1 Introduction

The spatial heterogeneity and homogeneity usually result in different effects on the invasion, extinction and coexistence of species [1, 4, 5, 14, 24, 30, 31]. In this paper, we are concerned about the effects of spatial heterogeneity and homogeneity on the following S-K-T competition system:

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\mu \Delta [(1+k(u+v))u]+u[a(x)-u-v], \quad \quad x\in \Omega ,~~~~t>0, \\&v_t=\mu \Delta [(1+k(u+v))v]+v[b(x)-u-v], \,\, \quad \quad x\in \Omega ,~~~~t>0, \\&\partial _{\nu }u=\partial _{\nu }v=0,\qquad \qquad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ~\,x\in \partial \Omega ,~~ t>0, \\ \end{aligned}\right. \end{aligned}$$
(1.1)

where u(x, t) and v(x, t) represent the population densities of the competing species, respectively. The habitat \(\Omega \) is a bounded domain in \({\mathbb {R}}^N\) with smooth boundary, \(\nu \) is the outward unit normal vector on \(\partial \Omega .\) Homogeneous Neumann boundary condition means that no individuals cross the boundary of the habitat \(\Omega .\) The positive constant \(\mu \) denotes the random diffusion rate, and the positive constant k can be regarded as both cross-diffusion and self-diffusion rates. The functions a(x) and b(x) are the intrinsic growth rates for u and v,  which are continuous functions on \(\overline{\Omega }\) with \(a(x), b(x)\ge \not \equiv 0.\)

In (1.1), the species are identical in all aspects except their intrinsic growth rates. Lou and Martínez [15] assumed that the competing species adopted different cross- and self-diffusion strategies, and showed that constant cross- and self-diffusion coefficients may be more beneficial for the evolution of species. Wang [25] considered that the competing species possessed different interspecific competitions, and revealed that spatially heterogeneous interspecific competitions can be either favorable or unfavorable for the coexistence of species. One can refer to [3, 8, 16] for the series work of classical Lotka-Volterra competition models without cross- and self-diffusion.

The goal of this paper is to characterize the dynamics and structure of the set of nontrivial steady states of (1.1). As \(a(x)=b(x),\) the result is clear and (1.1) has a curve of neutrally stable nontrivial steady states \(\{(s, 1-s)\theta :s\in [0, 1]\},\) where \(\theta \) is the unique positive steady state of

$$\begin{aligned} u_t=\mu \Delta [(1+ku)u]+u\left[ a(x)-u\right] ,\quad x\in \Omega , \quad \quad \partial _{\nu }u=0,\quad x\in \partial \Omega . \end{aligned}$$

It is also shown that the total mass \(\int _{\Omega }\theta (k)dx\) achieves its minimum at \(k=\infty .\) Moreover, different properties of \(\theta \) between \(k>0\) and \(k=0\) are revealed.

Whereas, if the intrinsic growth rates of the species are different, it is hard to give a complete characterization of the dynamics and structure of the set of positive steady states of (1.1). As the first step of achieving our aim, we consider the following special system:

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\mu \Delta \left[ (1+k(u+v))u\right] +u\left[ 1+\tau g(x)-u-v\right] , \quad \quad x\in \Omega ,~~~~t>0, \\&v_t=\mu \Delta \left[ (1+k(u+v))v\right] +v\left[ 1+\tau h(x)-u-v\right] , \, \quad \,\,\,\,~ x\in \Omega ,~~~~t>0, \\&\partial _{\nu }u=\partial _{\nu }v=0,\quad \,\,\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \qquad \quad \quad \quad \quad \quad \,\, ~~x\in \partial \Omega ,~~ t>0, \\ \end{aligned}\right. \nonumber \\ \end{aligned}$$
(1.2)

where \(\tau \) is a positive constant, g and h are continuous functions on \(\overline{\Omega }.\) A good understanding of the dynamics and structure of the set of positive steady states can be given as \(\tau >0\) is small enough.

Before showing the results, we introduce the following functions. For

$$\begin{aligned} W_\nu ^{2,p}(\Omega )=\left\{ w\in W^{2,p}(\Omega ): \partial _\nu w =0,x\in \partial \Omega \right\} ,\quad p>N, \end{aligned}$$

we define \(T: W_\nu ^{2,p}\rightarrow L^p(\Omega )\) by

$$\begin{aligned} Tf=-\mu (1+2k)\Delta f+f, \end{aligned}$$
(1.3)

and let

$$\begin{aligned} \alpha =T^{-1}(g-h), \quad \quad {\tilde{\alpha }}=T^{-1}g. \end{aligned}$$
(1.4)

When \(\int _{\Omega }(g-h)dx=0,\) let \(\beta \) with \(\int _{\Omega }\beta dx=0\) be the unique solution of

$$\begin{aligned} -\mu (1+k)\Delta \beta =g-h,\quad x\in \Omega ,\quad \quad \quad \partial _{\nu }\beta =0, \quad x\in \partial \Omega . \end{aligned}$$
(1.5)

Using Fourier series expansion, it is easy to see that as \(\int _{\Omega }(g-h)dx=0\) and \(g-h\) changes sign on \(\overline{\Omega },\) we have that

$$\begin{aligned} \int _{\Omega }[(g-h)\beta -(g-h-\beta )\alpha ]dx>0. \end{aligned}$$
(1.6)

Theorem 1.1

Suppose that \({\tilde{\alpha }}, \alpha \) and \(\beta \) are defined by (1.4) and (1.5), respectively.

\({(\textrm{i})}\) If \(\int _{\Omega }(g-h)dx\ne 0,\) then as \(\tau >0\) is small enough, (1.2) has no positive steady state; moreover, \(({\tilde{u}},0)\) is asymptotically stable and \((0, {\tilde{v}})\) is unstable when \(\int _{\Omega }(g-h)dx>0,\) and the stability is switched if \(\int _{\Omega }(g-h)dx<0.\)

\({(\textrm{ii})}\) If \(\int _{\Omega }(g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with

$$\begin{aligned} \int _{\Omega }[(g-h-\beta )\alpha -g\beta ]dx<\int _{\Omega }(g-h-\beta )\tilde{\alpha }dx< -\int _{\Omega }h\beta dx, \end{aligned}$$
(1.7)

then as \(\tau >0\) is small enough, (1.2) has a unique positive steady state, which is asymptotically stable; moreover, both \(({\tilde{u}},0)\) and \((0, {\tilde{v}})\) are unstable.

\({(\textrm{iii})}\) If \(\int _{\Omega }(g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with

$$\begin{aligned} \int _{\Omega }(g-h-\beta )\tilde{\alpha }dx<\int _{\Omega }[(g-h-\beta )\alpha -g\beta ]dx, \end{aligned}$$
(1.8)

then as \(\tau >0\) is small enough, (1.2) has no positive steady state; moreover, \(({\tilde{u}},0)\) is unstable and \((0, {\tilde{v}})\) is asymptotically stable.

\({(\textrm{iv})}\) If \(\int _{\Omega }(g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with

$$\begin{aligned} \int _{\Omega }(g-h-\beta )\tilde{\alpha }dx>-\int _{\Omega }h\beta dx, \end{aligned}$$
(1.9)

then as \(\tau >0\) is small enough, (1.2) has no positive steady state; moreover, \(({\tilde{u}},0)\) is asymptotically stable and \((0, {{\tilde{v}}})\) is unstable.

Now we examine the effects of spatial heterogeneity on the coexistence and extinction of the competing species. When both g and h are spatially heterogeneous, Theorem 1.1 asserts that if the total resources for u and v are different, that is, \(\int _{\Omega }gdx\ne \int _{\Omega }hdx\), then coexistence of the competing species is impossible, the species with higher total resource wins; whereas, if u and v have the same total resource, that is, \(\int _{\Omega }gdx=\int _{\Omega }hdx\), then the result becomes subtle and there exists a balance between g and h characterized by three integrals such that either coexistence or extinction is possible. It should be pointed out that all conditions in Theorem 1.1 may hold true for appropriate g and h. For example, let \((\phi _i, \lambda _i)(i\ge 0)\) with \(\int _{\Omega }\phi _i^2dx=1\) be the eigenfunctions and eigenvalues of \(-\Delta \) subject to homogeneous Neumann boundary condition. Then conditions (1.7), (1.8) and (1.9) hold true when \((g, h)=(g_i, h_i)\phi _i(i\ge 1)\) with appropriate coefficients \(g_i\) and \(h_i.\)

Theorem 1.2

Suppose that \(h=0.\) Let \(\alpha \) and \(\beta \) be defined by (1.4) and (1.5) with \(h=0,\) respectively.

\({(\textrm{i})}\) If \(\int _{\Omega } gdx\ne 0,\) then as \(\tau >0\) is small enough, (1.2) has no positive steady state; moreover, \(({\tilde{u}},0)\) is asymptotically stable and (0, 1) is unstable when \(\int _{\Omega }gdx>0,\) and the stability is switched if \(\int _{\Omega }gdx<0.\)

\({(\textrm{ii})}\) If \(\int _{\Omega } gdx=0,\) g changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }g\alpha dx<\int _{\Omega }\alpha \beta dx,\) then as \(\tau >0\) is small enough, (1.2) has a unique positive steady state, which is asymptotically stable; moreover, both \(({\tilde{u}},0)\) and (0, 1) are unstable.

\({(\textrm{iii})}\) If \(\int _{\Omega } gdx=0,\) g changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }g\alpha dx>\int _{\Omega }\alpha \beta dx,\) then as \(\tau >0\) is small enough, (1.2) has no positive steady state; moreover, \(({\tilde{u}},0)\) is asymptotically stable and (0, 1) is unstable.

Due to Theorem 1.2, the contest between spatial heterogeneity and homogeneity is observed. If the intrinsic growth rate for v is homogeneous, then the semitrivial solution \((0, {\tilde{v}})\) becomes (0, 1). Moreover, (0, 1) can never be stable when the total resources for u and v are the same. Hence, the result reveals that the species with spatially heterogeneous intrinsic growth rate has advantage over the species with spatially homogeneous intrinsic growth rate during the competition.

