Vector-Borne Pathogen and Host Evolution in a Structured Immuno-Epidemiological System

Abstract

Vector-borne disease transmission is a common dissemination mode used by many pathogens to spread in a host population. Similar to directly transmitted diseases, the within-host interaction of a vector-borne pathogen and a host’s immune system influences the pathogen’s transmission potential between hosts via vectors. Yet there are few theoretical studies on virulence–transmission trade-offs and evolution in vector-borne pathogen–host systems. Here, we consider an immuno-epidemiological model that links the within-host dynamics to between-host circulation of a vector-borne disease. On the immunological scale, the model mimics antibody-pathogen dynamics for arbovirus diseases, such as Rift Valley fever and West Nile virus. The within-host dynamics govern transmission and host mortality and recovery in an age-since-infection structured host-vector-borne pathogen epidemic model. By considering multiple pathogen strains and multiple competing host populations differing in their within-host replication rate and immune response parameters, respectively, we derive evolutionary optimization principles for both pathogen and host. Invasion analysis shows that the ${\mathcal {R}}_0$ maximization principle holds for the vector-borne pathogen. For the host, we prove that evolution favors minimizing case fatality ratio (CFR). These results are utilized to compute host and pathogen evolutionary trajectories and to determine how model parameters affect evolution outcomes. We find that increasing the vector inoculum size increases the pathogen ${\mathcal {R}}_0$, but can either increase or decrease the pathogen virulence (the host CFR), suggesting that vector inoculum size can contribute to virulence of vector-borne diseases in distinct ways.

Appendix

Analysis of Immunological Model: Virus Ultimately Dies Out

Notice that in the system (1), in the absence of immune response ($M_0=0, G_0=0$), the pathogen grows exponentially, in which case it is expected that the infected host dies since the parasite damages the host. However, when immune system is active, we establish the following result:

Theorem 4.1

If initial immune response is active ($M_0>0 \text { or } G_0>0$), then the pathogen eventually dies out ($\lim _{\tau \rightarrow \infty }P(\tau )=0$), the IgM immune response antibodies decay to zero after viral clearance, and subsequently, the IgG immune memory antibodies reach a steady state, i.e., $\ \lim _{\tau \rightarrow \infty }M(\tau )=0 \ \text {and} \ \lim _{\tau \rightarrow \infty }G(\tau )=G^*,$ where $G^*>0$.

Proof

Let $P(0)>0$. By the first equation in the immunological model (1), we obtain the following inequality:

$$\begin{aligned} P(\tau )\le P(0)e^{\int _0^\tau {\left[ r-\delta G(s)\right] }ds}. \end{aligned}$$

(8)

Also note that if $M_0>0 \text { or } G_0>0$, then by comparison principle (similar argument to above), after some time, namely $\tilde{\epsilon },$ we obtain

$$\begin{aligned} G(\tau ) \ge G(\tilde{\epsilon })e^{b\int _{\tilde{\epsilon }}^\tau {P(s)}ds}>0. \end{aligned}$$

(9)

By the way of contradiction assume that $\limsup _{\tau \rightarrow \infty }P(\tau )\ne 0$. Then, the right-hand side of the inequality (9) goes to infinity. Therefore, $\lim _{\tau \rightarrow \infty }G(\tau )= \infty$. Then, $\exists \tau ^*: \forall \tau >\tau ^*,$ $G(\tau )>\frac{r}{\delta }$. Then, as $\tau \rightarrow \infty$, the right-hand side of the inequality (8) goes to zero. It is a contradiction. Then, $\lim _{\tau \rightarrow \infty }P(\tau )=0,$ subsequently $\lim _{\tau \rightarrow \infty }M(\tau )=0$ and $\lim _{\tau \rightarrow \infty }G(\tau )=G^*$, for some $G^*>0$. $\square$

In summary, for any solution $(P(\tau ),M(\tau ),G(\tau ))$ of the immunological model (1), with nonzero initial condition (P(0), M(0), G(0)), we have $\lim _{\tau \rightarrow \infty } P(\tau )=0,\ \lim _{\tau \rightarrow \infty } M(\tau )=0, \ \lim _{\tau \rightarrow \infty } G(\tau )= G^*,$ where $G^*$ is a positive real number and depends on the initial condition.

