Upper and lower solutions
First, we give the definition of upper and lower solutions of (2.1).
Definition 2
The pair of continuous functions \((\bar{S}(\xi ),\bar{I}(\xi ), \bar{R}(\xi ))\) and \((\underline{S}(\xi ),\underline{I}(\xi ), \underline{R}(\xi ))\) is called a pair of upper and lower solutions for (2.1) if they satisfy
(3.1)
(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
except for finitely many points of ξ on \(\mathbb{R}\).
For \(\xi \in \mathbb{R}\), define the following nonnegative continuous functions:
$$\begin{aligned}& \bar{S}(\xi ):=S_{0}, \\& \underline{S}(\xi ):=\textstyle\begin{cases} S_{0}(1-\sigma _{2}^{-1}e^{\sigma _{2}\xi }), \quad \xi < \xi _{2}, \\ 0, \quad \xi \geq \xi _{2}, \end{cases}\displaystyle \\& \bar{I}(\xi ):=\textstyle\begin{cases} -L_{1} \xi e^{\lambda ^{*} \xi }, \quad \xi < \xi _{1}, \\ \frac{1}{\alpha } (\frac{\beta S_{0}}{\gamma }-1 ),\quad \xi \geq \xi _{1}, \end{cases}\displaystyle \\& \underline{I}(\xi ):=\textstyle\begin{cases} -L_{1} \xi e^{\lambda ^{*} \xi }-L_{2}(-\xi )^{\frac{1}{2}}e^{\lambda ^{*} \xi },\quad \xi < \xi _{3}, \\ 0, \quad \xi \geq \xi _{3}, \end{cases}\displaystyle \\& \bar{R}(\xi ):=L_{3}e^{\sigma _{1} \xi },\qquad \underline{R}(\xi ):=0, \end{aligned}$$
where \(\lambda ^{*}\) is defined in Lemma 1.1,
$$ \xi _{1}=-\frac{1}{\lambda ^{*}},\qquad \xi _{2}= \frac{\ln \sigma _{2}}{\sigma _{2}},\qquad \xi _{3}=-\frac{L_{2}^{2}}{L_{1}^{2}},\qquad L_{1}=\frac{e\lambda ^{*}}{\alpha } \biggl(\frac{\beta S_{0}}{\gamma }-1 \biggr), $$
and the constants \(\sigma _{1}\), \(\sigma _{2}\), \(L_{2}\), and \(L_{3}\in \mathbb{R^{+}}\) are to be determined later. In the next lemma, we will prove that \((\bar{S}(\xi ),\bar{I}(\xi ),\bar{R}(\xi ))\) and \((\underline{S}(\xi ),\underline{I}(\xi ),\underline{R}(\xi ))\) are a pair of upper and lower solutions of (2.1).
Lemma 3.1
The function
\(\bar{S}(\xi )\)satisfies the inequality
$$ d_{1} \bar{S}''(\xi )-c^{*}\bar{S}'(\xi )-\frac{\beta \bar{S}(\xi ) \underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*} \tau )}\leq 0 $$
and the function
\(\underline{R}(\xi )\)satisfies the inequality
$$ d_{3} \underline{R}''(\xi )-c^{*} \underline{R}'(\xi )+\gamma \underline{I}(\xi )\geq 0 $$
for all
\(\xi \in \mathbb{R}\).
Proof
By the definitions of \(\bar{S}(\xi )\), \(\underline{R}(\xi )\), and \(\underline{I}(\xi )\) on \(\mathbb{R}\), one can get
$$ d_{1} \bar{S}''(\xi )-c^{*}\bar{S}'(\xi )-\frac{\beta \bar{S}(\xi ) \underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*} \tau )}=- \frac{\beta S_{0}\underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*}\tau )}\leq 0 $$
(3.7)
and
$$ d_{3} \underline{R}''( \xi )-c^{*} \underline{R}'(\xi )+\gamma \underline{I}( \xi )=\gamma \underline{I}(\xi )\geq 0, $$
(3.8)
which completes the proof. □
Lemma 3.2
The function
\(\bar{I}(\xi )\)satisfies the inequality
$$ d_{2} \bar{I}''(\xi )-c^{*}\bar{I}'(\xi )+\frac{\beta \bar{S}(\xi ) \bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c^{*}\tau )}-\gamma \bar{I}(\xi )\leq 0 $$
for all
\(\xi \neq \xi _{1}\).
Proof
When \(\xi <\xi _{1}\), \(\bar{I}(\xi )=-L_{1}\xi e^{\lambda ^{*}\xi }\), then it follows that
$$\begin{aligned} &d_{2} \bar{I}''(\xi )-c^{*}\bar{I}'(\xi )+\frac{\beta \bar{S}(\xi ) \bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c^{*}\tau )}-\gamma \bar{I}(\xi ) \\ &\quad \leq d_{2} \bar{I}''(\xi )-c^{*}\bar{I}'(\xi )+\beta S_{0}\bar{I} \bigl( \xi -c^{*}\tau \bigr)-\gamma \bar{I}(\xi ) \\ &\quad =-L_{1} e^{\lambda ^{*} \xi } \biggl[\Delta \bigl(\lambda ^{*},c^{*}\bigr)+\frac{ \partial \Delta }{\partial \lambda }\bigl(\lambda ^{*},c^{*}\bigr) \biggr]=0,\quad \xi < \xi _{1}. \end{aligned}$$
When \(\xi >\xi _{1}\), \(\bar{I}(\xi -c^{*}\tau )<\bar{I}(\xi )=\frac{1}{ \alpha } (\frac{\beta S_{0}}{\gamma }-1 )\), we have that
$$\begin{aligned} &d_{2} \bar{I}''(\xi )-c^{*}\bar{I}'(\xi )+\frac{\beta \bar{S}(\xi ) \bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c^{*}\tau )}-\gamma \bar{I}(\xi ) \\ &\quad =\frac{\beta \bar{S}(\xi )\bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c^{*}\tau )}-\gamma \bar{I}(\xi ) \\ &\quad \leq \frac{\beta {S}_{0}\bar{I}(\xi )}{1+\alpha \bar{I}(\xi )}- \gamma \bar{I}(\xi )=0. \end{aligned}$$
This is the end of the proof. □
Lemma 3.3
The function
\(\bar{R}(\xi )\)satisfies the inequality
$$ d_{3} \bar{R}''(\xi )-c^{*}\bar{R}'(\xi )+\gamma \bar{I}(\xi )\leq 0 $$
for all
\(\xi \in \mathbb{R}\).
