# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA8 => Topic started by: Victor Ivrii on November 07, 2015, 07:19:01 AM

Title: HA8-P3
Post by: Victor Ivrii on November 07, 2015, 07:19:01 AM
Problem 3
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.3 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.3)
Title: Re: HA8-P3
Post by: Bruce Wu 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 $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{n}a^{-n}}{n}\sin(n\theta)$$
Title: Re: HA8-P3
Post by: Bruce Wu 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
$$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{-n}a^{n}}{n}\sin(n\theta)$$
Title: Re: HA8-P3
Post by: Victor Ivrii 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
Title: Re: HA8-P3
Post by: Bruce Wu 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.
Title: Re: HA8-P3
Post by: Rong Wei 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
Title: Re: HA8-P3
Post by: Emily Deibert 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.
Title: Re: HA8-P3
Post by: Emily Deibert 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
Title: Re: HA8-P3
Post by: Rong Wei 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.
Title: Re: HA8-P3
Post by: Emily Deibert 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.
Title: Re: HA8-P3
Post by: Victor Ivrii on December 04, 2015, 09:47:46 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.

Many things follows from formula. Several tools better than one, anyway