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C. The use of Green function for solving differential equations

Green's function $G\left(\mathbf{x},\mathbf{\xi}\right)$ is defined as a integral kernel of a linear operator which inverts a diffential operator. It can be defined thus as:

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

(C.1)

Where $\mathcal{L}$ is a linear differential operator, $\xi$ is an arbitrary point in

and $\delta$ is the Dirac Delta function. The Green function can be used to solve (weakly) a differential equation of the form:

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

(C.2)

If the differential equation is accompanied by appropriate boundary conditions, and if the green function corresponding to $\mathcal{L}$ is known the solution of C.2 has the following integral representation:

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

(C.3)

The demonstration is straight forward. Multiplying equation C.1 by $\psi(\mathbf{\xi})$ and integrating over the bounded space $\Omega$ with respect to $\xi$ we have:

$\displaystyle \int_\Omega \delta \left( \mathbf{x} - \xi \right) \psi\left( \xi... ... \int_\Omega \mathcal{L}G\left(\mathbf{x},\xi\right)\psi\left( \xi \right) d\xi$

(C.4)

Due to the Dirac delta function property, the first term is simply:

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

(C.5)

Or, recalling the differential equation definition C.2:

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

(C.6)

Hence, equation C.4 may be rewritten as:

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

(C.7)

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

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

(C.8)

Which is the integral representation presented above for . However, to evaluate this integral knowlagde of the explicit form of both and $\psi$ are required. Furthermore, even if both and $\psi$ 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 that in general, Green's functions are distributions rather than classical functions.

Emmanuel Branlard
http://emmanuel.branlard.free.fr/