Finite-time stability and optimal impulsive control for age-structured HIV model with time-varying delay and Lévy noise

Abstract

This paper investigates the finite-time stability and optimal impulsive control for stochastic age-structured HIV model with time-varying delay. A stochastic noise is introduced by using the Lévy process to characterize the phenomenon of discontinuous jumps in virus transmission, which cannot be described by a continuous stochastic process (e.g., Brownian motion). By employing the comparison theorem and the bounded impulsive interval method, we obtain the sufficient conditions of finite-time stability for a stochastic HIV system. The effects of impulse, delay and Lévy noise on the finite-time stability are considered in our sufficient conditions. Furthermore, optimal impulsive control is studied to seek the optimal and cost-effective strategy for HIV treatments, which shows that control strategies play an important role in HIV virus transmissions. Numerical simulations are performed to illustrate the validity of our results.

Appendices

Appendix A: The proof of Theorem 2.4

Proof

Since the coefficients of system (8) are locally Lipschitz continuous, for any given initial data $\{\Phi (t):-{\bar{\tau }}\le t\le 0\}\in {\mathcal {H}}$, there is a unique maximal local solution $\big (X(t),Y(\cdot ,t),V(t)\big )$ on $t\in [-{\bar{\tau }},\tau _e]$, where $\tau _e$ is the explosion time (see [22]). Let $n_0>0$ be sufficiently large for

$$\begin{aligned} \frac{1}{n_0}<\min \limits _{-{\bar{\tau }}\le t\le 0}|\Phi (t)|<\max \limits _{-{\bar{\tau }}\le t\le 0}|\Phi (t)|<n_0. \end{aligned}$$

For each integer $n\ge n_0$, define the stopping time

$$\begin{aligned} \tau _n= & {} \inf \{t\in [0,\tau _e]: \min \{X(t),Y(\cdot ,t),V(t)\}\\\le & {} \frac{1}{n} ~\text {or}~ \max \{X(t),Y(\cdot ,t),V(t)\}\ge n\}. \end{aligned}$$

Set $\inf \emptyset =\infty$ ($\emptyset$ represents the empty set). $\tau _n$ increases as $n\rightarrow \infty$. Let $\tau _\infty = \lim \limits _{n\rightarrow \infty }\tau _n$, then $\tau _\infty \le \tau _e$ a.s. In what follows, if we can show that $\tau _\infty =\infty$ a.s., then $\tau _e=\infty$ a.s. and $\big (X(t),Y(\cdot ,t),V(t)\big )\in {\mathscr {X}}$ a.s. Before showing this, we first show the boundedness of the global solution. Since $\big (X(t),Y(\cdot ,t),V(t)\big )$ is positive on $[0,\tau _n)$, and

$$\begin{aligned}&\big (x(t),y(a,t),v(t)\big )\\&\quad =\big (A_1(t)X(t),A_2(t)Y(a,t),A_3(t)V(t)\big ), \end{aligned}$$

according to (A1) and the define of $A_i(t)$ given by (10), it is obvious that $A_i(t)$ is positive and bounded. Thus, $\big (x(t),y(\cdot ,t),v(t)\big )$ is also positive on $[0,\tau _n)$. For any $t\ne t_k$, adding the first and second equation of (7), and it follows from (A2) that

