The Laplacian operator for a Scalar function
is defined by
![\begin{displaymath}
\nabla^2 \phi = {1\over h_1h_2h_3}\left[{{\partial \over \pa...
...h_2\over h_3}{\partial \over \partial u_3}}\right)}\right]\phi
\end{displaymath}](l1_741.gif) |
(1) |
in Vector notation, where the
are the Scale Factors
of the coordinate system. In Tensor notation, the Laplacian is written
where
is a Covariant Derivative and
![\begin{displaymath}
\Gamma^\lambda \equiv {\textstyle{1\over 2}}g^{\mu\nu}g^{\la...
...l x^\mu}-{\partial g_{\mu\nu}\over \partial x^\kappa}}\right).
\end{displaymath}](l1_746.gif) |
(3) |
The finite difference form is
For a pure radial function
,
Using the Vector Derivative identity
![\begin{displaymath}
\nabla\cdot (f{\bf A}) = f(\nabla\cdot{\bf A})+(\nabla f)\cdot(\bf A),
\end{displaymath}](l1_755.gif) |
(6) |
so
Therefore, for a radial Power law,
A Vector Laplacian can also be defined for a Vector A by
![\begin{displaymath}
\nabla^2{\bf A}=\nabla(\nabla\cdot {\bf A})-\nabla\times(\nabla\times{\bf A})
\end{displaymath}](l1_761.gif) |
(9) |
in vector notation. In tensor notation, A is written
, and the identity becomes
Similarly, a Tensor Laplacian can be given by
![\begin{displaymath}
\nabla^2 A_{\alpha \beta}= {A_{\alpha \beta ;\lambda}}^{;\lambda}.
\end{displaymath}](l1_766.gif) |
(11) |
An identity satisfied by the Laplacian is
![\begin{displaymath}
\nabla^2 \vert{\bf x}{\hbox{\sf A}}\vert = {{\vert{\hbox{\sf...
...sf A}}^{\rm T}\vert^2\over \vert{\bf x}{\hbox{\sf A}}\vert^3},
\end{displaymath}](l1_767.gif) |
(12) |
where
is the Hilbert-Schmidt Norm,
is a row Vector, and
is the
Matrix Transpose of A.
To compute the Laplacian of the inverse distance function
, where
, and integrate the
Laplacian over a volume,
![\begin{displaymath}
\int_V \nabla^2 \left({1\over \vert{\bf r}-{\bf r}'\vert}\right)d^3{\bf r}.
\end{displaymath}](l1_773.gif) |
(13) |
This is equal to
where the integration is over a small Sphere of Radius
. Now, for
and
, the integral
becomes 0. Similarly, for
and
, the integral becomes
. Therefore,
![\begin{displaymath}
\nabla^2\left({1\over \vert{\bf r}-{\bf r}'\vert}\right)= -4\pi \delta^3({\bf r}-{\bf r}'),
\end{displaymath}](l1_783.gif) |
(15) |
where
is the Delta Function.
See also Antilaplacian
© 1996-9 Eric W. Weisstein
1999-05-26