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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 $\sin(x_2)$...

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    I think you can prove that the term that involves $\sin(x_2)$ is smaller then the others so that the sum is always negative. notice me if it didn't work. – 2017-01-23
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    How can I do that? Do you have some thoughts? – 2017-01-23
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    Note that $$-2ax_1^3\sin x_2\leqslant2a|x_1|^3$$ and that $$x_1^2+\tfrac13x_1^4-|x_1|^3=(|x_1|-\tfrac12x_1^2)^2+\tfrac1{12}x_1^4$$ hence your RHS is at most $$-2a(|x_1|-\tfrac12x_1^2)^2\leqslant-\tfrac16ax_1^4-\tfrac83bx_2^2$$ which is strictly negative for every $(x_1,x_2)\ne(0,0)$. – 2017-01-24

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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 $$\left<\nabla V(x_1, x_2), (x_1', x_2')\right><0\quad\forall(x_1,x_2)\in B\setminus\{(0,0)\} $$ $$B=\{(x_1,x_2)\in I\times J,\ I=]x_1',x_1''[,\ J=]-\epsilon,+\epsilon[,\ \epsilon>0 \} $$

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    Expanding a square much simplifies and shortens this, see my comment on main. // At the end, note that the quantity is $<0$ on $B\setminus\{(0,0)\}$, not on $B$ (actually it is $<0$ everywhere except at $(0,0)$). – 2017-01-24
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    Thanks for the rectification. – 2017-01-24