The zero point $(0,0)$ of $x'=x+y,y'=xy-bx-y$ is stable when $b>1$

Asked Jul 02 '16 at 05:37

Active Jul 02 '16 at 06:31

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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

edited Jul 02 '16 at 06:31

asked Jul 02 '16 at 05:37

Liding Yao

2

Did you find the critical points, the Jacobian, and then analyze the Jacobian at each critical point for the parameter b? – Moo Jul 02 '16 at 05:58
@Moo It has zero real part and non-zero imag part. – Liding Yao Jul 02 '16 at 06:32
Related (see @Artem's detailed solution), here the perturbation term is $xy$. – Did Jul 02 '16 at 08:55

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