Pre-Quantum Electrodynamics

Suppose now that we have two solutions to the Poisson equation, \(V_1 ({\bf r})\) and \(V_2 ({\bf r})\). Defining \(U = V_1 - V_2\), we see that \(U\) manifestly obeys Laplace within \({\cal V}\), \({\boldsymbol \nabla}^2 U = 0\). We can now use Green's first identity (\ref{eq:GreensFirstIdentity}) to shed some light on the boundary problem for the electrostatic potential. Namely, put \(\phi = \psi = U\). This yields \[ \int_{\cal V} d\tau \left( U {\boldsymbol \nabla}^2 U + {\boldsymbol \nabla} U \cdot {\boldsymbol \nabla} U \right) = \oint_{\cal S} da ~U \frac{\partial U}{\partial n}. \] The first term on the left-hand side vanishes since \(U\) satisfies Laplace. The right-hand side can be made to vanish if \(U\) obeys either

\begin{align} &U|_{\cal S} = 0 &\mbox{({\bf Dirichlet})} \label{eq:Dirichlet}\\ \mbox{or}& & \nonumber \\ &\frac{\partial U}{\partial n}|_{\cal S} = 0 &\mbox{({\bf Neumann})} \label{eq:Newmann} \end{align}

boundary conditions on each individual boundary surface. In those cases, we are left with \[ \int_{\cal V} d\tau \left|{\boldsymbol \nabla} U \right|^2 = 0, \longrightarrow {\boldsymbol \nabla} U = 0. \] \(U\) is thus constant. For Dirichlet, \(U = 0\) throughout \({\cal V}\), and thus \(V_2 = V_1\) 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 'second case' above, in which the potential is specified on the boundaries, is the case of Dirichlet boundary conditions. The 'first case corollary', 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 our first case corollary involved only surface charges).

\paragraph{Note on Griffiths' presentation of uniqueness theorem(s):} we have used Green's identity to provide a general statement on uniqueness. Griffiths might mislead you into thinking that there are numerous cases and corollaries. Here (in italics) is his (confused) way of thinking about it (see Comment/warning below):

\subsubsection*{\it Boundary Conditions and Uniqueness Theorem} \paragraph{\it First uniqueness theorem:} {\it The solution to Laplace's equation in some volume \({\cal V}\) is uniquely determined if \(V\) is specified on the boundary surface \({\cal S}\). {\bf Corollary:} the potential in a volume \({\cal V}\) is uniquely determined if a) the charge density throughout the region and b) the value of \(V\) on all boundaries are specified.}

\subsubsection*{\it Conductors and the Second Uniqueness Theorem} \paragraph{\it Second uniqueness theorem:} {\it In a volume \({\cal V}\) surrounded by conductors and containing a specified charge density \(\rho\), the electric field is uniquely determined if the total charge on each conductor is given.}

{\it Comments: this is the same uniqueness as before, in view of the fact that conductors are equipotentials, and capacitance relates the charge to the potential. }