Proof. Let ’s denote

\begin{align*} u(\xi )\,:\!=\,e^{v_{1}\xi +\int _{0}^{\xi }\phi (y)dy}\ \mbox { with }v_{1}\,:\!=\,- {b}/{c} . \end{align*}

Then, $ u(\xi )$ satisfies the equation

(11) \begin{align} cu^{\prime}(\xi )=d\sum _{i\in \tilde{\Omega }}J(i)e^{iv_{1}} u(\xi -i)+b_{0}\int _{0}^{\tau }f(s)e^{csv_{1}} u(\xi -cs)ds\gt 0, \end{align}

which implies that $u(\xi )$ is strictly increasing on $\mathbb{R}$ . Thus, for any $p\in \tilde{\Omega }$ with $ p\gt 0$ , we have

(12) \begin{align} cu^{\prime}(\xi )\gt dJ(p)e^{-pv_{1}} u(\xi +p). \end{align}

Integrating the inequality (12) from $ \xi -\displaystyle \frac{p}{2}$ to $ \xi$ gives

\begin{align*} cu(\xi )-cu\!\left(\xi -\frac {p}{2}\right)\gt \frac {dp}{2} J(p) e^{-pv_{1}}u\!\left(\xi +\frac {p}{2}\right), \end{align*}

which implies

\begin{align*} u(\xi )\gt \frac {dp J(p) u\!\left(\xi +\frac {p}{2}\right)}{2ce^{pv_{1}}}\gt \frac {d^{2}p^{2}J^{2}(p)u(\xi +p)}{4c^{2}e^{2pv_{1}}}, \end{align*}

and hence

(13) \begin{align} \frac{u(\xi +p)}{u(\xi )}\lt \frac{4c^{2}e^{2pv_{1}}}{d^{2}p^{2}J^{2}(p)}. \end{align}

Since $u(\xi )$ is strictly increasing on $\mathbb{R}$ , it follows from (11) and (13) that ${u^{\prime}(\xi )}/{u(\xi )}\leq M_{0},$ for some $M_{0}\gt 0.$ Note that $\phi (\xi )={u^{\prime}(\xi )}/{u(\xi )}-v_{1}$ . Thus, $\phi (\xi )$ is uniformly bounded. Moreover, by (10), it is easy to see $ \phi ({\cdot}) \in C^{\infty } (\mathbb{R})$ . The proof is complete.

Proof. The proof of the case $b_{0}=0$ is similar to that of [Reference Guo and Lin12, Lemma 2.8], so we only consider the case $b_{0}\gt 0$ . Differentiating equation (10) gives

(14) \begin{align} c\phi ^{\prime}(\xi )= d\sum _{i\in \tilde{\Omega }}J(i) e^{\int ^{\xi -i}_{\xi }\phi (y)dy} (\phi (\xi -i)-\phi (\xi ) )+b_{0}\int _{0}^{\tau }f(s)e^{\int ^{\xi -cs}_{\xi }\phi (y)dy}(\phi (\xi -cs)-\phi (\xi ) )ds. \end{align}

Suppose that $ \phi (\xi )$ admits a global maxima at $\xi _*$ , then (14) gives

\begin{align*} 0=c\phi ^{\prime}(\xi _{\ast })= d\sum _{i\in \tilde {\Omega }}J(i) e^{\int ^{\xi _{\ast }-i}_{\xi _{\ast }}\phi (y)dy} (\phi (\xi _{\ast }-i)-\phi (\xi _{\ast }) )+b_{0}\int _{0}^{\tau }f(s)e^{\int ^{\xi _{\ast }-cs}_{\xi _{\ast }}\phi (y)dy}(\phi (\xi _{\ast }-cs)-\phi (\xi _{\ast }) )ds\leq 0, \end{align*}

which implies

\begin{align*} \phi (\xi _{\ast }-i)=\phi (\xi _{\ast }), \ \mbox {for } i\in \{-2,-1,1,2\}; \ \mbox {and}\ \phi (\xi _{\ast }-cs)=\phi (\xi _{\ast }), \ \mbox { for } s\in [0,\tau ]. \end{align*}

By induction arguments, we can conclude that $ \phi (\xi )=\phi (\xi _{\ast })$ for all $ \xi \in \mathbb{R}$ . Similarly, $\phi (\xi )$ is a constant function provided that it admits a global minima. The proof is complete.

Proof. (1) Let $\{y_{j}\}_{j=1}^\infty$ be a sequence such that

\begin{align*} \lim _{j\rightarrow +\infty } y_{j}=+\infty \ \mbox { and }\ \lim _{j\rightarrow +\infty } \phi (y_{j})= w^{\ast }\,:\!=\,\limsup _{\xi \rightarrow +\infty }\phi (\xi ). \end{align*}

Denote $\eta _j(\xi )\,:\!=\, \phi ( y_{j}+\xi )$ for $j\in \mathbb{N}$ . Clearly, $\{ \eta _j( \xi )\} _{j=1}^\infty$ is uniformly bounded and equicontinuous. Then it follows from the Arzela-Ascoli theorem that $\{ \eta _j( \xi )\} _{j=1}^\infty$ has a subsequence, still written as $\{ \eta _j( \xi )\} _{j=1}^\infty$ , such that $\eta _j(\xi )\rightarrow \eta (\xi )$ in $ C^{1}_{loc} (\mathbb{R})$ as $j\rightarrow +\infty$ . Hence, $\eta (\xi )$ is a solution of (10) with $\eta (0)=w^{\ast } =\max _{\xi \in \mathbb{R}} \eta (\xi )$ .

By Lemma 2.2, we have $ \eta (\xi )=w^{\ast }$ for all $ \xi \in \mathbb{R}$ , which implies

(15) \begin{align} \lim _{j\rightarrow +\infty }\max _{[y_{j}-2-c\tau, y_{j}+2+c\tau ]}| \phi (\xi )-w^{\ast }| =0. \end{align}

We first show that $\phi ({+}\infty )$ exists. Suppose $ w^{\ast } \gt w_{\ast }\,:\!=\,\liminf \limits _{\xi \rightarrow +\infty }\phi (\xi )$ , by (15), there exists a sufficiently large $ j$ such that

(16) \begin{align} \min _{\xi \in [y_{j}, y_{j+1}]}\phi (\xi )\lt \frac{w_{\ast }+w^{\ast }}{2}\ \mbox{ and }\ \phi (\xi )\gt \frac{w_{\ast }+w^{\ast }}{2},\ \mbox{for}\ \xi \in \!\left[y_{j},y_{j}+2+c\tau \right]\cup\! \left[y_{j+1}-2-c\tau, y_{j+1}\right]. \end{align}

Let $ \hat{y}$ be the point such that $ \phi (\hat{y})= \min \limits _{\xi \in [y_{j}, y_{j+1}]}\phi (\xi )$ . According to (16), we have $ \hat{y}\in (y_{j}+2+c\tau, y_{j+1}-2-c\tau )$ . Then, for $ s\in [0,\tau ]$ , it holds that $\phi ^{\prime}(\hat{y})=0$ ,

\begin{align*} \phi (\hat {y})\leq \min \!\left\{\phi (\hat {y}+1), \phi (\hat {y}+2)\right\},\ \phi (\hat {y})\lt \min\! \left\{\phi (\hat {y}-1), \phi (\hat {y}-2)\right\}\ \mbox { and }\ \phi (\hat {y})\leq \phi (\hat {y}-cs). \end{align*}

Taking $\xi = \hat{y}$ in (14) yields

\begin{align*} 0= d\sum _{i\in \tilde {\Omega }}J(i) e^{\int ^{\hat {y}-i}_{\hat {y}}\phi (y)dy} \!\left(\phi (\hat {y}-i)-\phi (\hat {y}) \right)+b_{0}\int _{0}^{\tau }f(s)e^{\int ^{\hat {y}-cs}_{\hat {y}}\phi (y)dy}\!\left(\phi (\hat {y}-cs)-\phi (\hat {y}) \right)ds\gt 0, \end{align*}

which leads to a contradiction. Hence $ w^{\ast }=w_{\ast }$ , i.e., the limit $\phi ({+}\infty )$ exists.

In the same way, we can obtain that $\phi ({-}\infty )$ exists. Letting $ \xi \rightarrow \pm \infty$ in (10), it follows that $ \phi (\pm \infty )$ satisfy the equation $ \Psi (d, b_{0},b,c, \phi (\pm \infty ))=0$ .

