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!
Best Answer
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.