In this section we will study the dynamic behavior of the solution of (2.4). First we give the following assumptions:

- (H
_{1}): -
\(d(t)\) is continuous differentiable decreasing positive function on \([0,+\infty )\) and satisfies

$$ \lim_{t\rightarrow +\infty }d(t)=d_{\infty }. $$

- (H
_{2}): -
\(r(t)\) is continuous differentiable function on \([0,+\infty)\) and satisfies

$$ r_{1}\leq \lim_{t\rightarrow +\infty }r(t)\leq r_{2}, $$

where \(r_{1}\), \(r_{2}\) are positive constants.

### Definition 3.1

A function \(\hat{u}\in C^{2,1}(\Omega (0)\times (0,\infty ))\cap C(\bar{ \Omega }(0)\times [0,\infty ))\) is called an upper solution of (2.4) if it satisfies

$$ \textstyle\begin{cases} \hat{u}_{t}-d(t)\Delta \hat{u}\geq f(t,\hat{u}),&t>0,x\in \Omega (0), \\ \hat{u}(t,x)\geq 0,&t>0,x\in \partial \Omega (0), \\ \hat{u}(0,x)\geq u_{0}(x),&x\in \Omega (0), \end{cases} $$

(3.1)

where \(f(t,u)=r(t)u(a-bu^{q})\). Similarly, *ǔ* is called a lower solution of (2.4) if it satisfies all the reversed inequalities in (3.1).

### Lemma 3.1

(Comparison principle)

*Let*
\(v(t,x)\)
*be a solution of* (2.4), \(\hat{v}(t,x)\)
*and*
\(\check{v}(t,x)\)
*are upper and lower solutions of* (2.4) *respectively*, *then*
\(\check{v}(t,x)\leq v(t,x)\leq \hat{v}(t,x)\)
*in*
\(\bar{\Omega }\times [0,+\infty )\).

### Lemma 3.2

*Let*
\(u(t,x)\)
*be a nonnegative nontrivial solution of the following problem*:

$$ \textstyle\begin{cases} u_{t}=d(t)\Delta u+r(t)u(a-bu^{q}),&t>0,x\in \Omega (0), \\ u(t,x)=0,&t>0,x\in \partial \Omega (0), \\ u(0,x)=u_{0}(x)\geq 0,& x\in \Omega (0). \end{cases} $$

(3.2)

*If*
\(u_{0}(x)\in C^{2}(\bar{\Omega }(0))\), \(u_{0}(x)=\Delta u(0,x)=0\)
*for*
\(x\in \partial \Omega (0)\)
*and*
\(\Delta u(0,x)\leq 0\)
*for*
\(x\in \bar{ \Omega }(0)\), *then*
\(u(t,x)\in C^{2,1}(\Omega (0)\times (0,\infty ))\)
*and*

$$ \Delta u(t,x)\leq 0\quad \textit{for } x\in \Omega (0),t>0. $$

### Proof

In view of \(u_{0}\) being smooth and for \(x\in \partial \Omega (0)\), \(d(t)u_{0}(x)+r(t)u_{0}(a-bu_{0}^{q})=0\), by the standard parabolic regularity theory it follows that the solution \(u(t,x)\in C^{2,1}( \Omega (0)\times (0,\infty ))\). Denote \(\omega =\Delta u\). For \(t>0\), \(x\in \Omega (0)\), we have

$$ \omega_{t}\leq d(t)\Delta \omega + \bigl(r(t)a-b(q+1)u^{q} \bigr)\omega . $$

From the condition \(\Delta u(x,0)\leq 0\) for \(x\in \Omega (0)\), we have

$$ \omega (0,x)\leq 0\quad \mbox{for } x\in \Omega (0). $$

By \(u(t,x)=0\) for \(x\in \partial \Omega (0)\), \(t>0\), we have

$$ \omega (t,x)=\frac{1}{d(t)}\bigl[u_{t}-r(t)u\bigl(a-bu^{q} \bigr)\bigr]=0,\quad x\in \partial\Omega (0),t>0. $$

By Lemma 3.1, we can obtain that

$$ \omega (t,x)\leq 0\quad \mbox{for } x\in \partial \Omega (0),\quad t>0, $$

which implies that \(\Delta u(t,x)\leq 0\) for \(x\in \partial \Omega (0)\), \(t>0\). The proof is completed. □

