 Solution

Write the Fourier coefficients of the two boundary functions as $a_{j,n}=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_j(\phi )e^{-in\phi }d\phi$. The Laplace equation in polar coordinates is $u_{rr}+r^{-1}{u}_r+r^{-2}{u}_{\phi \phi }=0$. A separated mode $R(r)e^{in\phi }$ satisfies $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. The ansatz $R=r^{\alpha }$ gives ${\alpha }^2=n^2$, so the solutions are $r^{|n|},r^{-|n|}$ for $n\ne 0$ and $1,\log r$ for $n=0$.

Set $s=\log (r/R_1)$ and $L=\log (R_2/R_1)$. Solving the two boundary equations for every mode yieldsFor real boundary data, the conjugate coefficients ensure this is real. On any compact subannulus $0<\delta \le s\le L-\delta$, the multipliers and all their differentiated versions are bounded by a polynomial in $|n|$ times $e^{-\delta |n|}$. The boundary coefficients are square summable by Parseval identity, so Cauchy-Schwarz inequality proves locally uniform convergence of the series and of all its derivatives. Thus it is smooth and satisfies the Laplace equation term by term.

Because the data are only $L^2$, the Dirichlet boundary condition is interpreted in $L^2$ on circles. As $s\downarrow 0$, the first multiplier tends to one and the second to zero; both lie between zero and one. Dominated convergence of the square-summable Fourier coefficients gives $u(r,\cdot )\to u_1$ in $L^2$, and similarly $u\to u_2$ as $s\uparrow L$. Finally, any harmonic solution with these $L^2$ traces has Fourier coefficients satisfying the radial equation above with the same two endpoint values. Uniqueness of those scalar boundary-value problems makes every coefficient equal to the displayed one. Hence this is the unique harmonic solution with the prescribed $L^2$ traces; pointwise values at every boundary angle are not required by the given data.