# Toronto Math Forum

## APM346-2016F => APM346--Tests => Q6 => Topic started by: XinYu Zheng on November 10, 2016, 08:54:19 PM

Title: Q6
Post by: XinYu Zheng on November 10, 2016, 08:54:19 PM
Solve
$$\begin{cases} u_{xx}+u_{yy}=0& r<a\\ u_r|_{r=a}=f(\theta) \end{cases}$$
Where
$$f(\theta)=\begin{cases} 1 & 0<\theta<\pi\\ -1 & \pi<\theta <2\pi \end{cases}$$

BONUS:
(a) What is a necessary and sufficient condition on $f(\theta)$ for solution to exist? Is this condition satisfied here?
(b) Is the solution unique?

Solution:
For Laplace's equation in the inner disk we know that the general solution takes the form
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1} r^n (A_n\sin(n\theta)+B_n\cos(n\theta))$$
Where we have dropped the logarithm and any terms with $r^{-n}$. Applying the boundary condition we have
$$f(\theta)=\sum_{n\geq 1}n a^{n-1}(A_n\sin(n\theta)+B_n\cos(n\theta))$$
At this point, the coefficients can be directly calculated:
$$A_n=\frac{a^{1-n}}{n\pi}\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta=\frac{a^{1-n}}{n\pi}\left(\int_0^{\pi}\sin(n\theta)\,d\theta+\int_\pi^{2\pi}-\sin(n\theta)\,d\theta\right)=\frac{a^{1-n}}{n^2\pi}(\cos(n\theta)|_{\pi}^0+\cos(n\theta)|_\pi^{2\pi})=\frac{4a^{1-n}}{n^2\pi}\,\,\,\text{n odd, 0 otherwise}$$
Similarly,
$$B_n=\frac{a^{1-n}}{n\pi}\int_0^{2\pi}f(\theta)\cos(n\theta)\,d\theta=\frac{a^{1-n}}{n\pi}\left(\int_0^{\pi}\cos(n\theta)\,d\theta+\int_\pi^{2\pi}-\cos(n\theta)\,d\theta\right)=0$$
Where the integrals are 0 because we will be evaluating sine functions at integer multiples of $\pi$. Thus we have our solution:
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1, n\,\,odd}\frac{4a^{1-n}}{n^2\pi}r^n\sin(n\theta)$$

BONUS:
(a) Note that in the fourier expansion of $f(\theta)$ we have no constant term. Thus for solution to exist we must demand
$$\int_0^{2\pi}f(\theta)\,d\theta=0$$
Which is fulfilled here.
(b) No. In Neumann problems the solution is defined up to a constant, in this case $A_0/2$.