Caspase-1-Mediated Pyroptosis of the Predominance for Driving CD4$^{+}$ T Cells Death: A Nonlocal Spatial Mathematical Model

Abstract

Caspase-1-mediated pyroptosis is the predominance for driving CD4$^{+}$ T cells death. Dying infected CD4$^{+}$ T cells can release inflammatory signals which attract more uninfected CD4$^{+}$ T cells to die. This paper is devoted to developing a diffusive mathematical model which can make useful contributions to understanding caspase-1-mediated pyroptosis by inflammatory cytokines IL-1$\beta$ released from infected cells in the within-host environment. The well-posedness of solutions, basic reproduction number, threshold dynamics are investigated for spatially heterogeneous infection. Travelling wave solutions for spatially homogeneous infection are studied. Numerical computations reveal that the spatially heterogeneous infection can make $\mathscr {R}_0>1$, that is, it can induce the persistence of virus compared to the spatially homogeneous infection. We also find that the random movements of virus have no effect on basic reproduction number for the spatially homogeneous model, while it may result in less infection risk for the spatially heterogeneous model, under some suitable parameters. Further, the death of infected CD4$^{+}$ cells which are caused by pyroptosis can make $\mathscr {R}_0<1$, that is, it can induce the extinction of virus, regardless of whether or not the parameters are spatially dependent.

Appendices

Appendix A

1.1 Proof of Theorem 2.1

For any $\phi =(\phi _1,\phi _2,\phi _3,\phi _4)\in [\mathbf 0 ,\mathbf M ]_\mathbb {C}$ and any $\acute{k}\ge 0$, we easily obtain that

$$\begin{aligned} \phi +\acute{k}\mathscr {F}(\phi )= & {} \left( \begin{array}{c} \displaystyle \phi _1+\acute{k}\Big (\xi (x)-d_U\phi _1(0,x)\Big ) -\acute{k}\frac{\beta (x)\phi _1(0,x) \phi _4(0,x)}{1+a\phi _4(0,x)}-\acute{k}\frac{q(x)\phi _1(0,x) \phi _3(0,x)}{1+b\phi _3(0,x)}\\ \displaystyle \phi _2+\acute{k}\int _{\varOmega } {\varGamma (\tau ,x,y)\frac{\beta (y)\phi _1(-\,\tau ,y) \phi _4(-\,\tau ,y)}{1+a\phi _4(-\,\tau ,y)}}\hbox {d}y-\acute{k}\Big (d_V(x)+\alpha _1(x)\Big )\phi _2(0,x)\\ \displaystyle \phi _3+\acute{k}\alpha _2(x)\phi _2(0,x)-\acute{k}d_M\phi _3(0,x)\\ \displaystyle \phi _4+\acute{k}k(x)\phi _2(0,x)-\acute{k}d_\omega \phi _4(0,x)\\ \end{array} \right) \\\ge & {} \left( \begin{array}{c} \displaystyle \Bigg (1-\Big (\acute{k}d_U+\frac{\acute{k}\overline{\beta }}{a}+\frac{\acute{k}\overline{q}}{b}\Big )\Bigg )\phi _1\\ \Big (1-\acute{k}(\overline{d}_V+\overline{\alpha }_1)\Big )\phi _2\\ \displaystyle (1-\acute{k}d_M)\phi _3\\ \displaystyle (1-\acute{k}d_\omega )\phi _4\\ \end{array} \right) \ge \left( \begin{array}{c} \displaystyle 0\\ \displaystyle 0\\ \displaystyle 0\\ \displaystyle 0\\ \end{array} \right) . \end{aligned}$$

For

$$\begin{aligned} 0<\acute{k}\le \min \left\{ \frac{1}{d_U+\frac{\overline{\beta }}{a}+\frac{\overline{q}}{b}},\ \frac{1}{\overline{d}_V+\overline{\alpha }_1},\ \frac{1}{d_M},\ \frac{1}{d_\omega }\right\} , \end{aligned}$$

we have that

$$\begin{aligned} \phi +\acute{k}\mathscr {F}(\phi )= & {} \left( \begin{array}{c} \displaystyle \phi _1+\acute{k}\Big (\xi (x)-d_U\phi _1(0,x)\Big ) -\frac{\acute{k}\beta (x)\phi _1(0,x) \phi _4(0,x)}{1+a\phi _4(0,x)}-\frac{\acute{k}q(x)\phi _1(0,x) \phi _3(0,x)}{1+b\phi _3(0,x)}\\ \displaystyle \phi _2+\acute{k}\int _{\varOmega } {\varGamma (\tau ,x,y)\frac{\beta (y)\phi _1(-\,\tau ,y) \phi _4(-\,\tau ,y)}{1+a\phi _4(-\,\tau ,y)}}\hbox {d}y-\acute{k}\Big (d_V(x)+\alpha _1(x)\Big )\phi _2(0,x)\\ \displaystyle \phi _3+\acute{k}\alpha _2(x)\phi _2(0,x)-\acute{k}d_M\phi _3(0,x)\\ \displaystyle \phi _4+\acute{k}k(x)\phi _2(0,x)-\acute{k}d_\omega \phi _4(0,x)\\ \end{array} \right) \\\le & {} \left( \begin{array}{c} \displaystyle \frac{\overline{\xi }}{d_U}\\ \displaystyle \frac{\acute{k}\overline{\beta }\ \overline{\xi }}{d_Ua}+\Big (1-\acute{k}(\underline{d}_V+\underline{\alpha }_1)\Big )\frac{\overline{\xi }\ \overline{\beta }}{ad_U(d_I+\underline{\alpha }_1)}\\ \displaystyle \acute{k}\overline{\alpha }_2\frac{\overline{\xi }\ \overline{\beta }}{ad_U(\underline{d}_V+\underline{\alpha }_1)}+(1-\acute{k}d_M)\frac{\overline{\alpha }_2\overline{\xi }\ \overline{\beta }}{d_Mad_U(\underline{d}_V+\underline{\alpha }_1)}\\ \displaystyle \acute{k}\overline{k}\frac{\overline{\xi }\ \overline{\beta }}{ad_U(\underline{d}_V+\underline{\alpha }_1)}+(1-\acute{k}d_\omega )\frac{\overline{k}\ \overline{\xi }\ \overline{\beta }}{d_\omega ad_U(\underline{d}_V+\underline{\alpha }_1)} \end{array} \right) \\= & {} \left( \begin{array}{c} \displaystyle \frac{\overline{\xi }}{d_U}\\ \displaystyle \frac{\overline{\xi }\ \overline{\beta }}{ad_U(\underline{d}_V+\underline{\alpha }_1)}\\ \displaystyle \frac{\overline{\alpha }_2\overline{\xi }\ \overline{\beta }}{d_Mad_U(\underline{d}_V+\underline{\alpha }_1)}\\ \displaystyle \frac{\overline{k}\ \overline{\xi }\ \overline{\beta }}{d_\omega ad_U(\underline{d}_V+\underline{\alpha }_1)}\\ \end{array} \right) =\mathbf M . \end{aligned}$$

