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Laplace's Equation

Laplace's equation is written

$\displaystyle \nabla^{\,2} \phi({\bf r}) = 0,$ (127)

where the function $ \phi({\bf r})$ is often referred to as a potential. Suppose that we wish to find a solution to this equation in some finite volume $ V$ , bounded by a closed surface $ S$ , subject to the boundary condition

$\displaystyle \phi({\bf r})= 0,$ (128)

when $ {\bf r}$ lies on $ S$ . Consider the vector identity

$\displaystyle \nabla\cdot (\phi\,\nabla \phi) \equiv \phi\,\nabla^{\,2} \phi + \nabla \phi\cdot \nabla \phi.$ (129)

Integrating this expression over $ V$ , making use of the divergence theorem, we obtain

$\displaystyle \int_S \phi\,\nabla\phi\cdot d{\bf S} = \int_V\left( \phi\,\nabla^{\,2} \phi + \nabla \phi\cdot \nabla \phi\right) dV.$ (130)

It follows from Equations (127) and (128) that

$\displaystyle \int_V \vert\nabla \phi\vert^{\,2} \,dV = 0,$ (131)

which implies that $ \nabla\phi={\bf0}$ throughout $ V$ and on $ S$ . Hence, Equation (128) yields

$\displaystyle \phi({\bf r}) = 0$ (132)

throughout $ V$ and on $ S$ . We conclude that the only solution to Laplace's equation, (127), subject to the boundary condition (128), is the trivial solution (132). Finally, if we let the surface $ S$ tend to infinity then we deduce that the only solution to Laplace's equation, (127), subject to the boundary condition

$\displaystyle \phi({\bf r}) \rightarrow 0$ as $\displaystyle \vert{\bf r}\vert\rightarrow \infty,$ (133)

is

$\displaystyle \phi({\bf r}) = 0$ (134)

for all $ {\bf r}$ .

Consider a potential $ \phi({\bf r})$ that satisfies Laplace's equation, (127), in some finite volume $ V$ , bounded by the closed surface $ S$ , subject to the boundary condition

$\displaystyle \phi({\bf r})= \phi_S({\bf r}),$ (135)

when $ {\bf r}$ lies on $ S$ . Here, $ \phi_S({\bf r})$ is a known surface distribution. We can demonstrate that this potential is unique. Let $ \phi_1({\bf r})$ and $ \phi_2({\bf r})$ be two supposedly different potentials that both satisfy Laplace's equation throughout $ V$ , as well as the previous boundary condition on $ S$ . Let us form the difference $ \phi_3({\bf r})=\phi_1({\bf r})-\phi_2({\bf r})$ . This function satisfies Laplace's equation throughout $ V$ , subject to the boundary condition

$\displaystyle \phi_3({\bf r}) = 0$ (136)

when $ {\bf r}$ lies on $ S$ . However, as we have already seen, this implies that $ \phi_3({\bf r})=0$ throughout $ V$ and on $ S$ . Hence, $ \phi_1({\bf r})$ and $ \phi_2({\bf r})$ are identical, and the potential $ \phi({\bf r})$ is therefore unique.


next up previous
Next: Poisson's Equation Up: Electrostatic Fields Previous: Introduction
Richard Fitzpatrick 2014-06-27