Proof
Choose a sufficiently large \(L_{3}>0\) and a sufficiently small \(\sigma _{1} \in (0,\min \{\lambda ^{*},c^{*}/d_{3}\})\) such that
$$ d_{3} \sigma _{1}^{2}-c^{*} \sigma _{1}-\gamma L_{1} {L_{3}}^{-1} \xi e ^{(\lambda ^{*}-\sigma _{1})\xi }< 0, \quad \xi < \xi _{1} $$
(3.9)
and
$$ d_{3} \sigma _{1}^{2}-c^{*} \sigma _{1}+ \frac{\gamma }{\alpha L_{3}} \biggl(\frac{\beta S_{0}}{\gamma }-1 \biggr)e^{-\sigma _{1}\xi }< 0, \quad \xi \geq \xi _{1}. $$
(3.10)
When \(\xi <\xi _{1}\), \(\overline{I}(\xi )=-L_{1}\xi e^{\lambda ^{*} \xi }\). Then, by using (3.9), we obtain that
$$\begin{aligned} &d_{3} \bar{R}''(\xi )-c^{*}\bar{R}'(\xi )+\gamma \bar{I}(\xi ) \\ &\quad =d_{3} {\sigma _{1}}^{2} L_{3} e^{\sigma _{1} \xi }-c^{*}L_{3}\sigma _{1} e^{\sigma _{1} \xi }+\gamma (-L_{1})\xi e^{\lambda ^{*}\xi } \\ &\quad =L_{3} e^{\sigma _{1} \xi }\bigl[d_{3} {\sigma _{1}}^{2}-c^{*}\sigma _{1}- \gamma L_{1} {L_{3}}^{-1}\xi e^{(\lambda ^{*}-\sigma _{1})\xi }\bigr] \\ &\quad \leq 0,\quad \xi < \xi _{1}. \end{aligned}$$
When \(\xi \geq \xi _{1}\), \(\bar{I}(\xi )=\frac{1}{\alpha } (\frac{ \beta S_{0}}{\gamma }-1 )\). Using (3.10), we get that
$$\begin{aligned} &d_{3} \bar{R}''(\xi )-c^{*}\bar{R}'(\xi )+\gamma \bar{I}(\xi ) \\ &\quad =d_{3} {\sigma _{1}}^{2} L_{3} e^{\sigma _{1} \xi }-c^{*}L_{3}\sigma _{1} e^{\sigma _{1} \xi }+\frac{\gamma }{\alpha } \biggl( \frac{\beta S _{0}}{\gamma }-1 \biggr) \\ &\quad =L_{3}e^{\sigma _{1} \xi } \biggl[d_{3} {\sigma _{1}}^{2}-c^{*}\sigma _{1}+ \frac{\gamma }{\alpha L_{3}} \biggl(\frac{\beta S_{0}}{\gamma }-1 \biggr)e^{-\sigma _{1}\xi } \biggr]\leq 0, \quad \xi \geq \xi _{1}. \end{aligned}$$
The proof of this lemma is finished. □
Lemma 3.4
The function
\(\underline{S}(\xi )\)satisfies the inequality
$$ d_{1} \underline{S}''(\xi )-c^{*}\underline{S}'(\xi )-\frac{\beta \underline{S}(\xi )\bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c ^{*}\tau )}\geq 0 $$
for all
\(\xi \neq \xi _{2}\).
Proof
Select \(\sigma _{2}\) to be small enough such that \(\sigma _{2}\in (0, \min \{\lambda ^{*},c^{*}/d_{1}\})\), \(\xi _{2} <\xi _{1} \), and
$$ -d_{1} \sigma _{2}+c^{*}+ \beta {L_{1}}\bigl(\xi -c^{*}\tau \bigr) \bigl(1-{\sigma _{2}} ^{-1}e^{\sigma _{2} \xi }\bigr)e^{(\lambda ^{*}-\sigma _{2})\xi -\lambda ^{*} c ^{*} \tau }\geq 0,\quad \xi < \xi _{2}. $$
(3.11)
If \(\xi <\xi _{2}\), then \(\underline{S}(\xi ) =S_{0}(1-\sigma _{2}^{-1}e ^{\sigma _{2}\xi })\) and \(\bar{I}(\xi )=-L_{1} \xi e^{\lambda ^{*} \xi }\). Utilizing (3.11), we deduce that
$$\begin{aligned} &d_{1} \underline{S}''(\xi )-c^{*}\underline{S}'(\xi )-\frac{\beta \underline{S}(\xi )\bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c ^{*}\tau )} \\ &\quad \geq d_{1} \underline{S}''(\xi )-c^{*}\underline{S}'(\xi )-\beta \underline{S}(\xi ) \bar{I}\bigl(\xi -c^{*}\tau \bigr) \\ &\quad =-d_{1} S_{0}\sigma _{2} e^{\sigma _{2} \xi }+c^{*}S_{0}e^{\sigma _{2} \xi }+L_{1} \beta S_{0}\bigl(1-\sigma _{2}^{-1}e^{\sigma _{2}\xi } \bigr) \bigl(\xi -c ^{*}\tau \bigr)e^{\lambda ^{*}(\xi -c^{*}\tau )} \\ &\quad =S_{0} e^{\sigma _{2} \xi } \bigl[-d_{1} \sigma _{2}+c^{*}+\beta {L _{1}}\bigl(\xi -c^{*}\tau \bigr) \bigl(1-{\sigma _{2}}^{-1}e^{\sigma _{2} \xi } \bigr)e^{( \lambda ^{*}-\sigma _{2})\xi -\lambda ^{*} c^{*} \tau } \bigr] \\ &\quad \geq 0, \quad \xi < \xi _{2}. \end{aligned}$$
If \(\xi >\xi _{2}\), then \(\underline{S}(\xi )=0\) and
$$ d_{1} \underline{S}''(\xi )-c^{*}\underline{S}'(\xi )-\frac{\beta \underline{S}(\xi )\bar{I}(\xi -c^{*}\tau )}{1+\alpha \bar{I}(\xi -c ^{*}\tau )}=0 $$
holds naturally. □
Lemma 3.5
The function
\(\underline{I}(\xi )\)satisfies the inequality
$$ d_{2} \underline{I}''(\xi )-c^{*}\underline{I}'(\xi )+\frac{\beta \underline{S}(\xi )\underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*}\tau )}- \gamma \underline{I}(\xi )\geq 0 $$
for all
\(\xi \neq \xi _{3}\).
Proof
Choosing large enough \(L_{2}>0\) such that \(\xi _{3} < \min \{\xi _{2},-c ^{*}\tau \}\) and recalling \(\xi _{2} <\xi _{1} \), then for \(\xi <\xi _{3}\) we have that
$$\begin{aligned}& S_{0}\bigl(1-{\sigma _{2}}^{-1}e^{\sigma _{2}\xi } \bigr)< S_{0}, \end{aligned}$$
(3.12)
$$\begin{aligned}& \underline{I}(\xi )=\overline{I}(\xi )-L_{2} (-\xi )^{\frac{1}{2}}e ^{\lambda ^{*}\xi }, \end{aligned}$$
(3.13)
$$\begin{aligned}& 1+c^{*}\tau \xi ^{-1}>0, \end{aligned}$$
(3.14)
and
$$ \frac{1}{16} L_{2}\bigl(c^{*} \tau \bigr)^{2}-\sigma _{2}^{-1}L_{1}(- \xi )^{ \frac{5}{2}}e^{\sigma _{2}\xi } -\alpha L_{1}^{2}(- \xi )^{\frac{7}{2}}e ^{\lambda ^{*}(\xi -c^{*}\tau )}\geq 0. $$
(3.15)
From (3.13), for \(\xi <\xi _{3}\), we can get that
