We use the approach developed by [
18] to find the proof. Let
\((u_{1}(t, x), u_{2}(t, x), u_{3}(t,x))\) be positive solution of (
1.1) and define the following Lyapunov function:
$$V_{1}(t)=\sum_{i=1}^{3} \beta_{i} \int_{\Omega} \biggl(u_{i}-k_{i}^{\ast}-k_{i}^{\ast} \ln \frac{u_{i}}{k_{i}^{\ast}} \biggr)\,dx, $$
where
\(\beta_{3} = 1\),
\(\beta_{1}\) and
\(\beta_{2}\) are positive constant to be determined. By calculating the derivative of
\(V_{1}(t)\) along positive solutions of system (
1.1), we obtain
$$\begin{aligned} \frac{dV_{1}(t)}{dt} =&\sum_{i=1}^{2} \beta_{i} \int_{\Omega}\frac {\partial u_{i}}{\partial t} \biggl(1-\frac{k_{i}^{\ast}}{u_{i}} \biggr) \,dx+ \int_{\Omega}\frac {\partial u_{3}}{\partial t} \biggl(1-\frac{k_{3}^{\ast}}{u_{3}} \biggr) \,dx \\ =&-\sum_{i=1}^{3}\beta_{i}d_{i}k_{i}^{\ast} \int_{\Omega}\frac {|\nabla u_{i}|}{u_{i}^{2}}\,dx+ \int_{\Omega}\beta_{1}\bigl(u_{1}-k_{1}^{\ast } \bigr) (r_{1}-a_{1}u_{2}-b_{1}u_{1}) \,dx \\ &{}+ \int_{\Omega}\beta_{2}\bigl(u_{2}-k_{2}^{\ast } \bigr) \biggl(r_{2}-b_{2}u_{2}-a_{2}u_{3}+a_{3} \int_{\Omega} \int_{-\infty}^{t} K_{1}(x,y,t-s)u_{1}(s,y) \,ds\,dy \biggr)\,dx \\ &{}+ \int_{\Omega}\bigl(u_{3}-k_{3}^{\ast} \bigr) \biggl(-\alpha-b_{3}u_{3}+a_{4} \int _{\Omega} \int_{-\infty}^{t} K_{2}(x,y,t-s)u_{2}(s,y) \,ds\,dy\biggr)\,dx. \end{aligned}$$
(2.2)
By using the inequality
\(ab\leq\frac{1}{2}\lambda a^{2}+\frac{1}{2\lambda}b^{2}\), we obtain
$$\begin{aligned} \frac{dV_{1}(t)}{dt} \leq &-\sum_{i=1}^{3} \beta_{i}d_{i}k_{i}^{\ast} \int_{\Omega}\frac {|\nabla u_{i}|}{u_{i}^{2}}\,dx-\beta_{1}b_{1} \int_{\Omega}\bigl(u_{1}-k_{1}^{\ast } \bigr)^{2}\,dx \\ &{}+\beta_{1}a_{1} \int_{\Omega} \biggl[\frac{1}{2}\lambda _{1} \bigl(u_{1}-k_{1}^{\ast}\bigr)^{2} + \frac{1}{2\lambda_{1}}\bigl(u_{2}-k_{2}^{\ast} \bigr)^{2} \biggr]\,dx-\beta _{2}b_{2} \int_{\Omega} \bigl(u_{2}-k_{2}^{\ast} \bigr)^{2}\,dx \\ &{}+\beta_{2}a_{2} \int_{\Omega} \biggl[\frac{1}{2}\lambda _{2} \bigl(u_{2}-k_{2}^{\ast}\bigr)^{2} + \frac{1}{2\lambda_{2}}\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2} \biggr]\,dx-b_{3} \int _{\Omega}\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2}\,dx \\ &{}+\frac{1}{2}\lambda_{3}a_{3}\beta_{2} \int_{\Omega} \int_{\Omega } \int_{-\infty}^{t} K_{1}(x,y,t-s) \bigl(u_{1}(s,y)-k_{1}^{\ast}\bigr)^{2}\,ds \,dy\,dx \\ &{}+\frac{1}{2\lambda_{3}}a_{3}\beta_{2} \int_{\Omega} \int_{\Omega } \int_{-\infty}^{t} K_{1}(x,y,t-s) \bigl(u_{2}(t,y)-k_{2}^{\ast}\bigr)^{2}\,ds \,dy\,dx \\ &{}+\frac{1}{2}\lambda_{4}a_{4} \int_{\Omega} \int_{\Omega} \int _{-\infty}^{t} K_{2}(x,y,t-s) \bigl(u_{2}(s,y)-k_{2}^{\ast}\bigr)^{2}\,ds \,dy\,dx \\ &{}+\frac{1}{2\lambda_{4}}a_{4} \int_{\Omega} \int_{\Omega} \int _{-\infty}^{t} K_{2}(x,y,t-s) \bigl(u_{3}(t,y)-k_{3}^{\ast}\bigr)^{2}\,ds \,dy\,dx. \end{aligned}$$
(2.3)
Employing the property of
\(K_{i}(x, y, t)\) (
\(i = 1, 2\)) as described in (
1.2), we obtain
$$\begin{aligned} \frac{dV_{1}(t)}{dt} \leq &-\sum_{i=1}^{3} \beta_{i}d_{i}k_{i}^{\ast} \int_{\Omega}\frac {|\nabla u_{i}|}{u_{i}^{2}}\,dx-\beta_{1}(b_{1}-a_{1} \lambda_{1}/2) \int_{\Omega }\bigl(u_{1}-k_{1}^{\ast} \bigr)^{2}\,dx \\ &{}-(\beta_{2}b_{2}-\beta_{1}a_{1}/2 \lambda_{1}-\lambda_{2}\beta_{2}a_{2}/2 -a_{3}\beta_{2}/2\lambda_{3}) \int_{\Omega}\bigl(u_{2}-k_{2}^{\ast } \bigr)^{2}\,dx \\ &{}-(b_{3}-\beta_{2}a_{2}/2\lambda_{2}-a_{4}/2 \lambda_{4}) \int _{\Omega}\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2}\,dx \\ &{}+\frac{1}{2}\lambda_{3}a_{3}\beta_{2} \int_{\Omega} \int_{\Omega } \int_{0}^{\infty} K_{1}(x,y,r) \bigl(u_{1}(t-r,y)-k_{1}^{\ast}\bigr)^{2}\,dr \,dy\,dx \\ &{}+\frac{1}{2}\lambda_{4}a_{4} \int_{\Omega} \int_{\Omega} \int _{0}^{\infty} K_{2}(x,y,r) \bigl(u_{2}(t-r,y)-k_{2}^{\ast}\bigr)^{2}\,dr \,dy\,dx. \end{aligned}$$
(2.4)
Define a new Lyapunov function
$$\begin{aligned} V(t) =& V_{1}(t)+\frac{1}{2} \lambda_{3}a_{3}\beta_{2} \int_{\Omega } \int_{\Omega} \int_{0}^{\infty} \int_{t-r}^{t}K_{1}(x,y,r) \bigl(u_{1}(l,y)-k_{1}^{\ast}\bigr)^{2}\,dl \,dr\,dy\,dx \\ &{}+\frac{1}{2}\lambda_{4}a_{4} \int_{\Omega} \int_{\Omega} \int _{0}^{\infty} \int_{t-r}^{t} K_{2}(x,y,r) \bigl(u_{2}(l,y)-k_{2}^{\ast}\bigr)^{2}\,dl \,dr\,dy\,dx. \end{aligned}$$
(2.5)
Then, combining (
2.4) and (
2.5), we get
$$\begin{aligned} \frac{dV_{1}(t)}{dt} \leq &-\sum_{i=1}^{3} \beta_{i}d_{i}k_{i}^{\ast} \int_{\Omega}\frac {|\nabla u_{i}|}{u_{i}^{2}}\,dx-\beta_{1}(b_{1}-a_{1} \lambda_{1}/2) \int_{\Omega }\bigl(u_{1}-k_{1}^{\ast} \bigr)^{2}\,dx \\ &{}-(\beta_{2}b_{2}-\beta_{1}a_{1}/2 \lambda_{1}-\lambda_{2}\beta_{2}a_{2}/2 -a_{3}\beta_{2}/2\lambda_{3}) \int_{\Omega}\bigl(u_{2}-k_{2}^{\ast } \bigr)^{2}\,dx \\ &{}-(b_{3}-\beta_{2}a_{2}/2\lambda_{2}-a_{4}/2 \lambda_{4}) \int _{\Omega}\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2}\,dx \\ &{}+\frac{1}{2}\lambda_{3}a_{3}\beta_{2} \int_{\Omega} \int_{\Omega } \int_{0}^{\infty} K_{1}(x,y,r) \bigl(u_{1}(t,y)-k_{1}^{\ast}\bigr)^{2}\,dr \,dy\,dx \\ &{}+\frac{1}{2}\lambda_{4}a_{4} \int_{\Omega} \int_{\Omega} \int _{0}^{\infty} K_{2}(x,y,r) \bigl(u_{2}(t,y)-k_{2}^{\ast}\bigr)^{2}\,dr \,dy\,dx. \end{aligned}$$
