The role of backward mutations on the within-host dynamics of HIV-1

Abstract

The quality of life for patients infected with human immunodeficiency virus (HIV-1) has been positively impacted by the use of antiretroviral therapy (ART). However, the benefits of ART are usually halted by the emergence of drug resistance. Drug-resistant strains arise from virus mutations, as HIV-1 reverse transcription is prone to errors, with mutations normally carrying fitness costs to the virus. When ART is interrupted, the wild-type drug-sensitive strain rapidly out-competes the resistant strain, as the former strain is fitter than the latter in the absence of ART. One mechanism for sustaining the sensitive strain during ART is given by the virus mutating from resistant to sensitive strains, which is referred to as backward mutation. This is important during periods of treatment interruptions as prior existence of the sensitive strain would lead to replacement of the resistant strain. In order to assess the role of backward mutations in the dynamics of HIV-1 within an infected host, we analyze a mathematical model of two interacting virus strains in either absence or presence of ART. We study the effect of backward mutations on the definition of the basic reproductive number, and the value and stability of equilibrium points. The analysis of the model shows that, thanks to both forward and backward mutations, sensitive and resistant strains co-exist. In addition, conditions for the dominance of a viral strain with or without ART are provided. For this model, backward mutations are shown to be necessary for the persistence of the sensitive strain during ART.

Appendix: Proofs of results

Proposition 1

(Boundedness of solutions) The closed positive 5-dimensional orthant, defined as $\mathbb{R }_+^5=\{x\in \mathbb{R }^5\backslash x\ge 0\}$ is positive invariant for System 1 and there exists $M>0$ such that all solutions satisfy $T(t),I_s(t),I_r(t),V_s(t),V_r(t)<M$ for all large $t$.

Proof

Positive invariance follows from the fact that all solutions are uniformly bounded in a proper subset $\varOmega$. To show that solutions of System 1 are bounded, let $\hat{T}$ be the steady-state of susceptible cells present before infection. In a healthy individual, the T-cell population dynamics are regulated by $f(T)=\lambda -\gamma T$, where, $f(T)$ is a smooth function and $f(T)>0$ for $0<T<\hat{T}$. Furthermore, $f(\hat{T})=0$ with $f^\prime (\hat{T})<0$ and $f(T)<0$ whenever $T>\hat{T}$.

From the first equation of System 1, we note that $\frac{dT}{dt}\le f(T)$. This means that there exists a $t_0>0$ such that $T(t)<\hat{T}+1$ for $t>t_0$. Let $S=\max \limits _{T\ge 0}f(T)$. Adding the first three equations of System 1, we obtain

$$\begin{aligned} \frac{dT}{dt}+\frac{dI_s}{dt}+\frac{dI_r}{dt}=f(T)-\delta (I_s+I_r)\le S-\delta {(I_s+I_r)}. \end{aligned}$$

Let $A,B>0$ be such that $\delta (A+B)>S+1$. Then as long as

$$\begin{aligned} T(t)+I_s(t)+I_r(t)\ge A+B+\hat{T}+1 \end{aligned}$$

and $t>t_0$, we have that

$$\begin{aligned} \frac{dT}{dt}+\frac{dI_s}{dt}+\frac{dI_r}{dt}<-1. \end{aligned}$$

Clearly, there exists $t_1>t_0$ such that

$$\begin{aligned} T(t)+I_s(t)+I_r(t)<A+B+\hat{T}+1 \end{aligned}$$

for all $t>t_1$.

Adding the last two equations, we get

$$\begin{aligned} \frac{dV_s}{dt}+\frac{dV_r}{dt}=aI_s+k_2 aI_r-c(V_s+V_r). \end{aligned}$$

The asymptotic bound for $I_s$ is $I_s(t)<A+\hat{T}+1$ while that of $I_r$ is $I_r(t)<B+\hat{T}+1$. Considering the asymptotic bounds for both $I_s$ and $I_r$ together with the differential inequality

$$\begin{aligned} \frac{dV_s}{dt}+\frac{dV_r}{dt}\le -c(V_s+V_r)+a(A+k_2B+(1+k_2)\hat{T}+1+k_2), \end{aligned}$$

which holds for large $t$, yields the asymptotic bound below;

$$\begin{aligned} c^{-1}a(A+k_2B+(1+k_2)\hat{T}+1+k_2). \end{aligned}$$

$\square$

Theorem 1

(Global stability of infection-free state) The infection-free equilibrium $E_0$ is globally asymptotically stable provided $R_0<1$.

