What is the Purpose of Green's Function?

[Greg Bernhardt]

Definition/Summary

Green's function G\left(\mathbf{x},\mathbf{\xi}\right) can be defined thus

\mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) + \delta\left(\mathbf{x} - \mathbf{\xi}\right) = 0\;\;\; \mathbf{x},\mathbf{\xi} \in \mathbb{R}^n

Where \mathcal{L} is a linear differential operator, \mathbf{\xi} is an arbitrary point in \mathbb{R}^n and \delta is the Dirac Delta function.

Equations



Extended explanation

The Green's function for a particular linear differential operator \mathcal{L}, is an integral kernel that can be used to solve differential equations with appropriate boundary conditions. Given a linear differential equation

\mathcal{L}u\left(\mathbf{x}\right) = \psi\left(\mathbf{x}\right)\ \ \ \ \ \ \left(*\right)

and the Green's function for the linear differential operator \mathcal{L}, one can derive an integral representation for the solution u\left(\mathbf{x}\right) as follows:

Consider the definition of Green's function above

\mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) + \delta\left(\mathbf{x} - \mathbf{\xi}\right) = 0 \Leftrightarrow \mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) = -\delta\left(\mathbf{x} - \mathbf{\xi}\right)

Now multiplying through by \psi\left(\mathbf{\xi}\right) and integrating over some bounded measure space \Omega with respect to \mathbf{\xi}

-\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \int_\Omega \mathcal{L}G \left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}\hspace{2cm}\left(**\right)

Noting that by the properties of the Dirac Delta function

\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \psi\left(\mathbf{x}\right)

And from (*)

\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \mathcal{L}u\left(\mathbf{x}\right)

Hence (**) may be rewritten thus

\mathcal{L}u\left(\mathbf{x}\right) = - \int_\Omega \mathcal{L}G \left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}

Since \mathcal{L} is a linear differential operator, which does not act on the variable of integration we may write:

u\left(\mathbf{x}\right) = - \int_\Omega G\left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}

Hence we have obtained an integral representation for u\left(x\right). However, to evaluate this integral knowlagde of the explicit form of both G\left(\mathbf{x},\mathbf{\xi}\right) and \psi\left(\mathbf{\xi}\right) are required. Furthermore, even if both G\left(\mathbf{x},\mathbf{\xi}\right) and \psi\left(\mathbf{\xi}\right) are known, the associated integral may not be a trivial exercise. In addition, every linear differential operator does not admit a Green's Function.

It would also be prudent to point out at this point that in general, Green's functions are distributions rather than classical functions.

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!

 

[fresh_42]

For short, a Green's function $G$ can be used to determine a given function $f$:

Loosely speaking, if such a function $G$ can be found for the operator $L$, then, if we multiply the equation
$$
LG(x,\xi) = \delta(\xi -x)
$$
for the Green's function by $f(\xi)$, and then integrate with respect to $\xi$, we obtain,
$$
\int LG(x,\xi)f(\xi)\,d\xi=\int \delta (x-\xi)f(\xi)\,ds=f(x)
$$