MAT244--2018F > Quiz-7

Q7 TUT 0301

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Michael Poon:

--- Quote from: Jingze Wang on November 30, 2018, 05:27:47 PM ---
--- Quote from: Michael Poon on November 30, 2018, 04:37:30 PM ---c) For the first critical point(s), the eigenvalues are $\pm i$. This means the phaseportrait is centre (CW).

--- End quote ---
Hi Michael, I think you should mention that phase portrait is counterclockwise for the first matrix since -1<0 rather than imaginary eigenvalues

--- End quote ---

From Prof. Ivrii's post: http://forum.math.toronto.edu/index.php?topic=1525.0

Given the matrix of the first critical point, b = 1, c = -1, and it seems the link above says, b >0, c<0 => clockwise?

Mengfan Zhu:
Hello, here is my solution.
There are two critical points:
(0, 2nπ) n=0,1,2,3... and (2, nπ) n=1,3,5...
By the way, I think there is no “±” for n here.
At (0, 2nπ), it's indeterminate center or spiral point.
At (2, nπ), it's an unstable saddle point.
The graph drawn by hand is also below.
Thank you ^_^

Victor Ivrii:
For general non-linear systems purely imaginary e.v. (non 0) coold mean both centers and "funny spiral points"
http://forum.math.toronto.edu/index.php?topic=1602.0
In this case those are centers because system is integrable (with integrable factor):
$$
dt =\frac{dx}{(1+x)\sin(y)}=\frac{dy}{1-x-cos(y)}\implies\\
[1-x-\cos(y)]dx -(1+x)\sin(y)dy=0\implies\\
\frac{1-x}{(1+x)^2}- \frac{\cos(y)}{(1+x)^2}dx-\frac{\sin(y)}{1+x}dy=0\implies\\
-\frac{2}{1+x}-\ln (1+x) + \frac{\cos(y)}{1+x}=C
$$
and since $1+x \ne 0$ for $x>0$ everything is fine.

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