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Let $$ \Delta u(r,\theta )=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta }. $$ Solve $$ \begin{cases} \Delta u(r,\theta)=0&\text{ for }0<r<3, -\pi<\theta<\pi\\ u(3,\theta)=f(\theta)=\cos\dfrac{\theta}{2}&\text{ for }-\pi\leq\theta\leq\pi \end{cases} $$ What I've tried:

We will take solutions given by $u(r,\theta)=\sum_{n=0}^\infty 3^n(A_n\cos n\theta +B_n\sin n\theta )$ And will use the boundary condition $f(\theta )=u(3,\theta )$. Thus we have $$ \begin{align} A_n=\frac{1}{\pi }\int_{-\pi }^{\pi }f(\theta )\cos(n\theta ) d\theta\\ B_n=\frac{1}{\pi }\int_{-\pi }^{\pi }f(\theta )\cos(n\theta ) d\theta \end{align} $$ Can anybody help me figure this out? And am I on doing it correctly? Thanks a lot

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  • $\begingroup$ There seems to be some information missing. You need boundary conditions on $r=0$, $r=3$, $\theta=-\pi$ and $\theta=\pi$. $\endgroup$ – Thales Sep 11 at 11:40
  • $\begingroup$ So it should be $r^n$ not $3^n$ in the general expression for $u$ but in terms of computing the coefficients this looks right (besides the typo in $B_n$). Now can you compute the integrals? $\endgroup$ – Ian Sep 11 at 12:30

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