Since the pioneering work by Shigesada et al. [21], the S-K-T system with cross-diffusion has been widely studied and cross-diffusion is shown to play an important role [2, 9,10,11,12,13, 17,18,19,20, 22, 26,27,29]. To show the effect of the cross- and self-diffusion coefficient k on the dynamics of (1.2), we set \((g, h)=(1, \ell )\phi _i(i\ge 1)\) and assume that \(\mu \lambda _i>1.\) Then as \(\ell <-[\mu (1+2k)\lambda _i+1]/[\mu (1+k)\lambda _i-1],\) the semitrivial steady state \((0,{\tilde{v}})\) is stable; as \(-[\mu (1+2k)\lambda _i+1]/[\mu (1+k)\lambda _i-1]<\ell <-[\mu (1+k)\lambda _i-1]/[\mu (1+2k)\lambda _i+1],\) \((0, {\tilde{v}})\) becomes unstable and there exists a unique stable positive steady state; as \(-[\mu (1+k)\lambda _i-1]/[\mu (1+2k)\lambda _i+1]<\ell <1,\) the unique positive steady state disappears and \(({\tilde{u}}, 0)\) is stable; as \(\ell >1,\) \((0, {\tilde{v}})\) again becomes stable. Noting the monotone property of \([\mu (1+2k)\lambda _i+1]/[1-\mu (1+k)\lambda _i]\) and \([1-\mu (1+k)\lambda _i]/[\mu (1+2k)\lambda _i+1]\) with respect to k,  one sees that if \(\mu \lambda _i>3,\) then the coexistence region becomes larger as k varies from small to large, which implies that k is beneficial for the coexistence of competing species; whereas, if \(1<\mu \lambda _i<3,\) then the coexistence region becomes smaller as k varies from small to large, and k is not favorable for the coexistence of competing species. Hence, the effect of k depends on the random diffusion \(\mu \) and the domain \(\Omega .\)

On the other hand, set \((g,h)=(\ell , 0)\phi _i(i\ge 1).\) Then as \(\mu (1+k)\lambda _i<1,\) (1.2) has a unique asymptotically stable positive steady state; once \(\mu (1+k)\lambda _i>1,\) the unique positive steady state vanishes and \(({\tilde{u}}, 0)\) becomes asymptotically stable. So for fixed \(\mu \lambda _i<1,\) the coexistence region disappears as k crosses the critical value \([1-\mu \lambda _i]/[\mu \lambda _i].\) The result reveals that the effect of k for constant h is different from that for nonconstant h. It should be pointed out that for the effects of diffusion and spatial heterogeneity of the classical Lotka-Volterra competition system without cross-diffusion, one can refer to the series work by He and Ni [6, 7].

The rest of this paper is organized as follows. In Sect. 2, we show the properties of positive solution of an elliptic equation. Sections 3 and 4 are devoted to the structure and stability of semitrivial and positive steady states of (1.2) for small \(\tau >0.\) Finally, Sect. 5 shows the asymptotic behavior of positive steady states of (1.2) as \(k\rightarrow \infty \) and gives nonexistence result of positive steady states for large k.

2 Properties of Positive Solution of an Elliptic Equation

In this section, we shall establish some properties of the positive solution to the following nonlinear elliptic equation:

$$\begin{aligned} -\mu \Delta [(1+ku)u]=u\left( a(x)-u\right) ,\quad x\in \Omega , \quad \quad \partial _{\nu }u=0,\quad x\in \partial \Omega , \end{aligned}$$
(2.1)

where \(a(x)\ge \not \equiv 0\) is a continuous function in \(\overline{\Omega }.\)

Lemma 2.1

For each fixed \(k\ge 0,\) (2.1) has a unique positive solution \(\theta (k),\) and it is asymptotically stable.

Proof

Let \(U=(1+ku)u.\) Then we obtain that (2.1) is equivalent to

$$\begin{aligned} -\mu \Delta U=g(U)\left[ a(x)-g(U)\right] ,\quad x\in \Omega , \quad \quad \partial _{\nu }U=0,\quad x\in \partial \Omega , \end{aligned}$$
(2.2)

where

$$\begin{aligned} u=g(U)=\frac{-1+\sqrt{1+4kU}}{2k}. \end{aligned}$$

Let \(\lambda _1<0\) be the principal eigenvalue of the following eigenvalue problem:

$$\begin{aligned} -\mu \Delta \varphi -a(x)\varphi =\lambda \varphi ,\quad x\in \Omega ,\quad \quad \partial _{\nu }\varphi =0,\quad x\in \partial \Omega , \end{aligned}$$

and let \(\varphi _1>0\) with \(\max _{\overline{\Omega }}\varphi _1=1\) be the corresponding eigenfunction. Set \({\underline{U}}=\varepsilon \varphi _1\) and \( {\overline{U}}=(1+kM)M\) with \(M>\max \{\Vert a\Vert _{\infty }, 1\}.\) Then some computations deduce that

$$\begin{aligned} -\mu \Delta {\underline{U}}= & {} \varepsilon \varphi _1\left[ \lambda _1+a(x)\right] \\= & {} g({\underline{U}})\left\{ a(x)-g({\underline{U}})+\lambda _1+g({\underline{U}})[1+k(\lambda _1+a(x))] \right\} ,\quad x\in \Omega . \end{aligned}$$

By choosing \(0<\varepsilon <1\) such that \(g({\underline{U}})[1+k(\lambda _1+a(x))]<-\lambda _1,\) we derive that \(({\underline{U}}, {\overline{U}})\) is a pair of sub-super solution of (2.2). Hence, the super-sub solution method yields that (2.2) has a positive solution in \([{\underline{U}}, {\overline{U}}].\)

Let \({\widetilde{U}}\) be the maximal positive solution of (2.2) in \([{\underline{U}}, {\overline{U}}]\) and let \(\Theta \) be any positive solution of (2.2). By the the maximum principle, we obtain that

$$\begin{aligned} \max _{\overline{\Omega }}\Theta =\Theta (x_0)\le (1+ka(x_0))a(x_0) \le (1+k\Vert a\Vert _{\infty })\Vert a\Vert _{\infty }<{\overline{U}}, \end{aligned}$$

which asserts that \(\Theta \le {\widetilde{U}}.\) Noting that \({\widetilde{U}}=(1+kg({\widetilde{U}}))g({\widetilde{U}})\) and \(\Theta =(1+kg(\Theta ))g(\Theta ),\) some computations deduce that

$$\begin{aligned} 0&=\int _{\Omega }\left[ {\widetilde{U}}g(\Theta )(a(x)-g(\Theta ))-\Theta g({\widetilde{U}})(a(x)-g({\widetilde{U}}))\right] dx\\&=\int _{\Omega }g(\Theta )g({\widetilde{U}})(g({\widetilde{U}})-g(\Theta ))(1+ka(x))dx\ge 0, \end{aligned}$$

as g is monotone increasing. So we obtain that \(\Theta ={\widetilde{U}}\) and (2.2) has a unique positive solution. Hence, (2.1) has a unique positive solution.

Next, let \(\theta \) be the unique positive solution of (2.1) and we prove the stability of \(\theta .\) It is clear that the linearized problem of (2.1) at \(\theta \) is given by

$$\begin{aligned} -\mu \Delta [(1+2k\theta )\psi ]-[a(x)-2\theta ]\psi =\sigma \psi ,\quad x\in \Omega , \quad \quad \partial _{\nu }\psi |_{\partial \Omega }=0. \end{aligned}$$
(2.3)

Let \(\Psi =(1+2k\theta )\psi .\) Then (2.3) is reduced to

$$\begin{aligned} -\mu \Delta \Psi +\frac{2\theta -a(x)}{1+2k\theta }\Psi =\sigma \frac{\Psi }{1+2k\theta },\quad x\in \Omega , \quad \quad \partial _{\nu }\Psi |_{\partial \Omega }=0, \end{aligned}$$
(2.4)

which has a real principal eigenvalue \(\sigma _1\) and a positive eigenfunction \(\Psi _1.\) So the stability of \(\theta \) is determined by the sign of \(\sigma _1.\) Multiplying both sides of (2.4) at \(\sigma =\sigma _1\) and \(\Psi =\Psi _1\) by \(\Psi _1\) and integrating over \(\Omega ,\) we obtain that

$$\begin{aligned} \sigma _1\int _{\Omega }\frac{\Psi _1^2}{1+2k\theta }dx= & {} \int _{\Omega }\left[ \mu |\nabla \Psi _1|^2+ \frac{2\theta -a(x)}{1+2k\theta }\Psi _1^2\right] dx\\> & {} \int _{\Omega }\left[ \mu |\nabla \Psi _1|^2+ \frac{\theta -a(x)}{1+k\theta }\Psi _1^2\right] dx. \end{aligned}$$

By the equation of \(\theta \) and the property of principal eigenvalue, we deduce that

$$\begin{aligned} 0=\inf _{\psi \in H^1\setminus \{0\}}\frac{\int _{\Omega }\left[ \mu |\nabla \psi |^2+ \frac{\theta -a(x)}{1+k\theta }\psi ^2\right] dx}{\int _{\Omega }\psi ^2dx}\le \frac{\int _{\Omega }\left[ \mu |\nabla \Psi _1|^2+ \frac{\theta -a(x)}{1+k\theta }\Psi _1^2\right] dx}{\int _{\Omega }\Psi _1^2dx}, \end{aligned}$$

which yields that \(\sigma _1>0.\) So the positive solution \(\theta \) is asymptotically stable and the proof of the lemma is complete. \(\square \)

It should be pointed out that if \(a(x)=a>0,\) then the unique positive solution \(\theta \) of (2.1) is constant and equal to a. Whereas, if a(x) is nonconstant, then \(\theta \) becomes nonconstant and depends on the diffusion coefficient k.

