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When considering the system

\begin{cases} x' = (A-By)x \\ y' = (C-Dx)y, & \end{cases}

($A,B,C,D > 0$)

I am trying to understand how to tell that the fixed points $(0,0)$ and $(C/D,A/B)$ are unstable. If I'm not mistaken, Lyapunov functions can be used to show that a point is stable but not that it is unstable. I was thinking of finding the eigenvalues (if we consider the system as $\bf{x'} = Ax$), but the eigenvalues will involve $x$ and $y$ and that hasn't happened before as I've been solving exercises, so I'm not sure if that is the way to go...

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  • $\begingroup$ Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function. $\endgroup$
    – user539887
    Commented Mar 14, 2019 at 8:07

1 Answer 1

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The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.

Let's compute the linearization of the right-hand side: $$ L(x,y) = \left[ \begin{array}{ll} A - By & -Bx\\ Dy & C - Dx\\ \end{array} \right] $$ Evaluating it at $(x,y) = (0,0)$, we get: $$ L(0,0) = \left[ \begin{array}{ll} A & 0\\ 0 & C\\ \end{array} \right], $$ so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.

Similarly inspect the eigenvalues of $L(C/D, A/B)$.

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