Now we are in a position to prove Theorem 1.3. Note that, by Theorem 1.2 and the assumptions of Theorem 1.3, problem (1.2) has a unique positive solution \((\tilde{u}(x), \tilde{u}(x))\). Then we will establish Theorem 1.3 by proving the following theorem without the assumption (1.10).

### Theorem 3.1

*Assume the hypotheses of Theorem *
1.1
*and that problem* (1.2) *has a unique positive solution*
\((\tilde{u}(x), \tilde{u}(x))\)
*in* Ω, *then*
\((\tilde{u}(x), \tilde{u}(x))\)
*is globally asymptotically stable in the following sense*. *Let*
\((u(x,t), v(x,t))\)
*be a solution of the initial boundary value problem* (1.1) *with both*
\(u^{0}, v^{0}\geq0,\not\equiv0\)
*in*
\(C^{\alpha}(\overline{\Omega})\), \(0<\alpha<1\), *and vanishing on*
*∂*Ω, *then*

$$\bigl(u(x,t), v(x,t)\bigr)\rightarrow\bigl(\tilde{u}(x), \tilde{u}(x)\bigr) \quad\textit{as } t\rightarrow\infty, $$

*uniformly in*
\(\overline{\Omega}\).

### Proof

For convenience, we introduce the following notation: If \(w\in C^{1}(\overline{\Omega})\), \(w(x)>0\) for all \(x\in\Omega\), and \(\partial w/\partial\nu<0\) everywhere on *∂*Ω, we write \(w\gg0\). If \(w, z\in C^{1}(\overline{\Omega})\), we write \(w\ll z\) if \(z-w \gg0\). We first prove the theorem under the additional conditions \(u^{0}, v^{0}\in C^{1}(\overline{\Omega})\),

$$ u^{0}\gg0, \qquad v^{0}\gg0, $$

(3.1)

and for all \(x\in\overline{\Omega}\),

$$ u^{0}\leq\bar{u}, \qquad v^{0}\leq\bar{v}, $$

(3.2)

where \(\bar{u}\) and \(\bar{v}\) are defined in (1.6).

Let \(\phi_{1}\) be the positive eigenfunction of the principal eigenvalue in (1.3). Choose \(\epsilon>0\) small such that

$$ \epsilon\phi_{1}(x)\leq u^{0}(x), \qquad \epsilon\phi_{1}(x)\leq v^{0}(x), $$

(3.3)

and

$$ \begin{aligned} &a(x)>\mu_{1} \lambda_{1}+c(x)\bar{v}+b(x)\epsilon\phi_{1}(x), \\ &d(x)>\mu_{2}\lambda_{1}+e(x)\bar{u}+f(x)\epsilon \phi_{1}(x), \end{aligned} $$

(3.4)

for all \(x\in\overline{\Omega}\). If we let \(\underline{u}=\underline {v}=\epsilon\phi_{1}\), then

$$\mu_{1}\Delta\bar{u}+\bar{u}\bigl[a(x)-b\bar{u}-c(x)\underline{v} \bigr]=\bar {u}\biggl[\biggl(1-\frac{b(x)}{\min_{x\in\overline{\Omega}}b(x)}\biggr)\theta_{\mu_{1}, a(x)}-c(x) \underline{v}\biggr]< 0, $$

for all \(x\in\Omega\); and from (3.4), we have

$$\mu_{2}\Delta\underline{v}+\underline{v}\bigl[d(x)-e\bar{u}-f(x) \underline {v}\bigr]=\underline{v}\bigl[d(x)-\mu_{2} \lambda_{1}-e(x)\bar{u}-f(x)\underline{v}\bigr]>0, $$

on Ω. Similarly, we have

$$\begin{aligned}& \mu_{1}\Delta\underline{u}+\underline{u}\bigl[a(x)-b(x) \underline{u}-c(x)\bar{v}\bigr]>0, \\& \mu_{2}\Delta\bar{v}+\bar{v}\bigl[d(x)-e(x)\underline{u}-f(x)\bar{v} \bigr]< 0. \end{aligned}$$

By Theorem 1.3 in [1] (also see Pao [11], Section 10.5), the conclusion of the theorem follows from the uniqueness assumption, the inequalities \(\underline {u}(x)\leq u^{0}\leq\bar{u}(x)\), \(\underline{v}(x)\leq v^{0}\leq\bar{v}(x)\), \(x\in\overline{\Omega}\), and a comparison with solutions of the differential system (1.1) with initial conditions replaced at the steady-state upper lower solutions \((\bar{u}(x), \underline{v}(x))\).

We next remove condition (3.2) on the initial functions \(u^{0}(x)\), \(v^{0}(x)\). First, observe that there exists large \(K>1\), such that

$$u^{0}(x)\leq K\bar{u}, \qquad V^{0}(x)\leq K\bar{v}, $$

on Ω. Define \((\overline{U}(x,t), \underline{V}(x,t))\) to be the solution of problem (1.1) with initial conditions replaced with

$$\bigl(\overline{U}(x,0), \underline{V}(x,0)\bigr)=(K\bar{u}, 0). $$

It is clear that \(\underline{V}\equiv0\), \(\overline{U}\) is non-negative in \(\Omega\times[0, \infty)\) and

$$ \lim_{t\rightarrow\infty}\overline{U}(x,t)= \overline{U}^{\star}(x) \quad \mbox{for } x\in\Omega, $$

