HIV inflicts its damage by infecting healthy CD4 + T cells (a type of white blood cell) that are necessary to fight infection. As the virus embeds in a T cell and the immune system produces more of these cells to fight the infection, the virus propagates in an opportunistic fashion. Normally T cells are produced at a rate $s$ and die at a rate $d$. The HIV virus is present in the bloodstream in the infected individual. These viruses in the bloodstream, called free viruses, infect healthy T cells at a rate $\beta$. Also, the viruses reproduce through the T cell multiplication process or otherwise at a rate $k$. Free viruses die at a rate c. Infected T cells die at a rate $\mu$.

A simple mathematical model to describe how the HIV/AIDS virus infects healthy cells is given by the following equations:

$$\begin{align}\frac{dT}{dt} &= s-dT-\beta T v \\ \frac{dT^*}{dt} &= \beta T v - \mu T^* \\ \frac{dv}{dt} &= kT^* -cv\end{align}$$

I am told that the system has two equilibrium points given by

$$(T_0, T_0^*,v_0) = (\frac{s}{d},0,0)$$ and $$(T_0, T_0^*,v_0) = (\frac{c\mu}{\beta k}, \frac{s}{\mu}-\frac{cd}{\beta k}, \frac{sk}{c\mu}-\frac{d}{\beta})$$

I am asked to prove that the equilibrium points are the ones shown above but I am having trouble proving this after setting the derivatives equal to $0$. I can obtain the first equilibirum point easily if $\mu$ is zero but otherwise not. How would I solve this system for the two equilbirium points? My differential equations knowledge is quite rusty.

  • $\begingroup$ How is $T^*$ defined? EDIT, is $T^*$ the infected cells and $T$ the healthy T cells? $\endgroup$ Aug 21, 2018 at 16:02
  • $\begingroup$ @AndrewAllen that is correct $\endgroup$
    – Blargian
    Aug 21, 2018 at 18:39

1 Answer 1


Setting the equations to zero gives:

$(1)\space 0 = s-dT-\beta T v $

$(2)\space0 = \beta T v - \mu T^*$

$(3)\space 0 = kT^* -cv$

My plan to solve the simultaneous equations here is to

  • use (3) to get $T^*$ in terms of $k, c, v$

  • plug this $T^*$ found into (2) to get $T$ in terms of $v, \beta, \mu, c, k$ when $v\ne0$

  • use (1) to get $v$ in terms of $s, d, \beta, \mu, c, k$

  • finally use the $v$ found back into our found $T$ and $T^*$ equations

This gives the second equilibrium point once simplified as $(T, T^*, v)$


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