Proof

To prove global stability of the disease free equilibrium $E_0$, we use Theorem 1 adopted from Castillo-Chávez et al. (2002). We can write System 1 in the form

$$\begin{aligned} X^\prime (t)&= F(X,Y)\\ Y^\prime (t)&= G(X,Y), \quad G(X,0)=0 \end{aligned}$$

where $X=(T)$ and $Y=(I_s,I_r,V_s,V_r)$ with $X\in \mathbb{R }_+$ and $Y\in \mathbb{R }_+^4$.

Taking $F(X,0)=[\lambda -\gamma T]$,

$$\begin{aligned} A=\left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} -\delta&0&\frac{\beta \lambda }{\gamma }&0\\ 0&-\delta&0&\frac{k_1\beta \lambda }{\gamma }\\ (1-q)a&zk_2a&-c&0\\ qa&(1-z)k_2a&0&-c \end{array}\right] \end{aligned}$$

and

$$\begin{aligned} \hat{G}(X,Y)=\left[ \begin{array}{c} \frac{\beta \lambda }{\gamma }V_s-\beta TV_s\\ \frac{k_1\beta \lambda }{\gamma }V_r-k_1\beta TV_r\\ 0\\ 0 \end{array}\right]. \end{aligned}$$

Since $\limsup \limits _{t\rightarrow \infty } T(t)\le \frac{\lambda }{\gamma },$ we have that $\hat{G}(X,Y)\ge 0$ for all $(X,Y)\in \varOmega$.

Therefore $G(X,Y)=AY-\hat{G}(X,Y), \text{ where} \hat{G}(X,Y)\ge 0$ for $(X,Y)\in \varOmega$, and applying Theorem 1 from Castillo-Chávez et al. (2002), the fixed point $E_0$ is globally asymptotically stable, provided, $R_0<1.$ $\square$

Theorem 2

(Uniqueness of endemic equilibrium) $E_1$ is in $\varOmega$ if and only if $R_0>1$. Moreover, $E_1$ is a unique endemic equilibrium in $\varOmega$ whenever it exists.

Proof

Substituting $E_1=(T^*,I_s^*,I_r^*,V_s^*,V_r^*)$ into System 1 shows that $E_1$ is a stationary point. From second and third equations of System 1, we obtain

$$\begin{aligned} I_s^*+I_r^*=\frac{1}{\delta }(\lambda -\gamma T^*), \end{aligned}$$

then

$$\begin{aligned} T^*+I_s^*+I_r^*=\frac{\lambda }{\gamma }\frac{1}{R_0}+\frac{1}{\delta }\left(\lambda -\gamma \frac{\lambda }{\gamma }\frac{1}{R_0}\right)=\frac{\lambda }{\gamma }\frac{1}{R_0}+\frac{\lambda }{\delta }\left(1-\frac{1}{R_0}\right)\le \frac{\lambda }{\delta } \end{aligned}$$

as long as $\delta \ge \gamma$.

Note that, as $k_1k_2<1$,

$$\begin{aligned} (1-q-z)-(1-q-k_1k_2z)<0. \end{aligned}$$

Thus,

$$\begin{aligned} (1-q-z)^2-(1-q-z)[(1-q)+k_1k_2(1-z)]<-k_1k_2(1-q-z) \end{aligned}$$

(under the assumption that $1-q-z>0$). Multiplying by $4$ and adding $[(1-q)+k_1k_2(1-z)]^2$ to both sides of the inequality leads to

$$\begin{aligned}&\{2(1-q-z)-[(1-q)+k_1k_2(1-z)]\}^2\\&\quad <[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z). \end{aligned}$$

Therefore,

$$\begin{aligned}&2(1-q-z)-[(1-q)+k_1k_2(1-z)]\\&\quad <\sqrt{[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z)}. \end{aligned}$$

and so

$$\begin{aligned} 2(1-q-z)&< \frac{1}{2}\left\{ (1-q)+k_1k_2(1-z)\right.\nonumber \\&+\left.\sqrt{[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z)}\right\} . \end{aligned}$$

Hence, multiplying by $\bar{R_0}=\frac{a\beta \lambda }{\delta \gamma c}$, we obtain $(1-q-z)\bar{R_0}<R_0$ (and then $(1-q-z)k_1k_2\bar{R_0}<R_0$). Therefore $I_s^*>0$ (and $I_r^*$) if and only if $R_0>1$, which implies $E_1\in \varOmega \,\text{ iff} R_0>1$

To show that $E_1$ is the only endemic equilibrium in $\varOmega$, assume that there is another such equilibrium $E_2=(\hat{T},\hat{I_s},\hat{I_r},\hat{V_s},\hat{V_r})$ in $\varOmega$. Note that by solving for stationary points, $\hat{T}$ must be a solution of the quadratic equation

$$\begin{aligned} \left(\frac{a\beta }{c}\right)^2(1-q-z)k_1k_2\hat{T}^2-[(1-q)+(1-z)k]\left(\frac{a\beta }{c}\right)\delta \hat{T}+\delta ^2=0. \end{aligned}$$

