First, by the same method as in [4], we can obtain the nonnegativity of the solutions of problem (2.2)-(2.4).

**Lemma 1** *If a nontrivial function* ${u}_{\epsilon}\in {C}_{T}({\overline{Q}}_{T})$ *solves* ${u}_{\epsilon}=G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+})$,

*then* ${u}_{\epsilon}(x,t)>0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}(x,t)\in {\overline{Q}}_{T}.$

In the following, we will show some *a priori* estimates for the upper bound of a nonnegative periodic solution of problem (2.2)-(2.4). Here and below, we denote by ${\parallel \cdot \parallel}_{p}$ ($1\le p\le \mathrm{\infty}$) the ${L}^{p}(\mathrm{\Omega})$ norm.

**Lemma 2** *For* $\lambda \in [0,1]$,

*let* $u(x,t)$ *be a nonnegative periodic solution which solves* ${u}_{\epsilon}=G(1,\lambda (m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+})$,

*then there exists a constant* *K* *independent of* *λ*,

*ε* *such that* ${\parallel u(t)\parallel}_{\mathrm{\infty}}<K,$

(3.1)

*where* $u(t)=u(\cdot ,t)$.

*Proof* Multiplying (2.2) by

${u}^{m+1}$ (

$m\ge 0$) and integrating over Ω, we have

$\frac{1}{m+2}\frac{d}{dt}{\parallel u(t)\parallel}_{m+2}^{m+2}+\frac{4(m+1)}{{(m+2)}^{2}}{\parallel \mathrm{\nabla}({|u(t)|}^{\frac{m}{2}}u(t))\parallel}_{2}^{2}\le {\parallel m(x,t)\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega}\times (0,T))}{\parallel u(t)\parallel}_{m+2}^{m+2},$

and hence

$\frac{d}{dt}{\parallel u(t)\parallel}_{m+2}^{m+2}+{C}_{1}{\parallel \mathrm{\nabla}({|u(t)|}^{\frac{m}{2}}u(t))\parallel}_{2}^{2}\le {C}_{2}(m+2){\parallel u(t)\parallel}_{m+2}^{m+2},$

(3.2)

where

${C}_{i}$ (

$i=1,2$) are positive constants independent of

*u* and

*m*. Assume that

${\parallel u(t)\parallel}_{\mathrm{\infty}}\ne 0$ and set

${u}_{k}(t)={|u(t)|}^{\frac{{m}_{k}}{2}}u(t),\phantom{\rule{1em}{0ex}}{m}_{k}={2}^{k}-2\phantom{\rule{0.25em}{0ex}}(k=1,2,\dots ),$

then

${m}_{k}=2{m}_{k-1}+2$. For convenience, we denote by

*C* a positive constant independent of

*k* and

*m*, which may take different values. From (3.2), we obtain

$\frac{d}{dt}{\parallel {u}_{k}(t)\parallel}_{2}^{2}+C{\parallel \mathrm{\nabla}{u}_{k}(t)\parallel}_{2}^{2}\le C(m+2){\parallel {u}_{k}(t)\parallel}_{2}^{2}.$

(3.3)

By using the Gagliardo-Nirenberg inequality, we have

${\parallel {u}_{k}(t)\parallel}_{2}\le C{\parallel \mathrm{\nabla}{u}_{k}(t)\parallel}_{2}^{\theta}{\parallel {u}_{k}(t)\parallel}_{1}^{1-\theta},$

(3.4)

with

$\theta =\frac{N}{N+2}\in (0,1).$

By inequalities (3.3), (3.4) and the fact that

${\parallel {u}_{k}(t)\parallel}_{1}={\parallel {u}_{k-1}(t)\parallel}_{2}^{2}$, we obtain the following differential inequality:

$\begin{array}{rl}\frac{d}{dt}{\parallel {u}_{k}(t)\parallel}_{2}^{2}& \le -C{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}}{\parallel {u}_{k}(t)\parallel}_{1}^{\frac{2(\theta -1)}{\theta}}+C({m}_{k}+2){\parallel {u}_{k}(t)\parallel}_{2}^{2}\\ \le -C{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}}{\parallel {u}_{k-1}(t)\parallel}_{2}^{\frac{4(\theta -1)}{\theta}}+C({m}_{k}+2){\parallel {u}_{k}(t)\parallel}_{2}^{2}.\end{array}$

Let

${\lambda}_{k}=max\{1,\underset{t}{sup}{\parallel {u}_{k}(t)\parallel}_{2}\},$

we have

$\begin{array}{rcl}\frac{d}{dt}{\parallel {u}_{k}(t)\parallel}_{2}^{2}& \le & {\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2({m}_{k}+1)}{{m}_{k}+2}}\{-C{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}-\frac{2({m}_{k}+1)}{{m}_{k}+2}}{\lambda}_{k-1}^{\frac{4(\theta -1)}{\theta}}\\ +C({m}_{k}+2){\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{{m}_{k}+2}}\}.\end{array}$

(3.5)

By Young’s inequality,

$ab\le \u03f5{a}^{{p}^{\prime}}+{\u03f5}^{-\frac{{q}^{\prime}}{{p}^{\prime}}}{b}^{{q}^{\prime}},$

where

${p}^{\prime}>1$,

${q}^{\prime}>1$,

$a>0$,

$b>0$,

$\u03f5>0$ and

$\frac{1}{{p}^{\prime}}+\frac{1}{{q}^{\prime}}=1$. Set

then we obtain

$({m}_{k}+2){\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{{m}_{k}+2}}\le \frac{1}{2}{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}-\frac{2({m}_{k}+1)}{{m}_{k}+2}}{\lambda}_{k-1}^{\frac{4(\theta -1)}{\theta}}+C{({m}_{k}+2)}^{\frac{{l}_{k}}{{l}_{k}-1}}{\lambda}_{k-1}^{4\frac{1-\theta}{\theta}\frac{1}{{l}_{k}-1}}.$

(3.6)

Here we have used the fact that

${p}^{\prime}={l}_{k}>r>1$ for some

*r* independent of

*k*. In fact, it is easy to verify that

$\underset{k\to \mathrm{\infty}}{lim}{l}_{k}=+\mathrm{\infty}.$

Denoting

${a}_{k}=\frac{{l}_{k}}{{l}_{k}-1},\phantom{\rule{2em}{0ex}}{b}_{k}=\frac{1-\theta}{\theta}\frac{4}{{l}_{k}-1},$

and combining (3.5) with (3.6), we have

$\frac{d}{dt}{\parallel {u}_{k}(t)\parallel}_{2}^{2}\le {\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2({m}_{k}+1)}{{m}_{k}+2}}\{-C{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}-\frac{2({m}_{k}+1)}{{m}_{k}+2}}{\lambda}_{k-1}^{\frac{4(\theta -1)}{\theta}}+C{({m}_{k}+2)}^{{a}_{k}}{\lambda}_{k-1}^{{b}_{k}}\}.$

(3.7)

Then

$({m}_{k}+2)\frac{d}{dt}{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{{m}_{k}+2}}\le -C{\parallel {u}_{k}(t)\parallel}_{2}^{\frac{2}{\theta}-\frac{2({m}_{k}+1)}{{m}_{k}+2}}{\lambda}_{k-1}^{\frac{4(\theta -1)}{\theta}}+C{({m}_{k}+2)}^{{a}_{k}}{\lambda}_{k-1}^{{b}_{k}}.$

(3.8)

From the periodicity of

${u}_{k}(t)$, we know that there exists

${t}_{0}$ at which

${\parallel {u}_{k}(t)\parallel}_{2}$ reaches its maximum and thus the left-hand side of (3.8) vanishes. Then we obtain