Lemma 2.2

Suppose that \(\theta =\theta (k)\) is the unique positive solution of (2.1). Then as \(k\rightarrow \infty ,\) we have that

$$\begin{aligned} \theta \rightarrow \frac{1}{|\Omega |}\int _{\Omega }a(x)dx \quad \quad {\text {uniformly in}}~~\overline{\Omega }. \end{aligned}$$
(2.5)

Proof

Let \(\{k_i\}\) be any sequence with \(\lim _{i\rightarrow \infty }k_i=\infty \) and \(\theta _i\) be the unique positive solution of (2.1) at \(k=k_i.\) Define \(V_i=\left( \frac{1}{k_i}+\theta _i\right) \theta _i.\) Then one sees that \(V_i\) satisfies

$$\begin{aligned} -\mu \Delta V_i=\frac{1}{k_i}\theta _i\left[ a(x)-\theta _i\right] ,\quad x\in \Omega , \quad \quad \partial _{\nu }V_i=0,\quad x\in \partial \Omega . \end{aligned}$$
(2.6)

Since \(\theta _i\) and \((1+k_i\theta _i)\theta _i\) achieve their maximums at the same point, the maximum principle deduces that \(\theta _i\le \Vert a\Vert _{\infty }.\) So both \(V_i\) and the right hand side of (2.6) are uniformly bounded. By the \(L^p\) estimates and Sobolev embedding theorems, we deduce that subject to a subsequence if necessary, \(V_i\rightarrow V\) in \(C^1\) as \(i\rightarrow \infty .\) Passing to the weak limit in (2.6), we obtain that V is a nonnegative weak solution of

$$\begin{aligned} -\mu \Delta V=0,\quad x\in \Omega , \quad \quad \partial _{\nu }V=0,\quad x\in \partial \Omega , \end{aligned}$$

which implies that V is a nonnegative constant C. So it follows that

$$\begin{aligned} \theta _i=\sqrt{V_i-\frac{\theta _i}{k_i}}\rightarrow \sqrt{C} \end{aligned}$$

uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\) Integrating the equation of \(\theta _i\) over \(\Omega ,\) we obtain that

$$\begin{aligned} \int _{\Omega }\theta _i\left[ a(x)-\theta _i\right] dx=0. \end{aligned}$$

Letting \(i\rightarrow \infty ,\) we have that either \(\sqrt{C}=0\) or

$$\begin{aligned} \sqrt{C}=\frac{1}{|\Omega |}\int _{\Omega }a(x)dx. \end{aligned}$$

In the following, we suppose that \(C=0.\) If \(k_i\Vert \theta _i\Vert _{\infty }\) is bounded, then \(w_i=k_i\theta _i\) is uniformly bounded and satisfies

$$\begin{aligned} -\mu \Delta \left[ (1+w_i)w_i\right] =w_i\left[ a(x)-\theta _i\right] ,\quad x\in \Omega , \quad \quad \partial _{\nu }w_i=0,\quad x\in \partial \Omega . \end{aligned}$$

A standard argument deduces that, subject to a subsequence if necessary, \((1+w_i)w_i\rightarrow W\) in \(C^1\) as \(i\rightarrow \infty \) and \(w_i\rightarrow w\) in \(C^1\) as \(i\rightarrow \infty .\) Moreover, \(W=(1+kw)w\) and w is a nonnegative solution of

$$\begin{aligned} -\mu \Delta \left[ (1+w)w\right] =a(x)w,\quad x\in \Omega , \quad \quad \partial _{\nu }w=0,\quad x\in \partial \Omega . \end{aligned}$$
(2.7)

By the maximum principle, it is clear that either \(w\equiv 0\) or \(w>0\) in \(\overline{\Omega }.\) If \(w=0,\) then \(\widehat{w_i}=w_i/\Vert w_i\Vert _{\infty }\) satisfies

$$\begin{aligned} -\mu \Delta (1+w_i)\widehat{w_i}=\widehat{w_i}(a(x)-\theta _i),\quad x\in \Omega , \quad \quad \partial _{\nu }\widehat{w_i}=0,\quad x\in \partial \Omega . \end{aligned}$$

By the standard argument, one sees that subject to a subsequence if necessary, \(\widehat{w_i}\rightarrow {\widehat{w}}\) in \(C^1\) as \(i\rightarrow \infty \) and \({\widehat{w}}\) is a nonnegative solution of

$$\begin{aligned} -\mu \Delta {\widehat{w}}={\widehat{w}}a(x),\quad x\in \Omega , \quad \quad \partial _{\nu }{\widehat{w}}=0,\quad x\in \partial \Omega . \end{aligned}$$

Since \({\widehat{w}}\) is nontrivial, we have that \({\widehat{w}}>0\) and

$$\begin{aligned} \int _{\Omega }a(x){\widehat{w}}dx=0, \end{aligned}$$

which is a contradiction. If \(w>0,\) then (2.7) yields that

$$\begin{aligned} \int _{\Omega }a(x)wdx=0, \end{aligned}$$

which is a contradiction.

If there exists a subsequence of \(\{k_i\},\) which is still denoted by itself, such that \(k_i\Vert \theta _i\Vert _{\infty }\rightarrow \infty \) as \(i\rightarrow \infty ,\) then \(\widehat{\theta _i}=\theta _i/\Vert \theta _i\Vert _{\infty }\) satisfies

$$\begin{aligned} -\mu \Delta \left[ \left( \frac{1}{k_i\Vert \theta _i\Vert _{\infty }}+\widehat{\theta _i}\right) \widehat{\theta _i}\right] =\frac{\widehat{\theta _i}(a(x)-\theta _i)}{k_i\Vert \theta _i\Vert _{\infty }},\quad x\in \Omega , \quad \quad \partial _{\nu }\widehat{\theta _i}=0,\quad x\in \partial \Omega . \end{aligned}$$

By the standard argument, we have that subject to a subsequence if necessary, \(\widehat{\theta _i}\rightarrow \widehat{\theta }\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty \) and \(\widehat{\theta }=1.\) Integrating the equation of \(\widehat{\theta _i}\) over \(\Omega ,\) we obtain that

$$\begin{aligned} \int _{\Omega }\widehat{\theta _i}\left[ a(x)-\theta _i\right] dx=0. \end{aligned}$$

Letting \(i\rightarrow \infty ,\) we have that

$$\begin{aligned} \int _{\Omega }a(x)dx=0, \end{aligned}$$

which is a contradiction. Thus, one sees that \(C>0\) and the proof of the lemma is complete. \(\square \)

It should be pointed out that the asymptotic behavior of positive solution of (2.1) as \(\mu \rightarrow \infty \) is the same to that as \(k\rightarrow \infty .\) By virtue of (2.1), we have that

$$\begin{aligned} \int _{\Omega }\theta dx=\int _{\Omega }a(x)dx+\mu \int _{\Omega }\frac{(1+2k\theta )|\nabla \theta |^2}{\theta ^2}dx >\int _{\Omega }a(x)dx. \end{aligned}$$

Hence, we obtain that

$$\begin{aligned} \int _{\Omega }\theta (k)dx>\int _{\Omega }\theta (\infty )dx. \end{aligned}$$

Lemma 2.3

Let \(\theta =\theta (k)\) be the positive solution of (2.1). Then we have that

$$\begin{aligned} \int _{\Omega }\theta ^3(k)dx<\int _{\Omega }\theta ^3(0)dx. \end{aligned}$$
(2.8)

Proof

It is clear that \(\theta (0)\not \equiv \theta (k),\) otherwise we have that \(-k\mu \Delta \theta ^2(k)=0\) and \(\theta (k)\) is constant. By the equation of \(\theta (0)\) and the property of principal eigenvalue, we get that

$$\begin{aligned} 0&\le \inf _{\varphi \in H^1\setminus \{0\}} \int _{\Omega }\left[ \mu |\nabla \varphi |^2+ (\theta (0)-a(x))\varphi ^2\right] dx \\&\quad <\int _{\Omega }\left[ \mu |\nabla \theta (k)|^2+ (\theta (0)-a(x))\theta ^2(k)\right] dx. \end{aligned}$$

By the equation of \(\theta (k),\) we obtain that

$$\begin{aligned} \mu \int _{\Omega }(1+2k\theta (k))|\nabla \theta (k)|^2dx= \int _{\Omega }(a(x)-\theta (k))\theta ^2(k)dx. \end{aligned}$$

Hence, it follows that

$$\begin{aligned} 0<\int _{\Omega }(\theta (0)-\theta (k))\theta ^2(k)dx-2k\mu \int _{\Omega }\theta (k)|\nabla \theta (k)|^2dx <\int _{\Omega }(\theta (0)-\theta (k))\theta ^2(k)dx. \end{aligned}$$

Then

$$\begin{aligned} \int _{\Omega }\theta ^3(k)dx<\int _{\Omega }\theta ^2(k)\theta (0)dx\le \left[ \int _{\Omega }(\theta ^2(k))^{3/2}dx\right] ^{2/3}\left[ \int _{\Omega }\theta ^3(0)dx\right] ^{1/3} \end{aligned}$$

and (2.8) is obtained. The proof of the lemma is complete. \(\square \)

3 Structure and Stability of Semitrivial Steady States

In this section, we shall establish the structure and stability of semitrivial steady states of (1.2). By virtue of Lemma 2.1, it is clear that there exists a small positive number \(\tau _0=\tau _0(k)\) such that for each fixed \(\tau \in [0, \tau _0),\) (1.2) has two semitrivial steady states \(({\tilde{u}}(\tau ), 0)\) and \((0, {\tilde{v}}(\tau )).\) Moreover, \({\tilde{u}}(\tau )\) and \({\tilde{v}}(\tau )\rightarrow 1\) as \(\tau \rightarrow 0.\) By the implicit function theorem and the asymptotic stability of positive solution of (2.1), we have that \({\tilde{u}}(\tau )\) and \({\tilde{v}}(\tau )\) are differentiable with respect to \(\tau >0.\) More precisely, we have the following lemma.

Lemma 3.1

Let \(({\tilde{u}}(\tau ), 0)\) and \((0, {\tilde{v}}(\tau ))\) be the semitrivial steady states of (1.2) for each fixed \(\tau \in [0, \tau _0).\) Then

$$\begin{aligned} {\tilde{u}}(\tau )=1+\tau u_1(\tau ),\quad {\tilde{v}}(\tau )=1+\tau v_1(\tau ),\quad u_1(0)=T^{-1}g,\quad v_1(0)=T^{-1}h, \end{aligned}$$

where T is defined by (1.3).

Lemma 3.2

Let \(\tilde{\alpha }(x)\) and \(\beta (x)\) be defined by (1.4) and (1.5), respectively.

\({(\textrm{i})}\):

If \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with (1.9), or \(\int _{\Omega } (g-h)dx>0,\) then as \(\tau >0\) is small enough, the semitrivial steady state \(({\tilde{u}}, 0)\) is asymptotically stable.

