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### Topics - Emily Deibert

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1
##### Textbook errors / Broken Link in S 10.2
« on: November 23, 2015, 11:36:59 AM »
Clicking the arrow to go to the next section at the bottom of this page:

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter10/S10.2.html

leads to a 404 missing page error.
Fixed

2
##### Website errors / Link to sections for Week 10
« on: November 18, 2015, 06:19:52 PM »
The link to section 9.1 for week 10 is broken.

http://www.math.toronto.edu/courses/apm346h1/20159/lectures.html

3
##### Textbook errors / HA7-P2
« on: November 15, 2015, 10:22:20 PM »

4
##### HA9 / HA9-P3
« on: November 13, 2015, 07:10:15 PM »

5
##### HA9 / HA9-P1
« on: November 13, 2015, 07:10:47 AM »

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##### Textbook errors / Equation 6.3.5'
« on: November 11, 2015, 12:47:44 PM »
Hi Professor, I think you made a mistake here by putting $dy^2$ twice in the equation.

Indeed

7
##### Website errors / Mistake in Quiz Statistics on Forum
« on: November 09, 2015, 10:23:02 AM »
Professor, I think the statistics you just posted for the quiz marks don't add up---there should be a category for 90-99 that you haven't included.

(http://forum.math.toronto.edu/index.php?topic=652.msg2612#new)

The best always fall between cracks

8
##### Textbook errors / Potential error in section 5.2?
« on: November 07, 2015, 05:57:46 PM »
I might have misunderstood, so maybe this isn't an error, but for the Fourier transform to be unitary, shouldn't we set $\kappa=\sqrt{2\pi{}}$, rather than $\kappa=\frac{1}{\sqrt{2\pi{}}}$, as mentioned in section 5.2 theorem 2.c? Thank you for the help!

Indeed

9
##### HA8 / HA8-P1
« on: November 06, 2015, 02:55:24 PM »
Problem 1. http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html

(a) We are asked to find the solutions of $\Delta u = u_{xx} + u_{yy} + u_{zz} = k^2u$ that depend only on $r$. So, we must turn this into spherical coordinates. By the methods of the textbook section 6.3.2, we arrive at:

\Delta u(r, \phi{}, \theta{}) = u_{rr} + \frac{2u_r}{r} + \frac{u_{\phi{} \phi{}}}{r^2} + \frac{\cot(\phi{})u_{\phi}}{r^2} + \frac{u_{\theta{} \theta{}}}{\sin^2(\phi{})r^2} = k^2u

Note that these are defined as in the textbook: $r$ is the radius, $\phi$ is the latitude, and $\theta$ is the longitude.

Now, the problem asks for a solution that depends only on r. In this case, any derivatives with respect to the other variables should be zero. So canceling these terms, we can write the equation as:

\Delta u(r) = u_{rr} + \frac{2}{r}u_r = k^2u

Now we can make use of the hint provided in the problem, namely that we should make the substitution $u(r) = \frac{v(r)}{r}$. Then $u''(r)$ and $u'(r)$ are:

\begin{cases}
u_r = \frac{v'(r)}{r} - \frac{v(r)}{r^2} \\
u_{rr} = \frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3}
\end{cases}

Now plugging these all into the problem, we have:

\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + \frac{2}{r} \Big( \frac{v'(r)}{r} - \frac{v(r)}{r^2} \Big) = k^2\frac{v(r)}{r} \longrightarrow

\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + 2\frac{v'(r)}{r^2} - 2\frac{v(r)}{r^3} = k^2\frac{v(r)}{r} \longrightarrow

\frac{v''(r)}{r} = k^2\frac{v(r)}{r} \longrightarrow

v''(r) = k^2v

We now have a simple ODE which can be solved by the usual methods. Recall that $k>0$ as defined in the problem. The solution to the ODE is:

v(r) = Ae^{kr} + Be^{-kr}

Now we put this back in terms of $u(r)$, where we recall that $u(r) = \frac{v(r)}{r}$. So:

u(r) = Ae^{kr}r^{-1} + Be^{-kr}r^{-1}

10
##### Quiz 4 / Will Quiz 4 be next week?
« on: October 22, 2015, 09:11:39 AM »
Just wondering if we will have quiz 4 on the 29th of October, or if we will have no quiz next week. Thank you!

11
##### Website errors / Wrong lecture room listed on website
« on: October 18, 2015, 05:13:41 PM »
Hi Professor,

I just noticed that on the website you have our lecture room listed as SF 1101, however our lecture is in Sid Smith! The room should be SS 2110.

http://www.math.toronto.edu/courses/apm346h1/20159/outlines.html

12
##### HA5 / HA5-P7
« on: October 17, 2015, 06:41:15 PM »
I will update my answers as I finish working through them. All of my answers are now uploaded.

(a) We consider the energy as given in the problem:
E(t) = \int_{J} u^2(x,t)dx

Now, take the time derivative to yield:
\frac{\partial}{\partial t}E(t) = \int_{J} \frac{\partial}{\partial t}u^2(x,t)dx = \int_{J} 2uu_tdx

Replacing $u_t$ with $ku_{xx}$ as per the heat equation, we get:
\int_J 2kuu_{xx}dx

Now, using integration by parts, we will get:
2k\Big[uu_x|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u_x^2dx\Big]

Here I assume that the function is fast-decaying, so the first term goes to 0 (professor, is this a valid assumption?) This will leave me with just:
-2k\int_{-\infty}^{\infty} u_x^2dx  \label{eq:integral}

Now consider \ref{eq:integral}. Since the integrand is positive (since it is $u_x^2$) and k is positive (by the definition of the heat equation we assume this), then this equation is $\leq 0$.

