# Liapunov Stable question

Show that the system $x^{'}=y-x^{3}$ and $y^{'}=-x-y^{3}$ has no closed orbits by constructing a Liapunov Function $V= ax^{2}+by^{2}$ with suitable a and b.

all i really know is that this is an offshoot on the idea of a gradient field and that i want to show that one side = 0 and the other side clearly isn't 0 to derive a contradiction but i am defiantly not well verse'd in the process.

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For $(x,y)\not=0$, $V>0$ follows from the stated form of $V$, provided $a,b>0$. But you also need $\dot{V}<0$.

To this end, compute \begin{align} \dot{V}&={\partial V\over \partial x}\cdot {dx\over dt}+{\partial V\over \partial y}\cdot{dy\over dt}\\ &=2ax(y-x^3)+2by(-x-y^3)\\ &=2(a-b)xy-2ax^4-2by^4. \end{align} Since you only need to find some Lyapunov function, make life easy for yourself and choose $a=b$ with $a>0$. Then, for $(x,y)\not=0$, $$\dot{V}=-2a(x^4+y^4)<0,$$ as desired.

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The condition that $V\gt0$ is not needed, is it? –  Did Mar 12 '13 at 6:20
–  JohnD Mar 12 '13 at 15:21
Sorry but maths is not theology, yet. If the condition $V\gt0$ is needed, where and why? –  Did Mar 12 '13 at 17:40
Requiring $V(0)=0$ and $V(x)>0$ $\forall x\in D/0$ simply guarantees there is only one minimum 'energy', and that the minimum is at the origin.