Ergodic stationary distribution and disease eradication in a stochastic SIR model with telegraph noises and Lévy jumps

Abstract

In this work, we present analysis of a SIR model where white noises, telegraph noises (Markov switching) and Lévy jumps serve as sources of environmental perturbations in the system. Based on the Feller property, we derive sufficient conditions for the existence of a unique stationary distribution with ergodic property using the mutually exclusive probabilities technique (Stenner in the existence and uniqueness of invariant measure for continuous-time Markov process, Tech. Report, pp 18-86, Brown University, Providence, RI, USA, 1986). Further, in a special case, we derived condition for disease eradication. Using numerical simulations, we were able to illustrate the analytical results obtained herein. Some of the results reveal that, in a Markovian switching regime, white noises and Lévy jumps could determine whether the disease is eradicated or not, with both sources of random perturbations affecting the degree of disease persistence.

Appendices

Appendix A: Proof to Lemma 2.1

Proof

Let $\phi (t)$ be the solution of

$$\begin{aligned} \begin{aligned} \mathrm{d}\phi (t)&= - [\mu (k) + \alpha (k)]\phi \mathrm{d}t + \sigma _1(k) \phi \mathrm{d}B_1(t)\\&\quad + \int \limits _Z G_1(z)\phi (t-){\tilde{N}}(\mathrm{d}t,\mathrm{d}z), \end{aligned} \end{aligned}$$

(5.1)

with $\phi (0) = 1$. Using the generalized Itô’s formula to $\ln \phi (t)$ and taking integrations, we have that

$$\begin{aligned} \begin{aligned} \phi (t)&= \exp \bigg [\int _0^t-\bigg (\bigg [\mu (k) + \alpha (k)\\&\quad +\frac{\sigma _1^2(k)}{2}\bigg ] + \int \limits _Z \big \{\ln [1+G_1(z)]-G_1(z)\big \}\nu (\mathrm{d}z)\bigg )\mathrm{d}s\\&\quad + \sigma _1(k)B_1(t) + \int _0^t \int \limits _Z \ln [1+G_1(z)]{\tilde{N}}(\mathrm{d}t,\mathrm{d}z)\bigg ]. \end{aligned} \end{aligned}$$

Applying the variation-of-constants formula, we have that $\theta (t) = \theta (0)f(t)+\Lambda (k)f(t)\int _0^t f^{-1}(s)\mathrm{d}s$, where

$$\begin{aligned} \begin{aligned} f(t)&= \exp \bigg (-\bigg (\bigg [\mu (k) + \alpha (k)+\frac{\sigma _1^2(k)}{2}\bigg ]t + \sigma _1(k)B_1(t)\bigg )\\&\quad \times \exp \bigg [\int _0^t \int \limits _Z \big \{\ln (1+G_1(z))-G_1(z)\big \}\nu (\mathrm{d}z)\mathrm{d}s\\&\quad + \int _0^t \int \limits _Z \ln [1+G_1(z)]{\tilde{N}}(\mathrm{d}s,\mathrm{d}z)\bigg ]. \end{aligned} \end{aligned}$$

Then,

$$\begin{aligned} \theta (t) \le \theta (0){\bar{f}}(t)+{\check{\Lambda }}{\bar{f}}(t)\int _0^t {\bar{f}}^{-1}(s)\mathrm{d}s, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&{\bar{f}}(t) = e^{\check{\sigma _1}B_1(t)} \times \exp \bigg [\int _0^t \int \limits _Z \big \{\ln [1+G_1(z)]-G_1(z)\big \}\nu (dz)ds\\&\quad + \int _0^t \int \limits _Z \ln [1+G_1(z)]{\tilde{N}}(\mathrm{d}t,\mathrm{d}z)\bigg ]. \end{aligned} \end{aligned}$$

Since, based on the results in [14], we have that

$$\begin{aligned}&\lim _{t\rightarrow \infty } \frac{\int _0^t \int \limits _Z \big \{\ln (1+G_1(z))-G_1(z)\big \}\nu (\mathrm{d}z)\mathrm{d}s + \int _0^t \int \limits _Z \ln (1+G_1(z)){\tilde{N}}(\mathrm{d}s,\mathrm{d}z)}{t} = 0\quad \text {a.s.}, \end{aligned}$$

it is easy to get the desired result that

$$\begin{aligned} \lim _{t\rightarrow \infty } \frac{\ln \theta (t)}{t} = 0 \quad \text {a.s.} \square \end{aligned}$$

