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### Messages - Bruce Wu

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31
##### Test 1 / Re: TT1-P1
« on: October 22, 2015, 01:06:07 AM »
One can look at the contour plot on this page http://www.wolframalpha.com/input/?i=x-arctan%28t%29 to see the characteristic curves

32
##### Test 1 / Re: TT1-P1
« on: October 21, 2015, 11:54:42 PM »
We have $$\frac{dt}{t^{2}+1}=\frac{dx}{1}=\frac{du}{u}$$
Xi Yue is right until the last line. Instead we need to proceed as follows: $$\frac{du}{u}=dx\Rightarrow \ln(u)=x+d\Rightarrow u=De^{x}=\phi(x-\arctan(t))e^{x}$$
So for part c): $$u(x,0)=\phi(x)e^x=\cos(x)\Rightarrow \phi(x)=\cos(x)e^{-x}$$ since $\arctan(0)=0$. Therefore, the solution to the IVP is:
$$u(x,t)=\cos(x-\arctan(t))e^{\arctan(t)-x}e^{x}=\cos(x-\arctan(t))e^{\arctan(t)}$$
The solution is fully determined because the original equation was first order in time, with no restrictions on x. Therefore one initial condition uniquely defines a solution.

33
##### Test 1 / Re: TT1-P3
« on: October 21, 2015, 10:43:19 PM »
Ah, I see. Our expressions are actually equivalent. I have just forgotten to collect the $e^{x-2t}$ terms. My apologies

34
##### Test 1 / Re: TT1-P3
« on: October 21, 2015, 10:23:52 PM »
Here we have an inhomogeneous Dirichlet boundary condition. Using http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.html#mjx-eqn-eq-2.6.10 I obtained the solution I gave above

35
##### Test 1 / Re: TT1-P3
« on: October 21, 2015, 09:50:24 PM »
I believe that for $0<x<2t$ the solution is actually \large
u(x,t)=e^{x-2t}-e^{-x-2t}+e^{-2t+x}

36
##### HA5 / Re: HA5-P3
« on: October 20, 2015, 01:09:01 AM »
Shouldn't the integral bounds in equation (6) be from $-\infty$ to $\infty$ instead of $0$ to $\infty$? Here the domain is not restricted yet. This is what is in Rong Wei's written solution also.

37
##### Textbook errors / 3.2 eqn 17
« on: October 19, 2015, 01:13:00 PM »

38
##### Textbook errors / 3.1 equation 12
« on: October 17, 2015, 08:47:21 PM »
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter3/S3.1.html#mjx-eqn-eq-3.1.12

I might be wrong, but based on the definition of the error function given, shouldn't it be equal to $$\frac{1}{4}+\frac{1}{4}erf\left(\frac{x}{\sqrt{4kt}}\right)$$?

39
##### HA4 / Quiz 3 Answers
« on: October 15, 2015, 10:56:49 PM »
Remember that it is just HA4 problem 2 part a): http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.P.2
With $c_{1}=1,c_{2}=2,\alpha=\beta=1$

For $x<-2t$: $$u(x,t)=0$$
For $-2t<x<0$: $$u(x,t)=\frac{4}{3}\phi\left(\frac{1}{2}x+t\right)-\frac{1}{3}\phi(0)$$
For $0<x<t$: $$u(x,t)=\phi(x+t)+\frac{1}{3}\phi(t-x)-\frac{1}{3}\phi(0)$$
For $x>t$: $$u(x,t)=\phi(x+t)$$

Visualization can be found here: http://forum.math.toronto.edu/index.php?topic=660.msg2418#new
Substitute the appropriate constants.

40
##### HA4 / Re: HA4-P3
« on: October 15, 2015, 12:41:21 AM »
Below is what I have so far. Please let me know if I am correct before I continue.
a) If $v>c$, then $x>ct$. In this region both $\phi(x+ct)$ and $\psi(x-ct)$ are fully defined. We do not need any boundary conditions, which is condition 1.
If $-c<v<c$, then $-ct<x<ct$. Here $\phi$ is fully defined but we need to define $\psi$ for negative arguments. Therefore we need 1 boundary condition, either 2. or 3. (need to work out which one)
If $v<-c$, then $x<-ct$. We need to define both $\phi$ and $\psi$ for negative arguments. We need 2 boundary conditions, so condition 4.

