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So I am working with the following set of non-dimensionalized differential equations:

$ dx/dτ = (xy - x) $

$dy/dτ= (-xy - ay + b)$

Where a is always positive and b can be positive or negative.

My next step is to find and classify fixed points. I have 2 fixed points: (0, b/a) and (b-a, 1).

I'm confused from here. I think my next step is to find the jacobian and evaluate eigenvalues, but I'm don't know how to go about it with a & b in my equation.

jacobian =

\begin{matrix} y-1&x\\ -y&-x-a \\ \end{matrix}

The jacobian evaluated at (0, b/a) gives eigenvalues λ = b/a-1 and λ = -a, how do I interpret these?

I don't know how to find the eigenvalues at (b-a, 1).

Any help would be appreciated!

umz
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1 Answers1

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Your fixed points seem to be correct. Assume that $a,b$ are real (and not complex). In the case of $(0,b/a)$ the eigenvalue $-a$ is strictly in the left half plane if $a$ is strictly positive. Therefore if $b<a$ then this fixed point would be a stable equilibrium. For the other equilibrium the Jacobian matrix evaluated at the fixed point gives,

$$ A= \left( \begin{array}{cc} 0 & b-a \\ -1 & -b \\ \end{array} \right) $$

Its eigenvalues can be computed as the roots of the determinant of $(A-\lambda I)$. They are:

$$ \frac{1}{2} \left(-b \pm \sqrt{4 a+b^2-4 b} \right) $$

Now find conditions on $a,b$ so that the real part of the eigenvalue is strictly in the left half plane.

ITA
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  • Very clear and helpful, thanks! So I can't definitely classify my points as sinks/sources/saddle points but I can set up stability conditions on a and b. How would I plot these conditions on a phase portrait? – umz Apr 10 '18 at 14:28