(2) Since $\phi (\xi )$ is a non-constant solution, according to Lemma 2.2, $\phi (\xi )$ cannot attain its global extrema. Thus, $ \phi ({-}\infty )\neq \phi ({+}\infty )$ and

\begin{align*} { \lambda }_{1}\,:\!=\,\min \!\left\{\phi ({-}\infty ), \phi ({+}\infty )\right\}\lt \phi (\xi )\lt \max\! \left\{\phi ({-}\infty ), \phi ({+}\infty )\right\}\,:\!=\,{ \lambda }_{2}.\end{align*}

We first claim that $\phi ({-}\infty )\lt \phi ({+}\infty )$ . If the claim is false, i.e., $\phi ({-}\infty )\gt \phi ({+}\infty )$ , one has $ \phi ({+}\infty )={ \lambda }_{1}$ . Thus, given any small $ \epsilon \gt 0$ , by translation if necessary, we may assume that

(17) \begin{align} \phi (\xi )\lt{ \lambda }_{1}+\epsilon, \ \mbox{ for } \xi \gt 0\ \mbox{ and }\,\,{ \lambda }_{1}+\epsilon \leq \phi (\xi ),\mbox{ for}\ \xi \leq 0. \end{align}

According to (10) and (17), we have

\begin{align*} c ({ \lambda }_{1}+\epsilon )=c\phi (0) =& \,d\sum _{i=-2}^{-1}J(i) e^{\int ^{-i}_{0}\phi (y)dy}+d\sum _{i=1}^{2}J(i) e^{\int ^{-i}_{0}\phi (y)dy}+b_{0}\int _{0}^{\tau }f(s)e^{\int ^{-cs}_{0}\phi (y)dy}ds+b\\[3pt] \leq &\, d\sum _{i=-2}^{-1}J(i) e^{-({ \lambda }_{1}+\epsilon )i}+d\sum _{i=1}^{2}J(i) e^{-({ \lambda }_{1}+\epsilon )i}+b_{0}\int _{0}^{\tau }f(s) e^{-({ \lambda }_{1}+\epsilon )cs}ds+b\\[3pt] =&\,d\sum _{i\in \tilde{\Omega }}J(i) e^{-({ \lambda }_{1}+\epsilon )i}+b_{0}\int _{0}^{\tau }f(s)e^{-({ \lambda }_{1}+\epsilon )cs}ds+b, \end{align*}

which implies

(18) \begin{align} \Psi (d, b_{0},b,c,{ \lambda }_{1}+\epsilon )=d\sum _{i\in \tilde{\Omega }}J(i) e^{-({ \lambda }_{1}+\epsilon )i}+b_{0}\int _{0}^{\tau }f(s)e^{-({ \lambda }_{1}+\epsilon )cs}ds+b-c ({ \lambda }_{1}+\epsilon )\geq 0. \end{align}

However, the inequality (18) contradicts to the fact that $ \Psi (d, b_{0},b,c,{ \lambda }_{1}+\epsilon )\lt 0$ for every small $ \epsilon \gt 0$ with $\lambda _1+\epsilon \lt \lambda _2$ . Therefore, ${ \lambda }_{1}=\phi ({-}\infty )\lt \phi ({+}\infty )={ \lambda }_{2}$ .

Next we show that $\displaystyle \int _{-\infty }^{0}(\phi (\xi )-{ \lambda }_{1})d\xi \lt +\infty$ . Combining the equations $ \Psi (d, b_{0},b,c,{ \lambda }_{1})=0$ and (10), we have

(19) \begin{align} c (\phi (\xi )-{ \lambda }_{1})= & d\sum _{i\in \tilde{\Omega }}J(i) (e^{\int ^{\xi -i}_{\xi }\phi (y)dy}-e^{-i{ \lambda }_{1}})+b_{0}\int _{0}^{\tau }f(s)(e^{\int ^{\xi -cs}_{\xi }\phi (y)dy}-e^{-cs{ \lambda }_{1}})ds\nonumber \\[3pt] =& d\sum _{i=-2}^{-1}J(i) e^{B_{1}(\xi, i)}\int ^{\xi -i}_{\xi }(\phi (y)-{ \lambda }_{1})dy+d\sum _{i=1}^{2}J(i) e^{B_{2}(\xi, i)}\int ^{\xi -i}_{\xi }(\phi (y)-{ \lambda }_{1})dy+\nonumber \\[3pt] &b_{0}\int _{0}^{\tau }f(s) e^{B_{3}(\xi, s)}\int ^{\xi -cs}_{\xi }(\phi (y)-{ \lambda }_{1})dyds, \end{align}

where $ B_{1}(\xi, i)\in [{-}i{ \lambda }_{1}, -i\max _{y\in [\xi, \xi -i]}{\phi (y)}] \mbox{ for } i\in \{-2,-1\}$ ;

\begin{align*} B_{2}(\xi, i)\in [{-}i\max _{y\in [\xi -i,\xi ]}{\phi (y)}, -i{ \lambda }_{1}]\ \mbox { for }\ i\in \{1,2\}; \ \mbox { and }\ B_{3}(\xi, s)\in [{-}cs\max _{y\in [\xi -cs,\xi ]}{\phi (y)}, -cs { \lambda }_{1}]. \end{align*}

Let’s denote $ R(\xi )\,:\!=\,\phi (\xi )-\lambda _{1}$ . Integrating (19) from $M$ to $ 0$ with $ M\lt -2-c\tau$ gives

(20) \begin{align} c \int _{M}^{0}R(\xi )d\xi = & d\sum _{i=-2}^{-1}J(i)\int _{M}^{0} \int ^{\xi -i}_{\xi }e^{B_{1}(\xi, i)}R(y)dyd\xi +d\sum _{i=1}^{2}J(i) \int _{M}^{0} \int ^{\xi -i}_{\xi }e^{B_{2}(\xi, i)}R(y)dyd\xi +\nonumber \\[3pt] &b_{0}\int _{0}^{\tau }f(s) \int _{M}^{0} \int ^{\xi -cs}_{\xi }e^{B_{3}(\xi, s)}R(y)dyd\xi ds. \end{align}

Changing the integration order in (20) yields

(21) \begin{align} 0=&-c \int _{M}^{0}R(\xi )d\xi +d\sum _{i=-2}^{-1}J(i) \int _{M}^{M-i}\int _{M}^{y} e^{B_{1}(\xi, i)}R(y)d\xi dy+d\sum _{i=-2}^{-1}J(i) \int _{M}^{0}\int _{y+i}^{y} e^{B_{1}(\xi, i)}R(y)d\xi dy\nonumber \\[3pt] &-d\sum _{i=-2}^{-1}J(i) \int _{M}^{M-i}\int _{y+i}^{y} e^{B_{1}(\xi, i)}R(y)d\xi dy+d\sum _{i=-2}^{-1}J(i) \int _{0}^{-i}\int _{y+i}^{0} e^{B_{1}(\xi, i)}R(y)d\xi dy\nonumber \\[3pt] &-d\sum _{i=1}^{2}J(i) \int _{M-i}^{M} \int ^{y+i}_{M}e^{B_{2}(\xi, i)}R(y)d\xi dy-d\sum _{i=1}^{2}J(i) \int _{M}^{0} \int ^{y+i}_{y}e^{B_{2}(\xi, i)}R(y)d\xi dy\nonumber \\[3pt] &+d\sum _{i=1}^{2}J(i) \int _{-i}^{0} \int ^{y+i}_{y}e^{B_{2}(\xi, i)}R(y)d\xi dy-d\sum _{i=1}^{2}J(i) \int _{-i}^{0} \int ^{0}_{y}e^{B_{2}(\xi, i)}R(y)d\xi dy\nonumber \\[3pt] &-b_{0}\int _{0}^{\tau }f(s) \int _{M-cs}^{M} \int ^{y+cs}_{M}e^{B_{3}(\xi, s)}R(y)d\xi dyds-b_{0}\int _{0}^{\tau }f(s) \int _{M}^{0} \int ^{y+cs}_{y}e^{B_{3}(\xi, s)}R(y)d\xi dyds\nonumber \\[3pt] &+b_{0}\int _{0}^{\tau }f(s) \int _{-cs}^{0} \int ^{y+cs}_{y}e^{B_{3}(\xi, s)}R(y)d\xi dyds-b_{0}\int _{0}^{\tau }f(s) \int _{-cs}^{0} \int ^{0}_{y}e^{B_{3}(\xi, s)}R(y)d\xi dyds. \end{align}

Let ${\varepsilon }\gt 0$ be small enough, by translation if necessary, we may assume that $\phi (\xi )\lt{ \lambda }_{1}+{\varepsilon }$ for $ \xi \lt \theta$ and ${ \lambda }_{1}+{\varepsilon }\leq \phi (\xi )$ for $ \xi \geq \theta$ , where $\theta =2+c\tau$ . Note that there exists some constant $K\gt 0$ such that $ |R(\xi )|\leq K$ for all $ \xi \in \mathbb{R}$ . Then we can obtain

(22) \begin{align} &d\sum _{i=-2}^{-1}J(i) \int _{M}^{M-i} \int ^{y}_{M} e^{B_{1}(\xi, i)}R(y)d\xi dy\nonumber \\[3pt] \leq & dK\sum _{i=-2}^{-1}J(i) \int _{M}^{M-i} \int ^{y}_{M} e^{-i ({ \lambda }_{1}+{\varepsilon })}d\xi dy =\frac{1}{2}dK\sum _{i=-2}^{-1}{J(i)i^{2}e^{-i ({ \lambda }_{1}+{ \varepsilon })}}\lt +\infty . \end{align}

It is also easy to verify that

\begin{align*} &d\sum _{i=-2}^{-1}J(i) \int _{0}^{-i}\int _{y+i}^{0} e^{B_{1}(\xi, i)}R(y)d\xi dy\lt +\infty, \quad d\sum _{i=1}^{2}J(i) \int _{-i}^{0} \int ^{y+i}_{y}e^{B_{2}(\xi, i)}R(y)d\xi dy\lt +\infty, \\[3pt] &\qquad \qquad \qquad b_{0}\int _{0}^{\tau }f(s) \int _{-cs}^{0} \int ^{y+cs}_{y}e^{B_{3}(\xi, s)}R(y)d\xi dyds\lt +\infty . \end{align*}

