Proof. We shall prove that $Q_t$ is the continuous in $\phi$ with respect to the compact open topology uniformly for $t\in [0,t_0]$ with $t_0\gt 0$ . In view of [Reference Weng and Zhao29, Lemma 3.1], the solution semigroup of the following linear nonlocal dispersal equation

(3.3) \begin{align} \begin{cases} \displaystyle \frac{\partial V(t,x)}{\partial t}=D(J\ast V-V)(t,x), \ \ t\gt 0, \ x\in \mathbb{R} \\[6pt] \displaystyle V(0,x)=\psi (x), \ \ \ x\in \mathbb{R}, \end{cases} \end{align}

is given by

(3.4) \begin{align} [P(t)\psi ](x)=e^{-Dt}\sum ^\infty _{k=0}\frac{(Dt)^k}{k!}a_k(\psi )(x), \ \ t\gt 0,\ x\in \mathbb{R} \end{align}

for any $\psi \in \mathbb{Y}$ , where $\mathbb{Y}$ is the set of all bounded and continuous functions from $\mathbb{R}$ to $\mathbb{R}$ , and

\begin{align*} a_0(\psi )(x)=\psi (x),\quad a_m(\psi )(x)=\int _{\mathbb{R}}J(x-y)a_{m-1}(\psi )(y)dy, \ \forall m\ge 1. \end{align*}

For any $\psi \in \mathbb{Y}$ , define $\|\cdot \|=\sup _{x\in \mathbb{R}}|\psi (x)|$ . It is easy to see that $\|a_0(\psi )\|=\|\psi \|$ , $\|a_1(\psi )(x)\|=\|\int _{\mathbb{R}}J(x-y)a_{0}(\psi )(y)dy\|\le \|\psi \|$ . By induction, we can obtain $\|a_k(\psi )(x)\|\le \|\psi \|$ for all $k=0,1,2,\cdots$ . By (3.4), we have

(3.5) \begin{align} \|P(t)\psi \|\le e^{-Dt}\sum ^\infty _{k=0}\frac{(Dt)^k}{k!}\|a_k(\psi )\|\le \|\psi \|. \end{align}

It is clear that the system (2.3) can be rewritten into the following integral system

(3.6) \begin{align} \begin{cases} \displaystyle U(t,x)= e^{-d_mt}\phi _1(0,x) +\int _0^te^{-d_m(t-s)}\mathcal{H}_1[U,V](s,x)ds, \\[6pt] \displaystyle V(t,x)= P(t)\phi _2(0,x) +\int _0^t P(t-s)\mathcal{H}_2[U,V](s,x)ds, \end{cases} \end{align}

where

(3.7) \begin{align} \begin{cases} \displaystyle \mathcal{H}_1[U,V](t,x) \,:\!=\alpha e^{-d_i\tau } \int _{\mathbb{R}}k(y)\frac{U(t{-}\tau,x-y)}{U(t{-}\tau,x-y)+\omega }(V(t{-}\tau,x-y)+1)dy,\\[6pt] \displaystyle \mathcal{H}_2[U,V](t,x)\,:\!=-V(1+V)+d\int _{\mathbb{R}}k(y)\frac{U(t,x-y)}{U(t,x-y)+\omega }(V(t,x-y)+1)dy. \end{cases} \end{align}

For $\phi ^1=(\phi _1^1,\phi ^1_2)$ , $\phi ^2=(\phi _1^2,\phi _2^2)$ $\in \mathcal{C}_{\mathbf{K}}$ , we define

\begin{align*} w(t,x)=(w^1(t,x),w^2(t,x)), \end{align*}

where

\begin{align*} w^1(t,x)=|U(t,x;\,\phi ^1)-U(t,x;\,\phi ^2)|, \ \ \ w^2(t,x)=|V(t,x;\,\phi ^1)-V(t,x;\,\phi ^2)|. \end{align*}

Choose $t_0\gt 0$ and for any $\varepsilon \gt 0$ , we let

\begin{equation*}\sigma =\frac {(\alpha e ^{-d_i\tau }+d)(v_++\omega )}{\omega }+(2v_+-1)\ \ \mbox {and} \ \ \varepsilon _1=\frac {\varepsilon }{2\sigma t_0e^{\sigma t_0}}.\end{equation*}

