Laplace operator: Definition from Answers.com

Wikipedia: Laplace operator

In mathematics and physics, the Laplace operator or Laplacian, denoted by $\Delta\,$ or $\nabla^2$ and named after Pierre-Simon Laplace, is a differential operator, specifically an important case of an elliptic operator, with many applications. In physics, it is used in modeling of wave propagation and heat flow, forming the Helmholtz equation. It is central in electrostatics, anchoring in Laplace's equation and Poisson's equation. In quantum mechanics, it represents the kinetic energy term of the Schrödinger equation. In mathematics, functions with vanishing Laplacian are called harmonic functions; the Laplacian is at the core of Hodge theory and the results of de Rham cohomology.

Definition

The Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence of the gradient:

$\Delta = \nabla^2 = \nabla \cdot \nabla.$

Equivalently, the Laplacian is the sum of all the unmixed second partial derivatives:

$\Delta = \sum_{i=1}^n \frac {\partial^2}{\partial x^2_i}.$

Here, it is understood that the $x i$ are Cartesian coordinates on the space; the equation takes a different form in spherical coordinates and cylindrical coordinates, as shown below.

In the three-dimensional space the Laplacian is commonly written as

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

As we shall see later, the Laplacian can be generalized to non-Euclidean spaces, where it may be elliptic or hyperbolic. For example, in the Minkowski space the Laplacian becomes the d'Alembert operator or d'Alembertian

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

The D'Alembert operator is often used to express the Klein-Gordon equation and the four-dimensional wave equation. The sign in front of the fourth term is negative, while it would have been positive in the Euclidean space. The additional factor of c is required because space and time are usually measured in different units; a similar factor would be required if, for example, the x direction were measured in inches, and the y direction were measured in centimeters. Indeed, physicists usually work in units such that c=1 in order to simplify the equation.

The Laplacian can operate on vector fields in addition to scalars. See vector Laplacian.

Motivation

One motivation for the Laplacian appearing in numerous areas of physics is that solutions to $Δ f = 0$ in a region U are functions that make the energy functional

$E(f) = \frac{1}{2} \int_U \Vert \nabla f \Vert^2 \mathrm{d}x$

stationary. To see this, suppose $f\colon U\to \mathbb{R}$ is a function, and $u\colon U\to \mathbb{R}$ is a function that vanishes on the boundary of U. Then

$\frac{d}{d\varepsilon}\Big|_{\varepsilon = 0} E(f+\varepsilon u) = \int_U \nabla f \cdot \nabla u \mathrm{d} x = -\int_U u \Delta f \mathrm{d} x$

where the last equality follows using Green's first identity. This calculation shows that if $Δ f = 0$ , then E is stationary around f. Conversely, if E is stationary around f, then $Δ f = 0$ by the fundamental theorem in calculus of variation.

Coordinate expressions

In three dimensions, it is common to work with the Laplacian in a variety of different coordinate systems. Given a function f, in cylindrical coordinates, one has:

$\Delta f = {1 \over r} {\partial \over \partial r} \left( r {\partial f \over \partial r} \right) + {1 \over r^2} {\partial^2 f \over \partial \theta^2} + {\partial^2 f \over \partial z^2 }.$

In spherical coordinates:

$\Delta f = {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial f \over \partial r} \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial f \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 f \over \partial \phi^2}.$

(here $θ$ represents the polar angle and $φ$ the azimuthal angle.) The term ${1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial f \over \partial r} \right)$ can be replaced by its equivalent ${1 \over r} {\partial^2 \over \partial r^2} \left( r f \right)$ as well. See also the article Del in cylindrical and spherical coordinates.

