Uniqueness Theorem

Consider a volume $V$ bounded by some surface $S$. Suppose that we are given the electric charge density $\rho$ throughout $V$, and the (not necessarily constant) value of the scalar potential $\phi_S$ on $S$. Is this sufficient information for Poisson's equation to uniquely specify the scalar potential throughout $V$? Suppose, for the sake of argument, that the solution is not unique. Let there be two different potentials $\phi_1$ and $\phi_2$ that both satisfy

$$\nabla^2 \phi_1 =- \frac{\rho}{\epsilon_0},$$ (2.100)

$$\nabla^2 \phi_2 =- \frac{\rho}{\epsilon_0}$$ (2.101)

throughout $V$, and

$$\phi_1 = \phi_S,$$ (2.102)

$$\phi_2 = \phi_S$$ (2.103)

on $S$. We can form the difference between these two potentials:

$$\phi_3 = \phi_1 - \phi_2.$$ (2.104)

The potential $\phi_3$ clearly satisfies

$$\nabla^2\phi_3 = 0$$ (2.105)

throughout $V$, and

$$\phi_3 =0$$ (2.106)

on $S$.

Now,

$$\nabla\cdot(\phi_3\,\nabla\phi_3) \equiv (\nabla \phi_3)^2 +\phi_3\nabla^2\phi_3.$$ (2.107)

(See Section A.24.) Thus, making use of the divergence theorem,

$$\int_V \left[ (\nabla\phi_3)^2 +\phi_3 \nabla^2\phi_3\right] dV = \oint_S \phi_3 \nabla \phi_3\cdot d{\bf S}.$$ (2.108)

(See Section A.20.) But, $\nabla^2\phi_3 = 0$ throughout $V$, and $\phi_3=0$ on $S$, so the previous equation reduces to

$$\int_V (\nabla \phi_3)^2\,dV = 0.$$ (2.109)

Note that $(\nabla \phi_3)^2$ is a positive definite quantity. The only way in which the volume integral of a positive definite quantity can be zero is if that quantity itself is zero throughout the volume. This is not necessarily the case for a non-positive definite quantity, because we could have positive and negative contributions from various regions inside the volume that cancel one another out. Thus, because $(\nabla \phi_3)^2\equiv \nabla\phi_3\cdot\nabla\phi_3$ is positive definite, it is zero throughout $V$. It follows that $\nabla\phi_3={\bf0}$ throughout $V$, and, hence, that

$$\phi_3 = {\rm constant}$$ (2.110)

throughout $V$. However, we know that $\phi_3=0$ on $S$, so we get

$$\phi_3 =0$$ (2.111)

throughout $V$. In other words,

$$\phi_1 = \phi_2$$ (2.112)

throughout $V$ and on $S$. Our initial supposition that $\phi_1$ and $\phi_2$ are two different solutions of Poisson's equation, satisfying the same boundary conditions, turns out to be incorrect. Hence, we deduce that the solutions to Poisson's equation in a volume bounded by a surface on which the electric potential is specified are unique. This important result is known as the uniqueness theorem.