We assume there are two functions $V_1$ and $V_2$ satisfying Poisson's equation and the same boundary conditions, and we define $U=V_1-V_2$, then: $$\nabla^2U=\nabla^2V_1-\nabla^2V_2=0$$ Now, lets say that the potential at every point of the boundary is given by a function $V_0 (\overset{\to}{r})$, where $\overset{\to}{r}$ ranges all over the boundary. So for any point in the boundary, with position vector $\overset{\to}{r}$ we have: $$U(\overset{\to}{r})=V_1 (\overset{\to}{r})-V_2(\overset{\to}{r})=V_0(\overset{\to}{r})-V_0(\overset{\to}{r})=0$$ So $U$ is a function satisfying Laplace's equation and so we know that every possible extremum must occur in the boundary but $U$ is also satisfying the condition of being zero all over the boundary, so it must be the case that $U$ is zero everywhere inside the region enclosed by the boundary. At this point I thought "so we're done because if $U=0$ then (because of the definition of $U$) $V_1=V_2$", why is this conclusion wrong? I mean, if it were not the case that $V_1=V_2$ then there would be some point, with vector position $\overset{\to}{r_0}$, inside the region enclosed by the boundary for which $V_1(\overset{\to}{r_0})\not=V_2(\overset{\to}{r_0})$ and hence $U(\overset{\to}{r_0})\not=0$, contradicting what we concluded before.

(All the proofs I've seen don't stop here, but keep going in order to get the result: $$(\nabla U)^2=0$$ and from there conclude that the fields due to potentials $V_1$ and $V_2$ are the same, which implies $V_1$ and $V_2$ can differ by at most a constant.)

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    $\begingroup$ and your question is ? $\endgroup$
    – Lelouch
    Commented Oct 1, 2016 at 11:58

1 Answer 1


There are two common types of boundary condition:

  1. Dirichlet boundary condition: $V$ is known
  2. Neumann boundary condition: $\mathbf{n} \cdot \nabla V$ is known

The boundary conditions for $U$ follow from the boundary conditions for $V$. If $V$ has a Dirichlet boundary condition everywhere, then $U=0$ along the entire boundary. In this case we can use your extremum argument. However, if $V$ has a Neumann boundary condition for some parts of the boundary, then we will have $\mathbf{n} \cdot \nabla U = 0$ instead of $U = 0$, so the extremum argument does not work.

On the other hand, the proof using $(\nabla U)^2$ works for an arbitrary mix of Dirichlet and Neumann boundary conditions, so it is more general.


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