Remark 2.1.

1. (1) Observing that $\lambda _{L}(\mathrm{Y})$ and $\phi _{L}(\sigma _{t}b)(x)$ are analogues of principal eigenvalue and principal eigenfunctions of elliptic problem, respectively. We call $\textrm {span}\{\phi _{L}(\sigma _{t}b)(x)\}_{t\in \mathbb{R}}$ as the principal Floquet bundle of (2.1) associate to the principal Lyapunov exponent.

2. (2) The principal eigenvalue of the following elliptic problem with mixed Dirichlet–Neumann boundary conditions,

   (2.3) \begin{equation} \begin{cases} -\Delta u-\sum ^{N}_{i=1}b_{i}(x)\frac {\partial u}{\partial x_{i}}-b_{0}(x)u=\lambda _{L}u,\quad x\in ({-}L,L)\times \omega ,\\[4pt] \partial _{\nu }u=0\quad \text{on }({-}L,L)\times \partial \omega ,\\[4pt] u({-}L,y)=u(L,y)=0\quad \text{on }y\in \overline {\omega } \end{cases} \end{equation}
   has been studied in [Reference Berestycki and Rossi3 , Theorem 3.1]. Their results show that for any $L\gt 0$ , there exists a unique principal eigenvalue $\lambda _L$ of (2.3) corresponding to a positive eigenfunction $\psi _{L}\in W^{2,p}(\Omega _{L})$ , for any $p\gt 1$ . Moreover, $\lambda _{L}$ converges to $\lambda _{1}(\Omega )$ as $L\rightarrow \infty$ , where $\lambda _{1}(\Omega )$ is the generalised Neumann principal eigenvalue defined by
   \begin{align*} \lambda _{1}(\Omega )\,:\!=\,\sup \{\lambda \in \mathbb{R}\,:\,\exists \phi \gt 0,\,\Delta \phi +\sum ^{N}_{i=1}b_{i}(x)\frac {\partial \phi }{\partial x_{i}}+b_{0}(x)\phi +\lambda u&\leq 0, \text{ a.e. in } \Omega ,\\ \partial _{\nu }\phi &\geq 0 \text{ on } \partial \Omega \}. \end{align*}
   We point out in $\lambda _{1}(\Omega )$ , the function $\phi$ is understood to belong to $W^{2,p}(({-}L,L)\times \omega )$ for some $p\gt N$ and every $L\gt 0$ .

Proof. For any $L\gt 0$ , let $\phi _{L}(\sigma _{t}b)(x)$ be as in Lemma 2.1. Let $0\lt L_{1}\lt L_{2}$ be fixed. Then, there exists $\epsilon _{0}\in (0,1)$ such that

\begin{equation*} \epsilon _{0}\phi _{L_{1}}(\sigma _{t}b)(x)\lt \phi _{L_{2}}(\sigma _{t}b)(x)\quad \text{on }[{-}L_{1},L_{1}]\times \omega . \end{equation*}

It then follows from the comparison principle that

\begin{equation*} \epsilon _{0}U_{L_{1}}(t,b)\phi _{L_{1}}(\sigma _{t}b)(x)\lt U_{L_{2}}(t,b)\phi _{L_{2}}(\sigma _{t}b)(x)\quad \text{on }[{-}L_{1},L_{1}]\times \omega . \end{equation*}

Then

\begin{align*} \lambda (L_{2})=\lim _{t\rightarrow \infty }\frac {\ln {\|U_{L_{2}}(t,b)\phi _{L_{2}}(\sigma _{t}b)\|_{\infty }}}{t} \geq \lim _{t\rightarrow \infty }\frac {\ln {\|\epsilon _{0}U_{L_{1}}(t,b)\phi _{L_{1}}(\sigma _{t}b)\|_{\infty }}}{t}=\lambda (L_{1}), \end{align*}

which implies that $\lambda (L)$ is non-decreasing in $L$ .

Let $b_{L}\,:\!=\,\sup _{t\in \mathbb{R},x\in \Omega _{L}}b_{0}(t,x)$ . Then, $e^{b_{L}t}$ is a supersolution of (2.1), and thus,

\begin{equation*} U_{L}(t,b)\phi _{L}(\sigma _{t}b)\leq e^{b_{L}t}\quad \text{in }\Omega _{L} \end{equation*}

for all $t\geq 0$ . It then follows that

\begin{equation*} \lambda (L)\leq \lim _{t\rightarrow \infty }\frac {\ln {e^{b_{L}t}}}{t}=b_{L}. \end{equation*}

Hence, $\lim _{L\rightarrow \infty }\lambda (L)$ exists. This completes the proof.

Proof. In the following proof, to emphasis the dependence of $\mathrm{Y}$ on $b$ , we write it as $\mathrm{Y}(b)$ . We divide the proof into two cases: $\lambda _{\infty }(\mathrm{Y}(b))\geq \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ and $\lambda _{\infty }(\mathrm{Y}(b))\lt \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ .

Firstly, we consider the case $\lambda _{\infty }(\mathrm{Y}(b))\geq \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ . By definitions of $\lambda _{\infty }(Y(b))$ and $\lambda (Y(b))$ , we have $\lambda _{\infty }(Y(b))\leq \lambda (Y(b))$ . Assume that $\lambda _{\infty }(Y(b))\lt \lambda (Y(b))$ . Without loss of generality, we may assume that

\begin{equation*} \lambda _{\infty }(Y(b))\lt 0\lt \lambda (Y(b)). \end{equation*}

In fact, we can take $\lambda _{0}\in (\lambda _{\infty }(Y(b)),\lambda (Y(b)))$ and replace $b_{0}(t,x)$ by $\overline {b}_{0}(t,x)=b_{0}(t,x)-\lambda _{0}$ . Then

\begin{equation*} \lambda _{\infty }(Y(\overline {b}))=\lambda _{\infty }(Y(b))-\lambda _{0}\lt 0\lt \lambda -\lambda _{0}=\lambda (Y(\overline {b})) \end{equation*}

and

\begin{equation*} \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }\overline {b}_{0}(t,x_{1},y)\lt 0. \end{equation*}

Consider

(2.5) \begin{equation} \begin{cases} \partial _{t}u=\Delta u+\sum ^{N}_{i=1}a_{i}(t,x)\frac {\partial u}{\partial x_{i}}+ug(t,x,u),\quad x\in \mathbb{R}\times \omega ,\\[4pt] \partial _{\nu }u=0\quad \text{on }\mathbb{R}\times \partial \omega . \end{cases} \end{equation}

Denote $v(t,x;\,v_{0})$ be the solution of (2.5) with initial value $v(t,x;\,v_{0})=v_{0}\in X^{+}$ . Choose $v^{*}_{0}\equiv 1$ , similar to that in Theorem 3.1 later, we have that $\lim _{t\rightarrow \infty }\|v(t,\cdot ;\,v^{*}_{0})\|_{\infty }=0$ for $\lambda _{\infty }(Y(b))\lt 0$ . Note that $\lambda (Y(b))\gt 0$ and $g$ is uniformly continuous on $\mathbb{R}\times \Omega \times [0,\delta ]$ for any $\delta \gt 0$ , then there exists large $T\gt 0$ such that

\begin{equation*} g(t,x,0)\leq g(t,x,v(t,x;\,v^{*}_{0}))+\frac {\lambda (Y(b))}{2} \end{equation*}

for any $t\geq T$ and $x\in \mathbb{R}\times \omega$ . By comparison principle,

(2.6) \begin{equation} U(t,\sigma _{T}b)v(T,\cdot ;\,v^{*}_{0})\leq e^{\frac {\lambda (Y(b))}{2}t}v(t+T,\cdot ;\,v^{*}_{0}),\quad \forall t\geq 0. \end{equation}

Due to that $v^{*}_{0}$ is uniform positive and $g$ is bounded on $\mathbb{R}\times \Omega \times [0,\delta ]$ for any $\delta \gt 0$ ,

\begin{equation*} \inf _{x\in \mathbb{R}\times \omega }v(T,\cdot ;\,v^{*}_{0})\gt 0. \end{equation*}

Then for any $v_{0}\in X^{+}$ with $\|v_{0}\|=1$ , there is a constant $M\gt 0$ such that

\begin{equation*} -Mv(T,\cdot ;\,v^{*}_{0})\leq v_{0}\leq Mv(T,\cdot ;\,v^{*}_{0}), \quad x\in \mathbb{R}\times \omega . \end{equation*}

Then the comparison principle implies that

\begin{equation*} -MU(t,\sigma _{T}b)v(T,\cdot ;\,v^{*}_{0})\leq U(t,\sigma _{T}v_{0})\leq MU(t,\sigma _{T}b)v(T,\cdot ;\,v^{*}_{0}),\quad t\geq 0, \end{equation*}

which implies that

\begin{equation*} \|U(t,\sigma _{T}b)v_{0}\|_{\infty }\leq M\|U(t,\cdot )v(T,\cdot ;\,v^{*}_{0})\|_{\infty },\quad \forall t\geq 0. \end{equation*}

Then by the definition of $\lambda (Y(b))$ and together with (2.6), we have that

\begin{equation*} 0\lt \lambda (Y(b))\leq \frac {\lambda (Y(b))}{2}, \end{equation*}

which is a contradiction. Therefore, $\lambda _{\infty }(Y(b))=\lambda (Y(b))$ and they have the same sign in the case $\lambda _{\infty }(\mathrm{Y}(b))\geq \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ .

Secondly, we consider the case $\lambda _{\infty }(\mathrm{Y}(b))\lt \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ . In this case, we show that if $\lambda _{\infty }(Y(b))\lt 0$ , then $\lambda (Y(b))\lt 0$ . Suppose that $\lambda (Y(b))\geq 0$ . Then there exists $\lambda _{0}\gt 0$ such that

\begin{equation*} \lambda _{\infty }(Y(b))+\lambda _{0}\lt \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)+\lambda _{0}\lt 0\lt \lambda (Y(b))+\lambda _{0}. \end{equation*}

Let $a(t,x)=g(t,x,0)+\lambda _{0}$ . Then,

\begin{equation*} \lambda _{\infty }(Y(b^{\prime }))\lt \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }a(t,x_{1},y)\lt 0\lt \lambda (Y(b^{\prime })). \end{equation*}

Similar to above, we have

\begin{equation*} \lambda _{\infty }(Y(b^{\prime }))\lt \frac {\lambda _{\infty }(Y(b^{\prime }))}{2}, \end{equation*}

which is a contradiction. Thus $\lambda (Y(b))\lt 0$ . This completes the proof.

Proof. Since $b_{i}(t,x)\ (i=1,\ldots ,N)$ and $b_{0}(t,x)$ are periodic in $t$ , by the definition of $\lambda _{L}$ , we know that $\lambda _{L}$ is the principal eigenvalue of (2.1) for any $L\gt 0$ . Thus, there exists a bounded positive time periodic eigenfunction $\phi _{L}$ of (2.1) corresponding to $\lambda _{L}$ , that is,

(2.7) \begin{equation} \begin{cases} \frac {\partial \phi _{L}}{\partial t}=\Delta \phi _{L}+\sum ^{N}_{i=1}b_{i}(t,x)\frac {\partial \phi _{L}}{\partial x_{i}}+b_{0}(t,x)\phi _{L}-\lambda _{L}\phi _{L},\quad t\gt 0,\,x\in ({-}L,L)\times \omega ,\\[5pt] \partial _{\nu }\phi _{L}=0,\quad t\gt 0,\,x\in ({-}L,L)\times \partial \omega ,\\[5pt] \phi _{L}(t,-L,y)=\phi _{L}(t,L,y)=0,\quad t\gt 0,\,y\in \omega ,\\[5pt] \phi _{L}\text{ is periodic in }t. \end{cases} \end{equation}

Normialised $\phi _{L}$ by $\max _{\mathbb{R}\times ({-}L,L)\times \omega }\phi _{L}(t,x_{1},y)=1$ and let $(t_{L},x_{1L},y_{L})\in [0,T]\times ({-}L,L)\times \omega$ be such that $\phi _{L}(t_{L},x_{1L},y_{L})=1$ . Extending by reflection the function $\phi _{L}$ to large cylinders, as done in [Reference Berestycki and Rossi3, Appendix A], then $\phi _{L}$ is uniformly bounded in $\mathbb{R}\times ({-}L,L)\times \omega$ . Hence, by standard parabolic regularity estimates, there exists $\phi _{\infty }$ such that $\phi _{L}(t,x)\rightarrow \phi _{\infty }(t,x)$ locally uniformly in $\mathbb{R}\times ({-}L,L)\times \omega$ as $L\rightarrow \infty$ , where $\phi _{\infty }$ is a bounded, nonnegative and time periodic function satisfying

(2.8) \begin{equation} \begin{cases} \frac {\partial \phi _{\infty }}{\partial t}=\Delta \phi _{\infty }+\sum ^{N}_{i=1}b_{i}(t,x)\frac {\partial \phi _{\infty }}{\partial x_{i}}+b_{0}(t,x)\phi _{\infty }-\lambda _{\infty }\phi _{\infty },\quad t\gt 0,\,x\in \mathbb{R}\times \omega ,\\[4pt] \partial _{\nu }\phi _{\infty }=0,\quad t\gt 0,\,x\in \mathbb{R}\times \partial \omega ,\\[4pt] \phi _{\infty }\text{ is periodic in }t. \end{cases} \end{equation}

Consider (2.7) at point $(t_{L},x_{1L},y_{L})$ , we have

\begin{equation*} 0=\Delta \phi _{L}(t_{L},x_{1L},y_{L})+b_{0}(t_{L},x_{1L},y_{L})\phi _{L}-\lambda _{L}\phi _{L}\leq (b_{0}(t_{L},x_{1L},y_{L})-\lambda _{L})\phi _{L}. \end{equation*}

Since $\lambda _{\infty }\gt \lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }b_{0}(t,x_{1},y)$ , we must have that $x_{1L}$ is bounded as $L\rightarrow \infty$ . Thus, $(t_{L},x_{1L},y_{L})\rightarrow (t^*,x^{*}_{1},y^{*})$ as $L\rightarrow \infty$ , and $\phi _{\infty }(t^*,x^{*}_{1},y^{*})=1$ . The strong maximum principal yields that $\phi _{\infty }\gt 0$ in $\mathbb{R}\times \Omega$ . Then, $\lambda _{\infty }$ is the principal eigenvalue of (2.8) with principal eigenfunction $\phi _{\infty }$ .

Proof. Before proving (3.2), we first show that

(3.3) \begin{equation} \lim _{t\rightarrow \infty ,\ |x_{1}|\rightarrow \infty }v(t,x_{1},y;\,v_{0})=0\quad \text{ uniformly for } y\in \omega . \end{equation}

For any fixed $v_{0}\in X^{+}$ , we assume that (3.3) does not hold. Then there exist $\epsilon _{0}\gt 0$ , $\{t_{n}\}\subset (0,\infty )$ and $\{x_{1,n}\}\subset \mathbb{R}$ with $t_{n}\rightarrow \infty$ and $|x_{1,n}|\rightarrow \infty$ , as $n\rightarrow \infty$ , such that

\begin{equation*} v(t_{n},x_{1,n},y;\,v_{0})\geq \epsilon _{0}\quad \text{for some }y\in \omega \text{ and all }n. \end{equation*}

Let $v^{n}(t,x_{1},y)=v(t+t_{n},x_{1}+x_{1,n},y;\,v_{0})$ . Then $v^{n}$ satisfies

(3.4) \begin{equation} \begin{cases} \partial _{t}v^{n}=\Delta v^{n}+c(t+t_{n})\partial _{1}v^{n}+f(t+t_{n},x_{1}+x_{1,n},y,v^{n}), \quad t\gt 0, (x_{1},y)\in \Omega ,\\ \partial _{\nu } v^{n}(t,x_{1},y)=0, \quad t\in \mathbb{R},(x_{1},y)\in \partial \Omega . \end{cases} \end{equation}

Then parabolic estimates and the almost periodicity of $c(t)$ and $f(t,x,v)$ imply that (up to subsequence) there are $v^{*}$ , $c^{*}$ and $f^{*}$ such that

\begin{align*} \lim _{n\rightarrow \infty }v^{n}(t,x_{1},y)=&v^{*}(t,x_{1},y)\quad \text{locally uniformly in }t\in \mathbb{R},(x_{1},y)\in \Omega ,\\ \lim _{n\rightarrow \infty }f(t+t_{n},x_{1}+x_{1,n},y,v^{n})=&f^{*}(t,x_{1},y,v^{*})\quad \text{locally uniformly in }t\in \mathbb{R},(x_{1},y)\in \Omega \end{align*}

and

\begin{align*} \lim _{n\rightarrow \infty }c(t+t_{n})=&c^{*}(t)\quad \text{uniformly in }t\in \mathbb{R}, \end{align*}

where $v^{*}(t,x_{1},y)$ is a nonnegative bounded solution of

(3.5) \begin{equation} \begin{cases} \partial _{t}v^{*}=\Delta v^{*}+c(t)\partial _{1}v^{*}+f(t,x_{1},y,v^{*}),\quad t\in \mathbb{R},(x_{1},y)\in \Omega ,\\ \partial _{\nu } v^{*}=0, \quad t\in \mathbb{R},(x_{1},y)\in \partial \Omega \end{cases} \end{equation}

and

\begin{equation*} v^{*}(0,0,y)=v(t_{n},x_{1,n},y;\,v_{0})\geq \epsilon _{0}. \end{equation*}

