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.
On
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.
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.