$$\begin{aligned}&\mathrm {d}\Big (x(t)+\int _0^Ay(a,t+\tau (t))\mathrm {d}a\Big )\\&\quad =\Big [\lambda -d'x(t)-\beta _1x(t)v(t)\\&\qquad -x(t)\int _0^A \beta (a)y(a,t)\mathrm {d}a\\&\qquad -\int _0^A\frac{\partial y(a,t+\tau (t))}{\partial a}\mathrm {d}a\\&\qquad -\int _0^A\mu _1'\mu _2(a) y(a,t+\tau (t))\mathrm {d}a\Big ]\mathrm {d}t\\&\qquad -\sigma _1'x(t)\mathrm {d}B_1(t)-x(t^-)\int _{{\mathbb {Y}}}\gamma _1(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&\qquad -\sigma _2'\int _0^A\mu _2(a)y(a,t+\tau (t))\mathrm {d}a\mathrm {d}B_2(t)\\&\qquad -\int _0^A\int _{{\mathbb {Y}}}\gamma _2(\nu ){\tilde{N}}(\mathrm {d}t, \mathrm {d}\nu )\mu _2(a)y(a,t^-+\tau (t))\mathrm {d}a\\&\quad =\Big [\lambda -d'x(t)-\beta _1x(t)v(t)\\&\qquad -x(t)\int _0^A\beta (a)y(a,t)\mathrm {d}a-y(a,t+\tau (t))|_0^A\\&\qquad -\int _0^A\mu _1'\mu _2(a) y(a,t+\tau (t))\mathrm {d}a\Big ]\mathrm {d}t\\&\qquad -\sigma _1'x(t)\mathrm {d}B_1(t)-x(t^-)\int _{{\mathbb {Y}}}\gamma _1(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&\qquad -\sigma _2'\int _0^A\mu _2(a)y(a,t+\tau (t))\mathrm {d}a\mathrm {d}B_2(t)\\&\qquad -\int _0^A\int _{{\mathbb {Y}}}\gamma _2(\nu ){\tilde{N}}(\mathrm {d}t, \mathrm {d}\nu )\mu _2(a)y(a,t^-+\tau (t))\mathrm {d}a\\&\quad \le \Big [\lambda -d'x(t)-\mu _1'{\underline{\mu }}_2\int _0^A y(a,t+\tau (t))\mathrm {d}a\Big ]\mathrm {d}t\\&\qquad -\sigma _1'x(t)\mathrm {d}B_1(t)-x(t^-)\int _{{\mathbb {Y}}}\gamma _1(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&\qquad -\sigma _2'{\underline{\mu }}_2\int _0^Ay(a,t+\tau (t))\mathrm {d}a\mathrm {d}B_2(t)\\&\qquad -{\underline{\mu }}_2\int _0^A\int _{{\mathbb {Y}}}\gamma _2(\nu ){\tilde{N}}(\mathrm {d}t, \mathrm {d}\nu )y(a,t^-+\tau (t))\mathrm {d}a\\&\quad =\Big [\lambda -\alpha \Big (x(t)+\int _0^A y(a,t+\tau (t))\mathrm {d}a\Big )\Big ]\mathrm {d}t+M(t), \end{aligned}$$

where $\alpha =\min \{d',\mu _1'{\underline{\mu }}_2\}$ and

$$\begin{aligned} \begin{aligned} M(t)=&-\sigma _1'x(t)\mathrm {d}B_1(t)-x(t^-)\int _{{\mathbb {Y}}}\gamma _1(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&-\sigma _2'{\underline{\mu }}_2\int _0^Ay(a,t+\tau (t))\mathrm {d}a\mathrm {d}B_2(t)\\&-{\underline{\mu }}_2\int _0^A\int _{{\mathbb {Y}}}\gamma _2(\nu ){\tilde{N}}(\mathrm {d}t, \mathrm {d}\nu )y(a,t^-+\tau (t))\mathrm {d}a. \end{aligned} \end{aligned}$$

Making use of the stochastic comparison theorem [41] and the method of variation of constant, we know that there exists a positive constant $K'$ such that

$$\begin{aligned} x(t)+\int _0^Ay(a,t+\tau )\mathrm {d}a\le K', \quad a.s. \end{aligned}$$

For the third equation of system (7), we have

$$\begin{aligned}&\mathrm {d}v(t+\tau (t))\\&\quad =\Big [\int _0^A k(a)e^{-m_2\tau (t)}y(a,t)\mathrm {d}a-c'v(t+\tau (t))\Big ]\mathrm {d}t\\&\qquad -\sigma _3'v(t+\tau (t))\mathrm {d}B_3(t)\\&\qquad -v(t^-+\tau (t))\int _{{\mathbb {Y}}}\gamma _3(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&\quad \le \Big [{\bar{k}}\int _0^A y(a,t)\mathrm {d}a-c'v(t+\tau (t))\Big ]\mathrm {d}t\\&\qquad -\sigma _3'v(t+\tau (t))\mathrm {d}B_3(t)\\&\qquad -v(t^-+\tau (t))\int _{{\mathbb {Y}}}\gamma _3(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\\&\quad \le \Big [{\bar{k}}K'-c'v(t+\tau (t))\Big ]\mathrm {d}t-\sigma _3'v(t+\tau (t))\mathrm {d}B_3(t)\\&\qquad -v(t^-+\tau (t))\int _{{\mathbb {Y}}}\gamma _3(\nu ){\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu ). \end{aligned}$$