Let \(\lambda_{1}\) be the principal positive eigenvalue of the problem

$$ \textstyle\begin{cases} -\Delta u=\lambda u,&y\in \Omega (0), \\ u(t,y)=0,&y\in \partial \Omega (0). \end{cases} $$

### Theorem 3.1

*If the assumptions* (H_{1}) *and* (H_{2}) *hold and*
\(a\leq \frac{d _{\infty }\lambda_{1}}{r_{2}}\), *then the solution of problem* (2.4) *satisfies*

$$ \lim_{t\rightarrow \infty }{u}(t,x)=0\quad \textit{for }x\in \bar{\Omega }(0). $$

### Proof

Obviously, \(\check{u}=0\) is a lower solution of (2.4). Next we construct a upper solution of (2.4). Let \(\hat{u}(t,x)\) be the unique solution of the problem:

$$ \textstyle\begin{cases} \hat{u}_{t}-d(t)\Delta \hat{u}=r(t)\hat{u}(a-b\hat{u}^{q}),&t>0,x\in \Omega (0), \\ \hat{u}(t,x)=0,& t>0,x\in \partial \Omega (0), \\ \hat{u}(0,x)=M\phi (x),&x\in \Omega (0), \end{cases} $$

(3.3)

where \(\phi (x)\) is the corresponding eigenfunction of \(\lambda_{1}\), *M* is a positive constant. We choose *M* so large that \(M\phi (x) \geq u_{0}(x)\), For any solution \(\hat{u}(t,x)\) of (3.3), we can see that \(\hat{u}(t,x)\) is an upper solution of (2.4). Let *u* be any solution of (2.4). It follows from Lemma 3.1 that

$$ 0\leq u(t,x)\leq \hat{u}(t,x),\quad t>0,x\in \Omega (0). $$

Since \(\Delta \hat{u}(0,x)=M\Delta \phi (x)=-\lambda_{1}M\phi (x) \leq 0\), it follows from Lemma 3.2 that

$$ \Delta \hat{u}(t,x)\leq 0\quad \mbox{for } x\in \Omega (0), t>0. $$

On the other hand, by assumption (H_{1}), \(d(t)\) tends decreasingly to \(d_{\infty }\) as \(t\rightarrow \infty \), then \(d(t)\geq d_{\infty }\) for \(t\geq 0\). By assumption (H_{2}), \(r_{1}< r(t)\leq r_{2}\) for \(t\geq 0\). Thus, \(\hat{u}(t,x)\) satisfies

$$ \hat{u}_{t}\leq d_{\infty }\Delta \hat{u}+r_{2} a \hat{u}-r_{1} b \hat{u}^{q+1},\quad t>0,x\in \Omega (0). $$

Now consider the following problem:

$$ \textstyle\begin{cases} \bar{u}_{t}=d_{\infty }\Delta \bar{u}+r_{2} a\bar{u}-r_{1} b\bar{u} ^{q+1},&t>0,x\in \Omega (0), \\ \bar{u}(t,x)=0,&t>0,x\in \partial \Omega (0), \\ \bar{u}(0,x)=M\phi (x),&x\in \Omega (0). \end{cases} $$

(3.4)

Let \(\bar{u}(t,x)\) be a solution of (3.4). By the comparison principle, for \(t>0\), \(x\in \Omega (0)\), we have

$$ \hat{u}(t,x)\leq \bar{u}(t,x) $$

and

$$ 0\leq u(t,x)\leq \hat{u}(t,x)\leq \bar{u}(t,x). $$

Since \(ar_{2}\leq d_{\infty }\lambda_{1}\), by Theorem 2.1, \(\lim_{t\rightarrow \infty }\bar{u}(t,x)=0\) for \(x\in \bar{\Omega }(0)\). Hence,

$$ \lim_{t\rightarrow \infty }{u}(t,x)=0\quad \mbox{for }x\in \bar{\Omega}(0). $$

□

### Theorem 3.2

*If the assumptions* (H_{1}) *and* (H_{2}) *hold and*
\(a> \frac{d_{ \infty }\lambda_{1}}{r_{1}}\), *then the solution of problem* (3.14) *satisfies*

$$ \check{u}^{*}(x)\leq u(x)\leq \bar{u}^{*}(x)\quad \textit{for }x\in \bar{\Omega }(0), $$

*where*
\(u^{*}(x)\)
*is a unique positive solution of* (3.8), \(\check{u} ^{*}(x)\)
*is a unique positive solution of* (3.15).

