# Laplace equation with boundary condition

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?

• Hint: There are values of $n$ that make both coefficients undefined. Extra hint: $\cos 2\theta = \cos^2\theta - \sin^2\theta$ – Ninad Munshi Jun 18 '20 at 5:09

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. .