The dynamical system in polar form is given by:

$\dot{r} = r(1 - r^2), \quad \dot{\theta} = 1.$

We are tasked with analyzing the stability of $r = 1$ and whether it is an attractor.

Step 1: Analyze the $\dot{r}$ equation

The equation for $\dot{r}$ is: $\dot{r} = r(1 - r^2).$

• At equilibrium points, $\dot{r} = 0$. Solving $r(1 - r^2) = 0$, we get: $r = 0 \quad \text{and} \quad r = \pm 1.$

Step 2: Stability analysis

To determine stability, we evaluate the derivative of $\dot{r}$ with respect to $r$ at the equilibrium points: $\frac{d\dot{r}}{dr} = \frac{d}{dr} \big[r(1 - r^2)\big] = (1 - r^2) - 2r^2 = 1 - 3r^2.$

Case 1: $r = 1$

$\frac{d\dot{r}}{dr} \bigg|_{r=1} = 1 - 3(1)^2 = 1 - 3 = -2.$ Since $\frac{d\dot{r}}{dr} < 0$, $r = 1$ is stable and acts as an attractor.

Case 2: $r = 0$

$\frac{d\dot{r}}{dr} \bigg|_{r=0} = 1 - 3(0)^2 = 1.$ Since $\frac{d\dot{r}}{dr} > 0$, $r = 0$ is unstable.

Case 3: $r = -1$

$\frac{d\dot{r}}{dr} \bigg|_{r=-1} = 1 - 3(-1)^2 = 1 - 3 = -2.$ $r = -1$ is stable, but it is not relevant here as the problem focuses on $r = 1$.

Step 3: Conclusion

From the analysis:

• $r = 1$ is an attractor.
• $r = 0$ is not an attractor.

Thus, the correct option is: $\text{D) } r = 1 \text{ is an attractor.}$

Let me know if you would like a detailed explanation of any specific step!

Here are 5 related questions for further understanding:

1. What is the significance of $\dot{\theta} = 1$ in the system?
2. How would the system behave for initial values $r > 1$?
3. What happens to the system as $r \to 0$?
4. Can the stability of $r = -1$ affect the system's behavior?
5. How do we graphically represent the stability of $r = 1$ in a phase portrait?

Tip: Stability analysis often involves linearization. For a quick test, check the sign of $\frac{d\dot{r}}{dr}$ at equilibrium points!