(m+1)m0n exp{u(m+1)m0n }

+g(xn, tm+1), n=2,. . . ,N, m=1,. . . ,M (19)

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Y. YANG ET AL.

Figure 1. Numerical simulation for test example.

which is a second-order scheme in the spatial direction and back Euler scheme in the temporal direction. The boundary conditionsare given by

um1 =umN+1=0, m=1,. . . ,M+1 (20)and the periodic condition is given by

u1n=uM+1n , n=1,. . . ,N+1 (21)The scheme (19) can be rewritten as

(1+k)U(m+1) = U(m)+kFm+1(U(m+1)), m=1,. . . ,MU(1) = U(M+1)

(22)

where

U(m) = (um2 ,. . . , umN )T, m=1,. . . ,M+1

(Fm(U(m)))n = 2umn

umn+12umn +umn1h2

+2(umn+1umn1

2h

)2+umm0n exp{umm0n }+g(xn, tm), n=2,. . . ,N, m=1,. . . ,M+1

and

umm0n =uM+mm0n , n=2,. . . ,N, m=1,. . . ,m0Step 2: The well-known solver for the nonlinear system (22) is the Newton method. However, at each iteration step it needs the

exact solution of the corresponding Newton equation, which is very costly in actual applications. There are many efficient tricks[15--19] to improve the Newton method, here, we use the so-called Newton-GMRES subspace method [20] to solve (22). We setthat the tolerance is 106 and the numerical result are shown by Figure 1.

Acknowledgements

This work is partially supported by the National Science Foundation of China, partially supported by a Specific Foundation for PhDSpecialities of Educational Department of China and partially supported by the Foundation for Post Doctor of China.

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