Finding a strict Liapunov function I need to show that the equilibrium point $(0,0)$ is asymptotically stable using Liapunov function. That means I shall find some strict Liapunov function. 
Its given me the non-linear system:
\begin{cases}
x_1' = -x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2) \\ 
x_2' = -x_2 -\frac{1}{3}x_2^3
\end{cases}
My attempt: $V(x_1, x_2) = ax_1^2 + bx_2^2$,  $a, b > 0$. I would like to determine $a$ and $b$.
Then, $V(0,0) = 0$ and $V(x_1, x_2) > 0$, for all $(x_1, x_2) \neq (0,0)$. 
In order to decide if $V(x_1, x_2)$ is a strict Liapunov function, I wish that $\left<\nabla V(x_1, x_2), (x_1', x_2')\right> < 0$.
\begin{align}
\left<\nabla V(x_1, x_2), (x_1', x_2')\right> &= \left<(2ax_1, 2bx_2) ( -x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2),  -x_2 -\frac{1}{3}x_2^3)\right>\\
&= 2ax_1(-x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2)) + 2bx_2(-x_2 -\frac{1}{3}x_2^3) \\
&= -2a(x_1^2 + \frac{1}{3}x_1^4) -2b(x_2^2 + \frac{1}{3}x_2^2) -2ax_1^3\sin(x_2)
\end{align}
I do not know what to do with the term that involves $\sin(x_2)$...

As i can remember, we have to prove that $\left<\nabla V(x_1, x_2), (x_1', x_2')\right> < 0$ over some neighborhood $B$ of $(0,0)$.
We call $sg(x)$ the sign function defined by:$$sg(x)=1\quad if\ x>0$$
$$sg(x)=-1\quad if\ x<0$$
First, we have \begin{align*}
-1\leq\sin(x_2)\leq 1\\
-2a\leq-2a\sin(x_2)\leq 2a\\
-2a|x_1^3|\leq-2ax_1^3\sin(x_2)\leq2a|x_1^3|\\
-2a(x_1^2 + \frac{1}{3}x_1^4)-2ax_1^3\sin(x_2)\leq2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)
\end{align*}
Now let's study the sign of the expression $(E)= 2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)$
\begin{align*}
 2a|x_1^3|-2a(x_1^2 + \frac{1}{3}x_1^4)&=2ax_1^2(sg(x_1)x_1-1+\frac{1}{3}x_1^2)\\
&=\frac{2}{3}2ax_1^3(x_1^2+3sg(x_1)x_1-3)
\end{align*}
$\Delta=((3sg(x_1))^2-4(-3))=9+12=21 \implies \sqrt{\Delta}\simeq4.58$ 
Then, $x_1'=\frac{-3sg(x_1^3)-\sqrt{\Delta}}{2}$ and $x_1''=\frac{-3sg(x_1^3)+\sqrt{\Delta}}{2}$
When $x_1\in I=]x_1',x_1''[$, $(E)<0$
 Notice that whether the sign of $x_1$ is $(+)$ or $(-)$,the interval I is always a neighborhood of $(0,0)$.
Now that we have $(E)<0$ we add to it the last term of our first equation $-2b(x_2^2+\frac{1}{3}x_2^2)$ which is obviously negative.
Thus we conclude that $$\left<\nabla V(x_1, x_2), (x_1', x_2')\right><0\quad\forall(x_1,x_2)\in B\setminus\{(0,0)\} $$
$$B=\{(x_1,x_2)\in I\times J,\ I=]x_1',x_1''[,\ J=]-\epsilon,+\epsilon[,\ \epsilon>0 \} $$  
