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Global Asymptotic Behavior of a Predator-Prey Diffusion System with Beddington-DeAngelis Function Response

2014-05-03 12:48:12MENGYijieandXIAOShiwu
Journal of Partial Differential Equations 2014年2期

MENG Yijieand XIAO Shiwu

School of Mathematics and Computer Science,Hubei University of Arts and Science, Xiangfan 441053,China.

Global Asymptotic Behavior of a Predator-Prey Diffusion System with Beddington-DeAngelis Function Response

MENG Yijie?and XIAO Shiwu

School of Mathematics and Computer Science,Hubei University of Arts and Science, Xiangfan 441053,China.

Received 14 April 2013;Accepted 14 October 2013

.Inthispaper,westudyaclassofreaction-diffusionsystemswith Beddington-DeAngelis function response.The global asymptotic convergence is established by using the comparison principle and the method of monotone iterations,which is via successive improvement of upper-lower solutions function.

Predator-prey diffusion system;asymptotic behavior;Beddington-DeAngelis function.

1 Introduction

It is the purpose of this paper to study the global asymptotic behavior of solutions to the predator-prey diffusion system with Beddington-DeAngelis function response and the homogeneous Neumann boundary condition,

where ??RN(N≥1)is a bounded domain with smooth boundary??,u and v represent the population densities of prey and predator,ν is the outward unit normal vector of the boundary??.The constant d1and d2,which are the diffusion coefficients,are positive.a, b,r,m and k are positive constants.The initial data u0(x),v0(x)are continuous functions.

It is known that there exist three equilibria(0,0),(1,0)and(?u,?v)provided that 0<k<(1+a)-1,where?u and?v are positive and satisfy

where

We note that(1.1)has a unique nonnegative global solution(u,v).In addition,if u06≡0,v06≡0,then the solution(u,v)is positive,i.e.,u(x,t)>0,v(x,t)>0 on ?,for all t>0.

In population dynamics,the prey-predator system with Beddington-DeAngelis function response has been extensively studied in[1-6].Reaction-diffusion systems with delays have been treated by many authors.However,most of the systems are mixed quasimonotone,and most of the discussions are in the framework of semi-group theory of dynamical systems[7-10].The method of upper and lower solutions and its associated monotone iterations have been used to investigate the dynamic property of the system, which is mixed quasimonotone with discrete delays[11-13].In[6],the author discussed the dissipation,persistence and the local stability of nonnegative constant steady states for(1.1).In this paper,we give sufficient conditions for the global asymptotic behavior of solutions of(1.1).The method of proof is via successive improvement of upper-lower solutions of some suitable systems,see[14,15].

2 Main results and proof

In thus section,we discuss the global asymptotic behavior of solutions by using the comparison principle and the method of monotone iterations.

Firstly,we give two results in[6].

Lemma 2.1.If k≥(1+a)-1,and b≤m,then

provided that u06≡0.

Lemma 2.2.If k<(1+a)-1and,then the positive constant solution(u?,v?)of(1.1)is locally stable.

Now,we discuss the global asymptotic stability of the solutions of(1.1)(?u,?v).Theorem 2.1.If(1+2a)-1<k<(1+a)-1and b<m,then

provided that u06≡0,v06≡0,where(?u,?v)is given by(1.2).

Proof.From(1.1),we know u satisfies

it follows by the comparison principle that

Thus,for any ?>0,there exists T1>0,such that

It then follows that v satisfies

Let w(t)be a solution of the following ordinary differential equation

Since k<(1+a)-1,for any ?>0,we have(1-k)(1+?)-ka>0.Then,

From the comparison principle,it follows that v(x,t)≤w(t).Thus,we get

From the arbitrariness of ?>0,we can get that

Thus,for any ?>0,there exists T2(≥T1),such that

Therefore,

thus,by the direct computation,we have

An application of the comparison principle gives

The arbitrariness of ? implies that

It is obvious that

thus,we have

By(2.3),for any sufficiently small ?>0,there exists T≥T3(≥T2),such that

Therefore,v satisfies

Since k<(1+a)-1and b<m,we have

The sufficiently small ? implies that

By the same comparison argument,we get

the arbitrariness of ? implies that

It is obvious that

Thus,we have

From(2.1 and(2.5),for any ?,0<??1,there exists T4(≥T3),such that

It follows that u satisfies

Since

thus,for sufficiently small ?>0,

So,by the comparison principle,we get

The arbitrariness of ? implies that

It is obvious that u2≤u1.

From(2.7),for any ?>0,there exists T5(≥T4),such that

It follows that v satisfies

Since

so,by the comparison principle,we get

The arbitrariness of ? implies that

Since

we get

From(2.1)and(2.5),For any ?,0<??1,there exists T6(≥T5),such that

It follows that u satisfies

Since u1≤u2≤u1.,and v1≤v2≤v1.,we have

so,for sufficiently small ?>0,

By by the comparison principle,we get

The arbitrariness of ? implies that

Sine

we have

Thus,we get

From(2.9),for any ?>0,there exists T7(≥T6),such that

It follows that v satisfies

Let Z(t)be a solution of the ordinary differential equation,

By u2(1-k)-ka≥u1(1-k)-ka>0,and the sufficiently small ?>0,we have

Thus,we get

The comparison principle gives that v(x.t)≥Z(t)for all x∈? and t≥T7,such that

The arbitrariness of ? implies that

Since

we have

Thus,we get

Define the sequences un,un,vn,vn(n≥1)as follows

Lemma 2.3.For the above defined sequences,we have

and the solution(u(x,t),v(x,t))satisfies

and

Proof.For n=1,2,we have shown that,and

and

Using induction and repeating the above process,we can complete the proof,and omit the detail.

Lemma2.3implies thatlimn→∞un,limn→∞un,limn→∞vnandlimn→∞vnexist,denoted as u,u,v,v,respectively.It is obvious that 0<u≤u and 0<v≤v.

Thus,u,u,v,and v satisfy that

and

and

Substituting the second equality into the third equality,and by straightforward computation,we have

Substituting the forth equality into the first equality,and by straightforward computation,we have

Let(2.19)minus(2.18),and by straightforward computation,we get

Since

and

from condition k>(1+2a)-1,so,we get u+u>1.Therefore,we can get(b-m)(1-k)+ mk(1-u-u)<0,such that u=u,from(2.20).

Now,from(2.17),we have

By(1-k)/mk>0 and u=u,it follows that v=v.Thus,we get u=u=?u,and v=v=?v, such that

The proof is complete.

Acknowledgments

This work is supported by the Science and Technology Research Plan of the Education Department of Hubei Province(Q20122504 and D20112605).

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10.4208/jpde.v27.n2.3 June 2014

?Corresponding author.Email addresses:yijie-meng@sina.com(Y.Meng),xshiwu@sina.com(S.Xiao)

AMS SubjectClassifications:35B35,35K51

Chinese Library Classifications:0193.26


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