It is easy to see that there exists $(t^*,x^*)\in [{-}\tau,t]\times \mathbb{R}$ such that

\begin{equation*}w_s(\theta,x)\leq \sup _{s\in [{-}\tau,t],x\in \mathbb {R}}(w^1(s,x),w^2(s,x))\leq w(t^*,x^*)+\left(\frac {\varepsilon }{8}e^{-\sigma t_{0}},\frac {\varepsilon }{8}e^{-\sigma t_{0}}\right)\end{equation*}

for $(s,\theta,x)\in [0,t]\times [{-}\tau,0]\times \mathbb{R}$ with $t\in [0,t_0]$ . Set

\begin{align*} \|\phi \|_{\Sigma _{M}(z)}=\sup _{(\theta,x)\in \Sigma _{M}(z)}|\phi _1(\theta,x)|+\sup _{(\theta,x)\in \Sigma _{M}(z)}|\phi _2(\theta,x)| \ \ \mbox{for} \ \phi =(\phi _1,\phi _2), \end{align*}

with

\begin{align*} \Sigma _M(z)=[{-}\tau,0]\times [z-M,z+M], \ \ M\gt 0, \ z\in \mathbb{R}. \end{align*}

Then, there exists $M=M(t_0,\varepsilon )$ such that

\begin{equation*}\int _{\mathbb {R}}k(y)\left [\frac {v_+}{\omega }w^1(s,x^*-y)+w^2(s,x^*-y)\right ]dy \leq \frac {v_++\omega }{\omega }\|w_s\|_{\Sigma _M(x^*)}+\varepsilon _1\end{equation*}

for $0\leq s\leq t$ . Hence, for above $\varepsilon \gt 0$ , choose $\delta =\displaystyle\frac{\varepsilon }{8}e^{-\sigma t_{0}}$ such that when $\|\phi ^1-\phi ^2\|_{\Sigma _M(x^*)}\lt \delta$ , by (3.5) and (3.6), we obtain

\begin{align*} & \|w_t(\theta,x)\|_{\Sigma _M(x^*)} \\ &\quad\leq w^1(t^*,x^*)+w^2(t^*,x^*)+\frac{\varepsilon }{4}e^{-\sigma t_{0}} \\ &\quad\leq e^{-d_mt}w^1(0,x^*)+w^2(0,x^*)+\frac{\varepsilon }{4}e^{-\sigma t_{0}} \\ &\qquad +\alpha e^{-d_i\tau }\int _0^{t^*}\int _{\mathbb{R}}e^{-d_m(t^*-s)}k(y)\left [\frac{v_+}{\omega }w^1(s{-}\tau,x^*-y) +w^2(s{-}\tau,x^*-y)\right ] dyds \\ &\qquad +\int _0^{t^*}(2v_+-1)w^2(s,x^*)ds+d\int _0^{t^*}\int _{\mathbb{R}}k(y)\left [\frac{v_+}{\omega }w^1(s,x^*-y) +w^2(s,x^*-y)\right ] dyds \\ &\quad\leq 2\|\phi ^1-\phi ^2\|_{\Sigma _M(x^*)}+\frac{\varepsilon }{4}e^{-\sigma t_{0}}+(2v_+-1)\int _0^{t}(\|w_s\|_{\Sigma _M(x^*)}+\varepsilon _1)ds \\ & \qquad +\frac{(\alpha e ^{-d_i\tau }+d)(v_++\omega )}{\omega }\int _0^{t}(\|w_s\|_{\Sigma _M(x^*)}+\varepsilon _1)ds \\ &\quad\leq 2\delta +\frac{\varepsilon }{4}e^{-\sigma t_{0}}+\varepsilon _1\sigma t+\sigma \int _0^{t}\|w_s\|_{\Sigma _M(x^*)}ds. \end{align*}

By Gronwall’s inequality, we further have

\begin{equation*}\|w_t(\theta,x)\|_{\Sigma _M(x^*)}\leq \left(\frac {\varepsilon }{2}e^{-\sigma t_{0}}+\varepsilon _1\sigma t\right)e^{\sigma t} \leq \left(\frac {\varepsilon }{2}e^{-\sigma t_{0}}+\varepsilon _1\sigma t_0\right)e^{\sigma t_0}=\varepsilon,\quad t\in [0,t_0].\end{equation*}

This shows that $Q_t$ is continuous in $\phi$ with respect to compact open topology uniformly for $t\in [0,t_0]$ , which, together with the continuity of $Q_t$ in $t$ from Lemma 2.1, implies that $Q_t$ is continuous in $(t,\phi )$ with respect to the compact open topology. The proof is complete.

