Laplace operator

In the physical theory of diffusion, the Laplace operator arises naturally in the mathematical description of equilibrium.[1] Specifically, if u is the density at equilibrium of some quantity such as a chemical concentration, then the net flux of u through the boundary ∂V of any smooth region V is zero, provided there is no source or sink within V:

$${\displaystyle \int _{\partial V}\nabla u\cdot \mathbf {n} \,dS=0,}$$

where n is the outward unit normal to the boundary of V. By the divergence theorem,

$${\displaystyle \int _{V}\operatorname {div} \nabla u\,dV=\int _{\partial V}\nabla u\cdot \mathbf {n} \,dS=0.}$$

Since this holds for all smooth regions V, one can show that it implies:

$${\displaystyle \operatorname {div} \nabla u=\Delta u=0.}$$

The left-hand side of this equation is the Laplace operator, and the entire equation Δu = 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.

Given a twice continuously differentiable function ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$, a point ${\displaystyle p\in \mathbb {R} ^{n}}$ and a real number ${\displaystyle h>0}$, we let ${\displaystyle {\overline {f}}_{B}(p,h)}$ be the average value of ${\displaystyle f}$ over the ball with radius ${\displaystyle h}$ centered at ${\displaystyle p}$, and ${\displaystyle {\overline {f}}_{S}(p,h)}$ be the average value of ${\displaystyle f}$ over the sphere (the boundary of a ball) with radius ${\displaystyle h}$ centered at ${\displaystyle p}$. Then we have:[2]

$${\displaystyle {\overline {f}}_{B}(p,h)=f(p)+{\frac {\Delta f(p)}{2(n+2)}}h^{2}+o(h^{2})\quad {\text{for}}\;\;h\to 0}$$

and

$${\displaystyle {\overline {f}}_{S}(p,h)=f(p)+{\frac {\Delta f(p)}{2n}}h^{2}+o(h^{2})\quad {\text{for}}\;\;h\to 0.}$$

If φ denotes the electrostatic potential associated to a charge distribution q, then the charge distribution itself is given by the negative of the Laplacian of φ:

$${\displaystyle q=-\varepsilon _{0}\Delta \varphi ,}$$

where ε₀ is the electric constant.

This is a consequence of Gauss's law. Indeed, if V is any smooth region with boundary ∂V, then by Gauss's law the flux of the electrostatic field E across the boundary is proportional to the charge enclosed:

$${\displaystyle \int _{\partial V}\mathbf {E} \cdot \mathbf {n} \,dS=\int _{V}\operatorname {div} \mathbf {E} \,dV={\frac {1}{\varepsilon _{0}}}\int _{V}q\,dV.}$$

where the first equality is due to the divergence theorem. Since the electrostatic field is the (negative) gradient of the potential, this gives:

$${\displaystyle -\int _{V}\operatorname {div} (\operatorname {grad} \varphi )\,dV={\frac {1}{\varepsilon _{0}}}\int _{V}q\,dV.}$$

Since this holds for all regions V, we must have

$${\displaystyle \operatorname {div} (\operatorname {grad} \varphi )=-{\frac {1}{\varepsilon _{0}}}q}$$

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.

Another motivation for the Laplacian appearing in physics is that solutions to Δf = 0 in a region U are functions that make the Dirichlet energy functional stationary:

$${\displaystyle E(f)={\frac {1}{2}}\int _{U}\lVert \nabla f\rVert ^{2}\,dx.}$$

To see this, suppose f : U → R is a function, and u : U → R is a function that vanishes on the boundary of U. Then:

$${\displaystyle \left.{\frac {d}{d\varepsilon }}\right|_{\varepsilon =0}E(f+\varepsilon u)=\int _{U}\nabla f\cdot \nabla u\,dx=-\int _{U}u\,\Delta f\,dx}$$

where the last equality follows using Green's first identity. This calculation shows that if Δf = 0, then E is stationary around f. Conversely, if E is stationary around f, then Δf = 0 by the fundamental lemma of calculus of variations.

The Laplace operator in two dimensions is given by:

In Cartesian coordinates,

$${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}}$$

where x and y are the standard Cartesian coordinates of the xy-plane.