Therefore, it then follows that $\phi +\acute{k}\mathscr {F}(\phi )\in [\mathbf 0 ,\mathbf M ]_{\mathbb {C}}$. This implies that

$$\begin{aligned} \mathop {\lim }\limits _{\acute{k} \rightarrow 0^{+} }\frac{1}{\acute{k}}\mathrm{dist}\Big (\phi +\acute{k}\mathscr {F}(\phi ),[\mathbf 0 ,\mathbf M ]_{\mathbb {C}}\Big )=0, \end{aligned}$$

for all $\phi \in [\mathbf 0 ,\mathbf M ]_{\mathbb {C}}.$ From Corollary 4 in Martin and Smith (1990) (see, also Wu 1996, Corollary 8.1.3), we obtain the conclusion. The proof is completed. $\square$

Appendix B

1.1 Proof of Lemma 4.1

For any fixed $\phi =(\phi _1,\phi _2,\phi _3,\phi _4) \in \mathbb {C_{+}}$, from the first equation of model (3), we obtain that

$$\begin{aligned} \frac{\partial U(x,t)}{\partial t}\le D_0\Delta U+\overline{\xi }-d_UU(x,t). \end{aligned}$$

Then, the comparison principle implies there exists $t_1(\phi )>0$ such that $U(x,t)\le \frac{2\overline{\xi }}{d_U}=:B_1$ for $t>t_1$.

From the second equation of model (3), it then follows that

$$\begin{aligned} \frac{\partial V(x,t)}{\partial t}\le D_0\Delta V+\frac{\overline{\beta }B_1}{a}-(\underline{d}_V+\underline{\alpha }_1) V(x,t). \end{aligned}$$

Again, the comparison principle implies there exists $t_2(\phi )>0$ such that $V(x,t)\le \frac{2\overline{\beta }B_1}{a(\underline{d}_V+\underline{\alpha }_1)}=:B_2$ for $t>t_2$.

From the third equation of model (3), we get that

$$\begin{aligned} \frac{\partial M(x,t)}{\partial t}\le D_1\Delta M+\overline{\alpha }_2B_2-d_MM(x,t). \end{aligned}$$

Again, there exists $t_3(\phi )>0$ such that $M(x,t)\le \frac{2\overline{\alpha }_2B_2}{d_M}=:B_3$ for $t>t_3$.

From the fourth equation of model (3), we obtain that

$$\begin{aligned} \frac{\partial \omega (x,t)}{\partial t}\le D_2\Delta \omega +\overline{k}B_2-d_\omega \omega (x,t). \end{aligned}$$

Again, there exists $t_4(\phi )>0$ such that $\omega (x,t)\le \frac{2\overline{k}B_2}{d_\omega } =:B_4$ for $t>t_4$.

Therefore, the solutions of model (3) are ultimately bounded with respect to the maximum norm. The solution semiflow $\varPhi (t)=u(t,.):\ \mathbb {C_{+}}\rightarrow \mathbb {C_{+}}$ is point dissipative. By Theorem 2.2.6 in Wu (1996), we get that $\varPhi (t)$ is compact for any $t > \tau$. Therefore, from Theorem 3.4.8 in Hale (1988), we know that $\varPhi (t)=u(t,.)$ has a compact global attractor in $\mathbb {C_{+}}$. The proof is completed. $\square$

1.2 Proof of Lemma 4.2

From model (3), we easily obtain that

$$\begin{aligned} \left\{ {\begin{array}{l} \displaystyle \frac{\partial V(x,t)}{\partial t} \ge D_0\Delta V- (\alpha _1(x)+d_V(x))V(x,t),\\ \displaystyle \frac{\partial M(x,t)}{\partial t} \ge D_1\Delta M- d_MM(x,t),\\ \displaystyle \frac{\partial \omega (x,t)}{\partial t} \ge D_2\Delta \omega -d_\omega \omega (x,t),\\ \displaystyle \frac{\partial V(x,t)}{\partial \nu }=\frac{\partial M(x,t)}{\partial \nu }=\frac{\partial \omega (x,t)}{\partial \nu }=0,\ x\in \partial \varOmega . \end{array}} \right. \end{aligned}$$

If $V(t_0,\cdot , \phi )\not \equiv 0$, $M(t_0,\cdot , \phi )\not \equiv 0$ and $\omega (t_0,\cdot , \phi )\not \equiv 0$, the comparison principle implies that $V(t,\cdot , \phi )>0$, $M(t,\cdot , \phi )>0$, and $\omega (t,\cdot , \phi )>0$ for $t> t_0$, $x\in \overline{\varOmega }$, that is, the conclusion (i) holds.

