Proof of Biot-Savart theorem in 3D using distributions and fourier transform

Question

While studying the book Vorticity and Incompressible Flow by A. Majda and A. Bertozzi I came across the following Biot-Savart theorem in 3D: The solution to: $div \ v=0$ $curl(v)=\omega$ is $v=\int K(x-y) \omega(y,t) dy$ where $K(x)\omega=\frac{1}{4 \pi} \frac{x \times \omega}{|x|^3}$

I tried to prove it using Fourier transform but I came across the following function for which I couldn't find the Inverse Fourier Transform: $\frac{h_1}{h_1^2+h_2^2+h_3^2}$ To my knowledge, there is no way to find the transform as this function is not in $L^2$ . After doing some research I found some hints that solving this set of equations by combining distributions and fourier transform is possible. The problem is that the math behind it is still pretty complicated and the calculations for this must've been done somewhere. So the question - does anyone know where can I find how this is calculated or alternatively - how to calculate the inverse transform?

score 4 · Answer 1 · edited Apr 13 '17 at 12:57

You can reduce your computation to the standard newtonian potential as follows (I denote by $\xi_i$ the dual coordinates and use the convention for the Fourier transform without the $\pi$ normalisation) $\widehat{x_i\over r^3} = \widehat{ -{1\over r^2} {\partial r \over \partial x_i}} = \widehat{{\partial \over \partial x_i} \Bigl({1\over r}\Bigr)}= i\xi_i \widehat{1\over r}$

Now $1/r$ is the Newtonian potential and the fundamental solution of the Laplacian in ${\bf R}^3$ , which implies, in the sense of distribution $\Delta({1\over r}) = -4\pi\delta_0$ Thus, taking the Fourier transform, $-|\xi|^2 \widehat{1\over r} = \widehat{\Delta({1\over r})} = -4\pi\widehat{\delta_0}=-4\pi$ A good reference is the book by Laurent Schwartz, "mathematics for the physical sciences". See also this post.

Carlo Beenakker · Accepted Answer · 2015-08-17T08:44:10.313

no distributions are needed to invert this Fourier transform:

$(2\pi)^{-3}\int_{-\infty}^\infty dk_x\int_{-\infty}^\infty dk_y\int_{-\infty}^\infty dk_z\, e^{-i\vec{k}\cdot\vec{r}} \,\frac{\vec{k}}{|\vec{k}|^2}\\ = (2\pi)^{-3}i\frac{\partial}{\partial\vec{r}}\int_{-\infty}^\infty dk_x\int_{-\infty}^\infty dk_y\int_{-\infty}^\infty dk_z\, e^{-i\vec{k}\cdot\vec{r}} \,|\vec{k}|^{-2}\\ =(2\pi)^{-3} i\frac{\partial}{\partial\vec{r}} \int_0^\infty k^2 dk\int_0^{2\pi}d\phi \int_0^\pi \sin\theta d\theta\, e^{-ikr\cos\theta}k^{-2}\\ =2(2\pi)^{-2} i\frac{\partial}{\partial\vec{r}}\,|\vec{r}|^{-1}\int_0^\infty dq\, \frac{\sin q}{q}\quad\quad({\rm with}\;q=kr)\\ =\frac{i}{4\pi} \frac{\partial}{\partial\vec{r}}\,|\vec{r}|^{-1} =-\frac{i}{4\pi}\frac{\vec{r}}{|\vec{r}|^3}$ so the inverse Fourier transform of $4\pi i\, \vec{k}\,|\vec{k}|^{-2}$ equals $\vec{r}\,|\vec{r}|^{-3}$ .

UPDATE AUGUST 2015

The OP asks how the similar calculation would proceed in two spatial dimensions. Here it goes (with a formal interchange of integration and differentiation):

$(2\pi)^{-2}\int_{-\infty}^\infty dk_x\int_{-\infty}^\infty dk_y\, e^{-i\vec{k}\cdot\vec{r}} \,\frac{\vec{k}}{|\vec{k}|^2}\\ = (2\pi)^{-2}i\frac{\partial}{\partial\vec{r}}\int_{-\infty}^\infty dk_x\int_{-\infty}^\infty dk_y\, e^{-i\vec{k}\cdot\vec{r}} \,|\vec{k}|^{-2}\\ =(2\pi)^{-2} i\frac{\partial}{\partial\vec{r}} \int_0^\infty k dk\int_0^{2\pi}d\phi \, e^{-ikr\cos\phi}k^{-2}\\ =(2\pi)^{-1} i\frac{\vec{r}}{|\vec{r}|}\int_0^\infty k^{-1}dk\, \frac{d}{dr}J_0(kr)\\ =-(2\pi)^{-1} i\frac{\vec{r}}{|\vec{r}|}\int_0^\infty dk\, J_1(kr)\\ =-\frac{i}{2\pi}\frac{\vec{r}}{|\vec{r}|^2}$

I don't exactly understand the calculations - from what I see you use a double integral (instead of a triple). Also did u denote $\overrightarrow{k}=[k_1, \ k_2, \ k_3]$ ? — Max, May 09 '15 at 13:45
I expanded the notation; the third integral is over $\phi$ (in spherical coordinates), which just contributes $2\pi$ ; and yes, $\vec{k}=(k_x,k_y,k_z)$ is a vector with magnitude $k=|\vec{k}|$ ; so this Fourier transform gives all three components, in your posting you wrote the first component — Carlo Beenakker, May 09 '15 at 14:07
While it is technically true that no distributions were used in the above calculation, the integrands here are not all absolutely integrable, so that Fubini's theorem, differentiation under the integral sign, change of variables, and the classical theory of the Fourier transform are not strictly applicable. This can be resolved using a distributional interpretation of all the integrals here, but one can also use other techniques (e.g. use of gaussians or other cutoff functions) to justify why this particular arrangement of integrals deserves to be called a Fourier transform. — Terry Tao, May 09 '15 at 19:19
In case of reversing the same integral in two dimensions similar trick cannot be applied (the integral diverges) - is there any other similar one that I might apply? — Max, Aug 17 '15 at 01:16

Proof of Biot-Savart theorem in 3D using distributions and fourier transform

2 Answers2

UPDATE AUGUST 2015