 Author Topic: FE-5  (Read 4763 times)

Victor Ivrii FE-5
« on: April 17, 2013, 03:05:12 PM »
For the system of differential equations
\begin{equation*}
\left\{\begin{aligned}
&x' =\tan (y) - \frac{1}{2}\tan (x)  \,,\\
&y' = \tan (x) - \frac{1}{2}\tan (y) \,.
\end{aligned}\right.
\end{equation*}

(a) Linearize the system at a critical point $(x_0 ,y_0)$ of your choice;

(b) Describe the type of the critical point $(x_0,y_0)$ of the linearized and of the original system;

(c) Sketch the phase portraits of the linearized and of the  original system near this critical point $(x_0,y_0)$.

Michal Staszewski

• Newbie
• • Posts: 3
• Karma: 1 Re: FE-5
« Reply #1 on: April 17, 2013, 08:22:58 PM »
Solution attached.

Victor Ivrii Re: FE-5
« Reply #2 on: April 17, 2013, 11:40:29 PM »
Solution attached.

Few remarks.
1. Field is obviously $\pi$-periodic with respect to both $x$ and $y$ and it is singular as $x=(m+\frac{1}{2})\pi$  or $y=(n+\frac{1}{2})\pi$ with $m,n\in \mathbb{Z}$ so one needs to consider only square $\{ -\frac{\pi}{2}<x < \frac{\pi}{2}, -\frac{\pi}{2}<y < \frac{\pi}{2}\}$ where $(0,0)$ is an only equilibrium point.

2. Missing: eigenvectors (so directions of separatrices have not been found.

3. Global phase portrait would be appreciated.

Michal Staszewski

• Newbie
• • Posts: 3
• Karma: 1 Re: FE-5
« Reply #3 on: April 18, 2013, 12:30:58 AM »
A follow-up.

Michal Staszewski

• Newbie
• • Posts: 3
• Karma: 1 Re: FE-5
« Reply #4 on: April 18, 2013, 12:32:24 AM »
And the computer-generated phase portrait.

Victor Ivrii Re: FE-5
« Reply #5 on: April 18, 2013, 01:08:39 AM »
Now it is OK. Note that vector field breaks (on the picture you see the change of direction)  at lines I mentioned