Editing Laplace operator (section)

== Motivation ==

=== 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.

=== Averages ===
Given a twice continuously differentiable function <math>f : \R^n \to \R </math> and a point <math>p\in\R^n</math>, the average value of <math>f </math> over the ball with radius <math>h</math> centered at <math>p</math> is:<ref>{{Cite journal | last=Ovall | first=Jeffrey S. | date=2016-03-01 | title=The Laplacian and Mean and Extreme Values|url=http://web.pdx.edu/~jovall/PDF/LaplaceMeanValue.pdf | journal=The American Mathematical Monthly | volume=123 | issue=3 | pages=287–291| doi=10.4169/amer.math.monthly.123.3.287 | s2cid=124943537 }}</ref>
<math display="block">\overline{f}_B(p,h)=f(p)+\frac{\Delta f(p)}{2(n+2)} h^2 +o(h^2) \quad\text{for}\;\; h\to 0</math>

Similarly, the average value of <math>f </math> over the sphere (the boundary of a ball) with radius <math>h</math> centered at <math>p</math> is:
<math display="block">\overline{f}_S(p,h)=f(p)+\frac{\Delta f(p)}{2n} h^2 +o(h^2) \quad\text{for}\;\; h\to 0.</math>

=== Density associated with a potential ===
If {{math|''φ''}} denotes the [[electrostatic potential]] associated to a [[charge distribution]] {{math|''q''}}, then the charge distribution itself is given by the negative of the Laplacian of {{math|''φ''}}:
<math display="block">q = -\varepsilon_0 \Delta\varphi,</math>
where {{math|''ε''<sub>0</sub>}} is the [[electric constant]].

This is a consequence of [[Gauss's law]]. Indeed, if {{math|''V''}} is any smooth region with boundary {{math|∂''V''}}, then by Gauss's law the flux of the electrostatic field {{math|'''E'''}} across the boundary is proportional to the charge enclosed:
<math display="block">\int_{\partial V} \mathbf{E}\cdot \mathbf{n}\, dS = \int_V \operatorname{div}\mathbf{E}\,dV=\frac1{\varepsilon_0}\int_V q\,dV.</math>
where the first equality is due to the [[divergence theorem]]. Since the electrostatic field is the (negative) gradient of the potential, this gives:
<math display="block">-\int_V \operatorname{div}(\operatorname{grad}\varphi)\,dV = \frac1{\varepsilon_0} \int_V q\,dV.</math>

Since this holds for all regions {{mvar|V}}, we must have
<math display="block">\operatorname{div}(\operatorname{grad}\varphi) = -\frac 1 {\varepsilon_0}q</math>

The same approach implies that the negative of the Laplacian of the [[gravitational potential]] is the [[mass distribution]]. Often the charge (or mass) distribution are given, and the associated potential is unknown. Finding the potential function subject to suitable boundary conditions is equivalent to solving [[Poisson's equation]].

=== Energy minimization ===
Another motivation for the Laplacian appearing in physics is that solutions to {{math|1=Δ''f'' = 0}} in a region {{math|''U''}} are functions that make the [[Dirichlet energy]] [[functional (mathematics)|functional]] [[stationary point|stationary]]:
<math display="block"> E(f) = \frac{1}{2} \int_U \lVert \nabla f \rVert^2 \,dx.</math>

To see this, suppose {{math|''f'' : ''U'' → '''R'''}} is a function, and {{math|''u'' : ''U'' → '''R'''}} is a function that vanishes on the boundary of {{mvar|U}}. Then:
<math display="block">\left. \frac{d}{d\varepsilon}\right|_{\varepsilon = 0} E(f+\varepsilon u) = \int_U \nabla f \cdot \nabla u \, dx = -\int_U u \, \Delta f\, dx </math>

where the last equality follows using [[Green's first identity]]. This calculation shows that if {{math|1=Δ''f'' = 0}}, then {{math|''E''}} is stationary around {{math|''f''}}. Conversely, if {{math|''E''}} is stationary around {{math|''f''}}, then {{math|1=Δ''f'' = 0}} by the [[fundamental lemma of calculus of variations]].