= Solution
For <Laplace's equation> in <polar coordinates>, put $\Phi=R(r)\Theta(\theta)$. <Separation of variables> gives $\Theta''+n^2\Theta=0$ and $r^2R''+rR'-n^2R=0$. Periodicity selects integers $n\geq0$. For $n>0$, the radial solutions are $r^n,r^{-n}$; for $n=0$, they are $1,\ln r$, while a term linear in $\theta$ is excluded by periodicity. Their superposition is
$$
\Phi=A_0+C_0\ln r+\sum_{n\geq1}[(A_nr^n+C_nr^{-n})\cos n\theta+(B_nr^n+D_nr^{-n})\sin n\theta].
$$
Inside the disk, boundedness at the origin removes the logarithm and all negative powers. The separately biased semicircular boundary values have mean $(V_1+V_2)/2$, no nonconstant cosine coefficients, and sine coefficients $2(V_1-V_2)/(\pi n)$ for odd $n$, zero for even $n$. The <disk potential with two semicircular boundary values> is therefore
$$
\Phi=\frac{V_1+V_2}{2}+\frac{2(V_1-V_2)}\pi\sum_{m\geq0}\frac{(r/a)^{2m+1}\sin((2m+1)\theta)}{2m+1}.
$$
Put $w=(r/a)e^{i\theta}$. The odd logarithmic series sums to $\sum_{m\geq0}w^{2m+1}/(2m+1)=\frac12\log[(1+w)/(1-w)]$. The quotient has positive real part for $|w|<1$, so its argument is the principal <arctangent> of $2(r/a)\sin\theta/[1-(r/a)^2]$. Taking imaginary parts gives
$$
\boxed{\Phi=\frac{V_1+V_2}{2}+\frac{V_1-V_2}\pi\arctan\frac{2ar\sin\theta}{a^2-r^2}.}
$$
The boundary limits away from the two voltage jumps are $V_1,V_2$. Along the diameter through a jump the value is the mean; incompatible adjacent boundary limits prevent a continuous extension at the junctions.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2-split-disk-potential.png]
{title=Harmonic potential inside a disk with two semicircular boundary voltages}
{description=Original normalized example with upper voltage $+1$ and lower voltage $-1$. White circles mark the two boundary jumps; the black curves are equipotentials.}
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