Can Vaccination Save a Zika Virus Epidemic?

Abstract

Zika virus (ZIKV) is a vector-borne disease that has rapidly spread during the year 2016 in more than 50 countries around the world. If a woman is infected during pregnancy, the virus can cause severe birth defects and brain damage in their babies. The virus can be transmitted through the bites of infected mosquitoes as well as through direct contact from human to human (e.g., sexual contact and blood transfusions). As an intervention for controlling the spread of the disease, we study a vaccination model for preventing Zika infections. Although there is no formal vaccine for ZIKV, The National Institute of Allergy and Infectious Diseases (part of the National Institutes of Health) has launched a vaccine trial at the beginning of August 2016 to control ZIKV transmission, patients who received the vaccine are expected to return within 44 weeks to determine if the vaccine is safe. Since it is important to understand ZIKV dynamics under vaccination, we formulate a vaccination model for ZIKV spread that includes mosquito as well as sexual transmission. We calculate the basic reproduction number of the model to analyze the impact of relatively, perfect and imperfect vaccination rates. We illustrate several numerical examples of the vaccination model proposed as well as the impact of the basic reproduction numbers of vector and sexual transmission and the effect of vaccination effort on ZIKV spread. Results show that high levels of sexual transmission create larger cases of infection associated with the peak of infected humans arising in a shorter period of time, even when a vaccine is available in the population. However, a high level of transmission of Zika from vectors to humans compared with sexual transmission represents that ZIKV will take longer to invade the population providing a window of opportunities to control its spread, for instance, through vaccination.

Appendices

Appendix A $\mathcal {R}_0$ Calculations

In this appendix, we calculate the basic reproduction of the ZIKV model using the Next Generation Operator (Van den Driessche and Watmough 2002); in our case, the population is divided into $n=5$ compartments, the infected compartments are i, $e_v$ and $i_v$ giving $m=3$. The ZFE is given by $\mathbf {x}_{0}=(0,0,0,\frac{ (\mu +\omega )}{\mu +\phi +\omega },\frac{\phi }{\mu +\phi +\omega })^t$ where $\mathbf {x}=(i,e_v,i_v,s,v)^t$. The progression of an infected individual through various compartments is given by:

$$\begin{aligned} \mathcal {F}(x)&= \left( \begin{array}{c} \beta s i_v + \alpha si \\ 0 \\ 0\\ 0\\ 0 \end{array} \right) \; \text {and}&\mathcal {V}(x)&= \left( \begin{array}{c} (\gamma +\mu ) i\\ -\beta _{v} s_{v}i+(\mu _{v} + \eta )e_{v}\\ -\eta e_{v} + \mu _{v} i_{v}\\ -\mu +\beta si_{v}+\alpha s i +\phi s - \omega v +\mu s\\ -\phi s + \omega v + \mu v \end{array} \right) . \end{aligned}$$

To calculate $\mathcal {R}_{0}$, we determine matrices F and V, given by

$$\begin{aligned} F(x_0) =\left( \begin{array}{ccc} \alpha s^* &{} 0 &{} \beta s^* \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) \; \text {and} \; V(x_0) =\left( \begin{array}{ccc} \mu +\gamma &{} 0 &{} 0 \\ -\beta _v &{} \mu _{v}+\eta &{} 0 \\ 0 &{} -\eta &{} \mu _{v} \\ \end{array} \right) . \end{aligned}$$

Using simple algebra to calculate $V^{-1}$, we obtain

$$\begin{aligned} V^{-1}(x_0) =\left( \begin{array}{ccc} \frac{1}{\mu +\gamma } &{} 0 &{} 0 \\ \frac{\beta _{v}}{(\mu +\gamma )(\mu _{v}+\eta )} &{} \frac{1}{\mu _{v}+\eta } &{} 0 \\ \frac{\eta \beta _{v}}{\mu _{v}(\mu +\gamma )(\mu _{v}+\eta )} &{} \frac{\eta }{\mu _{v}(\mu _{v}+\eta )} &{} \frac{1}{\mu _{v}} \\ \end{array}\right) \end{aligned}$$

and the next generation matrix of the model with vaccination is given by

$$\begin{aligned} FV^{-1}(x_0) =\left( \begin{array}{ccc} \frac{\alpha s^*}{(\mu +\gamma )}+\frac{\beta \beta _{v} \eta s^*}{\mu _{v}(\mu +\gamma )(\mu _{v}+\eta )} &{} \frac{\beta \eta s^* }{\mu _{v}(\mu _{v}+\eta )} &{} \frac{\beta s^*}{\mu _{v}} \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array}\right) \end{aligned}$$

where $\mathcal {R}_{0}=\rho (FV^{-1})$ and thus

$$\begin{aligned} \mathcal {R}_{0}= \frac{\alpha s^*}{(\mu +\gamma )}+\frac{\beta \beta _{v} \eta s^*}{\mu _{v}(\mu +\gamma )(\mu _{v}+\eta )}, \end{aligned}$$

