# Solution to Laplace's equation on the circle $0<r<1$

Consider Laplace's equation in polar coordinates $$\frac{\partial ^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial ^2u}{\partial \theta^2}=0$$ for $0<r<1$. I'm given the boundary condition $u(1,\theta)=\cos^2(\theta)$.

I know that the general solution is given by $$u(r,\theta)=b_0+\sum_{n=1}^{\infty}r^n\left(a_n\sin (n\theta) +b_n\cos(n\theta) \right)$$ with \begin{align*} b_0&=\frac{1}{2\pi}\int_0^{2\pi}\cos^2(\theta) \, d\theta \\ b_n&=\frac{1}{\pi}\int_0^{2\pi} \cos^2(\theta)\cos(n\theta) \, d\theta \\ a_n & = \frac{1}{\pi}\int_0^{2\pi}\cos^2(\theta)\sin(n\theta) \, d\theta \end{align*}

It is clear to me that here we have $a_n=0$ and $b_0=\frac{1}{2}$ but I'm not sure how to find an explicit form for $b_n$.

• You could probably use some addition formulas to pull the $\cos^2(\theta)\cos(n \theta)$ into a simpler expression involving sums of cosines. – Matt Apr 3 '17 at 16:51

Recall that $\cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$ and exploit orthogonality $\int_0^{2\pi}\cos(n\theta)\cos(m\theta)\,d\theta=0$ for $m\ne n$.
Use the double angle formula, $\cos^2 (\theta) = \frac{1 + \cos (2 \theta)}{2}$ and the orthogonality relations: $$\int\limits_0^{2 \pi} \cos (m x) \cos (n x) dx = 0,$$ if $m \neq n$. Then you get that the only non-zero term (other than $b_0$) is $b_2$, for which you get $$b_2 = \frac{1}{\pi} \int\limits_0^{2 \pi} \cos^2 (2 x) dx = 1.$$