From model (3), we easily obtain that

$$\begin{aligned} \begin{aligned} \displaystyle \frac{\partial U(x,t)}{\partial t}&\ge D_0\Delta U+\underline{\xi }-\Big (d_U+\frac{\overline{\beta }}{a}+\frac{\overline{q}}{b}\Big )U(x,t). \end{aligned} \end{aligned}$$

Let $v(x,t,\phi )$ be the solution of

$$\begin{aligned} \left\{ {\begin{array}{l} \displaystyle \frac{\partial v}{\partial t} \displaystyle = D_0\Delta v+\underline{\xi }-\Big (d_U+\frac{\overline{\beta }}{a}+\frac{\overline{q}}{b}\Big )v,\\ \displaystyle \frac{\partial v}{\partial \nu }=0,\ x\in \partial \varOmega ,\\ \displaystyle v(0,x)=\phi (0,x). \end{array}} \right. \end{aligned}$$

It then follows that $U(t,x,\phi )\ge v(t,x,\phi )>0$ for all $t>0$, and $x\in \overline{\varOmega }$. Moreover, by Lemma 3.1 and the comparison principle, we obtain that

$$\begin{aligned} \mathop {\liminf }\limits _{t \rightarrow \infty }U(t,\cdot , \phi ) \ge \frac{\underline{\xi }}{d_U+\frac{\overline{\beta }}{a}+ \frac{\overline{q}}{b}} \end{aligned}$$

uniformly for $x\in \overline{\varOmega }.$ The proof is completed. $\square$

1.3 Proof of Lemma 4.3

For any given $\phi \in \mathbb {C_{+}}$ with $\phi _2(0)\not \equiv 0$ and $\phi _4(0)\not \equiv 0$, let $u(t,x,\phi )=(U(t,x),V(x,t), M(t,x),\omega (t,x))$. In view of the parabolic maximum principle and Lemma 4.2, from model (3), we easily get that

$$\begin{aligned} \begin{aligned} V(t,x)>0,\ \omega (t,x)>0,\ \ \ \forall t>0,\ x\in \overline{\varOmega }. \end{aligned} \end{aligned}$$

By Lemma 3.3, there exists $\tau _1>0$ such that $\overline{\lambda }_0(\widehat{U}(x)-\tau _1)>0$. Suppose, by contradiction, we assume that there exists some $\phi \in \mathbb {C_{+}}$ with $\phi _2(0)\not \equiv 0$ and $\phi _4(0)\not \equiv 0$ such that

$$\begin{aligned} \mathop {\limsup }\limits _{t \rightarrow \infty }\parallel u(t,\cdot ,\phi )-\left( \widehat{U}(x),0,0,0\right) \parallel _{\mathbb {C_{+}}}< \tau _1. \end{aligned}$$

Then, there exists a sufficiently large positive number $T_1$ such that

$$\begin{aligned} \widehat{U}(x)-\tau _1<U(x,t)<\widehat{U}(x)+\tau _1,\ M(x,t)<\tau _1,\ \omega (x,t)<\tau _1,\ \ \ \forall t\ge T_1,\ x\in \overline{\varOmega }. \end{aligned}$$

Therefore, we obtain the following model

$$\begin{aligned} \left\{ {\begin{array}{ll} \displaystyle \frac{\partial v_2}{\partial t}&{}\ge D_0\Delta v_2 +\int _{\varOmega } {\varGamma (\tau ,x,y)\frac{\beta (y) \left( \widehat{U}(y)-\tau _1\right) v_4(y,t-\tau )}{1+a\tau _1}}\hbox {d}y\\ &{}\quad -\,\Big (\alpha _1(x)+d_V(x)\Big )v_2,\\ \displaystyle \frac{\partial v_4}{\partial t}&{}\ge D_2\Delta v_4+k(x)v_2-d_\omega v_4. \end{array}} \right. \end{aligned}$$

Let $\psi$ be the positive eigenfunction associated with $\overline{\lambda }_0(\widehat{U}(x)-\tau _1)$. Then, we obtain that

$$\begin{aligned} \left\{ {\begin{array}{ll} \displaystyle \frac{\partial u_2}{\partial t}&{}= D_0\Delta u_2+ \int _{\varOmega } {\varGamma (\tau ,x,y)\frac{\beta (y) \left( \widehat{U}(y)-\tau _1\right) u_4(y,t-\tau )}{1+a\tau _1}}\hbox {d}y\\ &{}\quad -\,\Big (\alpha _1(x)+d_V(x)\Big )u_2,\\ \displaystyle \frac{\partial u_4}{\partial t}&{}= D_2\Delta u_4+k(x)u_2-d_\omega u_4. \end{array}} \right. \end{aligned}$$

admits a solution $u(t,x)=e^{\overline{\lambda }_0(\widehat{U}(x)-\tau _1) t}(\psi _1(x),\ \psi _2(x))$. Since $u(t,x,\phi _0)\gg 0$ for all $t>0$ and $x\in \overline{\varOmega }$, there exists $\eta >0$ such that

$$\begin{aligned}\Big (V(x,t),\ \omega (x,t)\Big )\ge \eta \Big (u_2(x,t),\ u_4(x,t)\Big ),\ \ \ \ \ \forall t\in [T_1-\tau ,T_1],\ \ \ x\in \overline{\varOmega }.\end{aligned}$$

By the comparison principle, we have that

$$\begin{aligned}(V(x,t),\ \omega (x,t))\ge \eta e^{\overline{\lambda }_0(\widehat{U}(x)-\tau _1) t}(\psi _1(x),\ \psi _2(x)),\quad \forall t>T_1,\ \ \ x\in \overline{\varOmega }.\end{aligned}$$

Since $\lambda _0(\widehat{U}(x)-\tau _1)>0$, we obtain that $\mathop {\lim }\nolimits _{t \rightarrow +\infty }V(t,x)=\infty ,\ \mathop {\lim }\nolimits _{t \rightarrow +\infty }\omega (t,x)=\infty ,$ which is a contradiction. We complete the proof. $\square$

