Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 2

For a separated axisymmetric harmonic function $\Psi =R(r)\Theta (\theta )$, multiplying the Laplace equation by $r^2/(R\Theta )$ givesThe two sides depend on different variables, hence both equal a constant. Put $u=\cos \theta$ in the angular equation. Since $\sin \theta {\Theta }^{\prime }=-{\sin }^2\theta d\Theta /du$, it becomes the Legendre differential equationRegularity on both parts of the polar axis restricts $\lambda =\ell (\ell +1)$, with $\ell =0,1,\dots$, and selects the Legendre polynomial $P_{\ell }(u)$. The other angular solution is singular at an endpoint.

The radial Euler differential equation isTrying $R=r^s$ gives $s(s+1)=\ell (\ell +1)$, so $s=\ell$ or $s=-\ell -1$. Superposition of these separated harmonic functions givesThe Legendre polynomials form the angular expansion for regular data with rotational symmetry about the polar axis; the radial domain and the boundary conditions determine which coefficients survive and where the expansion converges.

With $u=\cos \theta$, the boundary data have the finite Legendre polynomial expansionInside the sphere, every $r^{-\ell -1}$ term is singular at the origin and must be absent. Matching at $r=a$ therefore givesIn the exterior, decay at infinity excludes all $r^{\ell }$ terms, including the constant term at $\ell =0$. ThusBoth harmonic functions have the prescribed Dirichlet boundary condition, and each is finite throughout its own domain. The interior expression is regular even at the origin because $r\cos \theta =z$ and $r^2(3{\cos }^2\theta -1)=2z^2-x^2-y^2$. Uniqueness of the Dirichlet problem identifies these as the respective solutions. Their normal derivatives need not match: the two Dirichlet problems do not require a single harmonic function across the bounding sphere.