# Positive Steady States of a Strongly Coupled Predator-Prey System with Holling-(n+1) Functional Response

Authors: Xiao-zhou Feng; Zhi-guo Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper discusses a predator-prey system with Holling-(n+1) functional response and the fractional type nonlinear diffusion term in a bounded domain under homogeneous
Neumann boundary condition. The existence and nonexistence results concerning nonconstant
positive steady states of the system were obtained. In particular, we prove that the positive constant solution (u~,v~) is asymptotically stable when the parameter k satisfies some conditions.


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## Body

## 1. Introduction

In this paper, we are interested in the positive steady states of the strongly coupled predator-prey system with Holling-(n+1) functional response. The specific system is as follows:
(1)-d1Δu=u(1-uk)-unva+uninΩ,-d2Δ(v+d3v1+u)=-bv+σunva+uninΩ,∂u∂ν=∂v∂ν=0on∂Ω,
where 1≤n<+∞; Ω is a bounded domain in RN with smooth boundary ∂Ω; ∂/∂ν is the outward directional derivative normal to ∂Ω; u and v stand for the densities of the prey and predator; the given coefficients di(i=1,2), a, b, k, and σ are positive constants. The term un/(a+un) is named Holling-(n+1) functional response [1, 2]. In the second equation, the fractional type nonlinear diffusion term Δd2(d3v/(1+u)) models a situation in which the population pressure of the predator species weakens in high-density areas of the prey species. For more precise details, we can refer to [3, 4]. Paper [3] discusses a strongly coupled predator-prey system with nonmonotonic functional response, the existence and nonexistence results concerning nonconstant positive steady states of the system were proved by degree theory. Paper [4] considers the positive steady states for a prey-predator model with some nonlinear diffusion terms, and the sufficient conditions for the existence of positive steady state solutions were obtained by bifurcation theory.In recent years, there has been considerable interest in the dynamics of strongly coupled reaction-diffusion systems with cross-diffusion. We point out that most efforts have concentrated on the Lotka-Volterra competition system which was proposed first by Shigesada et al. [5]. Since their pioneering work, many authors have studied population models with cross-diffusion terms from various mathematical viewpoints, for example, the global existence of time-depending solutions [6–11], the stability analysis for steady states [12–14], and the steady state problems [15–21]. In this paper, we mainly consider the existence of solutions of (1). The research method refers to [3, 22, 23].For convenience of the research, we write (1) as the following form:
(2)-ΔΦ(U)=G(U),U=[uv],
where
(3)Φ(U)=[ϕ1(U)ϕ2(U)]=[d1ud2v(1+d31+u)],G(U)=[g(u)-vp(u)-bv+σvp(u)],(4)g(u)=u(1-uk),p(u)=una+un,n∈N+.The main work of this paper is to study the effects of the fractional type nonlinear diffusion pressures on the existence of nonconstant positive steady states of (1). Here, a positive solution means a smooth solution (u,v) with both u and v being positive. We will demonstrate that the cross-diffusion pressure d3 may help forming more patterns. Obviously, for system (1), one notes that when σ≤b, there holds sups≥0{-b+σsn/(a+sn)}≤0, so that the only nonnegative solutions to (1) are U=(0,0) and U=(k,0). Consequently, (1) does not have any positive solution. On the other hand, when σ>b, the unique positive constant solution to (1) is (u~,v~); that is,
(5)u~=abσ-bn,v~=(1-u~/k)(a+u~n)u~n-1=(k-u~)σu~kb,ifk>u~.The organization of our paper is as follows. In Section2, we establish a priori upper and lower bounds for positive solutions of (1). In Section 3, we use a degree theory to develop a general result to enable one to conclude the existence or nonexistence of nonconstant steady-state solutions or patterns as the index of positive constant steady states changes. In Section 4, we establish the existence of nonconstant positive solutions to (1) for a large range of diffusion and cross-diffusion coefficients. Meanwhile, we prove that the positive constant solution (u~,v~) is asymptotically stable for different ranges of parameters.


