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:<math>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0.</math>
As before, since this is the volume integral of a positive quantity, we must have <math>\nabla \varphi = 0</math> at all points. Further, because the gradient of <math>\varphi</math> is everywhere zero within the volume <math>V</math>, and because the gradient of <math>\varphi</math> is everywhere zero on the boundary <math>S</math>, therefore <math>\varphi</math> must be constant---but not necessarily zero---throughout the whole region. Finally, since <math>\varphi = k</math> throughout the whole region, and since <math>\varphi = \varphi_2 - \varphi_1</math> throughout the whole region, therefore <math>\varphi_1 = \varphi_2 - k</math> throughout the whole region. This completes the proof that there is the unique solution up to an additive constant
[[Mixed 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 <math>(2)</math> still vanishes at large distances because the integrand decays faster than the surface area grows.
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