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Find all the fixed points of the system $f(x,y)=-x+x^3, g(x,y)=-2y$ and use linearization to classify them.

I have found the solutions to be : $x = 0$ or $x = ±1$ and $y = 0 \implies 3$ fixed points $(0, 0), (1, 0) $and$ (−1, 0)$

We then calculate the Jacobian matrix, which I did for each of the above fixed points. However, my only concern is that once I calculate the Jacobian matrix at each FP how do I know whether it is a stable, saddle, or unstable node?

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up vote 2 down vote accepted

Take the Jacobian and find its eigenvalues at the fixed points.

The Jacobian is: $\begin{pmatrix} -1+3x^2 & 0 \\ 0 & -2 \end{pmatrix}$

At (0,0): $\begin{vmatrix} -1-\lambda & 0 \\ 0 & -2 - \lambda \end{vmatrix} = (\lambda+1)(\lambda+2)$

So, we have two negative eigen values, ie, stable node at (0,0).

You do this for the rest of the fixed points, here is a sheet with the correspondence of eigenvalues to type of fixed point.

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thank u this clarifies my concern. – Saba Di Nov 20 '12 at 19:14

Well, theres a really simple method :) You could refer to Nonlinear Dynamics And Chaos by Steven H. Strogatz for the derivation; I'll just present the results here.
You have to create the Jacobian matrix and then calculate (a) the trace T of the matrix, (b) the Determinant D of the matrix, and (c) the quantity T^2-4D.
The results are,
(1) If D<0: The fixed point is a saddle point
(2) If T=0 and (1) is not satisfied: It is a center, i.e., with orbits around it - neutral stability.
(3) If T<0 and (1) is not satisfied:
(i) If T^2-4D<0: It is a stable spiral.
(ii) If T^2-4D>0: It is a stable node.
(4) If T>0 and (1) is not satisfied:
(i) If T^2-4D<0: It is an unstable spiral.
(ii) If T^2-4D>0: It is an unstable node.
For your example, (0,0) is a stable node and (1,0) & (-1,0) are both saddle points.
Hope that helped :)

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You have a stable node if the matrix is negative definite, unstable node if the matrix is positive definite and a saddle if it is neither of those two.

There are several conditions for positive definiteness and negative definiteness, one of the easiest being the sign of the eigenvalues. If all are positive, the matrix is positive definite, if all are negative then it is negative definite and otherwise it is neither.

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