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Green's function
In mathematics, a Green's function is the impulse response of an inhomogeneous differential equation defined on a domain, with specified initial conditions or boundary conditions. Via the superposition principle, the convolution of a Green's function with an arbitrary function f(x) on that domain is the solution to the inhomogeneous differential equation for f(x).
Green's functions are named after the British mathematician George Green, who first developed the concept in the 1830s. In the modern study of linear partial differential equations, Green's functions are studied largely from the point of view of fundamental solutions instead.
Under manybody theory, the term is also used in physics, specifically in quantum field theory, aerodynamics, aeroacoustics, electrodynamics and statistical field theory, to refer to various types of correlation functions, even those that do not fit the mathematical definition. In Quantum field theory, Green's functions take the roles of propagators.
Contents
Definition and uses
A Green's function, G(x, s), of a linear differential operator L = L(x) acting on distributions over a subset of the Euclidean space R^{n}, at a point s, is any solution of

<math>LG(x,s)=\delta(xs),</math>
<p style="margin:0; fontsize:4pt;"> </p>
(1)
</dl>

<math>Lu(x)=f(x).</math>
<p style="margin:0; fontsize:4pt;"> </p>
(2)
</dl>
 <math>LG(x,s)=\delta(xs).</math>
 <math>G(x,s)=G(xs).</math>
 <math>\int LG(x,s) f(s) \, ds = \int \delta(xs)f(s) \, ds = f(x).</math>
 <math>Lu(x)=\int LG(x,s) f(s) \, ds.</math>
 <math>Lu(x)=L\left(\int G(x,s) f(s) \,ds\right),</math>

<math>u(x)=\int G(x,s) f(s) \,ds.</math>
<p style="margin:0; fontsize:4pt;"> </p>
(3)
</dl>
 <math>L=\dfrac{d}{dx}\left[p(x) \dfrac{d}{dx}\right]+q(x)</math>
 <math>Du= \begin{cases}
 <math>\begin{align}
 <math> \begin{align}
 <math>u(x)=\int_0^\ell f(s) G(x,s) \, ds,</math>
 <math>G(x,s)</math> is continuous in <math>x</math> and <math>s</math>.
 For <math>x \ne s</math>, <math>L G(x, s)=0</math>.
 For <math>s \ne 0</math>, <math>D G(x, s)=0</math>.
 Derivative "jump": <math>G'(s_{+0}, s)G'(s_{0}, s)=1 / p(s)</math>.
 Symmetry: <math>G(x,s) = G(s, x)</math>.
 <math>\delta(xx')=\sum_{n=0}^\infty \Psi_n^\dagger(x) \Psi_n(x').</math>
 <math>G(x, x')=\sum_{n=0}^\infty \dfrac{\Psi_n^\dagger(x) \Psi_n(x')}{\lambda_n},</math>
 <math>\int_V \nabla \cdot \vec A\ dV=\int_S \vec A \cdot d\hat\sigma.</math>
 <math>\begin{align}
 <math>\int_V (\phi\nabla^2\psi\psi\nabla^2\phi) dV=\int_S (\phi\nabla\psi\psi\nabla\phi)\cdot d\hat\sigma.</math>
 <math>L G(x,x')=\nabla^2 G(x,x')=\delta(xx').</math>
 <math>\int_V \left[ \phi(x') \delta(xx')G(x,x') \nabla^2\phi(x')\right]\ d^3x' = \int_S \left[\phi(x')\nabla' G(x,x')G(x,x')\nabla'\phi(x')\right] \cdot d\hat\sigma'.</math>
 <math>\int\limits_V {\phi(x')\delta(xx')\ d^3x'}</math>
 <math>\phi(x)=\int_V G(x,x') \rho(x')\ d^3x'+\int_S \left[\phi(x')\nabla' G(x,x')G(x,x')\nabla'\phi(x')\right] \cdot d\hat\sigma'.</math>
 <math>\int_S \nabla' G(x,x') \cdot d\hat\sigma' = \int_V \nabla'^2 G(x,x') d^3x' = \int_V \delta (xx') d^3x' = 1</math>
 <math>\int_S \phi(x')\nabla' G(x,x')\cdot d\hat\sigma' = \langle\phi\rangle_S </math>
 <math>G(x,x')=\dfrac{1}{xx'}.</math>
 <math>\phi(x)=\int_V \dfrac{\rho(x')}{xx'} \, d^3x'.</math>
 <math>\begin{align}
 <math>g(x,s) + k^2 g(x,s) = \delta(xs).</math>
