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

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.
