What is a suitable Lyapunov function for this system?

Question

I have verified using the eigenvalue method that around $(0,0)$ the system

\begin{align}\dot x&=y - 3x - x^3 \\ \dot y &= 6x - 2y \end{align}

is stable. However, I have been trying to find a suitable Lyapunov function $V$ but from the expressions I have come up with so far, I cannot definitively deduce that the derivative is less than zero. I have tried the classical $V = x^2 + y^2$ and tried changing up the coefficients and exponents so that I can have some cancellations of the odd terms. It has been several hours now and still no luck. Any hints will be much appreciated.

Answer 1

You can try this: $$V(x,y)=\dfrac {y^2}2+ 3 x^2 \ge 0$$ Hence we have: $$V'(x,y)=\left(\dfrac {y^2}2+ 3 x^2\right)'$$ $$V'(x,y)=\left( {y'y}+ 6x'x\right)$$ $$V'(x,y)= {y(6x-2y)}+ 6x(y-3x-x^3)$$ Regroup some termes into a square: $$V'(x,y)=-2(y^2-6xy+9x^2)-6x^4 $$ $$V'(x,y)=-2(y-3x)^2-6x^4 \le 0$$ The derivative is zero only if: $$V'(x,y)=0 \implies (x,y)=(0,0)$$

Answer 2

Let $V(x,y)=6x^2+y^2$. Then \begin{align}\dot{V}&=12x(y - 3x - x^3)+2y(6x-2y)\\&= 12xy-36x^2-12x^4+12xy-4y^2\\&= -4(9x^2-6xy+y^2)-12x^4\\&= -4(3x-y)^2-12x^4\\&< 0 \end{align}

for $(x,y) \ne (0,0)$.