$\dfrac{dx_1}{dt} = x_2 - x_1^3 + x_1$
$\dfrac{dx_2}{dt} = -x_1 - x_2^3 + x_2$
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
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.