Existence and Uniqueness of Equilibrium

Proof

(Proof of Theorem (2.1)) The proof is omitted. However, a sketch of the proof is as follows. It can be shown that the system is dissipative and asymptotically compact. In addition, the system is uniformly persistent when ${\mathcal {R}}_0>1$ (it is not hard to show that DFE is unstable when ${\mathcal {R}}_0>1;$ then by a similar approach to Proposition 4.4 in Yang et al. (2012), it can be shown that the system is uniform persistent). Then by Zhao (2013), there exists at least one positive steady state. Subsequently, by the argument of Proposition 3.1 in Martcheva et al. (2016), it can be shown that this positive equilibrium has to be unique. The proof of local stability under the condition (5) is contained in the proof of ${\mathcal {R}}_0$ maximization in the next subsection. $\square$

Virus Evolution: ${\mathcal {R}}_0$ Maximization

Proof

(Proof of Theorem (3.1)) By taking $S_H(t)=S^*_H+x_H(t), \ i_{H_1}(\tau ,t)=i^*_{H_1}(\tau )+y_{H_1}(\tau ,t), \ i_{H_2}(\tau ,t)=y_{H_2}(\tau ,t), R_H(t)=R^*_H+z_H(t),S_V(t)=S^*_V+x_V(t), I_{V_1}(t)=I^*_{V_1}+y_{V_1}(t),$ and $I_{V_2}(t)=y_{V_2}(t),$ we linearize the one-host two-strain model (6) about the equilibrium ${\mathcal {E}}_1=(S^*_H, i^*_{H_1}(\tau ),0,R^*_H, S^*_V, I^*_{V_1},0)$ and look for eigenvalues of the linear operator, that is, we look for solutions of the form $x_H(t)=\overline{x}_H e^{\lambda t}, y_{H_1}(\tau ,t)=\overline{y}_{H_1}(\tau ) e^{\lambda t}, y_{H_2}(\tau ,t)=\overline{y}_{H_2}(\tau )e^{\lambda t}, z_H(t)=\overline{z}_H e^{\lambda t}, x_V(t) = \overline{x}_V e^{\lambda t}, y_{V_1}(t)=\overline{y}_{V_1}e^{\lambda t},$ and $y_{V_2}(t)=\overline{y}_{V_2} e^{\lambda t},$ where $\overline{x}_H, \overline{y}_{H_1},\overline{y}_{H_2},\overline{z}_H,\overline{x}_V,\overline{y}_{V_1}$ and $\overline{y}_{V_2}$ are arbitrary nonzero constants (a function of $\tau$ in the case of $y_{H_i}$), but the eigenvalue $\lambda$ is common. This process results in the following system (the bars have been omitted):

$$\begin{aligned} \left\{ \begin{array}{l} \lambda x_H=\tilde{T_0}\left( x_H+\sum _{i=1}^{2}{\displaystyle \int _0^\infty {(1+\frac{\gamma _i(\tau )}{(\lambda +d)})y_{H_i}(\tau )d\tau }}\right) \\ \qquad \qquad -\sum _{i=1}^{2}{\beta _{v_i}S_H^*y_{V_i}} - \beta _{v_1}x_HI_{V_1}^* -dx_H, \\ \displaystyle \frac{d y_{H_1}}{d\tau } +\lambda y_{H_1}(\tau ) =-\left( \alpha _1(\tau ) + \kappa _1(\tau )+\gamma _1(\tau )+d \right) y_{H_1}(\tau ), \\ y_{H_1}(0) = \beta _{v_1} S_H^*y_{V_1} +\beta _{v_1} x_H I_{V_1}^* ,\\ \displaystyle \frac{dy_{H_2}}{d\tau } +\lambda y_{H_2}(\tau ) =-\left( \alpha _2(\tau ) + \kappa _2(\tau )+\gamma _2(\tau )+d \right) y_{H_2}(\tau ), \\ y_{H_2}(0) = \beta _{v_2} S_H^*y_{V_2} ,\\ \lambda x_V = -S_V^* \sum _{i=1}^{2}{\displaystyle \int _0^\infty {\beta _{h_i}(\tau )y_{H_i}(\tau )d\tau }} - x_V \sum _{i=1}^{2}{\displaystyle \int _0^\infty {\beta _{h_i}(\tau )i_{H_i}^*(\tau )}d\tau } -\mu x_V ,\\ \lambda y_{V_1} = S_V^*\displaystyle \int _0^\infty {\beta _{h_1}(\tau )y_{H_1}(\tau )}d\tau + x_V\displaystyle \int _0^\infty {\beta _{h_1}(\tau )i_{H_1}^*(\tau )}d\tau -\mu y_{V_1} \\ \lambda y_{V_2} = S_V^*\displaystyle \int _0^\infty {\beta _{h_2}(\tau )y_{H_2}(\tau )}d\tau -\mu y_{V_2}, \\ \end{array}\right. \end{aligned}$$

