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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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2 Answers 2

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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
    
@Did: Yes; mathworld.wolfram.com/LyapunovFunction.html –  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
    
@Did: The link, although not theology, already answered that. Just read it. –  JohnD Mar 13 '13 at 2:00
    
(Amazing comment.) I did read the link and sorry but no, it does not answer that. The condition is mentioned but neither explained nor even, as far as I can see, used. –  Did Mar 13 '13 at 6:34

As i am unable to comment;

The intuition comes from physics, we think of a Lyapunov function as an energy function that looks like a potential well with a minimum at the origin.

If we can show this energy function is strictly decreasing with time (on some domain), then intuitively everything ends up at whatever value minimizes the energy.

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.

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