That is,

$$\begin{aligned} \mathcal {R}_{0}(\phi ,\omega ) = \frac{\alpha (\mu +\omega )}{(\mu +\phi +\omega )(\mu +\gamma )} + \frac{\beta \beta _{v} \eta (\mu +\omega ) }{ \mu _{v}(\mu +\gamma )(\mu _{v}+\eta )(\mu +\phi +\omega )}. \end{aligned}$$

Appendix B Proof of Theorem 1

In this appendix, we provide a proof of Theorem 1. In our case, we have that $\mathbf X =(s,v,r,s_v)^t$, $\mathbf Z =(i,e_v,i_v)^t$ thus: $\mathbf U _0(\mathbf X ^*,0)=(1-\rho ,\rho ,0,1,0,0,0)^t$,

$$\begin{aligned} H(\mathbf X ,\mathbf Z )=\left( \begin{array}{c} \mu -\beta s i_v - \alpha s i -\phi fs + \omega v -\mu s \\ \phi fs - \omega v - \mu v, \\ \gamma i -\mu r \\ \mu _{v} -\mu _{v}s_{v}-\beta _{v}s_{v}i \end{array} \right) \; \text {and} \; G(\mathbf X,Z )=\left( \begin{array}{c} \beta s i_v + \alpha s i- (\gamma +\mu ) i, \\ \beta _{v}s_{v}i-(\mu _{v} + \eta )e_{v},\\ \eta e_{v} - \mu _{v} i_{v} \\ \end{array} \right) . \end{aligned}$$

To show (H1), we calculate

$$\begin{aligned} H(\mathbf X ,0)=\left( \begin{array}{c} \mu -\phi fs + \omega v -\mu s \\ \phi fs - \omega v - \mu v, \\ -\mu r \\ \mu _{v} -\mu _{v}s_{v} \end{array} \right) \end{aligned}$$

and then the Jacobian matrix of H at $\mathbf X ^*$ to study if its eigenvalues have negative real part

$$\begin{aligned} J_H(\mathbf X* ,0)=\left( \begin{array}{c c c c} -(\phi f+\mu ) &{} \omega &{} 0 &{} 0 \\ \phi f&{} -(\omega +\mu ) &{} 0 &{} \\ 0 &{} 0 &{} -\mu &{} \\ 0 &{} 0 &{} 0 &{} -\mu _{v} \end{array} \right) . \end{aligned}$$

Therefore, it can be easily checked that all the eigenvalues of the matrix above have negative real part and then $\mathbf X ^*$ is l.a.s of the system $\frac{d\mathbf X }{dt} =H(\mathbf X ,0)$.

To show (H2), first we calculate the matrix A

$$\begin{aligned} A= D_\mathbf{Z }G(\mathbf X ^*,0)&= \left( \begin{array}{c c c} \alpha (1-\rho ) - (\gamma +\mu ) &{} 0 &{} \beta (1-\rho ) \\ \beta _{v} &{} -(\mu _{v} + \eta ) &{} 0\\ 0 &{} \eta &{} - \mu _{v} \end{array} \right) \\&= \left( \begin{array}{c c c} (\gamma + \mu ) (R_d(1-\rho ) - 1) &{} 0 &{} \beta (1-\rho ) \\ \beta _{v} &{} -(\mu _{v} + \eta ) &{} 0\\ 0 &{} \eta &{} - \mu _{v} \end{array} \right) , \end{aligned}$$

where $\rho =\frac{\phi f}{\mu +\omega +\phi f}$. Since $\mathcal {R}_0<1$, it follows that $(1-\rho )R_d<1$ as well and thus, the off elements of the diagonal of the matrix A have negative sign which means that A is an M-matrix. Second, the vector $\hat{G}$ can be expressed as $\hat{G}(\mathbf X,Z )=G(\mathbf X,Z )-A\mathbf Z$ and after some simple manipulations

$$\begin{aligned} \hat{G}(\mathbf X,Z )=\left( \begin{array}{c} (\alpha i+ \beta i_v)(1-\rho -s) \\ 0\\ 0 \end{array}\right) \end{aligned}$$

Since $0<s<1-\rho$, then the last condition holds with $\hat{G}(\mathbf X,Z )\ge 0$ and therefore $\mathbf U _0$ is g.a.s.