Then it follows from (21) that

(23) \begin{align} \int _{M}^{0}\Gamma (y) R(y)dy\leq \mathcal{O}(1), \end{align}

where $ \mathcal{O}(1)$ is uniformly bounded and

\begin{align*} \Gamma (y)\,:\!=\,c-d\sum _{i=-2}^{-1}J(i) \int _{y+i}^{y} e^{B_{1}(\xi, i)}d\xi +d\sum _{i=1}^{2}J(i) \int ^{y+i}_{y}e^{B_{2}(\xi, i)}d\xi +b_{0}\int _{0}^{\tau }f(s) \int ^{y+cs}_{y}e^{B_{3}(\xi, s)}d\xi ds. \end{align*}

Note that

\begin{align*} \Gamma (y)\geq &c+d\sum _{i=-2}^{-1}J(i) ie^{-i ({ \lambda }_{1}+{ \varepsilon })}+d\sum _{i=1}^{2}J(i) ie^{-i ({ \lambda }_{1}+{ \varepsilon })}+b_{0}\int _{0}^{\tau }f(s) cse^{-cs({ \lambda }_{1}+{ \epsilon })} ds\\[3pt] =&-\frac{\partial }{\partial \lambda } \Psi (d, b_{0},b,c,{ \lambda }_{1}+{ \varepsilon })\gt 0, \end{align*}

for small $\varepsilon \gt 0$ . Let $M\rightarrow -\infty$ in (23), we have

\begin{align*} -\frac{\partial }{\partial \lambda } \Psi (d, b_{0},b,c,{ \lambda }_{1}+{ \varepsilon })\int _{-\infty }^{0}R(y)dy\leq \mathcal{O}(1). \end{align*}

Thus, $\displaystyle \int _{-\infty }^{0}R(y)dy\lt +\infty$ , that is $\displaystyle \int _{-\infty }^{0}(\phi (\xi )-\phi ({-}\infty ))d\xi \lt +\infty$ . By the same way, we can obtain that $ \displaystyle \int _{0}^{+\infty }(\phi ({+}\infty )-\phi (\xi ))d\xi \lt +\infty$ . The proof is complete.

Proof. If $\phi (\xi )$ is constant solution of (10), by Lemma 2.3, we have $\phi (\xi )={ \lambda }_{1}$ or $\phi (\xi )={ \lambda }_{2}$ for $\xi \in \mathbb{R}$ , i.e., (24) holds with $l=1$ or $0$ respectively. Note that $\lambda _1$ may equal to $\lambda _2$ .

If $\phi (\xi )$ is a non-constant solution of (10), according to Lemma 2.3, we have ${ \lambda }_{1}=\phi ({-}\infty )\lt \phi (\xi )\lt \phi ({+}\infty ){ =\lambda }_{2}$ . Then we consider the functions

\begin{align*} w(\xi )\,:\!=\,e^{\int ^{\xi }_{0}\phi (z)dz}, w_{1}(\xi )\,:\!=\,{ l} e^{{ \lambda }_{1}\xi } \text{ and } w_{2}(\xi )\,:\!=\,w(\xi )-w_{1}(\xi ), \text{ with }{ l} \,:\!=\,e^{-\int ^{0}_{-\infty }(\phi (z)-{ \lambda }_{1})dz}\in (0,1). \end{align*}

It’s easy to verify that

(25) \begin{align} w^{\prime}(\xi )=\phi (\xi )w(\xi ),\ \ w_{2}(0)=1-l\ \mbox{ and }\ w_{2}(\xi )e^{-{ \lambda }_{1}\xi } =e^{\int ^{\xi }_{0}(\phi (z)-{ \lambda }_{1})dz}-e^{-\int ^{0}_{-\infty }(\phi (z)-{ \lambda }_{1})dz}. \end{align}

Note that $\displaystyle \int ^{\xi }_{0}(\phi (z)-{ \lambda }_{1})dz\gt -\int ^{0}_{-\infty }(\phi (z)-{ \lambda }_{1})dz$ for any $\xi \in \mathbb{R}$ . We have

(26) \begin{align} w_{2}(\xi )e^{-{ \lambda }_{1}\xi }\gt 0\ \mbox{ for }\xi \in \mathbb{R}\ \mbox{ and }\ w_{2}(\xi )e^{-{ \lambda }_{1}\xi }\rightarrow 0\ \mbox{ as}\ \xi \rightarrow -\infty . \end{align}

Based on (26), we further consider the function $\varphi (\xi )\,:\!=\,w^{\prime}_{2}(\xi )/w_{2}(\xi )$ . By simple computations, $\varphi (\xi )$ satisfies the equation (10). We claim that $\varphi (\xi )$ is a constant solution of (10). If false, that is $\varphi (\xi )$ is not a constant function, by Lemma 2.3, we have

\begin{align*} \displaystyle \int _{-\infty }^{0}(\varphi (\xi )-{ \lambda }_{1}))d\xi \lt \infty \ \mbox { and }\ \ln [w_{2}(\xi )e^{-{ \lambda }_{1}\xi }]-\ln [w_{2}(0)]=\int ^{\xi }_{0}\frac {(w_{2}(z)e^{-{ \lambda }_{1}z})^{\prime}}{(w_{2}(z)e^{-{ \lambda }_{1}z})}dz=\int ^{\xi }_{0}[\varphi (z)-{ \lambda }_{1}]dz. \end{align*}

As $\xi \rightarrow -\infty$ , it follows that

\begin{align*}\lim _{\xi \rightarrow -\infty }\ln [w_{2}(\xi )e^{-{ \lambda }_{1}\xi }]\lt \infty, \end{align*}

which contradicts to (26). Hence, by Lemma 2.3, $\varphi (\xi )$ is a constant function which equals to ${\lambda }_{1}$ or $\lambda _{2}$ for all $\xi \in \mathbb{R}$ . If $\varphi (\xi )={ \lambda }_{1}$ , we have $w(\xi )=ae^{{ \lambda }_{1}\xi }$ for some constant $a$ . Then $\phi (\xi )$ is constant function, which gives a contradiction. Therefore $\varphi (\xi )={ \lambda }_{2}$ . In view of the definition of $\varphi (\xi )$ , we have

\begin{align*} w_{2}(\xi )=w_{2}(0)e^{{ \lambda }_{2}\xi }=(1-{l})e^{{ \lambda }_{2}\xi }\ \mbox { and }\ w(\xi )={l}e^{{ \lambda }_{1}\xi }+(1-{l})e^{{ \lambda }_{2}\xi }. \end{align*}

Since $w^{\prime}(\xi )=\phi (\xi )w(\xi )$ , the solution form (24) holds obviously. According to (24), it is easy to verify that $\phi (\xi )$ is strictly increasing on $\mathbb{R}$ . The proof is complete.

Remark 1. From the proof of Lemma 2.4 , it can be easily seen that the conclusion of Lemma 2.3 (1) also holds for more general form of (10) by replacing the constant $b$ as any continuous function $b(\xi )$ whose limits $b_{\pm }\,:\!=\,b(\pm \infty )$ exist. That is, if $\phi (\xi )$ is a solution of (10) with $b$ replaced by such continuous function $b(\xi )$ , then $\phi (\pm \infty )$ exist and $ \Psi (d, b_{0},b_{\pm },c, \phi (\pm \infty ))=0$ .

Proof. The result can be proved in the following two cases.

Case 1: $\varphi (\xi _{0})=0$ for some $\xi _{0}\in \mathbb{R}$ . In this case, it can be easily deduced from (8) that $\varphi (\xi _{0}\pm 1)=0$ . An induction argument shows that $\varphi (\xi _{0}\pm k)=0$ for any $ k\in \mathbb{N_{+}}$ . On the other hand, by (8), one can see that

\begin{align*} c \varphi ^{\prime}(\xi )+|b| \varphi (\xi )\geq 0, \mbox{ for } \xi \in \mathbb{R}, \end{align*}

which implies that $ \varphi (\xi )e^{\frac{|b|}{c}\xi }$ is non-decreasing on $\mathbb{R}$ . Thus, it follows that $\varphi (\xi )=0$ on $\mathbb{R}$ .

Case 2: $\varphi (\xi )\gt 0$ for all $\xi \in \mathbb{R}$ . Dividing the equation (8) by $\varphi (\xi )$ , we have

(28) \begin{align} cq(\xi )= d\sum _{i\in \tilde{\Omega }}J(i) e^{\int ^{\xi -i}_{\xi }q(y)dy}+b_{0}\int _{0}^{\tau }f(s)e^{\int ^{\xi -cs}_{\xi }q(y)dy}ds+b, \mbox{ for } \xi \in \mathbb{R}. \end{align}

where $q(\xi )\,:\!=\,{\varphi ^{\prime}(\xi )}/{\varphi (\xi )}$ . By Lemma 2.4, $q(\xi )$ admits the form

\begin{align*}q(\xi )=\frac {\varphi ^{\prime}(\xi )}{\varphi (\xi )}=\frac {l\lambda _1e^{\lambda _1\xi }+(1- l)\lambda _2e^{ \lambda _2\xi }}{le^{ \lambda _1\xi }+(1- l)e^{ \lambda _2\xi }},\ \ l\in [0,1]. \end{align*}

Integrating the above equality gives

\begin{align*}\varphi (\xi )=p({ l}e^{{ \lambda }_{1}\xi }+(1-{ l})e^{{ \lambda }_{2}\xi })\end{align*}

for some $p\gt 0.$ Hence, (27) holds by letting $C_{1}=p{ l}$ and $C_{2}=p(1-{ l})$ .

