Consider the equation $\nabla^2f = 0$ on a smooth volume $V$ such that on the boundary $f(x) = u(x)$ for some specified function $u$ . This has a unique solution: Suppose that $f_1,f_2$ are solutions. Then let $F = f_1-f_2$ and note $\nabla^2 F = 0$ and $F = 0$ on the boundary . Also by Divergence theorem $$ 0 = \int_{\partial V}F\nabla F \cdot dS =\int_V \nabla\cdot(F\nabla F) \ dV = \int_VF\nabla^2F + |{\nabla F}|^2 \ dV = \int_V |{\nabla F}|^2 \ dV$$ So as $|{\nabla F}|^2 \geq 0$ we get that $\nabla F = 0$ everywhere so $F = 0$ so $f_1 = f_2$.
However if we let $V$ be the sphere of radius $1$ centred at the origin then the equation with the boundary condition $f = 0$ on surface with f spherically symmetric, then $\nabla^2f = 0$ has the solution $f = A/r-A$ for all real $A$. However this implies the solution is not unique. Any hints on what I am missing are much appreciated.