## 2. Upper and Lower Bounds for Positive Solutions

The main purpose of this section is to give a priori upper and lower positive bounds for positive solutions of (1). Firstly, we cite two known results.Harnack Inequality(see [24]). Let c∈C(Ω¯) and ω∈C2(Ω)⋂‍C1(Ω¯) be a positive classical solution to Δω(x)+c(x)ω(x)=0 in Ω subject to the homogeneous Neumann boundary condition. Then, there exists a positive constant C=C(Ω,∥c∥∞) such that maxΩ¯ω≤CminΩ¯ω.Maximum Principle(see [25]). Suppose that g∈C(Ω¯×R1). If w∈C2(Ω)∩C1(Ω¯) satisfies
(6)Δw(x)+g(x,w(x))≥0inΩ,∂w∂ν≤0on∂Ω,
and w(x0)=maxΩ¯w, then g(x0,w(x0))≥0.Theorem 1.
LetP=(σ,a,k,b)∈(0,∞)4 and d, d^>0 be fixed constants. Assume that d1, d2≥d and 0≤d3≤d^ there exists a positive constant C=C(P,d,d^,Ω), such that any positive solution (u,v) of (1) satisfies
(7)maxΩ¯u(x)+maxΩ¯v(x)+maxΩ¯u(x)minΩ¯u(x)+maxΩ¯v(x)minΩ¯v(x)≤C.Proof.
Letϕ1(U) and ϕ2(U) be defined as in (3), and denote that
(8)g1(U)=u(1-uk-un-1va+un),g2(U)=v(σuna+un-b),G(U)=(g1(U),g2(U))T.
Then, (1) becomes
(9)-ΔΦ(U)=G(U)inΩ,∂U∂ν=0on∂Ω.
For the first equation of (9), by the Maximum Principle, we have
(10)maxΩ¯u(x)≤1d1maxΩ¯ϕ1(U)<k.
The functionϕ2(U) satisfies
(11)Δϕ2(U)+c(x)ϕ2(U)=0inΩ,∂ϕ2(U)∂ν=0on∂Ω,
where c(x)=g2(U)/ϕ2(U)∈C(Ω¯). It is easy to verify that the norm
(12)∥c∥∞=∥v(σun/(a+un)-b)d2(v+d3v/(1+u))∥∞≤σkn/(a+kn)+bd2≤σkn+b(a+kn)d(a+kn);
then, ∥c∥∞ is bounded by a constant depending only on P and d(d2≥d). By the Harnack inequality, we have
(13)maxΩ¯ϕ2(U(x))≤C1minΩ¯ϕ2(U(x)),
where C1=C1(Ω,∥c∥∞)=C1(P,d,Ω). It follows that
(14)maxΩ¯vminΩ¯v≤maxΩ¯ϕ2(U(x))minΩ¯ϕ2(U(x))×maxΩ¯{1+d3(1+u)-1}minΩ¯{1+d3(1+u)-1}≤C1(1+d3)≤C1(1+d^).
Integrating the equations of (1) over Ω by parts and making use of the boundary conditions, we have
(15)σ∫Ω‍u(1-uk)dx=b∫Ω‍vdx.
For any f∈L1(Ω), we denote that f¯=(1/|Ω|)∫Ω‍f(x)dx. Then, from (15) and (10), we have
(16)v¯=σ|Ω|b∫Ω‍u(1-uk)dx<σbk.
Along with (14), we have
(17)maxΩ¯v≤C1(1+d^)σbk.
Similarly, consider the equation of ϕ1 as follows:
(18)∥c∥∞=1d1∥1-uk-un-1va+un∥∞≤1d1[1+1k∥u(x)∥∞+m∥v(x)∥∞],
where m=[(n-1)a](n-1)/n/na. Then, ∥c∥∞ is bounded by a constant depending only on P and d(d1≥d). By the Harnack Inequality, there exists a positive constant C2=C2(P,d,d^,Ω) such that maxΩ¯ϕ1(U(x))≤C2minΩ¯ϕ1(U(x)). Then,
(19)maxΩ¯uminΩ¯u=maxΩ¯ϕ1(U(x))minΩ¯ϕ1(U(x))≤C2.
Thus, along with (10), (14), and (17), we can complete the proof.Theorem 2.
Assume thatσkn≠b(a+kn). Let d, d^>0 be fixed constants. There exists a constant C_=C_(d,d^,P,Ω)>0, such that any positive solution (u,v) of (1) satisfies
(20)minΩ¯u(x)>C_,minΩ¯v(x)>C_,
provided that d1,d2≥d and d3≤d^.Proof.
Since∫Ω‍g2(U)dx=0, there exists x1∈Ω, such that g2(U(x1))=0; that is,
(21)σun(x1)=ba+bun(x1).
It follows that maxΩ¯u≥ab/σn. Consequently, by maxΩ¯u/minΩ¯u≤C in Theorem 1, we have
(22)minΩ¯u≥1CmaxΩ¯u≥ab/σnC∶=C3=C3(d,d^,P,Ω),
where  d3≤d^ and d1,d2≥d.
In the following, we mainly prove thatminΩ¯v(x)>C_ as σkn≠b(a+kn). Now, we suppose that claim is not true; then, there exists a sequence (d1,i,d2,i,d3,i)→(d1,d2,d3) with d1,i,d2,i≥d and d3,i≤d^. And the positive solution (ui,vi) of (1) corresponding to (d1,i,d2,i,d3,i) is such that
(23)ui⟶u,minΩ¯vi(x)⟶0,asi⟶∞.
By (14), we have maxΩ¯vi(x)/minΩ¯vi(x)≤C1(1+d^), and then, maxΩ¯vi(x)→0; furthermore, vi(x)→0  uniformly holds as i→∞. Then, the first equation of (1) becomes the following:
(24)-Δui=1d1,i(1-uik-uin-1via+uin)uiinΩ,∂ui∂ν=0on∂Ω.
According to (7), there exists C=C(d,d^,P,Ω) such that ∥ui∥∞≤C,  ∥(1-ui/k)-(uin-1vi/(a+uin))/d1,i(1+d3,ivi)∥∞≤C, where C is a positive constant which does not depend on i. For each i given in problem (24), it follows from Lp estimate that ∥ui∥W2,p(Ω)≤C2|Ω|1/p, where p>1. Let p>N; then, by Sobolev Imbedding Theorems, we get
(25)∥ui∥C1,α(Ω)≤C*∥ui∥W2,p(Ω)≤C*C2|Ω|1/p,
where α∈(0,1). We choose α<1-N/p such that Imbedding is compact, and along with elliptic equation regularity theory, there exists the subsequence of {ui}, which is still denoted by {ui}, and exists u such that ui→u uniformly holds in C2,α(Ω). On the other hand, vi→0, and when d1,i→d1∈[d,∞), the limit of (24) becomes the following problem:
(26)-Δu=1d1u(1-uk)inΩ,∂u∂ν=0on∂Ω.
Applying Maximum Principle to problem (26) and noting that minΩ¯ui≥C3, we have u=k.
However, whend1,i→∞, the limit of (24) becomes the following problem:
(27)-Δu=0inΩ,∂u∂ν=0on∂Ω,
which implies that u=C^ for some nonnegative constant C^. Since ∫Ω‍ui(1-ui/k-uin-1vi/(a+uin))dx=0, by letting i→∞ and noting that minΩ¯ui≥C3,  vi→0, and ui→C^, we also have u=k.
By a similar argument as that in (24), for the second equation of (1), we can prove that there exists a subsequence in C2,α(Ω), such that vi(x)/maxΩ¯vi(x)→v0. Since vi/maxΩ¯vi(x)≥minΩ¯vi(x)/maxΩ¯vi(x)≥1/C(1+d^), then v0≥1/C(1+d^). Dividing the second equation of (1) by maxΩ¯vi(x), and integrating over Ω, we have
(28)∫Ω‍vimaxΩ¯vi(x)(σuina+uin-b)dx=0.
Let i→∞, and note that vi/maxΩ¯vi(x)→v0 and ui→k; then, σkn/(a+kn)=b. This contradiction to the assumption completes the proof.


