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.

From
$$ \begin{cases}
     x x'=(1-x)x y+x^3\sin{(x)}& \\
     y y'=-(1-x)y x+y^3\sin{(y)}& 
   \end{cases}
$$
after addition
$$
\frac 12(x^2+y^2)' = x^3\sin x + y^3 y
$$
considering now small $x,y$ we have $\sin x\approx x$ and $\sin y\approx y$ hence
$$
\frac 12(x^2+y^2)' = x^4 + y^4
$$
which shows unstability at the origin.