Proof. Suppose the assertion is false, that is (5) admits a positive solution $ (S(\xi ),I(\xi ))$ satisfying (6) for $R_{0}\leq 1$ and some $c\in \mathbb{R}$ . We first claim that $ \displaystyle \int _{\mathbb{R}} I (\xi )d\xi \lt +\infty$ . Since $I(\pm \infty )=0$ , there exists $K\gt 0$ such that $I(\xi )\leq K$ for $\xi \in \mathbb{R}$ . Integrating the first equation of (5) from $y$ to $ x$ for any $ x,y\in \mathbb{R}$ , we have

\begin{align*} \Big | \int _{y}^{x}\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)dsd\xi \Big | \leq & d_{S}\Big |\sum _{i\in \tilde{\Omega }}J_{1}(i)i \int _{y}^{x}\int ^{1}_{0} S^{\prime}(\xi -ti) dt d\xi \Big | +\left | c(S(x)-S(y))\right | \\[3pt] \leq &d_{S}\Big |\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int ^{1}_{0} \left ( S(x-ti)- S(y-ti)\right ) dt\Big |+\left | c(S(x)-S(y))\right | \\[3pt] \leq & \big ( 2d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)|i|+2|c|\big ) S_{0}. \end{align*}

Then, it follows from the second equation of (5) that

\begin{align*} \Big | \gamma \int _{y}^{x} I (\xi )d\xi \Big | \leq & d_{I}\Big | \sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{y}^{x} \left (I(\xi -i)-I(\xi )\right )d\xi \Big |+ \Big | \int _{y}^{x}\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)dsd\xi \Big |\\[3pt] &+| c(I(x)-I(y))|\\[3pt] \leq &\big ( 2d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)|i|+ 2|c| \big ) K+\big ( 2d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)|i| + 2|c| \big ) S_{0}. \end{align*}

Thus, it follows that $ \displaystyle \int _{\mathbb{R}} I (\xi )d\xi \lt +\infty$ by the arbitrariness of $ x$ and $ y$ .

By the second equation of (5), we have

(29) \begin{align} \gamma \int _{\mathbb{R}}I(\xi )d\xi &\leq d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{\mathbb{R}}\left (I(\xi -i)-I(\xi )\right ) d\xi + \beta S_{0} \int _{0}^{\tau }f(s)\int _{\mathbb{R}}I(\xi -cs)d\xi ds \nonumber \\[3pt] &= d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{\mathbb{R}}\left (I(\xi -i)-I(\xi )\right ) d\xi + \beta S_{0} \int _{\mathbb{R}}I(\xi )d\xi = \beta S_{0} \int _{\mathbb{R}}I(\xi )d\xi . \end{align}

Note that $R_{0}={\beta S_{0} /}{\gamma }.$ If $ R_{0}\lt 1$ , by (29), we have

(30) \begin{align} \int _{\mathbb{R}}I(\xi )d\xi \leq \frac{\beta S_{0}}{\gamma } \int _{\mathbb{R}}I(\xi )d\xi \lt \int _{\mathbb{R}}I(\xi )d\xi, \end{align}

which gives a contradiction. If $ R_{0}= 1$ , the second equation of (5) also gives

(31) \begin{align} 0=&\,d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{\mathbb{R}}\left (I(\xi -i)-I(\xi )\right ) d\xi +\int _{\mathbb{R}}\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)ds d\xi - \gamma \int _{\mathbb{R}}I(\xi ) d\xi \nonumber \\[3pt] =&\,d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{\mathbb{R}}\left (I(\xi -i)-I(\xi )\right ) d\xi +\beta \int _{\mathbb{R}}\int _{0}^{\tau }f(s) S(\xi +cs) I(\xi ) dsd\xi - \beta S_{0} \int _{\mathbb{R}}I(\xi ) d\xi \nonumber \\[3pt] =&\,\beta \int _{\mathbb{R}} I(\xi ) \int _{0}^{\tau }f(s) \left ( S(\xi +cs)-S_{0} \right ) ds d\xi . \end{align}

Since $ 0\lt S(\xi )\leq S_{0}$ and $ I(\xi )\gt 0$ on $ \mathbb{R}$ , (31) implies that $ S(\xi )\equiv S_{0}$ on $ \mathbb{R}$ , which contradicts to $ S({+}\infty )=S_{\infty }\lt S_{0}$ . Therefore, $R_0\gt 1$ .

Next, we show that $c\gt 0$ . Suppose, by contradiction, that $(S(\xi ),I(\xi ))$ is a positive solution of (5) satisfying (6) with $c\leq 0$ . Since $R_{0}\gt 1$ and $ S({-}\infty )=S_{0}$ , there exists a $\hat{\xi }\lt 0$ such that

\begin{align*} S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)ds \gt \frac {\beta S_{0} +\gamma }{2\beta }\int _{0}^{\tau }f(s)I(\xi -cs)ds, \mbox { for }\ \xi \lt \hat {\xi }.\end{align*}

Then, for $\xi \lt \hat{\xi }$ , we have

\begin{align*} cI^{\prime}(\xi )=&d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\left ( I(\xi -i)- I(\xi )\right ) +\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)ds -\gamma I(\xi )\\[3pt] \gt & d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\left ( I(\xi -i)- I(\xi )\right ) +\frac{ \beta S_{0} +\gamma }{2 }\int _{0}^{\tau }f(s)\left ( I(\xi -cs) -I(\xi )\right ) ds +\frac{\beta S_{0}-\gamma }{2} I(\xi ). \end{align*}

Integrating the above inequality over $({-}\infty, \xi )$ with $\xi \lt \hat{\xi }$ , one can obtain

\begin{align*}\frac {\beta S_{0}-\gamma }{2} \Theta (\xi ) \lt cI(\xi )- d_{I}\sum _{i\in \tilde {\Omega }}J_{2}(i)\left ( \Theta (\xi -i)-\Theta (\xi )\right ) +\frac { \beta S_{0} +\gamma }{2 }\int _{0}^{\tau }f(s)\left ( \Theta (\xi )- \Theta (\xi -cs) \right )ds, \end{align*}

where $ \Theta (\xi ) \,:\!=\,\displaystyle \int ^{\xi }_{-\infty }I(y)dy\gt 0$ . Since

\begin{align*} \int ^{\xi }_{-\infty }\int _{0}^{\tau }f(s)(\Theta (\eta )-\Theta (\eta -cs)) ds d\eta &= \int ^{\xi }_{-\infty }\int _{0}^{\tau }f(s)cs\int ^{1}_{0} \Theta ^{\prime}(\eta -tcs)dt ds d\eta \\[3pt] &=\int _{0}^{\tau }f(s)cs\int ^{1}_{0}\Theta (\xi -tcs)dt ds\leq 0,\\[3pt] \sum _{i\in \tilde{\Omega }}J_{2}(i)\int ^{\xi }_{-\infty } \big ( \Theta (\eta -i)-\Theta (\eta )\big )d\eta &=\sum _{i\in \tilde{\Omega }}J_{2}(i) \int ^{\xi -i}_{\xi }\Theta (\eta )d\eta \\[3pt] &=\sum _{i =1}^{2}J_{2}(i)\Big ( \int ^{\xi +i}_{\xi }\Theta (\eta )d\eta -\int ^{\xi }_{\xi -i}\Theta (\eta )d\eta \Big )\gt 0, \end{align*}

it follows that

\begin{align*} 0\lt \frac{\beta S_{0}-\gamma }{2} \int ^{\xi }_{-\infty }\Theta (\eta )d\eta \lt c\Theta (\xi )\leq 0, \mbox{ for}\ \xi \lt \hat{\xi }, \end{align*}

which also gives a contradiction. Hence the assertion of the lemma holds.

Proof. Let’s prove the result by using the contradiction argument. Suppose, for some constant $ \epsilon \gt 0$ , there exists a subsequence of $\{ \xi _{j}\} _{j\in \mathbb{N}}$ , still written as $\{ \xi _{j}\}_{j\in \mathbb{N}}$ , such that $ S(\xi _{j})\geq \epsilon$ for each $ j\in \mathbb{N}$ . Let $d_{1}=d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)$ and $d_{2}=d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)$ . By the first equation of (5), we have $ S^{\prime}(\xi )\leq{d_{1}S_{0}}/{c}$ which implies $ S(\xi )\geq{\epsilon }/{2}$ for $ \xi \in [\xi _{j}- \xi _{0}, \xi _{j}]$ , where $ \xi _{0}\,:\!=\,{c\epsilon }/{(2d_{1}S_{0})}.$ Note that

\begin{align*} cI^{\prime}(\xi )+(d_{2}+\gamma ) I(\xi )=d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i) I(\xi -i) +\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -cs)ds\gt 0. \end{align*}

So, $I (\xi )e^{\frac{d_{2}+\gamma }{ c}\xi }$ is strictly increasing in $\mathbb{R}$ . Then,

(32) \begin{align} \int _{0}^{\tau }f(s)I(\xi -cs)ds\lt I(\xi )\int _{0}^{\tau }f(s) e^{( d_{2}+\gamma )s}ds \ \mbox{ and } \ \frac{I(\xi -x)}{I(\xi )}\lt e^{\frac{(d_{2}+\gamma )x}{ c}}, \mbox{ for } x\gt 0. \end{align}

Moreover, for any $p\in \tilde{\Omega }$ with $ p\gt 0$ , we have

\begin{align*} cI^{\prime}(\xi )+(d_{2}+\gamma ) I(\xi )\gt d_{I}J_{2}(p)I(\xi +p), \end{align*}

which implies

(33) \begin{align} \frac{d}{d\xi }\big ( e^{\frac{(d_{2}+\gamma )\xi }{c}}I(\xi )\big )\gt \frac{d_{I}J_{2}(p)}{c}e^{\frac{(d_{2}+\gamma )\xi }{c}}I(\xi +p). \end{align}

