In mathematics, Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace who first studied its properties. This is often written as:

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta\varphi = 0 \qquad\mbox{or}\qquad \nabla^2 \varphi = 0

where ∆ = ∇² is the Laplace operator and Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi

is a scalar function.  In general, ∆ = ∇² is the Laplace–Beltrami or Laplace–de Rham operator.

Laplace's equation and Poisson's equation are the simplest examples of elliptic partial differential equations. Solutions of Laplace's equation are called harmonic functions.

The general theory of solutions to Laplace's equation is known as potential theory. The solutions of Laplace's equation are the harmonic functions, which are important in many fields of science, notably the fields of electromagnetism, astronomy, and fluid dynamics, because they can be used to accurately describe the behavior of electric, gravitational, and fluid potentials. In the study of heat conduction, the Laplace equation is the steady-state heat equation.

Definition[link]

In three dimensions, the problem is to find twice-differentiable real-valued functions Failed to parse (Missing texvc executable; please see math/README to configure.): \scriptstyle f , of real variables x, y, and z, such that

In Cartesian coordinates

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta f = {\partial^2 f\over \partial x^2 } + {\partial^2 f\over \partial y^2 } + {\partial^2 f\over \partial z^2 } = 0.

In cylindrical coordinates,

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 \phi^2} + {\partial^2 f \over \partial z^2 } =0

In spherical coordinates,

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta f = {1 \over \rho^2}{\partial \over \partial \rho}\!\left(\rho^2 {\partial f \over \partial \rho}\right) \!+\!{1 \over \rho^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right) \!+\!{1 \over \rho^2\!\sin^2\theta}{\partial^2 f \over \partial \varphi^2} =0.

In Curvilinear coordinates,

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta f = {\partial \over \partial \xi^i}\!\left({\partial f \over \partial \xi^k}g^{ki}\right) \!+\!{\partial f \over \partial \xi^j}g^{jm}\Gamma^n_{mn} =0,

or

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta f = {1 \over \sqrt{|g|}}{\partial \over \partial \xi^i}\!\left(\sqrt{|g|}g^{ij}{\partial f \over \partial \xi^j}\right) =0, \quad (g=\mathrm{det}\{g_{ij}\}).

This is often written as

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla^2 \varphi = 0 \,

or, especially in more general contexts,

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta \varphi = 0,\,

where ∆ = ∇² is the Laplace operator or "Laplacian"

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta \varphi = \nabla^2 \varphi =\nabla \cdot \nabla \varphi =\operatorname{div}\; \operatorname{grad}\,\varphi,

where ∇ ⋅ = div is the divergence, and ∇ = grad is the gradient.

If the right-hand side is specified as a given function, f(x, y, z), i.e., if the whole equation is written as

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta\varphi = f\,

then it is called "Poisson's equation".

The Laplace equation is also a special case of the Helmholtz equation.

Boundary conditions[link]

Laplace's Equation on an annulus (r=2 and R=4) with Dirichlet Boundary Conditions: u(r=2)=0 and u(r=4)=4sin(5*θ)

The Dirichlet problem for Laplace's equation consists of finding a solution Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi

on some domain Failed to parse (Missing texvc executable; please see math/README to configure.): D
such that Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi
on the boundary of Failed to parse (Missing texvc executable; please see math/README to configure.): D
is equal to some given function. Since the Laplace operator appears in the heat equation, one physical interpretation of this problem is as follows: fix the temperature on the boundary of the domain according to the given specification of the boundary condition. Allow heat to flow until a stationary state is reached in which the temperature at each point on the domain doesn't change anymore. The temperature distribution in the interior will then be given by the solution to the corresponding Dirichlet problem.

The Neumann boundary conditions for Laplace's equation specify not the function Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi

itself on the boundary of Failed to parse (Missing texvc executable; please see math/README to configure.): D

, but its normal derivative. Physically, this corresponds to the construction of a potential for a vector field whose effect is known at the boundary of Failed to parse (Missing texvc executable; please see math/README to configure.): D

alone.

Solutions of Laplace's equation are called harmonic functions; they are all analytic within the domain where the equation is satisfied. If any two functions are solutions to Laplace's equation (or any linear homogeneous differential equation), their sum (or any linear combination) is also a solution. This property, called the principle of superposition, is very useful, e.g., solutions to complex problems can be constructed by summing simple solutions.

Laplace equation in two dimensions[link]

The Laplace equation in two independent variables has the form

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} \equiv \psi_{xx} + \psi_{yy} = 0.

