Here the surface S consists of S_r as well as S₁ , S₂, .....S₃.

We note that V_d =0 over the conducting boundaries, therefore the contribution to the surface integral from these surfaces are zero. For large surface S_r , we can think the surface to be a spherical surface of radius . In the surface integral, the term varies as whereas the area increases as r², hence the surface integral decreases as . For , the integral vanishes.

Therefore, . Since is non negative everywhere, the volume integral can be zero only if , which means V₁ = V₂. That is, our assumption that two solutions are different does not hold. So if a solution exists for Poisson’s (and Laplace’s) equation for a given set of boundary conditions, this solution is a unique solution.

Moreover, in the context of uniqueness of solution, let us look into the role of the boundary condition. We observe that, , which, can be written as . Uniqueness of solution is guaranteed everywhere if or =0 on S.