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### Messages - Emily Deibert

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31
##### Textbook errors / HA7-P2
« on: November 15, 2015, 10:22:20 PM »

32
##### HA9 / Re: HA9-P3
« on: November 13, 2015, 07:11:19 PM »
Professor, what is the question being asked here?

33
##### HA9 / HA9-P3
« on: November 13, 2015, 07:10:15 PM »

34
##### HA9 / Re: HA9-P1
« on: November 13, 2015, 07:36:12 AM »
(a) As in the previous home assignment, we're asked to find solutions of the 2-D Laplace equation that rely only on $r$. Recall from home assignment 8 that for solutions depending only on $r$, our problem in polar coordinates will reduce to:

u_{rr} + \frac{1}{r}u_r = 0 \longrightarrow ru_{rr} + u_r = 0

Notice that this is the derivative of $ru_r$, so our equation becomes:

\left( r u_r \right)' = 0

This is now a simple problem; the solution is as follows:

ru_r = c_1 \longrightarrow u_r = \frac{c_1}{r} \longrightarrow u = c_1\ln{r} + c_2

Thus the solution is:

u(r) = c_1\ln{r} + c_2

(b) This is a similar problem, but now we are dealing with a 3-D Laplace equation. Recall that when we switch to spherical coordinates, the Laplacian becomes:

\Delta = \partial_{\rho}^2 + \frac{2}{\rho}\partial_{\rho} + \frac{1}{\rho^2}\Lambda

Now, can we get rid of any of the terms in the Laplacian here, as we did in part one? Indeed, since we are looking for solutions that depend only on $\rho$ and $\Lambda$ by definition has no $\rho$-dependence. We proceed to reduce our problem to:

u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0

Proceeding as in the previous problem:

u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0 \longrightarrow \rho u_{\rho \rho} + 2u_{\rho} = 0

This is a Euler-Cauchy equation, so we assume a solution $\rho^m$. Plugging this in yields:

\rho(m)(m-1)\rho^{m-2} + 2m\rho^{m-1} = 0 \longrightarrow (m)(m-1)\rho^{m-1} + 2m\rho^{m-1} = 0

We look for the roots of the characteristic equation:

m(m-1) + 2m = 0 \longrightarrow m^2 - m + 2m = 0 \longrightarrow m^2 + m = 0 \longrightarrow m(m+1) = 0 \longrightarrow m = 0, -1

The roots are $m=0$ and $m= -1$. Plugging these into our solution $\rho^m$ and forming the general solution as a linear combination of these two solutions yields the final answer:

u(\rho) = \frac{c_1}{\rho} + c_2

35
##### HA9 / HA9-P1
« on: November 13, 2015, 07:10:47 AM »

36
##### HA8 / Re: HA8-P3
« on: November 12, 2015, 06:52:46 PM »
I think Fei Fan Wu just meant that the formulae are the same aside from those "replacements" he mentioned. But it is not an actual mathematical strategy to just replace these terms, he just didn't want to write it all out again.

37
##### HA7 / Re: HA7-P3
« on: November 12, 2015, 09:10:29 AM »
As far as I understand, there is technically a coefficient $\kappa$ in front of the FT and IFT---$\frac{\kappa}{2\pi}$ in front of the FT and $\frac{1}{\kappa}$ in front of the IFT. When I spoke to Professor Ivrii about this, he said that as long as we maintain the same value of $\kappa$ throughout our answers, we should get full marks. But perhaps Professor Ivrii can further clarify this.

In reference to problem one, we need the FT to be unitary, which would require us to choose a value of $\kappa=\sqrt{2\pi}$, I believe.

38
##### HA8 / Re: HA8-P4
« on: November 11, 2015, 02:23:09 PM »
By the way, we can only satisfy the problem if $h(\theta)$ is such that $\int_0^{2\pi}h(\theta)d\theta = 0$. This is clearly satisfied in this case.

39
##### 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

40
##### HA8 / Re: HA8-P3
« on: November 11, 2015, 09:44:40 AM »
To clarify, I am talking about the section Fei Fan Wu linked to, section 6.4: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11

41
##### HA8 / Re: HA8-P3
« on: November 11, 2015, 09:39:49 AM »
b) General solution is given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11. Fourier coefficients are the same as before, except we replace $A_n$ with $B_n$, $C_n$ with $D_n$, and $a^{-n}$ with $a^{n}$. The final solution is
$$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{-n}a^{n}}{n}\sin(n\theta)$$
I have question about itï¼Œso under this conditions given by Question 3, we can apply that replacing $A_n$ with $B_n$, $C_n$ with $D_n$, and $a^{-n}$ with $a^{n}$? where is it from?
Thank you! Fei Fan

@ Rong Wei, this is also described in the textbook. Eq. (10) and (11) in that section describe the solutions in the disk and outside of the disk, which are what we are looking for in this problem. The solutions differ in the way mentioned by Fei Fan Wu in the post.

42
##### 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

43
##### HA8 / Re: HA8-P4
« on: November 08, 2015, 12:07:45 PM »
(b) Part b is similar, but we consider equations outside the disk. This time the general solution is given by  http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11. Much of the solution is the same as part (a) so I will omit several steps. Note that the derivative in this case is:

u_r(r. \theta{}) = \sum_{n=1}^{\infty}-nr^{-n-1}(B_n\cos(n\theta{}) + D_n\sin(n\theta{}))

As with the $A_n$ terms in part (a), the $B_n$ terms will go to zero. The $D_n$ terms will become:

D_n = \frac{-a^{n+1}2(1-\cos(n\pi{}))}{\pi{}n^{2}}

Only odd-n terms survive; so:

D_n = \frac{-4a^{n+1}}{\pi{}n^2}

Plugging into the general solution, we get a final answer:

u(r, \theta{}) = -\frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{a^{n+1} \sin(n \theta{})}{n^2 r^n}

44
##### HA8 / Re: HA8-P4
« on: November 08, 2015, 11:51:00 AM »
As in problem three, we have a general solution (before imposing the boundary condition) given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.10. Now we must differentiate this general solution in order to make use of the boundary conditions. This yields:

u_r(r. \theta{}) = \sum_{n=1}^{\infty}nr^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{}))

Evaluating this at a, as per the boundary condition:

u_r(a, \theta{}) = \sum_{n=1}^{\infty}na^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{})) = f(\theta)

Using the Fourier method, we find the forms of the coefficients to be:

A_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\cos(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\cos(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\cos(n\theta{}')d\theta{}'\Big]  = 0

and

C_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\sin(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\sin(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\sin(n\theta{}')d\theta{}'\Big] \longrightarrow $$= \frac{1}{\pi{}na^{n-1}}\Big[\frac{1-\cos(n\pi)}{n}-\frac{\cos(n\pi)-\cos(2n\pi)}{n}\Big]= \frac{1}{\pi{}na^{n-1}} \frac{2(1-\cos(n\pi{}))}{n}$$

So:
C_n = \frac{2(1-\cos(n\pi{}))}{\pi{}n^{2}a^{n-1}}

As in the previous problem, only terms with odd n survive. This results in:

C_n = \frac{4}{\pi{}n^{2}a^{n-1}}

Now, plugging these coefficients back into the general solution, we arrive at the solution:

u(r, \theta{}) = \frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{r^n\sin(n\theta{})}{n^{2}a^{n-1}}

45
##### 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

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