Then

$$\begin{aligned} \hat{T}&= \frac{c\delta }{2k_1k_2a\beta (1-q-z)}\left[ (1-q)+k_1k_2(1-z)\right.\nonumber \\&+\left.\sqrt{[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z)}\right] \end{aligned}$$

(note that the other root of the above quadratic equation is $T^*$). As for $E_1,\,\hat{I_s}+\hat{I_r}=\lambda -\gamma \hat{T},$ so if $\lambda -\gamma \hat{T}<0$, then $E_2\notin \varOmega$. So assume that $\lambda -\gamma \hat{T}>0$. Thus

$$\begin{aligned} \hat{I_r}=\frac{q(\lambda -\gamma \hat{T})}{\delta -(1-q-z)\frac{a\beta }{c}\hat{T}}. \end{aligned}$$

Since $k_1k_2<1$ we have

$$\begin{aligned} k_1k_2+(1-q-z)<(1-q)+k_1k_2(1-z) \end{aligned}$$

and then

$$\begin{aligned} (2k_1k_2)^2-4k_1k_2[(1-q)+k_1k_2(1-z)]<-4k_1k_2(1-q-z). \end{aligned}$$

This implies that

$$\begin{aligned} \{2k_1k_2-[(1-q)+k_1k_2(1-z)]\}^2<[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z) \end{aligned}$$

and so

$$\begin{aligned} 2k_1k_2-[(1-q)+k_1k_2(1-z)]<\sqrt{[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z)}. \end{aligned}$$

Thus

$$\begin{aligned} 1&< \frac{1}{2k_1k_2}\left\{ [(1-q)+k_1k_2(1-z)]\right.\nonumber \\&+\left.\sqrt{[(1-q)+k_1k_2(1-z)]^2-4k_1k_2(1-q-z)}\right\} . \end{aligned}$$

This implies that $\delta -(1-q-z)\frac{a\beta }{c}\hat{T}<0$ and so $\hat{I_r}<0.$ Therefore $E_2\notin \varOmega$. $\square$

Theorem 3

(Local stability of endemic equilibrium) If $R_0>1$, the unique endemic equilibrium $E_1$ is locally asymptotically stable.

Using the standard linearization of the model to determine the local stability of $E_1$ is very laborious to track mathematically. For this reason, we employ the centre manifold theory (Carr 1981) as described in Castillo-Chávez and Song (2004) to establish the local asymptotic stability of $E_1$.

Making the following change of variables; $T=x_1,~I_s=x_2,~I_r=x_3,~V_s=x_4$ and $V_r=x_5$. Therefore, we get

$$\begin{aligned} \frac{dX}{dt}=F=(f_1,f_2,f_3,f_4,f_5)^T \end{aligned}$$

such that

$$\begin{aligned} x_1^\prime (t)&= f_1 = \lambda -\gamma x_1-\beta x_1x_4-k_1\beta x_1x_5 \\ x_2^\prime (t)&= f_2 = (1-q)\beta x_1x_4+zk_1\beta x_1x_5-\delta x_2\\ x_3^\prime (t)&= f_3 = q\beta x_1x_4+(1-z)k_1\beta x_1x_5-\delta x_3\\ x_4^\prime (t)&= f_4 = ax_2-cx_4\\ x_5^\prime (t)&= f_5 = k_2ax_3-cx_5 \end{aligned}$$

The corresponding Jacobian matrix at the disease free equilibrium is given by

$$\begin{aligned} J(E_0)=\left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} -\gamma&0&0&-\frac{\beta \lambda }{\gamma }&-\frac{k_1\beta \lambda }{\gamma }\\ 0&-\delta&0&\frac{(1-q)\beta \lambda }{\gamma }&\frac{zk_1\beta \lambda }{\gamma } \\ 0&0&-\delta&\frac{q\beta \lambda }{\gamma }&\frac{(1-z)k_1\beta \lambda }{\gamma } \\ 0&a&0&-c&0\\ 0&0&k_2a&0&-c \end{array} \right]. \end{aligned}$$

If $\beta$ is taken as the bifurcation point and we consider the case when $R_0=1$, then

$$\begin{aligned} \beta \!=\!\beta ^*\!=\!\frac{2\delta \gamma c}{a\lambda (k_1k_2(1\!-\!z)\!+\!(1\!-\!q)\!+\!\sqrt{[(1-q)\!+\!k_1k_2 (1\!-\!z)]^2-4k_1k_2(1\!-\!q\!-\!z)})}. \end{aligned}$$

