Construct a Liapunov function for this system Construct a Liapunov function for the system (Determine the stability of $x \equiv 0$):
I have an example:$$\begin{cases}
 & \mathrm { } \dot{x}= -x^3+xy^2\\ 
 & \mathrm { } \dot{y}= -2x^2y-y^3
\end{cases}  \tag{1}$$
Here's my solution:


*

*Let's try $V(x,y)=ax^2+by^2$.


Then we have: $\dot{V}(x,y)=-2ax^4+2(a-2b)x^2y^2-2by^4$


*

*When $a-2b<0$, for instance $a=b=1$. We have $V(x,y)=x^2+y^2$ such that: 
$$V(0,0)=0,V(x,y)>0, \forall (x,y) \ne (0,0)\ \text{and} \ \dot{V}(x,y)<0$$ 

*Hence, $x=y=0$ is asymptotically stable.
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What about the system:
$$\begin{cases}
 & \mathrm{  } \dot{x}= y-3x^3\\ 
 & \mathrm{  } \dot{y}= -x-7y^3
\end{cases} \tag{2}$$
How can we construct a Liapunov function for this system (Determine the stability of $x \equiv 0$ of the system).
I'm sorry I fixed it!

I doubt that there is a closed-form global Liapunov function.  Note that  besides the centre at $(0,0)$ you have critical points at $(\pm \sqrt{3}/3,\mp 7 \sqrt{3}/9)$, which I think are saddle points.
EDIT: It looks like e.g.
$$ V(x,y) = {x}^{2}+{y}^{2}+ 8.75\,x{y}^{3}+3\,{x}^{2}{y}^{2}+ 5.25\,{x}^{3}y$$
is a Liapunov function for $(x,y)$ near $(0,0)$.
EDIT: How could I find this? By symmetry, I wanted something invariant under
$ x \to -x,\; y \to -y$, so terms of even total degree in $x$,$y$.  So start
with $x^2 + y^2 + \sum_{i=0}^4 a_i x^i y^{4-i}$.  The lowest-order terms of
$\dot{V}$ are of total degree $4$ in $x,y$. I then substituted $x = \cos(s)$, $y = \sin(s)$ into those terms, and chose $a_0, \ldots, a_4$ so that the
result was negative for all $s \in [0,2\pi]$.  The one above is far from the only 
possible choice (and I don't remember exactly why I chose that particular one).