\({(\textrm{ii})}\):

If \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }\left[ \tilde{\alpha }(g-h-\beta )+ \beta h\right] dx<0,\) or \(\int _{\Omega } (g-h)dx<0,\) then as \(\tau >0\) is small enough, the semitrivial steady state \(({\tilde{u}}, 0)\) is unstable.

Proof

Linearizing (1.2) at its semitrivial steady state \(({\tilde{u}}, 0),\) we obtain the following eigenvalue problem:

$$\begin{aligned} \left\{ \begin{aligned}&-\mu \Delta [(1+k{\tilde{u}})\varphi ]-\mu k\Delta [{\tilde{u}}(\varphi +\psi )]-(1+\tau g-2{\tilde{u}})\varphi +{\tilde{u}}\psi =\lambda \varphi ,\quad x\in \Omega ,\\&-\mu \Delta [(1+k{\tilde{u}})\psi ]-(1+\tau h-{\tilde{u}})\psi =\lambda \psi , \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \,\,\,x\in \Omega ,\\&\partial _{\nu }\varphi =\partial _{\nu }\psi =0,\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \qquad \quad \,\,\, x\in \partial \Omega . \end{aligned}\right. \end{aligned}$$
(3.1)

It follows from [23] that if all eigenvalues of (3.1) have positive real parts, then \(({\tilde{u}},0)\) is asymptotically stable; however, if (3.1) possesses an eigenvalue with negative real part, then \(({\tilde{u}},0)\) is unstable.

If \(\psi \equiv 0,\) then \(\lambda \) must be an eigenvalue of

$$\begin{aligned} -\mu \Delta [(1+2k{\tilde{u}})\varphi ] -\left( 1+\tau g-2{\tilde{u}}\right) \varphi =\lambda \varphi , \quad x\in \Omega ,\quad \quad \partial _{\nu }\varphi |_{\partial \Omega }=0. \end{aligned}$$
(3.2)

Since \({\tilde{u}}\) is asymptotically stable for small \(\tau >0,\) we obtain that all eigenvalues of (3.2) have positive real parts.

If \(\psi \not \equiv 0,\) then \(\lambda \) is an eigenvalue of

$$\begin{aligned} -\mu \Delta [(1+k{\tilde{u}})\psi ]-\left( 1+\tau h-{\tilde{u}}\right) \psi =\lambda \psi ,\quad x\in \Omega ,\quad \quad \partial _{\nu }\psi |_{\partial \Omega }=0. \end{aligned}$$
(3.3)

By setting \(\Psi =(1+k{\tilde{u}})\psi ,\) we see that (3.3) has a real principal eigenvalue \(\lambda _1(\tau ),\) and the corresponding positive eigenfunction is denoted by \(\psi _1=\psi _1(\tau )\) with \(\max \psi _1(\tau )=1.\) If \(\lambda _1>0,\) then all eigenvalues of (3.1) have positive real parts, and \(({\tilde{u}}, 0)\) is asymptotically stable. However, if \(\lambda _1<0,\) then the first equation of (3.1) at \(\lambda =\lambda _1\) and \(\psi =\psi _1\) has a unique solution \(\varphi _1,\) which implies that \(\lambda _1<0\) is an eigenvalue of (3.1) with eigenfunction \((\varphi _1, \psi _1)\) and \(({\tilde{u}}, 0)\) is unstable.

Since \({\tilde{u}}(\tau )\rightarrow 1\) as \(\tau \rightarrow 0,\) we obtain that \(\lambda _1(\tau )\rightarrow 0\) and \(\psi _1(\tau )\rightarrow 1\) as \(\tau \rightarrow 0.\) Multiplying both sides of (3.3) at \(\lambda =\lambda _1\) and \(\psi =\psi _1\) by \([1+k{\tilde{u}}(\tau )]{\tilde{u}}(\tau )\) and integrating over \(\Omega ,\) we deduce that

$$\begin{aligned} \lambda _1(\tau )\int _{\Omega }[1+k{\tilde{u}}(\tau )]{\tilde{u}}(\tau )\psi _1(\tau )dx= \tau \int _{\Omega }[1+k{\tilde{u}}(\tau )]{\tilde{u}}(\tau )\psi _1(\tau )(g-h)dx. \end{aligned}$$
(3.4)

Let \({\tilde{u}}(\tau )=1+\tau u_1(\tau )\) and \(\psi _1(\tau )=1+\tau \tilde{\psi }_1(\tau ).\) Then (3.4) yields that as \(\tau \rightarrow 0,\)

$$\begin{aligned}&\lambda _1(\tau )\int _{\Omega }[1+k{\tilde{u}}(\tau )]{\tilde{u}}(\tau )\psi _1(\tau )dx\\&\quad =\tau (1+k) \int _{\Omega }(g-h)dx+ \tau ^2\\&\qquad \times \int _{\Omega }\left[ (1+k)\tilde{\psi }_1(\tau )+(1+2k) u_1(\tau )\right] (g-h)dx+o(\tau ^2). \end{aligned}$$

So it follows that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau }=\frac{\int _{\Omega } (g-h)dx}{|\Omega |}. \end{aligned}$$

Hence, as \(\tau >0\) is small enough, \(\lambda _1(\tau )<0\) for \(\int _{\Omega }(g-h)dx<0\) and \(\lambda _1(\tau )>0\) for \(\int _{\Omega }(g-h)dx>0.\)

Now, we assume that \(\int _{\Omega }(g-h)dx=0\) and \(g-h\) changes sign on \(\overline{\Omega }.\) Then one sees that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau ^2}=\frac{\int _{\Omega }\left[ (1+k)\tilde{\psi }_1(0)+(1+2k) u_1(0)\right] (g-h)dx}{(1+k)|\Omega |}. \end{aligned}$$
(3.5)

Since \(\lambda _1(\tau )/\tau \rightarrow 0\) as \(\tau \rightarrow 0,\) it is easy to check that \(\tilde{\psi }_1(0)\) satisfies

$$\begin{aligned} -\mu (1+k)\Delta \tilde{\psi }_1(0)=\mu k\Delta u_1(0)- u_1(0)+h, \quad x\in \Omega ,\quad \quad \partial _{\nu }\tilde{\psi }_1(0)|_{\partial \Omega }=0. \end{aligned}$$

Due to Lemma 3.1, we obtain that \(u_1(0)=T^{-1}g=\tilde{\alpha }(x)\) and

$$\begin{aligned} -\mu (1+k)\Delta [\tilde{\psi }_1(0)-\tilde{\alpha }(x)]=\mu (1+2k)\Delta \tilde{\alpha }(x) -\tilde{\alpha }(x)+h=h-g. \end{aligned}$$

So we have that \(\tilde{\psi }_1(0)=\tilde{\alpha }(x)-\beta (x)+C\) for some constant C. Then (3.5) yields that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau ^2}=\frac{\int _{\Omega }\left[ (1+k)(\tilde{\alpha }-\beta )+(1+2k) \tilde{\alpha }\right] (g-h)dx}{(1+k)|\Omega |}. \end{aligned}$$

Multiplying both sides of the equations of \(\tilde{\alpha }\) and \(\beta \) by \((1+k)\beta \) and \((1+2k)\tilde{\alpha }\) respectively, we obtain that

$$\begin{aligned} (1+k)\int _{\Omega }(g-\tilde{\alpha })\beta dx=(1+2k)\int _{\Omega }(g-h)\tilde{\alpha } dx. \end{aligned}$$

Thus, we derive that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau ^2}=\frac{\int _{\Omega }\left[ \tilde{\alpha }(g-h-\beta ) +\beta h\right] dx}{|\Omega |}. \end{aligned}$$

So the proof of the lemma is complete. \(\square \)

The stability analysis of the semitrivial steady state \((0, {\tilde{v}})\) for small \(\tau >0\) is the same to that of \(({\tilde{u}}, 0),\) which is given by the following lemma.

Lemma 3.3

Let \(\tilde{\alpha }(x), \alpha (x)\) and \(\beta (x)\) be defined by (1.4) and (1.5), respectively.

\({(\textrm{i})}\):

If \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with (1.8), or \(\int _{\Omega } (g-h)dx<0,\) then as \(\tau >0\) is small enough, the semitrivial steady state \((0, {\tilde{v}})\) is asymptotically stable.

\({(\textrm{ii})}\):

If \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }(g-h-\beta )\tilde{\alpha }dx>\int _{\Omega }[(g-h-\beta )\alpha -g\beta ]dx,\) or \(\int _{\Omega } (g-h)dx>0,\) then as \(\tau >0\) is small enough, the semitrivial steady state \((0, {\tilde{v}})\) is unstable.

Next we give the stability of the semitrivial steady states of (1.2) when h is constant. Without loss of generality, we suppose that \(h=0.\) In this case, (1.2) has two semitrivial steady states \(({\tilde{u}}, 0)\) and (0, 1). The stability of \(({\tilde{u}}, 0)\) for small \(\tau >0\) is given by Lemma 3.2. On the other hand, the linearized problem at (0, 1) shows that the stability of (0, 1) is determined by the sign of the principal eigenvalue of the following eigenvalue problem:

$$\begin{aligned} -\mu (1+k)\Delta \varphi -\tau g(x)\varphi =\lambda \varphi , \quad x\in \Omega ,\quad \quad \partial _{\nu }\varphi =0, \quad x\in \partial \Omega . \end{aligned}$$

Due to the variational formulations of eigenvalues in [1, Chapter 2], we have the following result.

