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### Messages - Michael Poon

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##### Quiz-7 / Re: Q7 TUT 0301
« 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?

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##### Quiz-7 / Re: Q7 TUT 0301
« 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!)

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##### Quiz-7 / Re: Q7 TUT 0301
« 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:

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##### Quiz-7 / Re: Q7 TUT 0301
« 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However, when we carry those constraints to the second equation, x = -1 is no longer valid as $\cos(y)$ is bound by -1 and 1.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}

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##### Term Test 2 / Re: TT2A-P4
« on: November 22, 2018, 05:06:53 PM »
Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically.

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##### Term Test 2 / Re: TT2A-P3
« on: November 20, 2018, 12:17:52 PM »
I think you could also characterise the phase portrait as a node? (unstable node)

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##### Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 12:12:59 PM »
Seems like no one has added a phase portrait yet.

I attached it below, it is a saddle, with eigenvalues real and opposite.

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##### Quiz-6 / Re: Q6 TUT 0401
« 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!

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##### Quiz-6 / Re: Q6 TUT 0401
« 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.

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##### Quiz-6 / Re: Q6 TUT 0401
« on: November 18, 2018, 11:18:17 PM »
Our graphs are all clockwise Isn't Guanyao's 2nd graph counterclockwise?

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##### Quiz-6 / Re: Q6 TUT 0401
« on: November 18, 2018, 04:25:25 PM »
I think the difference is the direction of rotation.

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##### Quiz-5 / Re: Q5 TUT 0501
« on: November 18, 2018, 09:16:18 AM »
Isolate the first equation for $x_2$:

$x_2 = 0.5x_1' + 0.25x_1$

Differentiating both sides we get:

$x_2' = 0.5x_1'' + 0.25x_1'$

Subbing in the above into the second equation:

$x_1'' + x_1 + 4.25x_1 = 0$

Solving for the characteristing eqn gives us:

$r = -0.5 \pm 2i$

So, we now know :

$x_1 = e^{-0.5t}(c_1cos(2t) + c_2sin(2t))$

Subbing the above and its derivative into eqn 2:

$x_2 = e^{-0.5t}(c_1(0.125)\cos(2t) + c_2(1.125)\sin(2t))$

By initial conditions:

$-2 = c_1$

$2 = 0.125c_2$

Final solution:

$x_1 = e^{-0.5t}(-2\cos(2t) + 16\sin(2t))$

$x_2 = e^{-0.5t}(-0.5\cos(2t) + 10\sin(2t))$

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##### Quiz-6 / Re: Q6 TUT 5102
« on: November 17, 2018, 04:42:29 PM »
Phase portraits attached below:

Top: Eigenvalues real & same sign (+ve), stable

Middle: Eigenvalues real & opposite sign, saddle

Bottom: Eigenvalues complex & negative, unstable spiral

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##### Quiz-6 / Re: Q6 TUT 5102
« on: November 17, 2018, 04:35:52 PM »
a) Finding the eigenvalues:

Set the determinant = 0

\begin{align}
(4 - \lambda)(-6 - \lambda) - 8\alpha &= 0\\
\lambda^2 + 2\lambda - 24 - 8\alpha &= 0\\
\lambda &= -1 \pm \sqrt{25 + 8\alpha}
\end{align}

b)

Case 1: Eigenvalues real and same sign
when: $\alpha$ > $\frac{-25}{8}$ + 1

Case 2: Eigenvalues real and opposite sign
when: $\frac{-25}{8}$ < $\alpha$ < $\frac{-25}{8} + 1$

Case 3: Eigenvalues complex
when: $\alpha$ < $\frac{-25}{8}$

critical points: $\alpha$ = $\frac{-25}{8}$, $\frac{-25}{8}$ + 1

c) will be posted below:

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##### Quiz-6 / Re: Q6 TUT 5101
« on: November 17, 2018, 04:08:43 PM »
Phase portrait attached

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