Wolbachia spreading dynamics in mosquitoes with imperfect maternal transmission

Abstract

Mosquitoes are primary vectors of life-threatening diseases such as dengue, malaria, and Zika. A new control method involves releasing mosquitoes carrying bacterium Wolbachia into the natural areas to infect wild mosquitoes and block disease transmission. In this work, we use differential equations to describe Wolbachia spreading dynamics, focusing on the poorly understood effect of imperfect maternal transmission. We establish two useful identities and employ them to prove that the system exhibits monomorphic, bistable, and polymorphic dynamics, and give sufficient and necessary conditions for each case. The results suggest that the largest maternal transmission leakage rate supporting Wolbachia spreading does not necessarily increase with the fitness of infected mosquitoes. The bistable dynamics is defined by the existence of two stable equilibria, whose basins of attraction are divided by the separatrix of a saddle point. By exploring the analytical property of the separatrix with some sharp estimates, we find that Wolbachia in a completely infected population could be wiped out ultimately if the initial population size is small. Surprisingly, when the infection shortens the lifespan of infected females that would impede Wolbachia spreading, such a reversion phenomenon does not occur.

Appendices

Appendices

A The trace of the Jacobian matrix

Lemma A.1

Let (1.6) hold. Then tr $\mathbf {J}<0$ at the interior equilibrium point of (1.4)–(1.5) if it is the unique one, or the one with a smaller y-coordinate.

Proof

In view of Lemma 2.2 and the labeling of equilibrium points following its proof, the current lemma assures tr $\mathbf {J}<0$ at $E^*(x^*, y^*)$ in Case (C2), or at $E_2^*(x_2^*, y_2^*)$ in Cases (C3), (C4) (i), or (C5). For notational simplicity, we let $(x_2,y_2)$ stand for $(x_2^*, y_2^*)$. By the second part of (3.3), it is equivalent to prove $H(y^*)>0$ or $H(y_2)>0$.

Case (C2): $\kappa < 1$, $\beta \mu <1$, and $(\kappa +\beta \mu )^2=4\beta \mu$. The system has a unique interior equilibrium point $E^*(x^*, y^*)$. By substituting $y=y^*$ into (3.4), changing $(1-\delta )\kappa$ to $-\delta \kappa$, and replacing $(\kappa +\beta \mu )^2$ with $4\beta \mu$ when applicable, we obtain

$$\begin{aligned} H(y^*)> & {} \frac{\kappa (\kappa +\beta \mu )^2}{4}-\left( \delta \kappa +2\right) \frac{\kappa (\kappa +\beta \mu )}{2}+(1+\delta )\kappa ^2\nonumber \\= & {} \kappa \beta \mu -\frac{\delta \kappa ^2(\kappa +\beta \mu )}{2}-\kappa ^2-\kappa \beta \mu +(1+\delta )\kappa ^2\\= & {} \frac{\delta \kappa ^2}{2}\left[ 2-(\kappa +\beta \mu )\right] >0\nonumber \end{aligned}$$

(A.1)

where the last inequality follows from the assumptions $\kappa <1$ and $\beta \mu <1$.

Case (C3): $\kappa <1$, $\beta \mu <1$, and $(\kappa +\beta \mu )^2>4\beta \mu$. By changing the equality in (A.1) to “>”, we see that $H(y^*)>0$ still holds. Instead of computing $H(y_2)$ directly, we proceed by showing that $H(y)-G(y)$ is positive at $y=y_2$ as $G(y_2)=0$. Subtracting (3.4) by (2.10) gives

$$\begin{aligned} H(y)-G(y)=c_1y+c_2 \end{aligned}$$

(A.2)

with the constant coefficients

$$\begin{aligned} c_1=\kappa +\beta \mu +(1-\delta )\kappa -2, \quad \mathrm{and}\quad c_2=(1+\delta )\kappa ^2-\beta \mu \kappa . \end{aligned}$$

(A.3)

We claim $c_2>0$. To confirm it, we first show that Condition (C3) implies

$$\begin{aligned} \beta \mu<\kappa \quad \mathrm{and}\quad \mu <\frac{1}{1+\delta }. \end{aligned}$$

(A.4)

Indeed, by using (2.2) we may rewrite $\kappa +\beta \mu$ as

$$\begin{aligned} \kappa +\beta \mu =\frac{\beta (1-\mu )}{\delta }+\beta \mu =\beta \mu (\gamma +1) \quad \mathrm{where}\quad \gamma =\frac{1-\mu }{\mu \delta }>0. \end{aligned}$$

Then by applying Condition (C3) we obtain

$$\begin{aligned} \frac{4}{(1+\gamma )^2}<\beta \mu<1 \, \Rightarrow \, (1+\gamma )^2>4 \, \Rightarrow \, \gamma>1 \, \Rightarrow \, 1-\mu >\mu \delta \, \Rightarrow \, \mu <\frac{1}{1+\delta }. \end{aligned}$$

