Basic reproduction number for complicated diseases model

26 Views Asked by At

For an epidemic model, the basic reproduction number is defined as the average number of new infections (e.g. infectious individuals) generated by one infectious individual in an otherwise completely susceptible population. I have a system (Basic HIV with treatment with intervention) like,

$$ \begin{align} S'&=\pi-\lambda S-\mu S\\ I_u'&=\lambda S-(\mu+\alpha+\epsilon_T\tau_u)I_u\\ A'&=\alpha I_u+\theta_{\alpha}I_T-(\epsilon_T \tau_A+\mu+\delta)A\\ I_T'&=\epsilon_T \tau_u I_u+\epsilon_T \tau_A I_u - (\mu+\theta_{\alpha})I_T \end{align} $$

Where the natural death, diseases induced death and treatment effectiveness are $\mu$,$\delta$ and $\epsilon_T$ respectively and $\lambda=\frac{\beta(I_u+\eta_A A+\eta_T I_T)}{N}$.

  • $I_u(t):$ untreated newly infected individuals
  • $A(t):$ infected individuals at the AIDS
  • $I_T(t):$ treated individuals

I get the next generation matrix,

$$ F-V=\pmatrix{\frac{\beta S}{N}&\frac{\beta \eta_A S}{N}&\frac{\beta \eta_T S}{N}\\0&0&0\\0&0&0}\pmatrix{T_u\\A\\I_T}-\pmatrix{k_1&0&0\\-\alpha&k_2&-\theta_{\alpha}\\-\epsilon_T \tau_u&-\epsilon_T \tau_T &k_3}\pmatrix{T_u\\A\\I_T} $$

where, $k_1=\alpha+\mu+\epsilon_T \tau_u$, $k_2=\epsilon_T\tau_u+\mu+\delta$ and $k_3=\theta_{\alpha}+\mu$

Now, to get $R_0$ from $\rho(FV^{-1})$ is very complicated and error prune. I give a try but the answer is not match with the solution. I was wondering how someone can get that in exam through that huge computation!