## 3. A Result on Degree Theory

In this section, we obtain nonexistence of nonconstant positive solutions to (1) as d3=0. Meanwhile, by degree theory, a general result to establish the existence of nonconstant positive solutions to (1) in the next section is proved.Denoted=(d1,d2,d3) and P=(σ,a,k,b). We will fix P∈(0,∞)4 and take d∈(0,∞)2×[0,∞) as bifurcation parameters, the dependence of P will often be suppressed. Define
(29)X={U∈[C2(Ω¯)]2∣∂U∂ν=0on∂Ω},X+={(u,v)∈X∣u>0,v>0onΩ¯},B(C)={(u,v)∈X∣1C<u,v<ConΩ¯}.
Since
(30)ΦU(U)=[d10-d2d3v(1+u)2d2[1+d3(1+u)]]
and detΦU(U) is positive for all nonnegative U, ΦU-1 exists. Hence, U is a positive solution to (1) if and only if
(31)F(d;U)∶=U-(I-Δ)-1×{ΦU(U)-1[G(U)+∇UΦUU∇U]+U}=0inX+,
where (I-Δ)-1 is the inverse of I-Δ in X. As F(d;·) is a compact perturbation of the identity operator I, the Leray-Schauder deg(F(d;·),0,B) is well defined if F(d;U)≠0 for all U∈∂B.By Theorems1 and 2, there exists a positive constant C0(d,P) such that if F(d;U)=0 in X+, then U∈B(C0). Note that C0(d,P) can be taken as a continuous function for σ≠ba/kn+b. By the invariance property of the Leray-Schauder degree, we then conclude that deg(F(d;·),0,B(C)) does not depend on d if C>C0(d,P), and it also does not depend on P, if P changes continuously in (0,∞)4 without touching the surface σ=ba/kn+b.For the caseσ>ba/kn+b, by degree invariance, we need only consider a special d; say d=(d,d,0) with large d. For this we can use the following nonexistence result. To compare the existence regions of (1) with and without cross-diffusion, we give a nonexistence result stronger than what is needed here.Theorem 3.
Let(d2,P)∈(0,∞)5 be given. Then, there exists a positive constant d^1(d2,P) such that when d=(d1,d2,0) and d1≥d^1(d2,P), (1) does not have any nonconstant positive solution.Proof.
First, by (7), there exists a positive constant C=C(P,d^2) such that a classical solution to (1) satisfies u(x), v(x)≤C provided that d2≥d^2. Now, we write f¯ as average of f over Ω, where
(32)f(u,v)=u(1-uk-un-1va+un),g(u,v)=σunva+un.
Multiplying the equation for u of (1) by (u-u¯) and integrating over Ω by parts, we have
(33)d1∫Ω‍∇|u-u¯|2dx=∫Ω‍f(u,v)(u-u¯)dx=∫Ω‍{f(u,v)-f(u¯,v¯)}(u-u¯)dx=∫Ω‍{fu(ξ,η)(u-u¯)2aaaaa+fv(ξ,η)(u-u¯)(v-v¯)fu(ξ,η)(u-u¯)2}dx≤∫Ω‍{(C1u-u¯)2+ɛ(v-v¯)2}dx
for some positive constant C1=C1(P,d^2) and ɛ=ɛ(P,d^2)≪1, where ξ and η lie between u and u¯, v and v¯, respectively. Similarly, we have
(34)d2∫Ω‍∇|v-v¯|2dx=∫Ω‍[-bv+g(u,v)](v-v¯)dx=∫Ω‍[-b(v-v¯)+g(u,v)-g(u¯,v¯)]aaaa×(v-v¯)dx=∫Ω‍{-b(v-v¯)2+gv(ξ,η)(v-v¯)2aaaaa+gu(ξ,η)(u-u¯)(v-v¯)gv(ξ,η)(v-v¯)2}dx≤∫Ω‍{(σ-b)(v-v¯)2+ɛ1(v-v¯)2aaaaaa+C2(u-u¯)2}dx
for some positive constant C2=C2(P,d^2) and ɛ1=ɛ1(P,d^2)≪1, where ξ and η are the same as in (33). Adding (33) and (34), we obtain
(35)∫Ω‍{d1∇|u-u¯|2+d2∇|v-v¯|2}dx≤∫Ω‍{(C1+C2)(u-u¯)2+(σ-b+ɛ+ɛ1)(v-v¯)2}dx.
It follows from the Poincaré Inequality that
