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While trying to solve the Poisson Equation by using Green's Function I have to Fourier transform the equation i.e

$$-\nabla^{2}\phi(r)=\rho(r).$$

In the book after Fourier transform, the solution is written as $$k^{2}\phi(k)=\rho(k).$$

I can understand the expression in the right hand side, but why there is $k^{2}$ in the left hand side?

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There really isnt any physics content in this question, so I suggest migration to math.SE (though a bit late..) –  Danu Jul 25 at 12:06
    
Cross-posted from math.stackexchange.com/q/877538/11127 –  Qmechanic Jul 26 at 0:50

2 Answers 2

up vote 7 down vote accepted

It's easiest to see if you start with the definition of the inverse fourier transform

$$ f(\mathbf x) = \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x} $$

and take a laplacian of both sides

$$ \nabla^2 f(\mathbf x) = \nabla^2 \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x} = \int d\mathbf k\, \hat f(\mathbf k) \nabla^2 e^{i \mathbf k \cdot \mathbf x} = \int d\mathbf k \, [ -k^2 \hat f(\mathbf k) ] e^{i \mathbf k \cdot \mathbf x}$$

So we say $$ \mathcal{F} \left[ \nabla^2 f(\mathbf x) \right] = -k^2 \hat f(\mathbf k)$$

Note: depending on your field and book, your mileage may vary when it comes to factors of $2\pi$ or signs

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i think $\nabla^{2}\phi(r)=\rho(r)$ is the same as $$ \frac{d^2}{dr^2} \phi(r) = \rho(r) $$. the convention i am used to in electrical engineering is to use large case letters for the Fourier Transform and use $\omega$ for angular frequency instead of "$k$". in that notation, Fourier transforming both sides is $$ (i \omega)^2 \Phi(\omega) = \mathrm{P}(\omega) $$

since $i^2 = -1$, i am curious if you're missing a minus sign with $k^2$.

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The angular frequency is the Fourier transform of the time: $t \Leftrightarrow \omega$, while $p$ or $k$ is the symbol conventionally used to denote the Fourier transform of the spatial components: $ x/r/q \Leftrightarrow k/p$... Not that it matters in a 1-D problem. –  Danu Jul 25 at 11:58

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