The resultant linearized system of the transformed model with $\beta =\beta ^*$ has a simple zero eigenvalue. This means that the centre manifold theory (Carr 1981) can be employed to analyze the dynamics of the model near the bifurcation parameter value $\beta ^*$. The Jacobian, $J(E_0)$ at $\beta ^*$ has a right eigenvalue associated with the zero eigenvalue given by $u=[u_1,u_2,u_3,u_4,u_5]$, where

$$\begin{aligned} u_1&= \frac{k_1\beta \lambda [(1-q-z)a\beta \lambda -\delta c\gamma ]}{\gamma ^2[\delta c\gamma -(1-q)a\beta \lambda ]}u_5,\\ u_2&= \frac{zk_1\beta \lambda c}{\delta c\gamma -(1-q)a\beta \lambda }u_5,\\ u_3&= \frac{c}{k_2a}u_5,\\ u_4&= \frac{zk_1\beta \lambda a}{\delta c\gamma -(1-q)a\beta \lambda }u_5,\\ u_5&= u_5>0. \end{aligned}$$

The left eigenvector for $J(E_0)$ associated with the zero eigenvalue is given by $v=[v_1,v_2,v_3,v_4,v_5]$, where

$$\begin{aligned} v_1&= 0 ,\\ v_2&= \frac{qk_2a^2\beta \lambda }{\delta [\delta c\gamma -(1-q)a\beta \lambda ]}v_5,\\ v_3&= \frac{k_2a}{\delta }v_5,\\ v_4&= \frac{qk_2a\beta \lambda }{\delta c\gamma -(1-q)a\beta \lambda }v_5,\\ v_5&= v_5>0. \end{aligned}$$

We state without proof, Theorem 4 as outlined in Castillo-Chávez and Song (2004).

Theorem 4

(Castillo-Chávez and Song) Consider the following general system of ordinary differential equations with a parameter $\phi$

$$\begin{aligned} \frac{dx}{dt}F(x,\phi ),~f:\mathbb{R }^n\times \mathbb{R } \text{ and} f\in \mathbb{R }^2(\mathbb{R }^n\times \mathbb{R }), \end{aligned}$$

(8)

where $0$ is the equilibrium of the system i.e, $f(0,\phi )=0$ for all $\phi$ and assume

• A1: $A=D_xf(0,0)=\left(\frac{\partial f_i}{\partial x_j}(0,0)\right)$ is the linearization of System 8 around the equilibrium $0$ evaluated with $\phi =0$. Then, zero is a simple eigenvalue of $A$ and other eigenvalues of $A$ have negative real parts.

• A2: Matrix $A$ has a right eigenvector $u$ and a left eigenvector $v$ corresponding to the zero eigenvalue.

Let $f_k$ be the $kth$ component of $f$ and

$$\begin{aligned} \hat{a}&= \sum \limits _{k,i,j=1}^{n}v_ku_iu_j\frac{\partial ^2f_k}{\partial x_i\partial x_j}(0,0),\\ \hat{b}&= \sum \limits _{k,i=1}^{n}v_ku_i\frac{\partial ^2f_k}{\partial x_i\partial \phi }(0,0). \end{aligned}$$

The local dynamics of 8 are completely governed by $\hat{a}$ and $\hat{b}$ as follows:

• (i) $\hat{a}>0, ~\hat{b}>0$. When $\phi <0$ with $|\phi |<<1,\,0$ is locally asymptotically stable and there exists a positive unstable equilibrium; when $0<\phi << 1,\,0$ is unstable and there exists a negative and locally asymptotically stable equilibrium.

• (ii) $\hat{a}<0, ~\hat{b}<0$. When $\phi <0$ with $|\phi |<<1,\,0$ is unstable; when $0<\phi << 1,\,0$ is locally asymptotically stable and there exists a positive unstable equilibrium.

• (iii) $\hat{a}>0, ~\hat{b}<0$. When $\phi <0$ with $|\phi |<<1,\,0$ is unstable and there exists a locally asymptotically stable negative equilibrium; when $0<\phi << 1,\,0$ is stable and a positive unstable equilibrium appears.

• (iv) $\hat{a}<0, ~\hat{b}>0$. When $\phi$ changes from negative to positive, $0$ changes its stability from stable to unstable. Correspondingly, a negative unstable equilibrium becomes positive and locally asymptotically stable.