 <math>g(x,s)=c_1 \cos kx+c_2 \sin kx.</math>
 <math>g(0,s)=c_1 \cdot 1+c_2 \cdot 0=0, \quad c_1 = 0</math>
 <math>g\left(\tfrac{\pi}{2k},s\right) = c_3 \cdot 0+c_4 \cdot 1=0, \quad c_4 = 0 </math>
 <math>g(x,s)= \begin{cases}
 <math>c_2 \sin ks=c_3 \cos ks</math>
 <math>c_3 \cdot \left(k \sin ks\right)c_2 \cdot \left( k \cos ks\right )=1</math>
 <math>c_2 = \frac{\cos ks}{k} \quad;\quad c_3 = \frac{\sin ks}{k}</math>
 <math>g(x,s)=\begin{cases}
 Let n = 1 and let the subset be all of R. Let L be d/dx. Then, the Heaviside step function H(x − x_{0}) is a Green's function of L at x_{0}.
 Let n = 2 and let the subset be the quarterplane {(x, y) : x, y ≥ 0} and L be the Laplacian. Also, assume a Dirichlet boundary condition is imposed at x = 0 and a Neumann boundary condition is imposed at y = 0. Then the Green's function is
 <math>\begin{align}
 Discrete Green's functions can be defined on graphs and grids.
 Feynman propagators
 Green's identities
 Impulse response, the analog of a Green's function in signal processing
 Keldysh formalism
 Spectral theory
 Green's function in manybody theory
 S. S. Bayin (2006), Mathematical Methods in Science and Engineering, Wiley, Chapters 18 and 19.
 Eyges, Leonard, The Classical Electromagnetic Field, Dover Publications, New York, 1972. ISBN 0486639479. (Chapter 5 contains a very readable account of using Green's functions to solve boundary value problems in electrostatics.)
 A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations (2nd edition), Chapman & Hall/CRC Press, Boca Raton, 2003. ISBN 1584882972
 A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1584882999
 G. B. Folland, Fourier Analysis and Its Applications, Wadsworth and Brooks/Cole Mathematics Series.
 K. D. Cole, J. V. Beck, A. HajiSheikh, and B. Litkouhi, Methods for obtaining Green's functions, Heat Conduction Using Green's Functions, Taylor and Francis, 2011, pp. 101–148. ISBN 9781439813546
 Green G, An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism (Nottingham, England: T. Wheelhouse, 1828). pages 1012
 ↑ some examples taken from Schulz, Hermann: Physik mit Bleistift. Frankfurt am Main: Deutsch, 2001. ISBN 3817116616 (German)
 Hazewinkel, Michiel, ed. (2001), "Green function", Encyclopedia of Mathematics, Springer, ISBN 9781556080104
 Weisstein, Eric W., "Green's Function", MathWorld.
 Green's function for differential operator at PlanetMath.org.
 Green's function at PlanetMath.org.
 GreenFunctionsAndConformalMapping at PlanetMath.org.
 Introduction to the Keldysh Nonequilibrium Green Function Technique by A. P. Jauho
 Green's Function Library
 Tutorial on Green's functions^{[dead link]}
 Boundary Element Method (for some idea on how Green's functions may be used with the boundary element method for solving potential problems numerically)
 At Citizendium
 MIT video lecture on Green's function^{[dead link]}
 Bowley, Roger. "George Green & Green's Functions". Sixty Symbols. Brady Haran for the University of Nottingham.
where <math>\delta</math> is the Dirac delta function. This property of a Green's function can be exploited to solve differential equations of the form
If the kernel of L is nontrivial, then the Green's function is not unique. However, in practice, some combination of symmetry, boundary conditions and/or other externally imposed criteria will give a unique Green's function. Also, Green's functions in general are distributions, not necessarily proper functions.
