Equilibrium Solutions to set of differential equations.

Question

In class today we were given this system of linear equations

$\frac{dS}{dt} = aS(1-\frac{S}{60})-cSF$

$\frac{dF}{dt} = -bF +dSF$

And we have to find the equilibrium solutions to this system of equations. I know that there is obviously one at (0,0) but I am a bit stuck at finding any more. Since this is a non-linear system do I have to use the Jacobian to linearise it first?

Any help would be much appreciated

Answer 1

$$ aS(1-\frac{S}{60}) - cSF = 0\\ -bF + d SF = F (-b + dS) = 0 $$

The second equation means either $F=0$ or $-b + d S = 0$.

First the $F=0$ case.

Plugging that into the first equation gives either $S=0$ or $S=60$. Those are the solutions already given in the question and comment.

Now the $S = \frac{b}{d}$ case. Again plug that into the first equation and solve for $F$.

$$ a \frac{b}{d} (1 - \frac{b}{60d}) - \frac{bc}{d} F = 0\\ F = a \frac{b}{d} (1 - \frac{b}{60d}) \frac{d}{bc}\\ = \frac{a}{c} (1 - \frac{b}{60d}) $$

So that is the last solution that you still had to find.

You were only asked to find the equilibria not about the linearizations.

Answer 2

In order to find all the equilibrium points of the system, you need to solve $$\frac{dS}{dt} = aS(1-\frac{S}{60})-cSF=0$$

$$\frac{dF}{dt} = -bF +dSF=0$$

After solving the system, then if you want to linearize your system about one the equilibrium points you find the Jacobian matrix at that equilibrium point.