Finding a Lyapunov Function for a system involving a trigonometric function I'm dealing with determining if $(0,0)$ is stable or not for the following system via constructing a Lyapunov function. The system is
$$ \begin{cases}
       x'(t)=(1-x)y+x^2\sin{(x)}& \\
       y'(t)=-(1-x)x+y^2\sin{(y)}& 
   \end{cases}
$$
My initial guess was to choose $V(x,y)=\frac{1}{2}x^2+\frac{1}{2}y^2$, however this leads to $\dot{V}=x^3 \sin{x}+y^3 \sin{y}$, which, for small $x$ and $y$, is positive. However, this does not agree with numerical evidence and also looking at the linearization method, which shows that the origin is indeed a stable node.
Might someone have a suggestion for a Lyapunov function, as well as the domain to choose for it? I suppose I might as well ask if it would be valid to approximate the sine terms by its argument since we would be looking at a small neighborhood around the origin. 
EDIT: Here is a streamplot of the system in a neighborhood of the origin,it appears that the origin is unstable, so I guess I was wrong with my initial assumption. I guess this agrees with my choosing of a Lyapunov function because the function is positive for all x and y(except at the origin), implying instability. 

EDIT2: After thinking for a little bit, the Lyapunov function
 $V(x,y)=\frac{1}{2}x^2+\frac{1}{2}y^2$ work, with the domain
$\Omega = \{ (x,y)\in \mathbb{R}^2 \vert -\pi < x < \pi$ and $-\pi < y <\pi \}$
so that $\dot{V}(x,y)>0$ in $\Omega$. This establishes instability of the origin.

I just want to point out that you can not use Lyapunov's second methods (Lyapunov functions) to show instability. The notion of Lyapunov's second method is not strong enough to do so. You can ONLY SHOW STABILITY. 
In a less ambiguous way: if you can show stability its fine and you are save that the system is stable. If you can not show stability with Lyapunov's second method you can not conclude instability from this. 
