Effect of Delayed Immune Response on the Dynamics of HIV Infection Under Multidrug Treatment

Abstract

This study considers a mathematical model that describes the interaction between HIV, CD4+ T-cells and immune response of the body. The immune response activates with a time delay. A combined antiretroviral drug therapy is applied to control the progression of HIV to AIDS. Conditions are obtained for the existence and stability of uninfected and infected steady states. The delay in immune response contributes to alter the stability of steady states. More precisely, this time delay may not affect the local asymptotical stability of uninfected steady state, but can destabilize the infected steady state, which further leads to Hopf bifurcations. A critical delay has been identified to classify the significance of immune system of the body in changing the HIV infection dynamics. The direction of Hopf bifurcation and the stability of bifurcating periodic solutions are obtained using the center manifold theory. Numerical simulations are computed and exhibited to illustrate the effects of delayed immune response on the growth or decay of infection in the presence of combined drug therapy.

Appendix A

Define a space of continuous real valued functions as $C=C([-1,0],R^5)$. Let $u_{1}(t)=T(t)-\overline{T}(t),$ $u_{2}(t)=T_{1}(t)-\overline{T_{1}}(t)$, $u_{3}(t)=T_{2}(t)-\overline{T_{2}}(t)$, $u_{4}(t)=V(t)-\overline{V}(t)$ and $u_{5}(t)=E(t)-\overline{E}(t)$. Define $x_{i}(t)=u_{i}(\tau t)$ for $i=1,2,3,4,5$, with $\tau =\tau _{j}+\nu ,$ $\nu \in R$, so that $\nu =0$ is Hopf bifurcation value for the system. Then, the delay system (2.5)–(2.9) transforms to the functional differential equation in C as follows:

$$\begin{aligned} \frac{dx}{dt}=L_\nu x_{t}+f(\nu , x_{t}), \end{aligned}$$

(7.1)

where $x(t)=(x_{1}(t),x_{2}(t), x_{3}(t), x_{4}(t), x_{5}(t))^{T}\in R^{5}$, $x_{t}(\theta )=x(t+\theta ),$ $\theta \in [-1,0]$. $L_{\nu }:C\rightarrow R^{5}$ and $f:C \times R \rightarrow R^{5}$ are given by

$$\begin{aligned} L_{\nu }\varphi =(\tau _{j}+\nu )[J_{1}\varphi (0)+J_{2}\varphi (-1)];~~\varphi =(\varphi _{1},\varphi _{2},\varphi _{3},\varphi _{4},\varphi _{5})^{T}\in C, \end{aligned}$$

(7.2)

and

$$\begin{aligned} f(\nu , \varphi )=(\tau _{j}+\nu )\left( \begin{array}{c} -k\varphi _{1}(0)\varphi _{4}(0)-r_{1}\varphi ^{2}_{1}(0)\\ k\varphi _{1}(0)\varphi _{4}(0)\\ -d_{x}\varphi _{3}(0)\varphi _{5}(0)\\ 0 \\ 0 \\ \end{array} \right) \end{aligned}$$

(7.3)

where the matrices $J_{1}=\mathbf J$ and $J_{2}=\mathbf K$ are defined in Sect. 4.2. By the Reisz representation theorem, there exists a function $\xi (\theta ,\nu )$ whose components are of bounded variation for $\theta \in [-1,0]$ such that

$$\begin{aligned} L_{\nu }\varphi =\int ^{0}_{-1}d\xi (\theta ,\nu )\varphi (\theta ). \end{aligned}$$

(7.4)

In relation to the Eq. (7.2), we choose

$$\begin{aligned} d\xi (\theta ,\nu )=(\tau _{j}+\nu )[J_{1}\overline{\delta }(\theta )-J_{2}\overline{\delta } (\theta +1)], \end{aligned}$$