$$\begin{aligned} d_{2}\underline{I}''( \xi ) =&d_{2}\overline{I}''(\xi )+d_{2}L_{2} e ^{\lambda ^{*}\xi } \biggl[\lambda ^{*}(-\xi )^{-\frac{1}{2}}-\bigl(\lambda ^{*} \bigr)^{2}(-\xi )^{\frac{1}{2}}+\frac{1}{4}(-\xi )^{-\frac{3}{2}} \biggr] \\ \geq &d_{2}\overline{I}''(\xi )+d_{2}L_{2} e^{\lambda ^{*}\xi } \bigl[ \lambda ^{*}(-\xi )^{-\frac{1}{2}}-\bigl(\lambda ^{*} \bigr)^{2}(-\xi )^{ \frac{1}{2}} \bigr] \end{aligned}$$
(3.16)
and
$$\begin{aligned} -c^{*}\underline{I}'(\xi )-\gamma \underline{I}(\xi ) = &-c^{*} \overline{I}'( \xi )-c^{*}L_{2} \biggl[\frac{1}{2}(-\xi )^{-\frac{1}{2}}- \lambda ^{*}(-\xi )^{\frac{1}{2}} \biggr]e^{\lambda ^{*}\xi } \\ &{}-\gamma \overline{I}(\xi )+\gamma L_{2} (-\xi )^{\frac{1}{2}}e^{ \lambda ^{*}\xi }. \end{aligned}$$
(3.17)
Using the inequality \(\frac{x}{1+\alpha x}\geq x(1-\alpha x)\) for \(x\geq 0\) and \(\alpha >0\), we obtain from (3.12) that
$$\begin{aligned} &\frac{\beta \underline{S}(\xi )\underline{I}(\xi -c^{*}\tau )}{1+ \alpha \underline{I}(\xi -c^{*}\tau )} \\ &\quad \geq \beta \underline{S}(\xi )\underline{I}\bigl(\xi -c^{*} \tau \bigr)\bigl[1- \alpha \underline{I}\bigl(\xi -c^{*}\tau \bigr) \bigr] \\ &\quad \geq \beta \underline{S}(\xi )\underline{I}\bigl(\xi -c^{*} \tau \bigr)-\alpha \beta \underline{S}(\xi ) \bigl(\bar{I}\bigl(\xi -c^{*}\tau \bigr)\bigr)^{2} \\ &\quad \geq \beta S_{0}\bigl(1-\sigma _{2}^{-1}e^{\sigma _{2}\xi } \bigr)\bigl[\bar{I}\bigl( \xi -c^{*}\tau \bigr)-L_{2} \bigl(-\xi +c^{*}\tau \bigr)^{\frac{1}{2}}e^{\lambda ^{*}( \xi -c^{*}\tau )}\bigr] \\ & \qquad {} -\alpha \beta S_{0}L_{1}^{2} \xi ^{2}e^{2\lambda ^{*}(\xi -c^{*} \tau )} \\ &\quad \geq \beta S_{0}\bigl[\bar{I}\bigl(\xi -c^{*} \tau \bigr)-L_{2}\bigl(-\xi +c^{*}\tau \bigr)^{ \frac{1}{2}}e^{\lambda ^{*}(\xi -c^{*}\tau )} \\ &\qquad {} +\sigma _{2}^{-1}L_{1}\bigl(\xi -c^{*}\tau \bigr)e^{\sigma _{2}\xi + \lambda ^{*}(\xi -c^{*}\tau )}-\alpha L_{1}^{2} \xi ^{2}e^{2\lambda ^{*}( \xi -c^{*}\tau )}\bigr] \end{aligned}$$
(3.18)
for \(\xi <\xi _{3}\). By Taylor’s formula, for \(\xi <\xi _{3} \), we have that
$$ \bigl(-\xi +c^{*}\tau \bigr)^{\frac{1}{2}}\leq (-\xi )^{\frac{1}{2}}+ \frac{1}{2}(-\xi )^{-\frac{1}{2}}c^{*} \tau -\frac{1}{8}(-\xi )^{- \frac{3}{2}}\bigl(c^{*}\tau \bigr)^{2}+\frac{1}{16}(-\xi )^{-\frac{5}{2}} \bigl(c^{*} \tau \bigr)^{3}. $$
(3.19)
From (3.12)–(3.19), (1.4), and (1.5), we deduce that
$$\begin{aligned} & d_{2} \underline{I}''(\xi )-c^{*}\underline{I}'(\xi )+\frac{\beta \underline{S}(\xi )\underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*}\tau )}- \gamma \underline{I}(\xi ) \\ &\quad \geq d_{2}\overline{I}''(\xi )+d_{2}L_{2} e^{\lambda ^{*}\xi }\bigl[ \lambda ^{*}(-\xi )^{-\frac{1}{2}}-\bigl(\lambda ^{*} \bigr)^{2}(-\xi )^{ \frac{1}{2}}\bigr] \\ &\qquad {} -c^{*}\overline{I}'(\xi )-c^{*}L_{2} \biggl[\frac{1}{2}(- \xi )^{-\frac{1}{2}}-\lambda ^{*}(-\xi )^{\frac{1}{2}} \biggr]e^{\lambda ^{*}\xi } \\ & \qquad {} -\gamma \overline{I}(\xi )+\gamma L_{2} (-\xi )^{\frac{1}{2}}e ^{\lambda ^{*}\xi }+\beta S_{0}\bigl[\bar{I}\bigl( \xi -c^{*}\tau \bigr)-L_{2}\bigl(-\xi +c ^{*} \tau \bigr)^{\frac{1}{2}}e^{\lambda ^{*}(\xi -c^{*}\tau )} \\ &\qquad {} +\sigma _{2}^{-1}L_{1}\bigl(\xi -c^{*}\tau \bigr)e^{\sigma _{2}\xi + \lambda ^{*}(\xi -c^{*}\tau )}-\alpha L_{1}^{2} \xi ^{2}e^{2\lambda ^{*}( \xi -c^{*}\tau )} \bigr] \\ &\quad \geq -L_{1} e^{\lambda ^{*}\xi }\bigl[\xi \bigl(d_{2}\bigl(\lambda ^{*}\bigr)^{2}-c^{*} \lambda ^{*}-\gamma +\beta S_{0}e^{-\lambda ^{*}c^{*}\tau } \bigr)+\bigl(2d_{2} \lambda ^{*}-c^{*}-\beta S_{0} c^{*}\tau e^{-\lambda ^{*}c^{*}\tau }\bigr)\bigr] \\ &\qquad {} +L_{1}\beta S_{0}\bigl(\xi -c^{*}\tau \bigr)e^{\lambda ^{*}(\xi -c^{*} \tau )}-L_{2} e^{\lambda ^{*}\xi }(-\xi )^{\frac{1}{2}}\bigl[d_{2}\bigl(\lambda ^{*}\bigr)^{2}-c^{*}\lambda ^{*}- \gamma +\beta S_{0}e^{-\lambda ^{*}c^{*} \tau }\bigr] \\ &\qquad {} +L_{2} \beta S_{0}(-\xi )^{\frac{1}{2}}e^{\lambda ^{*}(\xi -c ^{*}\tau )}+ \frac{1}{2}L_{2} e^{\lambda ^{*}\xi }(-\xi )^{-\frac{1}{2}} \bigl[2d_{2}\lambda ^{*}-c^{*}-\beta S_{0} c^{*}\tau e^{-\lambda ^{*}c ^{*}\tau } \bigr] \\ &\qquad {} +\frac{1}{2}L_{2} \beta S_{0}c^{*} \tau (-\xi )^{-\frac{1}{2}}e ^{\lambda ^{*}(\xi -c^{*}\tau )}+\beta S_{0} \bigl[ \bar{I}\bigl(\xi -c^{*} \tau \bigr)-L_{2}\bigl(-\xi +c^{*}\tau \bigr)^{\frac{1}{2}}e^{\lambda ^{*}(\xi -c^{*} \tau )} \\ &\qquad {} +\sigma _{2}^{-1}L_{1}\bigl(\xi -c^{*}\tau \bigr)e^{\sigma _{2}\xi + \lambda ^{*}(\xi -c^{*}\tau )}-\alpha L_{1}^{2} \xi ^{2}e^{2\lambda ^{*}( \xi -c^{*}\tau )} \bigr] \\ &\quad \geq -L_{1} e^{\lambda ^{*}\xi } \biggl[\Delta \bigl(\lambda ^{*},c^{*}\bigr)+\frac{ \partial \Delta }{\partial \lambda }\bigl(\lambda ^{*},c^{*}\bigr) \biggr]-L_{2} e ^{\lambda ^{*}\xi }(-\xi )^{\frac{1}{2}}\bigl[\Delta \bigl(\lambda ^{*},c^{*}\bigr)\bigr] \\ &\qquad {} +\frac{1}{2}L_{2} e^{\lambda ^{*}\xi }(-\xi )^{\frac{1}{2}} \biggl[\frac{\partial \Delta }{\partial \lambda }\bigl(\lambda ^{*},c^{*} \bigr) \biggr] \\ &\qquad {}+\beta S_{0} \biggl\{ L_{2}e^{\lambda ^{*}(\xi -c^{*}\tau )} \biggl[ \frac{1}{8}(-\xi )^{-\frac{3}{2}}\bigl(c^{*}\tau \bigr)^{2}-\frac{1}{16}(- \xi )^{-\frac{5}{2}} \bigl(c^{*}\tau \bigr)^{3} \biggr] \\ & \qquad {} +\sigma _{2}^{-1}L_{1}\xi e^{\sigma _{2}\xi +\lambda ^{*}(\xi -c ^{*}\tau )}-\alpha L_{1}^{2}\xi ^{2}e^{2\lambda ^{*}(\xi -c^{*}\tau )} \biggr\} \\ &\quad =\beta S_{0}(-\xi )^{-\frac{3}{2}}e^{\lambda ^{*}(\xi -c^{*}\tau )} \biggl[\frac{1}{16} L_{2}\bigl(c^{*}\tau \bigr)^{2}-\sigma _{2}^{-1}L_{1}(- \xi )^{ \frac{5}{2}}e^{\sigma _{2}\xi } -\alpha L_{1}^{2}(- \xi )^{\frac{7}{2}}e ^{\lambda ^{*}(\xi -c^{*}\tau )} \biggr] \\ &\qquad {} +\frac{1}{16}\beta S_{0}L_{2} \bigl(c^{*}\tau \bigr)^{2}(-\xi )^{- \frac{3}{2}}e^{\lambda ^{*}(\xi -c^{*}\tau )} \bigl(1+c^{*}\tau \xi ^{-1}\bigr) \\ &\quad \geq 0. \end{aligned}$$
If \(\xi >\xi _{3}\), then \(\underline{I}(\xi )=0\), and the inequality
$$ d_{2} \underline{I}''(\xi )-c^{*}\underline{I}'(\xi )+\frac{\beta \underline{S}(\xi )\underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*}\tau )}- \gamma \underline{I}(\xi )=\frac{ \beta \underline{S}(\xi )\underline{I}(\xi -c^{*}\tau )}{1+\alpha \underline{I}(\xi -c^{*}\tau )}\geq 0 $$
holds trivially. The proof of this lemma is finished. □
Application of Schauder’s fixed point theorem
Introduce a functional space
$$ B_{\mu }\bigl(\mathbb{R},\mathbb{R}^{3}\bigr):=\Bigl\{ \varphi (\xi )=\bigl(\varphi _{1}( \xi ),\varphi _{2}(\xi ),\varphi _{3}(\xi )\bigr)\in C\bigl(\mathbb{R},\mathbb{R} ^{3}\bigr):\sup_{\xi \in \mathbb{R}} \bigl\vert \varphi _{i}(\xi ) \bigr\vert e^{-\mu \vert \xi \vert }< \infty ,i=1,2,3\Bigr\} $$
equipped with the norm \(| \varphi | _{\mu }:=\max \{ \sup_{\xi \in \mathbb{R}}| \varphi _{i}(\xi )| e^{-\mu | \xi | },i=1,2,3\}\), where \(\mu \in (\sigma _{1},\mu _{0})\) is a constant and \(\mu _{0}\) is also a constant that will be specified later. Define a cone by
$$ \mathcal{S} := \left\{ \bigl(S(\xi ),I(\xi ),R(\xi )\bigr)\in B_{\mu }\bigl( \mathbb{R},\mathbb{R}^{3}\bigr)\left|\vphantom{\textstyle\begin{array}{l} \underline{S}(\xi )\leq S(\xi )\leq \bar{S}(\xi ), \\ \underline{I}(\xi )\leq I(\xi )\leq \bar{I}(\xi ), \\ \underline{R}(\xi )\leq R(\xi )\leq \bar{R}(\xi ) \end{array}\displaystyle }\right. \textstyle\begin{array}{l} \underline{S}(\xi )\leq S(\xi )\leq \bar{S}(\xi ), \\ \underline{I}(\xi )\leq I(\xi )\leq \bar{I}(\xi ), \\ \underline{R}(\xi )\leq R(\xi )\leq \bar{R}(\xi ) \end{array}\displaystyle \right\} . $$
It is easy to see that \(\mathcal{S}\) is nonempty, bounded, closed, and convex in \(B_{\mu }(\mathbb{R},\mathbb{R}^{3})\). Choosing a constant m to be satisfied \(m > \max \{\alpha ^{-1}\beta ,\beta \}\) and noting that \(\beta >\gamma \), one can get that:
$$ H_{1}[S,I,R](\xi ):=m S(\xi )-\frac{\beta S(\xi )I(\xi -c^{*}\tau )}{1+ \alpha I(\xi -c^{*}\tau )} $$
is increasing with respect to S and decreasing with respect to I;
$$ H_{2}[S,I,R](\xi ):=\frac{\beta S(\xi )I(\xi -c^{*}\tau )}{1+\alpha I( \xi -c^{*}\tau )}+(m-\gamma )I(z) $$
is increasing in both S and I;
$$ H_{3}[S,I,R](\xi ):=m R(\xi )+\gamma I(\xi ) $$
is increasing in both I and R. For any \((S,I,R)\in \mathcal{S}\), define a nonlinear operator \(\mathcal{M}:=(\mathcal{M}_{1}, \mathcal{M}_{2},\mathcal{M}_{3})\) on the space \(B_{\mu }(\mathbb{R}, \mathbb{R}^{3})\) by
$$ \mathcal{M}_{i}[S,I,R](\xi ):=\frac{1}{\varLambda _{i}} \biggl\{ \int _{- \infty }^{\xi }e^{r_{i1}(\xi -\eta )}H_{i}[S,I,R]( \eta )\,d\eta + \int _{\xi }^{\infty }e^{r_{i2}(\xi -\eta )}H_{i}[S,I,R]( \eta )\,d\eta \biggr\} , $$
where
$$ r_{i1}=\frac{c^{*}-\sqrt{(c^{*})^{2}+4md_{i}}}{2d_{i}},\qquad r_{i2}= \frac{c ^{*}+\sqrt{(c^{*})^{2}+4md_{i}}}{2d_{i}}, \qquad \varLambda _{i}=d_{i}(r _{i2}-r_{i1}) $$
for \(i=1,2,3\). Note that any fixed point of \(\mathcal{M}\) is a solution of (2.1).
Lemma 3.6
\(\mathcal{M}(\mathcal{S})\subset \mathcal{S}\).
Proof
Clearly, \((\mathcal{M}_{1}[S,I,R](\xi ),\mathcal{M}_{2}[S,I,R](\xi ), \mathcal{M}_{3}[S,I,R](\xi ))\in B_{\mu }(\mathbb{R},\mathbb{R}^{3})\) for any \((S,I,R)\in \mathcal{S}\). Then, by the monotonicity of \(H_{i}\) (\(i=1,2,3\)), we need to prove that
(3.20)
(3.21)
(3.22)
for any \((S,I,R)\in \mathcal{S}\).
