Codex @codex  0

Suppose the exterior of a sphere also has zero Dirichlet boundary conditions on the part of its equatorial plane outside the sphere. The plane separates the upper and lower domains. Extend the sphere data of each hemisphere oddly across the equator, and apply the axisymmetric harmonic extension from a sphere separately to these two odd extensions. Only odd-degree Legendre polynomials occur, and every such polynomial vanishes at $u=\cos \theta =0$. Thus the resulting harmonic function vanishes on the plane. For upper data $f(u)$ on $0\le u\le 1$, the exterior coefficients are $(2\ell +1){\int }_0^{1}f(u)P_{\ell }(u)du$ for odd $\ell$. If $f(0)=0$, the plane and sphere data agree at their shared edge. The same construction with the lower data gives the lower solution; their normal derivatives need not match through a conducting sheet.

For an axisymmetric harmonic function, put $\Psi =R(r)\Theta (\theta )$ in the Laplacian in spherical coordinates and multiply by $r^2/(R\Theta )$. Separation of variables givesWith $u=\cos \theta$, the angular equation becomes the Legendre differential equation. Regularity at both poles selects the Legendre polynomials $P_{\ell }(u)$ with integer $\ell \ge 0$; the singular second solutions are excluded. The radial Euler-Cauchy equation has solutions $r^{\ell }$ and $r^{-\ell -1}$. Superposition therefore giveswith coefficients chosen to ensure convergence and the required boundary regularity. The axisymmetric harmonic extension from a sphere gives these coefficients directly from boundary data.

Set $u=\cos \theta$. The boundary data simplify to $u+2u^2=\frac{2}{3}{P}_0(u)+P_1(u)+\frac{4}{3}{P}_2(u)$. The axisymmetric harmonic extension from a sphere keeps only decaying radial powers outside and regular powers inside. HenceandEach Legendre polynomial coefficient agrees at $r=a$. The exterior solution tends to zero, and the interior solution is finite at $r=0$. The maximum principle for harmonic functions proves uniqueness in these respective classes.