Analytic functions[link]

The real and imaginary parts of a complex analytic function both satisfy the Laplace equation. That is, if z = x + iy, and if

Failed to parse (Missing texvc executable; please see math/README to configure.): f(z) = u(x,y) + iv(x,y),\,

then the necessary condition that f(z) be analytic is that the Cauchy-Riemann equations be satisfied:

Failed to parse (Missing texvc executable; please see math/README to configure.): u_x = v_y, \quad v_x = -u_y.\,

where u_x is the first partial derivative of u with respect to x.

It follows that

Failed to parse (Missing texvc executable; please see math/README to configure.): u_{yy} = (-v_x)_y = -(v_y)_x = -(u_x)_x.\,

Therefore u satisfies the Laplace equation. A similar calculation shows that v also satisfies the Laplace equation.

Conversely, given a harmonic function, it is the real part of an analytic function, Failed to parse (Missing texvc executable; please see math/README to configure.): f(z)

(at least locally). If a trial form is

Failed to parse (Missing texvc executable; please see math/README to configure.): f(z) = \varphi(x,y) + i \psi(x,y),\,

then the Cauchy-Riemann equations will be satisfied if we set

Failed to parse (Missing texvc executable; please see math/README to configure.): \psi_x = -\varphi_y, \quad \psi_y = \varphi_x.\,

This relation does not determine ψ, but only its increments:

Failed to parse (Missing texvc executable; please see math/README to configure.): d \psi = -\varphi_y\, dx + \varphi_x\, dy.\,

The Laplace equation for φ implies that the integrability condition for ψ is satisfied:

Failed to parse (Missing texvc executable; please see math/README to configure.): \psi_{xy} = \psi_{yx},\,

and thus ψ may be defined by a line integral. The integrability condition and Stokes' theorem implies that the value of the line integral connecting two points is independent of the path. The resulting pair of solutions of the Laplace equation are called conjugate harmonic functions. This construction is only valid locally, or provided that the path does not loop around a singularity. For example, if r and θ are polar coordinates and

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi = \log r, \,

then a corresponding analytic function is

Failed to parse (Missing texvc executable; please see math/README to configure.): f(z) = \log z = \log r + i\theta. \,

However, the angle θ is single-valued only in a region that does not enclose the origin.

The close connection between the Laplace equation and analytic functions implies that any solution of the Laplace equation has derivatives of all orders, and can be expanded in a power series, at least inside a circle that does not enclose a singularity. This is in sharp contrast to solutions of the wave equation, which generally have less regularity.

There is an intimate connection between power series and Fourier series. If we expand a function f in a power series inside a circle of radius R, this means that

Failed to parse (Missing texvc executable; please see math/README to configure.): f(z) = \sum_{n=0}^\infty c_n z^n,\,

with suitably defined coefficients whose real and imaginary parts are given by

Failed to parse (Missing texvc executable; please see math/README to configure.): c_n = a_n + i b_n.\,

Therefore

Failed to parse (Missing texvc executable; please see math/README to configure.): f(z) = \sum_{n=0}^\infty \left[ a_n r^n \cos n \theta + b_n r^n \sin n \theta\right] + i \sum_{n=1}^\infty \left[ a_n r^n \sin n\theta + b_n r^n \cos n \theta\right],\,

which is a Fourier series for f. These trigonometric functions can themselves be expanded, using multiple angle formulae.

Fluid flow[link]

Let the quantities u and v be the horizontal and vertical components of the velocity field of a steady incompressible, irrotational flow in two dimensions. The condition that the flow be incompressible is that

Failed to parse (Missing texvc executable; please see math/README to configure.): u_x + v_y=0,\,

and the condition that the flow be irrotational is that

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla \times \mathbf{V}=v_x - u_y =0. \,

If we define the differential of a function ψ by

Failed to parse (Missing texvc executable; please see math/README to configure.): d \psi = v\, dx - u\, dy,\,

then the incompressibility condition is the integrability condition for this differential: the resulting function is called the stream function because it is constant along flow lines. The first derivatives of ψ are given by

Failed to parse (Missing texvc executable; please see math/README to configure.): \psi_x = v, \quad \psi_y=-u, \,

and the irrotationality condition implies that ψ satisfies the Laplace equation. The harmonic function φ that is conjugate to ψ is called the velocity potential. The Cauchy-Riemann equations imply that

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi_x=-u, \quad \varphi_y=-v. \,

Thus every analytic function corresponds to a steady incompressible, irrotational fluid flow in the plane. The real part is the velocity potential, and the imaginary part is the stream function.