Appendix C

1.1 Proof of Lemma 5.3

If $t>-\frac{1}{\varepsilon }\ln q_1$, then $\underline{\phi }(t)=0$. It is easy to obtain the result. If $t\le -\frac{1}{\varepsilon }\ln q_1$, it then follows that

$$\begin{aligned}&d_1\ddot{\underline{\phi }}(t)-c\dot{\underline{\phi }}(t)+ \frac{\beta \Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\gamma }(t)}{1+a \underline{\gamma }(t)} -d_U\underline{\phi }(t)+\frac{q\Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\psi }(t)}{1+b\underline{\psi }(t)}\\&\quad =d_1\ddot{\underline{\phi }}(t)-c\dot{\underline{\phi }}(t)+ \frac{\beta \Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\gamma }(t)}{1+a \underline{\gamma }(t)} -d_U\underline{\phi }(t)\\&\quad \quad +\,\frac{qU_0\underline{\psi }(t)(1+b\underline{\psi }(t))-qU_0b\underline{\psi }(t)\underline{\psi }(t)-q\underline{\phi }(t)\underline{\psi }(t)}{1+b\underline{\psi }(t)}\\&\quad \ge d_1\ddot{\underline{\phi }}(t)-c\dot{\underline{\phi }}(t)+ \frac{\beta \Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\gamma }(t)}{1+a \underline{\gamma }(t)}\\&\quad \quad -\,d_U\underline{\phi }(t)+qU_0\underline{\psi }(t)-qU_0b\underline{\psi }(t)^{2}-q\underline{\phi }(t)\underline{\psi }(t)\\&\quad =d_1\ddot{\underline{\phi }}(t)-c\dot{\underline{\phi }}(t) +\frac{\beta \xi \underline{\gamma }(t)}{d_U}\\&\quad \quad -\,d_U\underline{\phi }(t)-\frac{\frac{\beta \xi a}{d_U}\underline{\gamma }(t)\underline{\gamma }(t)}{1+a\underline{\gamma }(t)}-\frac{\beta \underline{\phi }(t)\underline{\gamma }(t)}{1+a \underline{\gamma }(t)}+qU_0\underline{\psi }(t)-qU_0b\underline{\psi }(t)^{2}-q\underline{\phi }(t)\underline{\psi }(t)\\&\quad \ge d_1\ddot{\underline{\phi }}(t)-c\dot{\underline{\phi }}(t) +\frac{\beta \xi \underline{\gamma }(t)}{d_U}\\&\quad \quad -\,d_U\underline{\phi }(t)-\frac{\beta \xi a}{d_U}\underline{\gamma }(t)\underline{\gamma }(t)-\beta \underline{\phi }(t)\underline{\gamma }(t)+qU_0\underline{\psi }(t)-qU_0b\underline{\psi }(t)^{2}-q\underline{\phi }(t)\underline{\psi }(t)\\&\quad \ge -q_1\Bigg (h_1(\lambda _c+\varepsilon ,c)\eta _1 +\frac{\beta \xi \eta _4}{d_U}\Bigg )e^{(\lambda _c+\varepsilon )t}-\Bigg (\frac{\beta \xi a\eta _4^{2}}{d_U}+\beta \eta _1\eta _4+qU_0b\eta _3^2+q\eta _1\eta _3\Bigg )e^{2\lambda _ct}\\&\quad \ge -\left[ q_1\Bigg (h_1(\lambda _c+\varepsilon ,c)\eta _1 +\frac{\beta \xi \eta _4}{d_U}\Bigg )+\frac{\beta \xi a\eta _4^{2}}{d_U}+\beta \eta _1\eta _4+qU_0b\eta _3^2+q\eta _1\eta _3\right] e^{(\lambda _c+\varepsilon )t}\ge 0. \end{aligned}$$

From (16) and Lemma 5.1, it then follows that $h_1(\lambda _c+\varepsilon ,c)\eta _1 +\frac{\beta \xi \eta _4}{d_U}<0$. Then for sufficiently large $q_1$, we obtain that the last inequality holds.

If $t>-\frac{1}{\varepsilon }\ln q_1$, then $\underline{\varphi }(t)=0$. We easily obtain the result. If $t\le -\frac{1}{\varepsilon }\ln q_1$, it then follows that

$$\begin{aligned} \begin{aligned} \displaystyle&d_2\ddot{\underline{\varphi }}(t)-c\dot{\underline{\varphi }}(t) +{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{ \beta \Big (\frac{\xi }{d_U}-\overline{\phi }(y,t-y-c\tau )\Big ) \underline{\gamma }(y,t-y-c\tau )}{1+a\underline{\gamma }(y,t-y-c\tau )}}\hbox {d}y\\&\quad \quad \displaystyle -\,d_V\underline{\varphi }(t)-\alpha _1\underline{\varphi }(t)\\&\quad \ge d_2\ddot{\underline{\varphi }}(t)-c\dot{\underline{\varphi }}(t)+ {\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{\beta \xi }{d_U}\underline{\gamma }(t-y-c\tau )}\hbox {d}y-(d_V+\alpha _1)\underline{\varphi }(t)\\&\quad \quad -{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{\frac{\beta \xi a}{d_U}\underline{\gamma }(t-y-c\tau )\underline{\gamma }(t-y-c\tau )}{1+a \underline{\gamma }(t-y-c\tau )}}\hbox {d}y\\&\quad \quad -\,{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{\beta \overline{\phi }(t-y-c\tau )\underline{\gamma }(t-y-c\tau )}{1+a \underline{\gamma }(t-y-c\tau )}}\hbox {d}y\\&\quad \ge -\Bigg [ q_1\Bigg (h_2(\lambda _c+\varepsilon ,c)\eta _2 +\frac{\beta \xi \eta _4e^{\tau (d_2(\lambda _c+\varepsilon )^{2}-c (\lambda _c+\varepsilon ))}}{d_U}\Bigg )\\&\quad \quad +\frac{\beta \xi a\eta _4^{2}e^{\tau (d_2(\lambda _c+\varepsilon )^{2}-c (\lambda _c+\varepsilon ))}}{d_U}+\beta \eta _1\eta _4e^{\tau (d_2(\lambda _c+\varepsilon )^{2}-c (\lambda _c+\varepsilon ))}\Bigg ]e^{(\lambda _c+\varepsilon )t}\ge 0. \end{aligned} \end{aligned}$$

Similarly, from (16) and Lemma 5.1, it then follows that $h_2(\lambda _c+\varepsilon ,c)\eta _2 +\frac{\beta \xi \eta _4e^{\tau (d_2(\lambda _c+\varepsilon )^{2}-c (\lambda _c+\varepsilon ))}}{d_U}<0$. Then for sufficiently large $q_1$, we obtain that the last inequality holds.