(36)μ1∫Ω‍{d1(u-u¯)2+d2(v-v¯)2}dx≤∫Ω‍{(C1+C2)(u-u¯)2+(σ-b+ɛ+ɛ1)(v-v¯)2}dx.
Since we can choose d2≥d^2 such that μ1d2>σ-b, we may also choose ɛ and ɛ1 sufficiently small such that μ1d2>σ-b+ɛ+ɛ1. Consequently, by (36),
(37)μ1d1∫Ω‍(u-u¯)2dx≤(C1+C2)∫Ω‍(u-u¯)2dx,
which implies that u=u¯= constant, and, hence, v=v¯= constant if d1≥d^1(d2,P):=(C1(P,d^2)+C2(P,d^2))/μ1. Thus, we complete the proof of the theorem.In the following, we only calculatedeg(F,0,B) when all solutions to F=0 are positive constant solutions in B(C).Let0=μ1<μ2<μ3<⋯ be the eigenvalues of the operator -Δ in Ω with zero flux boundary condition, E(μi) be the eigenspace corresponding to μi in C1(Ω¯),  ϕij,  j=1,2,3,…, dimE(μi) an orthonormal basis of E(μi), and Xij={Cϕij∣C∈R2}. Then, X=⊕i=1∞Xi, where Xi=⊕j=1dimE(μi)Xij. This decomposed method is similar to that of [22].LetU* be a positive root to G(U)=0. We can calculate
(38)DUF(d;U*)=I-(I-Δ)-1×{ΦU-1(U*)GU(U*)+I}inL(X,X),
where L(X,X) is a linear mapping from X to itself.Denote thatH(d,U*;μ)∶=det[μI-ΦU(U*)-1GU(U*)].Lemma 4.
LetU* be a positive root to G(·)=0, and assume that H(d,U*;μi)≠0 for all i. Then,
(39)index(F(d;·),U*)=(-1)r,r=∑i≥1,H(d,μi)<01.Proof.
IfDUF is invertible, then the index of F at U* is defined as index(F(d;·),U*)=(-1)r, where r is the number of eigenvalues of DUF with negative real parts. The deg(F(d;·),0,B) is then equal to summation of the indexes over all solutions to F=0 in B, provided that F≠0 on ∂B. Since for each integer i≥1,Xi is invariant under DU(F(d;U*)), and λ is an eigenvalue of DUF on Xi if and only if λ is an eigenvalue of following matrix:
(40)I-11+μi[ΦU(U*)-1GU(U*)+I]=11+μi[μiI-ΦU(U*)-1GU(U*)].
Since H(d,U*;μ)≠0,  DUF(d;U*) is invertible. Therefore, the number of eigenvalues with negative real parts of DUF(d;U*) on Xi is odd if and only if H(d,U*;μi)<0, and therefore
(41)index(F(d;·),U*)=(-1)r,r=∑i≥1,H(d,μi)<01.Theorem 5.
Assume thatP=(σ,a,k,b)∈(0,∞)4 with σkn≠b(a+kn). Then, for every d∈(0,∞)2×[0,∞), there exists a constant C0(d,P) such that for every C>C0(d,P),
(42)deg(F(d;·),0,B(C))={0,ifσ<bakn+b,1,ifσ<bakn+b.Proof.
Since (1) does not have any positive solution in B(C), when σ≤b, we then conclude that
(43)deg(F(d;·),0,B(C))=0,∀P∈{(σ,a,k,b)∈(0,∞)4∣σ<bakn+b},aaaaaaaaaaaaaaaaaaaaaaC>C0(d,P).
Next, to complete the proof of Theorem5. We need only to calculate the degree for the case σ>ba/kn+b. In this case, by Theorem 3, all positive solutions to F=0 are the unique positive constant solution to G(·)=0, denoted by U~, when d=(d,d,0),ΦU(U~)=dI so that H(d,U~;μ)=(1/d)det(dμI-GU(U~)),  det(GU(U~))=nb2(σ-b)v~/σ2u~>0. This implies that when d is sufficiently large, H(d,U~;μi)>0 for all i=1,2,3,…, and thus r=0. It follows from Lemma 4 that deg(F(d;·),0,B(C))=1. This completes the proof.Remark 6.
The change of degree whenσ passes the borderline σ=ba/kn+b is due to the appearance (disappearance) of a positive constant steady-state bifurcating from U≡(k,0).


