# Non-existence of closed orbits via construction of Liapunov function

Show that the system

\begin{aligned} x^{'} &= y-x^{3}\\ y^{'} &= -x-y^{3}\end{aligned}

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.

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.

• The condition that $V\gt0$ is not needed, is it?
– Did
Mar 12 '13 at 6:20
• 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.