Stability of an equilibrium point

By Carol Lord (T3480) on Saturday, April 28, 2001 - 03:06 pm :

dX/dt = eX - fXY

dY/dt = aY - bXY

( X & Y are fuctions of t )

At an equilibrium point (a/b , e/f), is there stability ?

By Kerwin Hui (Kwkh2) on Saturday, April 28, 2001 - 05:32 pm :

To investigate the stability, we form the Jacobian matrix. The Jacobian matrix of the system

dx / dt = f (x, y)

dy / dt = g (x, y)

(\begin{matrix} \partial f / \partial x & \partial f / \partial y \\ \partial g / \partial x & \partial g / \partial y \end{matrix})

So we have the Jacobian about the equilibrium point in question to be

(\begin{matrix} 0 & - a f / b \\ - b e / f & 0 \end{matrix})

Now we investigate its eigenvalues. We find that the eigenvalues are of opposite sign (

a

b

e

f > 0

), so it is a saddle and hence is unstable.

Kerwin

By Anonymous on Sunday, April 29, 2001 - 06:52 pm :

The Jacobian is a very useful trick for investigating stability that you don't often find out about - effectively it's the same as expanding the Taylor series about the point and then considering the sign of the second term (to see if, when you push it away, it keeps going away or comes back again), but it's a lot quicker and neater.

The sign of the evalues of the Jacobian always determines the type of equilibrium point. (Unfortunately I can't remember the others (ie both roots positive etc) so I'll have to look it up at some point.)