Lemma 3.4

Assume that \(h=0.\)

\({(\textrm{i})}\):

If \(\int _{\Omega } g(x)dx\ge 0,\) then the semitrivial steady state (0, 1) of (1.2) is unstable for any \(\tau >0.\)

\({(\textrm{ii})}\):

Assume that \(\int _{\Omega } g(x)dx< 0.\) If g(x) does not change sign on \(\overline{\Omega },\) then (0, 1) is asymptotically stable; if g(x) changes sign on \(\overline{\Omega },\) then (0, 1) is asymptotically stable for \(\tau <\mu (1+k)\eta _1\) and unstable for \(\tau >\mu (1+k)\eta _1,\) where \(\eta _1\) is the positive number defined by

$$\begin{aligned} \frac{1}{\eta _1}=\max _{\varphi \in H^{1}(\Omega )\setminus \{0\}}\frac{\int _{\Omega }g(x)\varphi ^2dx}{\int _{\Omega }|\nabla \varphi |^2dx}. \end{aligned}$$

It should be pointed out that the stability of the semitrivial steady state (0, 1) shown in Lemma 3.4 is not restricted to the case of small \(\tau .\) In addition, it is clear that the stability of (0, 1) may change once as k varies from 0 to \(\infty .\)

4 Structure and Stability of Positive Steady States

In this section, we shall investigate the existence, uniqueness and stability of positive steady states of (1.2) for small \(\tau >0.\) For the analytic framework, let \(X=W_\nu ^{2,p}(\Omega )\times W_\nu ^{2,p}(\Omega )\) and \(Y=L^p(\Omega )\times L^p(\Omega ).\) It is clear that the steady state problem of (1.2) is given by the following problem:

$$\begin{aligned} \left\{ \begin{aligned}&-\mu \Delta \left[ (1+k(u+v))u\right] =u\left[ 1+\tau g(x)-u-v\right] , \quad \quad x\in \Omega , \\&-\mu \Delta \left[ (1+k(u+v))v\right] =v\left[ 1+\tau h(x)-u-v\right] , \quad \quad x\in \Omega ,\\&\partial _{\nu }u=\partial _{\nu }v=0,\quad \,\,\quad \quad \quad \quad \quad \quad \quad \quad \qquad \qquad \quad \quad \quad \quad \quad \quad \,\, \,\,~x\in \partial \Omega . \\ \end{aligned}\right. \end{aligned}$$
(4.1)

By a similar argument to that of [25, Lemma 3.1], we obtain that as \(\tau >0\) is small enough, any positive solution of (4.1) is close to \((s,1-s)\) for \(s\in [0,1].\) So any positive solution (u, v) of (4.1) for small \(\tau >0\) is given by

$$\begin{aligned} (u,v)=(s,1-s)+(y,z), \end{aligned}$$

where \(s\in {\mathbb {R}},\) \((y,z)\in X_1\) is in a neighborhood of (0, 0) and

$$\begin{aligned} X_1=\left\{ (y,z)\in X: \int _{\Omega }(y-z)dx=0\right\} . \end{aligned}$$

In particular, the semitrivial solutions of (4.1) can be written by

$$\begin{aligned} ({\tilde{u}}(\tau ), 0)=(\sigma _1(\tau ), 1-\sigma _1(\tau ))+(\eta _1(\tau ), \varsigma _1(\tau )), \end{aligned}$$
(4.2)
$$\begin{aligned} (0, {\tilde{v}}(\tau ))=(\sigma _2(\tau ), 1-\sigma _2(\tau ))+(\eta _2(\tau ), \varsigma _2(\tau )), \end{aligned}$$
(4.3)

where \(\sigma _i\) and \((\eta _i, \zeta _i)(i=1,2)\) are smooth functions of \(\tau \) with \(\sigma _1(0)=1,\) \(\sigma _2(0)=0\) and \((\eta _i(0), \varsigma _i(0))=(0,0).\)

Theorem 4.1

(i) If \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with (1.7), then there exists a small positive number \(\delta \) such that for each fixed \(\tau \in (0, \delta ),\) (4.1) has a unique positive solution

$$\begin{aligned} (u(\tau ), v(\tau ))=(s(\tau ), 1-s(\tau ))+({\overline{y}}(\tau ),{\overline{z}}(\tau )). \end{aligned}$$
(4.4)

Moreover, \(({\overline{y}}(\tau ),{\overline{z}}(\tau ))\in X_1\) and \(s(\tau )\) are smooth functions with

$$\begin{aligned} ({\overline{y}}(0),{\overline{z}}(0))=(0,0),\quad \quad s(0)=s_0=\frac{\int _{\Omega } [(g-h-\beta )(\tilde{\alpha }-\alpha )+g\beta ]dx }{\int _{\Omega } \left[ \alpha \beta -(g-h)(\alpha -\beta ) \right] dx}. \end{aligned}$$
(4.5)

(ii) If \(\int _{\Omega }(g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with (1.8) or (1.9), then (4.1) has no positive solution for small \(\tau >0.\)

(iii) If \(\int _{\Omega } (g-h)dx\ne 0,\) then (4.1) has no positive solution for small \(\tau >0.\)

Proof

Let \(\delta _1\) be a positive constant and define \(F(y,z,s,\tau ):X_1\times (-\delta _1, 1+\delta _1)\times (-\delta _1, \delta _1)\rightarrow Y\) by

$$\begin{aligned} F(y,z,s,\tau ) ={\begin{pmatrix}\mu (1+k)\Delta y+\mu k\Delta \left[ (s+y) (y+z)\right] +(s+y)\left[ \tau g(x)-(y+z)\right] \\ \mu (1+k)\Delta z+\mu k\Delta \left[ (1-s+z) (y+z)\right] +(1-s+z)\left[ \tau h(x)-(y+z)\right] \end{pmatrix}.} \end{aligned}$$

Then \(F(y,z,s, \tau )\) is well defined and smooth. Moreover, \((u,v)=(s,1-s)+(y,z)\) is a solution of (4.1) if and only if \(F(y,z, s, \tau )=(0,0).\)

It is clear that \(F(0,0,s,0)=(0,0)\) and

$$\begin{aligned} F_{(y, z)}(0, 0, s, 0)[\varphi , \psi ]={\begin{pmatrix} \mu (1+k)\Delta \varphi +\mu ks\Delta (\varphi +\psi )-s(\varphi +\psi )\\ \mu (1+k)\Delta \psi +\mu k(1-s)\Delta (\varphi +\psi )-(1-s)(\varphi +\psi ) \end{pmatrix}.} \end{aligned}$$

Then it follows that

$$\begin{aligned} {\textrm{Ker}}F_{(y, z)}(0, 0, s, 0)= & {} {\text {span}}\{(1,-1)\}, \\ {\textrm{Range}}F_{(y, z)}(0, 0, s, 0)= & {} \left\{ (y,z)\in Y: \int _{\Omega }[(1-s)y-sz]dx=0\right\} . \end{aligned}$$

Define the projection

$$\begin{aligned} P_s(y,z)=\frac{\int _{\Omega } [(1-s)y-sz]dx}{|\Omega |}(1,-1). \end{aligned}$$

Then \(P_s^2=P_s,\) \(\mathrm{{Range}} P_s={\text {span}}\{(1,-1)\}\) and \(P_s F_{(y, z)}(0, 0, s, 0)=(0,0).\) Applying the Lyapunov-Schmidt procedure, \(F(y,z,s,\tau )=(0,0)\) is equivalent to

$$\begin{aligned} \left\{ \begin{aligned}&P_s F(y,z,s, \tau )=(0,0), \\&\left( I- P_s\right) F(y,z,s, \tau )=(0,0). \end{aligned}\right. \end{aligned}$$
(4.6)

Since \({\text {span}}\{(1,-1)\}\cap X_1= \{(0,0)\},\) the implicit function theorem and a compactness argument deduce that there exist a small positive number \(\delta _2\) and a neighborhood U of (0, 0) in \(X_1\) such that the second equation of (4.6) has a unique smooth solution \((y(s, \tau ), z(s, \tau ))\) in U,  which is a smooth function of \((s, \tau )\in (-\delta _2, 1+\delta _2)\times (-\delta _2, \delta _2)\) with

$$\begin{aligned} (y(s, 0), z(s,0))=(0,0). \end{aligned}$$
(4.7)

Hence, we obtain that \((y,z, s, \tau )\in U \times (-\delta _2, 1+\delta _2)\times (-\delta _2, \delta _2)\) satisfies \(F(y,z,s,\tau )=(0,0)\) if and only if \((y,z)=(y(s, \tau ), z(s, \tau ))\) and it solves the first equation of (4.6).

By the definition of \(P_s,\) there exists a smooth scalar function \(\xi (s, \tau )\) such that

$$\begin{aligned} P_sF(y(s,\tau ),z(s, \tau ),s,\tau )=\xi (s, \tau )(1,-1). \end{aligned}$$
(4.8)

Using (4.2), (4.3) and (4.7), it is clear that \(\xi (s, 0)=\xi (\sigma _1(\tau ), \tau )=\xi (\sigma _2(\tau ), \tau )=0,\) which asserts that

$$\begin{aligned} \xi (s, \tau )=\tau (\sigma _1(\tau )-s)(s-\sigma _2(\tau ))\xi _1(s, \tau ) \end{aligned}$$
(4.9)

for a certain smooth scalar function \(\xi _1(s, \tau ).\) Due to (4.8), (4.9) and the definition of \(P_s,\) it is easy to verify that

$$\begin{aligned} \xi _{\tau }(s, 0)=\frac{s(1-s)}{|\Omega |}\int _{\Omega } (g-h)dx =s(1-s)\xi _1(s, 0). \end{aligned}$$

Hence, we derive that

$$\begin{aligned} \xi _1(s, 0)=\frac{1}{|\Omega |}\int _{\Omega } (g-h)dx. \end{aligned}$$

(i) If \(\int _{\Omega } (g-h)dx\ne 0,\) by making \(\delta _2\) smaller if necessary, it can be seen that \(\xi _1(s, \tau )=0\) has no solutions for \((s,\tau )\in (-\delta _2, 1+\delta _2)\times (-\delta _2, \delta _2).\) Thus, the set of solutions of \(\xi (s,\tau )=0\) consists of \(\tau =0,\) \(s=\sigma _1(\tau )\) and \(s=\sigma _2(\tau ),\) which implies that (4.1) has no positive solutions except the trivial and semitrivial solutions for sufficiently small \(\tau >0.\)

(ii) Now we consider the case \(\int _{\Omega } (g-h)dx=0\) and \(g-h\) changes sign on \(\overline{\Omega }.\) Then it follows that \(\xi _1(s, 0)\equiv 0\) and there exists a smooth scalar function \(\xi _2(s, \tau )\) such that

$$\begin{aligned} \xi (s, \tau )=\tau ^2 (\sigma _1(\tau )-s)(s-\sigma _2(\tau ))\xi _2(s, \tau ). \end{aligned}$$

By (4.7), let

$$\begin{aligned} (y(s, \tau ), z(s, \tau ))=\tau \left( y_1(s, \tau ), z_1(s, \tau )\right) . \end{aligned}$$