Electrostatics[link]

According to Maxwell's equations, an electric field (u,v) in two space dimensions that is independent of time satisfies

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla \times (u,v) = v_x -u_y =0,\,

and

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla \cdot (u,v) = \rho,\,

where ρ is the charge density. The first Maxwell equation is the integrability condition for the differential

Failed to parse (Missing texvc executable; please see math/README to configure.): d \varphi = -u\, dx -v\, dy,\,

so the electric potential φ may be constructed to satisfy

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi_x = -u, \quad \varphi_y = -v.\,

The second of Maxwell's equations then implies that

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi_{xx} + \varphi_{yy} = -\rho,\,

which is the Poisson equation.

It is important to note that the Laplace equation can be used in three-dimensional problems in electrostatics and fluid flow just as in two dimensions.

Laplace equation in three dimensions[link]

Fundamental solution[link]

A fundamental solution of Laplace's equation satisfies

Failed to parse (Missing texvc executable; please see math/README to configure.): \Delta u = u_{xx} + u_{yy} + u_{zz} = -\delta(x-x',y-y',z-z'), \,

where the Dirac delta function Failed to parse (Missing texvc executable; please see math/README to configure.): \delta

denotes a unit source concentrated at the point Failed to parse (Missing texvc executable; please see math/README to configure.):  (x',\, y', \, z').

No function has this property, but it can be thought of as a limit of functions whose integrals over space are unity, and whose support (the region where the function is non-zero) shrinks to a point (see weak solution). It is common to take a different sign convention for this equation than one typically does when defining fundamental solutions. This choice of sign is often convenient to work with because Failed to parse (Missing texvc executable; please see math/README to configure.): -\Delta

is a positive operator. The definition of the fundamental solution thus implies that, if the Laplacian of u is integrated over any volume that encloses the source point, then

Failed to parse (Missing texvc executable; please see math/README to configure.): \iiint_V \nabla \cdot \nabla u \, dV =-1. \,

The Laplace equation is unchanged under a rotation of coordinates, and hence we can expect that a fundamental solution may be obtained among solutions that only depend upon the distance r from the source point. If we choose the volume to be a ball of radius a around the source point, then Gauss' divergence theorem implies that

Failed to parse (Missing texvc executable; please see math/README to configure.): -1= \iiint_V \nabla \cdot \nabla u \, dV = \iint_S \frac{du}{dr} \, dS = \left.4\pi a^2 \frac{du}{dr}\right|_{r=a}.\,

It follows that

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{du}{dr} = -\frac{1}{4\pi r^2},\,

on a sphere of radius r that is centered around the source point, and hence

Failed to parse (Missing texvc executable; please see math/README to configure.): u = \frac{1}{4\pi r}.\,

Note that, with the opposite sign convention (used in Physics), this is the potential generated by a point particle, for an inverse-square law force, arising in the solution of Poisson equation. A similar argument shows that in two dimensions

Failed to parse (Missing texvc executable; please see math/README to configure.): u = \frac{\log(r)}{2\pi}. \,

where Failed to parse (Missing texvc executable; please see math/README to configure.): \log(r)

denotes the natural logarithm.

Note that, with the opposite sign convention, this is the potential generated by a pointlike vortex (see point particle), which is the solution of Euler equations in two-dimensional incompressible flow.

Green's function[link]

A Green's function is a fundamental solution that also satisfies a suitable condition on the boundary S of a volume V. For instance, Failed to parse (Missing texvc executable; please see math/README to configure.): G(x,y,z;x',y',z')\,

may satisfy

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla \cdot \nabla G = -\delta(x-x',y-y',z-z') \quad \hbox{in} \quad V, \,

Failed to parse (Missing texvc executable; please see math/README to configure.): G = 0 \quad \hbox{if} \quad (x,y,z) \quad \hbox{on} \quad S. \,

Now if u is any solution of the Poisson equation in V:

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla \cdot \nabla u = -f, \,

and u assumes the boundary values g on S, then we may apply Green's identity, (a consequence of the divergence theorem) which states that

Failed to parse (Missing texvc executable; please see math/README to configure.): \iiint_V \left[ G \, \nabla \cdot \nabla u - u \, \nabla \cdot \nabla G \right]\, dV = \iiint_V \nabla \cdot \left[ G \nabla u - u \nabla G \right]\, dV = \iint_S \left[ G u_n -u G_n \right] \, dS. \,