Proof of (3.20). Using (3.1) and \(\bar{S}(\xi )=S_{0}\), we derive that
$$\begin{aligned} &\mathcal{M}_{1}[\bar{S},\underline{I},R](\xi ) \\ &\quad =\frac{1}{\varLambda _{1}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}H_{1}[ \bar{S},\underline{I},R](\eta )\,d\eta + \int _{\xi } ^{\infty }e^{r_{12}(\xi -\eta )}H_{1}[ \bar{S},\underline{I},R](\eta )\,d\eta \biggr\} \\ &\quad \leq \frac{1}{\varLambda _{1}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}\bigl[m \bar{S}(\eta )+c^{*}\bar{S}'(\eta )-d_{1} \bar{S}''( \eta )\bigr]\,d\eta \\ &\qquad {} + \int _{\xi }^{\infty }e^{r_{12}(\xi -\eta )}\bigl[m \bar{S}( \eta )+c ^{*}\bar{S}'(\eta )-d_{1} \bar{S}''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad = \frac{m S_{0}}{\varLambda _{1}} \biggl[ \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}\,d\eta + \int _{\xi }^{\infty }e^{r_{12}(\xi -\eta )}\,d\eta \biggr] \\ &\quad =S_{0}, \quad \xi \in \mathbb{R}. \end{aligned}$$
It follows from (3.4) that
$$\begin{aligned} &\mathcal{M}_{1}[\underline{S},\bar{I},R](\xi ) \\ &\quad =\frac{1}{\varLambda _{1}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}H_{1}[ \underline{S},\bar{I},R](\eta )\,d\eta + \int _{\xi } ^{\infty }e^{r_{12}(\xi -\eta )}H_{1}[ \underline{S},\bar{I},R](\eta )\,d\eta \biggr\} \\ &\quad \geq \frac{1}{\varLambda _{1}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}\bigl[m \underline{S}(\eta )+c^{*}\underline{S}'(\eta )-d_{1} \underline{S}''(\eta )\bigr]\,d\eta \\ & \qquad {} + \int _{\xi }^{\infty }e^{r_{12}(\xi -\eta )}\bigl[m \underline{S}( \eta )+c^{*}\underline{S}'(\eta )-d_{1} \underline{S}-''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad = \underline{S}(\xi )+\frac{e^{r_{11}(\xi -\xi _{2})}[\underline{S}'( \xi _{2}+0)-\underline{S}'(\xi _{2}-0)]}{r_{12}-r_{11}} \\ &\quad \geq \underline{S}(\xi ), \quad \xi \neq \xi _{2}. \end{aligned}$$
By the continuity of both \(\mathcal{M}_{1}[\underline{S},\bar{I},R]( \xi )\) and \(\underline{S}(\xi )\) on \(\mathbb{R}\), we get that
$$ \mathcal{M}_{1}[\underline{S},\bar{I},R](\xi )\geq \underline{S}( \xi ), \quad \xi \in \mathbb{R}. $$
Proof of (3.21). From (3.2) and (3.5), we get that
$$\begin{aligned} &\mathcal{M}_{2}[\bar{S},\bar{I},R](\xi ) \\ &\quad =\frac{1}{\varLambda _{2}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}H_{2}[ \bar{S},\bar{I},R](\eta )\,d\eta + \int _{\xi }^{\infty }e^{r_{22}(\xi -\eta )}H_{2}[ \bar{S},\bar{I},R](\eta )\,d\eta \biggr\} \\ &\quad \leq \frac{1}{\varLambda _{2}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}\bigl[m \bar{I}(\eta )+c^{*}\bar{I}'(\eta )-d_{2} \bar{I}''( \eta )\bigr]\,d\eta \\ &\qquad {} + \int _{\xi }^{\infty }e^{r_{22}(\xi -\eta )}\bigl[m \bar{I}( \eta )+c ^{*}\bar{I}'(\eta )-d_{2} \bar{I}''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad = \bar{I}(\xi )+\frac{e^{r_{21}(\xi -\xi _{1})}[\bar{I}'(\xi _{1}+0)- \bar{I}'(\xi _{1}-0)]}{r_{22}-r_{21}} \\ &\quad \leq \bar{I}(\xi ), \quad \xi \neq \xi _{1}, \end{aligned}$$
and
$$\begin{aligned} &\mathcal{M}_{2}[\underline{S},\underline{I},R](\xi ) \\ &\quad =\frac{1}{\varLambda _{2}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}H_{2}[ \underline{S},\underline{I},R](\eta )\,d\eta + \int _{\xi }^{\infty }e^{r_{22}(\xi -\eta )}H_{2}[ \underline{S}, \underline{I},R](\eta )\,d\eta \biggr\} \\ &\quad \geq \frac{1}{\varLambda _{2}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}\bigl[m \underline{I}(\eta )+c^{*}\underline{I}'(\eta )-d_{2} \underline{I}''(\eta )\bigr]\,d\eta \\ & \qquad {} + \int _{\xi }^{\xi _{3}}e^{r_{22}(\xi -\eta )}\bigl[m \underline{I}( \eta )+c^{*}\underline{I}'(\eta )-d_{2}\underline{I}''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad =\bar{I}(\xi )+\frac{e^{r_{21}(\xi -\xi _{3})}[\bar{I}'(\xi _{3}+0)- \bar{I}'(\xi _{3}-0)]}{r_{22}-r_{21}} \\ &\quad \geq \underline{I}(\xi ), \quad \xi \neq \xi _{3}. \end{aligned}$$
Using the continuity of both \(\mathcal{M}_{2}[\bar{S},\bar{I},R]( \xi )\), \(\mathcal{M}_{2}[\underline{S},\underline{I},R](\xi ) \), \(\bar{I}( \xi ) \) and \(\underline{I}(\xi )\) on \(\mathbb{R}\), we obtain
$$ \mathcal{M}_{2}[\bar{S},\bar{I},R](\xi )\leq \bar{I}(\xi ),\qquad \mathcal{M}_{2}[\underline{S},\underline{I},R](\xi )\geq \underline{I}(\xi ), \quad \xi \in \mathbb{R}. $$
Proof of (3.22). From (3.3), (3.6), and the expressions of \(\bar{R}(\xi )\) and \(\underline{R}(\xi )\), we deduce that
$$\begin{aligned} &\mathcal{M}_{3}[S,\bar{I},\bar{R}](\xi ) \\ &\quad =\frac{1}{\varLambda _{3}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{31}( \xi -\eta )}H_{3}[S, \bar{I},\bar{R}](\eta )\,d\eta + \int _{\xi }^{\infty }e^{r_{32}(\xi -\eta )}H_{3}[S, \bar{I},\bar{R}](\eta )\,d\eta \biggr\} \\ &\quad \leq \frac{1}{\varLambda _{3}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{31}( \xi -\eta )}\bigl[m \bar{R}(\eta )+c^{*}\bar{R}'(\eta )-d_{3} \bar{R}''( \eta )\bigr]\,d\eta \\ & \qquad {} + \int _{\xi }^{\infty }e^{r_{32}(\xi -\eta )}\bigl[m \bar{R}( \eta )+c ^{*}\bar{R}'(\eta )-d_{3} \bar{R}''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad = \bar{R}(\xi ),\quad \xi \in \mathbb{R}, \end{aligned}$$
and
$$\begin{aligned} &\mathcal{M}_{3}[S,\underline{I},\underline{R}](\xi ) \\ &\quad =\frac{1}{\varLambda _{3}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{31}( \xi -\eta )}H_{3}[S, \underline{I},\underline{R}](\eta )\,d\eta + \int _{\xi }^{\infty }e^{r_{32}(\xi -\eta )}H_{3}[S, \underline{I}, \underline{R}](\eta )\,d\eta \biggr\} \\ &\quad \geq \frac{1}{\varLambda _{3}} \biggl\{ \int _{-\infty }^{\xi }e^{r_{31}( \xi -\eta )}\bigl[m \underline{R}(\eta )+c^{*}\underline{R}'(\eta )-d_{3} \underline{R}''(\eta )\bigr]\,d\eta \\ & \qquad {} + \int _{\xi }^{\infty }e^{r_{32}(\xi -\eta )}\bigl[m \underline{R}( \eta )+c^{*}\underline{R}'(\eta )-d_{3}\underline{R}''(\eta )\bigr]\,d\eta \biggr\} \\ &\quad = \underline{R}(\xi ), \quad \xi \in \mathbb{R}. \end{aligned}$$
The proof of this lemma is finished. □
Lemma 3.7
The operator
\(\mathcal{M}:=(\mathcal{M}_{1},\mathcal{M}_{2}, \mathcal{M}_{3})\)is completely continuous with respect to the norm
\(|\cdot |_{\mu }\)in
\(B_{\mu }(\mathbb{R},\mathbb{R}^{3})\).
