Toronto Math Forum
MAT2442013S => MAT244 MathLectures => Easter and Semester End Challenge => Topic started by: Victor Ivrii 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.

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 counterclockwise, then the corresponding spiral points in $(b)$ is also directed counterclockwise 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.

Yes, correct. And saddles remain saddles.