(10)

Note that by fourth and fifth equation in (10), we have $y_{H_2}=y_{H_2}(0)e^{-\lambda \tau }\pi _2(\tau ),$ where $\pi _2(\tau )=e^{-\displaystyle \int _0^\tau {(\alpha _2(s)+\kappa _2(s)+\gamma _2(s)+d)}}$. Then, rearranging the last equation in (10), we obtain

$$\begin{aligned} y_{V_2}=\frac{S^*_V y_{H_2}(0)\displaystyle \int _0^\infty {\beta _{h_2}(\tau )e^{-\lambda \tau }\pi _2(\tau )d\tau }}{(\lambda +\mu )} \end{aligned}$$

(11)

Substituting (11) into the fifth equation (boundary condition for strain type 2), we obtain

$$\begin{aligned} 1=S_V^* \beta _{v_2}S_H^* \displaystyle \int _0^\infty {\beta _{h_2}(\tau )\frac{e^{-\lambda \tau }}{\lambda +\mu }\pi _2(\tau )d\tau }. \end{aligned}$$

(12)

Suppose that $y_{V_2} \ne 0$, then the eigenvalues of the system will be determined by the characteristic equation $G(\lambda ) =1$, where $G(\lambda )$ is the right-hand side of the equation (12).

Note by the equilibrium condition, we have

$$\begin{aligned} \displaystyle \frac{1}{S_V^*S_H^*}=\beta _{v_1}\displaystyle \int _0^\infty {\beta _{h_1}(\tau )\frac{1}{\mu }\pi _1(\tau )d\tau }, \end{aligned}$$

(13)

where $\pi _1(\tau )=e^{-\displaystyle \displaystyle \int _0^\tau {(\alpha _1(s)+\kappa _1(s)+\gamma _1(s)+d)ds}}$. Substituting (13) into (12), we obtain

$$\begin{aligned} G(\lambda )=\frac{\beta _{v_2}\displaystyle \int _0^\infty {\beta _{h_2}(\tau )\frac{e^{-\lambda \tau }}{\lambda +\mu }\pi _2(\tau )d\tau }}{\beta _{v_1}\displaystyle \int _0^\infty {\beta _{h_1}(\tau ) \frac{1}{\mu }\pi _1(\tau )d\tau }} \end{aligned}$$

(14)

Notice that $G(0)=\displaystyle \frac{{\mathcal {R}}^2_0}{{\mathcal {R}}^1_0},$ where ${\mathcal {R}}^i_0$ is reproduction number for strain type i. If ${\mathcal {R}}^2_0>{\mathcal {R}}^1_0,$ then $G(0)>1$. Since $G(\lambda )$ is a decreasing function of $\lambda$, where $\lambda$ is restricted to real numbers, and $\lim _{\lambda \rightarrow \infty } G(\lambda )=0,$ by intermediate value theorem, there exists a positive real number $\lambda ^*$ such that $G(\lambda ^*)=1$. Therefore, if ${\mathcal {R}}^2_0>{\mathcal {R}}^1_0,$ then the strain-one equilibrium ${\mathcal {E}}_1$ is unstable.

Assume ${\mathcal {R}}^2_0<{\mathcal {R}}^1_0,$ that is $G(0)<1$. Suppose that the system (10) has a solution $\lambda =a+ib$ such that $a\ge 0$. Then, by the equation (12), we have

$$\begin{aligned} |G(\lambda )|= & {} \left| S_V^* \beta _{v_2}S_H^* \displaystyle \int _0^\infty {\beta _{h_2}(\tau )\frac{e^{-\lambda \tau }}{\lambda +\mu }\pi _2(\tau )d\tau }\right| \nonumber \\< & {} \frac{S_V^*}{\mu } \beta _{v_2}S_H^* \displaystyle \int _0^\infty {\beta _{h_2}(\tau )\pi _2(\tau )d\tau }=G(0)<1, \end{aligned}$$

(15)

Hence, the characteristic equation $G(\lambda )=1$ does not have solutions with nonnegative real part.

