Proving stability of linear system equations

  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1716877322765.png

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!
 
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  • #2
You don't need to look at the Jacobian; the fact that the question tells you nothing about the partial derivatives of [itex]f[/itex] is a strong suggestion that this is the wrong way to proceed.

When I see scalar multiples of [itex](x,y)[/itex] and [itex](-y,x)[/itex] on the right hand side, I immediately think of polar coordinates [itex](x,y) = (r \cos \theta, r \sin \theta)[/itex]. This is because [tex]\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}[/tex] so that the coefficient of [itex](x,y)[/itex] tells you how [itex]r[/itex] behaves and the coefficient of [itex](-y,x)[/itex] tells you how [itex]\theta[/itex] behaves. If [itex]\dot r < 0[/itex] the origin is asymptotically stable; if [itex]\dot r > 0[/itex] the origin is unstable.
 
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  • #3
pasmith said:
You don't need to look at the Jacobian; the fact that the question tells you nothing about the partial derivatives of [itex]f[/itex] is a strong suggestion that this is the wrong way to proceed.

When I see scalar multiples of [itex](x,y)[/itex] and [itex](-y,x)[/itex] on the right hand side, I immediately think of polar coordinates [itex](x,y) = (r \cos \theta, r \sin \theta)[/itex]. This is because [tex]\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}[/tex] so that the coefficient of [itex](x,y)[/itex] tells you how [itex]r[/itex] behaves and the coefficient of [itex](-y,x)[/itex] tells you how [itex]\theta[/itex] behaves. If [itex]\dot r < 0[/itex] the origin is asymptotically stable; if [itex]\dot r > 0[/itex] 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!
 

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