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where ? 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 non-trivial, 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. Green's functions may be categorized, by the type of boundary conditions satisfied, by a Green's function number. Also, Green's functions in general are distributions, not necessarily functions of a real variable.
Loosely speaking, if such a function G can be found for the operator , then, if we multiply the equation (1) for the Green's function by f(s), and then integrate with respect to s, we obtain,
Because the operator is linear and acts only on the variable x (and not on the variable of integration s), one may take the operator outside of the integration, yielding
This means that
is a solution to the equation
Thus, one may obtain the function u(x) through knowledge of the Green's function in equation (1) and the source term on the right-hand side in equation (2). This process relies upon the linearity of the operator .
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 .
Not every operator admits a Green's function. A Green's function can also be thought of as a right inverse of . 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.
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 anti-causal. In these problems usually the causal part is the important one. These are frequently the solutions to the inhomogeneous electromagnetic wave equation.
Finding Green's functions
While it doesn't uniquely fix the form the Green's function will take, performing a dimensional analysis to find the units a Green's function must have is an important sanity check on any Green's function found through other means. A quick examination of the defining equation,
shows that the units depend not only on the units of but also on the number and units of the space of which the position vectors and are elements. This leads to the relationship:
If a differential operatorL admits a set of eigenvectors?n(x) (i.e., a set of functions ?n and scalars ?n such that L?n = ?n ?n ) that is complete, then it is possible to construct a Green's function from these eigenvectors and eigenvalues.
If the differential operator can be factored as then the Green's function of can be constructed from the Green's functions for and :
The above identity follows immediately from taking to be the representation of the right operator inverse of , analogous to how for the invertible linear operator, defined by , is represented by its matrix elements .
A further identity follows for differential operators that are scalar polynomials of the derivative, . The fundamental theorem of algebra, combined with the fact that commutes with itself, guarantees that the polynomial can be factored, putting in the form:
where are the zeros of . Taking the Fourier transform of with respect to both and gives:
The fraction can then be split into a sum using a Partial fraction decomposition before Fourier transforming back to and space. This process yields identities that relate integrals of Green's functions and sums of the same. For example, if then one form for its Green's function is:
While the example presented is tractable analytically, it illustrates a process that works when the integral is not trivial (for example, when is the operator in the polynomial).
Using this expression, it is possible to solve Laplace's equation ?2?(x) = 0 or Poisson's equation ?2?(x) = -?(x), subject to either Neumann or Dirichlet boundary conditions. In other words, we can solve for ?(x) everywhere inside a volume where either (1) the value of ?(x) is specified on the bounding surface of the volume (Dirichlet boundary conditions), or (2) the normal derivative of ?(x) is specified on the bounding surface (Neumann boundary conditions).
Suppose the problem is to solve for ?(x) inside the region. Then the integral
This form expresses the well-known 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.
If the problem is to solve a Dirichlet boundary value problem, the Green's function should be chosen such that G(x,x′) 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 G(x,x′) 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 1/S, where S is the surface area of the surface. The surface term in the solution becomes
where 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.
Supposing that the bounding surface goes out to infinity and plugging in this expression for the Green's function finally yields the standard expression for electric potential in terms of electric charge density as
Let , and all three are elements of the real numbers. Then, for any function from reals to reals, , with an th derivative that is integrable over the interval :
The Green's function in the above equation, , is not unique. How is the equation modified if is added to , where satisfies for all (for example, with )? Also, compare the above equation to the form of a Taylor series centered at .
^In technical jargon "regular" means that only the trivial solution () exists for the homogeneous problem ().
^some examples taken from Schulz, Hermann: Physik mit Bleistift. Frankfurt am Main: Deutsch, 2001. ISBN3-8171-1661-6 (German)
Bayin, S.S. (2006). Mathematical Methods in Science and Engineering. Wiley. Chapters 18 and 19.
Eyges, Leonard (1972). The Classical Electromagnetic Field. New York, NY: Dover Publications. ISBN0-486-63947-9. Chapter 5 contains a very readable account of using Green's functions to solve boundary value problems in electrostatics.
Polyanin, A.D. (2002). Handbook of Linear Partial Differential Equations for Engineers and Scientists. Boca Raton, FL: Chapman & Hall/CRC Press. ISBN1-58488-299-9.
Mathews, Jon; Walker, Robert L. (1970). Mathematical methods of physics (2nd ed.). New York: W. A. Benjamin. ISBN0-8053-7002-1.
Folland, G.B.Fourier Analysis and its Applications. Mathematics Series. Wadsworth and Brooks/Cole.
Cole, K.D.; Beck, J.V.; Haji-Sheikh, A.; Litkouhi, B. (2011). "Methods for obtaining Green's functions". Heat Conduction Using Green's Functions. Taylor and Francis. pp. 101-148. ISBN978-1-4398-1354-6.
Green, G (1828). An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism. Nottingham, England: T. Wheelhouse. pages 10-12.