Note that $f(t,x,u)\leq f_{u}(t,x,0)u=g(t,x,0)u$ and $\lim _{x_{1}\rightarrow \infty }\sup _{(t,y)\in \mathbb{R}\times \omega }g(t,x_{1},y,0)\lt 0$ . Choose $\alpha \gt 0$ such that

\begin{equation*} \lim _{x_{1}\rightarrow \infty }\sup _{(t,y)\in \mathbb{R}\times \omega }g(t,x_{1},y,0)\lt -\alpha . \end{equation*}

Recall that $S=\max \{M_{0},\|v_{0}\|_{L^{\infty }(\Omega )}\}$ . For any fixed $t_{0}\in \mathbb{R}$ , $w(t,x)=e^{-\alpha (t-t_{0})}S$ is a solution of

(3.6) \begin{equation} \begin{cases} \partial _{t}w= \Delta w+c(t)\partial _{1}w-\alpha w,\quad t\in \mathbb{R}, (x_{1},y)\in \Omega ,\\ \partial _{\nu } w=0, \quad t\in \mathbb{R}, (x_{1},y)\in \partial \Omega . \end{cases} \end{equation}

We also have that $v^{*}(t,x_{1},y)$ is a subsolution of (3.6). The comparison principle implies that

\begin{equation*} v^{*}(t,x_{1},y)\leq e^{-\alpha (t-t_{0})}S\quad \text{for any }t\gt t_{0}, (x_{1},y)\in \Omega , \end{equation*}

particularly, $v^{*}(0,x_{1},y)\leq e^{\alpha t_{0}S}$ for any $(x_{1},y)\in \Omega$ and $t_{0}\lt 0$ . Choose $t_{0}\ll -1$ such that $e^{\alpha t_{0}}S\leq \epsilon _{0}$ , then $v^{*}(0,x_{1},y)\lt \epsilon _{0}$ for $(x_{1},y)\in \Omega$ , which is a contradiction. Therefore, for any $\epsilon \gt 0$ , there exist constants $T_{\epsilon }\gt 0$ and $L(\epsilon )\gt 0$ such that

(3.7) \begin{equation} v(t,x_{1},y;\,v_{0})\lt \epsilon \quad \text{for $t\gt T(\epsilon )$ and $|x|\gt L_{\epsilon }$} \end{equation}

and (3.3) hold true.

Given $v_{0}\in X^{+}$ , $\epsilon \gt 0$ , let

\begin{equation*} v^{\epsilon }(t,x_{1},y)=v(t,x_{1},y;\,v_{0})-\epsilon . \end{equation*}

Then, $v^{\epsilon }(t,x_{1},y)$ satisfies

(3.8) \begin{equation} \begin{cases} \partial _{t}v^{\epsilon }= \Delta v^{\epsilon }+c(t)\partial _{1}v^{\epsilon }+f(t,x,v^{\epsilon }+\epsilon ),\quad t\in \mathbb{R},x\in \Omega ,\\[3pt] \partial _{\nu }v^{\epsilon }=0,\quad t\in \mathbb{R},x\in \partial \Omega . \end{cases} \end{equation}

By the assumption on $f$ , we have $f(t,x,v^{\epsilon }+\epsilon )\leq g(t,x,0)(v^{\epsilon }+\epsilon )$ . Then $v^{\epsilon }(t,x_{1},y)$ satisfies

(3.9) \begin{equation} \begin{cases} \partial _{t}v^{\epsilon }\leq \Delta v^{\epsilon }+c(t)\partial _{1}v^{\epsilon }+g(t,x,0)(v^{\epsilon }+\epsilon ),\quad t\in \mathbb{R},x\in \Omega ,\\[3pt] \partial _{\nu }v^{\epsilon }=0,\quad t\in \mathbb{R},x\in \partial \Omega . \end{cases} \end{equation}

Since $\lambda _{\infty }\lt 0$ , there exists $L\gt 0$ such that $\lambda _{L}\lt 0$ . Let $\eta \,:\,\mathbb{R}\rightarrow [0,1]$ be a smooth function with $\eta (x_{1})=1$ for $|x_{1}|\leq \frac {L}{2}$ and $\eta (x_{1})=0$ for $|x_{1}|\geq \frac {3L}{4}$ . Define

\begin{equation*} v^{\epsilon }_{0}(x_{1},y)=\eta (x_{1})\Big(v_{0}(x_{1},y)-\min _{[{-}L,L]\times \omega }v_{0}(x_{1},y)\Big),\quad (x_{1},y)\in [{-}L,L]\times \omega . \end{equation*}

Clearly, $v^{\epsilon }_{0}(x_{1},y)\in X^{+}$ . Let $v(t,x_{1},y;\,v^{\epsilon }_{0})$ be the unique solution of the following problem

\begin{equation*} \begin{cases} \partial _{t}v= \Delta v+c(t)\partial _{1}v+g(t,x,0)(v+\epsilon ),\quad t\gt 0, x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }v=0,\quad t\gt 0, x\in ({-}L,L)\times \partial \omega ,\\ v(t,-L,y)=v(t,L,y)=0,\quad t\gt 0, y\in \omega \end{cases} \end{equation*}

with initial value $v(0,x_{1},y;\,v^{\epsilon }_{0})=v^{\epsilon }_{0}(x_{1},y)$ , $(x_{1},y)\in [{-}L,L]\times \omega$ . Let $U_{L}(t,c)$ be the solution operator of

\begin{equation*} \begin{cases} \partial _{t}w= \Delta w+c(t)\partial _{1}w+f_{u}(t,x,0)w,\quad t\gt 0, x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }w=0,\quad t\gt 0, x\in ({-}L,L)\times \partial \omega ,\\ w(t,-L,y)=w(t,L,y)=0\quad t\gt 0, y\in \omega . \end{cases} \end{equation*}

Then by the variation of constants formula, we have

(3.10) \begin{equation} v(t,\cdot ;\,v^{\epsilon }_{0})=U_{L}(t,c)v^{\epsilon }_{0}+\epsilon \int ^{t}_{0}U_{L}(t-s,c\cdot s)g(s,x,0)ds. \end{equation}

Since $\lambda _{L}\lt 0$ , by Definition 2.1, there exists $M\gt 0$ such that

\begin{equation*} \epsilon \int ^{t}_{0}U_{L}(t-s,c\cdot s)g(s,x,0)ds\lt \epsilon M \end{equation*}

for $t\geq 0$ . Together with (3.10), we have

(3.11) \begin{equation} v(t,\cdot ;\,v^{\epsilon }_{0})\leq U_{L}(t,c)v^{\epsilon }_{0}+\epsilon M\quad \text{in }({-}L,L)\times \omega ,t\geq 0. \end{equation}

Note that

\begin{equation*} v^{\epsilon }(t,\pm L,y)=v(t,\pm L,y)-\epsilon \lt 0\quad \text{for all }t\gt 0,y\in \omega \end{equation*}

and

\begin{equation*} v^{\epsilon }(0,x_{1},y)=v_{0}(x_{1},y)-\epsilon \leq v^{\epsilon }_{0}(x_{1},y)\quad \text{for }x_{1}\in ({-}L,L), y\in \omega . \end{equation*}

Then, by (3.9) and the comparison principle, we obtain that

\begin{equation*} v^{\epsilon }(t,x_{1},y)\leq v(t,x_{1},y;\,v^{\epsilon }_{0})\quad \text{for }t\gt 0,x_{1}\in ({-}L,L),y\in \omega . \end{equation*}

By (3.11),

\begin{equation*} v^{\epsilon }(t,x_{1},y)\leq U_{L}(t,c)v^{\epsilon }_{0}+\epsilon M\quad \text{for }t\gt 0,x_{1}\in ({-}L,L),y\in \omega . \end{equation*}

Due to $\lambda _{L}\lt 0$ , Definition 2.1 ensures that there exists $T_{\epsilon }\gt 0$ such that

\begin{equation*} U_{L}(t,c)v^{\epsilon }_{0}\leq \varepsilon \quad \text{for all }t\geq T_{\epsilon }. \end{equation*}

Hence,

(3.12) \begin{equation} v^{\epsilon }(t,x_{1},y)\leq (1+M)\epsilon \end{equation}

for all $t\geq T_{\epsilon }$ , $x_{1}\in ({-}L,L)$ and $y\in \omega$ . By (3.9), choose large $T\gt 0$ and $L\gt 0$ , $v(t,x_{1},y;\,v_{0})\lt \epsilon$ for $t\geq T$ , $|x_{1}|\gt \frac {L}{2}$ and $y\in \omega$ . This, together with (3.12), implies that

\begin{equation*} v(t,x_{1},y;\,v_{0})\lt (2+M)\epsilon \quad \text{for all }t\gt \max \{T_{\epsilon },T\}, (x_{1},y)\in \Omega . \end{equation*}

By the arbitrariness of $\epsilon$ , we obtain $\lim _{t\rightarrow \infty }v(t,x_{1},y;\,v_{0})=0$ uniformly for $(x_{1},y)\in \Omega$ . This completes the proof.

