Given $dx/dt=\sin(y)+0.1\sin(x),dy/dt=-\sin(x)+0.2\sin(y)$ and $F(x,y)=\cos(x)+\cos(y)$ Determine if $F$ is a lyapunov function near $(0,0)$
I computed the jacobian matrix at $(0,0)$ as $\begin{pmatrix} 0.1&1\\-1&0.2\\\end{pmatrix}$
Which has eigenvalues $3/20 \pm i\sqrt{399}$
So both eigenvalues have positive real part, so this system should be asymptotically stable? Which I believe means it should also be lyapunov stable.
I have this lyapunov stability theorem,
Assume $U$ is an open neighbourhood of $x^*, F:U\to \mathbb{R}^n$ and $F(x^*)=0$. Further assume that there is a $C^1$-function $V:U\to\mathbb{R}$ satisfying $V(x^*)=0$, $V(x)>0$ for $x\neq x^*$. Then, V is called a lyapunov function for $x'=F(x)$ and:
a. If $V'(x)\leq 0$ for all $x\in U$, then $x^*$ is L-stable b. If $V'(x)<0$ for all $x\in U\{x^*\}$, then $x^*$ is asymptotically stable. c. If $V'(x)>0$ for all $x\in U\{x^*\}$ then $x^*$ is unstable.
I can show that $F'=-\sin(x)x'-\sin(y)y'=-\sin(x)\sin(y)-0.1\sin^2(x)+\sin(x)\sin(y)-0.2\sin^2(y)=0.1\sin^2(x)-0.2\sin^2(y)$
But I'm not sure how to relate that to the above theorem to say whether or not this is a lyapunov function.
You missed to transfer a minus sign, it should be $$ F'(x,y)=-0.1\sin^2x-0.2\sin^2y. $$ As $F$ has a local maximum at $(0,0)$, it would be more in tune with the traditional definition to call $-F$ a Lyapunov function.
Per the sign of the derivative, the solutions close to $(0,0)$ move downwards in the levels of $F$, which means away from the origin.
That the eigenvalues have a positive real part means that the origin is unstable, you need negative real parts for a stable fixed point.