The stochastic comparison theorem implies that

$$\begin{aligned} x(t)+\int _0^Ay(a,t)\mathrm {d}a+v(t)< \infty , \quad a.s. \end{aligned}$$

For $t=t_k$, based on (A1), the same result can be obtained. Therefore, there exists a positive constant K such that

$$\begin{aligned} X(t)+\int _0^AY(a,t)\mathrm {d}a+V(t)\le K, \quad a.s. \end{aligned}$$

(29)

Next we prove that $\tau _\infty = \lim \limits _{n\rightarrow \infty }\tau _n=\infty$ a.s. Let $W(t)=(X(t),Y(\cdot ,t),V(t))$, define a Lyapunov function

$$\begin{aligned} V(W(t))=X^2(t)+\int _0^AY^2(a,t)\mathrm {d}a+V^2(t), \end{aligned}$$

Let $T>0$ be arbitrary, for any $0\le t\le \tau _n\wedge T$, by Itô formula, we have

$$\begin{aligned} \mathrm {d}V(W(t))=&2X(t)\Big [\lambda A_1^{-1}(t)-(d'-\ln (1+I_{1k}))X(t) \\&-\beta _1X(t)A_3(t)V(t) \\&-X(t)\int _0^A \beta (a)A_2(t)Y(a,t)\mathrm {d}a\Big ]\mathrm {d}t \\&+(\sigma _1')^2X^2(t)\mathrm {d}t-2\sigma _1'X^2(t)\mathrm {d}B_1(t) \\&+\int _{{\mathbb {Y}}}\Big \{\big (X(t)-X(t)\gamma _1(\nu )\big )^2-X^2(t)\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu ) \\&+\int _{{\mathbb {Y}}}\Big \{\big (X(t)-X(t)\gamma _1(\nu )\big )^2-X^2(t) \\&+2X^2(t)\gamma _1(\nu )\Big \}\eta (\mathrm {d}\nu )\mathrm {d}t \\&+2\int _0^AY(a,t)\Big [-\frac{\partial Y(a,t)}{\partial a}-\big (\mu _1'\mu _2(a) \\&-\ln (1+I_{2k})\big ) Y(a,t)\Big ]\mathrm {d}a\mathrm {d}t \\&+\int _0^A(\sigma _2'\mu _2(a))^2Y^2(a,t)\mathrm {d}a\mathrm {d}t \\&-2\int _0^A\sigma _2'\mu _2(a)Y^2(a,t)\mathrm {d}a\mathrm {d}B_2(t) \\&+\int _{{\mathbb {Y}}}\Big \{\int _0^A\big (Y(a,t)-\mu _2(a)Y(a,t)\gamma _2(\nu )\big )^2\mathrm {d}a \\&-\int _0^AY^2(a,t)\mathrm {d}a\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu ) \\&+\int _{{\mathbb {Y}}}\Big \{\int _0^A\big (Y(a,t)-\mu _2(a)Y(a,t)\gamma _2(\nu )\big )^2\mathrm {d}a \\&-\int _0^AY^2(a,t)\mathrm {d}a \\&+2\int _0^AY(a,t)\mathrm {d}a\mu _2(a)Y(a,t)\gamma _2(\nu )\Big \}\eta (\mathrm {d}\nu )\mathrm {d}t \\&+2V(t)\Big [\int _0^A k(a)e^{-m_2\tau (t)}A_3^{-1}(t)A_2(t)Y(a,t\\&-\tau (t))\mathrm {d}a -(c'-\ln (1+I_{3k}))V(t)\Big ]\mathrm {d}t \\&+(\sigma _3')^2V^3(t)\mathrm {d}t-2\sigma _3'V^2(t)\mathrm {d}B_3(t) \\&+\int _{{\mathbb {Y}}}\Big \{\big (V(t)-V(t)\gamma _3(\nu )\big )^2-V^2(t)\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu ) \\&+\int _{{\mathbb {Y}}}\Big \{\big (V(t)-V(t)\gamma _3(\nu )\big )^2 \\&-V^2(t)+2V^2(t)\gamma _3(\nu )\Big \}\eta (\mathrm {d}\nu )\mathrm {d}t \end{aligned}$$