Then the definition of \(P_s\) and (4.8) yield that

$$\begin{aligned} \begin{aligned} \xi (s,\tau )&=\frac{\tau ^2}{|\Omega |}\left\{ \int _{\Omega } \left[ y_1(s,\tau )+z_1(s, \tau )\right] \left[ sz_1(s,\tau )-(1-s) y_1(s,\tau )\right] dx\right. \\&\quad \left. +(1-s)\int _{\Omega } y_1(s,\tau )g(x)dx-s\int _{\Omega } z_1(s,\tau )h(x)dx\right\} , \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned}&s(1-s)\xi _2(s,0)=\frac{1}{|\Omega |}\left[ \int _{\Omega } \left[ y_1(s,0)+z_1(s, 0)\right] \left[ sz_1(s,0)-(1-s) y_1(s,0)\right] dx\right. \\&\quad \left. +(1-s)\int _{\Omega } y_1(s,0)g(x)dx-s\int _{\Omega } z_1(s,\tau )h(x)dx\right] . \end{aligned} \end{aligned}$$
(4.10)

Since \((y(s, \tau ), z(s, \tau ))\) is the solution of the second equation of (4.6), we have that

$$\begin{aligned} F(y(s,\tau ),z(s,\tau ), s, \tau )=P_sF(y(s,\tau ),z(s,\tau ), s, \tau )=\xi (s,\tau )(1,-1). \end{aligned}$$

Then it is easy to check that \((y_1, z_1)=(y_1(s, 0), z_1(s,0))\) is a solution of

$$\begin{aligned} \left\{ \begin{aligned}&-\mu (1+k) \Delta y_1-\mu ks\Delta (y_1+z_1) +s(y_1+z_1)=sg,\\&-\mu (1+k) \Delta z_1-\mu k(1-s)\Delta (y_1+z_1) +(1-s)(y_1+z_1)=(1-s)h. \end{aligned}\right. \end{aligned}$$
(4.11)

By some computations, we obtain that for some constant C, 

$$\begin{aligned} y_1(s,0)+z_1(s,0)= & {} \tilde{\alpha }-(1-s)\alpha ,\quad \nonumber \\ (1-s)y_1(s,0)-sz_1(s,0)= & {} s(1-s)\beta +C. \end{aligned}$$
(4.12)

Then (4.10) and (4.12) yield that

$$\begin{aligned} \xi _2(s,0)=\frac{1}{|\Omega |}\left\{ s\int _{\Omega }[(g-h)(\alpha -\beta )-\alpha \beta ]dx +\int _{\Omega } [(g-h-\beta )(\tilde{\alpha }-\alpha )+g\beta ] dx\right\} , \end{aligned}$$

which implies that

$$\begin{aligned} \xi _2(0,0)= & {} \frac{1}{|\Omega |}\int _{\Omega }\left[ (g-h-\beta )\tilde{\alpha } +g\beta -(g-h-\beta )\alpha \right] dx, \\ \xi _2(1,0)= & {} \frac{1}{|\Omega |}\int _{\Omega }\left[ (g-h-\beta )\tilde{\alpha } +h\beta \right] dx. \end{aligned}$$

Due to (1.6), one sees that if (1.8) or (1.9) holds true, then \(\xi _2(\cdot , 0)=0\) has no solutions for \(s\in [0,1].\) By making \(\delta _2\) smaller if necessary, \(\xi _2(s, \tau )=0\) has no solutions for \((s,\tau )\in (-\delta _2, 1+\delta _2)\times (-\delta _2, \delta _2),\) which implies that (4.1) does not admit positive solutions for small \(\tau >0.\)

Whereas, if (1.7) holds true, then \(\xi _2(\cdot ,0)=0\) has a unique solution \(s=s_0\in (0,1),\) where \(s_0\) is given by (4.5). Moreover, (1.6) yields that

$$\begin{aligned} \frac{\partial \xi _2 }{\partial s}(s_0, 0)= \frac{1}{|\Omega |}\int _{\Omega }[(g-h)(\alpha -\beta )-\alpha \beta ]dx <0. \end{aligned}$$

Then the implicit function theorem yields that there exists a small positive number \(\delta _3\) such that \(\xi _2(s, \tau )=0\) has a unique solution \(s=s(\tau )\) for \(\tau \in (-\delta _3, \delta _3).\) Moreover, \(s(\tau )\) is a smooth function with \(s(0)=s_0.\) As a consequence, (4.1) has a unique positive solution \((u(\tau ), v(\tau ))\) for small \(\tau >0.\) Since \((u(\tau ),v(\tau ))=(s(\tau ), 1-s(\tau ))+(y(s(\tau ), \tau ),z(s(\tau ),\tau )),\) one sees that (4.4) follows by setting \(({\overline{y}}(\tau ),{\overline{z}}(\tau ))=(y(s(\tau ),\tau ),z(s(\tau ), \tau )).\) The proof of the theorem is complete. \(\square \)

Applying Theorem 4.1, the structure of positive solutions of (4.1) for \(h=0\) is given by the following result.

Corollary 4.2

Assume that \(h(x)=0.\) Let \(\alpha \) and \(\beta \) be the functions given by (1.4) and (1.5) for \(h=0\).

(i) If \(\int _{\Omega }gdx=0,\) g changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }g\alpha dx<\int _{\Omega }\alpha \beta dx,\) then there exists a small positive number \(\delta \) such that for \(0<\tau <\delta ,\) (4.1) has a unique positive solution \((u(\tau ), v(\tau ))\) near \((s_0, 1-s_0),\) where

$$\begin{aligned} s_0=\frac{\int _{\Omega }g\beta dx}{\int _{\Omega }(g\beta +\alpha \beta -g\alpha )dx}. \end{aligned}$$

(ii) If \(\int _{\Omega }gdx=0,\) g changes sign on \(\overline{\Omega }\) with \(\int _{\Omega }g\alpha dx>\int _{\Omega }\alpha \beta dx,\) or \(\int _{\Omega }g dx\ne 0,\) then (4.1) has no positive solution for small \(\tau >0.\)

In the following, we focus on the stability of positive steady state of (1.2) given by Theorem 4.1. Linearizing (1.2) at the positive steady state \((u,v)=(u(\tau ), v(\tau )),\) we obtain the following eigenvalue problem:

$$\begin{aligned} \left\{ \begin{aligned}&-\mu \Delta \left[ (1+k(u+v))\varphi \right] -\mu k \Delta [u(\varphi +\psi )]+A\varphi +B\psi =\lambda \varphi ,\quad x\in \Omega ,\\&\quad -\mu \Delta \left[ (1+k(u+v))\psi \right] -\mu k \Delta [v(\varphi +\psi )]+C\varphi +D\psi =\lambda \psi ,\quad x\in \Omega ,\\&\quad \partial _{\nu }\varphi =\partial _{\nu }\psi =0, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \qquad \quad \quad \quad \quad \quad \quad \quad \quad \,\, x\in \partial \Omega , \end{aligned}\right. \end{aligned}$$
(4.13)

where

$$\begin{aligned} \left\{ \begin{aligned}&A=A(u,v,\tau )=-(1+\tau g-2u-v),\quad B=B(u,v,\tau )=u,\\&C=C(u,v,\tau )=v,\quad \quad \quad \quad \quad \,\,\,\quad \quad \quad \, D=D(u,v,\tau )=-(1+ \tau h-u-2v). \end{aligned}\right. \end{aligned}$$

By virtue of Theorem 4.1, one sees that \((u(\tau ), v(\tau ))\rightarrow (s_0, 1-s_0)\) as \(\tau \rightarrow 0.\) Hence, (4.13) at \(\tau =0\) has a principal eigenvalue 0 and all other eigenvalues have positive real parts. In particular, the eigenfunction corresponding to the principal eigenvalue 0 can be chosen as \((1,-1).\) So for sufficiently small \(\tau >0,\) (4.13) has a real simple principal eigenvalue \(\lambda _1=\lambda _1(\tau )\) with \(\lambda _1(\tau )\rightarrow 0\) as \(\tau \rightarrow 0.\) Moreover, it can be supposed that the eigenfunction \((\varphi (\tau ), \psi (\tau ))\) corresponding to \(\lambda _1(\tau )\) tends to \((1,-1)\) as \(\tau \rightarrow 0.\) Therefore, the stability of the positive steady state \((u(\tau ), v(\tau ))\) of (1.2) is determined by the sign of \(\lambda _1(\tau ).\)

Theorem 4.3

Suppose that \(\int _{\Omega } (g-h)dx=0,\) \(g-h\) changes sign on \(\overline{\Omega }\) with (1.7). Then as \(\tau >0\) is sufficiently small, the unique positive solution \((u(\tau ), v(\tau ))\) of (4.1) is asymptotically stable.