Now assume that $y_{V_2}= 0$. Then, the stability of ${\mathcal {E}}_1$ depends on the eigenvalues of the following system:

$$\begin{aligned} \left\{ \begin{array}{l} \lambda x_H=\tilde{T_0}\left( x_H+\displaystyle \int _0^\infty {(1+\frac{\gamma _1(\tau )}{(\lambda +d)})y_{H_1}(\tau )d\tau }\right) \\ \qquad \qquad -\beta _{v_1}S_H^*y_{V_1} - \beta _{v_1}x_HI_{V_1}^* -dx_H, \\ \displaystyle \frac{d y_{H_1}}{ d\tau } +\lambda y_{H_1}(\tau ) = -\left( \alpha _1(\tau ) + \kappa _1(\tau )+\gamma _1(\tau )+d \right) y_{H_1}(\tau ), \\ y_{H_1}(0) = \beta _{v_1} S_H^*y_{V_1} +\beta _{v_1} x_H I_{V_1}^* ,\\ \lambda x_V = -S_V^* \displaystyle \int _0^\infty \beta _{h_1}(\tau )y_{H_1}(\tau )d\tau - x_V \displaystyle \int _0^\infty {\beta _{h_1}(\tau )i_{H_1}^*(\tau )}d\tau -\mu x_V ,\\ \lambda y_{V_1} = S_V^*\displaystyle \int _0^\infty {\beta _{h_1}(\tau )y_{H_1}(\tau )}d\tau + x_V\displaystyle \int _0^\infty {\beta _{h_1}(\tau )i_{H_1}^*(\tau )}d\tau -\mu y_{V_1}. \\ \end{array}\right. \end{aligned}$$

(16)

Since $y_{H_1}(\tau ) = y_{H_1}(0) e^{-\lambda \tau } \pi _1(\tau )$, we have

$$\begin{aligned}&\displaystyle \int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{(\lambda +d)}\right) y_{H_1}(\tau )d\tau }=(\beta _{v_1}S^*_Hy_{V_1}+\beta _{v_1}x_HI^*_{V_1}) \displaystyle \\&\quad \int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{(\lambda +d)}\right) e^{-\lambda \tau }\pi _1(\tau )d\tau }. \end{aligned}$$

Then, the first equation in (16) is an equality in the terms of $x_H, \ y_{V_1}$, and $\lambda$ obtained as follows:

$$\begin{aligned} x_H \left( \lambda +d-\tilde{T_0}+\beta _{v_1} I^*_{V_1}(1-\tilde{T_0}\displaystyle \int _0^\infty {(1+\frac{\gamma _1(\tau )}{\lambda +d})e^{-\lambda \tau }\pi _1(\tau )d\tau })\right) \\ +\,y_{V_1}\beta _{v_1} S^*_H \left( 1- \tilde{T_0}\displaystyle \int _0^\infty {(1+\frac{\gamma _1(\tau )}{\lambda +d})e^{-\lambda \tau }\pi _1(\tau )d\tau } \right) =0. \end{aligned}$$

Also from the last two equations in (16), we obtain an equation in the terms of $x_H, \ y_{V_1}$, and $\lambda$.

$$\begin{aligned} y_{V_1}\left( (\lambda +\mu +T_1)-\beta _{v_1}S^*_H S^*_V\displaystyle \int _0^\infty {\beta _{h_1}(\tau )e^{-\lambda \tau }\pi _1(\tau )d\tau }\right) \\ -\,x_H\left( S^*_V \beta _{v_1}I^*_{V_1}\displaystyle \int _0^\infty {\beta _{h_1}(\tau )e^{-\lambda \tau }\pi _1(\tau )d\tau }\right) =0, \end{aligned}$$

where $T_1=\displaystyle \int _0^\infty {\beta _{h_1}(\tau )i^*_{H_1}(\tau )d\tau } (>0)$. The second equality is also in the terms of $x_H, \ y_{V_1}$, and $\lambda$. Then, the characteristic equation is as follows:

$$\begin{aligned} \frac{\lambda +d+\beta _{v_1}I^*_{V_1}\hat{T}-\tilde{T_0}}{\lambda +d-\tilde{T_0}}=\frac{\beta _{v_1}S^*_H S^*_V \displaystyle \int _0^\infty {\beta _{h_1}e^{-\lambda \tau }\pi _1(\tau )d\tau }}{\lambda +\mu +T_1}, \end{aligned}$$

