A switching model for the impact of toxins on the spread of infectious diseases

Abstract

To study the effects of an environmental toxin, such as fine particles in hazy weather, on the spread of infectious diseases, we derive a toxin-dependent dynamic model that incorporates the birth rate with the toxin-dependent switching mode, the mortality rate, and infection rate with the toxin-dependent saturation effect. We analyze the model by showing the positive invariance, existence and stability of equilibria, and bifurcations. Numerical simulation is adopted to verify the mathematical results and exhibit transcritical and Hopf bifurcations. Our theoretical results show that there exists a threshold value of the environmental toxin: if the environmental toxin concentration is lower than the threshold, the system has a disease-free equilibrium and an interior equilibrium; if the environmental toxin concentration is higher than the threshold, the system has the extinction equilibrium. For the case where the disease-induced death is ignored, we show the global stability results. Numerical simulations clearly show that the environmental toxin facilitates the spread of infectious diseases. This study provides a theoretical basis for uncovering the impact of toxins on the spread of infectious diseases and for guiding the decision making by disease control agencies and governments.

Appendix: Proof of Theorem 4.2

Proof

The disease-free equilibrium is $E_{1} (S_{0},0)$, and the interior equilibrium is $E_{2}(S^{*},I^{*})$, where $S_{0}=\frac{(1-\frac{\alpha \sigma A}{\theta \xi })a}{\mu +\frac{k\sigma A}{\theta \xi }}, S^{*}=\frac{\mu +\frac{k\sigma A}{\theta \xi }+\gamma }{\lambda (1+\frac{m\sigma A}{b\theta \xi +\sigma A})}$, $I^{*}=\frac{(1-\frac{\alpha \sigma A}{\theta \xi })a}{\mu +\frac{k\sigma A}{\theta \xi }}-\frac{\mu +\frac{k\sigma A}{\theta \xi }+\gamma }{\lambda (1+\frac{m\sigma A}{b\theta \xi +\sigma A})}$.

In order to ensure $I^{*}>0$, we need $\frac{a\lambda (1-\frac{\alpha \sigma A}{\theta \xi })(1+\frac{m\sigma A}{b\xi \theta +\sigma A})}{(\mu +\frac{k\sigma A}{\xi \theta })(\mu +\gamma +\frac{k\sigma A}{\xi \theta })}>1$. Let $R_{0}=\frac{a\lambda (1-\frac{\alpha \sigma A}{\theta \xi })(1+\frac{m\sigma A}{b\xi \theta +\sigma A})}{(\mu +\frac{k\sigma A}{\xi \theta })(\mu +\gamma +\frac{k\sigma A}{\xi \theta })}$, which is called the basic reproductive number. We can obtain $I^{*}>0$, if $R_{0}>1$.

The Jacobian matrix at $E_{1}$ is

$$\begin{aligned} J(E_{1})= \left( \begin{array}{ccc} -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) &{}\quad -\lambda \frac{\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) +\gamma \\ 0&{}\quad \lambda \frac{\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \\ \end{array}\right) , \end{aligned}$$

and the eigenvalues are

$$\begin{aligned} \lambda _{1}= & {} -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) , \\ \lambda _{2}= & {} \lambda \frac{\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) . \end{aligned}$$

Obviously, $\lambda _{1}=-(\mu +\frac{k\sigma A}{\theta \xi })<0$.

If $R_{0}<1$, we have

$$\begin{aligned} \lambda \frac{\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) <0, \end{aligned}$$

thus $\lambda _{2}<0$.

If $R_{0}<1$, the equilibrium $E_{1}$ is locally asymptotically stable, and since the disease-free equilibrium $E_{1}$ is the only equilibrium under this condition, then $E_{1}$ is globally asymptotically stable.

