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