Proof

Let (u, v) be the positive solution of (4.1) given by Theorem 4.1, and let \((\varphi , \psi )\) be the eigenfunction corresponding to the principal eigenvalue \(\lambda _1(\tau ).\) Multiplying the first and second equation of (4.13) by v and u,  and integrating over \(\Omega \) respectively, we obtain that

$$\begin{aligned} \lambda _1(\tau ) \int _{\Omega }(\varphi v-\psi u)&=\mu k\left\{ \int _{\Omega }\varphi \Delta [v(u+v)]dx -\int _{\Omega }\psi \Delta [ u(u+v)]dx\right. \\&\quad \left. +\int _{\Omega }u\Delta [v(\varphi +\psi )]dx\right. \\&\quad \left. +\int _{\Omega }u\Delta [\psi (u+v)]dx -\int _{\Omega }\left[ v\left[ \Delta [u(\varphi +\psi )]\right. \right. \right. \\&\quad \left. \left. \left. +\Delta [\varphi (u+v)]\right] \right] dx\right\} \\&\quad -\tau \int _{\Omega }(g-h)(\varphi v+\psi u)dx \\&=I_1(\tau )+I_2(\tau ). \end{aligned}$$

Let

$$\begin{aligned} u(\tau )= & {} s_0+\tau U(\tau ), ~~ v(\tau )=1-s_0+\tau V(\tau ),~~\nonumber \\ \varphi (\tau )= & {} 1+\tau \tilde{\varphi }(\tau ), ~~ \psi (\tau )=-1+\tau \tilde{\psi }(\tau ), \end{aligned}$$
(4.14)

where \(s_0\) is given by Theorem 4.1, \(U, V, \tilde{\varphi }\) and \(\tilde{\psi }\) are smooth functions with respect to \(\tau .\) By (4.14) and some computations, we deduce that as \(\tau \rightarrow 0,\)

$$\begin{aligned} I_1(\tau )&=\mu k \tau ^2 \int _{\Omega }\left\{ (1-s_0)\tilde{\varphi }(0)\Delta \left( U(0)+V(0)\right) -s_0\tilde{\psi }(0) \Delta (U(0)+V(0))\right. \\&\quad \left. +U(0)\left[ (1-s_0)\Delta \left( \tilde{\varphi }(0)+ \tilde{\psi }(0)\right) -\Delta \left( U(0)+V(0)\right) \right] \right. \\&\quad \left. -V(0)\left[ s_0\Delta \left( \tilde{\varphi }(0)+\tilde{\psi }(0)\right) + \Delta \left( U(0)+V(0)\right) \right] \right\} dx+o(\tau ^2),\\ I_2(\tau )&=-\tau ^2\int _{\Omega }(g-h)\left[ V(0)-U(0)+(1-s_0)\tilde{\varphi }(0)+ s_0\tilde{\psi }(0)\right] dx+o(\tau ^2), \end{aligned}$$

where the assumption \(\int _{\Omega }(g-h)dx=0\) is used. Since \(\int _{\Omega }(\varphi v-\psi u)dx\rightarrow |\Omega |\) as \(\tau \rightarrow 0,\) it is clear that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau }=0. \end{aligned}$$
(4.15)

By (4.1) and the expansion (4.14), it is easy to verify that \((U,V)=(U(0), V(0))\) is a solution of

$$\begin{aligned} \left\{ \begin{aligned}&-\mu (1+k)\Delta U-\mu ks_0\Delta (U+V)+s_0(U+V)=s_0g, \\&-\mu (1+k)\Delta V-\mu k(1-s_0)\Delta (U+V)+(1-s_0)(U+V)=(1-s_0)h. \end{aligned}\right. \end{aligned}$$
(4.16)

Then it follows that for some constant \(C_1,\)

$$\begin{aligned} U(0)+V(0)= & {} \tilde{\alpha }-(1-s_0)\alpha ,\quad \quad \nonumber \\ (1-s_0)U(0)-s_0V(0)= & {} s_0(1-s_0)\beta +C_1. \end{aligned}$$
(4.17)

By (4.13), (4.14) and (4.15), we deduce that \((\tilde{\varphi }, \tilde{\psi })=(\tilde{\varphi }(0),\tilde{\psi }(0))\) satisfies

$$\begin{aligned} \left\{ \begin{aligned}&-\mu (1+k)\Delta \tilde{\varphi }-\mu ks_0\Delta \left( \tilde{\varphi }+\tilde{\psi }\right) +s_0\left( \tilde{\varphi }+\tilde{\psi }\right) =A_1,\\&-\mu (1+k)\Delta \tilde{\psi }-\mu k(1-s_0)\Delta \left( \tilde{\varphi }+\tilde{\psi }\right) +(1-s_0)\left( \tilde{\varphi }+\tilde{\psi }\right) =B_1, \end{aligned}\right. \end{aligned}$$
(4.18)

where

$$\begin{aligned} \left\{ \begin{aligned}&A_1=g+\mu k\Delta \left( U(0)+V(0)\right) -\left( U(0)+V(0)\right) ,\\&B_1=-h-\mu k\Delta \left( U(0)+V(0)\right) +\left( U(0)+V(0)\right) . \end{aligned}\right. \end{aligned}$$

By (4.17) and some computations, we obtain that for some constant \(C_2,\)

$$\begin{aligned} \tilde{\varphi }(0)+\tilde{\psi }(0)= & {} \alpha ,\quad \quad \nonumber \\ (1-s_0)\tilde{\varphi }(0)-s_0\tilde{\psi }(0)-(U(0)+V(0))= & {} (1-2s_0)\beta +C_2. \end{aligned}$$
(4.19)

Multiplying the first and second equation of (4.16) by \(\tilde{\psi }(0)\) and \(\tilde{\varphi }(0),\) and multiplying the first and second equation of (4.18) by V(0) and U(0) respectively, some detailed computations deduce that as \(\tau \rightarrow 0,\)

$$\begin{aligned} I_1(\tau )&=\tau ^2\int _{\Omega }\left\{ \left( U(0)+V(0)\right) \left[ (1-s_0)\tilde{\varphi }(0)-s_0 \tilde{\psi }(0)-\left( U(0)+V(0)\right) \right] \right. \\&\quad \left. +\left( \tilde{\varphi }(0)+\tilde{\psi }(0)\right) \left[ (1-s_0)U(0)-s_0V(0)\right] + g\left[ V(0)+s_0\tilde{\psi }(0)\right] \right. \\&\quad \left. +h\left[ U(0)-(1-s_0)\tilde{\varphi }(0)\right] \right\} dx+o(\tau ^2). \end{aligned}$$

Then it follows that as \(\tau \rightarrow 0,\)

$$\begin{aligned} \frac{I_1(\tau )+I_2(\tau )}{\tau ^2}&=\int _{\Omega }\left( U(0)+V(0)\right) \left[ (1-s_0)\tilde{\varphi }(0)-s_0 \tilde{\psi }(0)-\left( U(0)+V(0)\right) \right] dx\\&\quad +\int _{\Omega }\left( \tilde{\varphi }(0)+\tilde{\psi }(0)\right) \left[ (1-s_0)U(0)-s_0V(0)\right] dx\\&\quad +\int _{\Omega }\left\{ g\left[ U(0)-(1-s_0)\tilde{\varphi }(0)\right] +h\left[ V(0)+s_0\tilde{\psi }(0)\right] \right\} dx+O(\tau ). \end{aligned}$$

Due to (4.17) and (4.19), we obtain that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{I_1(\tau )+I_2(\tau )}{\tau ^2}&=(1-s_0)(3s_0-1)\int _{\Omega }[g\beta -(g-h-\beta )\alpha ]dx\\&\quad -s_0(2-3s_0)\int _{\Omega } h\beta dx\\&\quad +(1-2s_0)\int _{\Omega }\tilde{\alpha }[\beta -(g-h)] dx. \end{aligned}$$

By (1.6), (4.5) and some detailed computations, we deuce that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\frac{\lambda _1(\tau )}{\tau ^2}=\frac{s_0(1-s_0)}{|\Omega |} \int _{\Omega } [\alpha \beta +(g-h)(\beta -\alpha )] dx>0. \end{aligned}$$

Hence, \(\lambda _1(\tau )>0\) for small \(\tau >0.\) The proof of the theorem is complete. \(\square \)

Due to Theorem 4.3, the stability of the positive steady state of (1.2) for \(h=0\) is deduced.

Corollary 4.4

Suppose that \(h=0,\) \(\int _{\Omega } gdx=0,\) g changes sign on \(\overline{\Omega }\) with \(\int _{\Omega } g\alpha dx<\int _{\Omega } \alpha \beta dx,\) where \(\alpha \) and \(\beta \) are given by (1.4) and (1.5) for \(h=0,\) respectively. Then as \(\tau >0\) is sufficiently small, the unique positive solution \((u(\tau ), v(\tau ))\) of (4.1) is asymptotically stable.

5 Asymptotic Behavior of Positive Steady States

In this section, we are concerned about the asymptotic behavior of positive steady states of (1.2) as \(k\rightarrow \infty .\) In fact, this work can be done for the more general system (1.1). First, a priori estimate of positive steady states of (1.1) independent of k is needed, which is given by the following lemma.

Lemma 5.1

Let (u, v) be a positive steady state of (1.1). Then (u, v) satisfies

$$\begin{aligned} \Vert u\Vert _{\infty }\le & {} \max \left\{ \left( 1+\Vert a\Vert _{\infty } \right) \Vert a\Vert _{\infty }, \sqrt{\left( 1+ \Vert a\Vert _{\infty }\right) \Vert a\Vert _{\infty }}\right\} , \end{aligned}$$
(5.1)
$$\begin{aligned} \Vert v\Vert _{\infty }\le & {} \max \left\{ \left( 1+\Vert b\Vert _{\infty } \right) \Vert b\Vert _{\infty }, \sqrt{\left( 1+ \Vert b\Vert _{\infty }\right) \Vert b\Vert _{\infty }}\right\} . \end{aligned}$$
(5.2)

Proof

Let \(U=(1+k(u+v))u\) and \(V=(1+k(u+v))v.\) Set \(U(x_0)=\max _{\overline{\Omega }} U.\) By the maximum principle, we obtain that

$$\begin{aligned} u(x_0)+v(x_0)\le a(x_0)\le \Vert a\Vert _{\infty }. \end{aligned}$$

If \(k\le 1, \) then we have that

$$\begin{aligned} u\le U\le U(x_0)\le \left( 1+ \Vert a\Vert _{\infty } \right) \Vert a\Vert _{\infty }. \end{aligned}$$
(5.3)

Since \(v=uV/U,\) some computations deduce that

$$\begin{aligned} u=\frac{2U}{1+\sqrt{1+4k(U+V)}}. \end{aligned}$$

Hence, one sees that if \(k\ge 1,\) then

$$\begin{aligned} u<\sqrt{\frac{U}{k}}\le \sqrt{U(x_0)}\le \sqrt{\left( 1+ \Vert a\Vert _{\infty }\right) \Vert a\Vert _{\infty }}. \end{aligned}$$
(5.4)

Hence, (5.1) follows by (5.3) and (5.4). The proof of (5.2) is similar and the proof of the lemma is complete. \(\square \)

Theorem 5.2

Suppose that \(\int _{\Omega }a(x) dx=\int _{\Omega } b(x)dx\) and let (u(k), v(k)) be a positive steady state of (1.1). Then as \(k\rightarrow \infty ,\)

$$\begin{aligned} (u(k),v(k))\rightarrow \left( \frac{s}{K}, K-\frac{s}{K}\right) \quad \quad {\text { uniformly in}} ~\overline{\Omega }\times \overline{\Omega }, \end{aligned}$$
(5.5)

where \(s\in [0, K]\) and

$$\begin{aligned} K=\frac{1}{|\Omega |}\int _{\Omega }a(x)dx=\frac{1}{|\Omega |}\int _{\Omega }b(x)dx. \end{aligned}$$

