Let $u: \mathbb R^2 \to \mathbb R^2$ and let $\omega = \text{curl } u$ be the 2D vorticity of $u$, where $u, \omega \in L^2(\mathbb R^2)$ and $\nabla \cdot u = 0$. The classical Biot-Savart law states that $$ u_1(x) = -\frac1{2\pi}\int_{\mathbb R^2}\frac{x_2-y_2}{|x-y|^2}\omega(y) \, dy $$ and $$ u_2(x) = \frac1{2\pi}\int_{\mathbb R^2}\frac{x_1-y_1}{|x-y|^2}\omega(y) \, dy. $$

I am interested in a local version of the law in $B(x_o, r) \subset \mathbb R^2$. More precisely, suppose $u: \Omega \to \mathbb R^2$, where $\Omega \subset \mathbb R^2$ is bounded, and $u, \omega \in L_{loc}^2(\Omega)$ with $\nabla \cdot u = 0$. Is it true that $$ u_1(x) = -\frac1{2\pi}\int_{B(x_o,r)}\frac{x_2-y_2}{|x-y|^2}\omega(y) \, dy+A_1^{x_o}(x) $$ and $$ u_2(x) = \frac1{2\pi}\int_{B(x_o,r)}\frac{x_1-y_1}{|x-y|^2}\omega(y) \, dy+A_2^{x_o}(x). $$ for all $x \in B(x_o,r) \subset \Omega$ and for some smooth-ish function $A^{x_o}$? If not this, is there something similar which is true?

Finally, I am also interested in the general case of $\mathbb R^d$ for $d >2$, but for simplicity we may first restrict to the case $d=2$.

  • $\begingroup$ Probably you meant to assume that your $u$ is divergence free ? $\endgroup$
    – Hachino
    Jan 12, 2015 at 10:28

1 Answer 1


Remember how the classical Biot-Savart law is obtained : first, you take the curl of the defining equation $\omega = \nabla \wedge u$, and then, you use the fact that you know the solution of the Poisson equation in terms of the source. So, your problem boils down to computing explicitly some kernel.

Let's go through the proof in the 2D case, the necessary modifications for 3D and higher being easy.

First, extend $\omega$ by $0$ outside $\Omega$ and denote this extension by $\tilde{\omega}$. Now, $\tilde{\omega}$ is defined on the plane.

Let $\varphi$ be the unique solution in $L^2(\mathbb{R}^2)$ (actually also in $H^2$) of the Poisson equation $\Delta \varphi = \tilde{\omega}$. Define a new vector field $v := \nabla^{\perp} \varphi$, i.e.

$$v_1 = -\partial_2 \varphi \quad , \quad v_2 = \partial_1 \varphi .$$

Obviously, $\nabla \cdot v = 0$ and $\nabla \wedge v = \tilde{\omega}$. Thus, on $\Omega$, $u$ and $v$ agree up to a harmonic polynomial, which is smooth anyway.

Now, take a compact subset $K$ of $\Omega$ and choose a fattening $K'$ of $K$, that is, another compact subset of $\Omega$ whose interior contains $K$. $K'$ may very well be a set of the type $K_{\varepsilon} := \{x \in \Omega ; \text{dist}(x,K) \leq \varepsilon \}$ for some small $\varepsilon$.

Write $v = v_1 + v_2$ with $v_1 := \int_{K'} \nabla_x \left( \ln |x-y| \right) \omega (y) \: dy$ and $v_2$ is the same integral, but over the complement of $K'$.

As $v_2$ is harmonic over $K$, with the boundary of $K$ uniformly far from that of $K'$, $v_2$ enjoys both smoothness and integrability over $K$.

Up to some possible harmonic polynomial, this may satisfy you, I hope.

(On a side note : are you working with or around Navier-Stokes and fluid mechanics ?)

  • $\begingroup$ So $v$ and $u$ agree modulo a harmonic function, which would be my $A^{x_o}$ I suppose. And then one uses the Green's function in a ball to solve for $\varphi$. And then one may differentiate $\varphi$ to obtain $v_1$ and $v_2$. Seems doable although tedious. But now the Green's function should include a boundary term, right? So there should be a boundary term also in the Biot-Savart law in a bounded domain? $\endgroup$ Jan 12, 2015 at 16:43
  • $\begingroup$ And yes I have been studying Navier-Stokes theory and I am just trying to understand how some of these things work in a local setting. So eventually I would also like to understand this in a time-dependent setting, i.e. I would like to have some time regularity from my error function $A$, as well, if possible. $\endgroup$ Jan 12, 2015 at 16:56
  • $\begingroup$ I did some more calculations. At least in the case of a ball $B(0,1)$ the boundary term, indeed, disappears. However, the other integral seems to be $\int_{K'} \nabla_x \ln (|x||y-x/|x|^2|)\omega(y) \, dy$, which seems to have a singularity in the origin, if I did not do a mistake. On the other hand, I found the results as a Lemma in one of Serrin's papers, so I guess it must be true, but unfortunately Serrin did not provide a transparent proof. $\endgroup$ Jan 12, 2015 at 20:10
  • $\begingroup$ I have also studied Serrin's article for a few weeks and yes, he has a quite... concise style, at the very least. I suggest you follow his proof step by step, with more modern notations and more explicit integral splittings like I did above. Eventually, you will be a little bit more convinced than you seem to be now. :) Getting time-regularity is not something you do directly on your $A$, since it is only harmonic in space "locally uniformly in time", so you will not do better than $L^{\infty}_t$ in this way. (1/2) $\endgroup$
    – Hachino
    Jan 13, 2015 at 8:11
  • $\begingroup$ On the other hand, I have a more detailed proof of Serrin's result, which also includes an extension from Fabes, Jones and Rivière (equality in Serrin's condition and regularity up to the final time). Disclaimer : the style is quite informal, as it was some kind of writing exercise from my advisor and it's written in French (it was never intended to be used anywhere else). If you wish, you can give me your email and I will send it to you. $\endgroup$
    – Hachino
    Jan 13, 2015 at 8:16

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