Green's functions are also useful tools in solving wave equations and diffusion equations. In quantum mechanics, the Green's function of the Hamiltonian is a key concept with important links to the concept of density of states. As a side note, the Green's function as used in physics is usually defined with the opposite sign; that is,
This definition does not significantly change any of the properties of the Green's function.
If the operator is translation invariant, that is, when L has constant coefficients with respect to x, then the Green's function can be taken to be a convolution operator, that is,
In this case, the Green's function is the same as the impulse response of linear timeinvariant system theory.
Motivation
Loosely speaking, if such a function G can be found for the operator L, then if we multiply the equation (1) for the Green's function by f(s), and then perform an integration in the s variable, we obtain:
The righthand side is now given by the equation (2) to be equal to L u(x), thus:
Because the operator L = L(x) is linear and acts on the variable x alone (not on the variable of integration s), we can take the operator L outside of the integration on the righthand side, obtaining
which suggests
Thus, we can obtain the function u(x) through knowledge of the Green's function in equation (1) and the source term on the righthand side in equation (2). This process relies upon the linearity of the operator L.
In other words, the solution of equation (2), u(x), can be determined by the integration given in equation (3). Although f(x) is known, this integration cannot be performed unless G is also known. The problem now lies in finding the Green's function G that satisfies equation (1). For this reason, the Green's function is also sometimes called the fundamental solution associated to the operator L.
Not every operator L admits a Green's function. A Green's function can also be thought of as a right inverse of L. Aside from the difficulties of finding a Green's function for a particular operator, the integral in equation (3) may be quite difficult to evaluate. However the method gives a theoretically exact result.
This can be thought of as an expansion of f according to a Dirac delta function basis (projecting f over δ(x − s)) and a superposition of the solution on each projection.) Such an integral equation is known as a Fredholm integral equation, the study of which constitutes Fredholm theory.
Green's functions for solving inhomogeneous boundary value problems
The primary use of Green's functions in mathematics is to solve nonhomogeneous boundary value problems. In modern theoretical physics, Green's functions are also usually used as propagators in Feynman diagrams (and the phrase Green's function is often used for any correlation function).
Framework
Let L be the Sturm–Liouville operator, a linear differential operator of the form
and let D be the boundary conditions operator
\alpha_1 u'(0)+\beta_1 u(0) \\ \alpha_2 u'(l)+\beta_2 u(l).
\end{cases}</math>
Let f(x) be a continuous function in [0,l]. We shall also suppose that the problem
Lu &= f \\ Du &= 0
\end{align}</math>
is regular (i.e., only the trivial solution exists for the homogeneous problem).
Theorem
There is one and only one solution u(x) that satisfies
Lu & = f\\ Du & = 0,
\end{align}</math>
and it is given by
where G(x,s) is a Green's function satisfying the following conditions:
Advanced and retarded Green's functions
Sometimes the Green's function can be split into a sum of two functions. One with the variable positive (+) and the other with the variable negative (). These are the advanced and retarded Green's functions, and when the equation under study depends on time, one of the parts is causal and the other anticausal. In these problems usually the causal part is the important one.
Finding Green's functions
Eigenvalue expansions
If a differential operator L admits a set of eigenvectors <math>\Psi_n(x)</math> (i.e., a set of functions <math>\Psi_n</math> and scalars <math>\lambda_n</math> such that <math>L \Psi_n=\lambda_n \Psi_n</math>) that is complete, then it is possible to construct a Green's function from these eigenvectors and eigenvalues.
"Complete" means that the set of functions <math>\left\{ \Psi_n \right\}</math> satisfies the following completeness relation:
Then the following holds:
where <math>\dagger</math> represents complex conjugation.