(7.5)

where $\overline{\delta }(\theta )$ is the Dirac delta function. Through the relevant expressions for function $A(\nu )$ and $R(\nu )$, the system (7.1) can be rewritten as

$$\begin{aligned} \dot{x}_{t}=A(\nu )x_{t}+R(\nu )x_{t}, \end{aligned}$$

(7.6)

where $x_{t}(\theta )=x(t+\theta )$ for $\theta \in [-1,0]$. Let $q(\theta )=(\alpha _{1},~ \alpha _{2},~\alpha _{3},~\alpha _{4}, ~\alpha _{5})^{T}e^{i\omega _{0}\tau _{j}\theta }$ be the eigenvector of A(0) (referred hereafter as A) corresponding to the eigenvalue $i\omega _{0}\tau _{j}$ then

$$\begin{aligned} Aq(\theta )=i\omega _{0}\tau _{j}q(\theta ), \end{aligned}$$

(7.7)

where, for $\theta =0$, and $\alpha _{1}=1$, we get $\alpha _{2}=\frac{k\overline{V}-(i\omega _{0}+M)}{i\omega _{0}+\mu _{1}+(1-\eta )\alpha }$, $\alpha _{3}=\frac{(i\omega _{0}+\mu _{v})\alpha _{4}}{N(1-\gamma )\delta }$, $\alpha _{4}=\frac{(b+\eta \alpha )\alpha _{2}-(i\omega _{0}+M)}{k\overline{T}}$ and $\alpha _{5}=\frac{pe^{-i\omega _{0}\tau _{j}}\alpha _{3}}{i\omega _{0}+d_{E}}$.

Similarly, assume $q^{*}(s)=D(\alpha ^{*}_{1},~\alpha ^{*}_{2},~\alpha ^{*}_{3},~\alpha ^{*}_{4},~\alpha ^{*}_{5})^{T}e^{i\omega _{0}\tau _{j}s}$ to be the eigenvector of $A^{*}$ (the adjoint of A) corresponding to the eigenvalue $-i\omega _{0}\tau _{j}$, to get

$$\begin{aligned} A^{*}q^{*}(s)=-i\omega _{0}\tau _{j}q^{*}(s), \end{aligned}$$

(7.8)

with $\alpha ^{*}_{1}=1$, $\alpha ^{*}_{2}=\frac{-i\omega _{0}+M}{k\overline{V}}$, $\alpha ^{*}_{3}=\frac{(-i\omega _{0}+\mu _{1}+\alpha +b)\alpha ^{*}_{2}-(b+\eta \alpha )}{(1-\eta )\alpha }$, $\alpha ^{*}_{4}=\frac{k\overline{T}-k\overline{T}\alpha ^{*}_{2}}{i\omega _{0}-\mu _{v}}$, $\alpha ^{*}_{5}=\frac{d_{x}\overline{T_{2}}\alpha ^{*}_{3}}{i\omega _{0}-d_{E}}$.

The normalization condition (i.e., $\left\langle q^{*}(s),q(\theta )\right\rangle = 1)$ yields the value of $\bar{D}$ as

$$\begin{aligned} \bar{D}[1+\alpha _{2}\overline{\alpha ^{*}_{2}}+\alpha _{3}\overline{\alpha ^{*}_{3}}+\alpha _{4}\overline{\alpha ^{*}_{4}}+\alpha _{5}\overline{\alpha ^{*}_{5}}+\tau _{j}e^{-i\omega _{0}\tau _{j}}\overline{\alpha ^{*}_{5}}\alpha _{3}p]=1, \end{aligned}$$

(7.9)

where the inner product $< ,>$ is defined as follows.

$$\begin{aligned} \left\langle \varsigma , \varphi \right\rangle =\overline{\varsigma }(0)\varphi (0)-\int ^{0}_{\theta =-1}\int ^{\theta }_{\zeta =0}\overline{\varsigma }(\zeta -\theta )d\xi (\theta )\varphi (\zeta )d\zeta , \end{aligned}$$

(7.10)

where $\xi (\theta )=\xi (\theta ,0)$.

Following Hassard et al. [35], the coordinates are to be computed to describe the center manifold $C_{0}$ at $\nu =0$. Let $x_{t}$ denotes the solution of Eq. (7.6) for $\nu =0$. Define

$$\begin{aligned} z(t)=\left\langle q^{*}, x_{t}\right\rangle ,\quad W(t,\theta )=x_{t}(\theta )-2Re\left\{ z(t)q(\theta )\right\} . \end{aligned}$$

(7.11)

On the center manifold $C_{0}$, $W(t,\theta )$ is expressed as

$$\begin{aligned} W(t,\theta )=W(z,\overline{z},\theta ) =W_{20}(\theta )\frac{z^{2}}{2}+W_{11}(\theta )z\overline{z}+W_{02}(\theta )\frac{\overline{z^{2}}}{2}+\cdots , \end{aligned}$$

