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I am a bit unsure about the following problem:

Given:

$\dot{x} = y^2 - 2y + 1$

$\dot{y} = -x^2 + 2x -1$

Find and classify all equilibrium points of the system.

OK, så we know that equilibrium points occur when:

$y^2 - 2y + 1 = 0$

and

$-x^2 + 2x -1 = 0$

It is easy to see that this can only occur at $x = 1, y = 1$.

Now I find the Jacobi matrix for the system:

$J = \begin{bmatrix} 0 & 2y-2 \\ -2x+2 & 0 \end{bmatrix}$

By plotting $x = 1, y = 1$ into the matrix we are left with:

$J = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

I know that I am now supposed to find the eigenvalues, and from this deduct whether we are dealing with node, spiral, center, etc. But I have never before encoutered a zero matrix in these calcuations before. Basically, if I find the eigenvalue here, I get $\lambda^{2} = 0$, and I don't see how this can tell me anything about the nature of the equilibrium point.

Any help will be truly appreciated!

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    Tha$n$ks for the link! I will definitely check it out!2012-05-04

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You're right, you need more than the linearization to see what's going on here. Note that everywhere off the lines $x=1$ and $y=1$ we have $\dot{x}> 0$ and $\dot{y} < 0$. So, for example, if you start at $x=1+\epsilon$ and $y = 1-\epsilon$ with $\epsilon>0$ what will happen? What does this say about stability of the critical point?

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    Thanks for dispelling any possible confusion from my saddle point analogy.2012-05-04