### Proof

In view of assumption (H_{1}), we have \(\lim_{t\rightarrow \infty } {d}(t)=d_{\infty }\) and \(d(t)\) is decreasing for \(t>0\), there exists a \(T_{1}>0\) such that \(d_{\infty }\leq d(t)\leq d_{\infty }+\varepsilon \) for \(t>T_{1}\). Let \(\hat{u}(t,x)\) be the unique solution of the problem:

$$ \textstyle\begin{cases} \hat{u}_{t}-d(t)\Delta \hat{u}=r(t)\hat{u}(a-b\hat{u}^{q}),&t>T_{1},x\in \Omega (0), \\ \hat{u}(t,x)=0,&t>T_{1},x\in \partial \Omega (0), \\ \hat{u}(T_{1},x)=M\phi (x),&x\in \Omega (0), \end{cases} $$

(3.5)

where \(\phi (x)\) is the corresponding eigenfunction of \(\lambda_{1}\), *M* is a positive constant. We choose *M* so large that \(M\phi (x) \geq u_{0}(x)\), For any solution \(\hat{u}(t,x)\) of (3.5), we can see that \(\hat{u}(t,x)\) is an upper solution of (2.4). Let *u* be any solution of (2.4). It follows from Lemma 3.1 that

$$ 0\leq u(t,x)\leq \hat{u}(t,x),\quad t>T_{1}, x\in \Omega (0). $$

From \(\Delta \hat{u}(T_{1},x)\leq 0\) and Lemma 3.2, we have \(\Delta \hat{u}(t,x)\leq 0\) in \([T_{1},\infty )\times \Omega (0)\). Thus,

$$ \hat{u}_{t}\leq d_{\infty }\Delta \hat{u}+r_{2}a \hat{u}-r_{1}b\hat{u}^{q+1},\quad t>T_{1},x\in \Omega (0). $$

(3.6)

Now consider the following problem:

$$ \textstyle\begin{cases} {u}_{t}=d_{\infty }\Delta {u}+r_{2} a{u}-r_{1} b{u}^{q+1},&t>T_{1},x\in \Omega (0), \\ {u}(t,x)=0,& t>T_{1},x\in \partial \Omega (0), \\ {u}(T_{1},x)=M\phi (x),&x\in \Omega (0). \end{cases} $$

(3.7)

The problem (3.7) admits a unique solution \(\bar{u}(t,x)\). In view of \(a>\frac{d_{\infty }\lambda_{1}}{r_{1}}\) and \(r_{1}\leq r_{2}\), we have \(a>\frac{d_{\infty }\lambda_{1}}{r_{2}}\). Thus, the result of Theorem 2.1 shows that

$$ \bar{u}(t,x)\rightarrow \bar{u}^{*}(x)\quad \mbox{as }t\rightarrow \infty , $$

where \(\bar{u}^{*}(x)\) is the unique positive solution of the following problem:

$$ \textstyle\begin{cases} {u}_{t}=d_{\infty }\Delta {u}+r_{2} a{u}-r_{1} b{u}^{q+1},&t>T_{1},x\in \Omega (0), \\ {u}(x)=0,&x\in \partial \Omega (0). \end{cases} $$

(3.8)

Using (3.6), (3.7) and the comparison principle yields

$$ \hat{u}(t,x)\leq \bar{u}(t,x)\quad \mbox{for } t>T_{1},x\in \Omega (0). $$

This implies that

$$ \lim_{t\rightarrow \infty }\sup u(t,x)\leq u^{*}(x),\quad x\in \Omega (0). $$

(3.9)

On the other hand, let \(\check{u}(t,x)\) be the unique solution of the problem:

$$ \textstyle\begin{cases} \check{u}_{t}-d(t)\Delta \check{u}=r(t)\check{u}(a-b\check{u}^{q}),&t>T_{1},x\in \Omega (0), \\ \check{u}(t,x)=0,&t>T_{1},x\in \partial \Omega (0), \\ \check{u}(T_{1},x)=\delta \phi (x),&x\in \Omega (0), \end{cases} $$

