# Show that $L(x,y)=ax^2+by^2$ is a Lyapunov function for the equilibrium at the origin

I am looking at past paper questions and I'm a little stuck on this one.

I have the following system of ODEs:

$\dot{x}=(\epsilon x+2y)(x+1)$

$\dot{y}=(-x+\epsilon y)(x+1)$

where $\epsilon$ is a parameter.

a) Show that

$L(x,y)=ax^2+by^2$

is a Lyapunov function for the equilibrium at the origin of the system of the ODEs above if $\epsilon \leq 0$ and for suitable $a,b >0$. Give an example of suitable $a,b$.

b) What does the Lyapunov function tell us about the stability of the origin for $\epsilon <0$ and for $\epsilon =0$?

Okay, so my attempt:

$L(0,0)=a(0)^2+b(0)^2=0$

$L(x,y)>0$ when $a,b>0$

$\frac{dL}{dt}=\frac{dL}{dx} \frac{dx}{dt} + \frac{dL}{dy} \frac{dy}{dt}$

$=2xa(\epsilon x^2 +2yx +\epsilon x +2y) +2by(-x^2 + \epsilon yx -x + \epsilon y )$

But I'm not sure where to go from there, I feel like I'm overthinking it.

If anyone could help to find a solution to study I'd be really appreciative!

$$\frac{dL}{dt}= 2ax(\epsilon x+2y)(x+1)+2by(-x+\epsilon y)(x+1)$$ $$=2(x+1)(a\epsilon x^2+2axy-bxy+b\epsilon y^2).$$ If we take $a=1$, $b=2$, $\epsilon<0$ then $$\frac{dL}{dt}= 2(x+1)(\epsilon x^2+2\epsilon y^2)$$ is negative in the set $\{(x,y):\, x>-1\}\setminus (0,0)$, thus, the origin is asymptotically stable.
In the case when $\epsilon=0$ this Lyapunov function can tell us only that the origin is stable because in this case $\frac{dL}{dt}=0$ for all $x$ ,$y$.