Proof. It can be proved by properly modifying the arguments in [Reference Zhao32, Theorem 3.1]. For completeness, we provide the proof in the following.

Let $\mathcal{H}(c,f)$ be the closure of $\{(c\cdot s,f\cdot s)\,:\, s\in \mathbb{R}\}$ with the compact open topology. For any $(\widetilde {c},\widetilde {f})\in \mathcal{H}(c,f)$ , assumptions (H1)-(H4) imply that $w=M$ , $M\geq M_{0}$ , is an supersolution of (3.13). Hence, by the comparison principle and a priori estimates for parabolic equations, each solution $w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})$ of (3.13) exists globally on $[0,\infty )$ , and for any $t\gt 0$ , the set $\{w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})\,:\,t\gt 0\}$ is precompact in $X^{+}_{L}$ . Define the skew-product semiflow $\Pi _{t}\,:\,X^{+}_{L}\times \mathcal{H}(c,f)\rightarrow \Pi _{t}\,:\,X^{+}_{L}\times \mathcal{H}(c,f)$ by

\begin{equation*} \Pi _{t}\big(w_{0},\widetilde {c},\widetilde {f}\big)=(w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f}),\widetilde {c}_{t},\widetilde {f}_{t}). \end{equation*}

Then, the omega limit set $\omega (w_{0},\widetilde {c},\widetilde {f})$ of the forward orbit $\gamma ^{+}(w_{0},\widetilde {c},\widetilde {f})\,:\!=\,\{\Pi _{t}(w_{0},\widetilde {c},\widetilde {f})\,:\,t\geq 0\}$ is well defined, compact and invariant under $\Pi _{t}, t\geq 0$ , for each $(w_{0},\widetilde {c},\widetilde {f})\in X^{+}_{L}\times \mathcal{H}(c,f)$ .

We first show that there exists a $\delta \gt 0$ such that

\begin{equation*} \limsup _{t\rightarrow \infty }\|w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})\|_{X_L}\geq \delta , \quad \forall \big(w_{0},\widetilde {c},\widetilde {f}\big)\in X^{+}_{L}\times \mathcal{H}(c,f). \end{equation*}

We set $\lambda _{L}$ of (3.1) as $\lambda _{L}(g)$ . Since $\lambda _{L}(g)\gt 0$ , then we can choose a sufficient small $\varepsilon _{0}\gt 0$ such that

\begin{equation*} \lambda _{L}(f_{u}-\varepsilon _{0})\gt 0. \end{equation*}

Since $f$ is uniformly almost periodic in $t$ and $\mathcal{H}(f)$ is compact, there exists $\delta _{0}\gt 0$ such that

\begin{equation*} |\widetilde {f}(t,x_{1},y,u)-\widetilde {f}_{u}(t,x_{1},y,0)u|\lt \varepsilon _{0}, \quad (x_{1},y)\in \Omega ,\, t\in \mathbb{R},u\in (0,\delta _{0}], \widetilde {f}\in \mathcal{H}(f). \end{equation*}

Suppose on the contrary that there is $(w_{0},\widetilde {c},\widetilde {f})\in X^{+}_{L}\times \mathcal{H}(c,f)$ , such that

\begin{equation*} \limsup _{t\rightarrow \infty }\|w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f}\|_{X_L}\lt \delta . \end{equation*}

Then, there exists $t_{0}\gt 0$ such that

\begin{equation*} \|w(t,x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})\|_{X_L}\lt \delta ,\quad \forall t\geq t_{0}, (x_{1},y)\in \Omega . \end{equation*}

Since $w(t_{0},x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})\gt 0$ , then

\begin{equation*} \|w\big(t,x_{1},y;\,w_{*},\widetilde {c}_{t_{0}},\widetilde {f}_{t_{0}}\big)\|_{X_L}\lt \delta ,\quad \forall t\geq 0, (x_{1},y)\in \Omega , \end{equation*}

where $w_{*}=w(t_{0},x_{1},y;\,w_{0},\widetilde {c},\widetilde {f})$ , $\widetilde {c}_{t_{0}}=c(t+t_{0})$ and $\widetilde {f}_{t_{0}}=f(t+t_{0},x,w)$ . Then $w\big(t,x_{1},y;\,w_{*},\widetilde {c}_{t_{0}},\widetilde {f}_{t_{0}}\big)$ satisfies the following differential inequality

\begin{equation*} \begin{cases} \partial _{t}w\geq \Delta w+\widetilde {c}(t)\partial _{1}w+(f_{u}(t,x_{1},y,0)-\varepsilon _{0})w,\quad t\gt 0, x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }w=0,\quad t\gt 0, x\in ({-}L,L)\times \partial \omega ,\\ w(t,-L,y)=w(t,L,y)=0,\quad t\gt 0,y\in \omega . \end{cases} \end{equation*}

Similar to (2.2), $\lambda _{L}(f_{u}-\varepsilon _{0})=\lim _{t\rightarrow \infty }\frac {1}{t}\int ^{t}_{0}k_{L}(\sigma _{\tau }f_{u})d\tau$ and $e^{\int ^{t}_{0}k_{L}(\sigma _{\tau }f_{u})d\tau }\phi _{L}(\sigma _{t}f_{u})$ is a solution of

\begin{equation*} \begin{cases} \partial _{t}w= \Delta w+\widetilde {c}(t)\partial _{1}w+(f_{u}(t,x_{1},y,0)-\varepsilon _{0})w,\quad t\gt 0, x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }w=0,\quad t\gt 0, x\in ({-}L,L)\times \partial \omega ,\\ w(t,-L,y)=w(t,L,y)=0,\quad t\gt 0,y\in \omega . \end{cases} \end{equation*}

Choose a sufficiently small $\varepsilon \gt 0$ such that $w_{*}\geq \varepsilon \phi _{L}$ . By the standard comparison theorem, we have

\begin{equation*} w\big(t,x_{1},y;\,w_{*},\widetilde {c}_{t_{0}},\widetilde {f}_{t_{0}}\big)\geq \varepsilon e^{\int ^{t}_{0}k_{L}(\sigma _{\tau }f_{u})d\tau }\phi _{L}(\sigma _{t}f_{u}),\quad t\geq 0, (x_{1},y)\in \Omega . \end{equation*}

Letting $t\rightarrow \infty$ , we have

\begin{equation*} \lim _{t\rightarrow \infty }w\big(t,x_{1},y;\,w_{*},\widetilde {c}_{t_{0}},\widetilde {f}_{t_{0}}\big)=\infty ,\quad \forall (x_{1},y)\in \Omega , \end{equation*}

which is a contradiction.

Following from the invariance of $\omega (w_{0},\widetilde {c},\widetilde {f})$ and $\Pi _{t}((X^{+}_{L}\setminus \{0\})\times \mathcal{H}(c,f))\subset int{X^{+}_{L}\times \mathcal{H}(c,f)}$ , $\forall t\gt 0$ , we have

(3.15) \begin{equation} \omega \big(w_{0},\widetilde {c},\widetilde {f}\big)\subset int{X^{+}_{L}\times \mathcal{H}(c,f)},\quad \forall \big(w_{0},\widetilde {c},\widetilde {f}\big)\in (X^{+}_{L}\setminus \{0\})\times \mathcal{H}(c,f). \end{equation}

For each $(\widetilde {c},\widetilde {f},t)\in \mathcal{H}(c,f)\times (0,\infty )$ , for any $w^{\prime }_{0}(\cdot ),\, w^{\prime \prime }_{0}(\cdot )\in X^{+}_{L}$ , when $w^{\prime }_{0}(\cdot )\gt w^{\prime \prime }_{0}(\cdot )$ , by comparison principal, we have

\begin{equation*} w\big(t,\cdot ;\,w^{\prime }_{0},\widetilde {c},\widetilde {f}\big)\gt w\big(t,\cdot ;\,w^{\prime \prime }_{0},\widetilde {c},\widetilde {f}\big) \quad \text{for all }t\gt 0. \end{equation*}

Then, $w(t,\cdot ;\,\cdot ,\widetilde {c},\widetilde {f})$ is strongly monotone on $X^{+}_{L}$ for each $(\widetilde {c},\widetilde {f},t)\in \mathcal{H}(c,f)\times (0,\infty )$ . Moreover, by assumptions (H1)–(H4), $w(t,\cdot ;\,\cdot ,\widetilde {c},\widetilde {f})$ is also subhomongenous on $X^{+}_{L}$ for each $(\widetilde {c},\widetilde {f},t)\in \mathcal{H}(c,f)\times (0,\infty )$ . Let $w^{\prime }_{0}\in X^{+}_{L}\setminus \{0\}$ be fixed and $K_{0}=\omega (w^{\prime }_{0},\widetilde {c},\widetilde {f})$ . By [Reference Zhao32, Theorem 2.1] and (3.15), for every $w_{0}\in X^{+}_{L}\setminus \{0\}$ ,

\begin{equation*} \lim _{t\rightarrow \infty }\|w\big(t,\cdot ;\,w_{0},\widetilde {c},\widetilde {f}\big)-w\big(t,\cdot ;\,w^{*}_{0},\widetilde {c},\widetilde {f}\big)\|_{X_L}=0, \end{equation*}

where $(w^{*}_{0},\widetilde {c},\widetilde {f})\in K_{0}$ . Therefore, $w(t,\cdot ;\,w^{*}_{0},\widetilde {c},\widetilde {f})$ is a unique positive almost periodic solution of (3.13). This completes the proof.