(3.10)

where *δ* is a sufficiently small constant such that \(\delta \phi (x)\leq u_{0}\). It is easy to see that \(\check{u}(t,x)\) is a lower solution of (2.4) in \([T_{1},\infty )\times \bar{\Omega }(0)\). From \(\Delta \check{u}(T_{1},x)=-\delta \lambda_{1}\phi (x)\leq 0\) and Lemma 3.1, we have \(\Delta \check{u}(t,x)\leq 0\) in \([T_{1},\infty )\times \Omega (0)\). Thus,

$$ \check{u}_{t}\geq (d_{\infty }+\varepsilon )\Delta \check{u}+r_{1}a\hat{u}-r_{2}b\hat{u}^{q+1},\quad t>T_{1},x\in \Omega (0). $$

(3.11)

Now consider the following problem:

$$ \textstyle\begin{cases} \check{u}_{t}=(d_{\infty }+\varepsilon )\Delta \check{u}+r_{1}a\check{u}-r_{2}b\hat{u}^{q+1},&t>T_{1},x\in \Omega (0), \\ \check{u}(t,x)=0,& t>T_{1},x\in \partial \Omega (0), \\ \check{u}(T_{1},x)=\delta \phi (x),&x\in \Omega (0). \end{cases} $$

(3.12)

Clearly, (3.12) admits a unique positive solution, denoted by \(\check{u}_{\varepsilon }(t,x)\). Using the comparison principle yields that \(\check{u}_{\varepsilon }(t,x)\leq \check{u}(t,x)\). Since \(ar_{1}>d_{\infty }\lambda_{1}\), we can choose \(\varepsilon >0\) sufficiently small such that \(ar_{1}>(d_{\infty }+\varepsilon )\lambda _{1}\). Thus, by Theorem 2.1 we have

$$ \lim_{t\rightarrow \infty }\check{u}_{\varepsilon }(t,x)=\check{u} _{\varepsilon }^{*}(x),\quad x\in \bar{\Omega }(0), $$

where \(\check{u}_{\varepsilon }^{*}(x)\) is the unique positive solution of the problem

$$ \textstyle\begin{cases} -\check{u}_{t}=(d_{\infty }+\varepsilon )\Delta \check{u}+r_{1}a \check{u}-r_{2}b\hat{u}^{q+1},&x\in \Omega (0), \\ \check{u}(x)=0,&x\in \partial \Omega (0). \end{cases} $$

(3.13)

From (3.11) and (3.13), it follows that

$$ \lim_{t\rightarrow \infty }\inf u(t,x)\geq \check{u}_{\varepsilon } ^{*}(x),\quad x\in \Omega (0). $$

(3.14)

By the continuous dependence of \(\check{u}_{\varepsilon }^{*}(x)\) on *ε*, obviously,

$$ \check{u}_{\varepsilon }^{*}(x)\rightarrow \check{u}^{*}(x) \quad \mbox{as } \varepsilon \rightarrow 0^{+}, $$

where \(\check{u}^{*}(x)\) is a solution of the problem

$$ \textstyle\begin{cases} -d_{\infty }\Delta \check{u}=r_{1}a\check{u}-r_{2}b\hat{u}^{q+1},&x\in \Omega (0), \\ \check{u}(x)=0,&x\in \partial \Omega (0). \end{cases} $$

(3.15)

By (3.14), we have

$$ \lim_{t\rightarrow \infty }\inf u(t,x)\geq \check{u}^{*}(x),\quad x \in \Omega (0). $$

(3.16)

It follows from (3.9) and (3.16) that

$$ \check{u}^{*}(x)\leq u(x)\leq \bar{u}^{*}(x)\quad \mbox{for } x\in \bar{\Omega }(0). $$

This completes the proof. □

### Remark 3.1

If \(r(t)=r\) in (2.4) is a positive constant, then \(\check{u}^{*}(x)= \bar{u}^{*}(x):=u^{*}(x)\). Hence, for (2.4) there exists a unique positive steady state solution \(u^{*}(x)\). It is obvious that the results of [33] are special results of the present paper in the case of \(r(t)=r\) in (2.4) being a positive constant. On the other hand, by upper and lower solutions, we cannot find that for (2.4) there exists a unique positive steady state solution \(u^{*}(x)\) and we only obtain the scope of the solution to (2.4). We hope that some new methods can be developed by the researchers for obtaining an unique positive solution of (2.4).