(17)

Suppose that $\lambda =a+bi$ with $a\ge 0$ is a solution of the characteristic equation. Taking the absolute value of both sides of the equality above, we get

$$\begin{aligned} \left| \displaystyle \frac{\beta _{v_1} S_V^*S_H^* \displaystyle \int _0^\infty {\beta _{h_1}(\tau )e^{-\lambda \tau }\pi _1(\tau )d\tau }}{\lambda +\mu +T_1} \right|\le & {} \displaystyle \frac{\beta _{v_1} S_V^*S_H^* \displaystyle \int _0^\infty {\beta _{h_1}(\tau )e^{-a \tau }\pi _1(\tau )d\tau }}{\sqrt{(a+\mu +T_1)^2+b^2}} \\\le & {} \displaystyle \frac{\beta _{v_1} S_V^*S_H^* \displaystyle \int _0^\infty {\beta _{h_1}(\tau )e^{-a \tau }\pi (\tau )d\tau }}{\mu }\\\le & {} \beta _{v_1} \displaystyle \frac{S_V^*}{\mu }S_H^* \displaystyle \int _0^\infty {\beta _{h_1}(\tau )\pi _1(\tau )d\tau }=1\\ \end{aligned}$$

Moreover,

$$\begin{aligned} \left| \frac{ \lambda +d+\beta _{v_1}I^*_{v_1}\hat{T}-\tilde{T_0}}{\lambda +d-\tilde{T_0}}\right| =\frac{\sqrt{(a+d+\beta _{v_1}I_{V_1}^*\mathfrak {R}\,\hat{T}-\tilde{T_0})^2+(b+\beta _{v_1}I_{V_1}^*\mathfrak {I}\,\hat{T})^2}}{\sqrt{(a+d-\tilde{T_0})^2+b^2}}. \end{aligned}$$

Note that for $\lambda$ with nonnegative real part, when $\mathfrak {R}\,\hat{T}\ge 0$ and $\mathfrak {I}\,\hat{T}\ge 0$ the left-hand side remains strictly greater than one, while the right-hand side is strictly smaller than one. Thus, such $\lambda$’s cannot satisfy the characteristic equation (17). Hence the one strain endemic equilibrium ${\mathcal {E}}_1$ is locally asymptotically stable whenever it exists given the assumptions on $\hat{T}$. We notice that the conditions $\mathfrak {R}\,\hat{T}\ge 0$ and $\mathfrak {I}\,\hat{T}\ge 0$ for any $\lambda$ with $\mathfrak {R}\, \lambda \ge 0$ may not always hold. $\square$

Fig. 9

Host and parasite fitness landscape across distinct immune activation rate a and parasite growth rate r and the coevolutionary attractor $(a^*_{CoESS},r^*_{CoESS})$

Host Evolution: Case Fatality Ratio (${\mathcal {F}}$) Minimization

Proof

(Proof of Theorem (3.2))

Let $(x_1(t), y_1(\tau ,t),z_1(t),x_2(t), y_2(\tau ,t),z_2(t),u(t),v(t))$ be the small perturbations around the mutant-free equilibrium ${\mathcal {E}}_R$. Linearizing the system (7) (with two-host one-strain population) at the mutant-free equilibrium ${\mathcal {E}}_R$, we get:

$$\begin{aligned} \left\{ \begin{array}{l} \lambda x_1 =-\tilde{b}(\displaystyle \frac{\eta _1+\eta _2}{K})N^*_1 +\tilde{b}(1-\frac{N^*_1}{K})\eta _1-\beta _1(x_1 I^*_V+S^*_1 v)-dx_1,\\ \lambda y_{1}+\displaystyle \frac{ d y_1}{d\tau } = -\left( \gamma _1(\tau )+\alpha _1(\tau )+\kappa _1(\tau )+d \right) y_1(\tau ),\\ y_1(0) =\beta _1 (x_1 I^*_V +S^*_1 v)\\ \lambda z_1 =\displaystyle \int _0^\infty {\gamma _1(\tau )y_1(\tau )d\tau }-d z_1\\ \lambda u=-\displaystyle \int _0^\infty {\beta _{H_1}(\tau )(u i_{1}^*(\tau )+S^*_V y_1(\tau ))d\tau }-\displaystyle \int _0^\infty {\beta _{H_2}(\tau )S_{V}^*y_{2}(\tau )}d\tau -\mu u,\\ \lambda v=\displaystyle \int _0^\infty {\beta _{H_1}(\tau )(u i_{1}^*(\tau )+S^*_V y_1(\tau ))d\tau }+\displaystyle \int _0^\infty {\beta _{H_2}(\tau )S_{V}^*y_{2}(\tau )}d\tau -\mu v,\\ \lambda x_2 =\tilde{b}(1-\frac{N^*_1}{K})\eta _2-\beta _1x_2 I^*_V-dx_2,\\ \lambda y_{2}+\displaystyle \frac{ d y_{2}}{d\tau } = -\left( \gamma _2(\tau )+\alpha _2(\tau )+\kappa _2(\tau )+d \right) y_{2}(\tau ),\\ y_{2}(0) =\beta _1 x_2 I^*_V.\\ \lambda z_2 =\displaystyle \int _0^\infty {\gamma _2(\tau )y_{2}(\tau )d\tau }-d z_2, \end{array}\right. \end{aligned}$$

(18)

where $\eta _j=x_j+\displaystyle \displaystyle \int _0^\infty y_j(\tau ) d\tau +z_j$.

Solving the differential equations, we obtain

$$\begin{aligned} y_1(\tau )= & {} \beta _1(x_1 I^*_V +S^*_1 v)e^{-\lambda \tau }\pi _{H_1(\tau )}, \text { where } \nonumber \\ \pi _{H_1}(\tau )= & {} e^{-\displaystyle \int _0^\tau {(\gamma _1(s)+\alpha _1(s)+\kappa _1(s)+d)ds}} \end{aligned}$$

(19)

and

$$\begin{aligned} y_{2}(\tau )= & {} \beta _1x_2I_V^*e^{-\lambda \tau }\pi _{H_2(\tau )}, \text { where }\nonumber \\ \pi _{H_2}(\tau )= & {} e^{-\displaystyle \int _0^\tau {(\gamma _2(s)+\alpha _2(s)+\kappa _2(s)+d)ds}} \end{aligned}$$

(20)

Note that any eigenvalue of the system (18) is also an eigenvalue of the following decoupled subsystem

$$\begin{aligned} \left\{ \begin{array}{l} \lambda x_2 =\tilde{b}(1-\frac{N^*_1}{K})\eta _2-\beta _1x_2 I^*_V-dx_2,\\ \lambda y_{2}+\displaystyle \frac{d y_{2}(\tau )}{d\tau } = -\left( \gamma _2(\tau )+\alpha _2(\tau )+\kappa _2(\tau )+d \right) y_{2}(\tau ),\\ y_{2}(0) =\beta _1 x_2 I^*_V.\\ \lambda z_2 =\displaystyle \int _0^\infty {\gamma _2(\tau )y_{2}(\tau )d\tau }-d z_2. \end{array}\right. \end{aligned}$$

(21)

Rearranging the first equation in (21) and substituting $\displaystyle \displaystyle \int _0^\infty y_2(\tau ) d\tau$ and $z_2 = \displaystyle \frac{1}{\lambda +d}\displaystyle \int _0^\infty \gamma _2(\tau ) y_2(\tau ) d\tau$, we obtain

$$\begin{aligned} \left( \lambda -\tilde{b}\left( 1-\frac{N^*_1}{K}\right) +d+\beta _1 I^*_V\right) x_2=\tilde{b}\left( 1-\frac{N^*_1}{K}\right) \displaystyle \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{\lambda +d}\right) y_{2}(\tau )d\tau } \end{aligned}$$

(22)

Substituting the equation (20) into (22), we get

$$\begin{aligned}&\left( \lambda -\tilde{b}\left( 1-\frac{N^*_1}{K}\right) +d+\beta _1 I^*_V\right) \nonumber \\&\quad =\tilde{b}\left( 1-\frac{N^*_1}{K}\right) \beta _1 I_V^*\left( \displaystyle \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{\lambda +d}\right) e^{-\lambda \tau }\pi _{H_2(\tau ) d\tau }}\right) \end{aligned}$$