Since $I (\xi )e^{\frac{d_{2}+\gamma }{ c}\xi }$ is strictly increasing, integrating (33) from $ \xi -\displaystyle \frac{p}{2}$ to $ \xi$ gives

\begin{align*} e^{\frac{(d_{2}+\gamma )\xi }{c}}I(\xi )- e^{\frac{(d_{2}+\gamma )(\xi -\frac{p}{2}) }{c}}I(\xi -\frac{p}{2})& \gt \frac{d_{I}J_{2}(p)}{c}\int _{\xi -\frac{p}{2}}^{\xi }e^{\frac{(d_{2}+\gamma )\eta }{c}}I(\eta +p)d\eta \gt \frac{d_{I}J_{2}(p)p}{2c}e^{\frac{(d_{2}+\gamma )(\xi -\frac{p}{2}) }{c}}I(\xi +\frac{p}{2}). \end{align*}

It follows that

(34) \begin{align} \frac{I(\xi +\frac{p}{2})}{I(\xi )}\lt \frac{2c}{d_{I}J_{2}(p)p}e^{\frac{(d_{2}+\gamma )p}{2c}}\ \mbox{ and }\ \frac{I(\xi +p)}{I(\xi )}=\frac{I(\xi +p)}{I(\xi +\frac{p}{2})}\frac{I(\xi +\frac{p}{2})}{I(\xi )}\lt \frac{4c^{2}}{(d_{I}J_{2}(p)p)^{2}}e^{\frac{ (d_{2}+\gamma ) p }{c}}. \end{align}

Dividing the second equation of (5) by $I(\xi )$ , and using (32) and (34), we have

(35) \begin{align} \frac{I^{\prime}(\xi )}{I(\xi )}=&\frac{d_{I}}{c}\sum _{i\in \tilde{\Omega }}J_{2}(i)\frac{I(\xi -i)}{I(\xi )} - \frac{d_{2}}{c}+\frac{\beta }{c} S(\xi ) \int _{0}^{\tau }f(s)\frac{I(\xi -cs)}{I(\xi )}ds -\frac{\gamma }{c} \lt +\infty . \end{align}

Thus, there exists a constant $\rho \gt 0$ such that $ \left |{I^{\prime}(\xi )}/{I(\xi )} \right | \leq \rho$ , and hence

\begin{align*} \frac{I(\xi _{j})}{I(\xi )}=e^{\int _{\xi }^{\xi _{j}}\frac{I^{\prime}(\eta )}{I(\eta )} d\eta } \leq e^{\rho (c\tau + \xi _{0}) }, \mbox{ for } \xi \in [\xi _{j}- \xi _{0}-c\tau, \xi _{j}]. \end{align*}

Then it follows that

\begin{align*} \min _{\xi \in [\xi _{j}- \xi _{0}-c\tau, \xi _{j}]} I(\xi ) \geq I(\xi _{j})e^{-\rho (c\tau + \xi _{0}) }\rightarrow +\infty \ \mbox{ as }\ j\rightarrow +\infty . \end{align*}

Furthermore, by the first equation of (5), one can see

\begin{align*} \max _{\xi \in [\xi _{j}- \xi _{0}, \xi _{j}]} S^{\prime}(\xi )\leq & \frac{d_{1}S_{0}}{c}-\frac{\beta \epsilon }{2c} \min _{\xi \in [\xi _{j}- \xi _{0}, \xi _{j}]} \int _{0}^{\tau }f(s) I(\xi -cs)ds \leq \frac{d_{1}S_{0}}{c}-\frac{\beta \epsilon }{2c}I(\xi _{j})e^{-\rho (c\tau + \xi _{0}) } \rightarrow -\infty \end{align*}

as $j\rightarrow +\infty .$ Let $j$ be sufficiently large such that $ S^{\prime}(\xi )\leq{-2S_{0}}/{ \xi _{0}}$ for $ \xi \in [\xi _{j}- \xi _{0}, \xi _{j}]$ , then

\begin{align*} S(\xi _{j})-S(\xi _{j}- \xi _{0})\leq -2S_{0}, \end{align*}

which is impossible since $ 0\lt S(\xi )\leq S_{0}$ on $\mathbb{R}$ . Therefore, $S(\xi _{j})\to 0$ as $j\rightarrow +\infty$ .

Proof. Without loss of generality, we assume $\left \lbrace c_{k}\right \rbrace _{k=1}^{+\infty }$ is a strictly decreasing sequence with $ \lim _{k\rightarrow +\infty } c_{k}= c_0\gt 0$ . Let $ (S_{k}(\xi ),I_{k}(\xi ))$ be a positive solution of (5) satisfying (6) for $c= c_{k}$ . We first claim that the sequence $\{I_{k}(\xi )\} _{k\in \mathbb{N}}$ is uniformly bounded on $\mathbb{R}$ . If not, there exists a sequence $\{ z_{k}\}_{k\in \mathbb{N}}$ such that $\lim \limits _{k\rightarrow +\infty } I_{k}(z_{k})=+\infty$ . Then, it follows from Lemma 3.2 that $\lim \limits _{k\rightarrow +\infty } S_{k}(z_{k})=0$ . Since $ I_{k} (\pm \infty )=0$ and $ I_{k}(\xi )\gt 0$ , without loss of generality, we may assume that $ I_{k}(z_{k})=\max \limits _{\xi \in \mathbb{R}} I_{k}(\xi )$ . By the second equation of (5), we have

(36) \begin{align} 0=c_{k}I_{k}^{\prime}(z_{k})= & d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)\left ( I_{k}(z_{k}-i)-I_{k}(z_{k})\right ) +\beta S_{k}(z_{k}) \int _{0}^{\tau }f(s)I_{k}(z_{k}-c_{k}s)ds -\gamma I_{k}(z_{k})\nonumber \\[3pt] \leq &\beta S_{k}(z_{k}) \int _{0}^{\tau }f(s)I_{k}(z_{k}-c_{k}s)ds -\gamma I_{k}(z_{k}) \lt \left ( \beta S_{k}(z_{k})-\gamma \right ) I_{k}(z_{k}). \end{align}

However, the inequality (36) contradicts to the properties $\lim \limits _{k\rightarrow +\infty } I_{k}(z_{k})=+\infty$ and $\lim \limits _{k\rightarrow +\infty } S_{k}(z_{k})=0$ , when $k$ is large enough. Therefore, $\{I_{k}(\xi )\} _{k\in \mathbb{N}}$ is uniformly bounded on $\mathbb{R}$ .

Since $\{S_{k}(\xi )\} _{k\in \mathbb{N}}$ and $\{ I_{k}(\xi )\}_{k\in \mathbb{N}}$ are uniformly bounded on $\mathbb{R}$ , it follows from (5) that $||S_{k}||_{C^{2}(\mathbb{R})}$ and $||I_{k}||_{C^{2}(\mathbb{R})}$ are both uniformly bounded on $\mathbb{R}$ . Then, by Arzela-Ascoli theorem, there exists subsequences of $\{S_{k}(\xi )\}_{k\in \mathbb{N}}$ and $\{ I_{k}(\xi )\}_{k\in \mathbb{N}}$ , still written as $\{S_{k}(\xi )\}_{k\in \mathbb{N}}$ and $\{ I_{k}(\xi )\}_{k\in \mathbb{N}}$ , such that $ S_{k} (\xi ) \rightarrow S(\xi )$ and $ I_{k} (\xi ) \rightarrow I (\xi )$ in $ C^{1}_{loc} (\mathbb{R})$ as $k\rightarrow +\infty$ . By Lebesgue dominated convergence theorem, it is easy to see the limiting function $ ( S(\xi ), I(\xi ) )$ is a solution of (5) with $c= c_{0}$ , i.e., satisfies the following system

(37) \begin{align} \left \{ \begin{array}{ll} c_{0}S^{\prime}(\xi )=d_{S}\displaystyle \sum _{i\in \tilde{\Omega }}J_{1}(i)\left ( S(\xi -i)-S(\xi )\right ) -\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi - c_{0}s)ds,\\[3pt] c_{0}I^{\prime}(\xi )=d_{I}\displaystyle \sum _{i\in \tilde{\Omega }}J_{2}(i)\left (I(\xi -i)-I(\xi )\right ) +\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi - c_{0}s)ds-\gamma I(\xi ) . \end{array}\right . \end{align}

Since $\{I_{k}\} _{k\in \mathbb{N}}$ is uniformly bounded on $\mathbb{R}$ , there exists a constant $I_{0}\gt 0$ such that $ I(\xi )\leq I_{0}$ on $\mathbb{R}$ . Hence, the solution $ ( S(\xi ), I(\xi ) )$ satisfies $ 0\leq S(\xi )\leq S_{0}$ and $ 0\leq I(\xi )\leq I_{0}$ for all $\xi \in \mathbb{R}.$

Next, we claim that $ S(\xi )$ and $I(\xi )$ are non-trivial, and they satisfy the asymptotic boundary conditions of (6).