Proof. In view of Lemma 2.1, when $\phi \in \mathcal{C}_{\mathbf{K}}$ , the solution $(U(t,x;\,\phi ),V(t,x;\,\phi ))$ of (2.3) is bounded. More precisely, $\mathbf{0}\leq (U(t,x;\,\phi ),V(t,x;\,\phi )) \leq \mathbf{K}, \ \forall \ t\geq 0, \ x\in \mathbb{R}$ . It then follows from the first equation of (2.3) that

\begin{align*} |U_t(t,x;\,\phi )|\le & d_m|U|+\alpha e^{-d_i\tau }\int _{\mathbb{R}}k(x-y)\left |\frac{U(t{-}\tau,y;\,\phi )}{U(t{-}\tau,y;\,\phi )+\omega }\right ||V(t{-}\tau, y;\,\phi )+1|dy\\ \le & d_m \tilde u_++\alpha e^{-d_i\tau }\frac{\tilde u_+}{\omega }(\tilde v_++1)=\!:\,L, \end{align*}

which means that $U_t$ is bounded for $t\geq 0$ . Let $[a,b]\subseteq \mathbb{R}$ with $a\gt 0$ be any bounded interval, $I\subseteq \mathbb{R}$ be a compact interval and $\tilde K=\min \{K\in \mathbb{N}\,{:}\,I\subseteq [{-}K,K]\}$ . Then for any $t_1, t_2\in [a, b]$ and $x\in I$ , one has

\begin{align*} |U(t_1,x;\,\phi )-U(t_2,x;\,\phi )|\le L|t_1-t_2|. \end{align*}

Hence, for any $\varepsilon \gt 0$ , there exists $\delta =\frac{\varepsilon }{L}$ , such that for any $\phi \in \mathcal{C}_{\mathbf{K}}$ , any $x\in I$ , $s_1, s_2\in [{-}\tau,0]$ with $|s_1-s_2|\lt \delta$ , we obtain

\begin{align*} |Q_{t_0}^1[\phi ](s_1,x)-Q_{t_0}^1[\phi ](s_2,x)|\le |U(t_0+s_1,x;\,\phi )-U(t_0+s_2,x;\,\phi )|\lt \varepsilon, \end{align*}

where $t_0\gt \tau$ , which implies that $\{Q_t^1[\phi ](s,x)\,{:}\,\phi \in \mathcal{C}_{\mathbf{K}}, x\in \mathbb{R}\}$ is a family of equicontinuous functions of $s\in [{-}\tau,0]$ . By the Arzela-Ascoli Theorem, we obtain that $\{Q_t^1[\phi ](\cdot,x)\,{:}\,\phi \in \mathcal{C}_{\mathbf{K}}, x\in \mathbb{R}\}$ is precompact in $C([{-}\tau,0],\mathbb{R})$ if $t\geq \tau$ . Thus, $Q_t^1$ satisfies (A3)(a) for $t\geq \tau$ . On the other hand, if $t\lt \tau$ , we set $\varsigma\,{=}\,1$ . Then, for the $T$ -invariant set $\Pi$ defined in (A3), the set $\{S^1[\Pi ](\theta,0)\,{:}\,\theta \in [{-}\varsigma,0]\}$ is precompact in $C([{-}\varsigma,0],\mathbb{R})$ , where $S^1$ is the first component of the operator $S$ defined in (3.1). It is clear that $\{S^1[\Pi ](\theta,0)\,{:}\,\theta \in [{-}\tau,{-}\varsigma ]\}$ is an infinite set of constant functions in $C([{-}\tau,-\varsigma ],\mathbb{R})$ , and hence, it is precompact in $C([{-}\tau,-\varsigma ],\mathbb{R})$ . Therefore, $Q_t^1$ satisfies (A3)(b) for $t\lt \tau$ .

