Messages

16

Test 2 / Re: TT2-P2

« on: November 19, 2015, 01:04:42 AM »

I believe there may be a mistake in this one. As we did in HA7, we must change the given condition at $y=0$ by Fourier transform as well. So we will have (making use of the hint):

\begin{equation}
u|_{y=0} \longrightarrow \hat{u}|_{y=0} = \hat{\frac{1}{x^2 + 1}} = \frac{1}{2}e^{-|k|}
\end{equation}

We should be able to proceed similarly to what Xi Yue Wang has done, with the integral now being:

\begin{equation}
u(x,y)=\frac{1}{2} \int_{-\infty}^{\infty} e^{-|k|y}e^{-|k|}e^{ikx} dk
\end{equation}

It's actually a simple integral to solve; we split it up so that we have:

\begin{equation}
u(x,y)=\frac{1}{2} \left( \int_{-\infty}^{0} e^{ky}e^{k}e^{ikx} dk +  \int_{0}^{\infty} e^{-ky}e^{-k}e^{ikx} dk \right)
\end{equation}

So we get:

\begin{equation}
u(x,y) = \frac{1}{2} \left( \frac{e^{ky+k+ikx}|_{-\infty}^0}{y + 1 + ix} - \frac{e^{-ky-k+ikx}|_{0}^{\infty}}{y + 1 - ix} \right)
\end{equation}

Evaluating, we get:

\begin{equation}
u(x,y) = \frac{1}{2} \left( \frac{1}{y + 1 + ix} + \frac{1}{y + 1 - ix} \right)
\end{equation}

The final answer is then:

\begin{equation}
u(x,y) = \frac{y+1}{(y+1)^2 + x^2}
\end{equation}

17

Test 2 / Re: TT2-P4

« on: November 19, 2015, 12:34:51 AM »

We can go one step further by noting that all $a_n$ for even $n$ are zero. Then $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1,  odd}\frac{r^n\sin(n\theta)}{n(n+1)}$$

Of course! Thank you Fei Fan Wu!

18

Test 2 / Re: TT2-P4

« on: November 19, 2015, 12:25:56 AM »

Edit: Note, I accidentally wrote $\phi$ instead of $\theta$ throughout this problem. I do not wish to go back and change them all. Please consider $\phi$ as if I had written $\theta$!

As usual, let's separate variables. We let:

\begin{equation}
u(x,t) = \Phi(\phi)R(r)
\end{equation}

Plugging this into the given equation yields:

\begin{equation}
\Phi R'' + \frac{1}{r} \Phi R' + \frac{1}{r^2} \Phi'' R = 0 \longrightarrow \left( \frac{r^2R''}{R} + \frac{rR'}{R} \right) + \frac{\Phi''}{\Phi} = 0
\end{equation}

Now as usual, we have a term that depends on $R$ and a term that depends on $\Phi$. So we set the $\Phi$ term equal to $\lambda$ and solve the resulting ODE.

\begin{equation}
\Phi'' - \lambda\Phi = 0
\end{equation}

Let's first consider $\lambda = - \omega^2 < 0$. This yields a solution:

\begin{equation}
\Phi(\phi) = A\cos\omega\phi + B\sin\omega\phi
\end{equation}

Using the BCs, we show that:

\begin{equation}
\Phi(0) = A = 0
\end{equation}

\begin{equation}
\Phi(\pi) = B\sin\omega\pi = 0 \longrightarrow \omega\pi = n\pi \longrightarrow \omega = n
\end{equation}

So for negative eigenvalues we conclude that $\Phi_n(\phi) = B_n\sin(n\phi)$ with $\lambda = -n^2$.

Let's check the positive case ($\lambda = \omega^2 > 0$). Wellllll, this will yield:

\begin{equation}
\Phi(\phi) = C\cosh\omega\phi + D\sinh\omega\phi
\end{equation}

Using the boundary conditions here gives us:

\begin{equation}
\Phi(0) = C = 0
\end{equation}

\begin{equation}
\Phi(\pi) = D\sinh\omega\pi = 0 \longrightarrow \omega\pi = 0 \longrightarrow \omega = 0
\end{equation}

But we are checking the case where $\omega$ is a positive (i.e. greater than zero) eigenvalue, so this doesn't make sense. We conclude that there are no positive eigenvalues.

