Thread: Stability Analysis of two-dimensional system of DEs

1. Stability Analysis of two-dimensional system of DEs

Textbook: Mathematical Models in Population Biology and Epidemiology
Course: Self-study

I have this model with the hypothesis that $r > \mu$
\begin{align} S' &= r(S + I) - \beta S I - \mu S \\ I' &= \beta S I - (\mu + \alpha) I \end{align}

From the second equation the equilibria are given by either $I = 0$ or $S = \frac{\mu + \alpha}{\beta}$. For the first case, $I = 0$, the first equilibrium equation gives us $r S = \mu S$ which implies that $S = 0$ given our hypothesis. We then have an equilibrium $(S,I) = (0,0)$ and through linearizing the system I was able to show that this equilibrium is unstable.
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For the second case, $S = \frac{\mu + \alpha}{\beta}$, the first equation gives us $I = \frac{(r - \mu) (\mu + \alpha)}{\beta (\alpha + \mu - r)}$. We then have an equilibrium $(S,I) = \left(\frac{\mu + \alpha}{\beta}, \frac{(r - \mu) (\mu + \alpha)}{\beta (\alpha + \mu - r)}\right)$. For simplicity, let $\left(\frac{\mu + \alpha}{\beta}, \frac{(r - \mu) (\mu + \alpha)}{\beta (\alpha + \mu - r)}\right) = (S_\infty, I_\infty)$. Linearizing about this equilibrium we get this two-dimensional system
\begin{align} u' &= (r - \beta I_\infty - \mu) u + (r - \beta S_\infty) v \\ v' &= \beta I_\infty u + (\beta S_\infty - (\mu + \alpha)) v \end{align}

Finding the eigenvalues of the coefficient matrix we get $$(r - \beta I_\infty - \mu - \lambda)(\beta S_\infty - (\mu + \alpha) - \lambda) - \beta S_\infty (r - \beta S_\infty) = 0$$

This quadratic is really hard to solve by hand, as I am not familiar with any CAS. The authors of my textbook state that this equilibrium is stable, but they omit the verification. Should I take some other approach to determining the stability of this equilibrium?

EDIT: Another hypothesis is that $r < \mu + \alpha$

2. Re: Stability Analysis of two-dimensional system of DEs

I have figured out how to show that the equilibrium is stable. Starting from here,
For simplicity, let $\left(\frac{\mu + \alpha}{\beta}, \frac{(r - \mu) (\mu + \alpha)}{\beta (\alpha + \mu - r)}\right) = (S_\infty, I_\infty)$. Linearizing about this equilibrium we get this two-dimensional system
\begin{align} u' &= (r - \beta I_\infty - \mu) u + (r - \beta S_\infty) v \\ v' &= \beta I_\infty u + (\beta S_\infty - (\mu + \alpha)) v \end{align}
Our coefficient matrix is $A = \begin{pmatrix} r - \beta I_\infty - \mu & r - \beta S_\infty \\ \beta I_\infty & \beta S_\infty - (\mu + \alpha) \end{pmatrix} = \begin{pmatrix} r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & r - \mu - \alpha \\ \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & 0 \end{pmatrix}$

I had to use an approach that did not rely on eigenvalues. If the determinant of the matrix is positive and of the trace of the matrix is negative, then the equilibrium is stable.

$\begin{vmatrix} r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & r - \mu - \alpha \\ \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & 0 \end{vmatrix} = -(r - \mu - \alpha)\frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} = (r - \mu) (\mu + \alpha) > 0$

The trace is equal to $r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)}$. Using the fact that we are dealing with positive constants, $r > \mu$, and $r < \alpha + \mu$, we can show that the trace is negative.

\begin{align} \mu + \alpha &> \mu + \alpha - r \\[2 mm] \frac{\mu + \alpha}{\mu + \alpha - r} &> 1\\[2 mm] \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} &> r - \mu \end{align}

Therefore, $r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} < 0$. Hence our equilibrium is stable.