Appendix B: Proof to Lemma 2.2

Proof

Applying Itô’s formula to the addition of the first two equations in (1.2), we have that

$$\begin{aligned} \begin{aligned}&d[S(t)+I(t)] = [\Lambda (k)-\alpha (k)S(t)-\mu (k)(S(t)\\&\qquad +I(t))-(\gamma (k) + \delta (k))I(t)]\mathrm{d}t+ \sigma _1(k)S(t)\mathrm{d}B_1(t)\\&\qquad + \sigma _2(k)I(t)\mathrm{d}B_2(t) + \int \limits _Z (G_1(z)S(t-)\\&\qquad +G_2(z)I(t-)){\tilde{N}}(\mathrm{d}t,\mathrm{d}z)\\&\quad \le [{\check{\Lambda }}-{\hat{\mu }}(S(t)+I(t))]\mathrm{d}t\\&\qquad + \sigma _1(k)S(t)\mathrm{d}B_1(t)+\sigma _2(k)I(t)\mathrm{d}B_2(t)\\&\qquad + \int \limits _Z \max \{G_1(z),G_2(z)\}[S(t-)+I(t-)]{\tilde{N}}(\mathrm{d}t,\mathrm{d}z). \end{aligned} \end{aligned}$$

Now,

$$\begin{aligned} \begin{aligned}&\mathrm{d}[S(t)+I(t)]e^{\mu (k)t} = \mu (k)[S(t)+I(t)]e^{\mu (k)t}\mathrm{d}t\\&\qquad + e^{\mu (k)t}\mathrm{d}[S(t)+I(t)]\\&\quad \le {\check{\Lambda }}e^{{\check{\mu }}t}\mathrm{d}t + e^{{\check{\mu }}t}[\check{\sigma _1}S(t)\mathrm{d}B_1(t)+\check{\sigma _2}I(t)\mathrm{d}B_2(t)]\\&\qquad + e^{{\check{\mu }}t}\int \limits _Z \max \{G_1(z),G_2(z)\}[S(t-)+I(t-)]{\tilde{N}}(\mathrm{d}t,\mathrm{d}z). \end{aligned}\nonumber \\ \end{aligned}$$

(5.2)

Integrating (5.2), we have that

$$\begin{aligned} \begin{aligned}&S(t)+I(t) \le [S(0)+I(0)]e^{-{\hat{\mu }}t} + {\check{\Lambda }}\int _0^t e^{-{\hat{\mu }}(t-s)}\mathrm{d}s\\&\qquad + \int _0^t e^{-{\hat{\mu }}(t-s)}[\check{\sigma _1}S(s)\mathrm{d}B_1(s)+\check{\sigma _2}I(s)\mathrm{d}B_2(s)]\mathrm{d}s\\&\qquad + \int _0^t e^{-{\hat{\mu }}(t-s)}\int \limits _Z \max \{G_1(z),G_2(z)\}[S(s-)\\&\qquad +I(s-)]{\tilde{N}}(\mathrm{d}s,\mathrm{d}z)\mathrm{d}s. \end{aligned} \end{aligned}$$

Taking expectations, we have that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}[S(t)+I(t)]\le [S(0)+I(0)]e^{-{\hat{\mu }}t} + {\check{\Lambda }}\int _0^t e^{-{\hat{\mu }}(t-s)}\mathrm{d}s\\&\quad \le [S(0)+I(0)]e^{-{\hat{\mu }}t} + \frac{{\check{\Lambda }}}{{\hat{\mu }}}. \end{aligned} \end{aligned}$$

Now, we have that

$$\begin{aligned} \begin{aligned}&\lim _{t\rightarrow \infty }\sup \frac{1}{t}\int _0^t {\mathbb {E}}[S(s)+I(s)]\mathrm{d}s\\&\quad \le [S(0)+I(0)]\lim _{t\rightarrow \infty }\sup \frac{1}{t}\int _0^t e^{-{\hat{\mu }}s} ds + \frac{{\check{\Lambda }}}{{\hat{\mu }}} \\&\quad \le \frac{{\check{\Lambda }}}{{\hat{\mu }}}. \end{aligned} \end{aligned}$$

$\square$ .