The notations u_n and G_n denote normal derivatives on S. In view of the conditions satisfied by u and G, this result simplifies to

Failed to parse (Missing texvc executable; please see math/README to configure.): u(x',y',z') = \iiint_V G f \, dV + \iint_S G_n g \, dS. \,

Thus the Green's function describes the influence at Failed to parse (Missing texvc executable; please see math/README to configure.): \scriptstyle (x',y',z')\,

of the data f and g. For the case of the interior of a sphere of radius a, the Green's function may be obtained by means of a reflection (Sommerfeld, 1949): the source point P at distance ρ from the center of the sphere is reflected along its radial line to a point P' that is at a distance 

Failed to parse (Missing texvc executable; please see math/README to configure.): \rho' = \frac{a^2}{\rho}. \,

Note that if P is inside the sphere, then P' will be outside the sphere. The Green's function is then given by

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{4 \pi R} - \frac{a}{4 \pi \rho R'}, \,

where R denotes the distance to the source point P and R' denotes the distance to the reflected point P'. A consequence of this expression for the Green's function is the Poisson integral formula. Let ρ, θ, and φ be spherical coordinates for the source point P. Here θ denotes the angle with the vertical axis, which is contrary to the usual American mathematical notation, but agrees with standard European and physical practice. Then the solution of the Laplace equation inside the sphere is given by

Failed to parse (Missing texvc executable; please see math/README to configure.): u(P) = \frac{1}{4\pi} a^3\left( 1 - \frac{\rho^2}{a^2} \right) \iint \frac{g(\theta',\varphi') \sin \varphi' \, d\theta' \, d\varphi'}{(a^2 + \rho^2 - 2 a \rho \cos \Theta)^{3/2} }, \,

where

Failed to parse (Missing texvc executable; please see math/README to configure.): \cos \Theta = \cos \varphi \cos \varphi' + \sin\varphi \sin\varphi'\cos(\theta -\theta'). \,

A simple consequence of this formula is that if u is a harmonic function, then the value of u at the center of the sphere is the mean value of its values on the sphere. This mean value property immediately implies that a non-constant harmonic function cannot assume its maximum value at an interior point.

Electrostatics[link]

In free space the Laplace equation of any electrostatic potential must equal zero since Failed to parse (Missing texvc executable; please see math/README to configure.): \rho

(charge density) is zero in free space.

Taking the gradient of the electric potential we get the electrostatic field

Failed to parse (Missing texvc executable; please see math/README to configure.): E=-\nabla V

Taking the divergence of the electrostatic field, we obtain Poisson's equation, that relates charge density and electric potential

Failed to parse (Missing texvc executable; please see math/README to configure.): \nabla^2V = -\frac{\rho}{\varepsilon_0}

In the particular case of the empty space (Failed to parse (Missing texvc executable; please see math/README to configure.): \rho=0 ) Poisson's equation reduces to Laplace's equation for the electric potential.

Using a uniqueness theorem and showing that a potential satisfies Laplace's equation (second derivative of V should be zero i.e. in free space) and the potential has the correct values at the boundaries, the potential is then uniquely defined.

A potential that doesn't satisfy Laplace's equation together with the boundary condition is an invalid electrostatic potential.

See also[link]

• Spherical harmonic
• Quadrature domains
• Potential theory
• Potential flow
• Bateman transform
• Earnshaw's theorem uses the Laplace equation to show that stable static ferromagnetic suspension is impossible
• Vector Laplacian

References[link]

• Evans, L. C. (1998). Partial Differential Equations. Providence: American Mathematical Society. ISBN 0-8218-0772-2. 
• Petrovsky, I. G. (1967). Partial Differential Equations. Philadelphia: W. B. Saunders. 
• Polyanin, A. D. (2002). Handbook of Linear Partial Differential Equations for Engineers and Scientists. Boca Raton: Chapman & Hall/CRC Press. ISBN 1-58488-299-9. 
• Sommerfeld, A. (1949). Partial Differential Equations in Physics. New York: Academic Press. 

External links[link]

• Laplace Equation (particular solutions and boundary value problems) at EqWorld: The World of Mathematical Equations.
• Laplace Differential Equation on PlanetMath
• Example initial-boundary value problems using Laplace's equation from exampleproblems.com.
• Weisstein, Eric W., "Laplace’s Equation" from MathWorld.
• Module for Laplace’s Equation by John H. Mathews
• Find out how boundary value problems governed by Laplace's equation may be solved numerically by boundary element method