### Author Topic: HA8-P3  (Read 2767 times)

#### Victor Ivrii

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##### HA8-P3
« on: November 07, 2015, 07:19:01 AM »

#### Bruce Wu

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##### Re: HA8-P3
« Reply #1 on: November 08, 2015, 01:37:18 AM »
a) The general solution is given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.10, and Fourier coefficients by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.12.
We have $$A_{n}=\frac{a^{-n}}{\pi}\left[\int^{\pi}_{0}\cos(n\theta')d\theta'-\int^{2\pi}_{\pi}\cos(n\theta')d\theta'\right]=0$$
$$C_{n}=\frac{a^{-n}}{\pi}\left[\int^{\pi}_{0}\sin(n\theta')d\theta'-\int^{2\pi}_{\pi}\sin(n\theta')d\theta'\right]=\frac{a^{-n}}{\pi}\left[\frac{1-\cos(n\pi)}{n}-\frac{\cos(n\pi)-\cos(2n\pi)}{n}\right]=\frac{2a^{-n}[1-\cos(n\pi)]}{\pi n}$$
Only odd $n$ terms survive. In that case $$C_n=\frac{4a^{-n}}{n\pi}$$
The final solution is $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{n}a^{-n}}{n}\sin(n\theta)$$

#### Bruce Wu

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##### Re: HA8-P3
« Reply #2 on: November 08, 2015, 01:46:54 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
$$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{-n}a^{n}}{n}\sin(n\theta)$$

#### Victor Ivrii

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##### Re: HA8-P3
« Reply #3 on: November 08, 2015, 09:33:11 AM »
Yes, I wanted to ask to use Fourier method rather than the formulaâ€”and you did it

#### Bruce Wu

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##### Re: HA8-P3
« Reply #4 on: November 08, 2015, 09:46:49 AM »
To me the Fourier method looked easier than using the formula. Explicit forms are important but aren't always so nice to work with.

#### Rong Wei

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##### Re: HA8-P3
« Reply #5 on: November 10, 2015, 08:34:45 PM »
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

#### Emily Deibert

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##### Re: HA8-P3
« Reply #6 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.

#### Emily Deibert

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##### Re: HA8-P3
« Reply #7 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

#### Rong Wei

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##### Re: HA8-P3
« Reply #8 on: November 12, 2015, 03:56:06 PM »
Thank you Emily, I know (10) and (11), but it doesn't meantion the replacement.  . Maybe it from lectures.

#### Emily Deibert

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##### Re: HA8-P3
« Reply #9 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.