Now we prove that $Q^2_t$ satisfies (A3). By the second equation of (2.3), we have

\begin{align*} |V_t(t,x;\,\phi )|\le 2D\tilde v_++\tilde v_+(\tilde v_++1)+d\frac{\tilde u_+}{\tilde u_++\omega }(\tilde v_++1). \end{align*}

By a similar argument as that for $Q^1_t$ , we obtain that $\{Q_t^2[\phi ](\cdot,x)\,{:}\,\phi \in \mathcal{C}_{\mathbf{K}}, x\in \mathbb{R}\}$ is precompact in $C([{-}\tau,0],\mathbb{R})$ if $t\geq \tau$ . Thus, $Q_t^2$ satisfies (A3)(a) for $t\geq \tau$ . In the following, we verify that $Q_t^2$ satisfies (A3)(b) when $t\in [0,\tau ]$ . For any $\phi \in \mathcal{C}_{\mathbf{K}}$ , we fix $\tilde t\in (0,\tau ]$ and define

\begin{align*} S^2[\phi ](\theta,x)= \begin{cases} \phi _2(0,x), \ \ \ \ \ \ \ {-}\tau \leq \theta \leq -\tilde t, \\ Q^2_{\tilde t}[\phi ](\theta,x), \ \ {-}\tilde t\leq \theta \leq 0. \end{cases} \end{align*}

Let $\Pi \subset \mathcal{C}_{\mathbf{K}}$ be a $T$ -invariant set with $\Pi (0,\cdot )\,:\!=\{u(0,x)\,{:}\,u\in \Pi \}$ precompact in $X$ . We just need to show that for any given compact interval $I\in \mathbb{R}$ , $S^2(\Pi )$ is equicontinuous on $[{-}\tau,0]\times I$ .

When $(s,x)\in [{-}\tau,{-}\tilde t]\times I$ , one has $S^2[\phi ](s,x)=\phi _2(0,x)$ for all $\phi \in \Pi$ . Hence, by the precompactness of $\Pi (0,\cdot )$ in $X$ , we obtain that $S^2(\Pi )$ is equicontinuous on $[{-}\tau,{-}\tilde t]\times I$ .

Since $P(t)$ is uniformly continuous for $t$ in a bounded interval in the compact open topology with respect to the initial value, one can show that

\begin{equation*}\{P(t-s)\mathcal {H}_2[\phi ](s,x)\,{:}\, t\in [0,\tilde t],\ s \in [0,t],\ \phi \in \mathcal {C}_{\mathbf {K}} \}\end{equation*}

is bounded in $X_+$ , where $\mathcal{H}_2$ is defined in (3.7). Then there exists $M\gt 0$ such that

\begin{equation*}\|P(t-s)\mathcal {H}_2[\phi ](s,\cdot )\|_X\le M \ \mbox {for}\ t\in [0,\tilde t],\ s \in [0,t],\ \phi \in \mathcal {C}_{\mathbf {K}},\end{equation*}

where the norm

\begin{equation*}\|\phi \|_X=\sum _{k=1}^\infty \frac {\max _{|x|\le k}|\phi |}{2^k},\quad \forall \phi \in X,\end{equation*}

with $|\cdot |$ is the usual norm in $\mathbb{R}$ . Thus, we derive that $\sup _{|x|\le \tilde K}|P(t-s)\mathcal{H}_2[\phi ](x)|\le 2^{\tilde K}M$ . Hence, for any $\varepsilon \gt 0$ , there exists $\delta _1=\min \{\frac{\varepsilon }{2^{\tilde K}4M},\tilde t\}$ , such that for any $t\le \delta _1$ , $x\in I$ and $\phi \in \Pi$ , we have

(3.8) \begin{align} \left |\int _0^tP(t-s)\mathcal{H}_2[\phi ](s,x)ds\right |\le 2^{\tilde K}M\delta _1\lt \frac{\varepsilon }{4}. \end{align}

In [Reference Ignat and Rossi14, Section 2], Ignat and Rossi showed that the solution of (3.3) can also be written as $V(t,x)=[P(t)\psi ](x)=\int _{\mathbb{R}}G(t,y)\psi (x-y)dy$ , where $G(t,x)=e^{-Dt}\delta _0(x)+R(t,x)$ , $\delta _0(x)$ is the delta measure at zero and $R(t,x)=\frac{1}{2\pi }\int _{\mathbb{R}}(e^{D(\hat J(\xi )-1)t}-e^{-Dt})e^{jx\xi }d\xi$ with $j=\sqrt{-1}$ and $\hat J$ being the Fourier transform of $J$ . Moreover, it is proved that $|G(t,\cdot )|_{L^1(\mathbb{R})}\le 3$ for any $t\gt 0$ . Since $(\phi _1,\phi _2)\in \Pi (0, \cdot )$ and $\Pi (0,\cdot )$ is precompact in $X$ , then for the above $I$ , there exists $\delta _2\gt 0$ such that for any $x_1, x_2\in I$ satisfying $|x_1 {-}x_2|\lt \delta _2$ , we have $|\phi _2(x_1)-\phi _2(x_2)]\lt \frac{\varepsilon }{12}$ , and hence,

