Solving Laplace's Equation on unit sphere

Question

I have reduced a calculus of variations problem into solving laplaces equation in 3D in the unit sphere with boundary conditions $u=1$ on the boundary of the sphere and $\int\int\int u dV = 4\pi$ when integrated on the unit sphere. Is there a simple solution to these without delving into spherical coordinates/harmonics as its not a differential equation course so i suspect there should be. Thank yoiu for your help!

Answer 1

Since the boundary is direction-independent, it's reasonable to assume a radial solution, i.e. $u = u(r)$. The radial component of the 3D Laplacian is

$$\nabla^2 u = (r^2u_r)_r = 0 $$

which gives \begin{align} u_r &= \frac{C}{r^2} \\ \\ u &= \frac{c_1}{r} + c_2 \end{align}

The boundary condition gives

$$ u(1) = c_1 + c_2 = 1 $$

The radially-symmetric integral is

\begin{align} & \int_{\partial V}u dV = 4\pi \int_0^1 u(r) r^2 \ dr = 4\pi \\ &\quad \implies \int_0^1 u(r) r^2 dr = 1 \\ &\quad \implies \int_0^1 (c_1r + c_2r^2)\ dr = \frac{c_1}{2} + \frac{c_2}{3} = 1 \end{align}

Combining the two equations gives $c_1 = 4, \ c_2 = -3$, so the solution is $$ u(r) = \frac{4}{r} - 3 $$