= Solution
Write the <Fourier coefficients> of the two boundary functions as $a_{j,n}=(2\pi)^{-1}\int_0^{2\pi}u_j(\varphi)e^{-in\varphi}\,d\varphi$. The <Laplace equation> in polar coordinates is $u_{rr}+r^{-1}u_r+r^{-2}u_{\varphi\varphi}=0$. A separated mode $R(r)e^{in\varphi}$ satisfies $r^2R''+rR'-n^2R=0$. The ansatz $R=r^\alpha$ gives $\alpha^2=n^2$, so the solutions are $r^{|n|},r^{-|n|}$ for $n\ne0$ 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 yields
$$
\boxed{u(r,\varphi)=a_{1,0}\left(1-\frac sL\right)+a_{2,0}\frac sL
+\sum_{n\ne0}\left[a_{1,n}\frac{\sinh(|n|(L-s))}{\sinh(|n|L)}
+a_{2,n}\frac{\sinh(|n|s)}{\sinh(|n|L)}\right]e^{in\varphi}.}
$$
For 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 $s\downarrow0$, 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 \b[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.
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