|
In mathematics, 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(x − x0),
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.
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.
Motivation
Convolving with a Green's function gives solutions to inhomogeneous differentio-integral equations, most commonly a Sturm-Liouville problems . If g is the Green's
function of an operator L, then the solution for f of the equation Lf = h is the convolution of g with h. Namely,
-
This can be thought as an expansion of f according to Dirac delta function basis (projecting f over δ(x −
s)) and a superposition of the solution on each projection.
While in particle physics, Green's functions are usually used as
propagators in Feynman
diagrams (and the phrase "Green's function" is actually often used for any correlation
functions), one of GF main utilities in mathematics is to solve inhomogeneous boundary value problem.
Working Frame
Let L be a linear differential operator in the form of
-
and let D be the boundary conditions operator
-
Let f(x) be a continuous function in [0,l]. We
shall also suppose that the problem
-
is regular, i.e. only the trivial soluton exists for the homogenous problem.
Then there is one and only solution u(x) which satisfies
-
and it is given by
-
where g(x,s) is Green's function and satisfies the following demands:
- g(x,s) is continuous in x and s.
- For , Lg(x,s) = 0.
- For , Dg(x,s) = 0.
- Derivative "jump": g'(s +
0,s) - g'(s - 0,s) = 1 / p(s).
- Symmetry: g(x,s) = g(s,x).
Example
Given the problem
- Lu = u'' + u = f(x)
-
Find Green's function.
First step: From demand-2 we see that
-
For x < s we see from demand-3 that the c1(s) = 0, while for
x > s we see from demand-3 that the c2(s) = 0 (we leave to the
reader to fill in the in-between steps).
Summerize the results:
-
Second step: Now we shall determine a(s) and b(s).
Using demand-1 we get
- a(s)sin(s) = b(s)cos(s).
Using demand-4 we get
-
Using Cramer's rule or by intelligent guess solve for a(s)
and b(s) and obtain that .
Check that this automaticly satisfy demand-5 !
So, our Green's function to this problem is
-
Examples
-
|