(3.9) \begin{align} \left |P(t)[\phi _2](x_1)-P(t)[\phi _2](x_2)\right |=&\left |\int _{\mathbb{R}}G(t,y)[\phi _2(x_1-y)-\phi _2(x_2-y)]dy\right |\notag \\ \le &\int _{\mathbb{R}}|G(t,y)||\phi _2(x_1-y)-\phi _2(x_2-y)|dy\notag \\ \le &\|G(t,\cdot )\|_{L^1(\mathbb{R})}\frac{\varepsilon }{12}\le \frac{\varepsilon }{4},\ \forall t\in [0,\delta _1]. \end{align}

On the other hand, for all $t \gt 0$ , $x \in \mathbb{R}$ and $\phi \in \mathcal{C}_{\mathbf{K}}$ , we have $|[P(t)\phi _2](x)|\le \tilde v_+$ . It is easy to see that

\begin{align*} \left |\frac{\partial P(t)[\phi _2](x)}{\partial t}\right |=\left |D\left (\int _{\mathbb{R}}J(x-y)P(t)[\phi _2](y)dy-P(t)[\phi _2](x)\right )\right |\le 2D\tilde v_+. \end{align*}

Hence, for $t_1, t_2\in [0,\delta _1]$ , $(\phi _1, \phi _2)\in \Pi (0, \cdot )$ , there exists $\delta _3\,:\!=\displaystyle\frac{\varepsilon }{8D\tilde v_+}$ such that when $|t_1-t_2|\le \delta _3$ , we derive

(3.10) \begin{align} |P(t_1)[\phi _2](x)-P(t_2)[\phi _2](x)|=2D\tilde v_+|t_1-t_2|\lt \frac{\varepsilon }{4}. \end{align}

Combining (3.8)–(3.10), when $s_1, s_2\in [{-}\tilde t, \delta _1{-}\tilde t]$ and $x_1, x_2\in I$ satisfying $|s_1-s_2|\lt \delta _3$ and $|x_1-x_2|\lt \delta _2$ , for any $\phi \in \Pi (0, \cdot )$ , we obtain

\begin{align*} &|S^2(\phi )(s_1,x_2)-S^2(\phi )(s_2,x_2)|=|Q_{\tilde t}^2[\phi ](s_1,x_1)-Q_{\tilde t}^2[\phi ](s_2,x_2)|\\ &\quad=|V(\tilde t+s_1,x_1;\,\phi )-V(\tilde t+s_2,x_2;\,\phi )|\\ &\quad\le |P(\tilde t+s_1)[\phi _2](x_1)-P(\tilde t+s_2)[\phi _2](x_2)|\\ &\qquad +\left |\int _0^{\tilde t+s_1} P(\tilde t+s_1-s)\mathcal{H}_2[U,V](s,x_1)ds-\int _0^{\tilde t+s_2} P(\tilde t+s_2-s)\mathcal{H}_2[U,V](s,x_2)ds\right |\\ &\quad\le |P(\tilde t+s_1)[\phi _2](x_1)-P(\tilde t+s_1)[\phi _2](x_2)|+|P(\tilde t+s_1)[\phi _2](x_2)-P(\tilde t+s_2)[\phi _2](x_2)|\\ &\qquad +\left |\int _0^{\tilde t+s_1} P(\tilde t+s_1-s)\mathcal{H}_2[U,V](s,x_1)ds\right |+\left |\int _0^{\tilde t+s_2} P(\tilde t+s_2-s)\mathcal{H}_2[U,V](s,x_2)ds\right |\\ &\quad\lt \varepsilon, \end{align*}

which means $S^2(\Pi )$ is equicontinuous on $[{-}\tilde t,\delta _1{-}\tilde t]\times I$ .