Claim 1: $I(\pm \infty )=0.$

Integrating the first equation of (37) from $y$ to $ x$ for any $ x,y\in \mathbb{R}$ , we have

(38) \begin{align} \Big | \int _{y}^{x}\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \Big | \leq & d_{S}\Big |\sum _{i\in \tilde{\Omega }}J_{1}(i)i \int _{y}^{x}\int ^{1}_{0} S^{\prime}(\xi -ti) dt d\xi \Big | +c_{0}| S(x)-S(y)| \nonumber \\[3pt] \leq &d_{S}\Big |\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int ^{1}_{0} \left ( S(x-ti)- S(y-ti)\right ) dt\Big |+c_{0}\left | S(x)-S(y)\right | \nonumber \\[3pt] \leq & \big ( 2d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)|i|+2c_{0}\big ) S_{0}. \end{align}

Thus, by the second equation of (37), we can obtain

(39) \begin{align} \Big | \gamma \int _{y}^{x} I (\xi )d\xi \Big | \leq & d_{I}\Big | \sum _{i\in \tilde{\Omega }}J_{2}(i)\int _{y}^{x} \left (I(\xi -i)-I(\xi )\right )d\xi \Big |+ \Big | \int _{y}^{x}\beta S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \Big |\nonumber \\[3pt] &+c_{0}| I(x)-I(y)|\nonumber \\[3pt] \leq &\big ( 2d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i)|i|+ 2c_{0} \big ) I_{0}+\big ( 2d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)|i| + 2c_{0} \big ) S_{0}. \end{align}

Due to the arbitrariness of $ x$ and $ y$ , it follows from (39) that $ \displaystyle \int _{\mathbb{R}} I (\xi )d\xi \lt +\infty$ . In addition, it is easy to see from (37) that $ I^{\prime} (\xi )$ is bounded on $\mathbb{R}$ . Therefore, we have $ I (\pm \infty )=0$ .

Claim 2: $ S(\xi ) \gt 0$ on $\mathbb{R}$ and $S({-}\infty )=S_0$ .

Assume that $S({\xi }_0) =0$ for some ${\xi }_0\in \mathbb{R}$ . From the first equation of (37), we have $S({\xi }_0\pm 1) =0$ , and inductively that $S({\xi _0}\pm k) =0$ for any $k\in \mathbb{N}$ . Moreover, the first equation of (37) gives

\begin{align*} c_{0} S^\prime (\xi )+\delta S (\xi )\geq 0, \ \mbox{for }\xi \in \mathbb{R}, \end{align*}

where $\delta =(d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)+\beta I_{0})$ , which implies that $ S (\xi )e^{\frac{\delta }{ c_{0}}\xi }$ is non-decreasing over $\mathbb{R}$ . Then it follows that $ S(\xi ) \equiv 0$ on $ \mathbb{R}$ , and hence, $ S_{k}(\xi ) \rightarrow 0$ as $k\rightarrow +\infty$ for any $ \xi \in \mathbb{R}$ . On the other hand, let’s write

\begin{align*} &\sum _{i\in \tilde{\Omega }}J_{1}(i)\int _{-\infty }^{0} \left ( S_{k}(\xi -i)-S_{k}(\xi )\right )d\xi =-\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{-\infty }^{0}\int ^{1}_{0} S _{k}^{\prime}(\xi -ti)dt d\xi \\[3pt] =&-\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int ^{1}_{0}\int _{-\infty }^{0} S _{k}^{\prime}(\xi -ti) d\xi dt =\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}\left ( S _{0}-S _{k}({-}ti)\right )dt =-\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}S _{k}({-}ti)dt, \end{align*}

then the first equation of (5) gives

(40) \begin{align} c_{k}( S_{0}-S_{k}(0))=\int _{-\infty }^{0} \beta S_{k}(\xi ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi +d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}S _{k}({-}ti)dt. \end{align}

As $k\rightarrow +\infty$ , it follows from (40) that $ c_{0}S_{0}=0$ , which leads to a contradiction. Hence $ S(\xi ) \gt 0$ on $\mathbb{R}$ .

To prove $ S({-}\infty )=S_{0}$ , it is sufficient to show that $\underline{S}\,:\!=\, \liminf \limits _{\xi \rightarrow -\infty } S(\xi )= S_{0}$ . Suppose that $ \underline{S}\lt S_{0}$ , then there exists a sequence $\{ \zeta _{q}\}_{q\in \mathbb{N}}$ such

\begin{align*} \lim \limits _{q\rightarrow +\infty } \zeta _{q}=-\infty \ \text{ and } \ \lim \limits _{q\rightarrow +\infty } S( \zeta _{q})=\underline{S} . \end{align*}

Denote

\begin{align*} \tilde {S}_{q}(\xi ) \,:\!=\, S(\xi + \zeta _{q})\ \mbox { and }\ \tilde {I}_{q}(\xi ) \,:\!=\, I(\xi + \zeta _{q}),\ \mbox {for } \xi \in \mathbb {R}. \end{align*}

Obviously, $ \lim \limits _{q\rightarrow +\infty } \tilde{I}_{q}(\xi )=0$ locally uniformly on $ \mathbb{R}$ . Since $ ||\tilde{S}_{q} ||_{C^{2}(\mathbb{R})}$ is uniformly bounded on $ \mathbb{R}$ , there exists a subsequence of $\{\tilde{S}_{q}(\xi )\} _{q\in \mathbb{N}}$ , still denote as $\{\tilde{S}_{q}(\xi )\} _{q\in \mathbb{N}}$ , such that $\tilde{S}_{q}(\xi ) \rightarrow \tilde{S} (\xi )$ in $ C^{1}_{loc} (\mathbb{R})$ as $q\rightarrow +\infty$ . Hence, the first equation of (37) gives

(41) \begin{align} c_{0}\tilde{S}^{\prime}(\xi )=d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\big ( \tilde{S}(\xi -i)-\tilde{S}(\xi )\big ). \end{align}

It is easy to note that zero is the root of the characteristic equation of (41). Since $\tilde{S}(\xi )$ is nonnegative and bounded with $ \tilde{S}(0)=\underline{S}$ , it follows that $ \tilde{S}(\xi )=\underline{S}$ on $\mathbb{R}$ according to Proposition 2.5. Hence, we get that $ \tilde{ S}_{q}(\xi )\to \underline{S}$ in $ C^{1}_{loc} (\mathbb{R})$ as $q\rightarrow +\infty$ . Since

(42) \begin{align} c_{k}S^\prime _{k}(\xi )=d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\left ( S_{k}(\xi -i)-S_{k}(\xi )\right ) -\beta S_{k}(\xi ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)ds, \end{align}

we have

\begin{align*} c_{k}(S _{k}( \zeta _{q})-S_{0})=&d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\int _{-\infty }^{ \zeta _{q}} \left ( S_{k}(\xi -i)- S_{k}(\xi )\right )d\xi -\beta \int _{-\infty }^{ \zeta _{q}}S_{k}(\xi ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)ds d\xi . \end{align*}

Note that

\begin{align*} \sum _{i\in \tilde{\Omega }}J_{1}(i) \int _{-\infty }^{ \zeta _{q}} \left ( S_{k}(\xi -i)- S_{k}(\xi )\right )d\xi =-\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}S _{k}(\zeta _{q}-ti)dt. \end{align*}

As $k\rightarrow +\infty$ , one can see

(43) \begin{align} c_{0}(S ( \zeta _{q})-S_{0})=&-d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}S(\zeta _{q}-ti)dt -\beta \int _{-\infty }^{ \zeta _{q}} S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)ds d\xi . \end{align}

As $q\rightarrow +\infty$ , it follows that $ c_{0}(\underline{S} -S_{0})= 0$ , which contradicts to $ \underline{S} \lt S_{0}$ . Hence, $ S({-}\infty )=S_{0}$ .

Claim 3: $ I(\xi )\gt 0$ on $\mathbb{R}$ .

Assume that $I({\eta })=0$ for some ${\eta }\in \mathbb{R}$ , then it is easy to see from the second equation of (37) that $ I(\xi )\equiv 0$ on $\mathbb{R}$ . Then, the first equation of (37) gives

(44) \begin{align} c_{0}S^\prime (\xi )=d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\left ( S(\xi -i)-S(\xi )\right ). \end{align}

Since $S(\xi )$ is nonnegative and bounded, by Proposition 2.5, it follows that $S(\xi )$ is a constant function, i.e., $ S(\xi )\equiv \widehat{S}$ for some constant $ \widehat{S}\in [0,S_{0}]$ . Note that

\begin{align*} I_{k}(\pm \infty )=0\ \mbox { and }\ \int _{-\infty }^{+\infty } \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi = \int _{-\infty }^{+\infty } I_{k}(\xi )d\xi . \end{align*}

By the second equation of (5), we have

(45) \begin{align} 0=& \int _{-\infty }^{+\infty }\beta S_{k}(\xi ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi - \gamma \int _{-\infty }^{+\infty } I_{k} (\xi )d\xi \nonumber \\[3pt] =& \int _{-\infty }^{+\infty }\beta S_{k}(\xi ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi - \gamma \int _{-\infty }^{+\infty } \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi \nonumber \\[3pt] =& \int _{-\infty }^{+\infty }\Big (\beta S_{k}(\xi )-\gamma \Big ) \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)dsd\xi . \end{align}

Note that $ \displaystyle \int _{0}^{\tau }f(s)I_{k}(\xi -c_{k}s)ds\gt 0$ for any $\xi \in \mathbb{R}$ . There exists some ${\eta }_{k}$ such that $\gamma = \beta S_{k}({\eta }_{k}) .$ Since $ (S_{k}(\xi ), I_{k}(\xi ))$ is translation invariant, we may assume ${\eta }_{k}=0$ . As $k\to +\infty$ , we have

\begin{align*} \gamma = \beta S_{k}(0)\to \beta S(0)=\beta \widehat {S}, \end{align*}

which implies $\widehat{S}=\gamma /\beta$ . Since $R_{0}={\beta S_{0} }/{\gamma }\gt 1$ , then $ S (\xi )\equiv \widehat{S}\lt S_{0},$ which contradicts to $ S({-}\infty )=S_{0}$ . So, $ I(\xi )\gt 0$ on $\mathbb{R}$ .

