MHB Can you determine the stability of a fixed point in a system of ODEs?

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Here is this week's POTW:

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Consider the following system of ODE on $\Bbb R^2$.

\begin{align}
\dot{x} &= y\\
\dot{y} &= \lambda y(1 - x^2) - x
\end{align}

Determine a condition(s) on $\lambda$ such that the fixed point $(0,0)$ is asymptotically stable.

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No one answered this week's problem correctly. You can read my solution below.

Spoiler

The associated Jacobian of the zero solution of the system is

$$J(\lambda) = \begin{pmatrix}0 & 1\\-1 & \lambda\end{pmatrix}$$

The eigenvalues of $J(\lambda)$ are $t_1 = \frac{\lambda + \sqrt{\lambda^2 - 4}}{2}$ and $t_2 = \frac{\lambda - \sqrt{\lambda^2 - 4}}{2}$. If $\lambda \ge 0$ and $t_1$ is real, then $t_1 \ge \frac{\lambda}{2} \ge 0$, which implies $(0,0)$ is not asymptotically stable; $\lambda \ge 0$ and $t_1$ is non-real, then $\operatorname{Re}(t_1) = \frac{\lambda}{2} \ge 0$ and the conclusion the same. So assume $\lambda < 0$. If $t_1$ is real, then $\lambda = -|\lambda| < -\sqrt{\lambda^2 - 4}$ $\implies$ $\lambda + \sqrt{\lambda^2 - 4} < 0$ $\implies$ $t_1 < 0$. Hence also $t_2 < 0$ and $(0,0)$ is asymptotically stable. If $t_1$ is non-real, then since $\operatorname{Re}(t_1) = \operatorname{Re}(t_2) = \frac{\lambda}{2} < 0$, we deduce again that $(0,0)$ is asymptotically stable. Therefore, the origin is asymptotically stable if and only if $\lambda < 0$.