${\parallel {u}_{k}(t)\parallel}_{2}\le {\left\{C\left[{({m}_{k}+2)}^{{a}_{k}}{\lambda}_{k-1}^{{b}_{k}+\frac{4(1-\theta )}{\theta}}\right]\right\}}^{\frac{1}{{\alpha}_{k}}},$

where

${\alpha}_{k}=\frac{2}{\theta}-\frac{2({m}_{k}+1)}{{m}_{k}+2}=\frac{2{l}_{k}}{{m}_{k}+2}.$

Therefore, we conclude that

${\parallel {u}_{k}(t)\parallel}_{2}\le {\left\{C{({m}_{k}+2)}^{{a}_{k}}{\lambda}_{k-1}^{{b}_{k}+\frac{4(1-\theta )}{\theta}}\right\}}^{\frac{1}{{\alpha}_{k}}}={\left\{C{({m}_{k}+2)}^{{a}_{k}}\right\}}^{\frac{{m}_{k}+2}{2{l}_{k}}}{\lambda}_{k-1}^{\frac{2(1-\theta )({m}_{k}+2)}{({l}_{k}-1)\theta}}.$

Since

$\frac{{m}_{k}+2}{({l}_{k}-1)\theta}=\frac{1}{1-\theta}$ and

$\frac{{m}_{k}+2}{2{l}_{k}}$ and

${\alpha}_{k}$ are bounded, we get

${\parallel {u}_{k}(t)\parallel}_{2}\le C{A}^{k}{\lambda}_{k-1}^{2},$

where

$A>1$ is a positive constant independent of

*k*. Then we have

$ln{\parallel {u}_{k}(t)\parallel}_{2}\le ln{\lambda}_{k}\le lnC+klnA+2ln{\lambda}_{k-1},$

thus

$\begin{array}{rl}ln{\parallel {u}_{k}(t)\parallel}_{2}& \le lnC\sum _{i=0}^{k-2}{2}^{i}+{2}^{k-1}ln{\lambda}_{1}+lnA(\sum _{j=0}^{k-2}(k-j){2}^{j})\\ \le ({2}^{k-1}-1)lnC+{2}^{k-1}ln{\lambda}_{1}+f(k)lnA,\end{array}$

or

${\parallel {u}_{k}(t)\parallel}_{{m}_{k}+2}\le {\{{C}^{{2}^{k-1}-1}{\lambda}_{1}^{{2}^{k-1}}Af(k)\}}^{\frac{2}{{m}_{k}+2}},$

where

$f(k)=k-2(k+1)-{2}^{k-1}+{2}^{k+1}.$

Letting

$k\to \mathrm{\infty}$, we obtain

${\parallel u(t)\parallel}_{\mathrm{\infty}}\le C{\lambda}_{1}\le C(max\{1,\underset{t}{sup}{\parallel u(t)\parallel}_{2}\}).$

(3.9)

Now, we just need to show the estimate of

${\parallel u(t)\parallel}_{2}$. Multiplying (2.2) by

*u* and integrating by parts over

${Q}_{T}$, by the periodicity of

*u*, we have

${\iint}_{{Q}_{T}}{\lambda}_{1}{|\mathrm{\nabla}u|}^{2}+\epsilon {u}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\le {\iint}_{{Q}_{T}}\lambda {u}^{2}(m-\varphi [u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx,$

which implies that

${\iint}_{{Q}_{T}}{u}^{2}(m-\varphi [u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge 0.$