## 4. Existence of Nonconstant Positive Solutions

In this section, we establish the existence of positive nonconstant solutions for (1). In particular, we show that for certain ranges of parameters where (1) does not have any positive nonconstant steady state, our model can still produce patterns. The idea is as follows. First we calculate the index of F(d;·) at positive constant steady states. Suppose that the sum of all these indices is not equal to the degree stated in Theorem 5. Then, F(d;·)=0 in B(C) for C=C0(d,P) must have a nonconstant positive solution, which also solves (1).In the following, we always assume thatσ>b, and (1) has a unique positive constant solution (u~,v~), when U~=(u~,v~) is a positive solution to G(U)=0. We can get the following results at U~ by simply calculating(44)GU(U~)=[1σk{(n(σ-b)-2σ)u~-k[n(σ-b)-σ]}-bσnb(σ-b)v~σu~0]=:[g11-bσg210],ΦU(U~)=[d10-d2d3v~(1+u~)2d2[1+d3(1+u~)]]=∶[ϕ110ϕ21ϕ22],H(d,U~;μ)≡det(μI-ΦU-1(U~)GU(U~))=μ2+μ|ϕ21|b/σ-ϕ22g11ϕ11ϕ22+g21b/σϕ11ϕ22.To calculate the roots ofH(d,U~;·)=0, we will restrict our attention to large |d|. Note that
(45)limdi→∞H(d,U~;μ)=μ2-Λiμ,∀i=2,3,
where
(46)Λ2=Λ2(d1,d3;U)=(-d3bkv~+(1+u~+d3)(1+u~)×{(n(σ-b)-2σ)u~-k(n(σ-b)-σ)}Λ2(d1,d3;U))×(σkd1(1+u~+d3)(1+u~))-1,Λ3=Λ3(d1;U)=-bkv~+{(n(σ-b)-2σ)u~-k(n(σ-b)-σ)}(1+u~)σkd1(1+u~).The sign of the tracetr(GU)=g11 is determined by γ=(n(σ-b)-2σ)u~-k(n(σ-b)-σ) and the sign of the determinant det(GU)=nb2(σ-b)v/σ2u>0.Hence, we will discuss separately the following cases:(i)
n(σ-b)-2σ>0(n>2σ/(σ-b)), obviously k>u~>((n(σ-b)-2σ)/(n(σ-b)-σ))u~,   then  γ<0;(ii)
n(σ-b)-σ<0(n<σ/(σ-b)):
(iia)
ifu~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~, then γ<0;(iib)
ifk>((2σ-n(σ-b))/(σ-n(σ-b)))u~, then γ>0.