 

1.1 Computation of $\hat{a}$ and $\hat{b}$

For System 1, the associated nonzero partial derivatives of $F$ at the disease free equilibrium $E_0$ are given by

$$\begin{aligned}&\frac{\partial ^2f_1}{\partial x_1\partial x_4} =-\beta ,~~\frac{\partial ^2f_1}{\partial x_1 \partial x_5}=- k_1\beta ,~~\frac{\partial ^2f_2}{\partial x_1 \partial x_4}=(1- q)\beta ,~~\frac{\partial ^2f_2}{\partial x_1 \partial x_5} =zk_1\beta ,\\&\quad \frac{\partial ^2f_3}{\partial x_1\partial x_4} =q\beta ~~ \frac{\partial ^2f_3}{\partial x_1 \partial x_5}=(1- z)k_1\beta . \end{aligned}$$

Therefore, it follows that

$$\begin{aligned} \hat{a}&= -\beta ^* v_1u_1u_4-k_1 \beta ^* v_1u_1u_5+(1-q) \beta ^* v_2u_1u_4+zk_1 \beta ^* v_2u_1u_5 + q \beta ^* v_3u_1u_4+\,(1-z) k_1\beta ^* v_3u_1u_5\nonumber \\&= (1-q)\beta ^* v_2u_1u_4+zk_1 \beta ^* v_2u_1u_5+q \beta ^* v_3u_1u_4+\,(1-z) k_1\beta ^* v_3u_1u_5\nonumber \\&= \frac{k_1^2k_2a \beta ^{*2} \lambda [(1-q-z)a \beta ^*\lambda - \delta \gamma c] \theta }{\delta \gamma ^2[\delta c\gamma - (1-q)a \beta ^*\lambda ] ^3} \end{aligned}$$

(9)

where

$$\begin{aligned} \theta&= [z(1-q)qa^2\beta ^{*2}\lambda ^2+(za\beta ^*\lambda +qza\lambda )(\delta c\gamma -(1-q)a\beta ^*\lambda )\\&\quad +(1-z)(\delta c\gamma -(1-q)a\beta ^*\lambda )^2] \end{aligned}$$

It is observed that

$$\begin{aligned} \delta c\gamma -(1-q)a\beta ^*\lambda >0 \end{aligned}$$

for all possible parameter values. Assume that $\delta c\gamma -(1-q)a\beta ^*\lambda <0$, then, substituting for the value of $\beta ^*$, gives

$$\begin{aligned} k_1k_2(1-z)+(1-q)+\sqrt{[(1-q)\!+\!k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}\!-\!2(1\!-\!q). \end{aligned}$$

This simplifies to

$$\begin{aligned} k_1k_2qz<0 \end{aligned}$$

which is clearly not possible. Therefore, $\delta c\gamma -(1-q)a\beta ^*\lambda$ is always positive.

We note that $\hat{a}>0$ provided $\delta c\gamma -(1-q)a\beta ^* \lambda >0$ and $(1-q-z)a\beta ^*c-\delta c\gamma >0.$ Substituting for the value of $\beta ^*$, we get that $\hat{a}>0$ whenever;

$$\begin{aligned} k_1k_2(1-z)+\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)})>(1-q) \end{aligned}$$

and

$$\begin{aligned} k_1k_2(1-z)+\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)})+2z<(1-q). \end{aligned}$$

This means that

$$\begin{aligned} k_1k_2(1-z)+\Delta +2z<1-q<k_1k_2(1-z)+\Delta \end{aligned}$$

where $\Delta =\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)})$. This is clearly not possible. Therefore $\hat{a}<0$.

The value of the parameter $b$ is associated with following non-vanishing partial derivatives of $F$,

$$\begin{aligned} \frac{\partial ^2f_1}{\partial x_4\partial \beta ^*}&= -\frac{\lambda }{\gamma },\quad \frac{\partial ^2f_1}{\partial x_5\partial \beta ^*}=-\frac{k_1\lambda }{\gamma },\quad \frac{\partial ^2f_2}{\partial x_4\partial \beta ^*}=\frac{(1-q)\lambda }{\gamma },\quad \frac{\partial ^2f_2}{\partial x_5\partial \beta ^*}\\&= \frac{zk_1\lambda }{\gamma },\quad \frac{\partial ^2f_3}{\partial x_4\partial \beta ^*}=\frac{q\lambda }{\gamma },\quad \frac{\partial ^2f_3}{\partial x_5\partial \beta ^*}=\frac{(1-z)k_1\lambda }{\gamma }. \end{aligned}$$

Therefore, it follows that

$$\begin{aligned} \hat{b}&= \frac{(1-q)\lambda }{\gamma }v_2u_4+ \frac{zk_1\lambda }{\gamma }v_2u_5+\frac{q\lambda }{\gamma }v_3u_4+ \frac{(1-z)k_1\lambda }{\gamma }v_3u_5\nonumber \\&= \frac{qzk_1k_2a^2\beta ^*\lambda ^2cv_5u_5}{[\delta c\gamma -(1-q)a\beta ^*\lambda ]^2}+\frac{k_1k_2a\lambda [(1-z)\delta c\gamma -(1-q-z)a\beta ^*\lambda ]v_5u_5}{\gamma \delta [\delta c\gamma -(1-q)a\beta ^*\lambda ]} \end{aligned}$$

