HA8-P3

[Victor Ivrii]

Problem 3
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.3

[Bruce Wu]

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]

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]

Yes, I wanted to ask to use Fourier method rather than the formula—and you did it

[Bruce Wu]

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]

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 it，so 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]

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 it，so 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]

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]

Thank you Emily, I know (10) and (11), but it doesn't meantion the replacement.  . Maybe it from lectures.

[Emily Deibert]

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.

[Victor Ivrii]

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