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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 
Associated Topics:
College Calculus
College Linear Algebra

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