I've tried to solve a little problem that goes as follows:
Consider a system of ODEs: $$x'=y-x^3\text{ and } y'=-x^3-y^3$$ And the function $$L(x,y)=\frac{1}{2}y^2+\frac{1}{4}x^4.$$
Now I shall show that $(0,0)$ is not linearly stable, and that $L$ is a Lyapunov function.
I tried to investigate $\frac{dL}{dy}$ and $\frac{dL}{dx}$ and check if this is smaller or greater than $0$ at (0,0). However, these two terms are just $\frac{dL}{dy}=y,\frac{dL}{dx}=x^3$, so it seems we have to solve for $x,y$ explicitly? I'm confused because this looks like one of those exercise where brute-force can be avoided (And we did not deal with non-linear ODEs yet)?
yours, Marie
For $L(x,y)$ to be a Lyapunov function, what you want to do is consider $$\dfrac{dL}{dt} = y \dfrac{dy}{dt} + x^3 \dfrac{dx}{dt} = y (-x^3 - y^3) + x^3 (y - x^3)$$ Do you see why $\dfrac{dL}{dt} \le 0$, with equality only at $(0,0)$? And why $L(x,y) \ge 0$, with equality only at $(0,0)$?