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 $d=2$.



Local Biot-Savart law in $B(x_o,r) \subset \mathbb R^2$
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 ?)


