$x ^ { \prime } = y + \frac { x } { \sqrt { x ^ { 2 } + y ^ { 2 } } } \left( 16 – (x ^ { 2 } + y ^ { 2 }) \right)$
$y ^ { \prime } = – x + \frac { y } { \sqrt { x ^ { 2 } + y ^ { 2 } } } \left( 16 – (x ^ { 2 } + y ^ { 2 }) \right)$
There are four parts in this question. In the first part, I'm asked to show that there exists a periodic solution to the above system by using the Poincare-Bendixson Theorem. In the second part, I'm supposed to show that there exists a limit cycle of this system. In the third and the fourth part, I'm asked to sketch the phase diagram of this system and show that $(0,0)$ is attractor.
Here is my attempt:
Using polar coordinates, I have converted this system into the following one:
$r ^ { \prime } = 16 – r ^ { 2 }$
$\theta ^ { \prime } = – 1$
Since $r > 0$, the only root to $r'=0$ is $r=4$. $r=4$ gives us a stable limit cycle. And for $r = 0.5$, $r' > 0$, and for $r = 5$, $r' < 0$. I came up with the trapped region, $0.5 < r < 5$. Since there is no fixed point in this region, by the Poincare-Bendixson Theorem, we have $r=$ which is a stable limit cycle.
However, I couldn't understand why we are asked to show that there is a limit cycle in the second part. What I have done in the first part should be sufficient to both first and second parts. Is there something wrong?
Secondly, can you provide me any hint in showing $(0,0)$ is an attractor?
Best Answer
The answer to 1) is just "$r=4$, $θ=c-t$" and the computation that leads to it. All that follows about the trapping region is the answer to 2)
$(0,0)$ is a singular point, the vector field does not exist there. Close to the point it takes nonzero values pointing in all possible directions.