Pre-Quantum Electrodynamics

Suppose now that we have two solutions to the Poisson equation, \(\phi_1 ({\bf r})\) and \(\phi_2 ({\bf r})\). Defining \(\Phi = \phi_1 - \phi_2\), we see that \(\Phi\) manifestly obeys Laplace within \({\cal V}\), \({\boldsymbol \nabla}^2 \Phi = 0\). We can now use Green's first identity Green1 to shed some light on the boundary problem for the electrostatic potential. Namely, put \(\phi = \psi = \Phi\). This yields \[ \int_{\cal V} d\tau \left( \Phi {\boldsymbol \nabla}^2 \Phi + {\boldsymbol \nabla} \Phi \cdot {\boldsymbol \nabla} \Phi \right) = \oint_{\cal S} da ~\Phi \frac{\partial \Phi}{\partial n}. \] The first term on the left-hand side vanishes since \(\Phi\) satisfies Laplace. The right-hand side can be made to vanish if \(\Phi\) obeys either

\begin{equation} \left.\Phi\right|_{\cal S} = 0 \tag{Dirichlet}\label{Dirichlet} \end{equation}

or

\begin{equation} \left.\frac{\partial \Phi}{\partial n}\right|_{\cal S} = 0 \tag{Newmann}\label{Newmann} \end{equation}

boundary conditions on each individual boundary surface. In those cases, we are left with \[ \int_{\cal V} d\tau \left|{\boldsymbol \nabla} \Phi \right|^2 = 0, \longrightarrow {\boldsymbol \nabla} \Phi = 0. \] \(\Phi\) is thus constant. For Dirichlet, \(\Phi = 0\) throughout \({\cal V}\), and thus \(\phi_2 = \phi_1\) and the solution is unique. For Neumann, the solution is unique apart from an unimportant constant.

We can thus finally state the

Note that these types of boundary conditions can be mixed, i.e. Dirichlet on some surfaces, Neumann on others).

Existence of solutions: this is another matter. Intuitively, from our first case: the solution always exists for Dirichlet boundary conditions.

Link to earlier cases: the previous case in which the potential is specified on the boundaries, is thus the case of Dirichlet boundary conditions. The one where the normal derivative of the potential is given, is a subcase involving Neumann boundary conditions (subcase, because we could imagine other charges living outside volume \({\cal V}\), whereas the earlier example involved only surface charges).

Note on Griffiths' presentation of uniqueness theorem(s): we have used Green's identity to provide a general statement on uniqueness. Reading Griffiths, you might be misled into thinking that there are numerous cases and corollaries.