$$\begin{aligned} :=&{\mathcal {L}}V\mathrm {d}t-2\sigma _1'X^2(t)\mathrm {d}B_1(t)\nonumber \\&+\int _{{\mathbb {Y}}}\Big \{\big (X(t)-X(t)\gamma _1(\nu )\big )^2-X^2(t)\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\nonumber \\&-2\int _0^A\sigma _2'\mu _2(a)Y^2(a,t)\mathrm {d}a\mathrm {d}B_2(t)\nonumber \\&+\int _{{\mathbb {Y}}}\Big \{\big (V(t)-V(t)\gamma _3(\nu )\big )^2-V^2(t)\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )\nonumber \\&+\int _{{\mathbb {Y}}}\Big \{\int _0^A\big (Y(a,t)-\mu _2(a)Y(a,t)\gamma _2(\nu )\big )^2\mathrm {d}a\nonumber \\&-\int _0^AY^2(a,t)\mathrm {d}a\Big \}{\tilde{N}}(\mathrm {d}t,\mathrm {d}\nu )-2\sigma _3'V^2(t)\mathrm {d}B_3(t), \end{aligned}$$

(30)

where ${\mathcal {L}}V=J_1+J_2+J_3$ with

$$\begin{aligned} J_1=&2X(t)\Big [\lambda A_1^{-1}(t)-(d'-\ln (1+I_{1k}))X(t)\\&-\beta _1X(t)A_3(t)V(t)\\&-X(t)\int _0^A \beta (a)A_2(t)Y(a,t)\mathrm {d}a\Big ]\\&+(\sigma _1')^2X^2(t)+\int _{{\mathbb {Y}}}\Big \{\big (X(t)-X(t)\gamma _1(\nu )\big )^2\\&-X^2(t)+2X^2(t)\gamma _1(\nu )\Big \}\eta (\mathrm {d}\nu ),\\ J_2=&2\int _0^AY(a,t)\Big [-\frac{\partial Y(a,t)}{\partial a}\\&-\big (\mu _1'\mu _2(a)-\ln (1+I_{2k})\big )Y(a,t)\Big ]\mathrm {d}a\\&+\int _0^A(\sigma _2'\mu _2(a))^2Y^2(a,t)\mathrm {d}a\\&+\int _{{\mathbb {Y}}}\Big \{\int _0^A\big (Y(a,t)-\mu _2(a)Y(a,t)\gamma _2(\nu )\big )^2\mathrm {d}a\\&-\int _0^AY^2(a,t)\mathrm {d}a\\&+2\int _0^AY(a,t)\mathrm {d}a\mu _2(a)Y(a,t)\gamma _2(\nu )\Big \}\eta (\mathrm {d}\nu ),\\ J_3=&2V(t)\Big [\int _0^A k(a)e^{-m_2\tau (t)}A_3^{-1}(t)A_2(t\\&-\tau (t))Y(a,t-\tau (t))\mathrm {d}a\\&-(c'-\ln (1+I_{3k}))V(t)\Big ]\\&+(\sigma _3')^2V^2(t)+\int _{{\mathbb {Y}}}\Big \{\big (V(t)-V(t)\gamma _3(\nu )\big )^2\\&-V^2(t)+2V^2(t)\gamma _3(\nu )\Big \}\eta (\mathrm {d}\nu ). \end{aligned}$$