Proof. By the definition of $\lambda _{\infty }$ , due to $\lambda _{\infty }\gt 0$ , there is $L^{*}\gt 0$ such that $\lambda _{L}\gt 0$ for any $L\gt L^{*}$ . Then, by Proposition 3.1, there is a unique positive almost periodic solution $w^{*}(t,x)$ of (3.13). Note that any nonnegative solution of (1.2) is a supersolution of (3.13). Then,

\begin{equation*} v(t,x_{1},y;\,v_{0})\geq w(t,x_{1},y;\,v_{0})\quad \text{in }({-}L,L)\times \omega . \end{equation*}

Then by (3.14), we have

\begin{equation*} \liminf _{t\rightarrow \infty }\inf _{x_{1}\in [{-}L,L],y\in \omega }v(t,x_{1},y;\,v_{0})\gt 0. \end{equation*}

This completes the proof.

Proof. Let $M_{0}$ be given in (H3). Then $\overline {v}(t,x)=M_{0}$ is a supersolution of (4.1). Let $v(t,x_{1},y;\,\overline {v},c,f)$ be a solution of (1.2) with the initial condition $v(0,x_{1},y;\,\overline {v},c,f)=\overline {v}$ . By comparison principle, $v(t,x_{1},y;\,\overline {v},c,f)\leq \overline {v}(t,x)$ for $x\in \Omega$ and $t\gt 0$ . Since $c$ and $f$ are almost periodic function, there exists $t_{n}\rightarrow \infty$ such that $c\cdot t_{n}\rightarrow c$ and $f\cdot t_{n}\rightarrow f$ uniformly as $n\rightarrow \infty$ . By a priori estimates for parabolic equations, there exists a function $V(t,x)$ such that $v(t+t_{n},x_{1},y;\,\overline {v},c,f)$ converges to $V(t,x_{1},y)$ locally uniformly in $(t,x_{1},y)\in \mathbb{R}\times \Omega$ as $n\rightarrow \infty$ . Then, $V(t,x_{1},y)$ is a bounded entire solution satisfying (4.1) for $t\in \mathbb{R}$ and $(x_{1},y)\in \Omega$ . By Theorem 3.2, we have $\inf _{\mathbb{R}\times [{-}L,L]\omega }V(t,x_{1},y)\gt 0$ for all $L\gt 0$ .

Following from the arguments for (3.3), we also have $\lim _{|x_{1}|\rightarrow \infty }\sup _{t\in \mathbb{R},y\in \omega }V(t,x_{1},y)=0$ , that is, (4.2) holds true. In the following, we prove the uniqueness of $V(t,x_{1},y)$ . Suppose that there are two bounded positive solutions $V(t,x_{1},y)$ and $V^{\prime }(t,x_{1},y)$ of (4.1). We first show that $V(t,x_{1},y)\leq V^{\prime }(t,x_{1},y)$ . For any $\epsilon \gt 0$ , define

\begin{equation*} K_{\epsilon }\,:\!=\,\{k\gt 0\,:\,kV^{\prime }\geq V-\epsilon \quad \text{in }\mathbb{R}\times \overline {\Omega }\}. \end{equation*}

By (4.2), there exists $R(\epsilon )\gt 0$ such that

\begin{equation*} V(t,x_{1},y)-\epsilon \lt 0, \quad \forall \, |x_{1}|\gt R(\epsilon ),y\in \omega ,t\in \mathbb{R}. \end{equation*}

Together with $V^{\prime }\gt 0$ in $\mathbb{R}\times \overline {\Omega }$ , we find that the set $K_{\epsilon }$ is nonempty. Set $k(\epsilon )\,:\!=\,\inf \{k\,:\,k\in K_{\epsilon }\}$ . Then, $k\,:\,\mathbb{R}^{+}\rightarrow \mathbb{R}$ is a nonincreasing function and thus $\lim _{\epsilon \rightarrow 0^{+}}k(\epsilon )$ exists. Set $k^{*}=\lim _{\epsilon \rightarrow 0^{+}}k(\epsilon )$ . Assume that

(4.3) \begin{equation} k^{*}\gt 1. \end{equation}

For any $0\lt \epsilon \lt \sup _{\mathbb{R}\times \Omega }V$ small enough, we have $k(\epsilon )\gt 0$ ,

\begin{equation*} k(\epsilon )V^{\prime }-V+\epsilon \geq 0\quad \text{in }\mathbb{R}\times \overline {\Omega } \end{equation*}

and there exists a sequence $\{t^{\epsilon }_{n}\}\subset \mathbb{R}$ such that

\begin{equation*} \lim _{n\rightarrow \infty }\inf _{(x_{1},y)\in \mathbb{R}\times \omega }[k_{\epsilon }V^{\prime }(t^{\epsilon }_{n},x_{1},y)-V(t^{\epsilon }_{n},x_{1},y)+\epsilon ]=0. \end{equation*}

Without loss of generality, we assume that there are $V^{\prime }_{\epsilon }(t,x_{1},y)$ , $V_{\epsilon }(t,x_{1},y)$ and $(c^{\epsilon },f^{\epsilon })\in \mathcal{H}(c,f)$ such that

\begin{equation*} V^{\prime }_{\epsilon }(t,x_{1},y)=\lim _{n\rightarrow \infty }V^{\prime }\big(t+t^{\epsilon }_{n},x_{1},y\big), V_{\epsilon }(t,x_{1},y)=\lim _{n\rightarrow \infty }V\big(t+t^{\epsilon }_{n},x_{1},y\big) \end{equation*}

and

\begin{equation*} f^{\epsilon }(t,x,u)=\lim _{n\rightarrow \infty }f(t+t^{\epsilon }_{n},x_{1},y,u) \end{equation*}

locally uniformly in $(t,x_{1},y)\in \mathbb{R}\times \mathbb{R}\times \omega$ , $u\gt 0$ and

\begin{equation*} c^{\epsilon }(t)=\lim _{n\rightarrow \infty }c(t+t^{\epsilon }_{n})\quad \text{uniformly in }t\in \mathbb{R}. \end{equation*}

Then, $V_{\epsilon }$ and $V^{\prime }_{\epsilon }$ are bounded positive solution of (1.2) with $c$ and $f$ replaced by $c^{\epsilon }$ and $f^{\epsilon }$ , respectively. Moreover, there hold $k_{\epsilon }V^{\prime }_{\epsilon }\geq V_{\epsilon }-\epsilon$ and

\begin{equation*} \inf _{(x_{1},y)\in \mathbb{R}\times \omega }[k_{\epsilon }V^{\prime }_{\epsilon }(0,x)-V(0,x)+\epsilon ]=0. \end{equation*}

By (4.2) again, there is $(x^{\epsilon }_{1},y^{\epsilon })\in \mathbb{R}\times \omega$ such that

\begin{equation*} k_{\epsilon }V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)-V\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)+\epsilon =0. \end{equation*}

Then, the function $t\rightarrow k_{\epsilon }V^{\prime }_{\epsilon }(t,x^{\epsilon }_{1},y^{\epsilon })-V(t,x^{\epsilon }_{1},y^{\epsilon })+\epsilon$ attains its minimum value at $0$ and $k_{\epsilon }V^{\prime }_{\epsilon }(0,x_{1},y)-V(0,x_{1},y)+\epsilon$ attains its minimum value at $(x^{\epsilon }_{1},y^{\epsilon })$ .

