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**APM346--Misc / Thank You!**

« **on:**December 22, 2015, 06:35:01 PM »

Thank you for a great course professor! Happy holidays!

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Thank you for a great course professor! Happy holidays!

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Hi Professor,

On the course website, you say that studying HA1-10 is a good preparation for the final exam, but in class you said that HA10 will not appear on the exam. Could you clarify whether or not we should study this section?

On the course website, you say that studying HA1-10 is a good preparation for the final exam, but in class you said that HA10 will not appear on the exam. Could you clarify whether or not we should study this section?

3

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

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

leads to a 404 missing page error.

Fixed

4

The link to section 9.1 for week 10 is broken.

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

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

5

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter7/S7.P.html#problem-7.P.2

The link in this problem is broken.

Fixed

The link in this problem is broken.

Fixed

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7

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Hi Professor, I think you made a mistake here by putting $dy^2$ twice in the equation.

Indeed

Indeed

9

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

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

The best always fall between cracks

10

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

Indeed

11

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:

\begin{equation}

\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

\end{equation}

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:

\begin{equation}

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

\end{equation}

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{equation}

\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}

\end{equation}

Now plugging these all into the problem, we have:

\begin{equation}

\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

\end{equation}

\begin{equation}

\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

\end{equation}

\begin{equation}

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

\end{equation}

\begin{equation}

v''(r) = k^2v

\end{equation}

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:

\begin{equation}

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

\end{equation}

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

\begin{equation}

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

\end{equation}

(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:

\begin{equation}

\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

\end{equation}

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:

\begin{equation}

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

\end{equation}

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{equation}

\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}

\end{equation}

Now plugging these all into the problem, we have:

\begin{equation}

\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

\end{equation}

\begin{equation}

\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

\end{equation}

\begin{equation}

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

\end{equation}

\begin{equation}

v''(r) = k^2v

\end{equation}

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:

\begin{equation}

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

\end{equation}

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

\begin{equation}

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

\end{equation}

12

Just wondering if we will have quiz 4 on the 29th of October, or if we will have no quiz next week. Thank you!

13

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

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

14

(a) We consider the energy as given in the problem: \begin{equation}

E(t) = \int_{J} u^2(x,t)dx

\end{equation}

Now, take the time derivative to yield: \begin{equation}

\frac{\partial}{\partial t}E(t) = \int_{J} \frac{\partial}{\partial t}u^2(x,t)dx = \int_{J} 2uu_tdx \end{equation}

Replacing $u_t$ with $ku_{xx}$ as per the heat equation, we get: \begin{equation}

\int_J 2kuu_{xx}dx \end{equation}

Now, using integration by parts, we will get: \begin{equation}

2k\Big[uu_x|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u_x^2dx\Big] \end{equation}

Here I assume that the function is fast-decaying, so the first term goes to 0

-2k\int_{-\infty}^{\infty} u_x^2dx \label{eq:integral}\end{equation}

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: \begin{equation}

\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big] \label{eq:newint} \end{equation}

We consider the cases of both Dirichlet and Neumann boundary conditions. First, let's think about Dirichlet. This condition says that: \begin{equation}

u|_{x=0} = u|_{x=L} = 0 \end{equation}

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: \begin{equation}

u_x|_{x=0} = u_x|_{x=L} = 0. \end{equation}

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: \begin{equation}

\frac{\partial}{\partial t}E(t) = -2k\int_0^Lu_x^2dx \end{equation}

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: \begin{equation}

\frac{\partial}{\partial t}E(t) = 2k\Big[uu_x|_0^L - \int_0^Lu_x^2dx\Big]

\label{eq:c}

\end{equation}

This time, we have Robin boundary conditions. They are: \begin{equation}

\begin{cases}

u_x(0,t) - a_0u(0,t) = 0 \\

u_x(L,t) + a_Lu(L,t) = 0

\end{cases}

\end{equation}

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}: \begin{equation}

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)

\end{equation}

So, now let's put this back into \ref{eq:c} to yield: \begin{equation}

\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]

\end{equation}

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, \begin{equation}

-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

\end{equation}

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 $C\neq0$: \begin{equation}

\frac{d}{dx}C -a_0C = 0 - a_0C = -a_0C\neq 0 \end{equation}

(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 $E(t)$.

15

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*

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)

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