### Author Topic: Semester End Challenge 3  (Read 2274 times)

#### Victor Ivrii ##### Semester End Challenge 3
« on: April 04, 2013, 06:46:45 AM »
Draw phase portraits and explain them (using polar coordinates)
\begin{equation}
\left\{\begin{aligned}
&\frac{dx}{dt}=-y + x f(\sqrt{x^2+y^2}),\\
&\frac{dx}{dt}=  x + y f(\sqrt{x^2+y^2})
\end{aligned}\right.
\end{equation}
with
(a) $f(r) = \sin (\pi r)$,

(b) $f(r) = -\sin (\pi r)$,

(c) $f(r) = \sin^2 (\pi r)$,

(d) $f(r) = -\sin ^2 (\pi r)$,

(e) $f(r)=1/r$.
« Last Edit: April 04, 2013, 07:42:06 AM by Victor Ivrii »

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Semester End Challenge 3
« Reply #1 on: April 04, 2013, 09:10:59 AM »
In each case $(a)$ to $(e)$, there is one critical point $(0,0)$ that is a spiral point. Let us first make some generalizations. In cases $(a)$ to $(d)$, there is a set of limit cycles. It would seem that when $f(r) = 0$, a limit cycle occurs. Thus, the limit cycles are given by the equation $r = \ell$, where $\ell = 1,2,...,n$. In case $(e)$, there are no limit cycles.

The stream plot for $(a)$ tells us that the spiral point is unstable. We also see that the limit cycles with $r = 2k+1$, where $k = 0,1,...,n$, are asymptotically stable both internally and externally. Limit cycles with $r = 2k$, where $k = 1,2,...,n$, are unstable both internally and externally. In case $(b)$, the spiral point is stable, and the limit cycles behave in an opposite manner. This is because of the inclusion of a minus sign, which changes the direction of the trajectories.

In case $(c)$, the spiral point is unstable. In fact, all limit cycles are internally stable and externally unstable. The trajectories are only able to spiral outward from $(0,0)$--this corresponds to the fact that $\sin^2{x}$ is always positive. In case $(d)$, the spiral point is stable and the limit cycles behave in an opposite manner. This is because of presence of the minus sign.

Case $(e)$ is an outlier. No sinusoidal function appears, so no periodic behaviour is observed. As an aside, there is no trajectory that goes through the critical point, since $f(r)$ is discontinuous there.

The sine function has the following consequences in this situation:

• it generates a periodic limit cycle whenever $r = \ell$,
• the sign of its output determines the orbital stability of the limit cycle.

I attached stream plots and corresponding contour maps of $f(r)$. Light areas are positive and dark areas are negative. What I would like to continue working on is how the output of the sine function relates to the direction of the trajectories in the regions that are bounded by the limit cycles.
« Last Edit: April 04, 2013, 09:39:00 AM by Alexander Jankowski »

#### Victor Ivrii ##### Re: Semester End Challenge 3
« Reply #2 on: April 04, 2013, 09:54:50 AM »
Perfect. Almost: But (e) needs is different not only from (a)--(d), but from classical spiral with $f(r)=\pm \alpha1$. How? You can observe on the picture. Explanation?

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Semester End Challenge 3
« Reply #3 on: April 04, 2013, 11:01:31 AM »
I have plotted both spirals. It is now easier to see that in the case of $(e)$, the system has more circular character. The trajectories must conform to a missing spiral, which I think is the spiral for which $f(r)$ is discontinuous.

#### Victor Ivrii ##### Re: Semester End Challenge 3
« Reply #4 on: April 04, 2013, 11:13:48 AM »
Actually for classical spiral system is $\frac{d\theta}dt}=1$, $\frac{dr}{dt}=r$ and the trajectory makes an infinite number of rotations as $t\to -\infty$ and the curve "enters" an origin.

For (e) system is $\frac{d\theta}dt}=1$, $\frac{dr}{dt}=1$ and the trajectory makes a finite number of rotations as $t\to t_0+0$ and the curve "enters" an origin.