(7.12)

where z and $\overline{z}$ are local coordinates for center manifold $C_{0}$ in the direction of $q^{*}$ and $\overline{q^{*}}$. Considering only the real solution $x_{t}\in C_{0}$ of the Eq. (7.6), we get

$$\begin{aligned} \dot{z}=i\omega _{0}\tau _{j}z+\overline{q^{*}}(0).f(0,W(z,\overline{z},0)+2Re\left\{ zq(0)\right\} )=i\omega _{0}\tau _{j}z+g(z,\overline{z}), \end{aligned}$$

(7.13)

where

$$\begin{aligned} g(z,\overline{z})=\overline{q^{*}}(0).f_{0}(z,\overline{z}) =g_{20}\frac{z^{2}}{2}+g_{11}z\overline{z}+g_{02}\frac{\overline{z}^{2}}{2}+\cdots \end{aligned}$$

(7.14)

It follows from (7.11) and (7.12) that

$$\begin{aligned} x_{t}(\theta )= & {} W(z,\overline{z},\theta )+2Re\left\{ zq(\theta )\right\} \nonumber \\= & {} W_{20}(\theta )\frac{z^{2}}{2}+W_{11}(\theta )z\overline{z}+W_{02}(\theta )\frac{\overline{z}^{2}}{2}+z(\alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4},\alpha _{5})^{T}e^{i\omega _{0}\tau _{j}\theta }\nonumber \\&+ \overline{z}(\overline{\alpha _{1}},\overline{\alpha _{2}},\overline{\alpha _{3}}, \overline{\alpha _{4}}, \overline{\alpha _{5}})^{T}e^{-i\omega _{0}\tau _{j}\theta }+\cdots \end{aligned}$$

(7.15)

so that for $j=1,2,3,4,5$,

$$\begin{aligned} x_{jt}(\theta )=W_{20}^{(j)}(\theta )\frac{z^{2}}{2}+W_{11}^{(j)}(\theta )z\overline{z}+W_{02}^{(j)}(\theta )\frac{\overline{z}^{2}}{2}+z\alpha _{j}e^{i\omega _{0}\tau _{j}\theta }+\overline{z\alpha _{j}}e^{-i\omega _{0}\tau _{j}\theta }. \end{aligned}$$

(7.16)

Now from (7.14),

$$\begin{aligned} g(z,\overline{z})=\tau _{j}\overline{D}(\overline{\alpha ^{*}_{1}},\overline{\alpha ^{*}_{2}},\overline{\alpha ^{*}_{3}},\overline{\alpha ^{*}_{4}},\overline{\alpha ^{*}_{5}}) \left( \begin{array}{c} -kx_{1t}(0)x_{4t}(0)-r_{1}x_{1t}^{2}(0)\\ kx_{1t}(0)x_{4t}(0)\\ -d_{x}x_{3t}(0)x_{5t}(0)\\ 0\\ 0 \end{array}\right) . \end{aligned}$$

(7.17)

The Eq. (7.17) is simplified and compared with (7.14) to get

$$\begin{aligned} g_{20}= & {} 2\tau _{j}\overline{D}((-k\alpha _{4}-r_{1})+k\overline{\alpha _{2}^{*}}\alpha _{4}-d_{x}\overline{\alpha ^{*}_{3}}\alpha _{3}\alpha _{5}), \end{aligned}$$

(7.18)

$$\begin{aligned} g_{11}= & {} 2\tau _{j}\overline{D}((-kRe\alpha _{4}-r_{1})+k\overline{\alpha _{2}^{*}}Re\alpha _{4}-d_{x}\overline{\alpha _{3}^{*}}Re\alpha _{3}\overline{\alpha _{5}}),\end{aligned}$$

(7.19)

$$\begin{aligned} g_{02}= & {} 2\tau _{j}\overline{D}((-k\overline{\alpha _{4}}-r_{1})+k\overline{\alpha _{2}^{*}}\overline{\alpha _{4}}-d_{x}\overline{\alpha _{3}^{*}}\overline{\alpha _{3}}\overline{\alpha _{5}}),\end{aligned}$$

(7.20)

$$\begin{aligned} g_{21}= & {} 2\tau _{j}\overline{D}((-k+\alpha _{2}^{*}k)(2\alpha _{4}W^{(1)}_{11}(0)+2W^{(4)}_{11}(0)\nonumber \\&+\overline{\alpha _{4}}W_{20}^{(1)}(0)+W_{20}^{(4)}(0))-r_{1}2W_{20}^{(1)}(0)\nonumber \\&+4r_{1}W_{11}^{(1)}(0)-d_{x}\overline{\alpha _{3}^{*}}(\overline{\alpha _{3}}W_{20}^{(5)}(0)+\overline{\alpha _{5}}W_{20}^{(3)}(0)\nonumber \\&+2\alpha _{3}W_{11}^{5}(0)+2\alpha _{5}W_{11}^{(3)}(0))). \end{aligned}$$

