The uniqueness theorem for Poisson's equation states that, for a large class of boundary conditions, the equation may have many solutions, but the gradient of every solution is the same. In the case of electrostatics, this means that there is a unique electric field derived from a potential function satisfying Poisson's equation under the boundary conditions.

Proof

The general expression for Poisson's equation in electrostatics is where \varphi is the electric potential and \rho_f is the charge distribution over some region V with boundary surface S. The uniqueness of the solution can be proven for a large class of boundary conditions as follows. Suppose that we claim to have two solutions of Poisson's equation. Let us call these two solutions \varphi_1 and \varphi_2. Then It follows that is a solution of Laplace's equation, which is a special case of Poisson's equation that equals to 0. Subtracting the two solutions above gives By applying the vector differential identity we know that However, from we also know that throughout the region Consequently, the second term goes to zero and we find that By taking the volume integral over the region V, we find that By applying the divergence theorem, we rewrite the expression above as We now sequentially consider three distinct boundary conditions: a Dirichlet boundary condition, a Neumann boundary condition, and a mixed boundary condition. First, we consider the case where Dirichlet boundary conditions are specified as \varphi = 0 on the boundary of the region. If the Dirichlet boundary condition is satisfied on S by both solutions (i.e., if \varphi = 0 on the boundary), then the left-hand side of is zero. Consequently, we find that Since this is the volume integral of a positive quantity (due to the squared term), we must have at all points. Further, because the gradient of \varphi is everywhere zero and \varphi is zero on the boundary, \varphi must be zero throughout the whole region. Finally, since \varphi = 0 throughout the whole region, and since throughout the whole region, therefore throughout the whole region. This completes the proof that there is the unique solution of Poisson's equation with a Dirichlet boundary condition. Second, we consider the case where Neumann boundary conditions are specified as on the boundary of the region. If the Neumann boundary condition is satisfied on S by both solutions, then the left-hand side of is zero again. Consequently, as before, we find that As before, since this is the volume integral of a positive quantity, we must have at all points. Further, because the gradient of \varphi is everywhere zero within the volume V, and because the gradient of \varphi is everywhere zero on the boundary S, therefore \varphi must be constant---but not necessarily zero---throughout the whole region. Finally, since \varphi = k throughout the whole region, and since throughout the whole region, therefore throughout the whole region. This completes the proof that there is the unique solution up to an additive constant of Poisson's equation with a Neumann boundary condition. Mixed boundary conditions could be given as long as either the gradient or the potential is specified at each point of the boundary. Boundary conditions at infinity also hold. This results from the fact that the surface integral in still vanishes at large distances because the integrand decays faster than the surface area grows.