# Green's function

In mathematics, a Green's function is a type of function used to solve inhomogeneous differential equations subject to boundary conditions. Technically, a Green's function of a linear operator L acting on distributions over a manifold M, at a point x0, is any solution of (Lf)(x) = δ(xx0), where δ is the Dirac delta function. 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 would give us a unique Green's function. Also, please note Green's functions are distributions in general, not functions, meaning they can have discontinuities.

Not every operator L admits a Green's function. A Green's function can also be thought of as a one-sided inverse of L.

Green functions are also a useful tool in condensed matter theory - the Green function of the Hamiltonian is a key concept, with important links to the concept of density of states.

The Green's function was named after British mathematician George Green, who first developed the concept in the 1830s.

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## Motivation

Convolving with a Green's function gives solutions to inhomogeneous differentio-integral equations, most commonly a Sturm-Liouville problem . If g is the Green's function of an operator L, then the solution for f of the equation Lf = h is given by

[itex] f(x) = \int{ h(s) g(x,s) ds}. [itex]

This can be thought as an expansion of f according to Dirac delta function basis (projecting f over δ(xs)) and a superposition of the solution on each projection.

## Green's function for solving inhomogeneous boundary value problems

The primary use of Green's functions in mathematics is to solve inhomogeneous boundary value problems. In particle 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).

### Working frame

Let [itex] L [itex] be the Sturm-Liouville operator, a linear differential operator of the form

[itex] L = {d \over dx}\left( p(x) {d \over dx} \right) + q(x) [itex]

and let D be the boundary conditions operator

[itex] Du = \left\{\begin{matrix} \alpha _1 u'(0) + \beta _1 u(0) \\ \alpha _2 u'(l) + \beta _2 u(l) \end{matrix}\right. [itex]

Let [itex] f(x) [itex] be a continuous function in [itex][0,1][itex]. We shall also suppose that the problem

[itex] \begin{matrix}Lu = f \\ Du = 0 \end{matrix} [itex]

is regular, i.e. only the trivial solution exists for the homogeneous problem.

### Theorem

Then there is one and only one solution u(x) which satisfies

[itex] \begin{matrix}Lu = f \\ Du = 0 \end{matrix} [itex]

and it is given by

[itex] u(x) = \int_{0}^{l}{ f(s) g(x,s) ds} [itex]

where g(x,s) is Green's function and satisfies the following demands:

1. g(x,s) is continuous in x and s.
2. For [itex] x \ne s [itex], [itex] L g ( x, s ) = 0 [itex].
3. For [itex] s \ne 0, l [itex], [itex] D g ( x, s ) = 0 [itex].
4. Derivative "jump": [itex] g ' ( s_{ + 0}, s ) - g ' (s_{ - 0}, s ) = 1 / p(s) [itex].
5. Symmetry: g(x,s) = g(s,x).

## Example

Given the problem

[itex] \begin{matrix}Lu\end{matrix} = u ' ' + u = f( x ) [itex]
[itex] Du = u(0) = 0 \quad, \quad u( \pi / 2) = 0 [itex]

Find Green's function.

First step: From demand-2 we see that

[itex] g(x,s) = c_1 (s) \cdot \cos(x) + c_2 (s) \cdot \sin(x) [itex]

For x < s we see from demand-3 that the [itex] c_1 (s) = 0 [itex], while for x > s we see from demand-3 that the [itex] c_2 (s) = 0 [itex] (we leave it to the reader to fill in the in-between steps).

Summarize the results:

[itex] g(x,s)=\left\{\begin{matrix}

a(s) \sin(x), \;\; x < s \\ b(s) \cos(x), \;\; s < x \end{matrix}\right. [itex]

Second step: Now we shall determine a(s) and b(s).

Using demand-1 we get

[itex] a(s) \sin(s) = b(s) \cos(s) [itex].

Using demand-4 we get

[itex] b(s) \cdot ( - \sin(s) ) - a(s) \cdot cos (s) = 1/1 = 1 [itex]

Using Cramer's rule or by intelligent guess solve for a(s) and b(s) and obtain that [itex] a(s) = - \cos(s) \quad ; \quad b(s) = - \sin(s) [itex].

Check that this automatically satisfies demand-5.

So our Green's function for this problem is:

[itex] g(x,s)=\left\{\begin{matrix}

-1 \cdot \cos(s) \cdot \sin(x), \;\; x < s \\ -1 \cdot \sin(s) \cdot \cos(x), \;\; s < x \end{matrix}\right. [itex]

## Further examples

[itex]G(x, y, x_0, y_0)=\frac{1}{2\pi}\left[\ln\sqrt{(x-x_0)^2+(y-y_0)^2}-\ln\sqrt{(x+x_0)^2+(y-y_0)^2}\right][itex]
[itex]+\frac{1}{2\pi}\left[\ln\sqrt{(x-x_0)^2+(y+y_0)^2}-\ln\sqrt{(x+x_0)^2+(y+y_0)^2}\right].[itex]

## Bibliography

• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, 2002.de:Greensche Funktion

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