I have the system $x'=-x^3+2y^3$ and $y'=-2xy^2$. I need to prove that the point $(0,0)$ is asymptotically globally stable. Here's what I did: if we have a Lyapunov function $v(x,y)=ax^2+bxy+cy^2$, then
\begin{align}
\dot{v}(x,y)&=(2ax+by)(-x^3+2y^3)+(2cy+bx)(-2xy^2)\\
&= 2a(-x^4+2xy^3)+b(-yx^3+2y^4-2x^2y^2)+2c(-2xy^3)
\end{align}
Now, I can see that if we let $a=c=1$ and $b=0$, then $v(x,y)=x^2+y^2$ is positive definite and $\dot{v}(x,y)=-2x^4$ is negative semidefinite. However, this shows $(0,0)$ is stable and not asymptotically stable. How do I prove this stronger result? I know it's true because my textbook says the origin is "at least" stable.
The only idea I had was to write $\dot{v}(x,y)=-(x+y)^4+…$ but this hasn't gotten me very far.
Best Answer
With your current candidate Lyapunov function $v(x,y)=x^2+y^2$ you are already almost there in proving asymptotic stability. Namely, one can prove this with the aid of LaSalle's invariance principle.
In short LaSalle's invariance principle can show asymptotic stability for positive definite $v(x,y)$ and negative semidefinite $\dot{v}(x,y)$ by looking at trajectories where $\dot{v}(x,y)=0$. In your case $\dot{v}(x,y)=-2x^4=0$ implies $x=0$ and thus also $\dot{x}=0$. Substituting this into the system dynamics yields
\begin{align} \dot{x} &= -0^3 + 2 y^3 = 0, \\ \dot{y} &= -2\,0\,y^3. \end{align}
The only trajectory satisfying this is $(x,y)=(0,0)$ and from this one can conclude that the origin is asymptotically stable.