In polar coordinates,

$${\displaystyle {\begin{aligned}\Delta f&={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}\\&={\frac {\partial ^{2}f}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial f}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}},\end{aligned}}}$$

where r represents the radial distance and θ the angle.

In three dimensions, it is common to work with the Laplacian in a variety of different coordinate systems.

In Cartesian coordinates,

$${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}.}$$

In cylindrical coordinates,

$${\displaystyle \Delta f={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\left(\rho {\frac {\partial f}{\partial \rho }}\right)+{\frac {1}{\rho ^{2}}}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}},}$$

where ${\displaystyle \rho }$ represents the radial distance, φ the azimuth angle and z the height.

In spherical coordinates:

$${\displaystyle \Delta f={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}},}$$

or

$${\displaystyle \Delta f={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}(rf)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}},}$$

where φ represents the azimuthal angle and θ the zenith angle or co-latitude.

In general curvilinear coordinates (ξ¹, ξ², ξ³):

$${\displaystyle \Delta =\nabla \xi ^{m}\cdot \nabla \xi ^{n}{\frac {\partial ^{2}}{\partial \xi ^{m}\,\partial \xi ^{n}}}+\nabla ^{2}\xi ^{m}{\frac {\partial }{\partial \xi ^{m}}}=g^{mn}\left({\frac {\partial ^{2}}{\partial \xi ^{m}\,\partial \xi ^{n}}}-\Gamma _{mn}^{l}{\frac {\partial }{\partial \xi ^{l}}}\right),}$$

where summation over the repeated indices is implied, g^mn is the inverse metric tensor and Γ^l _mn are the Christoffel symbols for the selected coordinates.

In arbitrary curvilinear coordinates in N dimensions (ξ¹, …, ξ^N), we can write the Laplacian in terms of the inverse metric tensor, ${\displaystyle g^{ij}}$:

$${\displaystyle \Delta ={\frac {1}{\sqrt {\det g}}}{\frac {\partial }{\partial \xi ^{i}}}\left({\sqrt {\det g}}g^{ij}{\frac {\partial }{\partial \xi ^{j}}}\right),}$$

from the Voss-Weyl formula[3] for the divergence.

In spherical coordinates in N dimensions, with the parametrization x = rθ ∈ R^N with r representing a positive real radius and θ an element of the unit sphere S^N−1,

$${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial r^{2}}}+{\frac {N-1}{r}}{\frac {\partial f}{\partial r}}+{\frac {1}{r^{2}}}\Delta _{S^{N-1}}f}$$

where Δ_S^N−1 is the Laplace–Beltrami operator on the (N − 1)-sphere, known as the spherical Laplacian. The two radial derivative terms can be equivalently rewritten as:

$${\displaystyle {\frac {1}{r^{N-1}}}{\frac {\partial }{\partial r}}\left(r^{N-1}{\frac {\partial f}{\partial r}}\right).}$$

As a consequence, the spherical Laplacian of a function defined on S^N−1 ⊂ R^N can be computed as the ordinary Laplacian of the function extended to R^N∖{0} so that it is constant along rays, i.e., homogeneous of degree zero.

The Laplacian of any tensor field ${\displaystyle \mathbf {T} }$ ("tensor" includes scalar and vector) is defined as the divergence of the gradient of the tensor:

$${\displaystyle \nabla ^{2}\mathbf {T} =(\nabla \cdot \nabla )\mathbf {T} .}$$

For the special case where ${\displaystyle \mathbf {T} }$ is a scalar (a tensor of degree zero), the Laplacian takes on the familiar form.

If ${\displaystyle \mathbf {T} }$ is a vector (a tensor of first degree), the gradient is a covariant derivative which results in a tensor of second degree, and the divergence of this is again a vector. The formula for the vector Laplacian above may be used to avoid tensor math and may be shown to be equivalent to the divergence of the Jacobian matrix shown below for the gradient of a vector:

$${\displaystyle \nabla \mathbf {T} =(\nabla T_{x},\nabla T_{y},\nabla T_{z})={\begin{bmatrix}T_{xx}&T_{xy}&T_{xz}\\T_{yx}&T_{yy}&T_{yz}\\T_{zx}&T_{zy}&T_{zz}\end{bmatrix}},{\text{ where }}T_{uv}\equiv {\frac {\partial T_{u}}{\partial v}}.}$$

