# Classifying equilibrium points

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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Kind of like a saddle point. Have a look at the top right diagram on page 3 here. –  bgins May 4 '12 at 19:21
Thanks for the link! I will definitely check it out! –  Kristian May 4 '12 at 20:01

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?
No, it can't. In this case there is one trajectory that goes in to the critical point, on the line $x+y=2$ for $x < 1$, and one that goes out from the critical point, on the same line for $x > 1$. This is quite different from any of the phase portraits for linear systems. –  Robert Israel May 4 '12 at 19:52