Finally, we need to verify that $S^2(\Pi )$ is equicontinuous on $[\delta _1 {-}\tilde t, 0]\times I$ . Note that if $s\in [\delta _1 {-}\tilde t, 0]$ , then $\tilde t+ s \in [\delta _1, \tilde t]$ . Thus, we can prove the current case similar to that for (A3)(a). Therefore, $S^2(\Pi )$ is equicontinuous on $[{-}\tau,0]\times I$ . The proof is complete.

Proof. Clearly, the solution $(U(t,x),V(t,x))$ of (2.1) is a lower solution of (3.11), and hence, $Q_1(\phi )\leq N_1(\phi )$ for any $\phi \in \mathcal{C}_{\mathbf{K}}$ . It is easy to verify that $N_1$ and $\mathcal{B}_\mu ^1$ satisfies (C1)–(C6) in [Reference Liang and Zhao20]. By [Reference Liang and Zhao20, Theorem 3.10], it suffices to show that the principal eigenvalue $\gamma (0)$ is greater than $1$ , and the infimum of $\Phi (\mu )$ is attained at some $\mu ^*\gt 0$ .

When $\mu =0$ , it follows from (3.13) that

\begin{align*} \mathcal{P}(\lambda ) & =\left (\lambda +d_m-\frac{\alpha e^{-d_i\tau }}{\omega }e^{-\lambda \tau }\right )\left (\lambda +1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\right ) -\frac{d\alpha e^{-d_i\tau }\tilde{u}_+}{\omega (\tilde{u}_++\omega )}e^{-\lambda \tau } \\ & =\lambda ^2+\left (d_m+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\right )\lambda +\left (1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\right )d_m -\frac{\alpha e^{-d_i\tau }}{\omega }(\lambda +1)e^{-\lambda \tau }=0. \end{align*}

Let

\begin{align*} & f_1(\lambda )=\lambda ^2+\left (d_m+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\right )\lambda +\left (1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\right )d_m, \\ & f_2(\lambda,\tau )=\frac{\alpha e^{-d_i\tau }}{\omega }(\lambda +1)e^{-\lambda \tau }. \end{align*}

Since $d_m\lt \tilde{d}_m\,:\!=\frac{\alpha e^{-d_i\tau }}{\omega }$ by (H2), we obtain

\begin{equation*}f_1(0)=\left(1-\frac {d\tilde {u}_+}{\tilde {u}_++\omega }\right)d_m\lt f_2(0,\tau )=\frac {\alpha e^{-d_i\tau }}{\omega }.\end{equation*}

It is easy to see that $\frac{\partial f _2(\lambda, \tau )}{\partial \lambda }=\frac{\alpha e^{-d_i\tau }}{\omega }e^{-\lambda \tau }(1{-}\tau (\lambda +1))$ . Hence, if $\tau \geq 1$ , then $\frac{\partial f _2(\lambda, \tau )}{\partial \lambda }\leq 0$ for $\lambda \geq 0$ . If $\tau \lt 1$ , then $f_2(\lambda,\tau )$ reaches its unique local (thus global) maximum at $\lambda =\frac{1}{\tau }-1$ and tends to $0$ as $\lambda \rightarrow {+}\infty$ . Moreover, $f_1(\lambda )$ is convex for $\lambda \gt 0$ , while for any fixed $\tau \gt 0$ , $f_2(\lambda,\tau )$ has at most one reflection point for $\lambda \gt 0$ . Hence, there is a unique $\lambda ^*\gt 0$ such that $f_1(\lambda ^*)=f_2(\lambda ^*,\tau )$ no matter what value $\tau$ takes. This implies that $\lambda (0)=\lambda ^*\gt 0$ , and hence, $\gamma (0)=e^{\lambda (0)}\gt 1$ , i.e. the condition (C7) in [Reference Liang and Zhao20] is satisfied.

