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5.2 Resolution of poisson screened equation with the use of Fourier transform

The resolution of equation 5.6 is performed through a 3d spacial Fourier transfomation of function , this operation being possible due to the unbounded domain of definition $-\infty < x,y,z < \infty$ . The Fourier transform and its inverse are respectivly defined as:

$\displaystyle \hat{f}(\v{k})$	$\displaystyle = \mathcal{F}(f)$	$\displaystyle = \int_\Omega f(\v{r}) e^{- i \v{k}\cdot\v{r}} d^3 r$	(5.7)
$\displaystyle f(\v{r})$	$\displaystyle = \mathcal{F}^{-1}(f)$	$\displaystyle = \frac{1}{(2\pi)^{3}} \int_{\hat{\Omega}} \hat{f}(\v{k}) e^{ i \v{k}\cdot\v{r}} d^3 k$	(5.8)

Note: multiple conventions exists for Fourier transform, one could have used a coefficient $1/(2\pi)^{3/2}$ before the integral for both the Fourier transform and its inverse, or no coefficient at all but an exponential coefficient with $2\pi i$ . The choice made here is justified by the normalization convention $\mathcal{F}(\delta(\v{r}))=1$

The Fourier transform of equation 5.6, after two successive integrations and the use of boundary condition, reduces eventually to the algebraic equation:

$\displaystyle \left(k^2+\lambda^2 \right) \hat{u}(\v{k}) = \hat{f}(\v{k})$

(5.9)

Which yields the solution:

$\displaystyle \hat{u}(\v{k}) = \frac{\hat{f}(\v{k})}{k^2+\lambda^2 }$

(5.10)

The Fourier inverse transformation provides the desired solution:

$\displaystyle u(\v{r}) = \mathcal{F}^{-1}(u) = \frac{1}{(2\pi)^{3}} \int \frac{\hat{f}(\v{k})}{k^2+\lambda^2} e^{i \v{k}\cdot\v{r}} d^3 k$

(5.11)

The integrand is known because is known and it is straightforward to compute $\hat{f}$ . Thus one can be satisfied of this expression. Nevertheless it is interesting to develop the expression of $\hat{f}(\v{k})$ as a Fourier transform, leading to the double integral:

$\displaystyle u(\v{r}) = \frac{1}{(2\pi)^{3}} \int\int \frac{1}{k^2+\lambda^2} e^{i \v{k}\cdot \left(\v{r}-\v{r}'\right)} f(\v{r}') d^3 r' d^3 k$

(5.12)

Expression in which we recognize an integral representation for the Green's function

for equation 5.6:

$\displaystyle u(\v{r}) = \int G(\v{r},\v{r}') f(\v{r}') d^3 r'$

(5.13)

with

$\displaystyle G(\v{r},\v{r}') = \frac{1}{(2\pi)^{3}} \int \frac{1}{k^2+\lambda^2} e^{i \v{k} \cdot (\v{r}'-\v{r})} d^3 k$

(5.14)

The formalism of Green function is common in differential equation formalism and deserved attention. This approach will be thus studied in the next section.

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