Uniqueness Theorem vanishes on S vanishes in V There is a uniqueness theorem for Laplace's equation such that if a solution is found, by whatever means, it is the solution. The proof follows a proof by contradiction. Suppose that, in a given finite volume V bounded by the closed surface S, we have and f is given on the surface S. Now we assume that there are two different functions f1 and f2 satisfying these conditions. Then, Let us now calculate the integral This means that f must be a constant. However, since it vanishes on S, it must be zero! in V on S vanishes on S vanishes in V

A matter of terminology - specification of either the value or the derivative of the potential at the boundary of the region is an example of a boundary condition. A boundary value problem is a differential equation together with a set of additional restraints, called the boundary conditions. A solution to a boundary value problem is a solution to the differential equation which also satisfies the boundary conditions. To be useful in applications, a boundary value problem should be well posed. This means that given the input to the problem there exists a unique solution, which depends continuously on the input. If the boundary gives a value to the normal derivative of the problem then it is a Neumann boundary condition. If the boundary gives a value to the problem then it is a Dirichlet boundary condition. If the boundary has the form of a curve or surface that gives a value to the normal derivative and the problem itself then it is a Cauchy boundary condition. It corresponds to imposing both a Dirichlet and a Neumann boundary condition.

Uniqueness Theorem for Poisson Equation Suppose we know that some region contains a charge density r. Suppose we also know the value of the electrostatic potential f on the boundary of this region. The uniqueness theorem then guarantees that there is only one function f which describes the potential in that region. This means that no matter how we figure out f’s value — guessing, computer aided numerical computation, demonic invocation — the function φ we find is guaranteed to be the one we want. Let us assume that two functions, f1 and f2 both satisfy Poisson’s equation: Both of these functions must satisfy the same boundary condition. on boundary in V An electrostatic potential is completely determined within a region once its value is known on the region’s boundary

Example 1: Randomly shaped conductor

Example 2: Nested concentric spherical shells

Example 3: Point charge and flat plane conductor