SI model analysis

Question

We have an SI model with susceptible individuals $$\frac {dS}{dt} = bM\left(1 - \frac M K\right) - \mu S - \beta SI$$ and infected individuals $$ \frac {dI}{dt} = \beta SI - (\alpha+\mu)I.$$ In this case, show that the infection does not force the total popuation $M = S + I $ to die out.

So I tried to find the equilibria of the system by expressing $\frac {dM}{dt}=\frac {dS}{dt} + \frac {dI}{dt}$ and also using $ S = M - I$ in the second equation to get

$$\frac {dM}{dt} = bM\left(1 - \frac M K\right) - \mu M - \alpha I$$ $$ \frac {dI}{dt} = [\beta(M-I) - (\alpha+\mu)]I.$$

Both must be equal to $0$ so then I got equilibria $(0,0),(K ( 1 - \frac{\mu}{b}),0)$ and another quadratic equation. How do I proceed from here? Which case implies that the epidemic will not force the population to go extinct? Any help is appreciated.

Answer 1

With respect to the points of equilibrium, the points of stable equilibrium are those that give us an idea of ​​the values ​​achieved at steady state. Within the usual classification of source sink or center point, the sinks are associated with steady state stability. We will assume $\{\alpha,\beta,\mu, k\} > 0$

Considering the system

$$ \frac {dM}{dt} = \beta M\left(1 - \frac M k\right) - \mu M - \alpha I\\ \frac {dI}{dt} = [\beta(M-I) - (\alpha+\mu)]I $$

the equilibrium points are obtained by solving for $I,M$

$$ \beta M\left(1 - \frac M k\right) - \mu M - \alpha I = 0\\ \left(\beta(M-I) - (\alpha+\mu)\right)I = 0 $$

giving

$$ \left[ \begin{array}{cc} I & M \\ 0 & 0 \\ 0 & k(1-\frac{\mu }{\beta }) \\ \frac{k \beta-(k+2) (\alpha +\mu ) +\sqrt{k \left((k+4) \alpha ^2-2 \beta k \alpha +2 (k+2) \mu \alpha +k (\beta -\mu )^2\right)}}{2 \beta} & \frac{k \beta-k (\alpha +\mu ) +\sqrt{ k \left(k (\alpha -\beta +\mu )^2+4 \alpha (\alpha +\mu )\right)}}{2 \beta} \\ -\frac{-k \beta+(k+2) (\alpha +\mu ) +\sqrt{ k \left((k+4) \alpha ^2-2 \beta k \alpha +2 (k+2) \mu \alpha +k (\beta -\mu )^2\right)}}{2 \beta} & -\frac{-k \beta+k (\alpha +\mu ) +\sqrt{ k \left(k (\alpha -\beta +\mu )^2+4 \alpha (\alpha +\mu )\right)}}{2 \beta} \\ \end{array} \right] $$

Those points can be qualified with the associated jacobian eigenvalues . The jacobian is given by

$$ J = \left( \begin{array}{cc} -\alpha +\beta (M-2 I)-\mu & \beta I \\ -\alpha & -\frac{2 M \beta }{k}+\beta -\mu \\ \end{array} \right) $$

now assuming parametric values $\alpha = 1,\beta = \frac 12,\mu = 1,k = 1$ we have the equilibrium points

$$ \left( \begin{array}{cc} I & M \\ 0. & 0. \\ 0. & -1. \\ -2.29844 & 1.70156 \\ -8.70156 & -4.70156 \\ \end{array} \right) $$

and the associated eigenvalues

$$ \left( \begin{array}{lccl} 1&-2. & -0.5 &\mbox{sink}\\ 2&-2.5 & 0.5 &\mbox{saddle}\\ 3&-2.51518 & 1.46284 &\mbox{saddle}\\ 4&6.36336 & 2.18899 &\mbox{source}\\ \end{array} \right) $$

Attached a stream flow showing those points in red.

According to this plot, associated to the actual parameters, we have a total extinction at the sink $(0,0)$ and also we have a source point at $(-8.70156, -4.70156)$

Answer 2

It's a lot simpler than you think.

There are two simple ways to show that the infection cannot drive the population to extinction.

1. The first method is to show that the $(0,0)$ is an unstable steady state, which translates to no extinction is possible.

2. The second and fancier method is to prove the system is permanent, e.g. there exist positive constants that bound $S$ and $I$ from below. If these constants exist, it means the system cannot approach $(0,0)$, which again translates to no extinction is possible.

For the second method, just use inequality directly on the rate equations. Both methods are pretty straightforward mathematically.