According to (A1), (A2), (A3) and formula (29), we obtain

$$\begin{aligned} \begin{aligned} J_1\le&2X(t)\lambda A_1^{-1}(t)+2X^2(t)\ln (1+I_{1k})\\&+(\sigma _1')^2X^2(t)+X^2(t)\int _{{\mathbb {Y}}}\gamma _1^2(\nu )\eta (\mathrm {d}\nu )\\ \le&\lambda ^2 {\underline{A}}_1^{-2}+X^2(t)\Big [1+2\ln (1+I_{1k})+(\sigma _1')^2+C_0\Big ],\\ J_2\le&-Y^2(a,t)\Big |_0^A+2\ln (1+I_{2k})\int _0^AY^2(a,t)\mathrm {d}a\\&+(\sigma _2'{\bar{\mu }}_2)^2\int _0^AY^2(a,t)\mathrm {d}a\\&+{\bar{\mu }}_2^2\int _{{\mathbb {Y}}}\gamma _2^2(\nu )\eta (\mathrm {d}\nu )\int _0^AY^2(a,t)\mathrm {d}a\\ \le&\beta _1^2X^2(t-\tau (t))V^2(t-\tau (t))\\&+{\bar{\beta }}^2X^2(t-\tau (t))\int _0^AY^2(a,t-\tau (t))\mathrm {d}a\\&+\Big [2\ln (1+I_{2k})+(\sigma _2'{\bar{\mu }}_2)^2\\&+{\bar{\mu }}_2^2C_0\Big ]\int _0^AY^2(a,t)\mathrm {d}a,\\ J_3\le&V^2(t)+{\bar{k}}^2A_3^{-2}(t)\\&\Big [\int _0^AA_2(t-\tau (t))Y(a,t-\tau (t))\mathrm {d}a\Big ]^2\\&+V^2(t)2\ln (1+I_{3k})+(\sigma _3')^2X^2(t)+C_0V^2(t)\\ \le&{\bar{k}}^2{\underline{A}}_3^{-2}\Big [\int _0^AA_2(t-\tau (t))Y(a,t-\tau (t))\mathrm {d}a\Big ]^2\\&+V^2(t)\Big [1+2\ln (1+I_{3k})+(\sigma _3')^2+C_0\Big ]. \end{aligned} \end{aligned}$$

Now, for any $n\ge n_0$, we integrate the both sides of (30) from 0 to $\tau _n\wedge T$ and then take the expectations to derive

$$\begin{aligned}&{\mathbb {E}}V(W(\tau _n\wedge T))\\&\quad \le V(W(0))+\lambda ^2 {\underline{A}}_1^{-2}(\tau _n\wedge T)\\&\qquad +\Big [1+2\check{b}+(\sigma _1')^2+C_0\Big ]\int _0^{\tau _n\wedge T}X^2(t)\mathrm {d}t\\&\qquad +\beta _1^2\int _0^{\tau _n\wedge T}X^2(t-\tau (t))V^2(t-\tau (t))\mathrm {d}t\\&\qquad +{\bar{\beta }}^2\int _0^{\tau _n\wedge T}\int _0^AX^2(t-\tau (t))Y^2(a,t-\tau (t))\mathrm {d}a\mathrm {d}t\\&\qquad +\Big [2\check{b}+(\sigma _2'{\bar{\mu }}_2)^2+{\bar{\mu }}_2^2C_0\Big ]\\&\qquad \int _0^{\tau _n\wedge T}\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\&\qquad +{\bar{k}}^2{\underline{A}}_3^{-2} \\&\qquad \int _0^{\tau _n\wedge T}\Big [\int _0^AA_2(t-\tau (t))Y(a,t-\tau (t))\mathrm {d}a\Big ]^2\mathrm {d}t\\&\qquad +\Big [1+2\check{b}+(\sigma _3')^2+C_0\Big ]\int _0^{\tau _n\wedge T}V^2(t)\mathrm {d}t, \end{aligned}$$

where $\check{b}=\sup \{\ln (1+I_{ik})\}~(i=1,2,3)$. Since

$$\begin{aligned} \begin{aligned}&\beta _1^2\int _0^{\tau _n\wedge T}X^2(t-\tau (t))V^2(t-\tau (t))\mathrm {d}t\\&\quad \le \beta _1^2\frac{1}{1-\zeta }\int _{-\tau (t)}^{\tau _n\wedge T-\tau (t)}X^2(t)V^2(t)\mathrm {d}t\\&\quad \le \frac{\beta _1^2}{1-\zeta }K^4(\tau _n\wedge T), \end{aligned} \end{aligned}$$