If $u(x,t) =$ const., then the integrand will be zero (indeed, $\frac{d}{dx}C = 0$).

In all other cases, this is < 0. Therefore $E(t)$ does not increase; further, it decreases unless $u(x,t) =$ const.

(b) This part proceeds in the same way up to the integration by parts (except with J defined as the range from 0 to L rather than negative to positive infinity), so I will not type all of those steps again. Just recall that:
\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big] \label{eq:newint}

We consider the cases of both Dirichlet and Neumann boundary conditions. First, let's think about Dirichlet. This condition says that:
u|_{x=0} = u|_{x=L} = 0
Therefore the first term in \ref{eq:newint} goes to zero by the Dirichlet boundary condition on $u$.

Now let's consider the Neumann boundary condition. It says that:
u_x|_{x=0} = u_x|_{x=L} = 0.
Again, the first term in \ref{eq:newint} will go to zero, this time by the Neumann boundary condition on $u_x$.
In both cases, we see that the end result is:
\frac{\partial}{\partial t}E(t) = -2k\int_0^Lu_x^2dx
We proceed as in part (a), showing that $E(t)$ decreases unless $U(x,t) =$ const because the integrand and the constant k are both positive.

(c) This problem is identical to (a) and (b) up to the integration by parts (where J is defined as the region from 0 to L, as in part (b)), so I won't type up the first few steps again. We arrive at:
\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big]
\label{eq:c}

This time, we have Robin boundary conditions. They are:
\begin{cases}
u_x(0,t) - a_0u(0,t) = 0 \\
u_x(L,t) + a_Lu(L,t) = 0
\end{cases}

The first boundary condition says that at $x=0$, $u=\frac{u_x}{a_0}$. The second says that at $x=L$, $u=\frac{u_x}{-a_L}$. So now let's consider the first term in the integral in \ref{eq:c}:
uu_x|_0^L = u_x(L,t)\frac{u_x(L,t)}{-a_L} - u_x(0,t)\frac{u_x(0,t)}{a_0} = \frac{u_x^2}{-a_L} - \frac{u_x^2}{a_0} = -\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big)

So, now let's put this back into \ref{eq:c} to yield:
\frac{\partial}{\partial t}E(t) = -2k\Big[\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big) + \int_0^Lu_x^2dx\Big]

Since all functions and the integrand are to the power of 2, they are all positive; all constants ($a_L$, $a_0$, and k) are also defined to be positive. This implies that overall,
-2k\Big[\Big(\frac{u_x^2(L,t)}{a_L} + \frac{u_x^2(0,t)}{a_0}\Big) + \int_0^Lu_x^2dx\Big] \leq 0

Therefore the energy is decreasing or zero, and in fact the endpoints contribute to this decrease. Note that in this case the Robin boundary condition prevents a constant solution from being possible, since for $C\neq0$:
\frac{d}{dx}C -a_0C = 0 - a_0C  = -a_0C\neq 0
(And likewise for the other condition).
The only case where a constant solution is valid is the trivial solution $u(x,t) = 0$. In this case only will the equality to zero occur.
Therefore the endpoints contribute to the decrease of $E(t)$.

13
##### Textbook errors / Missing reference in HA5 (Problems to 3.2)
« on: October 17, 2015, 06:08:03 PM »
Hi Professor, I noticed that in Remark 1 in HA5 question 7, you have a link to an equation written as ().

Fixed. It is not a missing but broken reference. In LaTeX if \ref {mylabel} is written (without space) (and \ref has some variants like \pageref, \footref and some package-specific) but \label{mylabel} is missing (or misspelled) then \ref{mylabel}  appears.  MathJax has only \ref

PS On our forum \ref{x} appears as an emoticon (one needs to be careful with these things on forums, blogs, etc)

Oops, sorry about that! -Emily

14
##### HA4 / HA4-P1
« on: October 09, 2015, 05:07:44 PM »
Problem 1: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.1

(a) Here we have a Dirichlet boundary condition. In the domain ${t>0, x \geqslant ct}$, the solution is given by the D'Alembert formula:

u(x,t) = \frac{1}{2}[g(x+ct) + g(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy

So plugging in the initial conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi(x+ct)

In the domain ${0<x<ct}$, the solution is given by:

u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_{0}^{x+ct} h(y)dy + p\bigg(t - \frac{x}{c}\bigg) - \frac{1}{2}g(ct-x) - \frac{1}{2c} \int_0^{ct-x} h(y)dy

So plugging in the initial and boundary conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x)

So the solution to (a) is

u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}

(b) In this case we have a Neumann boundary condition. The solution for the domain ${t>0, x \geqslant ct}$ is the same as in part (a), namely:

u(x,t) = \phi(x+ct)

For the domain ${0<x<ct}$, the solution is given by:

u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_0^{x+ct} h(y)dy - c \int_0^{t-\frac{x}{c}} q(t')dt' + \frac{1}{2}g(ct-x) + \frac{1}{2c} \int_0^{ct-x} h(y)dy

Plugging in our initial and boundary conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x)

where

\int\chi(t')dt' = X(t')

So the solution to (b) is

u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}

15
##### Quiz 2 / Solution to Quiz 2
« on: October 08, 2015, 10:57:27 PM »
Here is my solution. The process is the same as HA3-Problem2 (http://forum.math.toronto.edu/index.php?topic=627.0) so I will not post that. Please let me know if you notice any errors or typos.

Professor, please excuse my ugly graph---I do not know how to make one as nice as you did in Problem 2! (http://forum.math.toronto.edu/index.php?topic=627.0) I also don't know how to put the image in the post as you did, so I will attach it here.

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