The Jacobian matrix at $E_{2}$ is

$$\begin{aligned}&J(E_{2}) \\&\quad = \left( \begin{array}{ccc} -\lambda I^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) &{}\quad -\lambda S^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) +\gamma \\ \lambda I^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) &{}\quad \lambda S^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \\ \end{array}\right) . \end{aligned}$$

We have

$$\begin{aligned} Tr(J(E_{2}))= & {} \lambda S^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \\&-\lambda I^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) \\= & {} \gamma -\lambda \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \frac{\left( 1-\frac{A\sigma \alpha }{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}. \end{aligned}$$

we know $\frac{\lambda (1-\frac{\alpha \sigma A}{\theta \xi })a(1+\frac{m\sigma A}{b\theta \xi +\sigma A})}{\gamma (\mu +\frac{k\sigma A}{\theta \xi })}>\frac{a\lambda (1-\frac{\alpha \sigma A}{\theta \xi })(1+\frac{m\sigma A}{b\xi \theta +\sigma A})}{(\mu +\frac{k\sigma A}{\xi \theta })(\mu +\gamma +\frac{k\sigma A}{\xi \theta })}$, and if $R_{0}>1$, we have $Tr(J(E_{2}))<0$.

On the other hand, we have

$$\begin{aligned} \det (J(E_{2}))= & {} \left( -\lambda I^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) \right) \left( \lambda S^{*} \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \right. \\&-\left. \left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \right) -\left( \lambda I^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \right) \\&\left( -\lambda S^{*}\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) +\gamma \right) \\= & {} \lambda \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \left( \mu +\frac{k\sigma A}{\theta \xi }\right) \left( \frac{\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{\mu +\frac{k\sigma A}{\theta \xi }}-\frac{2\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) }{\lambda \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) }\right) \\&+\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \left( \mu +\frac{k\sigma A}{\theta \xi }\right) \\= & {} \lambda \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a-\left( \mu +\frac{k\sigma A}{\theta \xi }+\gamma \right) \left( \mu +\frac{k\sigma A}{\theta \xi }\right) . \end{aligned}$$

Clearly, if $R_{0}>1$, we have $\det (J(E_{2}))>0.$

If $R_{0}>1$, the interior equilibrium $E_{2}$ is locally asymptotically stable.

Finally, we will prove that system (8) has no periodic orbits (Chauhan et al. 2015).

Now we choose the Dulac function

$$\begin{aligned} B(S,I)=\frac{\mu +S}{SI}. \end{aligned}$$

Let

$$\begin{aligned} f(S,I)= & {} \frac{dS}{dt}=\left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a-\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \lambda SI-\left( \mu +\frac{k\sigma A}{\theta \xi }\right) S+\gamma I, \\ g(S,I)= & {} \frac{dI}{dt}=\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \lambda SI-\left( \mu +\frac{k\sigma A}{\theta \xi }\right) I-\gamma I. \end{aligned}$$

We have

$$\begin{aligned} \frac{\partial }{\partial S}(Bf)+\frac{\partial }{\partial I}(Bg)= & {} \frac{\partial }{\partial S} \left( \frac{\mu +S}{SI}\left( \left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a-\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \lambda SI\right. \right. \\&\left. \left. -\left( \mu +\frac{k\sigma A}{\theta \xi }\right) S+\gamma I\right) \right) \\&+\frac{\partial }{\partial I}\left( \frac{\mu +S}{SI}\left( \left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \lambda SI\right. \right. \\&-\left. \left. \left( \mu +\frac{k\sigma A}{\theta \xi }\right) I-\gamma I\right) \right) \\= & {} -\frac{\mu \left( 1-\frac{\alpha \sigma A}{\theta \xi }\right) a}{IS^{2}}-\left( 1+\frac{m\sigma A}{b\theta \xi +\sigma A}\right) \lambda -\frac{\mu +\frac{k\sigma A}{\theta \xi }}{I}, \end{aligned}$$

then

$$\begin{aligned} \frac{\partial }{\partial S}(Bf)+\frac{\partial }{\partial I}(Bg)\le 0. \end{aligned}$$

According to the Bendixson–Dulac criterion, system (8) has no periodic orbits, hence the interior equilibrium $E_{2}$ is globally asymptotically stable. $\square$