Poisson kernel for the unit disk ($P_r(\theta )=(1-r^2)/(1-2r\cos \theta +r^2)$)

For $0\le r<1$, this Poisson kernel for the unit disk has the uniformly convergent Fourier seriesThe identity follows by taking the real part of the geometric series $1+2{\sum }_{n\ge 1}(r{e}^{i\theta }{)}^n$. For integrable $2\pi$-periodic boundary data $f$, its Poisson extension iswhere $a_n,b_n$ are the usual Fourier coefficients. Each term is a harmonic function in the open disk. For continuous boundary data, the extension solves the Dirichlet problem with those boundary values.

On a disk of radius $a$, prescribe $V_1$ on the upper open semicircle and $V_2$ on the lower one. The bounded harmonic extension has the formula in the title for $r<a$. The boundary data have only a constant term and odd sine Fourier coefficients, giving $u=(V_1+V_2)/2+[2(V_1-V_2)/\pi ]{\sum }_{m\ge 0}(r/a{)}^{2m+1}\sin ((2m+1)\theta )/(2m+1)$. The complex series is half of $\log [(1+w)/(1-w)]$, where $w=(r/a)e^{i\theta }$. Its quotient has positive real part, so taking the imaginary part yields the displayed principal arctangent without an argument ambiguity. At the two boundary jumps, the radial limit is the mean; the harmonic function has no continuous extension matching both adjacent constant arcs there.

For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.