Then, at point $\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)$ , by (H1)-(H4), we have

\begin{align*} 0 & = k_{\epsilon }\frac {\partial V^{\prime }_{\epsilon }}{\partial t}\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)-\frac {\partial V_{\epsilon }}{\partial t}\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\\ &=\left[k_{\epsilon }\Delta V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)+c^{\epsilon }(0)\partial _{1}\left(k_{\epsilon }V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\right)+f^{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon }V^{\prime }_{\epsilon }\right)\right]\\ & \quad -\left[\Delta V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)+c^{\epsilon }(0)\partial _{1}V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)+f^{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon },V_{\epsilon }\right)\right]\\ &=\left[k_{\epsilon }\Delta V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)-\Delta V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\right]+c^{\epsilon }(0)\left[\partial _{1}\left(k_{\epsilon }V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\right)-\partial _{1}V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\right]\\ &\quad +f^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon }V^{\prime }_{\epsilon })-f^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },V_{\epsilon })\\ &\geq f^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon }V^{\prime }_{\epsilon })-f^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },V_{\epsilon })\\ & \geq \left[k_{\epsilon } V^{\prime }_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)-V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)\right]\left[g^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon } V^{\prime }_{\epsilon })+V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)g^{\epsilon }_{u}\left(0,x^{\epsilon }_{1},y^{\epsilon },\eta _{\epsilon }\right)\right], \end{align*}

where $\eta _{\epsilon }$ is between $k_{\epsilon } V^{\prime }_{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon })$ and $V_{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon })$ . Since $k_{\epsilon } V^{\prime }_{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon })-V_{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon })\lt 0$ , we obtain that

\begin{equation*} g^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon } V^{\prime }_{\epsilon })+V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)g^{\epsilon }_{u}(0,x^{\epsilon }_{1},y^{\epsilon },\eta _{\epsilon })\geq 0. \end{equation*}

From (H2), $g^{\epsilon }_{u}\leq 0$ . Then

\begin{equation*} 0\leq g^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon } V^{\prime }_{\epsilon })+V_{\epsilon }\left(0,x^{\epsilon }_{1},y^{\epsilon }\right)g^{\epsilon }_{u}(0,x^{\epsilon }_{1},y^{\epsilon },\eta _{\epsilon })\leq g^{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon },k_{\epsilon } V^{\prime }_{\epsilon })\leq 0. \end{equation*}

By (H4), $\{x^{\epsilon }_{1}\}$ must be bounded. From $k_{\epsilon }V^{\prime }_{\epsilon }(0,x^{\epsilon }_{1},y^{\epsilon })-V(0,x^{\epsilon }_{1},y^{\epsilon })+\epsilon =0$ , we get that $\{k_{\epsilon }\}$ is bounded and then $k^{*}\in [1,\infty )$ . Choose a sequence $\{\epsilon _{n}\}$ in $\mathbb{R}^{+}$ such that

\begin{equation*} \lim _{n\rightarrow \infty }\epsilon _{n}=0,\quad \xi \,:\!=\,\lim _{n\rightarrow \infty }x^{\epsilon _{n}}_{1}\in \mathbb{R},\quad \eta \,:\!=\,\lim _{n\rightarrow \infty }y^{\epsilon _{n}}\in \overline {\omega }, \end{equation*}

and

\begin{equation*} V^{\prime *}(t,x_{1},y)=\lim _{n\rightarrow \infty }V^{\prime }_{\epsilon _{n}}(t,x_{1},y),\quad V^{*}(t,x_{1},y)=\lim _{n\rightarrow \infty }V_{\epsilon _{n}}(t,x_{1},y) \end{equation*}

locally uniformly in $(t,x)\in \mathbb{R}\times \Omega$ ,

\begin{equation*} c^{*}(t)=\lim _{n\rightarrow \infty }c^{\epsilon _{n}}(t)\quad \text{uniformly in }t\in \mathbb{R}, \end{equation*}

and

\begin{equation*} f^{*}(t,x_{1},y,u)=\lim _{n\rightarrow \infty }f^{\epsilon _{n}}(t,x,u)\quad \text{uniformly in }t\in \mathbb{R}, x\in \Omega , u\gt 0. \end{equation*}

Then $V^{*}$ and $V^{\prime *}$ are bounded positive entire solutions of (4.1) with $c$ and $f$ are replaced by $c^{*}$ and $f^{*}$ . Moreover, we have $k^{*}V^{\prime *}\geq V^{*}$ and $k^{*}V^{\prime *}(0,\xi ,\eta )-V^{*}(0,\xi ,\eta )=0$ . Set $W\,:\!=\,k^{*}V^{\prime *}-V^{*}$ satisfies $W(0,\xi ,\eta )=0$ . Then,

\begin{align*} W_{t}=&\Delta W+c^{*}(t)\partial _{1}W+k^{*}V^{\prime *}g(t,x,V^{\prime *})-V^{*}g(t,x,V^{*})\\ \geq &\Delta W+c^{*}(t)\partial _{1}W+[g(t,x,k^{*}V^{\prime *})+V^{*}g_{u}(t,x,\eta )]W,\quad \forall t\in \mathbb{R}, x\in \Omega \end{align*}

and

\begin{equation*} \partial _{\nu }W=0,\quad t\in \mathbb{R}, x\in \partial \Omega . \end{equation*}

Since $W$ is nonnegative in $\mathbb{R}\times \Omega$ , vanishes at $(\xi ,\eta )$ , the strong maximum principle and the Hopf lemma yield $W=0$ . Then

\begin{equation*} g(t,x,k^{*}V^{\prime *})=g(t,x,V^{\prime *})=g(t,x,V^{*}), \end{equation*}

which implies that $k^{*}=1$ and then $V(t,x_{1},y)\leq V^{\prime }(t,x_{1},y)$ . Similarly, we can also prove $V(t,x_{1},y)\geq V^{\prime }(t,x_{1},y)$ . Therefore, $V(t,x_{1},y)= V^{\prime }(t,x_{1},y)$ and $V(t,x_{1},y)$ is a unique bounded positive solution of (4.1). This completes the proof.

Proof. See [Reference Fink10, Theorems 1.17 and 2.10].

Proof. From Lemma 4.1, we need to show that $V(t,x_{1},y)$ is almost periodic in $t$ for $(x_{1},y)$ in bounded sets. We write $V(t,x_{1},y;\,c,f)$ for $V(t,x_{1},y)$ to indicate the dependence on $c$ and $f$ . The Schauder estimates ensure that $V(t,x_{1},y;\,c,f)$ is uniformly continuous in $(t,x_{1},y)\in \mathbb{R}\times \Omega$ . It then suffices to show that for each $x$ in bounded sets, $V(t,x_{1},y;\,c,f)$ is almost periodic in $t$ . To this end, let $\{\alpha _{n}\}$ and $\{\beta _{n}\}$ be any two sequence of $\mathbb{R}$ . Without loss of generality, we may assume that

\begin{equation*} c^{\prime }=\lim _{m\rightarrow \infty }c(t+\beta _{m}),\quad c^{\prime \prime }=\lim _{n\rightarrow \infty }c(t+\alpha _{n}+\beta _{n}), \end{equation*}

\begin{equation*} f^{\prime }(t,x,u)=\lim _{m\rightarrow \infty }f(t+\beta _{m},x,u),\quad f^{\prime \prime }(t,x,u)=\lim _{n\rightarrow \infty }f(t+\alpha _{n}+\beta _{n},x,u) \end{equation*}

locally uniformly in $(t,x,u)\in \mathbb{R}\times \Omega \times \mathbb{R}$ and

\begin{equation*} V^{\prime }(x)=\lim _{m\rightarrow \infty }V(\beta _{m},x;\,c,f)\quad V^{\prime \prime }(x)=\lim _{n\rightarrow \infty }V(\alpha _{n}+\beta _{n},x;\,c,f) \end{equation*}

locally uniformly in $x\in \Omega$ . Therefore,

(4.5) \begin{equation} V(t,x;\,v^{\prime },c^{\prime },f^{\prime })=\lim _{m\rightarrow \infty }V(t+\beta _{m},x;\,c,f),\quad V(t,x;\,v^{\prime \prime },c^{\prime \prime },f^{\prime \prime })=\lim _{n\rightarrow \infty }V(t+\alpha _{n}+\beta _{n},x;\,c,f) \end{equation}

locally uniformly in $(t,x)\in \mathbb{R}\times \Omega$ , where $V(t,x;\,v^{\prime },c^{\prime },f^{\prime })$ and $V(t,x;\,v^{\prime \prime },c^{\prime \prime },f^{\prime \prime })$ are bounded positive entire solution of (4.1) with $c$ and $f$ replaced by $c^{\prime },f^{\prime }$ and $c^{\prime \prime },f^{\prime \prime }$ , respectively. By Lemma 4.1,

\begin{equation*} \inf _{\mathbb{R}\times [{-}L,L]\times \omega }V(t,x;\,v^{\prime },c^{\prime },f^{\prime })\gt 0\quad \text{and}\quad \inf _{\mathbb{R}\times [{-}L,L]\times \omega }V(t,x;\,v^{\prime \prime },c^{\prime \prime },f^{\prime \prime })\gt 0 \end{equation*}

for all $L\gt 0$ .

