Finding a Lyapunov function for the differential system $x_1'=-8x_1^3-x_2$, $x_2'=-4x_2-4x_1^3$

Question

I've got the following system of equations: $$ x_1'=-8x_1^3-x_2 \qquad x_2'=-4x_2-4x_1^3 $$ I'm trying to check, if the equilibrium point in $(0,0)$ is stable or not. I am supposed to find so called Lyapunov function $L$, i.e. function which satisfies three following conditions:

1) $L(x_1,x_2)=0$ iff $(x_1,x_2)=(0,0)$

2) $\forall_{(x_1,x_2)\neq(0,0)}L(x_1,x_2)>0$

3) $\forall_{(x_1,x_2)}:\frac{dL}{dt}(x_1,x_2)\leqslant 0$ or $\frac{dL}{dt}(x_1,x_2)\geqslant 0$

The sign in 3) gives us the information about stability in the equilibrium point.

I was trying to find a polynomial in $\mathbb{R}[x,y]$ with only even exponents, but I was unable to find such function. Is there any way to do it without just "guessing"?

What is the Lyapunov function for this system of ODEs?

Answer 1

Hint: If $L(x_1,x_2)$ has a part $x_1^{2n}$, the stability condition involves some multiples of $x_1^{2n+2}$ and $x_1^{2n-1}x_2$. If $L(x_1,x_2)$ has a part $x_2^{2m}$, the stability condition involves some multiples of $x_2^{2m}$ and $x_2^{2m-1}x_1^3$. The mixed terms $x_1^{2n-1}x_2$ and $x_2^{2m-1}x_1^3$ are the same if $(n,m)=(2,1)$.

This suggest to check whether $L(x_1,x_2)=x_1^4+x_2^2$ is a Lyapunov function. If it was not (it is), one could have tried $L(x_1,x_2)=ax_1^4+x_2^2$ for some positive $a$, or even $L(x_1,x_2)=ax_1^4+bx_1^2x_2+x_2^2$ for some $(a,b)$ such that $b^2\lt4a$.

Answer 2

But differentiating your $L$, I get: $$\frac{dL}{dt}(x_1,x_2)=(x_1^4+x_2^2)'=4x_1^3 \cdot x_1'+2x_2\cdot x_2'=4x_1^3\cdot(-8x_1^3-x_2)+2x_2\cdot(-4x_2-4x_1^3)=-32x_1^6-4x_1^3x_2-8x_2^2-8x_1^3x_2=-4\cdot(8x_1^6+3x_1^3x_2+2x_2^2)$$ which doesn't have a constant sign. Differentiating $L=ax_1^4+x_2^2$ works only for $a=-2$, but it is not positive. I am going to try the remaining option.