(2.6)
Since
$$\begin{aligned}& \int_{\Omega} \int_{\Omega} \int_{0}^{\infty} K_{i}(x,y,r) \bigl(u_{i}(t,y)-k_{i}^{\ast}\bigr)^{2}\,dr \,dy\,dx \\& \quad = \int_{\Omega}\bigl(u_{i}(t,y)-k_{i}^{\ast} \bigr)^{2}\,dy\quad (i=1,2), \end{aligned}$$
it follows from (
2.6) that
$$\begin{aligned} \frac{dV_{1}(t)}{dt} \leq &-\sum_{i=1}^{3} \beta_{i}d_{i}k_{i}^{\ast} \int_{\Omega}\frac {|\nabla u_{i}|}{u_{i}^{2}}\,dx \\ &{}-\bigl(\beta_{1}(b_{1}-a_{1} \lambda_{1}/2)-\lambda_{3}a_{3} \beta_{2}/2\bigr) \int_{\Omega}\bigl(u_{1}-k_{1}^{\ast} \bigr)^{2}\,dx \\ &{}-(\beta_{2}b_{2}-\beta_{1}a_{1}/2 \lambda_{1}-\lambda_{2}\beta_{2}a_{2}/2 -a_{3}\beta_{2}/2\lambda_{3}-\lambda_{4}a_{4}/2) \int_{\Omega }\bigl(u_{2}-k_{2}^{\ast} \bigr)^{2}\, dx \\ &{}-(b_{3}-\beta_{2}a_{2}/2\lambda_{2}-a_{4}/2 \lambda_{4}) \int _{\Omega}\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2}\,dx. \end{aligned}$$
(2.7)
Integrating (
2.7) over
\([0, T]\) (
\(T > 0\)), we obtain
$$\begin{aligned}& \sum_{i=1}^{3} \beta_{i}d_{i}k_{i}^{\ast}\biggl\Vert \frac{|\nabla u_{i}|}{u_{i}}\biggr\Vert _{L^{2}_{(\Omega_{T})}}^{2} + \beta_{1}b_{1}\bigl\Vert u_{1}-k_{1}^{\ast} \bigr\Vert _{L^{2}_{(\Omega _{T})}}^{2}+\beta_{2}b_{2}\bigl\Vert u_{2}-k_{2}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2}+b_{3}\bigl\Vert u_{3}-k_{3}^{\ast} \bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} \\& \quad \leq V(0)+\frac{1}{2}(\beta_{1}a_{1} \lambda_{1}+\lambda_{3}\beta _{2}a_{3}) \bigl\Vert u_{1}-k_{1}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} +\frac{1}{2}(\beta_{1}a_{1}/ \lambda_{1}+\lambda_{2}\beta_{2}a_{2} +a_{3}\beta_{2}/\lambda_{3} \\& \qquad {}+\lambda_{4}a_{4})\bigl\Vert u_{2}-k_{2}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} +\frac{1}{2}(\beta_{2}a_{2}/\lambda_{2}+a_{4}/ \lambda_{4})\bigl\Vert u_{3}-k_{3}^{\ast} \bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2}. \end{aligned}$$
(2.8)
Taking
$$\lambda_{1}=\lambda_{3}=\frac{2\beta_{1}b_{1}}{a_{1}\beta _{1}+a_{3}} \quad \text{and} \quad \lambda_{2}=\lambda_{4}=\frac{\beta_{2}a_{2}+a_{4}}{2b_{3}}, $$
it is derived from (
2.8) that
$$\begin{aligned}& \sum_{i=1}^{3} \frac{\beta_{i}d_{i}k_{i}^{\ast}}{M_{i}}\bigl\Vert \vert \nabla u_{i}\vert \bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} +\beta_{2}b_{2}\bigl\Vert u_{2}-k_{2}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} \\& \quad \leq V(0)+\bigl((a_{1}\beta_{1}+a_{3} \beta_{2})^{2}/4\beta _{1}b_{1}+(a_{2} \beta_{2} +a_{4})^{2}/4b_{3}\bigr)\bigl\Vert u_{2}-k_{2}^{\ast}\bigr\Vert _{L^{2}_{(\Omega _{T})}}^{2}. \end{aligned}$$
