 Solution

A temperature on the full unit disk must be single valued, so its angular factor and angular derivative are $2\pi$-periodic. For positive separation constant $\lambda =k^2$, this requires $k=n$ to be a positive integer. For $\lambda =0$, periodicity removes $D\theta$. For $\lambda <0$, no nonzero linear combination of $e^{k\theta }$ and $e^{-k\theta }$ is $2\pi$-periodic, so that case disappears.

The origin belongs to the disk, so a classical harmonic function must remain bounded and extend smoothly there. This excludes $r^{-n}$ and $\log r$. The remaining individual separated solutions are constants and $r^n(A_n\cos (n\theta )+B_n\sin (n\theta ))$. By superposition, the regular harmonic expansion on a disk isThe admissible Fourier coefficients are fixed by the boundary conditions. Smoothness at the origin can also be seen from $r^n\cos (n\theta )$ and $r^n\sin (n\theta )$ being the real and imaginary parts of $(x+iy{)}^n$, hence harmonic polynomials.