It can be similarly shown that

$$\begin{aligned} d_3\ddot{\underline{\psi }}(t)-c\dot{\underline{\psi }}(t) +\alpha _2\underline{\varphi }(t)-d_M\underline{\psi }(t)\ge 0,\ \ \ d_4\ddot{\underline{\gamma }}(t)-c\dot{\underline{\gamma }}(t) +k\underline{\varphi }(t)-d_\omega \underline{\gamma }(t)\ge 0. \end{aligned}$$

We complete the proof. $\square$

1.2 Proof of Lemma 5.4

By Lemma 5.2 and simple computations, we easily obtain that

$$\begin{aligned} \begin{aligned}&\beta _1\phi (t)+\frac{\beta \Big (\frac{\xi }{d_U}-\phi (t)\Big )\gamma (t)}{1+a \gamma (t)}-d_U\phi (t)+\frac{q\Big (\frac{\xi }{d_U}-\phi (t)\Big )\psi (t)}{1+b\psi (t)}\\&\le \beta _1\overline{\phi }(t)+\frac{\beta \Big (\frac{\xi }{d_U}-\overline{\phi }(t)\Big )\overline{\gamma }(t)}{1+a \overline{\gamma }(t)} -d_U\overline{\phi }(t)+\frac{q\Big (\frac{\xi }{d_U}-\overline{\phi }(t)\Big )\overline{\psi }(t)}{1+b\overline{\psi }(t)}\\&\le \beta _1\overline{\phi }(t)-d_1\ddot{\overline{\phi }}(t) +c\dot{\overline{\phi }}(t). \end{aligned} \end{aligned}$$

For $t>\frac{1}{\lambda _c}\ln \frac{\xi }{d_U \eta _1}\triangleq t_1$, by using integration of parts twice, it then follows that

$$\begin{aligned} \begin{aligned} F_1(\phi ,\varphi ,\psi ,\gamma )(t)&=\frac{1}{\rho _1}\left[ \int _{-\infty }^t {e^{\lambda _{11}(t-s)}H(\phi ,\varphi ,\psi ,\gamma )(s)}\hbox {d}s\right. \\ {}&\left. \quad +\int _{t }^\infty {e^{\lambda _{12}(t-s)}H(\phi ,\varphi ,\psi ,\gamma )(s)}\hbox {d}s\right] \\&\le \frac{1}{\rho _1}\left[ \int _{-\infty }^{t_1} {e^{\lambda _{11}(t-s)}\Big (\beta _1\overline{\phi }(t)-d_1\ddot{\overline{\phi }} +c\dot{\overline{\phi }}\Big )}\hbox {d}s\right] \\&\ \ \ \ +\frac{1}{\rho _1}\left[ \int _{t_1 }^{t} {e^{\lambda _{11}(t-s)}\Big (\beta _1\overline{\phi }(t)-d_1\ddot{\overline{\phi }} +c\dot{\overline{\phi }}\Big )}\hbox {d}s\right] \\&\ \ \ \ +\frac{1}{\rho _1}\left[ \int _{t }^{+\infty } {e^{\lambda _{11}(t-s)}\Big (\beta _1\overline{\phi }(t)-d_1\ddot{\overline{\phi }} +c\dot{\overline{\phi }}\Big )}\hbox {d}s\right] \\&=\overline{\phi }(t)+\frac{1}{\lambda _{12}-\lambda _{11}}e^{\lambda _{11}(t-t_1)} \Big (\dot{\overline{\phi }}(t_1+0)-\dot{\overline{\phi }}(t_1-0)\Big )\le \overline{\phi }(t). \end{aligned} \end{aligned}$$

By employing the similar method above, we can prove $F_1(\phi ,\varphi ,\psi ,\gamma )(t)\le \overline{\phi }(t)$ for $t\le t_1$. Thus, for any $t\ge 0$, we have that $F_1(\phi ,\varphi ,\psi ,\gamma )(t)\le \overline{\phi }(t)$.

If $t<t^*$, it then follows that

$$\begin{aligned} \begin{aligned}&\beta _1\phi (t)+\frac{\beta \Big (\frac{\xi }{d_U}-\phi (t)\Big )\gamma (t)}{1+a \gamma (t)}-d_U\phi (t)+\frac{q\Big (\frac{\xi }{d_U}-\phi (t)\Big )\psi (t)}{1+b\psi (t)}\\&\quad \ge \beta _1 \underline{\phi }(t)-d_U\underline{\phi }(t)+\frac{\beta \Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\gamma }(t)}{1+a \underline{\gamma }(t)}+\frac{q\Big (\frac{\xi }{d_U}-\underline{\phi }(t)\Big )\underline{\psi }(t)}{1+b\underline{\psi }(t)}\\&\quad \ge \beta _1\underline{\phi }(t)-d_1\ddot{\underline{\phi }} +c\dot{\underline{\phi }}=D_1\underline{\phi }(t). \end{aligned} \end{aligned}$$

If $t\ge t^*$, it then follows that

$$\begin{aligned} \begin{aligned}&\beta _1\phi (t)+\frac{\beta \Big (\frac{\xi }{d_U}-\phi (t)\Big )\gamma (t)}{1+a \gamma (t)}-d_U\phi (t)+\frac{q\Big (\frac{\xi }{d_U}-\phi (t)\Big )\psi (t)}{1+b\psi (t)}\\&\quad \ge (\beta _1-d_U)\underline{\phi }(t)=0=D_1\underline{\phi }(t). \end{aligned} \end{aligned}$$

Therefore, by Lemma 3.2 in Wang et al. (2012a), it then follows that $F_1(\phi ,\varphi ,\psi ,\gamma )\ge D_1^{-1}(D_1\underline{\phi }(t))\ge \underline{\phi }(t).$ From the discussions above, for any $t\ge 0$, we easily obtain that $\underline{\phi }(t)\le F_1(\phi ,\varphi ,\psi ,\gamma )\le \overline{\phi }(t).$

Similarly, it then follows that

$$\begin{aligned} \begin{aligned}&\beta _2\varphi (t)+{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{ \beta \Big (\frac{\xi }{d_U}-\phi (t-y-c\tau )\Big ) \gamma (t-y-c\tau )}{1+a\gamma (t-y-c\tau )}}\hbox {d}y -(\alpha _1+d_V)\varphi (t)\\&\quad \le \beta _2 \overline{\varphi }(t)-(d_V+\alpha _1)\overline{\varphi }(t) +\frac{\beta \xi }{d_U}{\int }_{\mathbb {R} } {\varGamma (\tau ,y) \overline{\gamma }(t-y-c\tau )}\hbox {d}y\\&\quad =\beta _2\overline{\varphi }(t)-d_2\ddot{\overline{\varphi }} +c\dot{\overline{\varphi }}=D_2\overline{\varphi }(t). \end{aligned} \end{aligned}$$

Consequently, by Lemma 3.2 in Wang et al. (2012a), we have $F_2(\phi ,\varphi ,\psi ,\gamma )\le D_2^{-1}\Big (D_2\overline{\varphi }(t)\Big )=\overline{\varphi }(t)$.