Let

\begin{equation*} V^{\prime }(t,x)\,:\!=\,\lim _{n\rightarrow \infty }V(t+\alpha _{n},x;\,v^{\prime },c^{\prime },f^{\prime }) \end{equation*}

locally uniformly in $(t,x)\in \mathbb{R}\times \Omega$ . Since $\lim _{n\rightarrow \infty }c^{\prime }\cdot \alpha _{n}=c^{\prime \prime }$ , then $V^{\prime }$ is a bounded positive entire solutions of (1.2) with $c$ and $f$ are replaced by $c^{\prime }$ and $f^{\prime }$ , and $\inf _{\mathbb{R}\times [{-}L,L]\times \omega }V^{\prime }\gt 0$ . By the uniqueness of strictly positive solution, see Lemma 4.1, $V^{\prime }(t,x)=V(t+\alpha _{n},x;\,v^{\prime \prime },c^{\prime \prime },f^{\prime \prime })$ . It then follows from (4.5) that

\begin{align*} &\lim _{n\rightarrow \infty }\lim _{m\rightarrow \infty }V(t+\alpha _{n}+\beta _{m},x;\,c,f)\\ =&\lim _{n\rightarrow \infty }V(t+\alpha _{n},x;\,v^{\prime },c^{\prime },f^{\prime })=V(t+\alpha _{n},x;\,v^{\prime \prime },c^{\prime \prime },f^{\prime \prime })\\ =&\lim _{n\rightarrow \infty }V(t+\alpha _{n}+\beta _{n},x;\,c,f) \end{align*}

holds for all $t\in \mathbb{R}$ and uniformly for $x$ in bounded subsets. Hence, by Lemma 4.3, $V(t,x;\,c,f)$ is almost periodic in $t$ uniformly for $x$ in bounded sets.

Next, we show that (4.4) holds true. Let $v_{0}\in X^{+}\setminus \{0\}$ . Assume that the conclusion fails. Then there exist $\varepsilon _{0}\gt 0$ and $\{t_{n}\}\subset (0,\infty )$ satisfying $\lim _{n\rightarrow \infty }t_{n}=\infty$ and $\{(x^{n}_{1},y^{n})\}\subset \Omega$ such that

(4.6) \begin{equation} \left|v\left(t_{n},x^{n}_{1},y^{n};\,v_{0},c,f\right)-V\left(t_{n},x^{n}_{1},y^{n};\,c,f\right)\right|\geq \varepsilon _{0},\quad \forall n\in \Bbb {N}. \end{equation}

It is no loss of generality to assume that $y^{n}$ converges to some $\eta \in \overline {\omega }$ . We may assume without loss of generality that $c\cdot t_{n}\rightarrow c^{\prime }$ as $n\rightarrow \infty$ . When $\lim _{n\rightarrow \infty }|x^{n}_{1}|=\infty$ , by (3.3), we must have $\lim _{n\rightarrow \infty }V(t_{n},x^{n}_{1},y^{n};\,c,f)=0$ and $\lim _{n\rightarrow \infty }v(t_{n},x^{n}_{1},y^{n};\,v_{0},c,f)=0$ , which is a contradiction with (4.6). When $\{x^{n}_{1}\}$ is bounded, set $x^{n}_{1}\rightarrow x^{\prime }_{1}$ as $n\rightarrow \infty$ . By standard parabolic estimates, we may assume that

\begin{equation*} v\left(t+t_{n},x_{1},y;\,v_{0},c,f\right)\rightarrow v_{1}\left(t,x_{1},y;\,c^{\prime },f^{\prime }\right)\quad \text{and}\quad V(t+t_{n},x_{1},y;\,c,f)\rightarrow v_{2}(t,x_{1},y;\,c^{\prime },f^{\prime }) \end{equation*}

locally uniformly in $t\in \mathbb{R}$ and $(x_{1},y)\in \Omega$ , respectively. Then, $v_{i}(t,x_{1},y;\,c^{\prime },f^{\prime })\,(i=1,2)$ are bounded positive entire solution of (4.1) with $c$ and $f$ replaced by $c^{\prime }$ and $f^{\prime }$ . Moreover, $\inf _{\mathbb{R}\times [{-}L,L]\times \omega }v_{i}(\cdot ,\cdot ,\cdot ;\,c^{\prime },f^{\prime })\gt 0$ for all $L\gt 0$ . The uniqueness in Lemma 4.1 implies that $v_{1}(t,x_{1},y;\,c^{\prime },f^{\prime })=v_{2}(t,x_{1},y;\,c^{\prime },f^{\prime })$ for all $t\in \mathbb{R}$ and $(x_{1},y)\in \Omega$ . By (4.6) again, when $n\rightarrow \infty$ , there is $|v_{1}(0,x^{\prime }_{1},\eta ;\,c^{\prime })-v_{2}(0,x^{\prime }_{1},\eta ;\,c^{\prime })|\geq \varepsilon _{0}$ . This is a contradiction. Hence, (4.4) holds true. This completes the proof.

Proof. (1) When $A=0$ , by the Liouville transformation $v(t,x_{1},y)=u(t,x_{1},y)e^{\frac {c}{2}x_{1}}$ , problem (4.1) reduces to

(4.9) \begin{equation} \begin{cases} \partial _{t}v=\Delta v+f\left (t,x_{1},y,v(t,x_{1},y)e^{-\frac {c}{2}x_{1}}\right )e^{\frac {c}{2}x_{1}}-\frac {c^{2}}{4}v,\, t\gt 0,x\in \Omega ,\\[3pt] \partial _{\nu }v(t,x)=0,\quad t\gt 0,x\in \partial \Omega . \end{cases} \end{equation}

The associated linearised problem associated with (4.9) is

(4.10) \begin{equation} \begin{cases} \partial _{t}w=\Delta w+g(t,x,0)w-\frac {c^{2}}{4}w,\, t\gt 0,x\in \Omega ,\\[3pt] \partial _{\nu }w(t,x)=0,\quad t\gt 0,x\in \partial \Omega . \end{cases} \end{equation}

Let $\lambda ^{0}_{\infty }(0)$ is the principal Lyapunov exponent of

(4.11) \begin{equation} \begin{cases} \partial _{t}w=\Delta w+g(t,x,0)w,\, t\gt 0,x\in \Omega ,\\[3pt] \partial _{\nu }w(t,x)=0,\quad t\gt 0,x\in \partial \Omega . \end{cases} \end{equation}

Then

\begin{equation*} \lambda ^{0}_{\infty }(c)=\lambda ^{0}_{\infty }(0)-\frac {c^{2}}{4}. \end{equation*}

Define

\begin{equation*} c^{*}= \begin{cases} 2\sqrt {\lambda ^{0}_{\infty }}\quad \text{if }\lambda ^{0}_{\infty }\gt 0,\\[4pt] 0\quad \text{otherwise }. \end{cases} \end{equation*}

It then follows that $\lambda ^{0}_{\infty }(c)\gt 0$ for $c\in [0,c^{*})$ and $\lambda ^{0}_{\infty }(c)\lt 0$ for $c\gt c^{*}$ .

(2) Fix $A\gt 0$ . Note that $\sigma (t)$ and $f(t,x,u)$ are periodic in $t$ . Then $\lambda ^{A}_{L}$ is the principal eigenvalue of (4.10) on $({-}L,L)\times \Omega$ and $t\in \mathbb{R}$ . Then

(4.12) \begin{equation} \lambda ^{A}_{L}\geq \Lambda _{L}+\inf _{t\in \mathbb{R},x\in \Omega }g(t,x,0), \end{equation}

where $\Lambda _{L}$ is the principal eigenvalue of

\begin{equation*} \begin{cases} \partial _{t}w=\Delta w+[c+A\sigma (t)]\partial _{1}w,\quad x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }w=0,\quad x\in ({-}L,L)\times \partial \omega ,\\ w(t,L,y)=w(t,-L,y)=0,\quad y\in \omega \end{cases} \end{equation*}

and $\phi _{L}(t,x_{1},y)$ is the periodic eigenfunction corresponding to $\Lambda _{L}$ . Set

\begin{equation*} z=x_{1}+A\int ^{t}_{0}\sigma (s)ds \quad \text{and}\quad \Phi _{L}(t,z,y)=\phi _{L}(t,x_{1},y). \end{equation*}

Then $\Phi _{L}(t,z,x)$ is a periodic function defined on periodic domain

\begin{equation*} D_{T}=\cup _{t\in \mathbb{R}}\left (\{t\}\times [{-}L+A\int ^{t}_{0}\sigma (s)ds,L+A\int ^{t}_{0}\sigma (s)ds]\times \omega \right ) \end{equation*}

and $\Phi _{L}(t,z,x)$ also satisfies

\begin{equation*} \begin{cases} \partial _{t}\Phi _{L}=\Delta \Phi _{L}+c\partial _{1}\Phi _{L}-\Lambda _{L}\Phi _{L},\quad x\in ({-}L,L)\times \omega ,\\ \partial _{\nu }\Phi _{L}=0,\quad x\in ({-}L,L)\times \partial \omega ,\\ \Phi _{L}(t,L,y)=\Phi _{L}(t,-L,y)=0,\quad y\in \omega . \end{cases} \end{equation*}