For $\lambda = 0$, the solution will be $\Phi(\phi) = E\phi + F$. The first BC gives $\Phi(0) = F \longrightarrow F = 0$. The second gives $\Phi(\pi) = E\pi = 0 \longrightarrow E = 0$. We conclude that there is no zero eigenvalue.

Now we must consider the $R$ equation. Recall that we have:

\begin{equation}
\left( \frac{r^2R''}{R} + \frac{rR'}{R} \right) = - \lambda \longrightarrow r^2R'' + rR' + \lambda R = 0
\end{equation}

This is a typical Euler equation, so we assume a solution $R = r^m$. Plugging this in gives:

\begin{equation}
r^2m(m-1)r^{m-2} + rmr^{m-1} + \lambda r^m = 0 \longrightarrow m(m-1) + m + \lambda = 0 \longrightarrow m^2 - n^2 = 0 \longrightarrow m = \pm n
\end{equation}

The solutions are then:

\begin{equation}
R(r) = Gr^n + Hr^{-n}
\end{equation}

Well where are we? In the half-disk! So we discard the $H$ term, which will blow up at the origin. This leaves us with $R_n(r) = G_nr^n$.

Now let's combine the solutions, as we usually do. We can write our general solution as:

\begin{equation}
u(r, \phi) = \sum_1^{\infty}a_nr^n\sin(n\phi)
\end{equation}

Now we shall make use of this Robin boundary condition. We have that:

\begin{equation}
u(r, \phi) |_{r=1} = \sum_1^{\infty}a_n\sin(n\phi)
\end{equation}

\begin{equation}
u_r(r, \phi) |_{r=1} = \sum_1^{\infty}a_nn\sin(n\phi)
\end{equation}

Altogether, this is:

\begin{equation}
(u_r + u) |_{r=1} = \sum_1^{\infty} \left(a_nn\sin(n\phi) + a_n\sin(n\phi) \right) = 1 \longrightarrow \sum_1^{\infty} \left(a_n\sin(n\phi)(n+1) \right) = 1
\end{equation}

Now we'd better find out what $a_n$ is. So:

\begin{equation}
a_n = \frac{2}{\pi(n+1)} \int_0^{\pi}1\sin(n\phi)d\phi = \frac{2(1 - \cos(n\pi))}{\pi(n+1)n} = \frac{2(1 - (-1)^n)}{\pi(n+1)n}
\end{equation}

So we finally get that:

\begin{equation}
u(r, \phi) = \sum_1^{\infty} \frac{2(1 - (-1)^n)}{\pi(n+1)n} r^n\sin(n\phi)
\end{equation}

If anyone notices any errors, please let me know!

19

Website errors / Re: Link to sections for Week 10

« on: November 18, 2015, 11:28:31 PM »

Thank you professor!

20

Test 2 / Re: TT2-P5

« on: November 18, 2015, 11:22:32 PM »

Indeed, I got the same. Thank you for commenting Fei Fan Wu.

21

Test 2 / Re: TT2-P5

« on: November 18, 2015, 11:12:03 PM »

I think you accidentally added a factor of $\frac{1}{2}$ in your answer.

22

Website errors / Re: Link to sections for Week 10

« on: November 18, 2015, 06:26:05 PM »

Well, I suppose I should have posted this as a textbook error as when I check the table of contents for the online textbook it appears the link for 9.1 does not work there either!


http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/contents.html

The error seems to be that you have a comma before the html at the end of the link:

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter9/S9.1,html

23

Website errors / Re: Link to sections for Week 10

« on: November 18, 2015, 06:20:28 PM »

I just noticed the case is the same for section 9.2 as well.

24

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

25

Textbook errors / Re: Section 5.2 Theorem 3.c)

« on: November 18, 2015, 02:12:13 PM »

Great insight, Fei Fan Wu! Just for anyone else wondering, the link to this error is here:

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.2.html

26

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#6)

« on: November 18, 2015, 12:20:53 PM »

By the way, this is only part a. I will respond with part b later!

27

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#6)

« on: November 18, 2015, 12:20:03 PM »

Can you please clarify what you mean by $c_{1/2}$ in the line above equation (6)?

