# Fourier Transform of Poisson Equation

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?

• There really isnt any physics content in this question, so I suggest migration to math.SE (though a bit late..) – Danu Jul 25 '14 at 12:06
• Cross-posted from math.stackexchange.com/q/877538/11127 – Qmechanic Jul 26 '14 at 0:50

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

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 $k^2$.

• 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 '14 at 11:58