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Messages - Victor Ivrii

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1
MAT244--Announcements / TT1 papers
« on: Today at 07:28:30 AM »
TT1 grades are on BlackBoard

Papers will be returned at Lecture sections. We planned to do it on Tutorial sections, but too many students failed to write properly their Tutorial sections (time slot instead of code does not cut).



Term Test 2 papers will be returned at Tutorial sections first, and only one or two weeks later––on Lecture sections.

2
MAT244--Misc / Re: Test Paper
« on: February 23, 2018, 08:57:18 PM »
Because too many students failed to write properly their tutorial sections, papers will be distributed at lecture sections.

3
MAT244--Misc / Re: TT1 grade
« on: February 23, 2018, 03:10:09 PM »
The obvious answer: when TAs finish grading and upload marks

4
Term Test 1 / Re: P3 Night
« on: February 22, 2018, 07:38:07 AM »
(a) Solution to equation is
\begin{equation}
u(x,t)=\phi(x+3t)+\psi(x-3t)
\label{eq-3-5}
\end{equation}
with unknown functions $\phi$ and $\psi$. Plugging into initial conditions we get
\begin{gather*}
\phi(x)+\psi(x)=0, \quad 2\phi'(x)-2\psi'(x)=0\implies \phi(x)=\psi(x)=0 \qquad x>0
\end{gather*}
and $u(x,t)=\sin(x+3t)$ as $x>t$.

(b) Plugging into boundary condition  we get $\psi (2-2\sqrt{t+1}-t)= t$ as $t>0$. For $t>0$
$2-2\sqrt{t+1}-t$ is a monotone decreasing function, from $0$ to $-\infty$. Solve:
$$
x=-2-2\sqrt{t+1}-t\implies (\sqrt{t+1}+1)^2 = -x\implies  \sqrt{t+1}=\sqrt{1-x}-1\implies
t=-1+(\sqrt{1-x}-1)^2,
$$
and therefore $\psi(x)=-1+(\sqrt{1-x}-1)^2$ for $x<0$ and finally
$$
u(x,t)=-1+(\sqrt{1-x+t}-1)^2\qquad x<t.
$$



5
Term Test 1 / Re: P2 Night
« on: February 22, 2018, 07:32:19 AM »
A bit more polished:

$\newcommand{\erf}{\operatorname{erf}}$
By D'Alembert formula
\begin{equation}
u(x,t)=
3\iint_{\Delta (x,t)}\, e^{-\xi^2}\, d\xi d\tau,
\label{2-3}
\end{equation}
where $\Delta (x,t)$ is bounded by $\tau=0$, $x-\xi-3(t-\tau)=0$, $x-\xi +3(t-\tau)=0$.

Then the double integral becomes
\begin{align*}
&3 \int_{x-3t}^x \Bigl(\int_0^{t+\frac{1}{3}(\xi-x)}d\tau \Bigr)e^{-\xi^2}\,d\xi+
3 \int^{x+3t}_x \Bigl(\int_0^{t-\frac{1}{3}(\xi-x)}d\tau \Bigr)e^{-\xi^2}\,d\xi=\\[2pt]
&\int_{x-3t}^x (3t-x+\xi )e^{-\xi^2}\,d\xi + \quad \qquad \int^{x+3t}_x (3t+x-\xi)e^{-\xi^2}\,d\xi=\\
&(3t-x) \int_{x-3t}^x e^{-\xi^2}\,d\xi -\frac{1}{2}e^{-\xi^2}\bigr|_{\xi= x-3t}^{\xi=x} +
(3t+x) \int_x^{x+3t} e^{-\xi^2}\,d\xi +\frac{1}{2}e^{-\xi^2}\bigr|^{\xi= x-3t}_{\xi=x} =\\[4pt]
&(3t-x)\Bigl(\erf(x)-\erf(x-3t)\Bigr)+ (3t+x)\Bigl(\erf(x+3t)-\erf(x)\Bigr)+\\[4pt]
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\frac{1}{2}\Bigl(e^{-(x-3t)^2}+e^{-(x+3t)^2}-2e^{-x^2}\Bigr).
\end{align*}

6
Term Test 1 / Re: P1 Night
« on: February 22, 2018, 07:26:38 AM »
Crucial error

7
Term Test 1 / Re: P5
« on: February 22, 2018, 07:01:40 AM »
Tristan
$u$ is not split , but the integral, expressinfg it is

8
Term Test 1 / Re: P4
« on: February 22, 2018, 06:50:03 AM »
This problem describes the string with the masses on its ends (forces are $\pm c^2u_x$ on the left/right ends and acceleration $u_{tt}$)

9
Term Test 1 / Re: P3
« on: February 22, 2018, 06:47:10 AM »
OK

10
Term Test 1 / Re: P2
« on: February 22, 2018, 06:42:46 AM »
Definitely change of variables makes calculations easier; but one could  make calculations without it. To do so one needs to know the primitive of $]\arctan(x)$:
$$
\int \arctan(x)\,dx = x\arctan(x)-\int x\,d\arctan(x)= x\arctan(x) -\int \frac{x\,dx}{x^2+1}= x\arctan(x) -\frac{1}{2}\ln (x^2+1).
$$

11
Term Test 1 / Re: P1
« on: February 22, 2018, 06:33:57 AM »
Jilong
In the last line of (a) should be $D=x-t^3 + 3t$. Correct from here

12
Term Test 1 / Re: P3-Morning
« on: February 21, 2018, 09:22:08 AM »
Vivian
errors, in part (b)

13
Term Test 1 / Re: P2-Morning
« on: February 21, 2018, 09:11:30 AM »
Cheryl
Almost unreadable.

Meng
(9) is OK, but (10) is not

14
Term Test 1 / Re: P1-Day
« on: February 21, 2018, 08:17:00 AM »
(4) is correct but then integrated with an error

15
Quiz-3 / Re: Q3-T0301
« on: February 21, 2018, 07:48:37 AM »
use \cos \sin etc

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