Furthermore, $\gamma >1$ gives $(1-\mu )/\delta >\mu$, and so $\kappa =\beta (1-\mu )/\delta >\beta \mu$. This completes the proof of (A.4). Consequently,

$$\begin{aligned} c_2= & {} \frac{\beta \kappa }{\delta } \left[ (1+\delta )(1-\mu )-\mu \delta \right] = \frac{\beta \kappa }{\delta } \left[ 1+\delta -(1+2\delta )\mu )\right] \\> & {} \frac{\beta \kappa }{\delta } \left( 1+\delta -\frac{1+2\delta }{1+\delta }\right) =\frac{\beta \kappa \delta }{1+\delta }>0. \end{aligned}$$

Of course, if $c_1\ge 0$, then $H(y)>G(y)$ for all $y>0$ and there remains nothing to prove. We note that $c_1<0$ is possible as shown by an example with $\delta =1$, $\kappa +\beta \mu =\beta <2$, and $c_1=\beta -2<0$. Since $y_2$ is the smaller root between the two distinct roots of G(y) in $(0,\kappa )$, $y_2$ is less than the critical point of G where it takes the minimum value. Hence

$$\begin{aligned} y_2<y^*\quad \mathrm{and}\quad G(y^*)<0. \end{aligned}$$

Recall that $H(y^*)>0$. Now, when $c_1<0$, we have

$$\begin{aligned} H(y_2)=H(y_2)-G(y_2)>H(y^*)-G(y^*)>H(y^*)>0. \end{aligned}$$

Cases (C4) (i) and (C5): By changing $(1-\delta )\kappa$ to $-\delta \kappa$ in (3.4), and then replacing $y_2$ with $\kappa -x_2$, we find

$$\begin{aligned} H(y_2)>\frac{(\kappa -x_2)^2}{\kappa }-\left( \delta \kappa +2\right) (\kappa -x_2)+(1+\delta )\kappa ^2 =\frac{x_2^2}{\kappa }+\delta \kappa x_2+\kappa ^2-\kappa . \end{aligned}$$

Since $\kappa \ge 1$ in these cases, it follows immediately that $H(y_2)> \kappa (\kappa -1)\ge 0$. $\square$

B Proof of Lemma 3.1

Proof

In Case (C2), $E^*(x^*, y^*)\in \Gamma _y$. In Case (C3), the intersection at a point $(x,y^*)$ with $x_1^*<x<x_2^*$ is also clear in view of (2.13). To show the uniqueness of the intersection, we first prove

$$\begin{aligned} y^*=\frac{\kappa (\kappa +\beta \mu )}{2}>\beta \mu \end{aligned}$$

(B.1)

when either (C2) or (C3) holds. Recall from (A.4) that $\kappa >\beta \mu$ in Case (C3). By checking the proof of (A.4) it is easy to see that it again goes through for Case (C2) and that $\kappa >\beta \mu$ remains valid. Now we have

$$\begin{aligned} 4\kappa ^2=(\kappa +\kappa )^2>(\kappa +\beta \mu )^2\ge 4\beta \mu \quad \Rightarrow \quad \kappa ^2>\beta \mu \quad \Rightarrow \quad \kappa >\sqrt{\beta \mu }. \end{aligned}$$

As $(\kappa +\beta \mu )^2\ge 4\beta \mu$ implies $\kappa +\beta \mu \ge 2 \sqrt{\beta \mu }$, (B.1) follows at once.

Let $(x, y^*)\in \Gamma _y$. Then $g(x, y^*)=0$, or

$$\begin{aligned} (\beta \mu -y^*) x+ \frac{(y^*)^2}{x+y^*}- (y^*)^2=0. \end{aligned}$$

We may rewrite it as a quadratic equation of x in the form

$$\begin{aligned} (\beta \mu -y^*) x^2+ y^*(\beta \mu -2y^*)x+(y^*)^2- (y^*)^3=0 \end{aligned}$$

with the leading coefficient $\beta \mu -y^*<0$ by (B.1), and $(y^*)^2- (y^*)^3>0$ as $y^*<\kappa <1$. Hence it has at least one negative root and no more than one positive root. $\square$

C Proof of Lemma 3.2

Proof

Let $a\ge \delta$. When $x>x^*$, it is apparently true by (3.10) that $y=h_a(x)>y^*$ in Case (C2). For Case (C3), we see from (2.13) that $x_1^*<x^*$, $y_1^*>y^*$, and therefore

$$\begin{aligned} y=h_a(x)>h_a(x^*)>h_a(x_1^*)=y_1^*>y^*. \end{aligned}$$

This and (B.1) help us derive

$$\begin{aligned} F_a(x, h_a(x))>\frac{a-\delta }{a}\, x-\frac{\beta \mu x}{y}-1>\left( 1-\frac{\beta \mu }{y^*}-\frac{\delta }{a} \right) x-1>0, \end{aligned}$$

provided that

$$\begin{aligned} a>{\bar{a}}=\frac{2\delta y^*}{y^*-\beta \mu },\quad \mathrm{and}\quad x>\frac{{\bar{a}}}{\delta }=\frac{2y^*}{y^*-\beta \mu }. \end{aligned}$$

(C.1)