(7.21)

$W_{20}(\theta )$ and $W_{11}(\theta )$, required to compute $g_{21}$, are derived as follows:

Using the Eqs. (7.11) and (7.13), $\dot{W}$ is written as

$$\begin{aligned} \dot{W}= & {} \dot{x}_{t}-\dot{z}q-\dot{\overline{z}}~\overline{q}= \left\{ \begin{array}{ll} AW-2Re\left\{ \overline{q^{*}}(0).f_{0}q(\theta )\right\} ,\quad \theta \in [-1,0), &{} \hbox {} \\ AW-2Re\left\{ \overline{q^{*}}(0).f_{0}q(\theta )\right\} +f_{0},\quad \theta =0, &{}\hbox {}\\ \end{array} \right\} \end{aligned}$$

(7.22)

$$\begin{aligned}\equiv & {} AW + H(z,\overline{z},\theta );~~~ H(z,\overline{z},\theta )=H_{20}(\theta )\frac{z^{2}}{2}+H_{11}(\theta )z\overline{z}+H_{02}\frac{\overline{z}^{2}}{2}+\cdots \qquad \qquad \end{aligned}$$

(7.23)

Using chain rule $\dot{W}=W_{z}\dot{z}+W_{\overline{z}}\dot{\overline{z}}$, the Eqs. (7.13) and (7.23) yields

$$\begin{aligned} (A-2i\omega _{0}\tau _{j})W_{20}=-H_{20},\quad AW_{11}=-H_{11}. \end{aligned}$$

(7.24)

For $\theta \in [-1,0)$, we have

$$\begin{aligned} H(z,\overline{z},\theta )= & {} -\overline{q^{*}}(0).f_{0}q(\theta )-q^{*}(0).\overline{f_{0}}\overline{q}(\theta ) =-g(z,\overline{z})q(\theta )-\overline{g}(z,\overline{z})\overline{q}(\theta ),\nonumber \\= & {} -(g_{20}q(\theta )+\overline{g_{02}}~\overline{q}(\theta ))\frac{z^{2}}{2}-(g_{11}q(\theta )+\overline{g_{11}}~\overline{q}(\theta ))z\overline{z}+\cdots , \end{aligned}$$

(7.25)

which, on comparing the coefficients with (7.23), gives

$$\begin{aligned} H_{20}(\theta )= & {} -g_{20}q(\theta )-\overline{g_{02}}~\overline{q}(\theta ), \end{aligned}$$

(7.26)

$$\begin{aligned} H_{11}(\theta )= & {} -g_{11}q(\theta )-\overline{g_{11}}\overline{q}(\theta ). \end{aligned}$$

(7.27)

From (7.24), (7.26) and the definition of A,

$$\begin{aligned} W_{20}^{'}(\theta )=2i\omega _{0}\tau _{j}W_{20}(\theta )+g_{20}q(\theta )+\overline{g_{02}}\overline{q}(\theta ). \end{aligned}$$

(7.28)

Noting $q(\theta )=q(0)e^{i\omega _{0}\tau _{j}\theta },$ hence

$$\begin{aligned} W_{20}(\theta )=\frac{ig_{20}}{\omega _{0}\tau _{j}}q(\theta )+\frac{i\overline{g_{02}}}{3\omega _{0}\tau _{j}}\overline{q}(\theta )+E_{1}e^{2i\omega _{0}\tau _{j}\theta }. \end{aligned}$$

(7.29)

Similarly, we have

$$\begin{aligned} W_{11}(\theta )=-\frac{ig_{11}}{\omega _{0}\tau _{j}}q(\theta )+\frac{i\overline{g_{11}}}{\omega _{0}\tau _{j}}\overline{q}(\theta )+E_{2}, \end{aligned}$$

(7.30)

where $E_{1}=(E_{1}^{(1)}, E_{1}^{(2)},E_{1}^{(3)},E_{1}^{(4)},E_{1}^{(5)})$ and $E_{2}=(E_{2}^{(1)}, E_{2}^{(2)},E_{2}^{(3)},E_{2}^{(4)},E_{2}^{(5)})$ $\in R^{5}$ are constant vectors, to be determined. It follows from the definition of A and (7.24) that

$$\begin{aligned} \int _{0}^{-1}d\xi (\theta )W_{20}(\theta )= & {} 2i\omega _{0}\tau _{j}W_{20}(0)-H_{20}(0), \end{aligned}$$

