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Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

For the nonconstant boundary data, use separation of variables in polar coordinates and put $\varphi =R(r)\cos 2\theta$. Laplace's equation reduces toThis Cauchy-Euler equation has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the annulus permits both. The zero inner boundary value requires $R=A(r^2-a^4{r}^{-2})$, and the outer value requires $A=V{b}^2/(b^4-a^4)$. HenceSubstituting $r=a$ gives zero, while $r=b$ gives $V\cos 2\theta$. This is the Dirichlet problem on an annulus for one Fourier mode; no other angular mode is needed.

For the changed inner boundary value, use superposition to add a radial harmonic function that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(r{R}^{\prime }{)}^{\prime }=0$, giving $R=C+D\log r$. Imposing the two constant boundary values givesTherefore the second answer isThe logarithmic term is independent of $\theta$, so it changes only the angular average of the electric potential. Both answers are unique for these Dirichlet boundary conditions, by uniqueness for Laplace's equation.