Finding Lyapunov function for a given system of differential equations

Question

I am being introduced to the Lyapunov functions in order to determine the stability of a given system. I know that finding a Lyapunov function is not easy, so I would like to ask for any trick or hint in order to find a Lyapunov function for $$ \left\{\begin{array}{l}x'=-4y+x^2,\\y'=4x+y^2\end{array}\right. $$ at $(0,0)$. I have tried combinations of $x^{2n}$ and $y^{2m} $ and also products of $x$ and $y$ but got nothing clear. Also, I've searched the phase plot for the system and it is clear that $(0,0)$ is a stable point (not asymptotically stable). Thanks in advance.

Answer 1

It is not quite clear why you put a bounty on this question, since @Evgeny answered it in the best possible way. However, if you are looking for a Lyapunov function, here it is (up to an additive constant): $$ L(x,y)=\frac{2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right) \left(\left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \log \left(2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right)\right)+\left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right) \log \left(2\ 2^{2/3} \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right)\right)-3\right)}{9 \left(-\frac{\left(x^2-12 x+8 y+48\right)^3}{2 \left(x^2-4 y\right)^3}+\frac{3 x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right)}-\frac{x (x+4) \left(4 \log \left(x^2-4 x+16\right)+x\right)}{18 \sqrt[3]{2} \sqrt[3]{x^3 (x+4)^3}} $$