Claim 4: $ S({+}\infty )=S_{\infty }\lt S_{0}$ .

We first show that $ \bar \alpha \,:\!=\,\liminf \limits _{\xi \rightarrow +\infty } S (\xi )= \limsup \limits _{\xi \rightarrow +\infty } S (\xi )=:\hat \alpha$ . If $\bar{\alpha }\lt \hat \alpha$ , one can find two sequences $\{ \xi _{k}\}_{k\in \mathbb{N}}$ and $\{ \eta _{k}\}_{k\in \mathbb{N}}$ , with $\xi _{k}\lt \eta _{k}$ , $\lim \limits _{k\rightarrow +\infty } \xi _{k}=+\infty$ and $\lim \limits _{k\rightarrow +\infty } \eta _{k}=+\infty$ , such that

\begin{align*} \lim _{k\rightarrow +\infty } S (\xi _{k})=\bar \alpha, \ S^{\prime}(\xi _{k})=0; \mbox { and } \lim _{k\rightarrow +\infty } S (\eta _{k})=\hat \alpha, \ S^{\prime}(\eta _{k})=0.\end{align*}

Denote

\begin{align*} \hat{S}_{k}(\xi ) \,:\!=\, S(\xi +\xi _{k})\ \text{ and } \ \hat{I}_{k}(\xi ) \,:\!=\, I (\xi + \xi _{k}), \text{ for } \xi \in \mathbb{R} . \end{align*}

Obviously, $ \hat{I}_{k}(\xi )\to 0$ locally uniformly on $ \mathbb{R}$ as $k\rightarrow +\infty .$ Since $ ||\hat{S}_{k} ||_{C^{2}(\mathbb{R})}$ is uniformly bounded on $ \mathbb{R}$ , there exists a subsequence of $\{ \hat{S}_k(\xi )\}_{k\in \mathbb{N}}$ , still written as $\{ \hat{S}_k(\xi )\}_{k\in \mathbb{N}}$ , such that $\hat{S}_{k}(\xi ) \rightarrow \hat{S} (\xi )$ in $ C^{1}_{loc} (\mathbb{R})$ as $k\rightarrow +\infty$ . Hence, the first equation of (37) yields

(46) \begin{align} c_{0} \hat{S}^{\prime}(\xi )=d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\big ( \hat{S}(\xi -i)-\hat{S}(\xi )\big ). \end{align}

By Proposition 2.5, it follows that $ \hat{S}(\xi )$ is constant function. Since $ \hat{S}(0)=\bar{\alpha }$ , we have $ \hat{S}(\xi )\equiv \bar{\alpha }$ on $\mathbb{R}$ . Hence, we get that $ S(\xi +\xi _{k})\rightarrow \bar{\alpha }$ in $ C^{1}_{loc} (\mathbb{R})$ as $ k\rightarrow +\infty$ . By the same way, we also obtain that $ \lim _{k\rightarrow +\infty } S(\xi +\eta _{k})=\hat{\alpha }$ in $ C^{1}_{loc} (\mathbb{R})$ . Then, integrating the first equation of (37) from $ \xi _{k}$ to $ \eta _{k}$ , we have

\begin{align*} c_{0}(\hat{\alpha }-\bar{\alpha })= & \lim _{k\rightarrow +\infty } d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\int _{\xi _{k}}^{\eta _{k}}\left ( S(\xi -i)-S(\xi )\right ) d\xi -\lim _{k\rightarrow +\infty }\beta \int _{\xi _{k}}^{\eta _{k}}S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \\[3pt] \leq &\lim _{k\rightarrow +\infty } d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\int _{\xi _{k}}^{\eta _{k}}\left ( S(\xi -i)-S(\xi )\right ) d\xi \\[3pt] =&d_{S}\lim _{k\rightarrow +\infty }\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{0}^{1}\left ( S(\xi _{k}-ti)-S(\eta _{k}-ti)\right ) dt =0, \end{align*}

which contradicts to $\bar{\alpha }\lt \hat \alpha$ . Thus, $\hat{\alpha }=\bar{\alpha }$ , and then $S_{\infty }\,:\!=\, S ({+}\infty )$ exists.

Next, we further show that $S_{\infty }\lt S_{0}$ by proving $ \liminf \limits _{\xi \rightarrow +\infty } S (\xi )\lt S_{0} .$ Suppose, by contradiction, that $ \liminf \limits _{\xi \rightarrow +\infty } S (\xi )= S_{0}$ , then $ S ({+}\infty ) =S_{\infty }=S_{0}$ and the first equation of (37) gives

\begin{align*} 0&= d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)\int _{-\infty }^{+\infty }\left ( S(\xi -i)-S(\xi )\right ) d\xi -\beta \int _{-\infty }^{+\infty } S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \\[3pt] &= -d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int _{-\infty }^{+\infty }\int ^{1}_{0} S^{\prime}(\xi -ti)dt d\xi -\beta \int _{-\infty }^{+\infty } S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \\[3pt] &=-d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i)i\int ^{1}_{0}\int _{-\infty }^{+\infty } S^{\prime}(\xi -ti) d\xi dt-\beta \int _{-\infty }^{+\infty } S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \\[3pt] &=-\beta \int _{-\infty }^{+\infty } S(\xi ) \int _{0}^{\tau }f(s)I(\xi -c_{0}s)dsd\xi \lt 0, \end{align*}

which gives a contradiction. Thus, $ S_{\infty }\lt S_{0}$ . The proof is complete.

Proof. The idea of proof is similar to that of [Reference Xu, Tan and Hsu28, Proposition 3.1]. However, due to the consideration of distributed latent period, the computations are more complicated than those of [Reference Xu, Tan and Hsu28].

Firstly, for large $Y\gt 0$ , we define the set

\begin{align*} \Phi _{Y}\,:\!=\,&\big \{ (\omega ({\cdot}),\sigma ({\cdot}))\in C ([{-}Y,Y],\mathbb{R}^{2}): S_{-}(\xi )\leq \omega (\xi )\leq S_{0}, \,I_{-}(\xi )\leq \sigma (\xi )\leq I_{+}(\xi ), \,\xi \in [{-}Y,Y] \big \}\\[3pt] &{\cap } \big \{ (\omega ({\cdot}),\sigma ({\cdot}))\in C ([{-}Y,Y],\mathbb{R}^{2}): \omega ({-}Y)=S_{-}({-}Y),\ \sigma ({-}Y)=I_{-}({-}Y) \big \}. \end{align*}

Obviously, $ \Phi _{Y}$ is a closed and convex set. In addition, we extend any function $ (\tilde{\Omega }({\cdot}),\sigma ({\cdot}))$ to $ ( \tilde{\omega }({\cdot}), \tilde{\sigma } ({\cdot}))\in C(\mathbb{R},\mathbb{R}^{2})$ in the way

\begin{align*} (\tilde {\omega }(\xi ),\tilde {\sigma }(\xi ))=\left \{ \begin {array}{r@{\quad}l} (S_{-}(\xi ),I_{-}(\xi )),&\mbox {for }\xi \lt -Y;\\[3pt] (\omega (\xi ),\sigma (\xi )),&\mbox {for }|\xi |\leq Y;\\[3pt] (\omega (Y),\sigma (Y)),&\mbox {for }\xi \gt Y. \end {array} \right . \end{align*}

For the convenience of statement, without loss of generality, we next assume that $\sum _{i\in \tilde{\Omega }}J_{k}(i)=1$ ( $k=1,2$ ). Define an operator $\Gamma$ on $ \Phi _{Y}$ by

\begin{align*} \Gamma \,:\!=\,(\Gamma _{1}, \Gamma _{2}):(\omega ({\cdot}),\sigma ({\cdot}))\in \Phi _{Y}\rightarrow (S({\cdot}),I({\cdot}))\in C^{1} ([{-}Y,Y]),\end{align*}

where $(S(\xi ),I(\xi ))$ satisfies the following initial value problem of ODE:

(48) \begin{align} cS^{\prime}(\xi )= &- (d_{S}+l)S(\xi )+d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i) \tilde{\omega }(\xi -i)+l\omega (\xi )-\beta \omega (\xi )\displaystyle \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds, \\[-32pt] \nonumber \end{align}

(49) \begin{align} cI^{\prime}(\xi )=&-(d_{2}+\gamma ) I(\xi )+d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i) \tilde{\sigma }(\xi -i) +\beta \omega (\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds, \\[-32pt] \nonumber \end{align}

(50) \begin{align} S({-}Y)=&S_{-}({-}Y),\, \, I({-}Y)=I_{-}({-}Y), \end{align}

where $l$ is a constant satisfying $l\geq \beta e^{\lambda _{1}Y}\displaystyle \int _{0}^{\tau }f(s)e^{-c\lambda _{1}s}ds$ . Then we claim that $\Gamma$ is completely continuous which maps from $ \Phi _{Y}$ to $ \Phi _{Y}$ .

