Liapunov's Function and kind of Stability Problem :

Using the right Liapunov's Function V find the kind of stability of the non-linear system:
  $$ x'=\sin(x+y) $$
$$y'=-\sin(x-y)$$
given the Liapunov's function: $$V=x\cdot y $$ 

So, I don't have any experience on this type of exercices and I don't know even how to start the exercice.I know that i have to differentiate  the liapunov's function $$V=x\cdot y$$ so i take :$$V'=Vx\cdot x'+Vy\cdot y'$$ but i don't find the reason and i don't know if it's right too.
I would really appreciate a thorough solution and explanation, since I've just started working on Liapunov's functions and Liapunov's stability and I have to clear my mind on them.I have to say that my dynamical's system book doesn't have problems of these types so to proceed.Thanks in advance!
 A: First of all the $V$ you mentioned is not a Lyapunov function. Define the Lyapunov function $V(x,y)=x^2+y^2$. It is an exercise for you to verify that this is indeed a Lyapunov function for determining the stability of the point $(0,0)$. 
If we consider $V(x(t),y(t))$ then this function is a function of $t$. Moreover the derivative w.r.t. $t$ is $\dot V(x(t),y(t))= \partial _x V(x(t),y(t))\cdot x'(t)+\partial_y V(x(t),y(t)) \cdot y'(t)$. Lyapunov's second theorem about stability is the following.

Theorem (Lyapunov's second method). Let $U\subset \mathbb{R}^n$ be a neighbourhood of the origin. Let $V:U\to \mathbb{R}$ be a Lyapunov function. Then the following is true: 
  
  
*
  
*If $\dot V \leq 0$ on $U\setminus \textbf{0}$, then $\textbf{0}$ is stable.
  
*If $\dot V < 0$ on $U\setminus \textbf{0}$, then $\textbf{0}$ is asymptotic stable.
  
*If $\dot V > 0$ on $U\setminus \textbf{0}$, then $\textbf{0}$ is unstable.
  

Let's go back to our $V$ that we have defined there. We calculate the derivative w.r.t. $t$ but do not use $t$ in our notation because that makes it a little bit messy. We just write $x$ instead of $x(t)$. 
\begin{align}
\dot V(x,y) &= 2x \cdot x'+2y \cdot y'\\
&= 2x \sin(x+y)  - 2y\sin(x-y)
\end{align}
Now we must find out what the sign is of this expression in a neighbourhood of $(0,0)$. A natural thing to do is to take the Taylorpolynomial of the sine function around zero, so:
\begin{align}
\dot V(x,y) = 2x(x+y) -2y(x-y) + O(||(x,y)||^3) = 2x^2+2y^2 + O(||(x,y)||^3)
\end{align}
Since $O(||(x,y)||^3)$ has a much smaller magnitude than $2x^2+2y^2$ around $(x,y)=(0,0)$ we can actually find a neighbourhood of $(0,0)$ where $\dot V$ is positive in that neighbourhood excluding $(0,0)$. So $(0,0)$ is an unstable stationary point of the given differential equation.
A: You should take the gradient of the Liapunov function
$$\nabla V = \left(\frac{\partial}{\partial x} V, \frac{\partial}{\partial y} V\right)$$
and consider what angle this makes with the vector field of the given differential equation.
Can you show the cosine of this angle is always positive or always negative, in certain regions of the plane?
This will tell you something about how the trajectory specified by the differential equation will be constrained by the level sets of the Liapunov function $V(x,y) = c$.
A: Set $V(x,y)=\sin^2(\frac{x+y}2)+\sin^2(\frac{x-y}2)$ then
\begin{align}
(\dot V f)(x,y)&=(\sin(x+y)+\sin(x-y))\sin(x+y)-(\sin(x+y)-\sin(x-y))\sin(x-y)
\\
&=\sin^2(x+y)+\sin^2(x-y)
\end{align}
which shows that the point $(x,y)=(0,0)$ is unstable. 

Using the half-angle formulas one can further transform
\begin{align}
(\dot V f)(x,y)&=4V(x,y)-4\sin^4(\frac{x+y}2)-4\sin^4(\frac{x-y}2)
\\
&=4V(x,y)-2V(x,y)^2-2\left(\sin^2(\frac{x+y}2)-\sin^2(\frac{x-y}2)\right)^2
\end{align}