Proof

Let \(\{k_i\}\) be a sequence with \(\lim _{i\rightarrow \infty }k_i=\infty ,\) and let \((u_i, v_i)\) be a positive steady state of (1.1) at \(k=k_i.\) Then Lemma 5.1 yields that both \(u_i\) and \(v_i\) are uniformly bounded. Let

$$\begin{aligned} U_i=\left( \frac{1}{k_i}+u_i+ v_i\right) u_i,\quad \quad V_i=\left( \frac{1}{k_i}+u_i+ v_i\right) v_i. \end{aligned}$$

Then we obtain that both \(U_i\) and \(V_i\) are uniformly bounded. By a similar argument to that in the proof of Lemma 2.2, we obtain that subject to a subsequence if necessary, \((U_i, V_i)\rightarrow (C_1, C_2)\) in \(C^1\times C^1\) as \(i\rightarrow \infty \) for some nonnegative constants \(C_1\) and \(C_2.\) It follows that \(u_i(u_i+v_i)\rightarrow C_1\) and \(v_i(u_i+v_i)\rightarrow C_2\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty ,\) which implies that \(u_i+v_i\rightarrow \sqrt{C_1+C_2}\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\)

We claim that \(C_1+C_2>0.\) If not, then \(C_1=C_2=0\) and \(u_i,v_i\rightarrow 0\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\) If \(k_i\Vert u_i+v_i\Vert _{\infty }\) is uniformly bounded, then \(\overline{u_i}=k_i u_i\) and \(\overline{v_i}=k_iv_i\) are uniformly bounded. Since \((\overline{u_i}, \overline{v_i})\) satisfies

$$\begin{aligned} \left\{ \begin{aligned}&-\mu \Delta \left[ (1+\overline{u_i}+\overline{v_i})\overline{u_i}\right] =\overline{u_i}\left[ a(x)-u_i-v_i\right] ,\quad \quad x\in \Omega , \\&-\mu \Delta \left[ (1+\overline{u_i}+\overline{v_i})\overline{v_i}\right] =\overline{v_i} \left[ b(x)-u_i-v_i\right] , \, \quad \quad x\in \Omega , \end{aligned}\right. \end{aligned}$$
(5.6)

the standard argument yields that subject to a subsequence if necessary, \((1+\overline{u_i}+\overline{v_i})\overline{u_i}\rightarrow {\overline{U}}\) and \((1+\overline{u_i}+\overline{v_i})\overline{v_i}\rightarrow {\overline{V}}\) in \(C^1\) as \(i\rightarrow \infty \) for some nonnegative functions \({\overline{U}}\) and \({\overline{V}}.\) Then one sees that \(\overline{u_i}\rightarrow {\bar{u}}\) and \(\overline{v_i}\rightarrow {\bar{v}}\) in \(C^1\) as \(i\rightarrow \infty ,\) where \({\bar{u}}\) and \({\bar{v}}\) are nonnegative functions with \({\overline{U}}=(1+{\bar{u}}+{\bar{v}}){\bar{u}}\) and \({\overline{V}}=(1+{\bar{u}}+{\bar{v}}){\bar{v}}.\) Passing to the weak limit in (5.6), we have that \({\bar{u}}\) is a nonnegative solution of

$$\begin{aligned} -\mu \Delta \left[ (1+{\bar{u}}+{\bar{v}}){\bar{u}}\right] ={\bar{u}}a(x),\quad x\in \Omega , \quad \quad \partial _{\nu }{\bar{u}}=0,\quad x\in \partial \Omega . \end{aligned}$$

Applying the maximum principle, one sees that \({\bar{u}}>0\) or \({\bar{u}}=0\) in \(\overline{\Omega }.\) If \({\bar{u}}>0,\) then

$$\begin{aligned} \int _{\Omega }{\bar{u}}a(x) dx>0, \end{aligned}$$

which is impossible. If \({\bar{u}}=0,\) then the standard argument deduces that subject to a subsequence if necessary, \(\widehat{u_i}=u_i/\Vert u_i\Vert _{\infty }\rightarrow {\hat{u}}\) in \(C^1\) as \(i\rightarrow \infty \) for some nonnegative function \({\hat{u}},\) which satisfies

$$\begin{aligned} -\mu \Delta \left[ (1+{\bar{v}}){\hat{u}}\right] ={\hat{u}}a(x),\quad x\in \Omega , \quad \quad \partial _{\nu }{\hat{u}}=0,\quad x\in \partial \Omega . \end{aligned}$$

As \({\hat{u}}\) is nontrivial, the maximum principle asserts that \({\hat{u}}>0\) in \(\overline{\Omega }.\) Then a contraction is derived.

If there exists a subsequence of \(\{k_i\},\) which is still denoted by itself, such that \(k_i\Vert u_i+v_i\Vert _{\infty }\rightarrow \infty \) as \(i\rightarrow \infty .\) Set

$$\begin{aligned} \widetilde{u_i}=\frac{u_i}{\Vert u_i+v_i\Vert _{\infty }},\quad \quad \widetilde{v_i}=\frac{v_i}{\Vert u_i+v_i\Vert _{\infty }}. \end{aligned}$$

Then \((\widetilde{u_i}, \widetilde{v_i})\) satisfies

$$\begin{aligned} \left\{ \begin{aligned} -\mu \Delta \left[ \left( \frac{1}{k_i\Vert u_i+v_i\Vert _{\infty }}+\widetilde{u_i}+\widetilde{v_i}\right) \widetilde{u_i}\right]&=\frac{\widetilde{u_i}(a(x)-u_i-v_i)}{k_i\Vert u_i+v_i\Vert _{\infty }},\quad \quad x\in \Omega , \\ -\mu \Delta \left[ \left( \frac{1}{k_i\Vert u_i+v_i\Vert _{\infty }}+\widetilde{u_i}+\widetilde{v_i}\right) \widetilde{v_i}\right]&=\frac{\widetilde{v_i}(b(x)-u_i-v_i)}{k_i\Vert u_i+v_i\Vert _{\infty }}, \, \quad \quad x\in \Omega . \end{aligned}\right. \end{aligned}$$

Then subject to a subsequence if necessary, \((\widetilde{u_i}+\widetilde{v_i})\widetilde{u_i}\rightarrow C_3\) and \((\widetilde{u_i}+\widetilde{v_i})\widetilde{v_i}\rightarrow C_4\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty \) for some nonnegative constants \(C_3\) and \(C_4,\) which implies that \(\widetilde{u_i}+\widetilde{v_i}\rightarrow \sqrt{C_3+C_4}\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\) As \(\Vert \widetilde{u_i}+\widetilde{v_i}\Vert _{\infty }=1,\) we have that \(C_3+C_4>0.\) So \(C_3>0\) or \(C_4>0.\) If \(C_3>0,\) then one sees that \(\widetilde{u_i}\rightarrow C_3/\sqrt{C_3+C_4}>0\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\) As

$$\begin{aligned} \int _{\Omega }\widetilde{u_i}(a(x)-u_i-v_i)dx=0, \end{aligned}$$

a contradiction can be derived by letting \(i\rightarrow \infty \) in the above equation. Similarly, a contradiction can be derived as \(C_4>0.\) Therefore, we have that \(C_1+C_2>0.\)

Since \(C_1+C_2>0,\) one sees that \(i\rightarrow \infty ,\)

$$\begin{aligned} (u_i,v_i)\rightarrow \left( \frac{C_1}{\sqrt{C_1+C_2}},\frac{C_2}{\sqrt{C_1+C_2}} \right) \quad \quad {\text {uniformly in}}~ \overline{\Omega }\times \overline{\Omega }. \end{aligned}$$
(5.7)

If \(C_1>0,\) letting \(i\rightarrow \infty \) in

$$\begin{aligned} \int _{\Omega }u_i(a(x)-u_i-v_i)dx=0, \end{aligned}$$

we have that

$$\begin{aligned} \sqrt{C_1+C_2}=\frac{1}{|\Omega |}\int _{\Omega }a(x)dx. \end{aligned}$$
(5.8)

Similarly, if \(C_2>0,\) it holds that

$$\begin{aligned} \sqrt{C_1+C_2}=\frac{1}{|\Omega |}\int _{\Omega }b(x)dx. \end{aligned}$$
(5.9)

Hence, one sees that (5.5) holds true by setting \(s=C_1\) and \(K=\sqrt{C_1+C_2}.\) The proof of the theorem is complete. \(\square \)

Corollary 5.3

If \(\int _{\Omega }a(x) dx\ne \int _{\Omega } b(x)dx,\) then (1.1) has no positive steady state as k is large enough.

Proof

If the conclusion is not true, then there exists a sequence \(\{k_i\}\) with \(\lim _{i\rightarrow \infty }k_i=\infty \) such that (1.1) at \(k=k_i\) has a positive steady state \((u_i, v_i).\) Then the proof of Theorem 5.2 shows that (5.7) holds true.

If \(C_1>0\) and \(C_2>0,\) then it is easy to see that (5.8) and (5.9) hold true at the same time, which is impossible. If \(C_1>0,\) \(C_2=0,\) then (5.8) with \(C_2=0\) holds true. Let \(\widehat{v_i}=v_i/\Vert v_i\Vert _{\infty }.\) By a standard argument, we deduce that subject to a subsequence if necessary, \((u_i+v_i)\widehat{v_i}\rightarrow C_5\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty \) for some nonnegative constant \(C_5.\) Hence, it follows that \(\widehat{v_i}\rightarrow C_5/\sqrt{C_1}\) uniformly in \(\overline{\Omega }\) as \(i\rightarrow \infty .\) As \(\Vert \widehat{v_i}\Vert _{\infty }=1,\) we have that \(C_5>0.\) Since

$$\begin{aligned} \int _{\Omega }\widehat{v_i}(b(x)-u_i-v_i)dx=0, \end{aligned}$$

letting \(i\rightarrow \infty ,\) one sees that (5.9) with \(C_2=0\) holds true at the same time, which is a contradiction. Similarly, a contradiction can be derived as \(C_1=0\) and \(C_2>0.\) So the proof of the corollary is complete. \(\square \)