Definitions

# Laplace's equation

In mathematics, Laplace's equation is a partial differential equation named after Pierre-Simon Laplace who first studied its properties. The solutions of Laplace's equation are important in many fields of science, notably the fields of electromagnetism, astronomy, and fluid dynamics, because they describe the behavior of electric, gravitational, and fluid potentials. The general theory of solutions to Laplace's equation is known as potential theory. In the study of heat conduction, the Laplace equation is the steady-state heat equation.

## Definition

In three dimensions, the problem is to find twice-differentiable real-valued functions, $scriptstylevarphi$ of real variables, x, y, and z, such that


{partial^2 varphiover partial x^2 } + {partial^2 varphiover partial y^2 } + {partial^2 varphiover partial z^2 } = 0.

This is often written as

$nabla^2 varphi = 0 ,$

or

$operatorname\left\{div\right\},operatorname\left\{grad\right\},varphi = 0,$

$Delta varphi = 0,,$

where Δ is the Laplace operator.

Solutions of Laplace's equation are called harmonic functions.

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

$Delta varphi = f,$

then the equation is called "Poisson's equation." Laplace's equation and Poisson's equation are the simplest examples of elliptic partial differential equations. The partial differential operator, $scriptstylenabla^2$, or $scriptstyleDelta$, (which may be defined in any number of dimensions) is called the Laplace operator, or just the Laplacian.

## Boundary conditions

The Dirichlet problem for Laplace's equation consists of finding a solution $varphi$ on some domain $D$ such that $varphi$ on the boundary of $D$ 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 and wait until the temperature in the interior 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 $varphi$ itself on the boundary of $D$, 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 $D$ 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 homogenous 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

The Laplace equation in two independent variables has the form

$varphi_\left\{xx\right\} + varphi_\left\{yy\right\} = 0.,$

### Analytic functions

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

$f\left(z\right) = u\left(x,y\right) + iv\left(x,y\right),,$

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

$u_x = v_y, quad v_x = -u_y.,$

It follows that

$u_\left\{yy\right\} = \left(-v_x\right)_y = -\left(v_y\right)_x = -\left(u_x\right)_x.,$

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, $f\left(z\right)$ (at least locally). If a trial form is

$f\left(z\right) = varphi\left(x,y\right) + i psi\left(x,y\right),,$

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

$psi_x = -varphi_y, quad psi_y = varphi_x.,$

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

$d psi = -varphi_y, dx + varphi_x, dy.,$

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

$psi_\left\{xy\right\} = psi_\left\{yx\right\},,$

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

$varphi = log r, ,$

then a corresponding analytic function is

$f\left(z\right) = log z = log r + itheta. ,$

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

$f\left(z\right) = sum_\left\{n=0\right\}^infty c_n z^n,,$

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

$c_n = a_n + i b_n.,$

Therefore

$f\left(z\right) = sum_\left\{n=0\right\}^infty left\left[a_n r^n cos n theta - b_n r^n sin n thetaright\right] + i sum_\left\{n=1\right\}^infty left\left[a_n r^n sin ntheta + b_n r^n cos n thetaright\right],,$

which is a Fourier series for f.

### Fluid flow

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

$u_x + v_y=0,,$

and the condition that the flow be irrotational is that

$nabla times mathbf\left\{V\right\}=v_x - u_y =0. ,$

If we define the differential of a function ψ by

$d psi = v, dx - u, dy,,$

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

$psi_x = v, quad psi_y=-u, ,$

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

$varphi_x=-u, quad varphi_y=-v. ,$

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

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

$nabla times \left(u,v\right) = v_x -u_y =0,,$

and

$nabla cdot \left(u,v\right) = rho,,$

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

$d varphi = -u, dx -v, dy,,$

so the electric potential φ may be constructed to satisfy

$varphi_x = -u, quad varphi_y = -v.,$

The second of Maxwell's equations then implies that

$varphi_\left\{xx\right\} + varphi_\left\{yy\right\} = -rho,,$

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

### Fundamental solution

A fundamental solution of Laplace's equation satisfies

$Delta u = u_\left\{xx\right\} + u_\left\{yy\right\} + u_\left\{zz\right\} = -delta\left(x-x\text{'},y-y\text{'},z-z\text{'}\right), ,$

where the Dirac delta function $delta$ denotes a unit source concentrated at the point $\left(x\text{'},, y\text{'}, , z\text{'}\right).$ 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). 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

$iiint_V operatorname\left\{div\right\} nabla u , dV =-1. ,$

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

$-1= iiint_V operatorname\left\{div\right\} nabla u , dV = iint_S u_r dS = 4pi a^2 u_r\left(a\right).,$

It follows that

$u_r\left(r\right) = -frac\left\{1\right\}\left\{4pi r^2\right\},,$

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

$u = frac\left\{1\right\}\left\{4pi r\right\}.,$

A similar argument shows that in two dimensions

$u = frac\left\{-log r\right\}\left\{2pi\right\}. ,$

### Green's function

A Green's function is a fundamental solution that also satisfies a suitable condition on the boundary S of a volume V. For instance, $scriptstyle G\left(x,y,z;x\text{'},y\text{'},z\text{'}\right),$ may satisfy

$nabla cdot nabla G = -delta\left(x-x\text{'},y-y\text{'},z-z\text{'}\right) quad hbox\left\{in\right\} quad V, ,$

$G = 0 quad hbox\left\{if\right\} quad \left(x,y,z\right) quad hbox\left\{on\right\} quad S. ,$

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

$nabla cdot nabla u = -f, ,$

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

$iiint_V left\left[G , nabla cdot nabla u - u , nabla cdot nabla G right\right], dV = iiint_V nabla cdot left\left[G nabla u - u nabla G right\right], dV = iint_S left\left[G u_n -u G_n right\right] , dS. ,$

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

$u\left(x\text{'},y\text{'},z\text{'}\right) = iiint_V G f , dV + iint_S G_n g , dS. ,$

Thus the Green's function describes the influence at $scriptstyle \left(x\text{'},y\text{'},z\text{'}\right),$ 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

$rho\text{'} = frac\left\{a^2\right\}\left\{rho\right\}. ,$

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

$frac\left\{1\right\}\left\{4 pi R\right\} - frac\left\{a\right\}\left\{4 pi rho R\text{'}\right\}, ,$

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

$u\left(P\right) = frac\left\{1\right\}\left\{4pi\right\} a^3left\left(1 - frac\left\{rho^2\right\}\left\{a^2\right\} right\right) iint frac\left\{g\left(theta\text{'},varphi\text{'}\right) sin varphi\text{'} , dtheta\text{'} , dvarphi\text{'}\right\}\left\{\left(a^2 + rho^2 - 2 a rho cos Theta\right)^\left\{3/2\right\} \right\}, ,$

where

$cos Theta = cos varphi cos varphi\text{'} + sinvarphi sinvarphi\text{'}cos\left(theta -theta\text{'}\right). ,$

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.