1
$\begingroup$

Consider the differential system $$ x'=x+y $$ $$y'=x-y+xy$$

What would be a Lyapunov function for this system at $(0,0)$?

I have considered functions $V(x,y)=ax^{2n}+by^{2m}$ but none of them seems to be Lyapunov.

$\endgroup$
  • $\begingroup$ Yes $(0,0)$ is unstable solution but I want to investigate it by LF . $\endgroup$ – Finish Jan 13 '15 at 14:17
  • 1
    $\begingroup$ As far as I'm aware, you construct a Lyapunov function to show that a critical point is stable when linearising the system doesn't work. If a critical point is unstable, then you won't ever be able to find a Lyapunov function, so you can't investigate the critical point in the manner you want to. You can use the second derivative test (Hessian) to determine whether $\frac{d}{dt}L \le 0$ though. $\endgroup$ – mattos Jan 13 '15 at 14:36
  • $\begingroup$ @Mattos The critical point $(0,0)$ is unstable for the system $x'=-(x^2+y^2)x$, $y'=(x^2+y^2)y$, which admits the Lyapunov function $L:(x,y)\mapsto x^2-y^2$. $\endgroup$ – Did Jan 13 '15 at 17:29
  • 1
    $\begingroup$ @Did The function $L(x,y) = x^{2} - y^{2}$ doesn't satisfy the condition that $L(x,y) > 0, \ \forall (x,y) \ne (0,0)$. For example, choose $\lvert x \rvert < \lvert y \rvert \implies x^{2} - y^{2} < 0$. Or am I forgetting something? $\endgroup$ – mattos Jan 13 '15 at 17:37
  • 2
    $\begingroup$ @Mattos No, no, you are right, I guess I am talking too much to people using a relaxed definition of Lyapunov function. Thanks for putting me back in the right path. :-) $\endgroup$ – Did Jan 13 '15 at 18:16
2
$\begingroup$

As it is indicated in comments Lyapunov function is usually used to infer the stability of the equilibrium. There is a useful theorem by Chetaev, which allows one to build a function (often called Chetaev's function in Russian-language literature) to prove instability. In your case you can take $$ V(x,y)=xy, $$ which is positive in the first quadrant, zero on its boundary, and the origin also belong to the boundary.

I have $$ \dot V=y^2+x^2+x^2y>0, $$ for $x,y>0$. Hence, by the aforementioned theorem, the origin is unstable.

$\endgroup$
  • $\begingroup$ Nice theorem! I'll have to remember that one. $\endgroup$ – hasnohat Jan 15 '15 at 7:17
1
$\begingroup$

We refer to Liapunov's technique to determine the stability only when the linearized system is not hyperbolic at the equilibrium point.

Now ,bearing this in mind, let's just check the linearized system, which can be derived by simply dropping the non-linear terms : $\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array} \right)$

This system has a positive eigenvalue ($\sqrt{2}$) and a negative one ($-\sqrt{2}$), so it's hyperbolic, and from the linearization we can state that the origin is a saddle (hence not stable).

Bottom line, search for a Liapunov function only when the equilibrium is not hyperbolic. If it is hyperbolic, such as your case, then you're a lucky boy! (or a lucky girl if you're a girl)

;)

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.