### 4.1. The Casen(σ-b)-2σ>0  (n>2σ/(σ-b))

In this subsection, we consider local stability of the constant steady stateU≡U~ for evolution dynamics
(47)Ut=ΔΦ(U)+G(U)inΩ×(0,∞),∂U∂ν=0on∂Ω,
where U~=(u~,v~) is a positive constant solution to G(·)=0 by (5).Theorem 7.
Letk>u~>((n(σ-b)-2σ)/(n(σ-b)-σ))u~  (n>2σ/(σ-b)). Then the positive constant solution U(x,t)≡(u~,v~) is asymptotically stable with respect to the dynamics (47). Consequently, in a small neighborhood of  U~, (1) does not have any nonconstant positive solution.Proof.
The linearization of (47) at U~ takes the form
(48)Vt=ΦU(U~)ΔV+GU(U~)VinΩ×(0,∞),∂V∂ν=0on∂Ω.
Denote that the corresponding linear operator L∶=ΦU(U~)Δ+GU(U~). For each eigen-pair (μi,ϕi)  (i=1,2,3,…) of -Δ on Ω with no flux boundary condition, V=Ceλtϕ is a solution to (48) if and only if (λ,C) is an eigen-pair of the matrix Ai:=-μiΦU(U~)+GU(U~). Xi is invariant under the operator L. From (44), we have
(49)g11<0(k>n(σ-b)-2σn(σ-b)-σu~),g21>0(σ>b).
Then
(50)detAi∶=ϕ11ϕ22μi2-(bσϕ21+g11ϕ22)μi+bσg21>0,trAi∶=-(ϕ11+ϕ22)μi+g11<0.
Hence, we conclude that the two eigenvalues λi+ and λi- of Ai have negative real parts. More specifically, we have the following:
(i) Fori=1, μ1=0, if g112-4g21b/σ≤0, then Reλi±=(1/2)g11<0; if g112-4g21b/σ>0, then
(51)Reλi+=12[g11+g112-4g21bσ]≤0,Reλi-=12[g11-g112-4g21bσ]<0.
(ii) Fori≥2, as μi is increasing with respect to i and μi→∞ as i→∞, it follows that if (trAi)2-4detAi≤0, then
(52)Reλi±=12trAi=12[-(ϕ11+ϕ22)μi+g11]≤12[-(ϕ11+ϕ22)μ2+g11]<0;
if (trAi)2-4detAi>0, since detAi>0 and trAi<0, then
(53)Reλi-=12(trAi-(trAi)2-4detAi)≤12trAi≤12[-(ϕ11+ϕ22)μ2+g11]<0;Reλi+=12(trAi+(trAi)2-4detAi)=2detAitrAi-(trAi)2-4detAi≤detAitrAi<-ɛ~.
for some positive constant ɛ~ that does not depend on i. The previous arguments show that there exists a positive constant ɛ, which does not depend on i, such that Reλi<-ɛ, ∀i. Consequently, the spectrum of L, which consists of eigenvalues, lies in {Re(λ)<-ɛ}. It then follows from Theorem 5.1.1 of [26, page 98] that the constant steady-state U(x,t)≡U~ is asymptotically stable to (47).


### 4.2. The Casen(σ-b)-σ<0(n<σ/(σ-b))

#### 4.2.1.(γ<0)u~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~

Theorem 8.
Letu~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~. Then the positive constant solution U(x,t)≡(u~,v~) is asymptotically stable with respect to the dynamics (47).Proof.
Similar to the proof of Theorem7.


#### 4.2.2.(γ>0)k>((2σ-n(σ-b))/(σ-n(σ-b)))u~

In this case,U~=(u~,v~) is the only positive constant solution to G(·)=0. By fixing the diffusion coefficients d1 (for prey) and using the diffusion coefficients d2 and d3 (for predator) as bifurcation parameters, we will show that (1) can create nonconstant positive solutions. We want to emphasize that it is caused by the presence of cross-diffusion which has a more complex role than that of the diffusion coefficients d1 and d2.Theorem 9 (existence with suitabled2 and d3).
Assume thatσ>ba/kn+b and k>((2σ-n(σ-b))/(σ-n(σ-b)))u~, define Λ2(d1,d3;U) and Λ3(d1;U) as (46).
(i) Suppose thatd1 and d3 are given such that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j. There exists a positive constant D2 such that if d2≥D2, then (1) has at least one nonconstant positive solution.
(ii) Suppose thatd1 is given such that Λ3(d1;U)∈(μj,μj+1) for some positive even integer j. Then, for any given d2>0, there exists a positive constant D3 such that if d3≥D3, (1) has at least one nonconstant positive solution.Proof.
Denote byμ±(d,U~), with Re(μ-)≤Re(μ+), the two roots to H(d,U~;μ)=0, and then
(54)μ+(d,U~)μ-(d,U~)=[det(ΦU(U~))]-1det(GU(U~))=g21b/σϕ11ϕ22>0.
From (45), we see that
(55)limd2→∞μ-(d,U~)=0,limd2→∞μ+(d,U~)=Λ2(d1,d3;U~),limd3→∞μ-(d,U~)=0,limd3→∞μ+(d,U~)=Λ3(d1;U~).
Suppose that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j; then, there exists a positive constant D2≫1 such that d2≫D2, and we have
(56)0=μ1<μ-(d,U~)<μ2,μ+(d,U~)∈(μj,μj+1).
Hence, if d2≫1, H(d,U~;μi)<0 is equivalent to i∈{2,3,…,j}, since j is even. It follows from Lemma 4 that
(57)index(F(d;·),U~)=(-1)j-1=-1.
Consequently, F(d;U)=0 has at least one nonconstant positive solution that is different from the constant function U=U~. Otherwise, the degree of F=0 in B(C) would be -1 for all large enough C, which would contradict Theorem 5. This proves, the first assertion of the theorem, and the second assertion is similarly proved.Remark 10.
ForΛ3(d1;U) to be positive, it is necessary and sufficient to have
(58)k>(k+1)σu~(1+u~)[σ-n(σ-b)]+u~.
When this inequality holds, Λ2(d1,d3;U~) is also positive provided that d3 is large, and we then can adjust to d1 and make the assumptions in (i) or (ii) of theorem hold.Remark 11.
In fact, ifσ/(σ-b)<n<2σ/(σ-b), then γ<0. This case is similar to case (i). Furthermore, the positive constant solution U(x,t)≡(u~,v~) is also asymptotically stable with respect to the dynamics (47).


