Author Topic: Q6 TUT 0401  (Read 2841 times)

Victor Ivrii

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Q6 TUT 0401
« 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.

$$\mathbf{x}' =\begin{pmatrix}
\alpha  &1\\
-1 &\alpha
\end{pmatrix}\mathbf{x}.$$

Guanyao Liang

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Re: Q6 TUT 0401
« Reply #1 on: November 17, 2018, 03:54:53 PM »
This is my answer.

Jingze Wang

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Re: Q6 TUT 0401
« Reply #2 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
« Last Edit: November 25, 2018, 09:27:37 AM by Victor Ivrii »

Jiacheng Ge

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Re: Q6 TUT 0401
« Reply #3 on: November 18, 2018, 12:50:58 PM »
My answer to c is slightly different.

Jingze Wang

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Re: Q6 TUT 0401
« Reply #4 on: November 18, 2018, 03:10:49 PM »
I am sorry, but what's the difference? I cannot find it. :)

Michael Poon

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Re: Q6 TUT 0401
« Reply #5 on: November 18, 2018, 04:25:25 PM »
I think the difference is the direction of rotation.

Jingze Wang

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Re: Q6 TUT 0401
« Reply #6 on: November 18, 2018, 07:39:14 PM »
Our graphs are all clockwise :)

Victor Ivrii

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Re: Q6 TUT 0401
« Reply #7 on: November 18, 2018, 10:59:34 PM »

Nikita Dua

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Re: Q6 TUT 0401
« Reply #8 on: November 18, 2018, 11:10:01 PM »
The potraits are clockwise, Since b > 0 and  c < 0

Michael Poon

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Re: Q6 TUT 0401
« Reply #9 on: November 18, 2018, 11:18:17 PM »
Our graphs are all clockwise :)

Isn't Guanyao's 2nd graph counterclockwise?

Michael Poon

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Re: Q6 TUT 0401
« Reply #10 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.

Victor Ivrii

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Re: Q6 TUT 0401
« Reply #11 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

Michael Poon

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Re: Q6 TUT 0401
« Reply #12 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!