41
##### HA4 / Re: HA4-P4
« on: October 12, 2015, 04:09:18 PM »
a) For $x>3t$ the solution is given by D'Alembert's formula: $$u(x,t)=\frac{1}{6}[\sin(x+3t)-\sin(x-3t)]=\frac{1}{3}\cos(x)\sin(3t)$$
Here we have Dirichlet boundary condition. We use http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.html#mjx-eqn-eq-2.6.31. Then for $0<x\leq3t$:$$u(x,t)=\frac{1}{6}[\sin(x+3t)-\sin(3t-x)]=\frac{1}{3}\cos(3t)\sin(x)$$
b) For $x>3t$ the solution is the same as in part a)
Here we have Neumann boundary condition. We use http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.html#mjx-eqn-eq-2.6.31-. Then for $0<x\leq3t$:$$u(x,t)=\frac{1}{6}[\sin(x+3t)+\sin(3t-x)]=\frac{1}{3}\cos(x)\sin(3t)$$
c) For $x>3t$: $$u(x,t)=\frac{1}{6}[-\cos(x+3t)+\cos(x-3t)]=-\frac{1}{3}\sin(x)\sin(-3t)=\frac{1}{3}\sin(x)\sin(3t)$$
Dirichlet boundary condition. For $0<x\leq3t$:$$u(x,t)=\frac{1}{6}[-\cos(x+3t)+\cos(3t-x)]=\frac{1}{3}\sin(x)\sin(3t)$$
d) For $x>3t$ the solution is the same as in part c)
Neumann boundary condition. For $0<x\leq3t$:$$u(x,t)=\frac{1}{6}[-\cos(3t-x)+1+-\cos(x+3t)+1]=-\frac{1}{6}[\cos(3t-x)+\cos(x+3t)-2]=-\frac{1}{3}[\cos(3t)\cos(x)-1]$$

42
##### Textbook errors / Section 2.6 errors
« on: October 11, 2015, 03:34:15 PM »
In text: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.html#mjx-eqn-eq-2.6.23
The middle of paragraph b. it should be $u_{x}|_{x=b}=q_{r}(t)$ instead of $u_{x}|_{x=a}=q_{r}(t)$

In problem 2 part a.: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.P.2
At the bottom it is missing some letters. The different regions should be $x>c_{1}t,0<x<c_{1}t,-c_{2}t<x<0$ and $x<-c_{2}t$

43
##### Textbook errors / Section 2.7 problem 3
« on: October 09, 2015, 10:44:10 PM »
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.7.P.html#problem-2.7.P.3

This is bonus problem 4 of week 4. I believe that the boundary conditions are missing their defining functions.

Right. All $0$ (and there were several similar misprints). V.I.

44
##### HA3 / Re: HA3-P2
« on: October 08, 2015, 01:32:57 PM »
Sorry I forgot about the yellow diamond. My solution above has been fixed.

45
##### HA3 / Re: HA3-P2
« on: October 08, 2015, 11:44:58 AM »
I will do the last part.
$$g(x) = 0, h(x) = \begin{cases} \cos(x) &\mbox{if } |x| < \frac{\pi}{2} \\ 0 & \mbox{if } |x| \geq \frac{\pi}{2}. \end{cases}$$

The solution is $$u=\frac{1}{2c}\int^{x+ct}_{x-ct}h(y)dy$$
Let $H$ be the primitive of $h$, i.e $H'(y)=h(y)$
Then $$H(y)=\begin{cases} 0 &\mbox{if } y \leq -\frac{\pi}{2} \\ \sin(y)+1 & \mbox{if } -\frac{\pi}{2} < y < \frac{\pi}{2} \\ 2 & \mbox{if } y \geq \frac{\pi}{2}. \end{cases}$$
And we have $$u=\frac{1}{2c}[H(x+ct)-H(x-ct)]$$
Since g and h here are also even with respect to x, we apply the same logic as the professor and only worry about $x>0$ and $t>0$.
So we have $$u(x,t)=\begin{cases} \frac{1}{c} &\mbox{if } x-ct \leq -\frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ \frac{1}{2c}[1-\sin(x-ct)] & \mbox{if } -\frac{\pi}{2} < x-ct < \frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ 0 & \mbox{if } x-ct \geq \frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ \frac{1}{2c}[\sin(x+ct)-\sin(x-ct)] & \mbox{if } -\frac{\pi}{2} < x-ct < \frac{\pi}{2}, -\frac{\pi}{2} < x+ct < \frac{\pi}{2} \end{cases}$$
Then we can use the symmetry of u to find its value in other regions.

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