Hi Fei Fan Wu,

Thank you for your excellent question. By $c_{1/2}$, I meant that the relation applies for either $c_1$ or $c_2$.

28

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#6)

« on: November 18, 2015, 11:55:53 AM »

The energy that we want to conserve is given by:

\begin{equation}
E(t) = E_1(t) + E_2(t) = \frac{m_1}{2}\int_0^{\infty}(u_t^2 + c_1^2u_x^2)dx + \frac{m_2}{2}\int_{-\infty}^{0}(u_t^2 + c_2^2u_x^2)dx
\end{equation}

To make the energy conserved, we take the time derivative of $E(t)$ and set it equal to zero. So:

\begin{equation}
\frac{dE}{dt} = \frac{m_1}{2}\int_0^{\infty}(2u_tu_{tt} + c_1^22u_xu_{xt})dx + \frac{m_2}{2}\int_{-\infty}^0(2u_tu_{tt} + c_2^2u_xu_{xt})dx
\label{eq:timeder}
\end{equation}

Doing integration by parts on the second term of the first integral (where the result will extend to the second term of the second integral), we get:

\begin{equation}
\int_0^{\infty}u_xu_{xt}dx = u_xu_t|_0^{\infty}-\int_0^{\infty}u_{xx}u_tdx
\label{eq:parts}
\end{equation}

Plugging \ref{eq:parts} into \ref{eq:timeder}, we get:

\begin{equation}
\frac{dE}{dt} = m_1 \left( \int_0^{\infty}u_tu_{tt}dx + c_1^2u_xu_t|_{-\infty}^0 - c_1^2\int_0^{\infty}u_{xx}u_tdx \right) + m_2 \left( \int_{-\infty}^0u_tu_{tt}dx + c_2^2u_xu_t|_0^{\infty} - c_2^2\int_{-\infty}^0u_{xx}u_tdx \right)
\label{eq:terms}
\end{equation}

For each term, we can combine the integrals. For example, for the first term we have:

\begin{equation}
m_1 \int_0^{\infty} \left( u_tu_{tt}dx - c_1^2u_{xx}u_tdx \right) = m_1 \int_0^{\infty} \left( u_t \left(u_{tt} - c_1^2u_{xx} \right) dx \right)
\label{eq:rearrange}
\end{equation}

Recall from the problem that $u_{tt} - c_{1/2}^2u_{xx} = 0$. Therefore we get:

\begin{equation}
\frac{dE}{dt} = m_1c_1^2u_xu_t|_0^{\infty} + m_2c_2^2u_xu_t|_{-\infty}^0
\end{equation}
\begin{equation}
\frac{dE}{dt} = m_1c_1^2u_x(\infty)u_t(\infty) - m_1c_1^2u_x(0)u_t(0) + m_2c_2^2u_x(0)u_t(0) - m_2c_2^2u_x(-\infty)u_t(-\infty)
\end{equation}

Here we must assume that the solutions are fast-decaying, so the terms at infinity go to zero. We then get:

\begin{equation}
\frac{dE}{dt} = - m_1c_1^2u_x(0)u_t(0) + m_2c_2^2u_x(0)u_t(0)
\end{equation}

Since we wanted the energy to be conserved, we set this to zero. We can then solve for the required relation between $m_1$ and $m_2$.

\begin{equation}
- m_1c_1^2u_x(0)u_t(0) + m_2c_2^2u_x(0)u_t(0) = 0 \longrightarrow m_1c_1^2u_x(0)u_t(0) = m_2c_2^2u_x(0)u_t(0)
\end{equation}
\begin{equation}
\longrightarrow m_1c_1^2 = m_2c_2^2 \longrightarrow m_1 = \frac{m_2c_2^2}{c_1^2}
\end{equation}

Therefore we can conclude that for the energy to be conserved, $m_1:m_2 = \frac{c_2^2}{c_1^2}$.

29

HA7 / Re: HA7-P6

« on: November 16, 2015, 08:22:12 AM »

@Chi Ma, thank you!

30

HA7 / Re: HA7-P6

« on: November 15, 2015, 11:13:35 PM »

For the solution to part a, at the very end, why did we put $(x-x')$? Thank you!