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Messages - Victor Ivrii

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1
FE / Re: FE6
« on: December 20, 2016, 06:22:30 AM »
Indeed, occasionally (remnant of the previous version) misleading "use separation of variables" appeared. Because everybody followed and none got a correct solution, this problem was removed and FE was normalized 90->40.

There is a correct solution:

Solution is spherically symmetric because the problem is. Then
\begin{equation}
u_{tt}- \bigl(u_{rr}+\frac{2}{r}u_r\bigr)=0\qquad r>0, t>0.
\label{6-4A}
\end{equation}
Multiplying by $r$ and using (\ref{6-3}) we arrive to the first equation below:
\begin{align}
&v_{tt}-v_{rr}=0\qquad r>0,\label{6-5}\\
&v(0,t)=0,
\label{6-6}\\
&v(r,0)=g(r)=\left\{\begin{aligned} &r
\quad &&r<1,\\
&0 &&r\ge 1,\end{aligned}\right.  && v_t(r,0)=0.
\label{6-7}
\end{align}

Continuing $g(r)$ as and odd function  $\tilde{g}(r)=\left\{\begin{aligned} &r
\quad &&|r|<1,\\
&0 &&|r|\ge 1,\end{aligned}\right.$ and solving Cauchy problem we get
\begin{equation}
v(r,t)=\frac{1}{2}\bigl( \tilde{g}(r+t)+\tilde{g}(r-t)\bigr)=\left\{\begin{aligned}
&0  &&r>t+1,\\
&\frac{1}{2}(r-t) \qquad&&1-t<r<t+1,\\
&r &&0<r<1-t,\\
&0 && 0< r<t-1
\end{aligned}\right.
\end{equation}
and finally
\begin{equation}
u(r,t)=r^{-1}v(r,t)=\left\{\begin{aligned}
&0  &&r>t+1,\\
&\frac{1}{2r}(r-t) \qquad&&1-t<r<t+1,\\
&1 &&0<r<1-t,\\
&0 && 0< r<t-1
\end{aligned}\right.
\end{equation}


2
APM346--Announcements / FE and FM
« on: December 20, 2016, 06:14:23 AM »
Final Exam:
Count   53
Minimum Value   17.33
Maximum Value   40.00
Range   22.67
Average   27.37
Median   28.00
Standard Deviation   6.15


This is what I submitted

3
APM346--Announcements / Re: Bonuses
« on: December 19, 2016, 07:12:36 AM »
Exam bonuses


Luyu CEN =4
Sajjan Heerah= .5
Shaghayegh A =1
XinYu Zheng=.3
Shentao YANG=.3
Bruce Wu=1

4
FE / Re: FE5
« on: December 18, 2016, 01:14:35 PM »
Shaghayegh, the last correct line is the one I denoted by $(\checkmark)$; I also put (  )  where they should be. Please proceed with the correct solution after.

Ziling Also not completely correct solution

Indeed, both solutions are correct! Sorry... Shaghayegh, unwritten rule: red is mine (admin/moderator).

5
FE / Re: FE7
« on: December 18, 2016, 10:11:50 AM »
Bruce's solution is correct

6
FE / Re: FE6
« on: December 18, 2016, 10:10:06 AM »
Wrong solution removed. Please post a correct one

7
FE / Re: FE5
« on: December 18, 2016, 08:25:07 AM »
Minor errors. Still I expect a correct solution to be posted.

8
FE / Re: FE4
« on: December 17, 2016, 12:44:42 PM »
Further, even if the shape was a disk, three cases are different:

(1) Circular membrane; then $P''=-\lambda P$ for all $\theta$, $P$ is $2\pi$-periodic;
(2) Circular membrane, with a  cut along $\theta=\pi$,  and it is fixed there; then $P''=-\lambda P$ for  $-\pi<\theta<\pi$, $P(-\pi)=P(\pi)=0$;
(3) Circular membrane, with a  cut along $\theta=\pi$,  and it is free there; then $P''=-\lambda P$ for  $-\pi<\theta<\pi$, $P'(-\pi)=P'(\pi)=0$.

9
FE / Re: FE4
« on: December 17, 2016, 11:03:05 AM »
Correct. Can you simplify the answer?

Also, one can simplify calculations observing that the problem is symmetric with respect to $y=0$ (that means $\theta=0$), in particular, because $U|_{r=4}=1$ is an even function. Therefore from the very beginning we need to consider $\cos(\omega \theta)$, s.t. $\cos(\omega \times 4\pi/6)=0$, that is $5\pi\omega /6= (2m+1)\pi /2\implies \omega=\omega_m= =3(2m+1) /5$, $\lambda_m= 9(2m+1)/25$.

Many students thought of $2\pi$-periodic $P(\theta)$ which is not the case as a membrane is not disk, but a sector.

10
FE / Re: FE3
« on: December 17, 2016, 11:02:04 AM »
Correct; better to plug $n=2m+1$, $m=0,1,\ldots$ in the end

11
FE / Re: FE3
« on: December 16, 2016, 08:38:33 AM »
Wrong solution removed

Someone, who solved it correctly (including one, who posted the wrong solution), post a correct one -- and it better to be typed!

Most found correctly the general form of solution, but then rather frequent error: tried to decompose $f(x)$ into Fourier series on $[0,1]$ and $[1,2]$ separately. No, we do not have two separate $f(x)$ but a single one defined above:

12
FE / Re: FE2
« on: December 15, 2016, 07:42:07 PM »
Indeed, $u=\frac{x}{(t+1)^{\frac{3}{2}}}e^{-\frac{x^2}{(t+1)}}$ and there is a way to get it even almost without calculations. First, as mentioned, since $f(x)=xe^{-x^2}$ is an odd function, it is sufficient to solve Cauchy problem.

We know that $v(x,t)=t^{-\frac{1}{2}}e^{-\frac{x^2}{t}}$ satisfies our equation (\ref{2-1}).
Plugging $t:=t+1$ we conclude that $w(x,t)=(t+1)^{-\frac{1}{2}} e^{-\frac{x^2}{t+1}}$ also satisfies (\ref{2-1}) (since coefficients do not depend on $t$). Further, $w(x,0)=g(x):=e^{-x^2}$.

Next, $-\frac{1}{2}w_x$ also satisfies (\ref{2-1}) (since coefficients do not depend on $x$) and $-\frac{1}{2}w_x(x,0)=-\frac{1}{2}g'(x)$ which is $f(x)$. Bingo! $u=-\frac{1}{2}w_x=\frac{x}{(t+1)^{\frac{3}{2}}}e^{-\frac{x^2}{(t+1)}}$.

13
FE / Re: FE2
« on: December 15, 2016, 08:00:35 AM »
Wrong solution removed. Several students got correct solutions, including calculations. I hope that  one of them posts (or I will).

14
FE / Re: FE1
« on: December 14, 2016, 01:08:42 PM »
Luyu CEN
Correct


PS. In LaTeX we write \cos to get $\cos$ (upright, with a proper space after), and so on

15
FE / Re: FE1
« on: December 14, 2016, 11:08:16 AM »
Sajjan

This is incorrect solution. Obviously it does not satisfy the boundary condition.

I finished to grade this problem. One third of those who wrote got  it right, and another third made a single error (forgot $C$ in the definition of $\psi$, which is selected to make it continuous). There is no point to post a wrong solution, especially the scan of poor handwriting.

If nobody posts a correct solution, I will post one, Dec 19 or 20

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