And, in the same manner, a dot product, which evaluates to a vector, of a vector by the gradient of another vector (a tensor of 2nd degree) can be seen as a product of matrices:

$${\displaystyle \mathbf {A} \cdot \nabla \mathbf {B} ={\begin{bmatrix}A_{x}&A_{y}&A_{z}\end{bmatrix}}\nabla \mathbf {B} ={\begin{bmatrix}\mathbf {A} \cdot \nabla B_{x}&\mathbf {A} \cdot \nabla B_{y}&\mathbf {A} \cdot \nabla B_{z}\end{bmatrix}}.}$$

This identity is a coordinate dependent result, and is not general.

An example of the usage of the vector Laplacian is the Navier-Stokes equations for a Newtonian incompressible flow:

$${\displaystyle \rho \left({\frac {\partial \mathbf {v} }{\partial t}}+(\mathbf {v} \cdot \nabla )\mathbf {v} \right)=\rho \mathbf {f} -\nabla p+\mu \left(\nabla ^{2}\mathbf {v} \right),}$$

where the term with the vector Laplacian of the velocity field ${\displaystyle \mu \left(\nabla ^{2}\mathbf {v} \right)}$ represents the viscous stresses in the fluid.

Another example is the wave equation for the electric field that can be derived from Maxwell's equations in the absence of charges and currents:

$${\displaystyle \nabla ^{2}\mathbf {E} -\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}=0.}$$

This equation can also be written as:

$${\displaystyle \Box \,\mathbf {E} =0,}$$

where

$${\displaystyle \Box \equiv {\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\nabla ^{2},}$$

is the D'Alembertian, used in the Klein–Gordon equation.

The Laplacian also can be generalized to an elliptic operator called the Laplace–Beltrami operator defined on a Riemannian manifold. The Laplace–Beltrami operator, when applied to a function, is the trace (tr) of the function's Hessian:

$${\displaystyle \Delta f=\operatorname {tr} {\big (}H(f){\big )}}$$

where the trace is taken with respect to the inverse of the metric tensor. The Laplace–Beltrami operator also can be generalized to an operator (also called the Laplace–Beltrami operator) which operates on tensor fields, by a similar formula.

Another generalization of the Laplace operator that is available on pseudo-Riemannian manifolds uses the exterior derivative, in terms of which the "geometer's Laplacian" is expressed as

$${\displaystyle \Delta f=\delta df.}$$

Here δ is the codifferential, which can also be expressed in terms of the Hodge star and the exterior derivative. This operator differs in sign from the "analyst's Laplacian" defined above. More generally, the "Hodge" Laplacian is defined on differential forms α by

$${\displaystyle \Delta \alpha =\delta d\alpha +d\delta \alpha .}$$

This is known as the Laplace–de Rham operator, which is related to the Laplace–Beltrami operator by the Weitzenböck identity.

The Laplacian can be generalized in certain ways to non-Euclidean spaces, where it may be elliptic, hyperbolic, or ultrahyperbolic.

In Minkowski space the Laplace–Beltrami operator becomes the D'Alembert operator ${\displaystyle \Box }$ or D'Alembertian:

$${\displaystyle \square ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-{\frac {\partial ^{2}}{\partial x^{2}}}-{\frac {\partial ^{2}}{\partial y^{2}}}-{\frac {\partial ^{2}}{\partial z^{2}}}.}$$

It is the generalization of the Laplace operator in the sense that it is the differential operator which is invariant under the isometry group of the underlying space and it reduces to the Laplace operator if restricted to time-independent functions. The overall sign of the metric here is chosen such that the spatial parts of the operator admit a negative sign, which is the usual convention in high-energy particle physics. The D'Alembert operator is also known as the wave operator because it is the differential operator appearing in the wave equations, and it is also part of the Klein–Gordon equation, which reduces to the wave equation in the massless case.

The additional factor of c in the metric is needed in physics if space and time are measured in different units; a similar factor would be required if, for example, the x direction were measured in meters while the y direction were measured in centimeters. Indeed, theoretical physicists usually work in units such that c = 1 in order to simplify the equation.

The d'Alembert operator generalizes to a hyperbolic operator on pseudo-Riemannian manifolds.