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Set $\varphi =R(r)\Theta (\theta )$ in Laplace's equation. Separation gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Single-valued angular periodicity requires integer $n\ge 0$. For $n>0$ the radial factors are $r^n,r^{-n}$; for $n=0$ they are $1,\ln r$. A term linear in $\theta$ is excluded by periodicity. Thus the harmonic Fourier expansions in planar concentric domains have the general formFor $0<a<b$, the specified boundaries contain only a constant mode and a first cosine mode. The two-boundary annular harmonic interpolation givesAt $r=a$, the cosine term vanishes and the logarithmic term is $U$; at $r=b$, the first term vanishes and the cosine amplitude is $V$. Each component is harmonic, and the maximum principle ensures uniqueness for these continuous Dirichlet boundary conditions.

Let $s(\theta )$ be the $2\pi$-periodic ramp represented by $\theta$ on $(-\pi ,\pi )$. The Fourier series of a periodic sawtooth givesBoth boundary data are multiples of this odd function, so their angular means and cosine coefficients vanish. Consequently $A_0=C_0=A_n=C_n=0$. Put $h_n=2(-1{)}^{n+1}/n$. Matching the sine coefficients at the two radii gives $B_n{a}^n+D_n{a}^{-n}=h_n{\mu }_a$ and $B_n{b}^n+D_n{b}^{-n}=h_n{\mu }_b$. Solving this two-by-two linear system yieldsEquivalently, two-boundary annular harmonic interpolation expresses the same solution asEvery term is a harmonic function, and the radial factors enforce the specified boundary values. For a radius strictly between $a$ and $b$, their geometric decay makes the series smooth. At the boundary discontinuity $\theta =\pi$, the limiting Fourier value is the mean of the two one-sided values, zero. The boundary data are imposed away from that exceptional point, as stated.