Solve Delta u=0 for r≤ R with u|_(r=R)=f(varphi)

Solve $\Delta u=0$ for $r\le R$ with $u|_{r=R}=f(\varphi)$.

Step 1 · Separate variables

In polar coordinates, $u_{rr}+\frac1ru_r+\frac{1}{r^2}u_{\varphi\varphi}=0$; with $u=R(r)\Phi(\varphi)$, $2\pi$-periodicity forces $\lambda=n^2$, $\Phi_n=a_n\cos n\varphi+b_n\sin n\varphi$.

Step 2 · Radial (Euler) equation

$r^2R''+rR'-n^2R=0$ has solutions $r^n,r^{-n}$ (and $1,\ln r$ for $n=0$); boundedness at $r=0$ eliminates the singular branch, leaving $R_n=r^n$.

Step 3 · Apply the boundary condition and sum the series

\[ u(r,\varphi)=\frac{a_0}{2}+\sum_{n=1}^\infty r^n(a_n\cos n\varphi+b_n\sin n\varphi), \] with $a_n,b_n$ the Fourier coefficients of $f$ scaled by $R^{-n}$. Summing the resulting geometric series (via $\mathrm{Re}\frac{1+\rho e^{i\theta}}{1-\rho e^{i\theta}}$, $\rho=r/R$) gives the Poisson kernel.

Verification

Each $r^n\cos n\varphi$, $r^n\sin n\varphi$ is harmonic (real/imaginary part of $z^n$). \checkmark At $r=R$ the series reduces to the Fourier series of $f(\varphi)$. \checkmark At $r=0$, (II) gives the mean-value property $u(0)=\frac{1}{2\pi}\int_0^{2\pi}f\,d\psi$.