### Author Topic: Semester End Challenge 1  (Read 1657 times)

#### Victor Ivrii ##### Semester End Challenge 1
« on: April 04, 2013, 06:05:54 AM »
Draw phase portraits :

\begin{align}
&\left\{\begin{aligned}
&x'=-\sin(y),\\
&y'=   \sin (x);
\end{aligned}\right. \tag{a}\\
&\left\{\begin{aligned}
&x'=-\sin(y)-\alpha \sin (x),\\
&y'=   \sin (x)-\alpha \sin(y);
\end{aligned}\right. \tag{b}
\end{align}
where (a) was part of the Easter challenge and consider (b) for $\alpha=\pm 1$.

Observe the differences between these  phase portraits. Explain them.
« Last Edit: April 04, 2013, 06:10:48 AM by Victor Ivrii »

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Semester End Challenge 1
« Reply #1 on: April 04, 2013, 10:43:33 AM »
Attached are the stream plots of the three systems. One can refer to Easter challenge topic to see how system $(a)$ can be characterized. When we look at system $(b)$ for both cases $\alpha = \pm 1$, we find that it has the same critical points as system $(a)$.

If $\alpha = 1$, then the stable centers in $(a)$ are spiral points in $(b)$. If in $(a)$ the closed trajectories are directed counter-clockwise, then the corresponding spiral points in $(b)$ is also directed counter-clockwise and are stable. This is observed at $(0,0)$. If the closed trajectories in $(a)$ are directed clockwise, then the corresponding spiral points in $(b)$ are also directed clockwise but are unstable. This is observed at $(\pi,\pi)$.

If $\alpha = -1$, then the same analysis applies. However, the stabilities are inverted because we have changed the sign of the coefficient.

In both systems, saddle points remain saddle points, although they are rotated.
« Last Edit: April 04, 2013, 10:52:43 AM by Alexander Jankowski »

#### Victor Ivrii ##### Re: Semester End Challenge 1
« Reply #2 on: April 04, 2013, 10:53:07 AM »