Proof
First, we show that \(\mathcal{M}\) is continuous with respect to the norm \(|\cdot |_{\mu }\) in \(B_{\mu }(\mathbb{R},\mathbb{R}^{3})\). For any \(\varPsi _{1}=(S_{1},I_{1},R_{1})\in \mathcal{S}\) and \(\varPsi _{2}=(S_{2},I _{2},R_{2})\in \mathcal{S}\), we derive that
$$\begin{aligned} & \bigl\vert H_{1}(S_{1},I_{1},R_{1}) (\xi )-H_{1}(S_{2},I_{2},R_{2}) (\xi ) \bigr\vert e^{- \mu \vert \xi \vert } \\ &\quad \leq \biggl(m+\frac{\beta }{\alpha } \biggr) \bigl\vert S_{1}(\xi )-S_{2}(\xi ) \bigr\vert e ^{-\mu \vert \xi \vert }+\beta S_{0} \bigl\vert I_{1}\bigl( \xi -c^{*}\tau \bigr)-I_{2}\bigl(\xi -c^{*} \tau \bigr) \bigr\vert e^{-\mu \vert \xi \vert } \\ &\quad \leq \biggl(m+\frac{\beta }{\alpha } \biggr) \vert S_{1}-S_{2} \vert _{\mu }+ \beta S_{0} e^{\mu c^{*}\tau } \vert I_{1}-I_{2} \vert _{\mu } \\ &\quad \leq l \vert \varPsi _{1}-\varPsi _{2} \vert _{\mu }, \\ & \bigl\vert H_{2}(S_{1},I_{1},R_{1}) (\xi )-H_{2}(S_{2},I_{2},R_{2}) ( \xi ) \bigr\vert e^{- \mu \vert \xi \vert } \\ &\quad \leq \frac{\beta }{\alpha } \vert S_{1}-S_{2} \vert _{\mu }+\bigl(m+\beta S_{0}e^{\mu c^{*}\tau }- \gamma \bigr) \vert I_{1}-I_{2} \vert _{\mu } \\ &\quad \leq l \vert \varPsi _{1}-\varPsi _{2} \vert _{\mu }, \end{aligned}$$
and
$$\begin{aligned} & \bigl\vert H_{3}(S_{1},I_{1},R_{1}) (\xi )-H_{3}(S_{2},I_{2},R_{2}) ( \xi ) \bigr\vert e^{- \mu \vert \xi \vert } \\ &\quad \leq m \vert R_{1}-R_{2} \vert _{\mu }+\gamma \vert I_{1}-I_{2} \vert _{\mu } \\ &\quad \leq l \vert \varPsi _{1}-\varPsi _{2} \vert _{\mu }, \end{aligned}$$
where \(l=m+\frac{\beta }{\alpha }+\gamma +\beta S_{0} e^{\mu c^{*} \tau }\). Then, choosing \(\mu \in (\sigma _{2},-r_{i1})\), we have that
$$\begin{aligned} & \bigl\vert \mathcal{M}_{i}[S_{1},I_{1},R_{1}]( \xi )-\mathcal{M}_{i}[S_{2},I _{2},R_{2}]( \xi ) \bigr\vert e^{-\mu \vert \xi \vert } \\ &\quad \leq \frac{1}{\varLambda _{i}} \bigl\vert H_{i}(S_{1},I_{1},R_{1})-H_{i}(S_{2},I _{2},R_{2}) \bigr\vert _{\mu } \biggl[ \int _{-\infty }^{\xi }e^{r_{i1}(\xi -\eta )}e ^{\mu \vert \eta \vert -\mu \vert \xi \vert }\,d\eta \\ & \qquad {} + \int _{\xi }^{\infty }e^{r_{i2}(\xi -\eta )}e^{\mu \vert \eta \vert - \mu \vert \xi \vert }\,d\eta \biggr] \\ &\quad \leq \frac{l}{\varLambda _{i}} \vert \varPsi _{1}-\varPsi _{2} \vert _{\mu } \biggl[ \int _{-\infty }^{\xi }e^{r_{i1}(\xi -\eta )}e^{\mu \vert \eta -\xi \vert }\,d\eta + \int _{\xi }^{\infty }e^{r_{i2}(\xi -\eta )}e^{\mu \vert \eta -\xi \vert }\,d\eta \biggr] \\ &\quad =\frac{l(2\mu +r_{i1}-r_{i2})}{d_{i}(r_{i2}-r_{i1})(r_{i2}-\mu )(r _{i1}+\mu )} \vert \varPsi _{1}-\varPsi _{2} \vert _{\mu }, \quad i=1,2,3, \end{aligned}$$
which implies that \(\mathcal{M}\) is continuous with respect to the norm \(|\cdot |_{\mu }\) in \(B_{\mu }(\mathbb{R},\mathbb{R}^{3})\).
Now we turn to proving that \(\mathcal{M}\) is compact with respect to the norm \(|\cdot |_{\mu }\) in \(B_{\mu }(\mathbb{R},\mathbb{R}^{3})\). For any \((S,I,R)\in \mathcal{S}\), we deduce for \(\xi \in \mathbb{R}\) that
$$\begin{aligned} & \biggl\vert \frac{d\mathcal{M}_{1}[S,I,R](\xi )}{d\xi } \biggr\vert \\ &\quad = \biggl\vert \frac{r _{11}}{\varLambda _{1}} \int _{-\infty }^{\xi }e^{r _{11}(\xi -\eta )}H_{1}[S,I,R]( \eta )\,d\eta + \frac{r_{12}}{\varLambda _{1}} \int _{\xi }^{\infty }e^{r_{12}(\xi -\eta )}H _{1}[S,I,R](\eta )\,d\eta \biggr\vert \\ &\quad \leq -\frac{r_{11}}{\varLambda _{1}} \int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}H_{1}[S,I,R]( \eta )\,d\eta +\frac{r_{12}}{\varLambda _{1}} \int _{\xi }^{\infty }e^{r_{12}(\xi -\eta )}H_{1}[S,I,R]( \eta )\,d\eta \\ &\quad \leq -\frac{r_{11}m S_{0}}{\varLambda _{1}} \int _{-\infty }^{\xi }e^{r _{11}(\xi -\eta )}\,d\eta + \frac{r_{12}m S_{0}}{\varLambda _{1}} \int _{ \xi }^{\infty }e^{r_{12}(\xi -\eta )}\,d\eta \\ &\quad =\frac{2m S_{0}}{\varLambda _{1}}, \end{aligned}$$
(3.23)
$$\begin{aligned} & \biggl\vert \frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } \biggr\vert = \biggl\vert \frac{r _{21}}{\varLambda _{2}} \int _{-\infty }^{\xi }e^{r_{21}(\xi -\eta )}H_{2}[S,I,R]( \eta )\,d\eta \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |=}{} +\frac{r_{22}}{\varLambda _{2}} \int _{\xi }^{\infty }e^{r_{22}( \xi -\eta )}H_{2}[S,I,R]( \eta )\,d\eta \biggr\vert \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |}\leq -\frac{r_{21}}{\varLambda _{2}} \int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}H_{2}[S,I,R]( \eta )\,d\eta \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |=}{} +\frac{r_{22}}{\varLambda _{2}} \int _{\xi }^{\infty }e^{r_{22}( \xi -\eta )}H_{2}[S,I,R]( \eta )\,d\eta \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |}\leq -\frac{r_{21}(m+\beta S_{0}-\gamma )\bar{I}}{\varLambda _{2}} \int _{-\infty }^{\xi }e^{r_{21}(\xi -\eta )}\,d\eta \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |=}{} +\frac{r_{22}(m+\beta S_{0}-\gamma )\bar{I}}{\varLambda _{2}} \int _{\xi }^{\infty }e^{r_{22}(\xi -\eta )}\,d\eta \\ &\hphantom{ |\frac{d\mathcal{M}_{2}[S,I,R](\xi )}{d\xi } |}=\frac{2(m+\beta S_{0}-\gamma )\bar{I}}{\varLambda _{2}}, \end{aligned}$$
