# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:06:33 PM

Title: Q7 TUT 0301
Post by: Victor Ivrii on November 30, 2018, 04:06:33 PM
(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

(d)  Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = (1 + x) \sin (y), \\ &\frac{dy}{dt} = 1 - x - \cos (y). \end{aligned}\right.

Bonus: Computer generated picture
Title: Re: Q7 TUT 0301
Post by: Michael Poon on November 30, 2018, 04:33:58 PM
a) To find the critical points, we need to set x' = 0 and y' = 0

In the first equation, that is only satisfied when x = -1 or y = n$\pi$, where n is an integer.

However, when we carry those constraints to the second equation, x = -1 is no longer valid as $\cos(y)$ is bound by -1 and 1.

So the critical points are (0, 2n$\pi$), where n is an integer and (2, n$\pi$), where n is $\pm1, \pm3, \pm5, ...$ or 0

b) To get the corresponding linear system we take the Jacobian matrix and substitute the critical points in:

For (0, 2n$\pi$), where n is an integer:

\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}

For (2, n$\pi$), where n is $\pm1, \pm3, \pm5, ...$:

\begin{pmatrix} 0 & 3\\ -1 & 0 \end{pmatrix}

Title: Re: Q7 TUT 0301
Post by: 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).

For the next critical point(s), the eigenvalues are $\pm \sqrt{3}$, this means the phaseportrait is a saddle.

Phaseportrait coming below:
Title: Re: Q7 TUT 0301
Post by: Michael Poon on November 30, 2018, 04:48:48 PM
Phaseportrait computer generated using Wolfram Alpha:

Includes a centre centred at (0,0) and saddles centred at (2, $\pi$), (2, $-\pi$).

(Click to enlarge!)
Title: Re: Q7 TUT 0301
Post by: Jingze Wang on November 30, 2018, 05:27:47 PM
c) For the first critical point(s), the eigenvalues are $\pm i$. This means the phaseportrait is centre (CW).
Hi Michael, I think you should mention that phase portrait is counterclockwise for the first matrix since -1<0 rather than imaginary eigenvalues
Title: Re: Q7 TUT 0301
Post by: Michael Poon on November 30, 2018, 06:34:02 PM
c) For the first critical point(s), the eigenvalues are $\pm i$. This means the phaseportrait is centre (CW).
Hi Michael, I think you should mention that phase portrait is counterclockwise for the first matrix since -1<0 rather than imaginary eigenvalues

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?
Title: Re: Q7 TUT 0301
Post by: Mengfan Zhu on December 01, 2018, 03:09:51 AM
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 ^_^
Title: Re: Q7 TUT 0301
Post by: Victor Ivrii on December 01, 2018, 04:36:49 AM
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 (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.