In spherical coordinates in $N$ dimensions, i.e. if $x \in {\mathbb R}^N$ is parametrized by $x = r θ$ with $r \in [0,+\infty)$ and $\theta \in S^{N-1}$ , one has:

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

where $\Delta_{S^{N-1}}$ is the Laplace-Beltrami operator on the $N - 1$ dimensional sphere $S N - 1$ . One can also write the term ${\partial^2 f \over \partial r^2} + \frac{N-1}{r} \frac{\partial f}{\partial r}$ equivalently as Failed to parse (unknown function\Bigl): \frac{1}{r^{N-1}} \frac{\partial}{\partial r} \Bigl(r^{N-1} \frac{\partial f}{\partial r} \Bigr)

Identities

The Laplacian of a function is the trace of the function's Hessian.

If f and g are functions, then the Laplacian of the product is given by

$\Delta(fg)=(\Delta f)g+2((\nabla f)\cdot(\nabla g))+f(\Delta g).$

Note the special case where f is a radial function $f (r)$ and g is a spherical harmonic, $Y l m (θ,φ)$ . One encounters this special case in numerous physical models. The gradient of $f (r)$ is a radial vector and the gradient of an angular function is tangent to the radial vector, therefore

$2(\nabla f(r))\cdot(\nabla Y_{lm}(\theta,\phi))=0.$

In addition, the spherical harmonics have the special property of being eigenfunctions of the angular part of the Laplacian in spherical coordinates.

$\Delta Y_{\ell m}(\theta,\phi) = -\frac{\ell(\ell+1)}{r^2} Y_{\ell m}(\theta,\phi).$

Therefore,

$\Delta( f(r)Y_{\ell m}(\theta,\phi) ) = \left(\frac{d^2f(r)}{dr^2} + \frac{2}{r} \frac{df(r)}{dr} - \frac{\ell(\ell+1)}{r^2} f(r)\right)Y_{\ell m}(\theta,\phi).$

Laplace-Beltrami operator

The Laplacian can be extended to functions defined on surfaces, or more generally, on Riemannian and pseudo-Riemannian manifolds. This more general operator goes by the name Laplace-Beltrami operator. One defines it, just as the Laplacian, as the divergence of the gradient. To be able to find a formula for this operator, one will need to first write the divergence and the gradient on a manifold.

If $g$ denotes the (pseudo)-metric tensor on the manifold, one finds that the volume form in local coordinates is given by

$\mathrm{vol}_n := \sqrt{|g|} \;dx^1\wedge \ldots \wedge dx^n$

where the $d x i$ are the 1-forms forming the dual basis to the basis vectors

$\partial_i := \frac {\partial}{\partial x^i}$

for the local coordinate system, and $\wedge$ is the wedge product. Here $| g | : = | det g i j |$ is the absolute value of the determinant of the metric tensor. The divergence of a vector field X on the manifold can then be defined as

$(\mbox{div} X) \; \mathrm{vol}_n := \mathcal{L}_X \mathrm{vol}_n$

where $L X$ is the Lie derivative along the vector field X. In local coordinates, one obtains

$\mbox{div} X = \frac{1}{\sqrt{|g|}} \partial_i \left(\sqrt {|g|} X^i\right).$

Here (and below) we use the Einstein notation, so the above is actually a sum in i. The gradient of a scalar function f may be defined through the inner product $\langle\cdot,\cdot\rangle$ on the manifold, as

$\langle \mbox{grad} f(x) , v_x \rangle = df(x)(v_x)$

for all vectors $v x$ anchored at point x in the tangent bundle $T x M$ of the manifold at point x. Here, df is the exterior derivative of the function f; it is a 1-form taking argument $v x$ . In local coordinates, one has

$\left(\mbox{grad} f\right)^i = \partial^i f = g^{ij} \partial_j f.$

Combining these, the formula for the Laplace-Beltrami operator applied to a scalar function f is, in local coordinates

$\Delta f = \mbox{div grad} \; f = \frac{1}{\sqrt {|g|}} \partial_i \left(\sqrt{|g|} g^{ij} \partial_j f \right).$

Here, $g i j$ are the components of the inverse of the metric tensor $g$ , so that $g^{ij}g_{jk}=\delta^i_k$ with $\delta^i_k$ the Kronecker delta.