(3.5)

where \(\overline{U}^{\star}(x)\) is the unique positive solution of the problem

$$ \mu_{1}\Delta z+z\bigl[a(x)-b(x)z\bigr]=0 \quad\mbox{in } \Omega, \qquad z=0 \quad\mbox{on } \partial\Omega. $$

(3.6)

Moreover, the convergence above is monotone, because \(\overline {U}(x,0)\), \(\underline{V}(x,0)\) satisfies

$$\begin{aligned}& \mu_{1}\Delta\overline{U}(x,0)+\overline{U}(x,0)\bigl[a(x)-b(x) \overline {U}(x,0)-c(x)\underline{V}(x,0)\bigr]\\& \quad=\bigl(\overline{U}(x,0) \bigr)^{2}\biggl[\frac{\min_{x\in\overline{\Omega}}b(x)}{K}-b\biggr]< 0, \\& \mu_{2}\Delta\underline{V}(x,0)+\underline{V}(x,0)\bigl[d(x)-e(x) \overline {U}(x,0)-f(x)\underline{V}(x,0)\bigr]=0. \end{aligned}$$

The convergence in (3.5) is also in \(C^{1}(\overline {\Omega})\) norm by using the \(W^{2,p}\) estimates, compact embedding, and (1.1). Similarly, define \((\underline{U}(x,t), \overline{V}(x,t))\) to be the solution of problem (1.1) with initial conditions replaced with

$$\bigl(\underline{U}(x,0), \overline{V}(x,0)\bigr)=(0, K\bar{v}). $$

We have \(\underline{U}\equiv0\), \(\overline{V}\) is non-negative in \(\Omega\times[0, \infty)\), and we have monotone \(C^{1}(\overline{\Omega })\) convergence,

$$ \lim_{t\rightarrow\infty}\overline{V}(x,t)= \overline{V}^{\star}(x), $$

(3.7)

where \(\overline{V}^{\star}(x)\) is the unique positive solution of the problem

$$ \mu_{2}\Delta z+z\bigl[d(x)-f(x)z\bigr] \quad\mbox{in } \Omega, \qquad z|_{\partial\Omega}=0. $$

(3.8)

On the other hand, one readily verifies that the functions \(\underline {U}(x,t)\), \(\overline{U}(x,t)\), \(\underline{V}(x,t)\), \(\overline {V}(x,t)\) satisfy

$$ \begin{aligned} &\mu_{1}\Delta\overline{U}+ \overline{U}\bigl[a(x)-b(x)\overline {U}-c(x)\underline{V}\bigr]-\partial \overline{U}/\partial t< 0, \\ &\mu_{2}\Delta\underline{V}+\underline{V}\bigl[d(x)-e(x)\overline {U}-f(x)\underline{V}\bigr]-\partial\underline{V}/\partial t\geq0, \\ &\mu_{2}\Delta\overline{V}+\overline{V}\bigl[d(x)-e(x)\underline {U}-f(x)\overline{V}\bigr]-\partial\overline{V}/\partial t< 0, \\ &\mu_{1}\Delta\underline{U}+\underline{U}\bigl[a(x)-b(x)\underline {U}-c(x)\overline{V}\bigr]-\partial\underline{U}/\partial t\geq0, \end{aligned} $$

(3.9)

for \((x,t)\in\Omega\times(0, \infty)\), and

$$ \begin{aligned} &0=\underline{U}(x,0)\leq u^{0}(x)\leq \overline{U}(x,0)=K\bar{u}, \\ &0=\underline{V}(x,0)\leq v^{0}(x)\leq\overline{V}(x,0)=K\bar{v}, \end{aligned} $$

(3.10)

for \(x\in\overline{\Omega}\). From the comparison theorems, we assert that

$$ \begin{aligned} &0=\underline{U}(x,t)\leq u^{0}(x,t)\leq\overline{U}(x,t), \\ &0=\underline{V}(x,t)\leq v^{0}(x,t)\leq\overline{V}(x,t), \end{aligned} $$

(3.11)

for \((x,t)\in\Omega\times[0,\infty)\). We next observe that \(\mu _{1}\Delta\bar{u}+\bar{u}[a(x)-b(x)\bar{u}]<0\) in Ω, \(\bar {u}|_{\partial\Omega}=0\), thus \(\bar{u}=\theta_{\mu_{1},a(x)}/\min_{x\in\overline{\Omega}}b(x)\) is a strict upper solution of the problem (3.6). Similarly, \(\bar{v}\) is a strict upper solution of the problem (3.8).

By monotone iteration and comparison, we obtain

$$ \overline{U}^{\star}\ll\bar{u}, \qquad \overline{V}^{\star} \ll\bar{v}. $$

(3.12)

For \(s>0\), let \(u^{s}(x)=u(x,s)\), \(v^{s}(x)=v(x,s)\) for \(x\in\overline {\Omega}\). We obtain from (3.5), (3.7), (3.11), and (3.12) for \(s>0\) sufficiently large

$$ u^{s}(x)\leq\bar{u}, \qquad v^{s}(x)\leq \bar{v}, $$

(3.13)

for \(x\in\overline{\Omega}\). On the other hand for \(s>0\), we find from the theory of parabolic equations and the strong maximum principle that \(u^{s}\), \(v^{s}\) are in \(C^{1}(\overline{\Omega})\) and

$$ u^{s}(x)\gg0,\qquad v^{s}(x)\gg0. $$

(3.14)

Comparing (3.13) and (3.14), respectively, with (3.2) and (3.1), we obtain the conclusion of this theorem by using the first part of the proof. □