Applying the operator L to each side of this equation results in the completeness relation, which was assumed true.
The general study of the Green's function written in the above form, and its relationship to the function spaces formed by the eigenvectors, is known as Fredholm theory.
There are several other methods for finding Green's functions, including the method of images, separation of variables, and Laplace transforms (Cole 2011).
Table of Green's functions
The following table gives an overview of Green's functions of frequently appearing differential operators, where <math>\theta(t)</math> is the Heaviside step function, <math>r=\sqrt{x^2+y^2+z^2}</math> and <math>\rho=\sqrt{x^2+y^2}</math>.^{[1]}
Differential Operator <math>L</math>  Green's Function <math>G</math>  Example of application 

<math>\partial_t + \gamma </math>  <math>\theta(t)\mathrm e^{\gamma t}</math>  
<math>\left(\partial_t + \gamma \right)^2</math>  <math>\theta(t)t\mathrm e^{\gamma t}</math>  
<math>\partial_t^2 + 2\gamma\partial_t + \omega_0^2</math>  <math>\theta(t)\mathrm e^{\gamma t}\frac{1}{\omega}\sin(\omega t)</math> with <math>\omega=\sqrt{\omega_0^2\gamma^2}</math>  onedimensional damped harmonic oscillator 
<math>\Delta_\text{2D}</math><math>=\partial_x^2 + \partial_y^2</math>  <math>\frac{1}{2 \pi}\ln \rho</math>  
<math>\nabla^2</math><math>=\partial_x^2 + \partial_y^2 + \partial_z^2= \Delta</math>  <math>\frac{1}{4 \pi r}</math>  Poisson equation 
Helmholtz operator <math>\Delta + k^2</math>  <math>\frac{\mathrm e^{ikr}}{4 \pi r}</math>  stationary 3D Schrödinger equation for free particle 
D'Alembert operator <math>\square = \frac{1}{c^2}\partial_t^2\Delta</math>  <math>\frac{\delta(t\frac{r}{c})}{4 \pi r}</math>  3D wave equation 
<math>\partial_t  k\Delta</math>  <math>\theta(t)\left(\frac{1}{4\pi kt}\right)^{3/2}\mathrm e^{r^2/4kt}</math>  diffusion 
Green's functions for the Laplacian
Green's functions for linear differential operators involving the Laplacian may be readily put to use using the second of Green's identities.
To derive Green's theorem, begin with the divergence theorem (otherwise known as Gauss's theorem):
Let <math>\vec A=\phi\nabla\psi\psi\nabla\phi</math> and substitute into Gauss' law. Compute <math>\nabla\cdot\vec A</math> and apply the product rule for the <math>\nabla</math> operator:
\nabla\cdot\vec A &=\nabla\cdot(\phi\nabla\psi \;\; \psi\nabla\phi)\\ &=(\nabla\phi)\cdot(\nabla\psi) \;+\; \phi\nabla^2\psi \;\; (\nabla\phi)\cdot(\nabla\psi) \;\; \psi\nabla^2\phi\\ &=\phi\nabla^2\psi \;\; \psi\nabla^2\phi.
\end{align}</math>
Plugging this into the divergence theorem produces Green's theorem:
Suppose that the linear differential operator L is the Laplacian, <math>\nabla^2</math>, and that there is a Green's function G for the Laplacian. The defining property of the Green's function still holds:
Let <math>\psi=G</math> in Green's theorem. Then:
Using this expression, it is possible to solve Laplace's equation <math>\nabla^2\phi(x)=0</math> or Poisson's equation <math>\nabla^2\phi(x)=\rho(x)</math>, subject to either Neumann or Dirichlet boundary conditions. In other words, we can solve for <math>\phi(x)</math> everywhere inside a volume where either (1) the value of <math>\phi(x)</math> is specified on the bounding surface of the volume (Dirichlet boundary conditions), or (2) the normal derivative of <math>\phi(x)</math> is specified on the bounding surface (Neumann boundary conditions).
