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I am a bit unsure about the following problem:


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

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

Find and classify all equilibrium points of the system.

OK, så we know that equilibrium points occur when:

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


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

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
up vote 3 down vote accepted

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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Well, the phase paths will move away from the equilibrium point, as far as I can tell. So obviously this is not a stable point. But can it be classified as a clear example of a "saddle" or "unstable node"? – Kristian May 4 '12 at 19:43
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
Thanks a lot! I understand this now. Much appreciated! – Kristian May 4 '12 at 20:01
Thanks for dispelling any possible confusion from my saddle point analogy. – bgins May 4 '12 at 20:19

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