(2.9)
Using the conditions
\(b_{2}b_{3} > 2a_{2}a_{4}\),
\(b_{1}b_{2} > 2a_{1}a_{3}\), one can choose
\(\beta_{1},\beta_{2}>0\) such that
$$\beta_{2}b_{2}>\frac{1}{4}\bigl[(a_{1} \beta_{1}+a_{3}\beta _{2})^{2}/ \beta_{1}b_{1} +(a_{2}\beta_{2}+a_{4})^{2}/b_{3} \bigr], $$
because the inequalities
\(2b_{1}b_{2}\beta_{1}\beta_{2}>a_{1}^{2}\beta_{1}^{2} +2a_{1}a_{3}\beta_{1}\beta_{2}+a_{2}^{2}\beta_{2}^{2}\) and
\(2b_{2}b_{3}\beta_{2}>a_{2}^{2}\beta_{2}^{2} +2a_{2}a_{4}\beta_{2}+a_{4}^{2}\) hold for certain
\(\beta_{1}\) and
\(\beta_{2}\).
Therefore, we obtain
$$ \bigl\Vert \vert \nabla u_{i}\vert \bigr\Vert _{L^{2}_{(\Omega_{T})}}\leq C_{1}, \qquad \bigl\Vert u_{2}-k_{2}^{\ast} \bigr\Vert _{L^{2}_{(\Omega_{T})}}\leq C_{1}, $$
(2.10)
where
\(C_{1}\) is constant independent of
T. In a similar way, by taking
$$\lambda_{1}=\lambda_{3}=\frac{\beta_{1}a_{1}+a_{3}\beta_{2}}{\beta _{2}b_{2}} \quad \text{and} \quad \lambda_{2}=\lambda_{4}=\frac{\beta_{2}b_{2}}{a_{2}\beta_{2}+a_{4}}, $$
it is derived from (
2.8) that
$$\begin{aligned}& \sum_{i=1}^{3} \frac{\beta_{i}d_{i}k_{i}^{\ast}}{M_{i}}\bigl\Vert \vert \nabla u_{i}\vert \bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} +\beta_{1}b_{1}\bigl\Vert u_{1}-k_{1}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} +b_{3}\bigl\Vert u_{3}-k_{3}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} \\& \quad \leq V(0)+\bigl((\beta_{1}a_{1}+a_{3} \beta_{2})^{2}/2\beta_{2}b_{2}\bigr)\bigl\Vert u_{1}-k_{1}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2} \\& \qquad {}+\bigl((\beta_{2}a_{2}+a_{4})^{2}/2 \beta_{2}b_{2}\bigr)\bigl\Vert u_{3}-k_{3}^{\ast} \bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2}. \end{aligned}$$
(2.11)
Using the conditions
\(b_{2}b_{3} > 2a_{2}a_{4}\),
\(b_{1}b_{2} > 2a_{1}a_{2}\), one can choose
\(\beta_{1}, \beta_{2}>0 \) again such that
$$\beta_{1}b_{1}>\frac{(a_{1}\beta_{1}+a_{3}\beta_{2})^{2}}{2\beta _{2}b_{2}} \quad \text{and} \quad b_{3}>\frac{(a_{2}\beta_{2}+a_{4})^{2}}{2\beta_{2}b_{2}}. $$
Therefore, we see that
$$ \bigl\Vert u_{1}-k_{1}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2}\leq C_{2}\quad \text{and} \quad \bigl\Vert u_{3}-k_{3}^{\ast}\bigr\Vert _{L^{2}_{(\Omega_{T})}}^{2}\leq C_{2}, $$
(2.12)
where
\(C_{2}\) is constant independent of
T.
Choosing
$$\lambda_{1}=\lambda_{3}=\frac{\beta_{1}a_{1}+a_{3}\beta_{2}}{\beta _{2}b_{2}}+\varepsilon, \qquad \lambda_{2}=\lambda_{4}=\frac{\beta_{2}b_{2}}{a_{2}\beta_{2}+a_{4}}, $$
where
ε is sufficiently small positive constant. Using the conditions of Theorem
2.1 and (
2.7), one can easily verify that there exists a positive constant
δ (
\(\delta>0\)) such that
$$ \begin{aligned} &\frac{dV}{dt}\leq -\delta \int_{\Omega}\bigl[\bigl(u_{1}-k_{1}^{\ast} \bigr)^{2}+\bigl(u_{2}-k_{2}^{\ast } \bigr)^{2}+\bigl(u_{3}-k_{3}^{\ast} \bigr)^{2}\bigr]\,dx, \\ & \frac{dV}{dt}\leq0, \qquad (u_{1}, u_{2}, u_{3})\neq \bigl(k_{1}^{\ast}, k_{2}^{\ast}, k_{3}^{\ast}\bigr). \end{aligned} $$
(2.13)
Using integration by parts, the Hölder inequality, (
2.1), and (
2.10), one can easily check that
\(\frac{d}{dt}\int_{\Omega}[(u_{1}-k_{1}^{\ast})^{2} + (u_{2}-k_{2}^{\ast})^{2} + (u_{3}-k_{3}^{\ast})^{2}]\,dx \) is bounded from above. Then, using Lemma
2.1, (
2.10), and (
2.13), we see that
$$\begin{aligned}& \bigl\Vert u_{1}(t,\cdot)-k_{1}^{\ast} \bigr\Vert _{L_{(\Omega)}^{2}}\rightarrow0, \\& \bigl\Vert u_{2}(t, \cdot)-k_{2}^{\ast}\bigr\Vert _{L_{(\Omega)}^{2}}\rightarrow0, \\& \bigl\Vert u_{3}(t,\cdot)-k_{3}^{\ast}\bigr\Vert _{L_{(\Omega)}^{2}}\rightarrow0. \end{aligned}$$
(2.14)
Obviously,
$$ \bigl\Vert u(t,x)\bigr\Vert _{L_{(\Omega)}^{\infty}}\leq C_{3}\|u\|_{W_{2}^{1}}^{\frac{1}{2}}\|u\|_{L^{2}(\Omega)}. $$
(2.15)
It follows from (
2.10), (
2.12), (
2.14), (
2.15), and (
2.1) that
$$\begin{aligned}& \bigl\Vert u_{1}(t,\cdot)-k_{1}^{\ast}\bigr\Vert _{L_{(\Omega)}^{\infty}}\rightarrow 0, \\& \bigl\Vert u_{2}(t, \cdot)-k_{2}^{\ast}\bigr\Vert _{L_{(\Omega)}^{\infty}}\rightarrow 0, \\& \bigl\Vert u_{3}(t,\cdot)-k_{3}^{\ast}\bigr\Vert _{L_{(\Omega)}^{\infty}}\rightarrow 0. \end{aligned}$$
Namely,
\((u_{1}, u_{2}, u_{3})\) converges uniformly to
\((k_{1}^{\ast}, k_{2}^{\ast}, k_{3}^{\ast})\). Using the fact that
\(V(u_{1}, u_{2}, u_{3})\) is decreasing for
t, one can derive that
\((k_{1}^{\ast}, k_{2}^{\ast}, k_{3}^{\ast})\) is globally stable. This completes the proof. □