Similarly, we have

$$\begin{aligned} \begin{aligned}&{\bar{\beta }}^2\int _0^{\tau _n\wedge T}\int _0^AX^2(t-\tau (t))Y^2(a,t-\tau (t))\mathrm {d}a\mathrm {d}t\\&\quad = {\bar{\beta }}^2\frac{1}{1-\zeta }\int _{-\tau (t)}^{\tau _n\wedge T-\tau (t)}\int _0^AX^2(t)Y^2(a,t)\mathrm {d}a\mathrm {d}t\\&\quad \le \frac{{\bar{\beta }}^2K^2}{1-\zeta }\int _{-{\bar{\tau }}}^0\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\&\qquad +\frac{{\bar{\beta }}^2K^2}{1-\zeta }\int _0^{\tau _n\wedge T}\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&{\bar{k}}^2{\underline{A}}_3^{-2}\int _0^{\tau _n\wedge T}\Big [\int _0^AA_2(t-\tau (t))Y(a,t-\tau (t))\mathrm {d}a\Big ]^2\mathrm {d}t\\&\quad \le \frac{1}{1-\zeta }{\bar{k}}^2{\underline{A}}_3^{-2}{\bar{A}}_2^2\int _{-{\bar{\tau }}}^{\tau _n\wedge T-\tau (t)}\Big [\int _0^AY(a,t)\mathrm {d}a\Big ]^2\mathrm {d}t\\&\quad \le \frac{{\bar{k}}^2{\underline{A}}_3^{-2}{\bar{A}}_2^2K^2}{1-\zeta }\big (\tau _n\wedge T\big ). \end{aligned} \end{aligned}$$

Thus, we obtain

$$\begin{aligned}&{\mathbb {E}}V(W(\tau _n\wedge T))\\&\quad \le V(W(0))+\lambda ^2 {\underline{A}}_1^{-2}(\tau _n\wedge T)\\&\qquad +\Big [1+2\check{b}+(\sigma _1')^2+C_0\Big ]\int _0^{\tau _n\wedge T}X^2(t)\mathrm {d}t\\&\qquad +\frac{1}{1-\zeta }\beta _1^2K^4\big (\tau _n\wedge T\big )+\frac{1}{1-\zeta }{\bar{k}}^2{\underline{A}}_3^{-2}{\bar{A}}_2^2K^2\big (\tau _n\wedge T\big )\\&\qquad +\frac{1}{1-\zeta }{\bar{\beta }}^2K^2\int _{-{\bar{\tau }}}^0\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\&\qquad +\frac{1}{1-\zeta }{\bar{\beta }}^2K^2\int _0^{\tau _n\wedge T}\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\&\qquad +\Big [2\check{b}+(\sigma _2'{\bar{\mu }}_2)^2+{\bar{\mu }}_2^2C_0\Big ]\int _0^{\tau _n\wedge T}\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\&\qquad +\Big [1+2\check{b}+(\sigma _3')^2+C_0\Big ]\int _0^{\tau _n\wedge T}V^2(t)\mathrm {d}t\\&\quad \le C_1+C_2\int _0^{\tau _n\wedge T}\Big (X^2(t)\\&\qquad +\int _0^AY^2(a,t)\mathrm {d}a+V^2(t)\Big )\mathrm {d}t\\&\quad =C_1+C_2\int _0^{\tau _n\wedge T}V(Z(t))\mathrm {d}t, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} C_1=&V(W(0))+\lambda ^2 {\underline{A}}_1^{-2}(\tau _n\wedge T)\\&+\frac{1}{1-\zeta }\beta _1^2K^4\big (\tau _n\wedge T\big )\\&\frac{1}{1-\zeta }{\bar{k}}^2{\underline{A}}_3^{-2}{\bar{A}}_2^2K^2\big (\tau _n\wedge T\big )\\&+\frac{1}{1-\zeta }{\bar{\beta }}^2K^2\int _{-{\bar{\tau }}}^0\int _0^AY^2(a,t)\mathrm {d}a\mathrm {d}t\\ <&\infty , \end{aligned} \end{aligned}$$

and according to (A2), we choose

$$\begin{aligned} \begin{aligned} C_2=1+2\check{b}+({\hat{\sigma }}')^2+C_0+\frac{1}{1-\zeta }{\bar{\beta }}^2K^2. \end{aligned} \end{aligned}$$

Using Gronwall inequality, we obtain

$$\begin{aligned} {\mathbb {E}}V(W(\tau _n\wedge T))\le C_1e^{C_2T}, \quad \forall ~n\ge n_0. \end{aligned}$$

(31)

Define

$$\begin{aligned} \mu _n=\inf \limits _{|W|\ge n, 0\le t<\infty }V(W(t)),~~~\text {for any} ~n\ge n_0. \end{aligned}$$

Then from (31), we know

$$\begin{aligned} \mu _n{\mathbb {P}}(\tau _n\le T)\le C_1e^{C_2T}. \end{aligned}$$

(32)

