ODEs - Stability

We have the system of ODEs

$y_1'=y_2-y_1 \cdot f((y_1,y_2)^T)$

$y_2'=-y_1-y_2 \cdot f((y_1,y_2)^T)$,

where $f \in C^1(\mathbb{R}^2, \mathbb{R})$.

Prove the following statement:

If $f((y_1,y_2)^T)>0$ in a neighbourhood of $(0,0)$, then $(0,0)$ is asymptotically stable, but if $f((y_1,y_2)^T)<0$ in a neighbourhood of $(0,0)$, then $(0,0)$ is instable.

Ordinary Differential Equations Differential Equations
Ichbinanonym Ichbinanonym
116
Report

Answer

Answers can only be viewed under the following conditions:
  1. The questioner was satisfied with and accepted the answer, or
  2. The answer was evaluated as being 100% correct by the judge.
View the answer
Matchmaticians Alumnus Matchmaticians Alumnus
328
  • Ichbinanonym Ichbinanonym
    0

    Is it not possible that f(0,0)=0 although f is positive (or negative) in a neighborhood of (0,0), so tr J_G = 0? If so, does the argument still hold?

  • Ichbinanonym Ichbinanonym
    0

    Nevermind, stupid mistake on my part

  • Matchmaticians Alumnus Matchmaticians Alumnus
    0

    I took a "neighborhood" to include (0,0) as well. If this is not the case, then I can try to adjust the answer in a follow up.

The answer is accepted.
Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.