Proving stability of linear system equations

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My solution is to find the characteristic equation of the system by putting the system into a matrix. This gives $\lambda^2 + 2f \lambda + f^2 + 1 = 0$
Then each eigenvalue is $\lambda_1 = -f - i$ and $\lambda_2 = -f + i$

I then want to find the Jacobian, however, I would need to find the partial derivatives (with respect to x and y) of $F(x,y) = y - xf(x,y)$ and $G(x,y) = - x - yf(x,y)$, however, I'm not sure how to do that with the $f(x,y)$ in there.

Does anybody please know what I should do?

Thanks!

 

[pasmith]

You don't need to look at the Jacobian; the fact that the question tells you nothing about the partial derivatives of $f$ is a strong suggestion that this is the wrong way to proceed.

When I see scalar multiples of $(x,y)$ and $(-y,x)$ on the right hand side, I immediately think of polar coordinates $(x,y) = (r \cos \theta, r \sin \theta)$. This is because $$\begin{split} r\frac{dr}{dt} &= x\frac{dx}{dt} + y\frac{dy}{dt} \\ r^2\frac{d\theta}{dt} &= x\frac{dy}{dt} - y\frac{dx}{dt} \end{split}$$ so that the coefficient of $(x,y)$ tells you how $r$ behaves and the coefficient of $(-y,x)$ tells you how $\theta$ behaves. If $\dot r < 0$ the origin is asymptotically stable; if $\dot r > 0$ the origin is unstable.

 

[member 731016]

You don't need to look at the Jacobian; the fact that the question tells you nothing about the partial derivatives of $f$ is a strong suggestion that this is the wrong way to proceed.

When I see scalar multiples of $(x,y)$ and $(-y,x)$ on the right hand side, I immediately think of polar coordinates $(x,y) = (r \cos \theta, r \sin \theta)$. This is because $$\begin{split} r\frac{dr}{dt} &= x\frac{dx}{dt} + y\frac{dy}{dt} \\ r^2\frac{d\theta}{dt} &= x\frac{dy}{dt} - y\frac{dx}{dt} \end{split}$$ so that the coefficient of $(x,y)$ tells you how $r$ behaves and the coefficient of $(-y,x)$ tells you how $\theta$ behaves. If $\dot r < 0$ the origin is asymptotically stable; if $\dot r > 0$ the origin is unstable.

Thank you for your reply @pasmith!

That is a interesting idea that I have not seen before. They have taught me to me so far to methods to find the stability of a non-linear system of equations, to either use the Jacobian matrix for linearization to find a equivalent linear DE system to the non-linear DE system or use the direct method.

If I think about for the later method, we could also try to solve this problem by using a Liapunov function of the form $V(x,y) = dx^2 + dy^2 = d(x^2 + y^2)$? If that does not work then I could generalize to more coefficients, $V(x,y) = dx^2 + gy^2$?

Thanks!