Next we prove (3.11) for x close to $x^*$ and $x>x^*$. First let (C3) hold. In this case, we have $y=h_a(x)>y_1^*$ for all $x>x_1^*$. In the domain where $x>x_1^*$ and $y>y_1^*$, we have $f(x,y)<0$ and therefore for all $a\ge \delta$

$$\begin{aligned} F_a(x,y)=\frac{f(x,y)}{a x}-\frac{g(x,y)}{y}\ge \frac{f(x,y)}{\delta x}-\frac{g(x,y)}{y}= z'(t)=-\,\frac{G(y_s)}{y_s}. \end{aligned}$$

When $x\in (x_1^*, x_2^*]$, the shadow point $(x_s,y_s)$ of $(x,h_a(x))$ satisfies $y_s\in (y_2^*, y_1^*)$ in which $G<0$. It follows that $F_a(x, h_a(x))>0$ for $x\in (x_1^*, x_2^*]$ which includes $(x^*, x_2^*]$ as a subinterval. This verifies (3.11) for all $a\ge \delta$ and $x\in (x^*, x_2^*]$ in Case (C3). Now let (C2) hold. We need a more delicate approach since $G(y)>0$ for all $y>y^*$ now. As $f(x^*,y^*)=g(x^*,y^*)=0$, we have $F_a(x^*,y^*)=0$ and so

$$\begin{aligned} F_a(x,y)= & {} \frac{a-\delta }{a}\, (x-x^*+y-y^*)-\frac{\beta \mu x}{y}+\frac{\beta \mu x^*}{y^*} - \frac{ y}{x+y}+\frac{ y^*}{\kappa } \\= & {} \frac{a-\delta }{a}\, (x-x^*+y-y^*)-\frac{\beta \mu y^*(x-x^*)-\beta \mu x^*(y-y^*)}{yy^*}\\&+\,\frac{y^*(x-x^*)-x^*(y-y^*)}{\kappa (x+y)}\\= & {} \left[ \frac{a-\delta }{a}-\frac{\beta \mu }{y}+\frac{y^*}{\kappa (x+y)}\right] (x-x^*)\\&+\left[ \frac{a-\delta }{a}+\frac{\beta \mu x^*}{yy^*}-\frac{x^*}{\kappa (x+y)}\right] (y-y^*). \end{aligned}$$

In Case (C2), $\kappa +\beta \mu =2\sqrt{\beta \mu }$, $y^*=\kappa (\kappa +\beta \mu )/2=\kappa \sqrt{\beta \mu }$, and therefore

$$\begin{aligned} \lim _{(x,y)\rightarrow (x^*,y^*)} \left( \frac{y^*}{\kappa (x+y)}-\frac{\beta \mu }{y}\right) =\frac{y^*}{\kappa ^2}-\frac{\beta \mu }{y^*}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{(x,y)\rightarrow (x^*,y^*)} \left( \frac{\beta \mu x^*}{yy^*}-\frac{x^*}{\kappa (x+y)}\right) =-\frac{x^*}{y^*}\left( \frac{y^*}{\kappa ^2}-\frac{\beta \mu }{y^*}\right) =0. \end{aligned}$$

Let $a\ge 2\delta$, there is ${\bar{x}}>x^*$ such that for all $x\in (x^*, {\bar{x}}]$,

$$\begin{aligned} \frac{y^*}{\kappa (x+h_a(x))}-\frac{\beta \mu }{h_a(x)}>-\frac{1}{4}, \quad \mathrm{and}\quad \frac{\beta \mu x^*}{h_a(x)y^*}-\frac{x^*}{\kappa (x+h_a(x))}>-\frac{1}{4},\qquad \end{aligned}$$

(C.2)

and consequently

$$\begin{aligned} F_a(x,h_a(x))>(x-x^*)/4+(h(x)-y^*)/4>0. \end{aligned}$$

Finally, let $I_x=[{\bar{x}}, {{\bar{a}}}/\delta ]$ in Case (C2), or $I_x=[x_2^*, {{\bar{a}}}/\delta ]$ in Case (C3), where ${\bar{a}}$ is given in (C.1). For all $x\in I_x$ and $y\in [y^*, h_\delta (x)]$, $f(x, y)<0$ as $x+y>\kappa$, and $g(x, y)<0$ by Lemma 3.1. Let

$$\begin{aligned} M_f=\max _{I_x\times [y^*, \, h_\delta (x)]} \frac{|f(x, y)|}{x} \quad \mathrm{and}\quad m_g=\min _{I_x\times [y^*, \,h_\delta (x)]} \frac{|g(x, y)|}{y}. \end{aligned}$$

Then for any $a> M_f/m_g$ and $x\in I_x$, we have

$$\begin{aligned} F_a(x, h_a(x))=\frac{|g(x,h_a(x))|}{h_a(x)}-\frac{|f(x,h_a(x))|}{a x}\ge m_g-\frac{M_f}{a}>0. \end{aligned}$$

In conclusion, we find that (3.11) is valid if $a>\max \{{\bar{a}}, 2\delta , M_f/m_g\}$. $\square$