Please can you explain me why
$$ \nabla^2 G_{\omega}(R) = f(R) $$
is equivalent to
$$ \frac{1}{R} \frac{d^2(RG_\omega(R))}{d R^2} = f(R) $$
Here:
- $G_\omega(R)$ is the Fourier transform of $G$
- $G$ is the Green's function
- We suppose that $G$ and $f$ are spherically symmetric
- $R = \sqrt{x^2 + y^2 + z^2}$
Thanks!
$\endgroup$1 Answer
$\begingroup$If $H(R)$ is spherically symmetric, we can write the Laplacian of $H$ in spherical coordinates:
$\nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg)$
(see wikipedia).
Then, $\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = R^2\frac{d^2H}{dR^2} + 2R \frac{dH}{dR}$
On the other way, $\frac{d}{dR}(RH(R)) = R\frac{dH}{dR} + H$ and thus
$\frac{d^2}{dR^2}(RH(R)) = R\frac{d^2H}{dR^2} + 2\frac{dH}{dR}$
$\Rightarrow \frac{1}{R^2}\frac{d}{dR}\bigg(R\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$
$\Rightarrow \nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$.
Now, apply it to $G_{\omega}$.
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