We first show that $ \Gamma [\Phi _{Y}]\subseteq \Phi _{Y}$ , i.e.,

(51) \begin{align} S_{-}(\xi )\leq \Gamma _{1}[ \omega, \sigma ](\xi )\leq S_{0}\ \mbox{ and }\ I_{-}(\xi )\leq \Gamma _{2}[ \omega, \sigma ](\xi )\leq I_{+}(\xi ), \ \mbox{for} \ \xi \in [{-}Y,Y]. \end{align}

It is easy to see that $0$ is a lower solution of (48). By comparison principle, we have

\begin{align*} \Gamma _{1}[ \omega, \sigma ](\xi ) \geq 0,\ \mbox {for } \xi \in [{-}Y,Y]. \end{align*}

Since $0\leq \tilde{\omega }(\xi )\leq S_{0}$ on $ \mathbb{R}$ , for $ \xi \in [{-}Y,Y]$ , one can obtain

\begin{align*} & - (d_{S}+l)S_{0} +d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i) \tilde{\omega }(\xi -i)+l\omega (\xi )-\beta \omega (\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds \\[3pt] \leq &-\beta \omega (\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds \leq 0, \end{align*}

which implies that $ S_{0}$ is an upper solution of (48). Then, by comparison principle again,

\begin{align*} \Gamma _{1}[ \omega, \sigma ](\xi ) \leq S_{0}, \text{ for } \xi \in [{-}Y,Y]. \end{align*}

Hence, the left part of (51) holds.

Now we prove the right part of (51). According to the choice of $l$ , it is easy to see that

\begin{align*}h(\omega )\,:\!=\,l\omega (\xi )-\beta \omega (\xi ) \int _{0}^{\tau }f(s)\tilde {\sigma }(\xi -cs)ds\end{align*}

is non-decreasing in $\omega$ . Then, for $ \xi \in [{-}Y,\displaystyle \frac{1}{\nu }\ln \frac{S_{0}}{\delta })$ , one can see $S_{-}(\xi )= S_{0}-\delta e^{\nu \xi }$ and

(52) \begin{align} & cS_{-}^{\prime}(\xi )-d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i) \tilde{\omega }(\xi -i)+(d_{S}+l)S_{-}(\xi ) -l\omega (\xi )+\beta \omega (\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds\nonumber \\[3pt] \leq & cS_{-}^{\prime}(\xi )-d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i) S_{-}(\xi -i)+(d_{S}+l)S_{-}(\xi )-lS_{-}(\xi )+\beta S_{-}(\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }(\xi -cs)ds\nonumber \\[3pt] \leq & cS_{-}^{\prime}(\xi )-d_{S}\sum _{i\in \tilde{\Omega }}J_{1}(i) S_{-}(\xi -i)+d_{S}S_{-}(\xi )+\beta S_{-}(\xi ) \int _{0}^{\tau }f(s)I_{+}(\xi -cs)ds. \end{align}

Since $I_{+}(\xi )=e^{\lambda _{1}\xi }$ and $S_{-}(\xi )\geq S_{0}-\delta e^{\nu \xi }$ for $\xi \in \mathbb{R}$ , direct calculation gives

\begin{align*} cS_{-}^{\prime}(\xi )-d_{S}\sum _{i\in \tilde {\Omega }}J_{1}(i) S_{-}(\xi -i)+d_{S}S_{-}(\xi )+\beta S_{-}(\xi ) \int _{0}^{\tau }f(s)I_{+}(\xi -cs)ds \leq 0, \end{align*}

for small $\nu$ and large $\delta$ . Then it follows from (52) that $ S_{-}(\xi )$ is a lower solution of (48). By comparison principle, one can obtain

\begin{align*} S_{-}(\xi )\leq \Gamma _{1}[ \omega, \sigma ] (\xi ), \text{ for } \xi \in [{-}Y,\displaystyle \frac{1}{\nu }\ln \frac{S_{0}}{\delta }). \end{align*}

On the other hand, when $ \xi \in [\displaystyle \frac{1}{\nu }\ln \frac{S_{0}}{\delta },Y],$ we have $S_{-}(\xi )=0$ . Thus,

\begin{align*} S_{-}(\xi )\leq \Gamma _{1}[ \omega, \sigma ] (\xi )\leq S_{0},\ \text{ for } \xi \in [{-}Y,Y] . \end{align*}

Similarly, we can obtain

\begin{align*} I_{-}(\xi )\leq \Gamma _{2}[ \omega, \sigma ] (\xi )\leq I_{+}(\xi ),\text{ for } \xi \in [{-}Y,Y] . \end{align*}

Hence, the inequalities of (51) hold.

Next, we show that $ \Gamma [\cdot, \cdot ]$ is continuous on $ \Phi _{Y}$ . Assume that $ (\omega _{i}(\xi ),\sigma _{i}(\xi ))\in \Phi _{Y}(i=1,2)$ and $ \Gamma _{2}[\omega _{i}, \sigma _{i}](\xi )=I_{i}(\xi ) (i=1,2)$ for $ \xi \in [{-}Y,Y]$ , one can verify that

\begin{align*} &c(I^\prime _{1}(\xi )-I^{\prime}_{2}(\xi ))+(d_{I}+\gamma )(I_{1}(\xi )-I_{2}(\xi ))=h(\xi ), \end{align*}

where

\begin{align*} h(\xi )\,:\!=\,&d_{I}\sum _{i\in \tilde{\Omega }}J_{2}(i) \left ( \tilde{\sigma }_{1}(\xi -i)- \tilde{\sigma }_{2}(\xi -i)\right ) +\\[3pt] &\beta \Big ( \omega _{1}(\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }_{1}(\xi -cs)ds - \omega _{2}(\xi ) \int _{0}^{\tau }f(s)\tilde{\sigma }_{2}(\xi -cs)ds \Big ) . \end{align*}

Thus,

(53) \begin{align} I _{1}(\xi )-I _{2}(\xi ) = \frac{1}{c}\int _{-Y}^{\xi } e^{\frac{d_{I}+\gamma }{c}(x-\xi )} h(x)dx, \end{align}

Note that for $ x\in [{-}Y,Y]$ , one has

\begin{align*} \left |\tilde{\sigma }_{1}(x-i)- \tilde{\sigma }_{2}(x-i)\right | \leq \max _{x\in [{-}Y,Y]}\left | \sigma _{1}(x)-\sigma _{2}(x)\right |\ \end{align*}

and

\begin{align*} & \Big | \omega _{1}(x) \int _{0}^{\tau }f(s)\tilde{\sigma }_{1}(x-cs)ds - \omega _{2}(x) \int _{0}^{\tau }f(s)\tilde{\sigma }_{2}(x-cs)ds\Big | \\[3pt] \leq &\left | \omega _{1}(x) - \omega _{2}(x) \right | \int _{0}^{\tau }f(s)\tilde{\sigma }_{1}(x-cs)ds + \omega _{2}(x) \int _{0}^{\tau }f(s)\left | \tilde{\sigma }_{1}(x-cs) - \tilde{\sigma }_{2}(x-cs)\right |ds \\[3pt] \leq & e^{\lambda _{1}Y} \max _{x\in [{-}Y,Y]}\left | \omega _{1}(x)-\omega _{2}(x)\right | + S_{0}\max _{x\in [{-}Y,Y]}\left | \sigma _{1}(x)-\sigma _{2}(x)\right |. \end{align*}

Then, for $ x\in [{-}Y,Y]$ , it follows that

(54) \begin{align} \left | h(x)\right | \leq \beta e^{\lambda _{1}Y} \max _{x\in [{-}Y,Y]}\left | \omega _{1}(x)-\omega _{2}(x)\right | + ( \beta S_{0}+d_{I})\max _{x\in [{-}Y,Y]}\left | \sigma _{1}(x)-\sigma _{2}(x)\right | . \end{align}

Combine (53) and (54), we have

\begin{align*} |I _{1}(\xi )-I _{2}(\xi )| \leq \frac {1}{c}\int _{-Y}^{\xi } e^{\frac {d_{I}+\gamma }{c}(x-\xi )} |h(x)|dx\leq \frac {\kappa }{d_{I}+\gamma }(1-e^{-\frac {d_{I}+\gamma }{c}(Y+\xi )})\leq \frac {\kappa }{d_{I}+\gamma }, \end{align*}

where

\begin{align*} \kappa \,:\!=\,\beta e^{\lambda _{1}Y} \max _{x\in [{-}Y,Y]}\left | \omega _{1}(x)-\omega _{2}(x)\right | + ( \beta S_{0}+d_{I})\max _{x\in [{-}Y,Y]}\left | \sigma _{1}(x)-\sigma _{2}(x)\right | . \end{align*}

Thus, $\Gamma _{2}$ is continuous on $ \Phi _{Y}$ . Similarly, $\Gamma _{1}$ is continuous on $ \Phi _{Y}$ . Furthermore, in view of (48) and (49), $ S^{\prime}(\xi )$ and $ I^{\prime}(\xi )$ are bounded on $ [{-}Y,Y]$ . Then, it can be deduced by the Arzela-Ascoli theorem that $\Gamma$ is compact. So, $\Gamma$ is completely continuous which maps from $ \Phi _{Y}$ to $ \Phi _{Y}$ .

Finally, by Schauder’s fixed point theorem, one can see that there exists a fixed point $ (S(\xi ),I(\xi ))\in \Phi _{Y}$ such that

\begin{align*} (S(\xi ),I(\xi ))=\Gamma [S,I] (\xi ),\ \mbox {for } \xi \in [{-}Y,Y]. \end{align*}

Clearly, $(S(\xi ),I(\xi ))$ satisfies (47) with $\mathcal{X}=Y-2-c\tau$ ,

\begin{align*}0\leq S_{-}(\xi )\leq S(\xi )\leq S_{0}\ \mbox { and }\ 0\leq I_{-}(\xi )\leq I(\xi )\leq I_{+}(\xi ).\end{align*}

The proof is complete.