Editing Laplace operator (section)

=== Diffusion ===
In the [[physics|physical]] theory of [[diffusion]], the Laplace operator arises naturally in the mathematical description of [[Diffusion equilibrium|equilibrium]].<ref>{{harvnb|Evans|1998|loc=§2.2}}</ref> Specifically, if {{math|''u''}} is the density at equilibrium of some quantity such as a chemical concentration, then the [[net flux]] of {{math|''u''}} through the  boundary {{math|∂''V''}} (also called {{math|''S''}}) of any smooth region {{math|''V''}} is zero, provided there is no source or sink within {{math|''V''}}:
<math display="block">\int_{S} \nabla u \cdot \mathbf{n}\, dS = 0,</math>
where {{math|'''n'''}} is the outward [[unit normal]] to the boundary of {{math|''V''}}. By the [[divergence theorem]],
<math display="block">\int_V \operatorname{div} \nabla u\, dV = \int_{S} \nabla u \cdot \mathbf{n}\, dS = 0.</math>

Since this holds for all smooth regions {{math|''V''}}, one can show that it implies:
<math display="block">\operatorname{div} \nabla u = \Delta u = 0.</math>
The left-hand side of this equation is the Laplace operator, and the entire equation {{math|1=Δ''u'' = 0}} is known as [[Laplace's equation]]. Solutions of the Laplace equation, i.e. functions whose Laplacian is identically zero, thus represent possible equilibrium densities under diffusion.

The Laplace operator itself has a physical interpretation for non-equilibrium diffusion as the extent to which a point represents a source or sink of chemical concentration, in a sense made precise by the [[diffusion equation]]. This interpretation of the Laplacian is also explained by the following fact about averages.