(3.24)
and
$$\begin{aligned} \biggl\vert \frac{d\mathcal{M}_{3}[S,I,R](\xi )}{d\xi } \biggr\vert =& \biggl\vert \frac{r _{31}}{\varLambda _{3}} \int _{-\infty }^{\xi }e^{r_{31}(\xi -\eta )}H_{3}[S,I,R]( \eta )\,d\eta \\ &{} +\frac{r_{32}}{\varLambda _{3}} \int _{\xi }^{\infty }e^{r_{32}( \xi -\eta )}H_{3}[S,I,R]( \eta )\,d\eta \biggr\vert \\ \leq& -\frac{r_{31}}{\varLambda _{3}} \int _{-\infty }^{\xi }e^{r_{31}( \xi -\eta )}\bigl(m L_{3} e^{\sigma _{1}\eta }+\gamma \bar{I}\bigr)\,d\eta \\ &{} +\frac{r_{32}}{\varLambda _{3}} \int _{\xi }^{\infty }e^{r_{32}( \xi -\eta )}\bigl(m L_{3} e^{\sigma _{1}\eta }+\gamma \bar{I}\bigr)\,d\eta \\ \leq& \frac{m L_{3} \vert 2r_{31}r_{32}-\sigma _{1}(r_{31}+r_{32}) \vert }{\varLambda _{3} \vert (\sigma _{1}-r_{31})(\sigma _{1}-r_{32}) \vert }e ^{\sigma _{1} \xi }+\frac{2\gamma \bar{I}}{\varLambda _{3}}. \end{aligned}$$
(3.25)
It follows from Lemma 3.6 that \(|\mathcal{M}_{1}[S,I,R](\xi ) |+ |\mathcal{M}_{2}[S,I,R](\xi ) |+ |\mathcal{M}_{3}[S,I,R]( \xi ) |\leq S_{0}+\bar{I}+L_{3} e^{\sigma _{1}\xi }\) on \(\mathbb{R}\). Recall that \(\mu >\sigma _{1}\). Then, for any \(\varepsilon >0\), there is a sufficiently large number \(N>0\) such that
$$\begin{aligned} & \bigl\{ \bigl\vert \mathcal{M}_{1}[S,I,R]( \xi ) \bigr\vert + \bigl\vert \mathcal{M}_{2}[S,I,R]( \xi ) \bigr\vert + \bigl\vert \mathcal{M}_{3}[S,I,R](\xi ) \bigr\vert \bigr\} e^{-\mu \vert \xi \vert } \\ & \quad \leq \bigl(S_{0}+\bar{I}+L_{3} e^{\sigma _{1}\xi }\bigr)e^{-\mu \vert \xi \vert } \\ &\quad < (S_{0}+\bar{I})e^{-\mu N}+L_{3} e^{(\sigma _{1}-\mu )N} \\ & \quad < \varepsilon , \quad \vert \xi \vert >N. \end{aligned}$$
(3.26)
Utilizing (3.23)–(3.25) and Arzerà–Ascoli theorem, we can select finite elements in \(\mathcal{M}(\mathcal{S})\) such that they are a finite ε-net of \(\mathcal{M}(\mathcal{S})( \xi )\) on \([-N,N]\) with the supremum norm, a finite ε-net of \(\mathcal{M}(\mathcal{S})(\xi )\) on \(\mathbb{R}\) with the norm \(|\cdot |_{\mu }\) (see (3.26)). Thus \(\mathcal{M}\) is compact with respect to the norm \(|\cdot |_{\mu }\) in \(B_{\mu }(\mathbb{R}, \mathbb{R}^{3})\). The proof of this lemma is completed. □
By Lemma 3.6, Lemma 3.7, and Schauder’s fixed point theorem, we deduce that the operator \(\mathcal{M}\) has a fixed point \((S(\xi ),I(\xi ),R( \xi ))\in \mathcal{S}\), which is a solution of the system
$$ \textstyle\begin{cases} d_{1} S''(\xi )-c^{*}S'(\xi )-\frac{\beta S(\xi )I(\xi -c^{*}\tau )}{1+ \alpha I(\xi -c^{*}\tau )}=0, \\ d_{2} I''(\xi )-c^{*}I'(\xi )+\frac{\beta S(\xi )I(\xi -c^{*}\tau )}{1+ \alpha I(\xi -c^{*}\tau )}-\gamma I(\xi )=0, \\ d_{3} R''(\xi )-c^{*}R'(\xi )+\gamma I(\xi )=0. \end{cases} $$
(3.27)
Based on the above analysis, we have the following results.
Proposition 3.1
If
\(\mathcal{R}_{0} >1\)and
\(c=c^{*}\), then system (1.1) admits a traveling wave solution
\((S(\xi ),I(\xi ),R(\xi ))\)such that
$$ \underline{S}(\xi )\leq S(\xi )\leq \bar{S}(\xi ),\qquad \underline{I}(\xi )\leq I(\xi )\leq \bar{I}(\xi ),\qquad \underline{R}(\xi )\leq R(\xi )\leq \bar{R}(\xi ), \quad \xi \in \mathbb{R}. $$
(3.28)
Properties of the critical traveling wave solutions
In this section, we focus on some properties of the critical traveling wave solution of (2.1), that is, the proof of the four properties in Theorem 2.1.
Proof
(1) By contradiction, suppose that \(I(\hat{\xi })=0\) for some \(\hat{\xi }\in \mathbb{R}\). Then there are two constants \(a,b\in \mathbb{R}\) such that \(a<\xi _{3}\leq b\) and \(a<\hat{\xi }<b\), which implies that \(I(\xi )\) attains its minimum in \((a,b)\). It follows from the second equation in (3.27) that \(-d_{2} I''(\xi )+c^{*}I'( \xi )+\gamma I(\xi )\geq 0\) for \(\xi \in [a,b]\). By the strong maximum principle, we deduce that \(I(\xi )\equiv 0\) for \(\xi \in [a,b]\), which contradicts the fact that \(I(\xi )\geq I_{-}(\xi )>0\) for \(\xi \in [a, \xi _{3})\). Thus, \(I(\xi )>0\) on \(\mathbb{R}\). Similarly, one can obtain \(S(\xi )>0\) on \(\mathbb{R}\). Assume that \(R(\tilde{\xi })=0\) for some \(\tilde{\xi }\in \mathbb{R}\), then \(R'(\tilde{\xi })=0\) and \(R''(\tilde{\xi })\geq 0\). We infer from the third equation in (3.28) that \(I(\tilde{\xi })\leq 0\), which contradicts the positiveness of \(I(\xi )\) on \(\mathbb{R}\). This implies that \(R(\xi )>0\) on \(\mathbb{R}\).
(2) From (3.28), we get
$$ \begin{aligned} &S_{0}\bigl(1-\sigma _{2}^{-1}e^{\sigma _{2}\xi } \bigr)\leq S(\xi )\leq S_{0}, \\ &\bigl[-L_{1} \xi -L_{2}(-\xi )^{\frac{1}{2}} \bigr]e^{\lambda ^{*} \xi }\leq I( \xi )\leq -L_{1} \xi e^{\lambda ^{*}\xi }, \\ &0\leq R(\xi )\leq L_{3}e^{\sigma _{1} \xi },\quad \xi \in \mathbb{R}. \end{aligned} $$
Then using the squeeze rule yields that
$$ S(-\infty )=S_{0}, \qquad I(-\infty )=0, \qquad R(-\infty )=0\quad \text{and} \quad I(\xi )=O\bigl(-\xi e^{\lambda ^{*}\xi } \bigr) $$
(3.29)
as \(\xi \rightarrow -\infty \).