If $t<t^*$, it then follows that

$$\begin{aligned} \begin{aligned}&\beta _2\varphi (t)+{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{ \beta \Big (\frac{\xi }{d_U}-\phi (t-y-c\tau )\Big ) \gamma (t-y-c\tau )}{1+a\gamma (t-y-c\tau )}}\hbox {d}y -(\alpha _1+d_V)\varphi (t)\\&\quad \ge \beta _2\underline{\varphi }(t)+{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{ \beta \Big (\frac{\xi }{d_U}-\overline{\phi }(t-y-c\tau )\Big ) \underline{\gamma }(t-y-c\tau )}{1+a\underline{\gamma }(t-y-c\tau )}}\hbox {d}y -d_V\underline{\varphi }(t) -\alpha _1\underline{\varphi }(t)\\&\quad \ge \beta _2\underline{\varphi }(t)-d_2\ddot{\underline{\varphi }} +c\dot{\underline{\varphi }}=D_2\underline{\varphi }(t). \end{aligned} \end{aligned}$$

If $t\ge t^*$, it then follows that

$$\begin{aligned} \begin{aligned}&\beta _2\varphi (t)+{\int }_{\mathbb {R} } {\varGamma (\tau ,y)\frac{ \beta \Big (\frac{\xi }{d_U}-\phi (t-y-c\tau )\Big ) \gamma (t-y-c\tau )}{1+a\gamma (t-y-c\tau )}}\hbox {d}y -(\alpha _1+d_V)\varphi (t)\\&\quad \ge \Big (\beta _2-(d_V+\alpha _1)\Big )\underline{\varphi }(t)\\&\quad \ge 0=D_2\underline{\varphi }(t). \end{aligned} \end{aligned}$$

A combination of the above two inequalities yields $F_2(\phi ,\varphi ,\psi ,\gamma )\ge D_2^{-1}(D_2\underline{\varphi }(t))\ge \underline{\varphi }(t).$ Similarly, we can show that $\underline{\psi }\le F_3(\phi ,\varphi ,\psi ,\gamma )\le \overline{\psi },$ $\underline{\gamma }\le F_4(\phi ,\varphi ,\psi ,\gamma )\le \overline{\gamma }.$ We complete the proof. $\square$

1.3 Proof of Lemma 5.5

We firstly check that $H=(H_1,H_2,H_3,H_4):\varPi \rightarrow \varPi$ is continuous with respect to the norm $|{\cdot }|_{\mu }$ in $B_{\mu }(\mathbb {R},\mathbb {R}^{4}).$ In fact, for any $\varPhi _{11}, \varPsi _{11} \in C(\mathbb {R},\mathbb {R}^{4})$, it then follows that

$$\begin{aligned} \begin{aligned}&\mid H_1(\varPhi _{11})(t)-H_1(\varPsi _{11})(t)\mid e^{-\mu |t|}\\&\quad \le (d_U+\beta _1)|\phi _1(t)-\phi _2(t)|e^{-\mu |t|}\\ {}&\qquad + \left| \frac{\beta \Big (\frac{\xi }{d_U}-\phi _1(t)\Big )\gamma _1(t)}{1+a \gamma _1(t)}-\frac{\beta \Big (\frac{\xi }{d_U}-\phi _2(t)\Big )\gamma _2(t)}{1+a \gamma _2(t)}\right| e^{-\mu |t|}\\&\quad \quad + \left| \frac{\beta \Big (\frac{q}{d_U}-\phi _1(t)\Big )\psi _1(t)}{1+b \psi _1(t)}-\frac{\beta \Big (\frac{q}{d_U}-\phi _2(t)\Big )\psi _2(t)}{1+b \psi _2(t)}\right| e^{-\mu |t|}\\&\quad \le \Big (d_U+\beta _1+\frac{\beta }{a}+\frac{q}{b}\Big )|\phi _1(t)-\phi _2(t)|e^{-\mu |t|} + \frac{2\beta \xi }{d_U}|\gamma _1(t)-\gamma _2(t)|e^{-\mu |t|}\\&\quad \quad +\frac{2q\xi }{d_U}|\psi _1(t)-\psi _2(t)|e^{-\mu |t|}\\&\quad =\Big (d_U+\beta _1+\frac{\beta }{a}+\frac{q}{b}+\frac{2\beta \xi }{d_U}+\frac{2q\xi }{d_U}\Big ) |\varPhi _{11}(t)-\varPsi _{11}(t)|e^{-\mu |t|}.\\ \end{aligned} \end{aligned}$$

For any given $\varepsilon >0$, choose $0<\delta _1<\displaystyle \frac{\varepsilon }{d_U+\beta _1+\frac{\beta }{a}+\frac{q}{b}+\frac{2\beta \xi }{d_U}+\frac{2q\xi }{d_U}}$, as $|\varPhi _{11}-\varPsi _{11}|_{\mu }<\delta _1$, then we have $|H_1(\varPhi _{11})(t)-H_1(\varPsi _{11})(t)|e^{-\mu |t|}<\varepsilon .$ Similarly, we have that

$$\begin{aligned} \begin{aligned}&\mid H_2(\varPhi _{11})(t)-H_2(\varPsi _{11})(t)\mid e^{-\mu |t|}\le e^{-\mu |t|}\int _{\mathbb {R} } \varGamma (\tau ,y)\frac{\beta }{a}|\phi _1(t-y-c\tau )\\&\qquad -\phi _2(t-y-c\tau )|\hbox {d}y\\&\quad \quad +e^{-\mu |t|}\int _{\mathbb {R} } {\varGamma (\tau ,y)\frac{2\beta \xi }{d_U}|\gamma _1(t-y-c\tau ) -\gamma _2(t-y-c\tau )|}\hbox {d}y\\&\quad \quad + (\beta _2+d_V+\alpha _1)|\varphi _{1}(t)-\varphi _{2}(t)|e^{-\mu |t|}\\&\quad \le \Big (e^{\tau (d_2\mu ^{2}+c\mu )}\Big (\frac{\beta }{a}+\frac{2\beta \xi }{d_U}\Big )+\beta _2+d_V+\alpha _1\Big ) |\varPhi _{11}(t)-\varPsi _{11}(t)| e^{-\mu |t|}.\\ \end{aligned} \end{aligned}$$

