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:<math>\mathbf{\nabla}^2 \varphi = -\frac{\rho_f}{\epsilon_0},</math>
where <math>\varphi</math> is the [[electric potential]] and <math>\rho_f</math> is the [[charge density|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.
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:<math>\nabla \cdot (\varphi \, \nabla \varphi )= \, (\nabla \varphi )^2.</math>
By taking the volume integral over the region $V$, we find that
:<math>\int_V \mathbf{\nabla}\cdot(\varphi \, \mathbf{\nabla}\varphi) \, \mathrm{d}V= \int_V \mathbf{(\nabla}\varphi)\cdot( \mathbf{\nabla}\varphi) \, \mathrm{d}V= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V.</math>
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:<math>\int_{S} (\varphi \, \mathbf{\nabla}\varphi) \cdot \mathrm{d}\mathbf{S}= \int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V. \qquad (2)</math>
If the Dirichlet boundary condition is
▲If the Dirichlet boundary condition is satisified by both solutions (ie, <math>\varphi = 0</math> on the boundary) then the left-hand side of <math>(2)</math> is zero thus
:<math>\int_V (\mathbf{\nabla}\varphi)^2 \, \mathrm{d}V = 0</math>
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