## 4.1. The Casen(σ-b)-2σ>0  (n>2σ/(σ-b))

In this subsection, we consider local stability of the constant steady stateU≡U~ for evolution dynamics
(47)Ut=ΔΦ(U)+G(U)inΩ×(0,∞),∂U∂ν=0on∂Ω,
where U~=(u~,v~) is a positive constant solution to G(·)=0 by (5).Theorem 7.
Letk>u~>((n(σ-b)-2σ)/(n(σ-b)-σ))u~  (n>2σ/(σ-b)). Then the positive constant solution U(x,t)≡(u~,v~) is asymptotically stable with respect to the dynamics (47). Consequently, in a small neighborhood of  U~, (1) does not have any nonconstant positive solution.Proof.
The linearization of (47) at U~ takes the form
(48)Vt=ΦU(U~)ΔV+GU(U~)VinΩ×(0,∞),∂V∂ν=0on∂Ω.
Denote that the corresponding linear operator L∶=ΦU(U~)Δ+GU(U~). For each eigen-pair (μi,ϕi)  (i=1,2,3,…) of -Δ on Ω with no flux boundary condition, V=Ceλtϕ is a solution to (48) if and only if (λ,C) is an eigen-pair of the matrix Ai:=-μiΦU(U~)+GU(U~). Xi is invariant under the operator L. From (44), we have
(49)g11<0(k>n(σ-b)-2σn(σ-b)-σu~),g21>0(σ>b).
Then
(50)detAi∶=ϕ11ϕ22μi2-(bσϕ21+g11ϕ22)μi+bσg21>0,trAi∶=-(ϕ11+ϕ22)μi+g11<0.
Hence, we conclude that the two eigenvalues λi+ and λi- of Ai have negative real parts. More specifically, we have the following:
(i) Fori=1, μ1=0, if g112-4g21b/σ≤0, then Reλi±=(1/2)g11<0; if g112-4g21b/σ>0, then
(51)Reλi+=12[g11+g112-4g21bσ]≤0,Reλi-=12[g11-g112-4g21bσ]<0.
(ii) Fori≥2, as μi is increasing with respect to i and μi→∞ as i→∞, it follows that if (trAi)2-4detAi≤0, then
(52)Reλi±=12trAi=12[-(ϕ11+ϕ22)μi+g11]≤12[-(ϕ11+ϕ22)μ2+g11]<0;
if (trAi)2-4detAi>0, since detAi>0 and trAi<0, then
(53)Reλi-=12(trAi-(trAi)2-4detAi)≤12trAi≤12[-(ϕ11+ϕ22)μ2+g11]<0;Reλi+=12(trAi+(trAi)2-4detAi)=2detAitrAi-(trAi)2-4detAi≤detAitrAi<-ɛ~.
for some positive constant ɛ~ that does not depend on i. The previous arguments show that there exists a positive constant ɛ, which does not depend on i, such that Reλi<-ɛ, ∀i. Consequently, the spectrum of L, which consists of eigenvalues, lies in {Re(λ)<-ɛ}. It then follows from Theorem 5.1.1 of [26, page 98] that the constant steady-state U(x,t)≡U~ is asymptotically stable to (47).


## 4.2. The Casen(σ-b)-σ<0(n<σ/(σ-b))

### 4.2.1.(γ<0)u~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~

Theorem 8.
Letu~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~. Then the positive constant solution U(x,t)≡(u~,v~) is asymptotically stable with respect to the dynamics (47).Proof.
Similar to the proof of Theorem7.