(10)

Since $\delta c\gamma -(1-q)a\beta ^*\lambda >0$, in order to show that $\hat{b}>0$, it is enough to show that $(1-z)\delta c\gamma -(1-q-z)a\beta ^*\lambda >0$. If we assume that $(1-z)\delta c\gamma -(1-q-z)a\beta ^*\lambda <0$, then

$$\begin{aligned} (1\!-\!z)\sqrt{[(1\!-\!q)\!+\!k_1k_2 (1\!-\!z)]^2\!-\!4k_1k_2(1\!-\!q\!-\!z)}<(1\!-\!q\!-\!z)\!-\!k_1k_2(1\!-\!z)^2\!-\!qz. \end{aligned}$$

This reduces to

$$\begin{aligned} 2qz(1-q-z)<0 \end{aligned}$$

which is not possible since $q,z<<1.$ Therefore $\hat{b}>0$.

Thus $\hat{a}<0$ and $\hat{b}>0$ and from Theorem 4 item (iv), if $R_0>1$, the unique endemic equilibrium $E_1$ is locally asymptotically stable.

Theorem 5

(Sensitive strain dominance) The sensitive strain is dominant at endemic equilibrium $E_1$ (i.e., $V_s^*>V_r^*$) whenever

$$\begin{aligned} \frac{z}{q}>\frac{k_2\left(1-(1-q-z)\frac{\bar{R_0}}{R_0}\right)}{\left(1-(1-q-z)k_1k_2\frac{\bar{R_0}}{R_0}\right)}. \end{aligned}$$

Proof

The sensitive strain will dominate dynamics when $V_s^*>V_r^*$ This happens when

$$\begin{aligned} \frac{a}{c}I_s^*>\frac{k_2a}{c}I_r^*. \end{aligned}$$

Substituting for $I_s^*$ and $I_r^*$ and simplifying, we get

$$\begin{aligned} z\left(1-(1-q-z)k_1k_2\frac{\bar{R_0}}{R_0}\right)>k_2q\left(1-(1-q-z)\frac{\bar{R_0}}{R_0}\right). \end{aligned}$$

Therefore, for the sensitive strain to be dominant, it is required that

$$\begin{aligned} \frac{z}{q}>\frac{k_2\left(1-(1-q-z)\frac{\bar{R_0}}{R_0}\right)}{\left(1-(1-q-z)k_1k_2\frac{\bar{R_0}}{R_0}\right)}. \end{aligned}$$

$\square$

Theorem 6

(Effect of forward mutations on $R_0$) Given positive forward mutations, $q>0$, and positive fitness cost of resistant strain, $k_1,k_2<1$, we have $R_0\le \bar{R_0}.$

Proof

It is enough to show that

$$\begin{aligned} (1-q)+k_1k_2(1-z) +\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}\le 2. \end{aligned}$$

When $q=z=0$ and $k_1=k_2=1$, we have

$$\begin{aligned} (1-q)+k_1k_2(1-z) +\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}= 2 \end{aligned}$$

and $R_0=\bar{R_0}.$ This is the case when we have no drug resistance but a single sensitive strain. Let $q\not =0,\,z\not =0$ and $k_1,k_2\in (0,1)$, and suppose that

$$\begin{aligned} (1-q)+k_1k_2(1-z) +\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}> 2. \end{aligned}$$

This means that

$$\begin{aligned} \sqrt{[(1-q)+k_1k_2 (1\!-\!z)]^2-4k_1k_2(1-q-z)}> 2\!-\!k_1k_2(1\!-\!z)\!-\!(1\!-\!q). \end{aligned}$$

(11)

Since $k_1,k_2\in (0,1),\,q\not =0,\,z\not =0$ and $q,z<<1$, then $0<k_1k_2(1-z)+(1-q)<2.$ Squaring both sides of (4), we get

$$\begin{aligned}&[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)>4\\&\quad +k_1^2k_2^2(1-z)^2+(1-q)^2-4k_1k_2(1-z)-4(1-q)+2k_1k_2(1-z)(1-q). \end{aligned}$$

On simplification,

$$\begin{aligned} k_1k_2z(1-q)-k_1k_2(1-q)>1-k_1k_2(1-z)-(1-q). \end{aligned}$$

Collecting like terms,

$$\begin{aligned} q[k_1k_2(1-z)-1]>0 \end{aligned}$$

which is a contradiction since $k_1k_2(1-z)-1<0$. Therefore,

$$\begin{aligned} R_o<\bar{R_0}. \end{aligned}$$

$\square$

Theorem 7

(Effect of backward mutations on ${R_0}$) Whenever $q>0$ and $k_1,k_2<1,$ the presence of backward mutations $z>0$ increases the basic reproductive number (i.e., $R_0>R_0^\prime$).

