Similar to when deriving the weak form, multiply Laplace’s equation with a test function υ and integrate by parts:

Repeat with the roles of u and υ reversed:

with integration over x:

For the second term of the volume integral, since u is assumed to be a solution to Laplace’s equation −∇ · (∇u) = 0:

The resulting relationship is known as the representation of u in terms of boundary integrals:

The value of u over the entire domain Ω is thus determined completely by the values of u and the flux n · ∇u on the boundary. The integrals are not singular anywhere inside Ω since the integral for each y is taken over x on the boundary where x ≠ y. Furthermore, note that this relationship is not an equation but merely a representation of u when we already know the solution and the flux on the boundary. Indeed, this representation is used to reconstruct the solution anywhere inside Ω once the field and flux are solved for and known on the boundary.

The representation of u can be written more compactly as:

where ∂nu = n · ∇u.