Note that the above definition is, by construction, valid only for scalar functions $f:M\rightarrow \mathbb{R}$ . One may want to extend the Laplacian even further, to differential forms; for this, one must turn to the Laplace-deRham operator, defined in the next section. One may show that the Laplace-Beltrami operator reduces to the ordinary Laplacian in Euclidean space by noting that it can be re-written using the product and chain rule as

$\Delta f = \partial_i \partial^i f + (\partial^i f) \partial_i \ln \sqrt{|g|}.$

When $| g | = 1$ , such as in the case of Euclidean space with Cartesian coordinates, one then easily obtains

Δ f = \partial i \partial i f

which is the ordinary Laplacian. Using the Minkowski metric with signature (+++-), one regains the D'Alembertian given previously. Under local parametrization $u 1, u 2$ , the Laplace-Beltrami operator can be expanded in terms of the metric tensor and Christoffel symbols as follows:

$\Delta f = g^{ij}\left(\frac{\partial^2 f}{\partial u^i\, \partial u^j} - \Gamma_{ij}^k \frac{\partial f}{\partial u^k} \right).$

Note that by using the metric tensor for spherical and cylindrical coordinates, one can similarly regain the expressions for the Laplacian in spherical and cylindrical coordinates. The Laplace-Beltrami operator is handy not just in curved space, but also in ordinary flat space endowed with a non-linear coordinate system.

Note also that the exterior derivative d and -div are adjoint:

$\int_M df(X) \;\mathrm{vol}_n = - \int_M f \mbox{div} X \;\mathrm{vol}_n$ (proof)

where the last equality is an application of Stokes theorem. Note also, the Laplace-Beltrami operator is symmetric:

$\int_M f\Delta h \;\mathrm{vol}_n = \int_M \langle \mbox{grad} f, \mbox{grad} h \rangle \;\mathrm{vol}_n = \int_M h\Delta f \;\mathrm{vol}_n$

for functions f and h.

Laplace-de Rham operator

In the general case of differential geometry, one defines the Laplace-de Rham operator as the generalization of the Laplacian. It is a differential operator on the exterior algebra of a differentiable manifold. On a Riemannian manifold it is an elliptic operator, while on a Lorentzian manifold it is hyperbolic. The Laplace-de Rham operator is defined by

$\Delta= \mathrm{d}\delta+\delta\mathrm{d} = (\mathrm{d}+\delta)^2,\;$

where d is the exterior derivative or differential and δ is the codifferential. When acting on scalar functions, the codifferential may be defined as δ = −∗d∗, where ∗ is the Hodge star; more generally, the codifferential may include a sign that depends on the order of the k-form being acted on.

One may prove that the Laplace-de Rham operator is equivalent to the previous definition of the Laplace-Beltrami operator when acting on a scalar function f; see the Laplace operator article proofs for details. Notice that the Laplace-de Rham operator is actually minus the Laplace-Beltrami operator; this minus sign follows from the conventional definition of the properties of the codifferential. Unfortunately, Δ is used to denote both; which can sometimes be a source of confusion.

Properties

Given scalar functions f and h, and a real number a, the Laplace-de Rham operator has the following properties:

$\Delta(af + h) = a\Delta f + \Delta h\!$
$Δ(f h) = f Δ h + 2(\partial i f)(\partial i h) + h Δ f$ (proof)

References

Charles W. Misner, Kip S. Thorne, John Archibald Wheeler, Gravitation, (1970) W.H. Freeman, New York; ISBN 0-7167-0344-0. (Provides a basic review of differential geometry in the special case of four-dimensional space-time.)
Jürgen Jost, Riemannian Geometry and Geometric Analysis, (2002) Springer-Verlag, Berlin ISBN 3-540-42627-2 . (Provides a general introduction to curved surfaces).

External links

M.A. Shubin (2001), "Laplace operator", in Hazewinkel, Michiel, Encyclopaedia of Mathematics, Kluwer Academic Publishers, ISBN 1-55608-010-7
MathWorld: Laplacian
Derivation of the Laplacian in Spherical coordinates by Swapnil Sunil Jain

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Laplace operator