= Solution
Suppose $\Psi_1$ and $\Psi_2$ have the same boundary data, and put $u=\Psi_1-\Psi_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the <product rule> for the <divergence> gives
$$
\nabla\cdot(u\nabla u)=|\nabla u|^2+u\Delta u=|\nabla u|^2.
$$
The <divergence theorem> therefore yields
$$
\int_V|\nabla u|^2dV=\int_Su\,\partial_nu\,dS=0.
$$
The nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{\Psi_1=\Psi_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla\cdot(u^*\nabla u)$ and its real part.
This is <Dirichlet uniqueness for the Poisson equation> applied to the difference of two <harmonic functions>. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.
Back to article page