For any given $\varepsilon >0$, choose $0<\delta _2=\displaystyle \frac{\varepsilon }{e^{\tau (d_2\mu ^{2}+c\mu )}\Big (\frac{\beta }{a}+\frac{2\beta \xi }{d_U}\Big )+\beta _2+d_V+\alpha _1}$, as $|\varPhi _{11}-\varPsi _{11}|_{\mu }<\delta _2$, then we have $|H_2(\varPhi _{11})(t)-H_2(\varPsi _{11})(t)|e^{-\mu |t|}<\varepsilon .$ Hence, $H_2$ is continuous with respect to the norm $|{\cdot }|_{\mu }$ in $B_{\mu }(\mathbb {R},\mathbb {R}^{4}).$ Similarly, we can show that $H_3$ and $H_4$ is continuous with respect to the norm $|{\cdot }|_{\mu }$ in $B_{\mu }(\mathbb {R},\mathbb {R}^{4}).$

Applying the method similar to Ma (2001), Lemma 2.4 (see, also, Li et al. (2006) Lemma 3.4), we show that F is continuous with respect to the norm $B_{\mu }(\mathbb {R},\mathbb {R}^{4}).$ By simple computations, we have that

$$\begin{aligned}&\left| F_1(\varPhi _{11}({\cdot }))(t)-F_1(\varPsi _{11}({\cdot }))(t)\right| e^{-\mu |t|} \le \frac{e^{-\mu \mid t\mid }}{\rho _1}\Bigg (\int _{-\infty }^t e^{\lambda _{11}(t-s)}|H_1(\varPhi _{11})(s)\\&\quad \quad -H_1(\varPsi _{11})(s)|\hbox {d}s+\int _{t }^\infty {e^{\lambda _{12}(t-s)}|H_1(\varPhi _{11})(s)-H_1(\varPsi _{11})(s)|}\hbox {d}s\Bigg )\\&\quad \le \frac{Pe^{-\mu \mid t\mid }}{\rho _1}\left[ \int _{-\infty }^t {e^{\lambda _{11}(t-s)+\mu |s|}}\hbox {d}s +\int _{t }^\infty {e^{\lambda _{12}(t-s)+\mu |s|}}\hbox {d}s\right] \\&\qquad |\varPhi _{11}(t)-\varPsi _{11}(t)|e^{-\mu |t|}, \end{aligned}$$

where $P=d_U+\beta _1+\frac{\beta }{a}+\frac{q}{b}+\frac{2\beta \xi }{d_U}+\frac{2q\xi }{d_U}.$ If $t<0$, it then follows that

$$\begin{aligned} \begin{aligned}&\left| F_1\big (\varPhi _{11}({\cdot })\big )(t)-F_1(\varPsi _{11}({\cdot }))(t)\right| e^{-\mu |t|}\\&\quad \le \frac{Pe^{\mu t}}{\rho _1}\Big (e^{\lambda _{11}t}\int _{-\infty }^t {e^{-(\lambda _{11}+\mu )s}}\hbox {d}s+e^{\lambda _{12}t}\int _{t }^0 {e^{-(\lambda _{12}+\mu )s}}\hbox {d}s+e^{\lambda _{12}t}\int _{0 }^\infty {e^{(\mu -\lambda _{12})s}}\hbox {d}s\Big )\\&\quad \quad |\varPhi _{11}(t)-\varPsi _{11}(t)|\\&\quad \le \frac{Pe^{\mu t}}{\rho _1}\left( \frac{1}{-\lambda _{11}-\mu } +\frac{1}{\lambda _{12}+\mu } +\frac{1}{\lambda _{12}-\mu }\right) |\varPhi _{11}(t)-\varPsi _{11}(t)|.\\ \end{aligned} \end{aligned}$$

If $t\ge 0$, we have that

$$\begin{aligned} \begin{aligned}&\left| F_1(\varPhi _{11}({\cdot }))(t)-F_1(\varPsi _{11}({\cdot }))(t)\right| e^{-\mu |t|}\\&\le \frac{Pe^{-\mu t}}{\rho _1}\Bigg (e^{\lambda _{11}t}\int _{-\infty }^0 {e^{-(\lambda _{11}+\mu )s}}\hbox {d}s+e^{\lambda _{11}t}\int _{0 }^t {e^{(\mu -\lambda _{11})s}}\hbox {d}s+e^{\lambda _{12}t}\int _{t }^\infty {e^{(\mu -\lambda _{12})s}}\hbox {d}s\Bigg )\\&\qquad |\varPhi _{11}(t)-\varPsi _{11}(t)|\\&\le \frac{Pe^{-\mu t}}{\rho _1}\left( \frac{1}{-\lambda _{11}-\mu } +\frac{1}{\mu -\lambda _{11}} +\frac{1}{\lambda _{12}-\mu }\right) |\varPhi _{11}(t)-\varPsi _{11}(t)|.\\ \end{aligned} \end{aligned}$$