(7.31)

$$\begin{aligned} \int _{-1}^{0}d\xi (\theta )W_{11}(\theta )= & {} -H_{11}(0). \end{aligned}$$

(7.32)

From Eqs. (7.22) and (7.23),

$$\begin{aligned} H_{20}(0)=-g_{20}q(0)-\overline{g}_{02}~\overline{q}(0)+2\tau _{j} \left( \begin{array}{c} -k\alpha _{4}-r_{1}\\ k\alpha _{4}\\ -d_{x}\alpha _{3}\alpha _{5}\\ 0\\ 0 \end{array}\right) \end{aligned}$$

(7.33)

and

$$\begin{aligned} H_{11}(0)=-g_{11}q(0)-\overline{g}_{11}~\overline{q}(0)+2\tau _{j} \left( \begin{array}{c} -kRe\alpha _{4}-r_{1}\\ kRe\alpha _{4}\\ -d_{x}Re\alpha _{3}\overline{\alpha _{5}}\\ 0\\ 0 \end{array}\right) . \end{aligned}$$

(7.34)

Using (7.29) and (7.33) in (7.31) and noting that $q(\theta )$ is eigenvector of A, we have

$$\begin{aligned}&\left( \begin{array}{ccccc} 2i\omega _{0}+M &{}\quad -(\eta \alpha +b) &{}\quad 0 &{}\quad k\overline{T} &{}\quad 0\\ -k\overline{V}&{}\quad 2i\omega _{0}+\mu _{1}+\alpha +b &{}\quad 0&{}\quad -k\overline{T} &{}\quad 0\\ 0 &{}\quad -(1-\eta )\alpha &{}\quad 2i\omega _{0}+\delta +d_{x}\overline{E} &{}\quad 0&{}\quad d_{x}\overline{T_{2}}\\ 0&{}\quad 0&{}\quad -N(1-\gamma )\delta &{}\quad 2i\omega _{0}+\mu _{v} &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -e^{-2i\omega _{0}\tau _{j}}p &{}\quad 0 &{}\quad 2i\omega _{0}+d_{E} \end{array}\right) \left( \begin{array}{c} E^{(1)}_{1}\\ E^{(2)}_{1}\\ E^{(3)}_{1}\\ E^{(4)}_{1}\\ E^{(5)}_{1} \end{array}\right) \nonumber \\&\quad = 2 \left( \begin{array}{c} -k\alpha _{4}-r_{1}\\ k\alpha _{4}\\ -d_{x} \alpha _{3} \alpha _{5}\\ 0\\ 0 \end{array}\right) . \end{aligned}$$

(7.35)

Similarly, using (7.30) and (7.34) in (7.32), we have

$$\begin{aligned} \left( \begin{array}{ccccc} M &{}\quad -(\eta \alpha +b) &{}\quad 0 &{}\quad k\overline{T} &{}\quad 0\\ -k\overline{V}&{}\quad \mu _{1}+\alpha +b &{}\quad 0&{}\quad -k\overline{T} &{}\quad 0\\ 0 &{}\quad -(1-\eta )\alpha &{}\quad 2i\omega _{0}+\delta +d_{x}\overline{E} &{}\quad 0&{}\quad d_{x}\overline{T_{2}}\\ 0&{}\quad 0&{}\quad -N(1-\gamma )\delta &{}\quad \mu _{v} &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -p &{}\quad 0 &{}\quad 2d_{E} \end{array}\right) \left( \begin{array}{c} E^{(1)}_{2}\\ E^{(2)}_{2}\\ E^{(3)}_{2}\\ E^{(4)}_{2}\\ E^{(5)}_{2} \end{array}\right) = 2 \left( \begin{array}{c} -kRe \alpha _{4}-r_{1}\\ kRe \alpha _{4}\\ -d_{x} Re\alpha _{3}\overline{ \alpha _{5}}\\ 0\\ 0 \end{array}\right) . \end{aligned}$$

(7.36)

The systems (7.35) and (7.36) are solved for $E_{1}$ and $E_{2}$, respectively. Then, $g_{21}$ is calculated from (7.21), through the expressions (7.29) for $W_{20}$ and (7.30) for $W_{11}$.