(23)

(assuming $x_2 \ne 0$). Susceptible host type 1 when at equilibrium satisfies

$$\begin{aligned} \tilde{b} N_1^* \left( 1-\displaystyle \frac{N_1^*}{K}\right) -\beta _1 S_1^* I_V^* - { dS }_V^* =0 \end{aligned}$$

(24)

where $N_1^* = S_1^* + i_1^*(0)\displaystyle \int _0^\infty \pi _{H_1}(\tau ) d\tau + i_1^*(0)\displaystyle \int _0^\infty \displaystyle \frac{\gamma _1(\tau )}{d}\pi _{H_1}(\tau ) d\tau$.

$$\begin{aligned} -\tilde{b}\left( 1-\frac{N^*_{1}}{K}\right) +\beta _1 I^*_V+d=\frac{\tilde{b}(1-\frac{N^*_{1}}{K})}{S^*_{1}}\displaystyle \int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{d}\right) i^*_{1}(0)\pi _{H_1}(\tau )d\tau }. \end{aligned}$$

(25)

Substituting the right-hand side of the equation into (23), we obtain the characteristic equation

$$\begin{aligned}&\left( \lambda +\frac{\tilde{b}(1-\frac{N^*_{1}}{K})}{S^*_{1}} \displaystyle \int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{d}\right) i^*_{1}(0)\pi _{H_1}(\tau )d\tau }\right) \nonumber \\&\quad =\tilde{b}\left( 1-\frac{N^*_1}{K}\right) \beta _1 I_V^*\left( \displaystyle \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{\lambda +d}\right) e^{-\lambda \tau }\pi _{H_2}(\tau ) d\tau }\right) \end{aligned}$$

(26)

Claim 4.1

Assume that ${\mathcal {F}}_1<{\mathcal {F}}_2$. Then, all eigenvalues of the system (18) have negative real part.

Proof

By the way of contradiction, suppose that the system (18) has an eigenvalue $\lambda$ with nonnegative real part. Then, $\lambda$ also is an eigenvalue of the decoupled subsystem (21). Then,

$$\begin{aligned}&\left| \lambda +\frac{\tilde{b}\left( 1-\frac{N^*_{1}}{K}\right) }{S^*_{1}}\int _0^\infty {(1+\frac{\gamma _1(\tau )}{d})i^*_{1}(0)\pi _{H_1}(\tau )d\tau }\right| \nonumber \\&\quad \ge \frac{\tilde{b}\left( 1-\frac{N^*_{1}}{K}x\right) }{S^*_{1}}\int _0^\infty {(1+\frac{\gamma _1(\tau )}{d})i^*_{1}(0)\pi _{H_1}(\tau )d\tau } \end{aligned}$$

(27)

Also for the right-hand side of the equation (26), we get

$$\begin{aligned}&\left| \tilde{b}\left( 1-\frac{N^*_1}{K}\right) \beta _1 I_V^*\left( \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{\lambda +d}\right) e^{-\lambda \tau }\pi _{H_2}(\tau ) d\tau }\right) \right| \nonumber \\&\quad \le \tilde{b}\left( 1-\frac{N^*_1}{K}\right) \beta _1 I_V^*\left( \int _0^\infty {(1+\frac{\gamma _2(\tau )}{d})\pi _{H_2}(\tau ) d\tau }\right) . \end{aligned}$$

(28)

Then, by (27) and (28) if

$$\begin{aligned}&\int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{d}\right) i^*_{1}(0)\pi _{H_1}(\tau )d\tau } \le \beta _1 I_V^*S^*_{1} \left( \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{d}\right) \pi _{H_2(\tau ) d\tau }}\right) ,\nonumber \\&\quad \text { where } i^*_{1}(0)=\beta _1 I_V^*S^*_{1}. \end{aligned}$$

(29)

Therefore,

$$\begin{aligned} \int _0^\infty {\left( \frac{d+\gamma _1(\tau )}{d}\right) \pi _{H_1}(\tau )d\tau } \le \int _0^\infty {\left( \frac{d+\gamma _2(\tau )}{d}\right) \pi _{H_2}(\tau ) d\tau }. \end{aligned}$$

(30)

