|
|
|
Green's function
|
Definition/Summary
|
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. |
|
Recent forum threads on Green's function
|
|
|
|
|
Breakdown
|
|
Mathematics
> Calculus/Analysis
>> Differential Equations
|
|
|
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. |
Commentary
|
Physorg.com Science News Partner