Hence, it then follows that

$$\begin{aligned} \left| F_1(\varPhi _{11}({\cdot }))(t)-F_1(\varPsi _{11}({\cdot }))(t)\right| _{\mu } \le P_1 |\varPhi _{11}-\varPsi _{11}|_{\mu }, \end{aligned}$$

where

$$\begin{aligned} P_1=\frac{P}{\rho _1}\max \left\{ \frac{1}{-\lambda _{11}-\mu } +\frac{1}{\lambda _{12}+\mu } +\frac{1}{\lambda _{12}-\mu },\ \ \ \frac{1}{-\lambda _{11}-\mu } +\frac{1}{\mu -\lambda _{11}} +\frac{1}{\lambda _{12}-\mu }\right\} . \end{aligned}$$

Similarly, we can also prove that $F_2,\ F_3$, and $F_4$ are continuous with respect to the norm |.| in $B_{\mu }(\mathbb {R},\mathbb {R}^{4})$. The proof is completed. $\square$

1.4 Proof of Lemma 5.6

By employing Arzela–Ascoli theorem, we show that the operator $F:\varPi \rightarrow \varPi$ is compact with respect to the norm |.| in $B_{\mu }(\mathbb {R},\mathbb {R}^{4})$. Let $I_\mathbf k =[-\,\mathbf k ,\mathbf k ]$ with $\mathbf k \in N$ be a compact interval on $\mathbb {R}$. We regard $\varPi$ as a bounded subset of $C(I_\mathbf k ,\mathbb {R}^{4})$ equipped with the maximum norm. From Lemma 5.4, it then follows that F is uniformly bounded equipped with the maximum norm. In the following, we employ the following equalities to show that F is equi-continuous. In terms of the definition of $F_i$ in (18) and integral representation for the derivative of $D_i^{-1}\ (i=1,2,3,4)$, for any $(\phi ,\varphi ,\psi ,\gamma )\in \varPi$, it then follows that

$$\begin{aligned} \begin{aligned} \left| [F_1(\phi ,\varphi ,\psi ,\gamma )]'(t)\right|&\le \frac{-\lambda _{11}}{\rho _{ 1}}\int _{-\infty }^t {e^{\lambda _{ 11}(t-s)}|H_1(s)|}\hbox {d}s+ \frac{\lambda _{12}}{\rho _{ 1}}\int _{t }^\infty {e^{\lambda _{ 12}(t-s)}|H_1(s)|}\hbox {d}s\\&\le \frac{\xi }{d_U}\Big (\beta _1-d_U+\frac{\beta }{a}+\frac{q}{b}\Big ). \end{aligned} \end{aligned}$$

Similarly, we also have that

$$\begin{aligned} \begin{aligned} \left| [F_2(\phi ,\varphi ,\psi ,\gamma )]'(t)\right|&\le \frac{-\lambda _{21}}{\rho _2}\int _{-\infty }^t {e^{\lambda _{21}(t-s)}|H_2(s)|}\hbox {d}s+ \frac{\lambda _{22}}{\rho _2}\int _{t }^\infty {e^{\lambda _{22}(t-s)}|H_2(s)|}\hbox {d}s\\&\le A\Big (\frac{-\lambda _{21}}{\rho _2}\int _{-\infty }^t {e^{\lambda _{21}(t-s)}e^{\lambda _cs}}\hbox {d}s+ \frac{\lambda _{22}}{\rho _2}\int _{t }^\infty {e^{\lambda _{22}(t-s)}e^{\lambda _cs}}\hbox {d}s\Big )\\&=\frac{e^{\lambda _ct}A}{\rho _2} \Big (\frac{-\lambda _{21}}{\lambda _c-\lambda _{21}} +\frac{\lambda _{22}}{\lambda _{22}-\lambda _c}\Big ), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} A=\Big (\beta _2-(d_V+\alpha _1)\Big )\eta _2 +\frac{\beta \xi \eta _4e^{\tau (d_2\lambda _c^{2}-c\lambda _c)}}{d_U}. \end{aligned}$$

Let $\mathbf{u _n}$ be a sequence of $\varPi$, which is viewed as a bounded subset of $C(I_\mathbf k )$ with $I_\mathbf k :=[-\,\mathbf k ,\mathbf k ]$. From the discussions above, we conclude that F is uniformly bounded and equi-continuous. According to the Arzela–Ascoli theorem, there exists a subsequence $\mathbf{u _{n_\mathbf k }}$ such that $\mathbf v _{n_\mathbf k }=F\mathbf u _{n_\mathbf k }$ converges in $C(I_\mathbf k )$ for any $\mathbf k \in N$. Let $\mathbf v$ be the limit of $\mathbf v _{n_\mathbf k }$. It is easily observe that $\mathbf v \in C(\mathbb {R},\mathbb {R}^{4})$. Furthermore, from Lemma 5.4 and $\varPi$ is closed, we have that $\mathbf v \in \varPi$. Since $\mu>\lambda _c>0$, it can be obtained that $e^{-\mu \mid t\mid }\overline{\phi }(t)$, $e^{-\mu \mid t\mid }\overline{\varphi }(t)$, $e^{-\mu \mid t\mid }\overline{\psi }(t)$ and $e^{-\mu \mid t\mid }\overline{\gamma }(t)$ are uniformly bounded on $\mathbb {R}$. Consequently, $\varPi$ is uniformly bounded with respect to the norm $|{\cdot }|_{\mu }$. Therefore, it follows that the norm $|\mathbf v _{n_\mathbf k }-\mathbf v |_{\mu }$ is uniformly bounded for all $\mathbf k \in N$. For any $\varepsilon >0$, there exists an integer $Z>0$, independent of $\mathbf v _{n_\mathbf k }$ such that $e^{-\mu \mid t\mid }|\mathbf v _{n_\mathbf k }(x)-\mathbf v (x)|<\varepsilon$ for $|t|>Z$ and $\mathbf k \in N$. Due to the fact that $\mathbf v _{n_\mathbf k }$ converges to $\mathbf v$ on the compact interval $[-\,Z,Z]$ with respect to the maximum norm, it then follows that there exists $K\in N$, such that $e^{-\mu \mid t\mid }|\mathbf v _{n_\mathbf k }(x)-\mathbf v (x)|<\varepsilon$ for $|t|\le Z$ and $\mathbf k > K$. Consequently, it can be concluded that $\mathbf v _{n_\mathbf k }$ converges to $\mathbf v$ with respect to the norm $|{\cdot }|_{\mu }$. The proof is completed. $\square$