Check Equilibrium Point for Stability

Check Equilibrium Point for Stability
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$\dot x = -y^2$
$\dot y = x^3y$

The task is to draw global phase portrait (I guess I understand how to manage this using isoclines) and to check equilibrium point for stability. I found out that equilibrium points are (x; 0) and using Jacobian matrix found out, that if x > 0 one of eigenvalues is always positive so points (x, 0) where x > 0 are not stable at all. (Correct me, if I am wrong, please). And I don't know how to check the solutions where x <= 0.

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1 Answer

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Note that $x^4 + 2 y^2$ is constant on trajectories of this system. The curves $x^4 + 2 y^2 = c$ are shaped like distorted ellipses. Any solution in the left half plane will approach the negative $x$ axis along one of these curves. In particular, the equilibrium points $(x,0)$ for $x \le 0$ are stable (but not asymptotically stable).

EDIT:

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David Miller

David Miller brings 15 years of experience in global economics, personal finance strategy, and market dynamics. He specializes in turning complex economic trends into actionable insights for everyday readers.