Proof when $b>1$, the zero point $(0,0)$ of ODE $\left\{\begin{align}&x'=x+y\\&y'=xy-bx-y\end{align}\right.$ is stable.

I couldn't find a proper Lyapunov V function.

Let $v=x'$ we has 1st order ODE $v\frac{dv}{dx}=xv-x^2-bx$, but it has no suitable expressions

  • 2
    $\begingroup$ Did you find the critical points, the Jacobian, and then analyze the Jacobian at each critical point for the parameter b? $\endgroup$ – Moo Jul 2 '16 at 5:58
  • $\begingroup$ @Moo It has zero real part and non-zero imag part. $\endgroup$ – yaoliding Jul 2 '16 at 6:32
  • $\begingroup$ Related (see @Artem's detailed solution), here the perturbation term is $xy$. $\endgroup$ – Did Jul 2 '16 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.