Subtracting and adding $\nu _1(\tau )$ on the left-hand side of the equation above and $\nu _2(\tau )$ on the right side of the equation, we obtain

$$\begin{aligned} 1- \int _0^\infty {\nu _1(\tau )\pi _{H_1}(\tau )d\tau } \le 1-\int _0^\infty {\nu _2(\tau ) \pi _{H_2}(\tau ) d\tau }. \end{aligned}$$

Then,

$$\begin{aligned} \int _0^\infty {\nu _1(\tau )\pi _{H_1}(\tau )d\tau } \ge \int _0^\infty {\nu _2(\tau ) \pi _{H_2}(\tau ) d\tau }, \end{aligned}$$

which implies that ${\mathcal {F}}_1 \ge {\mathcal {F}}_2$. It is a contradiction. This completes the proof for the case $x_2 \ne 0$. Now assume that $x_2=0$. Then, $\eta _2=0,$ in which case we obtain the subsystem

$$\begin{aligned} \left\{ \begin{array}{l} \lambda x_1 =-\tilde{b}(\frac{\eta _1N^*_{1} }{K})+\tilde{b}(1-\frac{N^*_{1}}{K})\eta _1-\beta _1(x_1 I^*_V+S^*_1 v)-dx_1,\\ \lambda y_{1}(\tau )+\frac{ d y_{1}(\tau )}{d\tau } = -\left( \gamma _1(\tau )+\alpha _1(\tau )+\kappa _1(\tau )+d \right) y_{1}(\tau ),\\ y_{1}(0) =\beta _1 (x_1 I^*_V +S^*_1 v)\\ \lambda z_1 =\int _0^\infty {\gamma _1(\tau )y_{1}(\tau )d\tau }-d z_1\\ \lambda u=-\int _0^\infty {\beta _{H_1}(\tau )(u i^*_{1}(\tau )+S^*_V y_{1}(\tau ))d\tau }-\mu u,\\ \lambda v=\int _0^\infty {\beta _{H_1}(\tau )(u i^*_{1}(\tau )+S^*_V y_{1}(\tau ))d\tau } -\mu v.\\ \end{array}\right. \end{aligned}$$

(31)

This is exactly the linearized system determining the local stability of ${\mathcal {E}}$, the endemic equilibrium for the single-host single-strain model. Thus, by Theorem 2.1, the conditions (5) imply that the eigenvalues of subsystem (31) have negative real part. Then, the case $x_2=0$ also contradicts the assumption that the system (18) has an eigenvalue with nonnegative real part. $\square$

Then, when ${\mathcal {F}}_1<{\mathcal {F}}_2$ and boundary conditions hold, the mutant-free equilibrium is locally asymptotically stable.

Claim 4.2

(Local invasion) If ${\mathcal {F}}_2 < {\mathcal {F}}_1$. Then, the mutant-free equilibrium ${\mathcal {E}}^*_R$ is unstable; i.e., mutant population invades resident population.

Proof

Let define the left-hand side of the equation (26) as $g(\lambda )$ and the right-hand side as $f(\lambda )$. Note that any $\lambda$ solution of this system must be an eigenvalue of the decoupled subsystem (31). Now we want to show that the equation (26) has a positive real root $\lambda ,$ whenever ${\mathcal {F}}_2 < {\mathcal {F}}_1$. First note that $g(\lambda )$ is an increasing function of $\lambda$ and $f(\lambda )$ is a decreasing function of $\lambda$ for $\lambda \in \mathbb R$. Therefore, if $g(0)< f(0),$ then the equality (26) has a positive real root $\lambda$. Next we want to show that $g(0)< f(0) \Leftrightarrow {\mathcal {F}}_2 < {\mathcal {F}}_1$. Note that

$$\begin{aligned} g(0)< & {} f(0)\\\Leftrightarrow & {} \\ \int _0^\infty {\left( 1+\frac{\gamma _1(\tau )}{d}\right) i^*_{1}(0)\pi _{H_1}(\tau )d\tau }< & {} \beta _1 I_V^*S^*_{1} \left( \int _0^\infty {\left( 1+\frac{\gamma _2(\tau )}{d}\right) \pi _{H_2(\tau ) d\tau }}\right) , \end{aligned}$$

where $i^*_{1}(0)=\beta _1 I^*_VS^*_{1}$. Then, the rest of the proof follows similar steps to the proof of Claim (4.1) after the inequality (29). $\square$

$\square$