Suppose the problem is to solve for <math>\phi(x)</math> inside the region. Then the integral
reduces to simply <math>\phi(x)</math> due to the defining property of the Dirac delta function and we have:
This form expresses the wellknown property of harmonic functions that if the value or normal derivative is known on a bounding surface, then the value of the function inside the volume is known everywhere.
In electrostatics, <math>\phi(x)</math> is interpreted as the electric potential, <math>\rho(x)</math> as electric charge density, and the normal derivative <math>\nabla\phi(x')\cdot d\hat\sigma'</math> as the normal component of the electric field.
If the problem is to solve a Dirichlet boundary value problem, the Green's function should be chosen such that <math>G(x,x')</math> vanishes when either x or x′ is on the bounding surface. Thus only one of the two terms in the surface integral remains. If the problem is to solve a Neumann boundary value problem, the Green's function is chosen such that its normal derivative vanishes on the bounding surface, as it would seem to be the most logical choice. (See Jackson J.D. classical electrodynamics, page 39). However, application of Gauss's theorem to the differential equation defining the Green's function yields
meaning the normal derivative of <math>G(x,x')</math> cannot vanish on the surface, because it must integrate to 1 on the surface. (Again, see Jackson J.D. classical electrodynamics, page 39 for this and the following argument). The simplest form the normal derivative can take is that of a constant, namely <math>1/S</math>, where S is the surface area of the surface. The surface term in the solution becomes
where <math>\langle\phi\rangle_S </math> is the average value of the potential on the surface. This number is not known in general, but is often unimportant, as the goal is often to obtain the electric field given by the gradient of the potential, rather than the potential itself.
With no boundary conditions, the Green's function for the Laplacian (Green's function for the threevariable Laplace equation) is:
Supposing that the bounding surface goes out to infinity, and plugging in this expression for the Green's function, this gives the familiar expression for electric potential in terms of electric charge density as
Example
Example. Find the Green function for the following problem:
Lu & = u + k^2 u = f(x)\\ u(0)& = 0, \quad u\left(\tfrac{\pi}{2k}\right) = 0.
\end{align}</math>
First step: The Green's function for the linear operator at hand is defined as the solution to
If <math>x\ne s</math>, then the delta function gives zero, and the general solution is
For <math>x<s</math>, the boundary condition at <math>x=0</math> implies
if <math>x < s</math> and <math>s \ne \tfrac{\pi}{2k}</math>.
For <math>x>s</math>, the boundary condition at <math>x=\tfrac{\pi}{2k}</math> implies
The equation of <math>g(0,s)=0</math> is skipped for similar reasons.
To summarize the results thus far:
c_2 \sin kx, & \text{for }x<s\\ c_3 \cos kx, & \text{for }s<x \end{cases}</math>
Second step: The next task is to determine <math>c_{2}</math> and <math>c_{3}</math>.
Ensuring continuity in the Green's function at <math>x=s</math> implies
One can ensure proper discontinuity in the first derivative by integrating the defining differential equation from <math>x=s\epsilon</math> to <math>x=s+\epsilon</math> and taking the limit as <math>\epsilon</math> goes to zero:
The two (dis)continuity equations can be solved for <math>c_{2}</math> and <math>c_{3}</math> to obtain
So the Green's function for this problem is:
\frac{\cos ks}{k} \sin kx, & x<s\\ \frac{\sin ks}{k} \cos kx, & s<x \end{cases}</math>
Further examples
G(x, y, x_0, y_0) =\dfrac{1}{2\pi} &\left[\ln\sqrt{(xx_0)^2+(yy_0)^2}\ln\sqrt{(x+x_0)^2+(yy_0)^2} \right. \\ &\left.  \ln\sqrt{(xx_0)^2+(y+y_0)^2}+ \ln\sqrt{(x+x_0)^2+(y+y_0)^2}\right] \end{align}</math>
See also
References
External links
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