Proof

We need to show that

$$\begin{aligned} (1\!-\!q)<\frac{1}{2}\left[ (1\!-\!q)\!+\!k_1k_2(1\!-\!z)\!+\! \sqrt{[(1-q)\!+\!k_1k_2 (1\!-\!z)]^2\!-\!4k_1k_2(1\!-\!q\!-\!z)} \right] \end{aligned}$$

which is equivalent to

$$\begin{aligned} (1-q)-k_1k_2(1-z)< \sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}. \end{aligned}$$

We proceed by assuming the opposite, i.e.,

$$\begin{aligned} (1-q)-k_1k_2(1-z)\ge \sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)} \end{aligned}$$

then, squaring both sides

$$\begin{aligned}{}[(1-q)-k_1k_2(1-z)]^2\ge [(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z) \end{aligned}$$

which reduces to

$$\begin{aligned} (1-q)(1-z)\le 1-q-z \end{aligned}$$

or equivalently,

$$\begin{aligned} qz\le 0 \end{aligned}$$

which contradicts our assumptions. $\square$

Theorem 8

(Rate of change of $R_0$ with respect to mutations) Given $R_0$ as described in 3,

$$\begin{aligned} \frac{\partial R_0}{\partial q}<0 \quad \text{ and} \frac{\partial R_0}{\partial z}>0. \end{aligned}$$

Proof

$$\begin{aligned} \frac{\partial R_0}{\partial q}=\frac{a\beta \lambda }{2\delta \gamma c}\left[-1+\frac{[k_1k_2z+(k_1k_2+q-1)]}{\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}} \right] \end{aligned}$$

and

$$\begin{aligned} \frac{\partial R_0}{\partial z}=\frac{k_1k_2a\beta \lambda }{2\delta \gamma c}\left[-1+\frac{[k_1k_2z+(1+q-k_1k_2)]}{\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}} \right]. \end{aligned}$$

$\frac{\partial R_0}{\partial q}<0$ iff

$$\begin{aligned}{}[k_1k_2z+(k_1k_2+q-1)]<{\sqrt{[(1-q)+k_1k_2 (1-z)]^2-4k_1k_2(1-q-z)}}. \end{aligned}$$

Since the term inside the square root is positive ($k_1,k_2\in (0,1)$ and $q,z<<1$), it is enough to show that

$$\begin{aligned}{}[k_1k_2z+(k_1k_2+q-1)]^2<k_1^2k_2^2z^2+2k_1k_2z(1+q-k_1k_2)+(k_1k_2+q-1)^2. \end{aligned}$$

This reduces to

$$\begin{aligned} 2k_1k_2z(k_1k_2-1)<2k_1k_2z(1-k_1k_2) \quad \text{ or} \quad k_1k_2<1 \end{aligned}$$

which is satisfied in our region of interest, i.e., when $k_1,k_2\in (0,1)$.

Similarly, $\frac{\partial R_0}{\partial z}>0$ iff

$$\begin{aligned}{}[k_1k_2z+(1+q-k_1k_2)]>{\sqrt{k_1^2k_2^2z^2+2k_1k_2z(1+q-k_1k_2)+(k_1k_2+q-1)^2}}. \end{aligned}$$

It is enough to show that

$$\begin{aligned}{}[k_1k_2z+(1+q-k_1k_2)]^2>k_1^2k_2^2z^2+2k_1k_2z(1+q-k_1k_2)+(k_1k_2+q-1)^2 \end{aligned}$$

which reduces to

$$\begin{aligned} 4q>4qk_1k_2 \quad \mathrm {or}\quad 1>k_1k_2 \end{aligned}$$

which is satisfied in our region of interest, i.e., when $k_1,k_2\in (0,1).$ Therefore,

$$\begin{aligned} \frac{\partial R_0}{\partial q}<0 \quad \text{ and}\quad \frac{\partial R_0}{\partial z}>0. \end{aligned}$$

$\square$

Theorem 9

(Sensitive strain dominance during ART) The sensitive strain dominates dynamics at the endemic equilibrium during treatment if

$$\begin{aligned} \frac{z}{q}>\frac{(1-p_2\varepsilon _{pi})}{(1-\varepsilon _{pi})}\frac{k_2\left(1-(1-q-z)\varepsilon _{s}\frac{\bar{R_0}}{R_0}\right)}{\left(1-(1-q-z)\varepsilon _{r}k_1k_2\frac{\bar{R_0}}{R_0}\right)}. \end{aligned}$$

Proof

The proof follows same argument as for Theorem 5. $\square$

Theorem 10

(Rate of change of $R_0$ with respect to mutations during ART) Given $\hat{R_0}$ as described in 6,

$$\begin{aligned} \frac{\partial \hat{R_0}}{\partial q}>0\quad \text{ and} \quad \frac{\partial \hat{R_0}}{\partial z}<0 \end{aligned}$$

whenever $k_1k_2\varepsilon _r>\varepsilon _s$.

