$ODEs - Stability$

Question

$ODEs - Stability$

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

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Define
$$G(y_1, y_2) = (y_2 - y_1 f(y_1, y_2), -y_1 - y_2f(y_1, y_2)).$$
Let us compute the Jacobian of $G$ at $(0, 0)$. It is
$$\mathbf{J}_G = \begin{bmatrix}-f(0,0) & 1\\ -1 & - f(0,0)\end{bmatrix}.$$
Then $\det \mathbf{J}_G = f(0,0)^2 + 1 > 0$. Thus the system is stable if $\operatorname{tr} \mathbf{J}_G < 0$ and unstable if $\operatorname{tr} \mathbf{J}_G > 0$. We easily compute that $\operatorname{tr} \mathbf{J}_G = -2f(0,0)$.

Then if $f$ is positive in a neighborhood of $(0,0)$, one must have $\operatorname{tr} \mathbf{J}_G < 0$ so the system is stable. If $f$ is negative in a neighborhood of $(0,0)$, then $\operatorname{tr} \mathbf{J}_G > 0$ so the system is unstable.

Matchmaticians Alumnus

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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

0

Nevermind, stupid mistake on my part
Matchmaticians Alumnus

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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.

$ODEs - Stability$

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