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MAT244--Misc / FE Marks
« Last post by Victor Ivrii on April 20, 2018, 05:22:07 PM »
I removed this topic since it does not carry any useful information but also some students exposed some personal info. Not much, but better to be on the safe side.
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Final Exam / FE-P5 Comments
« Last post by Victor Ivrii on April 19, 2018, 01:16:12 PM »
a. Some students missed some stationary points and/or reported wrong points. All further analysis in the wrong points was ignored as irrelevant. For all three correct points found I gave 10pts, with a reduction of 3 pts for each missed points, and 1pts for extra points (only for those who found 3 correct points).

b. Linearization was easy, but some borked it. Finding eigenvalues was supposed to be a breeze but ...
One does not need to solve any equations to find eigenvalues of the diagonal or triangular matrices (some students wrote wrong equations and found wrong eigenvalues). Also eigenvectors of the diagonal matrices are obvious, and of the triangular are easy.

Not everyone found correctly eigenvalues of the matrix at $(2,3)$. T recall, that they are $1\pm i\sqrt{5}$. One does not need to look for eigenvectors, however one should look at the sign of bottom left element of the matrix and conclude what is the direction of rotation (I have not subtracted points for missing justification that it is counter-clockwise)

c. Drawing of the local pictures. At (0,0) mant=y draw incoming and outgoing lines like X instead of +, others indicated the wrong directions. Point $(-1,0)$ was more difficult. And in $(2,3)$ some draw "hairy monsters"

d. Even when all local pictures were drawn correctly, some students draw intersecting lines (trajectories do not intersect!) and not everyone observed that $x=0$ and $y=0$ consist of trajectories (see "skeleton" in my post above)



In some papers with no calculations or with calculations, leading to wrong conclusions) there are "miraculously" correct pictures. Those were discarded because "only solutions (not just answers) are evaluated". 
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Final Exam / FE-P6 Comments
« Last post by Victor Ivrii on April 19, 2018, 11:45:12 AM »
Observe that Hessian of $H(x,y)$ is
$$
\begin{pmatrix}
2x &2y\\
2y &2x
\end{pmatrix};
$$
compare with the  Jacobi matrix (Jacobian is its determinant). In this particular case (of exact system) sometimes it is called skew-Hessian.

I attach the Contour plot of $H(x,y)$; note that $(-1,0)$ is the local maximum and $(1,0)$ is the local minimum, while $(0,\pm 1)$ are two saddle points
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Final Exam / Re: FE-P5--solution
« Last post by Victor Ivrii on April 18, 2018, 06:48:47 AM »
a.  Solving $x(x-y+1)=0$, $y(x-2)=0$ we get cases
\begin{align*}
&x=y=0  &&\implies A_1=(0,0),\\
&x=x-2=0  &&\implies \text{impossible}\\
&y=x-y+1=0 &&\implies A_2=(-1,0),\\
&x-y+1=x-2=0 &&\implies A_3=(2,3).
\end{align*}
b. Linearizations at these points have matrices
\begin{align*}
&
\begin{pmatrix}
1 &\ \ 0\\
0 &-2
\end{pmatrix}
&&
\begin{pmatrix}
-1 &\ \ 1\\
0 &-3
\end{pmatrix}
&&
\begin{pmatrix}
2 &-2\\
3 &0
\end{pmatrix}
\\[5pt]
\text{with eigenvalues    }&\{1,-2\} &&\{-1,-3\} && \{1-\sqrt{5}i,1+\sqrt{5}i \}
\end{align*}
and therefore
* $A_1$ is a saddle,
* $A_2$ is a stable node, and
* $A_3$ is unstable focal point and since left bottom number is $3>0$ it is counterclockwise oriented.


c. Axis are:
in $A_1$:  $\mathbf{e}_1=(1,0)^T$ unstable ($\lambda_1=1$), $\mathbf{e}_2=(0,1)^T$ stable ($\lambda_2=-2$).

in $A_2$: $\mathbf{f}_1=(1,0)^T$ ($\lambda_1=-1$), $\mathbf{f}_2=(1,-2)^T$ ($\lambda_1=-3$). Since $\lambda_1 >\lambda_2$, all trajectories have an entry directions $\pm \mathbf{f}_1$ (except two, which have entry directions $\pm \mathbf{f}_2$). Then we draw trajectories near critical points (See attachment  P5-loc.png).

d. One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a "skeleton'' of the phase portrait (see attachment), impose local pictures on it and finally draw a global portrait
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Final Exam / Re: FE-P4
« Last post by Victor Ivrii on April 15, 2018, 03:26:19 AM »
Solution is complete, but the answer must be written.

Also a simpler form $\ln(\tan(t))$ is preferable.

Finally \sin, \cos, and so on must be escaped by \ to provide upright letters and a proper spacing
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Final Exam / Re: FE-P3
« Last post by Victor Ivrii on April 15, 2018, 03:15:29 AM »
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
The only thing which was missing in the solution, is the final answer, but it warrants neither such claim, nor uploading your solution.

General remark:
It would be better to denote "parameters" by uppercase letters $C_1(x)$, $C_2(x)$,... and constants by lowercase letters $c_1$, $c_2$,...
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Final Exam / Re: FE-P6
« Last post by Victor Ivrii on April 14, 2018, 02:25:56 PM »
There is nothing missing.
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Web Bonus Problems / Re: Exam Week
« Last post by Andrew Hardy on April 14, 2018, 02:12:26 PM »
Writing the change of coordinates Sheng applied to the boundary conditions I found,  $ w|_{\pi=0}=0$ and $  w|_{r=1}=2\sin^2\theta $ I can apply separation of variables to the function $ w = P(\theta)R(r) $ which I can solve for this half disk as
 $$ w = \sum A r^n \sin(n\theta) $$ and therefore
 $$ A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta  =\frac{ (4 (2 \cos(π n) - 2))}{(π (n^3 - 4 n)) }$$
and this is $ \frac{-8}{(π (n^3 - 4 n)) } $ when n is odd. So I have
$$ w =  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$
and therefore I can write $$ u = w + v =  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta) - r^2 sin^2(\theta) $$ which can be written most concisely as
$$ u = r^2 \cos(2 \theta) +  \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$

 
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Final Exam / Re: FE-P6
« Last post by Andrew Hardy on April 14, 2018, 01:46:20 PM »
If this is still open for karma, I will go into more detail missing from the solution?

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Final Exam / Re: FE-P6
« Last post by Victor Ivrii on April 14, 2018, 01:19:45 PM »
 .
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