3
$\begingroup$

Consider the nonlinear system:

\begin{cases} x' = x^2 + y \\\\ y' = x - y + a \\\\ \end{cases} where $ a $ is a parameter.

$ a) $ Find all equilibrium points and compute the linearized equation at each.

For this question I solve \begin{cases} x^2 + y = 0 \\\\ x - y + a = 0 \\\\ \end{cases} to give me $ \displaystyle x = \frac{-1 \pm \sqrt{1 - 4a}}{2} $ and $ \displaystyle y = \frac{-1 \pm \sqrt{1 - 4a}}{2} + a $ as equilibrium points. The linearized system I got is \begin{cases} x' = y \\\\ y' = x - y \\\\ \end{cases}

$ b) $ Describe the behavior of the linearized system at each equilibrium point?

Can someone help me with this one? The linearized system doesn't depend on $ a $ and so why does it ask for the behavior at each equilibrium point?

$\endgroup$
2
$\begingroup$

The linearization does depend on $a $. Your linearization is missing a term. Setting up the Jacobian, you get: \begin{bmatrix} 2x & 1\\ 1 & -1\\ \end{bmatrix}
evaluating this at the fixed points gives two separate Jacobians for each fixed point of the form: \begin{bmatrix} -1 \pm \sqrt{1-4a} & 1\\ 1 & -1\\ \end{bmatrix}

In general, to describe the behavior of linearization, you need to find the eigenvalues and eigenvectors of the matrix. The eigenvalues will tell you whether the flow is going inwards or outwards from the fixed point as well as the magnitude. The eigenvectors will tell you the direction of local flow around the fixed point. If the eigenvalues are complex with nonzero real part, it will spiral (the direction determined by the sign of the real part). If it is entirely complex, linearization will predict a center. However, linearization often fails in this scenario, as centers are very delicate. Hartman-Grobman Theorem tells us that linearization is accurate in the neighborhood of a hyperbolic fixed point (which a center is not).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.