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What is Green's function

  1. Jul 23, 2014 #1

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

    [tex]\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[/tex]

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


    Extended explanation

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

    [tex]\mathcal{L}u\left(\mathbf{x}\right) = \psi\left(\mathbf{x}\right)\ \ \ \ \ \ \left(*\right)[/tex]

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

    Consider the definition of Green's function above

    [tex]\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)[/tex]

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

    [tex]-\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)[/tex]

    Noting that by the properties of the Dirac Delta function

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

    And from (*)

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

    Hence (**) may be rewritten thus

    [tex] \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}[/tex]

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

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

    Hence we have obtained an integral representation for [itex]u\left(x\right)[/itex]. However, to evaluate this integral knowlagde of the explicit form of both [itex]G\left(\mathbf{x},\mathbf{\xi}\right)[/itex] and [itex]\psi\left(\mathbf{\xi}\right)[/itex] are required. Furthermore, even if both [itex]G\left(\mathbf{x},\mathbf{\xi}\right)[/itex] and [itex]\psi\left(\mathbf{\xi}\right)[/itex] 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!
  2. jcsd
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