Let $\widetilde {L}\gt 0$ large such that $\mathbb{R}\times [{-}\widetilde {L}-1,\widetilde {L}+1]\times \omega \subset D_{T}$ . Moreover, $\widetilde {L}\rightarrow \infty$ as $L\rightarrow \infty$ . Then on $\mathbb{R}\times [{-}\widetilde {L}-1,\widetilde {L}+1]\times \omega$ , we find that $\Phi _{L}$ is periodic in $t$ and $\inf \Phi _{L}\gt 0$ and

\begin{equation*} \begin{cases} \partial _{t}\Phi _{L}=\Delta \Phi _{L}+c\partial _{1}\Phi _{L}-\Lambda _{L}\Phi _{L},\quad [{-}\widetilde {L}-1,\widetilde {L}+1]\times \omega ,\\ \partial _{\nu }\Phi _{L}=0,\quad x\in ({-}L,L)\times \partial \omega . \end{cases} \end{equation*}

Set $\widetilde {\Lambda }_{L}$ is the principal eigenvalue of

\begin{equation*} \begin{cases} \Delta \phi +c\partial _{1}\phi =\widetilde {\Lambda }\phi ,\quad x\in ({-}\widetilde {L},\widetilde {L})\times \omega ,\\ \partial _{\nu }\phi _{L}=0,\quad x\in ({-}\widetilde {L},\widetilde {L})\times \partial \omega ,\\ \phi (\widetilde {L},y)=\phi ({-}\widetilde {L},y)=0,\quad y\in \omega . \end{cases} \end{equation*}

Then the comparison principal implies that $\Lambda _{L}\geq \widetilde {\Lambda }_{L}$ . By (4.12), we get

\begin{equation*} \lambda ^{A}_{L}\geq \widetilde {\Lambda }_{L}+\inf _{t\in \mathbb{R},x\in \Omega }g(t,x,0). \end{equation*}

Following from [Reference Berestycki and Rossi3, Proposition 1], we have that $\lim _{L\rightarrow \infty }\widetilde {\lambda }_{L}$ exists and $\lim _{L\rightarrow \infty }\widetilde {\lambda }_{L}=-\lambda (\Omega )$ . Thus, we obtain that $\lambda ^{A}_{\infty }\geq -\lambda (\Omega )+\inf _{t\in \mathbb{R},x\in \Omega }g(t,x,0).$ This completes the proof.

Proof. The proof is similar to that in [Reference Berestycki and Rossi3]. For sake of completeness, we include the proof here. By the hypotheses on $v$ , there exist $\varepsilon \gt 0$ and $L\gt 0$ such that

\begin{equation*} \partial _{t}v-\Delta v\lt ({-}\gamma -2\varepsilon )v \end{equation*}

for a.e. $t\in \mathbb{R}$ and $(x_{1},y)\in \Omega \setminus \overline {\Omega }_{L}$ . We list the following functions as that in [Reference Berestycki and Rossi3, Lemma 3.2]. For $a\gt 0$ , let $\theta _{a}\,:\,[L,L+a]\rightarrow \mathbb{R}$ be the solution to

(4.14) \begin{equation} \begin{cases} \theta ^{\prime \prime }=(\gamma +\varepsilon )\theta ,\quad \text{in }(L,L+a),\\ \theta (L)=ke^{\sqrt {\gamma }L},\\ \theta (L+a)=ke^{\sqrt {\gamma }(L+a)}. \end{cases} \end{equation}

Hence, $\theta _{a}(\rho )=A_{a}e^{-\sqrt {\gamma +\varepsilon }\rho }+B_{a}e^{\sqrt {\gamma +\varepsilon }\rho }$ , $\rho \in [L,L+a]$ with

\begin{equation*} A_{a}=ke^{(\sqrt {\gamma }+\sqrt {\gamma +\varepsilon })L}\left (1-\frac {e^{\sqrt {\gamma }a}-e^{-\sqrt {\gamma +\varepsilon }a}}{e^{\sqrt {\gamma +\varepsilon }a}-e^{-\sqrt {\gamma +\varepsilon }a}}\right ) \end{equation*}

and

\begin{equation*} B_{a}=ke^{(\sqrt {\gamma }-\sqrt {\gamma +\varepsilon })L}\left (1-\frac {e^{\sqrt {\gamma }a}-e^{-\sqrt {\gamma +\varepsilon }a}}{e^{\sqrt {\gamma +\varepsilon }a}-e^{-\sqrt {\gamma +\varepsilon }a}}\right ). \end{equation*}

Fix $\tau \in \mathbb{R}$ and define

\begin{equation*} \theta _{a}(t,x_{1},y)\,:\!=\,\theta _{a}(|x_{1}|)+(\tau -t)\delta _{a}, \end{equation*}

where

\begin{equation*} \delta _{a}=\frac {\varepsilon }{2}\min _{[L,L+a]\times \omega }\theta _{a}\gt 0. \end{equation*}

By the similar computation in [Reference Berestycki and Rossi2, Lemma 3.2],

(4.15) \begin{equation} \begin{cases} \partial _{t}\theta _{a}\geq \Delta \theta _{a}-(\gamma +2\varepsilon )\theta _{a},\quad t\in ({-}\infty ,\tau ), (x_{1},y)\in (L,L+a)\times \omega ,\\ \partial _{\nu }\theta _{a}=0,\quad t\in ({-}\infty ,\tau ), (x_{1},y)\in (L,L+a)\times \partial \omega . \end{cases} \end{equation}

Note that $v$ is a subsolution of (4.15) and $v\leq \theta _{a}$ on $\{\pm L,\pm (L+a)\}\times \omega$ , $t\in ({-}\infty ,\tau )$ . Then the comparison principle implies that

\begin{equation*} v\leq \theta _{a}\quad \text{in } (x_{1},y)\in \Omega _{L+a}\setminus \widetilde {\Omega }_{L}, t\lt \tau ,\quad \text{for any }a\gt 0. \end{equation*}

For $|x_{1}|\gt L$ and $y\in \omega$ , we have

\begin{equation*} v(\tau ,x_{1},y)\leq \lim _{a\rightarrow \infty }\theta _{a}(\tau ,x_{1},y)=ke^{\sqrt {\gamma }+\sqrt {\gamma +\varepsilon }L}e^{-\sqrt {\gamma +\varepsilon }|x_{1}|}. \end{equation*}

Due to the arbitrarily of $\tau$ , we obtain that (4.13) holds true. This completes the proof.

Proof. The function $v(t,x_{1},y)=V(t,x_{1},y)e^{\frac {c}{2}x_{1}}$ is a solution of (4.9). Hence,

\begin{equation*} \frac {\partial _{t}v-\Delta v(t,x_{1},y)}{v(t,x_{1},y)}=\zeta (t,x_{1},y)-\frac {c^{2}}{4} \end{equation*}

for $t\in \mathbb{R}$ and $(x_{1},y)\in \Omega$ , where $\zeta (t,x_{1},y)\leq g(t,x_{1},y,0)$ .

Set $\zeta =-\lim _{r\rightarrow \infty }\sup _{|x_{1}|\gt r,t\in \mathbb{R},y\in \omega }g(t,x_{1},y,0)$ and (H4) implies $\zeta \gt 0$ . Then for $\gamma \in (\frac {c^{2}}{4},\frac {c^{2}}{4}+\zeta )$ ,

\begin{equation*} \lim _{L\rightarrow \infty }\sup _{|x_{1}|\gt L,y\in \Omega }\frac {\partial _{t}v-\Delta v(t,x_{1},y)}{v(t,x_{1},y)}\leq \lim _{L\rightarrow \infty }\sup _{|x_{1}|\gt L,y\in \Omega }g(t,x_{1},y,0)-\frac {c^{2}}{4}\leq -\zeta -\frac {c^{2}}{4}\lt -\gamma . \end{equation*}

On the other hand, $v(t,x_{1},y)e^{-\sqrt {\gamma }|x_{1}|}\leq v(t,x_{1},y)e^{-\frac {c}{2}|x_{1}|}$ , which is bounded. Then by Lemma 4.3, there exist two constants $h$ and $\beta \gt 0$ such that

\begin{equation*} V(t,x_{1},y)=v(t,x_{1},y)e^{-\frac {c}{2}x_{1}}\leq he^{-\sqrt {\gamma }|x_{1}|-\frac {c}{2}|x_{1}|}\leq he^{-\beta |x_{1}|}, \forall (t,x_{1},y)\in \mathbb{R}\times \Omega . \end{equation*}

This completes the proof.