Let

$M={max}_{(x,t)\in {\overline{Q}}_{T}}m(x,t)$, by assumption (A2), we have

$\begin{array}{rl}0& \le {\iint}_{{Q}_{T}}{u}^{2}(m-\varphi [u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\le M{\iint}_{{Q}_{T}}{u}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-{\iint}_{{Q}_{T}}{u}^{2}\varphi [u]\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ \le M{\iint}_{{Q}_{T}}{u}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-C{\int}_{0}^{T}{\parallel u\parallel}_{2}^{2}\phantom{\rule{0.2em}{0ex}}dt,\end{array}$

that is,

${\int}_{0}^{T}{\parallel u\parallel}_{2}^{4}\phantom{\rule{0.2em}{0ex}}dt\le C{\int}_{0}^{T}{\parallel u\parallel}_{2}^{2}\phantom{\rule{0.2em}{0ex}}dt,$

where

*C* is a positive independent of

*λ*. By Young’s inequality, we have

${\int}_{0}^{T}{\parallel u\parallel}_{2}^{2}\phantom{\rule{0.2em}{0ex}}dt\le {\int}_{0}^{T}\frac{1}{4{\epsilon}^{2}}+{\epsilon}^{2}{\parallel u\parallel}_{2}^{4}\phantom{\rule{0.2em}{0ex}}dt.$

Combining with the above inequality, we have

${\parallel {u}_{k}(t)\parallel}_{2}\le C,$

which together with (3.9) implies (3.1), and thus the proof is complete. □

**Corollary 1**
*There exists a positive constant R independent of*
*ε*
*such that*
$deg(I-G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),{B}_{R},0)=1,$

*where* ${B}_{R}$ *is a ball centered at the origin with radius* *R* *in* ${L}^{\mathrm{\infty}}({Q}_{T})$.

*Proof* It follows from Lemma 2 that there exists a positive constant

*R* independent of

*λ*,

*ε* such that

$u\ne G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in \partial {B}_{R},\lambda \in [0,1].$

So, the degree is well defined on

${B}_{R}$. From the homotopy invariance of the Leray-Schauder degree and the existence and uniqueness of the solution of

$G(1,0)$, we can see that

$\begin{array}{rl}deg(1-G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),{B}_{R},0)& =deg(1-G(1,\lambda (m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),{B}_{R},0)\\ =deg(1-G(1,0),{B}_{R},0)\\ =1.\end{array}$

The proof is completed. □

**Lemma 3** *There exist constants* ${r}_{0}>0$ *and* $\epsilon >0$ *such that for any* $r<{r}_{0}$,

$\epsilon <{\epsilon}_{0}$,

$u=G(\tau ,(m-\mathrm{\Phi}[u]){u}^{+}+(1-\tau ))$ *admits no nontrivial solution* *u* *satisfying* $0<{\parallel u\parallel}_{{L}^{\mathrm{\infty}}({Q}_{T})}\le r,$

*where* *r* *is a positive constant independent of* *ε*.

*Proof* By contradiction, let

*u* be a nontrivial fixed point of

$u=G(\tau ,(m-\mathrm{\Phi}[u]){u}^{+}+1-\tau )$ satisfying

$0<{\parallel u\parallel}_{{L}^{\mathrm{\infty}}({Q}_{T})}\le r$. For any given

$\varphi (x)\in {C}_{0}^{\mathrm{\infty}}({B}_{\delta}({x}_{0}))$, multiplying (2.2) by

$\frac{{\varphi}^{2}}{u}$ and integrating over

${Q}_{T}^{\ast}={B}_{\delta}({x}_{0})\times (0,T)$, we have

By the periodicity of

*u*, the first term on the left-hand side is zero. The second term on the left-hand side can be rewritten as

The third term of the left-hand side of equation (

3.11) can be rewritten as

Then from (3.10), we obtain

From assumption (A1), we can see that

Since

$\tau \in [0,1]$, we have

${\iint}_{{Q}_{T}}\gamma {|D\varphi |}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-{\iint}_{{Q}_{T}}{\varphi}^{2}(m-\epsilon -\mathrm{\Phi}[u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge 0.$

