Find one domain of attraction for this system Assume the system:
 \begin{align}
  \begin{pmatrix} 
           x \\
           y \\
         \end{pmatrix}'   &= \begin{pmatrix} 
           -(1-y)x \\
           -(1-x)y \\
         \end{pmatrix}
  \end{align}
and the Lyapunov function
$$
V(x,y)=x^2+y^2$$
To begin with, the origin is a point of equilibrium and: 
$\bullet\,$$V$ is positive definite since $$V(x,y) >0, \quad \forall x,y \in \mathbb{R^2}\setminus(0,0)$$ and $V(0,0)=0$.
$\bullet\,$
$$
\dot{V}(x,y)=-2\left[x^2-(x^2y+xy^2)+y^2\right]=-2\left[x^2(1-y)+y^2(1-x)\right]$$
For $\dot{V}$ to be negative definite, we must have
$$
1-y>0 \iff y<1 \quad\text{and}\quad 1-x<0 \iff x<1
$$
Moreover, for the circular disks $x^2+y^2=c$ to lie inside the region:
$$
R=\Big\{ (x,y)\in \mathbb{R^2}: x<1,y<1\Big\}$$
we must have $c<1$. Therefore, one of the system's attraction domains is:
$$
S=\Big\{(x,y)\in \mathbb{R^2}: x^2+y^2 < 1\Big\}
$$
Question: It's very likely that this domain of attraction is not the maximum possible. Are there any ways to obtain the maximum one from the inequality $\dot{V}(x,y)<0$? Is it necessary to find the maximum domain or finding a smaller one, which can be derived easily, does the job?
 A: This particular system has closed form solutions.  The integral curves are
$x-\ln(|x|)-y+\ln(|y|) =$constant.  In particular, the separatrices forming the 
boundary of the "maximal" basin of attraction of $(0,0)$  are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - \ln(x) - y + \ln(y) = 0$ that are not on $x=y$.

A: In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction. 
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$\qquad \qquad \qquad$