We now prove that $\Phi (\mu )$ attains its infimum at some $\mu ^*\gt 0$ , which can be obtained by proving that $\lim \limits _{\mu \rightarrow {+}\infty }\Phi (\mu )={+}\infty$ . By (3.13), we have

(3.14) \begin{align} \mathcal{P}(\lambda ) = &\lambda ^2+\left (d_m-D\tilde{J}(\mu )+D+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\tilde{k}(\mu )\right )\lambda \notag \\ & +\left (-D\tilde{J}(\mu )+D+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\tilde{k}(\mu )\right )d_m \notag \\ & -\frac{\alpha e^{-d_i\tau }}{\omega }\tilde{k}(\mu )(\lambda -D\tilde{J}(\mu )+D+1)e^{-\lambda \tau } = 0. \end{align}

Let

\begin{align*} f_3(\lambda )&= \lambda ^2+\left (d_m-D\tilde{J}(\mu )+D+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\tilde{k}(\mu )\right )\lambda \\[2pt] &\quad +\left (-D\tilde{J}(\mu )+D+1-\frac{d\tilde{u}_+}{\tilde{u}_++\omega }\tilde{k}(\mu )\right )d_m, \\[2pt] f_4(\lambda,\tau )&=\frac{\alpha e^{-d_i\tau }}{\omega }\tilde{k}(\mu )(\lambda -D\tilde{J}(\mu )+D+1)e^{-\lambda \tau }. \end{align*}

It is easy to compute that $\tilde J'(\mu )\gt 0$ for $\mu \gt 0$ and $\lim _{\mu \rightarrow {+}\infty }\tilde J(\mu )={+}\infty$ . Hence, for any large $\mu$ , we have

\begin{align*} &f_3\left(D\tilde{J}(\mu )-D-1+\frac{1}{\tau }\right)\lt 0,\\ &f^{\prime}_3(\lambda )\gt 0,\quad \forall \lambda \gt D\tilde{J}(\mu )-D-1+\frac{1}{\tau },\\ &\lim \limits _{\lambda \rightarrow {+}\infty }f_3(\lambda )={+}\infty, \end{align*}

and

\begin{align*} &f_4\left(D\tilde{J}(\mu )-D-1+\frac{1}{\tau },\tau \right)\gt 0,\\ &\frac{\partial f_4(\lambda,\tau )}{\partial \lambda }\lt 0,\quad \forall \lambda \gt D\tilde{J}(\mu )-D-1+\frac{1}{\tau },\\ &\lim \limits _{\lambda \rightarrow {+}\infty }f_4(\lambda,\tau )=0. \end{align*}

Thus, (3.14) has a unique positive root $\lambda (\mu )\gt D\tilde{J}(\mu )-D-1+\frac{1}{\tau }$ . Hence,

\begin{equation*}\lim \limits _{\mu \rightarrow {+}\infty }\Phi (\mu )=\lim \limits _{\mu \rightarrow {+}\infty }\frac {\lambda (\mu )}{\mu } \ge \lim \limits _{\mu \rightarrow {+}\infty }\frac {D\tilde {J}(\mu )-D-1+\frac {1}{\tau }}{\mu }={+}\infty .\end{equation*}

The proof is complete.

Proof. Choose any small $\varepsilon \gt 0$ . Let $P^\varepsilon _t$ be the solution operator of the following linear system:

(3.15) \begin{align} \begin{cases} \displaystyle \frac{\partial U}{\partial t}=-d_m U+\alpha e^{-d_i\tau }\int _{\mathbb{R}}k(y)\frac{1}{\omega +\varepsilon }U(t{-}\tau,x-y)dy, \\[8pt] \displaystyle \frac{\partial V}{\partial t}=D\int _{\mathbb{R}}J(y)[V(t,x-y)-V(x,t)]dy -(1+\varepsilon )V \\[8pt] \displaystyle \, \, \, \, \, \, \, \ +d\int _{\mathbb{R}}k(y)\frac{1}{\omega +\varepsilon }U(t,x-y)dy, \end{cases} \end{align}

where $t\gt 0$ , $x\in \mathbb{R}$ . By a similar argument as that in the proof of Proposition 3.6, we can obtain that $P^\varepsilon _t$ satisfies (C1)–(C7) in [Reference Liang and Zhao20]. Moreover, for any given $\varepsilon \in (0,1)$ , there exists $\delta =(\delta _1,\delta _2)$ such that the solution $(U,V)$ of (3.15) satisfying

\begin{equation*}0\lt U(t,x;\,\phi )\lt \varepsilon, \ \ 0\lt V(t,x;\,\phi )\lt \varepsilon, \ \ t\in [0,1],\end{equation*}

