# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:54:02 PM

Title: Q6 TUT 0401
Post by: Victor Ivrii on November 17, 2018, 03:54:02 PM
The coefficient matrix contains a parameter $\alpha$.

(a) Determine the eigenvalues in terms of $\alpha$.
(b) Find the critical value or values of  $\alpha$  where the qualitative nature of the phase portrait for
the system changes.
(c) Draw a phase portrait for a value of  $\alpha$ slightly below, and for another value slightly above,
each critical value.

$$\mathbf{x}' =\begin{pmatrix} \alpha &1\\ -1 &\alpha \end{pmatrix}\mathbf{x}.$$
Title: Re: Q6 TUT 0401
Post by: Guanyao Liang on November 17, 2018, 03:54:53 PM
This is my answer.
Title: Re: Q6 TUT 0401
Post by: Jingze Wang on November 17, 2018, 03:56:11 PM
First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} \alpha&1\\ -1&\alpha\\ \end{bmatrix}$

$det(A-rI)=(\alpha-r)(\alpha-r)+1=0$

$r^2-2{\alpha}r+\alpha^2+1=0$

$r=\frac{2\alpha\pm\sqrt{-4}}{2}$

$r=\alpha\pm2i$       $\color{red}{r_\pm =\alpha \pm i\; V.I.}$

Notice there are always complex eigenvalues, and $\alpha=0$ is critical value since $\alpha=0, \alpha>0, \alpha<0$ have different phase portraits

When $\alpha=0$ , real parts of eigenvalues are 0

When value of $\alpha$ is slightly below 0
Then $\alpha<0$ , real parts of eigenvalues are negative

When value of $\alpha$ is slightly above 0
Then $\alpha>0$ , real parts of eigenvalues are positive
Title: Re: Q6 TUT 0401
Post by: Jiacheng Ge on November 18, 2018, 12:50:58 PM
My answer to c is slightly different.
Title: Re: Q6 TUT 0401
Post by: Jingze Wang on November 18, 2018, 03:10:49 PM
I am sorry, but what's the difference? I cannot find it. :)
Title: Re: Q6 TUT 0401
Post by: Michael Poon on November 18, 2018, 04:25:25 PM
I think the difference is the direction of rotation.
Title: Re: Q6 TUT 0401
Post by: Jingze Wang on November 18, 2018, 07:39:14 PM
Our graphs are all clockwise :)
Title: Re: Q6 TUT 0401
Post by: Victor Ivrii on November 18, 2018, 10:59:34 PM
Our graphs are all clockwise :)
Why?
Title: Re: Q6 TUT 0401
Post by: Nikita Dua on November 18, 2018, 11:10:01 PM
The potraits are clockwise, Since b > 0 and  c < 0
Title: Re: Q6 TUT 0401
Post by: Michael Poon on November 18, 2018, 11:18:17 PM
Our graphs are all clockwise :)

Isn't Guanyao's 2nd graph counterclockwise?
Title: Re: Q6 TUT 0401
Post by: Michael Poon on November 18, 2018, 11:30:36 PM
Our graphs are all clockwise :)
Why?

If we choose a unit vector $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and do matrix vector multiplication with the matrix $\begin{bmatrix} \alpha > 0 & 1 \\ -1 & \alpha > 0 \end{bmatrix}$, the vector $\begin{bmatrix}\alpha > 0 \\ -1\end{bmatrix}$ follows the phaseportrait CW. $\alpha$ > 0 means the phaseportrait points outward and is unstable.

If we choose a unit vector $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and do matrix vector multiplication with the matrix $\begin{bmatrix} \alpha < 0 & 1 \\ -1 & \alpha < 0 \end{bmatrix}$, the vector $\begin{bmatrix}\alpha < 0 \\ -1\end{bmatrix}$ also follows the phaseportrait CW. $\alpha$ < 0 means the phaseportrait points inward and is stable.
Title: Re: Q6 TUT 0401
Post by: Victor Ivrii on November 19, 2018, 02:15:43 AM
Michael, I asked why it is clockwise. Not about stability.

I especially made an announcement. For not providing explanation about clockwise/counter-clockwise rotation on Test (and Exam) the mark will be reduced
Title: Re: Q6 TUT 0401
Post by: Michael Poon on November 19, 2018, 03:29:21 AM
Thank you for posting the announcement about the explanation of CW/CCW. I found the textbook a little confusing to read and myself and a few others found this video helpful: https://www.youtube.com/watch?v=dpbRUQ-5YWc

At time 19:42 they display a technique to determine CW vs CCW using generic vectors and matrix A. I think it might be more intuitive but not as rigorous as the explanation you gave. And you explanation you gave in the announcement was very helpful!