Benford's law

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:

$\Delta \varphi =0\qquad {\mbox{or}}\qquad \nabla ^{2}\varphi =0$

where ∆ = ∇² is the Laplace operator and φ 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

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

In Cartesian coordinates

$\Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}=0.$

In cylindrical coordinates,

$\Delta f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \phi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}=0$

In spherical coordinates,

$\Delta f={\frac {1}{\rho ^{2}}}{\frac {\partial }{\partial \rho }}\left(\rho ^{2}{\frac {\partial f}{\partial \rho }}\right)+{\frac {1}{\rho ^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{\rho ^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}=0.$

In Curvilinear coordinates,

$\Delta f={\frac {\partial }{\partial \xi ^{j}}}\left({\frac {\partial f}{\partial \xi ^{k}}}g^{ki}\right)+{\frac {\partial f}{\partial \xi ^{j}}}g^{jm}\Gamma _{mn}^{n}=0,$

or

$\Delta f={\frac {1}{\sqrt {|g|}}}{\frac {\partial }{\partial \xi ^{i}}}\!\left({\sqrt {|g|}}g^{ij}{\frac {\partial f}{\partial \xi ^{j}}}\right)=0,\qquad (g=\mathrm {det} \{g_{ij}\}).$

This is often written as

$\nabla ^{2}f=0$

or, especially in more general contexts,

$\Delta f=0,$

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

$\Delta f=\nabla ^{2}f=\nabla \cdot \nabla f=\operatorname {div} \operatorname {grad} f,$

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

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

$\Delta f=h$

then it is called "Poisson's equation".

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

Boundary conditions

Laplace's Equation on an annulus (inner radius r=2 and outer radius 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 φ on some domain D such that φ 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 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 φ 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 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

The Laplace equation in two independent variables has the form

${\frac {\partial ^{2}\psi }{\partial x^{2}}}+{\frac {\partial ^{2}\psi }{\partial y^{2}}}\equiv \psi _{xx}+\psi _{yy}=0.$

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

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

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

$u_{x}=v_{y},\quad v_{x}=-u_{y}.$

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

$u_{yy}=(-v_{x})_{y}=-(v_{y})_{x}=-(u_{x})_{x}.$

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, f(z) (at least locally). If a trial form is

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

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:

$\psi _{xy}=\psi _{yx},$

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

$\varphi =\log r,$

then a corresponding analytic function is

$f(z)=\log z=\log r+i\theta .$

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

$f(z)=\sum _{n=0}^{\infty }c_{n}z^{n},$

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

$c_{n}=a_{n}+ib_{n}.$

Therefore

$f(z)=\sum _{n=0}^{\infty }\left[a_{n}r^{n}\cos n\theta -b_{n}r^{n}\sin n\theta \right]+i\sum _{n=1}^{\infty }\left[a_{n}r^{n}\sin n\theta +b_{n}r^{n}\cos n\theta \right],$

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

Fluid flow

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

$\nabla \times \mathbf {V} =v_{x}-u_{y}=0.$

If we define the differential of a function ψ by

$d\psi =vdx-udy,$

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

$\nabla \times (u,v)=v_{x}-u_{y}=0,$

and:$\nabla \cdot (u,v)=\rho ,$ 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

$\varphi _{xx}+\varphi _{yy}=-\rho ,$

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

Fundamental solution

A fundamental solution of Laplace's equation satisfies

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

where the Dirac delta function δ denotes a unit source concentrated at the point (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 −Δ 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

$\iiint _{V}\nabla \cdot \nabla u\,dV=-1.$

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

$-1=\iiint _{V}\nabla \cdot \nabla u\,dV=\iint _{S}{\frac {du}{dr}}\,dS=\left.4\pi a^{2}{\frac {du}{dr}}\right|_{r=a}.$

It follows that

${\frac {du}{dr}}=-{\frac {1}{4\pi r^{2}}},$

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

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

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

$u=-{\frac {\log(r)}{2\pi }}.$

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

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,

$G(x,y,z;x',y',z')$

may satisfy

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

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

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

$\iiint _{V}\left[G\,\nabla \cdot \nabla u-u\,\nabla \cdot \nabla G\right]\,dV=\iiint _{V}\nabla \cdot \left[G\nabla u-u\nabla G\right]\,dV=\iint _{S}\left[Gu_{n}-uG_{n}\right]\,dS.\,$

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

$u(x',y',z')=\iiint _{V}Gf\,dV+\iint _{S}G_{n}g\,dS.\,$

Thus the Green's function describes the influence at (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

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

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

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

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

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

where

$\cos \Theta =\cos \varphi \cos \varphi '+\sin \varphi \sin \varphi '\cos(\theta -\theta ').$

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

In free space the Laplace equation of any electrostatic potential must equal zero since ρ (charge density) is zero in free space.

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

$E=-\nabla V$

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

$\nabla ^{2}V=-{\frac {\rho }{\varepsilon _{0}}}$

In the particular case of the empty space (ρ = 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

• 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

• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534

External links

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• 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.
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• 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