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Solve Laplace's equation in polar coordinates $$ \frac {1}{r} \frac {\partial u} {\partial r} + \frac {\partial^2 u} {\partial r^2} + \frac {1} {r^2} \frac {\partial^2 u} {\partial \theta^2} = 0$$ on the disk $$ {{(r, \theta) | 0 \leq r \leq R , 0 \leq \theta \leq 2}} $$ subject to the boundary condition $ u (R, \theta) = Tsin^2 (\theta) $

I got $ u (r, \theta) = \sum_{n=0}^{\infty}r^n [a_ncos (n\theta)+b_nsin (n\theta)] $ for $ n \in \mathbb{N} $

And solvin for the condition using

$$ a_n= 1/\pi \int_{0}^{2\pi} Tsin^2 (\theta)cos (n\theta) d\theta $$and

$$ b_n= 1/\pi \int_{0}^{2\pi} Tsin^2 (\theta)sin(n\theta) d\theta $$

I get $$ a_n = \frac{2Tsin (2\pi n)}{4n \pi-n^3 \pi} $$ and $$ b_n = \frac{4Tsin^2(\pi n)}{4n \pi-n^3 \pi} $$

But $ sin (n\pi)=0 $ for $n \in \mathbb{N} $ so i would get $ u (r, \theta) = 0 $

What should i consider for solving it?

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  • $\begingroup$ Hint: There are values of $n$ that make both coefficients undefined. Extra hint: $\cos 2\theta = \cos^2\theta - \sin^2\theta$ $\endgroup$ – Ninad Munshi Jun 18 '20 at 5:09
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You have to work a little harder to write down the exact values of $a_n$ and $b_n$.

$\int_0^{2\pi} sin ^{2}\theta \cos (n\theta) d\theta=\frac 1 2\int_0^{2\pi} (1-\cos (2\theta) \cos (n\theta) d\theta$ using the identity $cos (2\theta) \cos (n\theta)=\frac 1 2 (\cos (n+2) \theta -\cos (n-2) \theta$ this becomes $\pi \delta_{0,n} -\frac {\pi} 2 \delta_{2,n} -\frac {\pi} 2 \delta_{-2,n}$ wheer $\delta_{i.j}=1$ if $i=j$ and $0$ if $i \neq j$.

I will let you handle $b_n$ by a similar method. .

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