(3) Since \(S(\xi )\) and \(I(\xi )\) are uniformly bounded on \(\mathbb{R}\), we have from the first two equations in (3.27) that
$$ \textstyle\begin{cases} S(\xi )=\frac{1}{\varLambda _{1}} \{\int _{-\infty }^{\xi }e^{r_{11}( \xi -\eta )}H_{1}[S,I,R](\eta )\,d\eta +\int _{\xi }^{\infty }e^{r_{12}( \xi -\eta )}H_{1}[S,I,R](\eta )\,d\eta \}, \\ I(\xi )=\frac{1}{\varLambda _{2}} \{\int _{-\infty }^{\xi }e^{r_{21}( \xi -\eta )}H_{2}[S,I,R](\eta )\,d\eta +\int _{\xi }^{\infty }e^{r_{22}( \xi -\eta )}H_{2}[S,I,R](\eta )\,d\eta \}, \end{cases} $$
(3.30)
where
$$ H_{1}[S,I,R](\eta )=m S(\eta )-\frac{\beta S(\eta )I(\eta -c^{*} \tau )}{1+\alpha I(\eta -c^{*}\tau )} $$
and
$$ H_{2}[S,I,R](\eta )=\frac{\beta S(\eta )I(\eta -c^{*}\tau )}{1+\alpha I(\eta -c^{*}\tau )}+(m-\gamma )I(\eta ). $$
Using L’Hôpital rule in (3.30) gives
$$ S'(\pm \infty )=0 \quad \text{and}\quad I'(\pm \infty )=0. $$
(3.31)
Integrating the first equation in (3.27) from −∞ to ξ and using (3.29) and (3.31), we have that
$$ \beta \int _{-\infty }^{\xi }\frac{ S(\eta )I(\eta -c^{*}\tau )}{1+ \alpha I(\eta -c^{*}\tau )}\,d\eta =c^{*}\bigl[S(\xi )-S_{0}\bigr]-d_{1} S'( \xi ) \leq c^{*}S_{0}-d_{1} S'(\xi ), \quad \xi \in \mathbb{R}. $$
(3.32)
Again, integrating the second equation in (3.27) from −∞ to ξ and utilizing (3.29), (3.31), and (3.32), we get that
$$\begin{aligned} \gamma \int _{-\infty }^{\xi }I(\eta )\,d\eta =&d_{2} I'(\xi )-c^{*} I( \xi )+\beta \int _{-\infty }^{\xi }\frac{ S(\eta )I(\eta -c^{*}\tau )}{1+ \alpha I(\eta -c^{*}\tau )}\,d\eta \\ \leq& d_{2} I'(\xi )+\beta \int _{-\infty }^{\xi }\frac{ S(\eta )I( \eta -c^{*}\tau )}{1+\alpha I(\eta -c^{*}\tau )}\,d\eta \\ \leq& d_{2} I'(\xi )+c^{*}S_{0}-d_{1} S'(\xi ),\quad \xi \in \mathbb{R}. \end{aligned}$$
Then, by the virtue of (3.31), we further obtain \(\int _{\mathbb{R}}I(\xi )\,d\xi <\infty \), which together with the boundedness of \(I'(\xi )\) on \(\mathbb{R}\) (see (3.31)) implies that
$$ I(\infty )=0. $$
(3.33)
It follows from the first equation in (3.27) that
$$ \bigl[e^{-\frac{c^{*}}{d_{1}}\xi }S'(\xi ) \bigr]'= \frac{\beta }{d_{1}}e^{-\frac{c^{*}}{d_{1}}\xi } \frac{S(\xi )I(\xi -c ^{*}\tau )}{1+\alpha I(\xi -c^{*}\tau )}. $$
(3.34)
Integrating (3.34) from ξ to ∞, utilizing \(S'(\infty )=0\) and \(S(\xi ),I(\xi )>0\) on \(\mathbb{R}\), we deduce
$$ S'(\xi )=-\frac{\beta }{d_{1}} \int _{\xi }^{\infty }e^{\frac{c^{*}}{d _{1}}(\xi -\eta )} \frac{S(\eta )I(\eta -c^{*}\tau )}{1+\alpha I( \eta -c^{*}\tau )}\,d\eta < 0, $$
(3.35)
which means that \(S(\xi )\) is strictly decreasing on \(\mathbb{R}\). This together with \(S(\xi )>0\) on \(\mathbb{R}\) gives that the limit \(S(\infty )\) exists and \(S(\infty ):=\varepsilon_{0} < S_{0}\). Moreover, an integration of the first equation in (3.27) over \(\mathbb{R}\) gives
$$ \beta \int _{-\infty }^{\infty }\frac{ S(\xi )I(\xi -c^{*}\tau )}{1+ \alpha I(\xi -c^{*}\tau )}\,d\xi =c^{*}( S_{0}-\varepsilon_{0} ), $$
(3.36)
where we have used (3.29) and (3.31). Another integration of the second equation in (3.27) over \(\mathbb{R}\) yields
$$ \gamma \int _{-\infty }^{\infty }I(\xi )\,d\xi =\beta \int _{-\infty } ^{\infty }\frac{ S(\xi )I(\xi -c^{*}\tau )}{1+\alpha I(\xi -c^{*} \tau )}\,d\xi , $$
(3.37)
since \(I(\pm \infty )=I'(\pm \infty )=0\). Solving the third equation in (3.27) and using \(R(-\infty )=0\) lead to
$$ R(\xi )=Ce^{\frac{c^{*}}{d_{3}}\xi }+\frac{\gamma }{c^{*}} \int _{- \infty }^{\xi }I(\eta )\,d\eta + \frac{\gamma }{c^{*}} \int _{\xi }^{+ \infty }e^{\frac{c^{*}}{d_{3}}(\xi -\eta )}I(\eta )\,d\eta , $$
where C is a constant of integration. Since \(R(\xi )\leq L_{3} e ^{\sigma _{1}\xi }\) and \(\sigma _{1}< c^{*}/d_{3}\) (see the proof of Lemma 3.4), we obtain
$$ R(\xi )=\frac{\gamma }{c^{*}} \int _{-\infty }^{\xi }I(\eta )\,d\eta + \frac{ \gamma }{c^{*}} \int _{\xi }^{\infty }e^{\frac{c^{*}}{d_{3}}(\xi - \eta )}I(\eta )\,d\eta . $$
(3.38)
We infer from (3.36)–(3.38) and L’Hôpital’s rule that
$$ R(\infty )=\frac{\gamma }{c^{*}} \int _{-\infty }^{\infty }I(\xi )\,d\xi =S_{0}- \varepsilon_{0} . $$
(3.39)
Differentiating (3.38) with respect to ξ and using \(I(\xi )>0\) on \(\mathbb{R}\), we have
$$ R'(\xi )=\frac{\gamma }{d_{3}} \int _{\xi }^{\infty }e^{\frac{c^{*}}{d _{3}}(\xi -\eta )}I(\eta )\,d\eta >0, $$
(3.40)
which means that \(R(\xi )\) is strictly increasing on \(\mathbb{R}\). Combining (3.40), \(I(\pm \infty )=0\), and L’Hôpital’s rule yields
$$ R'(\pm \infty )=0. $$
(3.41)
Note from (3.29), (3.31), (3.32), (3.39), and (3.41) that
$$ S''(\pm \infty )=0, \qquad I''(\pm \infty )=0,\quad \text{and}\quad R''(\pm \infty )=0. $$
(3.42)
(4) Since \(S(\xi )\) is strictly decreasing and \(R(\xi )\) is strictly increasing on \(\mathbb{R}\), we obtain \(S(\xi )< S_{0}\) and \(R(\xi )< S _{0}\) for \(\xi \in \mathbb{R}\). Now we claim that \(I(\xi )<\frac{1}{ \alpha } (\frac{\beta S_{0}}{\gamma }-1 )\) on \(\mathbb{R}\). For contradiction, we assume that \(I(\acute{\xi })=\frac{1}{\alpha } (\frac{\beta S_{0}}{\gamma }-1 )\) for some \(\acute{\xi } \in \mathbb{R}\), which results in \(I'(\acute{\xi })=0\) and \(I''( \acute{\xi })\leq 0\). By the second equation in (3.27) and \(S(\acute{\xi })< S_{0}\), we deduce that
$$\begin{aligned} 0 =&d_{2} I''(\acute{\xi })-c^{*}I'(\acute{\xi })+\frac{\beta S( \acute{\xi })I(\acute{\xi }-c^{*}\tau )}{1+\alpha I(\acute{\xi }-c ^{*}\tau )}-\gamma I(\acute{\xi }) \\ \leq& \frac{\beta S(\acute{\xi })I(\acute{\xi }-c^{*}\tau )}{1+ \alpha I(\acute{\xi }-c^{*}\tau )}-\gamma I(\acute{\xi }) \\ < &\frac{\frac{\beta S_{0}}{\alpha } (\frac{\beta S_{0}}{\gamma }-1 )}{1+ (\frac{\beta S_{0}}{\gamma }-1 )}- \frac{ \gamma }{\alpha } \biggl(\frac{\beta S_{0}}{\gamma }-1 \biggr) \\ =&0, \end{aligned}$$
leading to a contradiction. Thus \(I(\xi )<\frac{1}{\alpha } (\frac{ \beta S_{0}}{\gamma }-1 )\) on \(\mathbb{R}\). The proof is completed. □