Poincaré-Bendixson Theorem and Limit Cycle:find the trapping region $\dfrac{dx_1}{dt} = x_2 - x_1^3 + x_1$
$\dfrac{dx_2}{dt} = -x_1 - x_2^3 + x_2$
Hi, I have been given the non-linear system as above and I need to show if there exists limit cycle in the annular region $1 \leq r \leq \sqrt{2}$
I need help in applying the Poincare Bendixsson theorem. 
I have applied the linearisation theorem and established that there is a unstable critical point meaning a limit cycle can exist. 
I know the next step is to use polar-coordinates. What equations do I have to create and what should I obtain?
Many thanks
 A: Dynamical System
Find the periodic solution for the dynamical system:
$$
%
\begin{align}
%
 \dot{x} &= y - x^{3} + x \\
%
 \dot{y} &= -x - y^{3} + y \\
%
\end{align}
\tag{1}
%
$$
Fixed points
The single fixed point is at the origin:
$$
\left[ \begin{array}{c}
 \dot{x} \\ \dot{y}
\end{array} \right]_{(0,0)}
=
\left[ \begin{array}{c}
 0 \\ 0
\end{array} \right]
$$
Compute $\dot{r}$
The polar coordinate transform
$$
%
\begin{align}
%
 x &= r \cos \theta \\
%
 y &= r \sin \theta \\
%
\tag{2}
\end{align}
%
$$
implies
$$
  r^{2} = x^{2} + y^{2}
\tag{3}
$$
Differentiate (3) with respect to time:
$$
2r\dot{r} = 2x \dot{x} + 2y \dot{y}
$$
Therefore
$$
\dot{r} = \frac{x \dot{x} + y \dot{y}} {r}
\tag{4}
$$
Transform $\dot{x}$ and $\dot{y}$ to $r$ and $\theta$ using (2):
$$
%
\begin{align}
%
 \dot{x} 
&= y - x^{3} + x 
= r\left(\sin \theta - r^{2}\cos^{3}\theta + \cos \theta \right) \\
%
 \dot{y} 
&= -x - y^{3} + y
= r\left(-\cos \theta - r^{2}\sin^{3}\theta + \sin \theta \right) \\
%
\end{align}
%
$$
Inserting these identities in $(4)$ produces the final differential equation:
$$
\dot{r} = -\frac{1}{4} r \left(r^{2} \left(\cos(4\theta)+3 \right)- 4 \right)
\tag{5}
$$
Trapping region
Identify regions where $\dot{r}$ is expanding $\left(\dot{r}>0 \right)$ or shrinking $\left(\dot{r}<0 \right)$. Examine the bounding values 
$$-1 \le \cos 4\theta \le 1$$
Case 1: $\cos 4\theta = 1$
Equation (4) becomes
$$
\dot{r} = r \left( 1 -r^{2} \right)
$$
The zones of $\dot{r}$ $\color{blue}{increasing}$ and $\color{red}{decreasing}$ are
$$
\begin{cases}
  \color{blue}{\dot{r} > 0} &  \color{blue}{r < 1} \\ 
  \color{red}{\dot{r} < 0} &  \color{red}{r > 1} \\ 
\end{cases}
$$
Case 2: $\cos 4\theta = -1$
Equation (4) is now
$$
\dot{r} = r \left( 1 - \frac{r^{2}}{2} \right)
$$
The two zones are
$$
\begin{cases}
  \color{blue}{\dot{r} > 0} &  \color{blue}{r < \sqrt{2}} \\ 
  \color{red}{\dot{r} < 0} &  \color{red}{r > \sqrt{2}} \\ 
\end{cases}
$$
The zones are shown in the figure below. Case 1 on the left, case 2 on the right. The third case combines the first two. Red regions are where the flow is inward; blue regions mark outward flow. You can think of the process as adding the first two figures and using the rules red + red = red, blue + blue = blue, and red + blue = gray.

The trapping region is $1 < r <\sqrt{2}$. When $r<1$, $\dot{r}>0$ and $r$ will $\color{blue}{increase}$. When $r > \sqrt{2}$, $\dot{r} < 0$, and $r$ will $\color{red}{decrease}$. But in the trapping region the sign of $\dot{r}$ $\color{gray}{oscillates}$.
Results
The flow field is plotted with the trapping region shown as the shaded annulus and the null clines as red, dashed lines.