### 4.2.2.(γ>0)k>((2σ-n(σ-b))/(σ-n(σ-b)))u~

In this case,U~=(u~,v~) is the only positive constant solution to G(·)=0. By fixing the diffusion coefficients d1 (for prey) and using the diffusion coefficients d2 and d3 (for predator) as bifurcation parameters, we will show that (1) can create nonconstant positive solutions. We want to emphasize that it is caused by the presence of cross-diffusion which has a more complex role than that of the diffusion coefficients d1 and d2.Theorem 9 (existence with suitabled2 and d3).
Assume thatσ>ba/kn+b and k>((2σ-n(σ-b))/(σ-n(σ-b)))u~, define Λ2(d1,d3;U) and Λ3(d1;U) as (46).
(i) Suppose thatd1 and d3 are given such that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j. There exists a positive constant D2 such that if d2≥D2, then (1) has at least one nonconstant positive solution.
(ii) Suppose thatd1 is given such that Λ3(d1;U)∈(μj,μj+1) for some positive even integer j. Then, for any given d2>0, there exists a positive constant D3 such that if d3≥D3, (1) has at least one nonconstant positive solution.Proof.
Denote byμ±(d,U~), with Re(μ-)≤Re(μ+), the two roots to H(d,U~;μ)=0, and then
(54)μ+(d,U~)μ-(d,U~)=[det(ΦU(U~))]-1det(GU(U~))=g21b/σϕ11ϕ22>0.
From (45), we see that
(55)limd2→∞μ-(d,U~)=0,limd2→∞μ+(d,U~)=Λ2(d1,d3;U~),limd3→∞μ-(d,U~)=0,limd3→∞μ+(d,U~)=Λ3(d1;U~).
Suppose that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j; then, there exists a positive constant D2≫1 such that d2≫D2, and we have
(56)0=μ1<μ-(d,U~)<μ2,μ+(d,U~)∈(μj,μj+1).
Hence, if d2≫1, H(d,U~;μi)<0 is equivalent to i∈{2,3,…,j}, since j is even. It follows from Lemma 4 that
(57)index(F(d;·),U~)=(-1)j-1=-1.
Consequently, F(d;U)=0 has at least one nonconstant positive solution that is different from the constant function U=U~. Otherwise, the degree of F=0 in B(C) would be -1 for all large enough C, which would contradict Theorem 5. This proves, the first assertion of the theorem, and the second assertion is similarly proved.Remark 10.
ForΛ3(d1;U) to be positive, it is necessary and sufficient to have
(58)k>(k+1)σu~(1+u~)[σ-n(σ-b)]+u~.
When this inequality holds, Λ2(d1,d3;U~) is also positive provided that d3 is large, and we then can adjust to d1 and make the assumptions in (i) or (ii) of theorem hold.Remark 11.
In fact, ifσ/(σ-b)<n<2σ/(σ-b), then γ<0. This case is similar to case (i). Furthermore, the positive constant solution U(x,t)≡(u~,v~) is also asymptotically stable with respect to the dynamics (47).


## 4.2.1.(γ<0)u~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~

Theorem 8.
Letu~<k<((2σ-n(σ-b))/(σ-n(σ-b)))u~. Then the positive constant solution U(x,t)≡(u~,v~) is asymptotically stable with respect to the dynamics (47).Proof.
Similar to the proof of Theorem7.


## 4.2.2.(γ>0)k>((2σ-n(σ-b))/(σ-n(σ-b)))u~

In this case,U~=(u~,v~) is the only positive constant solution to G(·)=0. By fixing the diffusion coefficients d1 (for prey) and using the diffusion coefficients d2 and d3 (for predator) as bifurcation parameters, we will show that (1) can create nonconstant positive solutions. We want to emphasize that it is caused by the presence of cross-diffusion which has a more complex role than that of the diffusion coefficients d1 and d2.Theorem 9 (existence with suitabled2 and d3).
Assume thatσ>ba/kn+b and k>((2σ-n(σ-b))/(σ-n(σ-b)))u~, define Λ2(d1,d3;U) and Λ3(d1;U) as (46).
(i) Suppose thatd1 and d3 are given such that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j. There exists a positive constant D2 such that if d2≥D2, then (1) has at least one nonconstant positive solution.
(ii) Suppose thatd1 is given such that Λ3(d1;U)∈(μj,μj+1) for some positive even integer j. Then, for any given d2>0, there exists a positive constant D3 such that if d3≥D3, (1) has at least one nonconstant positive solution.Proof.
Denote byμ±(d,U~), with Re(μ-)≤Re(μ+), the two roots to H(d,U~;μ)=0, and then
(54)μ+(d,U~)μ-(d,U~)=[det(ΦU(U~))]-1det(GU(U~))=g21b/σϕ11ϕ22>0.
From (45), we see that
(55)limd2→∞μ-(d,U~)=0,limd2→∞μ+(d,U~)=Λ2(d1,d3;U~),limd3→∞μ-(d,U~)=0,limd3→∞μ+(d,U~)=Λ3(d1;U~).
Suppose that Λ2(d1,d3;U~)∈(μj,μj+1) for some positive even integer j; then, there exists a positive constant D2≫1 such that d2≫D2, and we have
(56)0=μ1<μ-(d,U~)<μ2,μ+(d,U~)∈(μj,μj+1).
Hence, if d2≫1, H(d,U~;μi)<0 is equivalent to i∈{2,3,…,j}, since j is even. It follows from Lemma 4 that
(57)index(F(d;·),U~)=(-1)j-1=-1.
Consequently, F(d;U)=0 has at least one nonconstant positive solution that is different from the constant function U=U~. Otherwise, the degree of F=0 in B(C) would be -1 for all large enough C, which would contradict Theorem 5. This proves, the first assertion of the theorem, and the second assertion is similarly proved.Remark 10.
ForΛ3(d1;U) to be positive, it is necessary and sufficient to have
(58)k>(k+1)σu~(1+u~)[σ-n(σ-b)]+u~.
When this inequality holds, Λ2(d1,d3;U~) is also positive provided that d3 is large, and we then can adjust to d1 and make the assumptions in (i) or (ii) of theorem hold.Remark 11.
In fact, ifσ/(σ-b)<n<2σ/(σ-b), then γ<0. This case is similar to case (i). Furthermore, the positive constant solution U(x,t)≡(u~,v~) is also asymptotically stable with respect to the dynamics (47).



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*Source: 851028-2013-04-04.xml*