Determining and Checking Saddle Point for Autonomous System
Date: 06/18/2004 at 16:05:00 From: Dennis Subject: How to see if autonomous system has a saddle point If you have a autonomous system in the form: dx/dt = x + x^2 + y^2 dy/dt = 2y - xy Then you can find the critical points (0,0) and (-1,0). The Jacobian matrix of the system is: | 1+2x 2y | | -y 2-x | If you fill in the point (-1,0) you get the matrix | -1 0 | | 0 3 | How can you see that this point is a saddle point? I think it has something to do with the fact that Fx is negative and Fy is positive but I'm not sure. I searched a lot on the net about Jacobian but I can't find anything related to this problem. I hope you can help me.
Date: 06/18/2004 at 16:23:37 From: Doctor Douglas Subject: Re: How to see if autonomous system has a saddle point Hi Dennis. Thanks for writing to the Math Forum. When you linearize the system by inserting -1 for x and 0 for y into the Jacobian, you arrive at a matrix which has one positive eigenvalue (3) and one negative eigenvalue (-1). The fact that these two eigenvalues have different signs identify the equilibrium point as a saddle. If the two eigenvalues were both positive, then the equilibrium would be unstable (a repeller). If the two eigenvalues were complex conjugates with negative real parts, then the fixed point would be a spiral sink. If they were complex conjugates with positive real parts, the point would be a spiral repellor. And so on. The nature of the point depends on the type of eigenvalues. This is true in higher dimensions as well. I hope that helps. Please write back if you need more help with this. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum