Hello there!
I have a question regarding linear stability analysis in MatLab.
given is the ODE
x' = x - x^3
of which I have calculated the fixed points x1 = 0 x2 = 1 x3 = -1
Now these points have to be checked for stability, both graphically and by means of linear stability analysis. I started doing that, by doing a linearization of the given differential equation and trying to set up a Jacobian.
But: since I've only got one variable and one equation, the Jacobian is reduced to a skalar, or am I seeing this wrong?
I'd need the Jacobian to get the eigenvalues and further the stability, so I can plot it in MatLab, but I don't see how I could do it with this equation.
So I would really be grateful for some help. Greetings, Tanja :-)

댓글 수: 1

Sagar Doshi
Sagar Doshi 2017년 6월 20일
You are correct about the Jacobian for single equation with one variable is a scalar. This can also be seen in the Wikipedia link for liner stability here .

댓글을 달려면 로그인하십시오.

답변 (0개)

카테고리

질문:

2017년 6월 17일

댓글:

2017년 6월 20일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by