It is straightforward to drive $\lim \limits _{|Z|\rightarrow \infty }\inf \limits _{0\le t<\infty }V(W(t))=\infty$, thus $\lim \limits _{n\rightarrow \infty }\mu _n=\infty$. Letting $n\rightarrow \infty$ in the inequality (32), we obtain ${\mathbb {P}}(\tau _n\le T)=0$, namely,${\mathbb {P}}(\tau _n>T)=1.$

Thus, the required assertion is obtained. Hence, we complete the proof. $\square$

Appendix B: Numerical algorithm of control

where

$$\begin{aligned} \begin{aligned} \text {Term}_1=&\Big [\lambda -d'x_j-(1-(u_1)_j)\beta _1x_jv_j\\&-x_j\sum _{i=1}^{A/\Delta a} \beta (i\Delta a)y_{i,j}\Delta a\Big ]\Delta t\\&-\sigma _1'x_j\sqrt{\Delta t}\xi _j\\&-\frac{{\sigma _1'}^2}{2}x_j(\xi _j^2-1)\Delta t-\gamma _1 x_j\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }}, \\ \text {Term}_2=&\Big [-\mu _1'\mu _2(i\Delta a)y_{i,j}\Big ]\Delta t\\&-\sigma _2'\mu _2(i\Delta a)y_{i,j}\sqrt{\Delta t}\xi _j\\&-\frac{{\sigma _2'}^2\mu _2^2(i\Delta a)}{2}y_{i,j}(\xi _j^2-1)\Delta t\\&-\gamma _2 \mu _2(i\Delta a) y_{i,j}\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }},\\ \text {Term}_3=&\Big [(1-u_{2j})\sum _{i=1}^{N} k(i\Delta a)\\&e^{-m \tau (j\Delta t)}y_{i,j-\tau (j\Delta t)/\Delta t}\Delta a-c'v_j\Big ]\Delta t\\&-\sigma _3'v_j\sqrt{\Delta t}\xi _j\\&-\frac{{\sigma _3'}^2}{2}v_j(\xi _j^2-1)\Delta t-\gamma _3 v_j\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }},\\ \text {Project}_1=&\Big [(p_1)_n\big (d'+(1-(u_1)_j)\beta _1v_j\\&+\sum _{i=1}^{A/\Delta a} \beta (i\Delta a)y_{i,j}\big )+\gamma _1r_1+q_1\sigma _1'+F_1\\&-\chi _{[0,T-\tau ]}\big ((1-(u_1)_j)\beta _1v_{j-m}\\&+\sum _{i=1}^{N} \beta (i\Delta a)y_{i,j-m}\Delta a\big )e^{-m_1\tau }(p_2)_{1,n+m}\Big ]\Delta t\\&+q_1\sqrt{\Delta t}\xi _j+r_1\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }}, \end{aligned} \end{aligned}$$

$$\begin{aligned} \text {Project}_2=&\Big [-\chi _{[0,T-\tau ]}x_{j-m}e^{-m_1\tau }\beta (i\Delta a)(p_2)_{1,j+m}\\&+(p_1)_nx_{j}\beta (i\Delta a)+(p_2)_{i,n}\mu _1'\mu _2(i\Delta a)\\&+\mu _2(i\Delta a)\gamma _2r_2+q_2\sigma _2'\mu _2(i\Delta a)\\&-\chi _{[0,T-\tau ]}(1-(u_2)_j)e^{-m_2\tau }(p_3)_{n+m}k(i\Delta a)\Big ]\Delta t\\&+q_2\sqrt{\Delta t}\xi _j+r_2\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }},\\ \text {Project}_3=&\Big [(p_1)_n(1-(u_1)_j)\beta _1x_j+c'(p_3)_n\\&-\chi _{[0,T-\tau ]}(1-(u_1)_j)\beta _1x_{j-m}e^{-m_1\tau }(p_2)_{1,n+m}\\&+q_3\sigma _3'+\gamma _3r_3-F_2\Big ]\Delta t\\&+q_3\sqrt{\Delta t}\xi _j+r_3\sum _{k=1}^\infty V_k\Gamma _k^{-\frac{1}{\alpha }}, \end{aligned}$$

with $\alpha \in (0,2)$, $\{V_k\}_{k\ge 1}$ is an independent and identically distributed (i.i.d.) random sequence such that ${\mathbb {P}}(V_k=\pm 1)=\frac{1}{2}$, which is independent of the random sequence $\{\Gamma _k\}_{k\ge 1}$.