By an approaching process, we choose

$\varphi ={\varphi}_{1}$, where

${\varphi}_{1}$ is the eigenvector of the first eigenvalue

${\lambda}_{1}$ in (A3), and then we obtain

$\begin{array}{rcl}0& \le & {\iint}_{{Q}_{T}}\gamma {|D{\varphi}_{1}|}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-{\iint}_{{Q}_{T}}{\varphi}_{1}^{2}(m-\epsilon -\mathrm{\Phi}[u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ =& -{\iint}_{{Q}_{T}}\gamma {\varphi}_{1}\mathrm{\u25b3}{\varphi}_{1}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-{\iint}_{{Q}_{T}}{\varphi}_{1}^{2}(m-\epsilon -\mathrm{\Phi}[u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ =& {\iint}_{{Q}_{T}}\gamma {\lambda}_{1}{\varphi}_{1}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx-{\iint}_{{Q}_{T}}{\varphi}_{1}^{2}(m-\epsilon -\mathrm{\Phi}[u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\\ =& {\int}_{\mathrm{\Omega}}{\varphi}_{1}^{2}{\int}_{0}^{T}(\gamma {\lambda}_{1}-m+\epsilon +\mathrm{\Phi}[u])\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx.\end{array}$

(3.11)

Thus, there exists

${x}_{0}\in \mathrm{\Omega}$ such that

${\int}_{0}^{T}({\lambda}_{1}-m({x}_{0},t)+\mathrm{\Phi}[u({x}_{0},t)])\phantom{\rule{0.2em}{0ex}}dt\ge 0$, then

$\frac{1}{T}{\int}_{0}^{T}m({x}_{0},t)\phantom{\rule{0.2em}{0ex}}dt\le \gamma {\lambda}_{1}+\epsilon +\frac{1}{T}{\int}_{0}^{T}\mathrm{\Phi}[u({x}_{0},t)]\phantom{\rule{0.2em}{0ex}}dt.$

From assumption (A2), we can see that

$\frac{1}{T}{\int}_{0}^{T}m({x}_{0},t)\phantom{\rule{0.2em}{0ex}}dt\le \gamma {\mu}_{1}+\epsilon +C{r}^{2}|\mathrm{\Omega}|$

holds for any sufficiently small *r* and *ε*, which is a contradiction to assumption (A3). The proof is complete. □

**Corollary 2** *There exists a small positive constant* $r<R$ *which is independent of* *ε*,

*τ* *such that* $deg(I-G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),{B}_{r},0)=0,$

*where* ${B}_{r}$ *is a ball centered at the origin with radius* *r* *in* ${L}^{\mathrm{\infty}}({Q}_{T})$.

*Proof* Similar to Lemma 3, we can see that there exists a positive constant

$0<r<R$ independent of

*ε* such that

${u}_{\epsilon}\ne G(\tau ,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}+1-\tau ),\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in \partial {B}_{r},\lambda \in [0,1].$

So, the degree is well defined on

${B}_{r}$. By Lemma 3.3, we can easily see that

$u=G(0,(m-\mathrm{\Phi}[u]){u}^{+}+1)$ admits no solution in

${B}_{r}$. Then, by the homotopy invariance of the Leray-Schauder degree, we have

$\begin{array}{rl}deg(I-G(1,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}),{B}_{r},0)& =deg(1-G(0,(m-\mathrm{\Phi}[{u}_{\epsilon}]){u}_{\epsilon}^{+}+1),{B}_{r},0)\\ =0.\end{array}$

The proof is completed. □

Now, we show the proof of the main result of this paper.

*Proof of Theorem 1*

Using Corollaries 1 and 2, we have

$deg(1-G(f(\cdot )),{B}_{R}\setminus {B}_{r},0)=1,$

where *R* and *r* are positive constants and $R>r$. Problem (2.2)-(2.4) admits a nonnegative nontrivial solution ${u}_{\epsilon}$ with $r\le {\parallel {u}_{\epsilon}\parallel}_{\mathrm{\infty}}\le R$. Combining with the regularity results [9] and a similar argument as in [10], we can prove that the limit function of ${u}_{\epsilon}$ is a nonnegative nontrivial periodic solution of problem (1.1)-(1.3). □