Proof

$$\begin{aligned} \frac{\partial \hat{R_0}}{\partial q}=\frac{a\beta \lambda \varepsilon _s }{2\delta \gamma c}\left[-1+\frac{-\varepsilon _s(1-q)+k_1k_2\varepsilon _r(1+z)}{\sqrt{[\varepsilon _s(1-q)+\varepsilon _r k_1k_2(1-z)]^2-4\varepsilon _s\varepsilon _rk_1k_2(1-q-z)}} \right] \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \hat{R_0}}{\partial z}\!=\!\frac{k_1k_2a\beta \lambda \varepsilon _r}{2\delta \gamma c}\left[-1\!+\!\frac{\varepsilon _s(1+q)\!-\!k_1k_2\varepsilon _r(1\!-\!z)}{\sqrt{[\varepsilon _s(1-q)+\varepsilon _r k_1k_2(1-z)]^2-4\varepsilon _s\varepsilon _rk_1k_2(1\!-\!q\!-\!z)}} \right]. \end{aligned}$$

$\frac{\partial \hat{R_0}}{\partial q}>0$ iff

$$\begin{aligned} -\varepsilon _s(1\!-\!q)\!+\!k_1k_2\varepsilon _r(1\!+\!z)>{\sqrt{[\varepsilon _s(1\!-\!q)\!+\!\varepsilon _r k_1k_2(1\!-\!z)]^2\!-\!4\varepsilon _s\varepsilon _rk_1k_2(1\!-\!q\!-\!z)}}. \end{aligned}$$

Since the term inside the square root is positive $(k_1,k_2\in (0,1)\,\text{ and}\,q,z<<1)$, it is enough to show that

$$\begin{aligned}{}[-\varepsilon _s(1-q)+k_1k_2\varepsilon _r(1+z)]^2&> [\varepsilon _s(1-q)+\varepsilon _r k_1k_2(1-z)]^2\\&-4\varepsilon _s\varepsilon _rk_1k_2(1-q-z). \end{aligned}$$

This reduces to

$$\begin{aligned} k_1k_2\varepsilon _rz+\varepsilon _s(1-q-z)>\varepsilon _s(1-q) \quad \text{ or} \quad k_1k_2\varepsilon _r>\varepsilon _s \end{aligned}$$

which is satisfied depending on the parameters $k_1$ and $k_2$ and the treatment parameters $p_1,p_2,\varepsilon _{rt}$ and $\varepsilon _{pi}$.

Similarly, $\frac{\partial \hat{R_0}}{\partial z}<0$ iff

$$\begin{aligned} \varepsilon _s(1+q)\!-\!k_1k_2\varepsilon _r(1-z)<{\sqrt{[\varepsilon _s(1-q)+\varepsilon _r k_1k_2(1\!-\!z)]^2\!-\!4\varepsilon _s\varepsilon _rk_1k_2(1\!-\!q\!-\!z)}}. \end{aligned}$$

It is enough to show that

$$\begin{aligned}{}[\varepsilon _s(1+q)-k_1k_2\varepsilon _r(1-z)]^2<[\varepsilon _s(1\!-\!q)\!+\!\varepsilon _r k_1k_2(1\!-\!z)]^2-4\varepsilon _s\varepsilon _rk_1k_2(1\!-\!q\!-\!z) \end{aligned}$$

which reduces to

$$\begin{aligned} q\varepsilon _s+\varepsilon _rk_1k_2(1-q-z)<\varepsilon _rk_1k_2(1-z) \quad \text{ or} \quad \varepsilon _s<k_1k_2\varepsilon _r \end{aligned}$$

which is satisfied depending on the parameters $k_1$ and $k_2$ and the treatment parameters $p_1,p_2,\varepsilon _{rt}$ and $\varepsilon _{pi}$. Therefore,

$$\begin{aligned} \frac{\partial \hat{R_0}}{\partial q}>0 \quad \text{ and}\quad \frac{\partial \hat{R_0}}{\partial z}<0 \end{aligned}$$

whenever $k_1k_2\varepsilon _r>\varepsilon _s$. $\square$