28  The uniqueness theorem

The uniqueness theorem for Poisson’s equation states that, for a large class of boundary conditions, the equation may have many solutions, but the gradient of every solution is the same. In the case of electrostatics, this means that there is a unique electric field derived from a potential function satisfying Poisson’s equation under the boundary conditions.

28.1 Proof

Consider the general case of Poisson’s equation in electrostatics over some region \(\Omega\) with boundary \(\partial\Omega\):

\[ -\Delta \varphi = \frac{\rho_E}{\varepsilon_0} \]

Suppose that there are two solutions, \(\varphi_1\) and \(\varphi_2\). Then

\[ \begin{align} -\Delta \varphi_1 = \frac{\rho_E}{\varepsilon_0} \\ -\Delta \varphi_2 = \frac{\rho_E}{\varepsilon_0} \end{align} \]

It follows that \(\varphi = \varphi_2 - \varphi_1\) is a solution of the Laplace equation.

Subtracting these equations gives

\[ \Delta \varphi = \Delta \varphi_1 - \Delta \varphi_2 = 0. \]

We know that

\[ \div (\varphi \grad \varphi) = \grad\varphi\cdot\grad\varphi + \varphi \Delta \varphi. \]

Since \(\Delta\varphi=0\), we find that

\[ \div (\varphi \grad \varphi) = \grad\varphi\cdot\grad\varphi \]

Integrate over \(\Omega\): \[ \int_\Omega \div (\varphi \grad \varphi) \,\dd V = \int_\Omega (\grad\varphi)^2\,\dd V \]

We rewrite using the divergence theorem as

\[ \int_{\partial\Omega} (\varphi \grad \varphi) \cdot \dd {\vb S} = \int_\Omega \grad\varphi\cdot\grad\varphi\,\dd V \tag{28.1}\]

We now consider the boundary conditions.

First, consider a Dirichlet boundary condition as \(\varphi=0\) on the boundary of the region.

Since this condition is satisfied on the boundary, i.e., on \(\partial\Omega\), the left-hand is zero.

We find that

\[ \int_\Omega (\grad\varphi)^2\,\dd V = \int_\Omega \grad\varphi\cdot\grad\varphi\,\dd V = 0 \]

It follows that \(\grad\varphi = 0\) at all points in \(\Omega\). Further, since the gradient is zero and \(\varphi\) is zero on the boundary, \(\varphi\) must be zero everywhere throughout the region. Therefore, \(\varphi_1 = \varphi_2\) throughout the whole region. \(\square\)

Second, we consider Neumann boundary conditions specified as \(\grad\varphi = 0\) on \(\partial\Omega\).

The left-hand side of (28.1) is zero. We find that

\[ \int_\Omega \grad\varphi\cdot\grad\varphi\,\dd V = 0 \]

Again, we must have \(\grad\varphi = 0\) at all points in \(\Omega\). The gradient of \(\varphi\) is zero within the volume, and because the gradient is zero on the boundary, it follows that \(\varphi\) must be constant throughout the whole region. Therefore, since \(\varphi = \varphi_2 - \varphi_1 = k\),

\[ \varphi_1 = \varphi_2 - k \]

The solution is unique up to an additive constant. \(\square\)

28.2 Summary

In all cases, the gradient \(\grad\varphi\) is unique. This is the only relevant quantity as seen from the physics perspective.