for any initial $\phi =(\phi _1,\phi _2)$ with $0\leq \phi _1\leq \delta _1$ , $0\leq \phi _2\leq \delta _2$ . Hence, $(U(t,x;\,\phi ),V(t,x;\,\phi ))$ satisfies

\begin{align*} \frac{\partial U}{\partial t} & =-d_{m}U+\alpha e^{-d_i\tau }\int _{\Omega }k(y)\frac{U(t{-}\tau, x-y)}{U(t{-}\tau, x-y)+\omega }(V(t{-}\tau, x-y)+1)dy\\ & \geq -d_m U+\alpha e^{-d_i\tau }\int _{\mathbb{R}}k(y)\frac{1}{\omega +\varepsilon }U(t{-}\tau,x-y)dy, \ \ \ t\in [0,1], \end{align*}

and

\begin{align*} \frac{\partial V}{\partial t} & =D\int _{\mathbb{R}}J(y)[V(t, x-y)-V(t, x)]dy-V(1+V) \\ &\ \ \ +d\int _{\Omega }k(y)\frac{U(t, x-y)}{U(t, x-y)+\omega }(V(t, x-y)+1)dy \\ &\geq D\int _{\mathbb{R}}J(y)[V(t,x-y)-V(t,x)]dy -(1+\varepsilon )V \\ &\ \ \ +d\int _{\mathbb{R}}k(y)\frac{1}{\omega +\varepsilon }U(t,x-y)dy, \ \ \ t\in [0,1]. \end{align*}

By the comparison principle, we obtain that $P^\varepsilon _t[\phi ]\leq Q_t[\phi ]$ for $t\in [0,1]$ . In particular, $P^\varepsilon _1[\phi ]\leq Q_1[\phi ]$ for $0\leq \phi _1\leq \delta _1$ and $0\leq \phi _2\leq \delta _2$ . It then follows from [Reference Liang and Zhao20, Theorem 3.10] that the spreading speed of $P^\varepsilon _t$ can be attained by the infimum of $\Psi ^\varepsilon (\mu )\,:\!=\frac{\Lambda ^\varepsilon (\mu )}{\mu }$ , where $\Lambda ^\varepsilon (\mu )$ is the principle eigenvalue of

\begin{align*} \left (\Lambda +d_m-\frac{\alpha e^{-d_i\tau }\tilde{k}(\mu )}{\omega +\varepsilon }e^{-\Lambda \tau }\right )\left (\Lambda -D\tilde{J}(\mu )+D+1+\varepsilon \right )=0, \end{align*}

which is the characteristic equation for the equation of $\eta$ corresponding to (3.15). It is easy to verify that the statements in Lemma 3.5 also hold for $\Psi ^\varepsilon (\mu )$ . Then we obtain

\begin{equation*}c^*\geq \inf \limits _{\mu \gt 0}\Psi ^\varepsilon (\mu ).\end{equation*}

Since $\varepsilon \gt 0$ can be chosen arbitrarily, one further has

\begin{equation*}c^*\geq \inf \limits _{\mu \gt 0}\Psi (\mu )=\inf \limits _{\mu \gt 0}\frac {\Lambda (\mu )}{\mu },\end{equation*}

where $\Lambda (\mu )$ is the principal eigenvalue of

\begin{align*} \left (\Lambda +d_m-\frac{\alpha e^{-d_i\tau }\tilde{k}(\mu )}{\omega }e^{-\Lambda \tau }\right )\left (\Lambda -D\tilde{J}(\mu )+D+1\right )=0. \end{align*}

The proof is complete.

Remark 3.8. For a fixed $\mu \gt 0$ , we can compute that $\frac{\partial \mathcal{L}(\Lambda,\mu )}{\partial \Lambda }=1+\tau \frac{\alpha e^{-d_i\tau }\tilde{k}(\mu )}{\omega }e^{-\Lambda \tau }\gt 0$ and $\mathcal{L}(0,\mu )=d_m-\frac{\alpha e^{-d_i\tau }\tilde{k}(\mu )}{\omega }$ . Since $d_m-\frac{\alpha e^{-d_i\tau }}{\omega }\lt 0$ by (H2) and $\tilde{k}(\mu )\gt 1$ for $\mu \gt 0$ by (H1), we have that $\mathcal{L}(0,\mu )\lt 0$ . Hence, the existence and uniqueness of $\Lambda _2(\mu )$ can be easily obtained.