Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 2 / Solution

For Laplace's equation in polar coordinates, put $\Phi =R(r)\Theta (\theta )$. Separation of variables gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Periodicity selects integers $n\ge 0$. 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 isInside 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 thereforePut $w=(r/a)e^{i\theta }$. The odd logarithmic series sums to ${\sum }_{m\ge 0}{w}^{2m+1}/(2m+1)=\frac{1}{2}\log [(1+w)/(1-w)]$. 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 givesThe 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.