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Let $\mathbf F$ be a smooth vector field in $\mathbb R^3$, which is null outside a finite compact domain $V$. By the Helmholtz decomposition theorem, there exist a scalar field $\Phi$ and a vector field $\mathbf A$ such that $${\mathbf F} = \nabla \Phi + \nabla \times {\mathbf A}.$$

Do there always exist smooth fields $\Phi$ and $\mathbf A$, which are null outside $V$ and which satisfy this identity?

EDIT: this question can be reformulated in the following way: Assume that $\bf F$ is a smooth vector field with compact support. Does there exist a scalar smooth solution $f$ to $$\nabla^2 f = \nabla \cdot \bf F$$ whose support is included inside the support of $\bf F$?

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    $\begingroup$ I assume not because of the Aharonov–Bohm effect in quantum mechanics. The magnetic field can be expressed as $\mathbf{B}=\nabla\times\mathbf{A}$ because of Gauss law and the movement of a charged particle is governed by this vector potential $\mathbf{A}$. We can now observe a phase shift of electrons passing nearby a magnetic field in a region where it vanishes. The explanation is, that the vector potential does not. $\endgroup$
    – Samuel Adrian Antz
    Commented May 8, 2022 at 14:36
  • $\begingroup$ Nevertheless, even for magnetic fields, there are possible harmonic functions $\Phi$ and possible $\bf A$ such that $B = \nabla \times {\bf A} + \nabla \Phi$. The question is if such fields exist that are null outside $V$. $\endgroup$
    – MikeTeX
    Commented May 8, 2022 at 15:10
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    $\begingroup$ @SamuelAdrianAntz --- the Aharonov-Bohm effect would be an obstacle to a vanishing vector potential if the compact region V encloses a nonzero flux (as in a solenoid), but since it is assumed that the magnetic field vanishes outside of V the enclosed flux is zero so I don't think this obstacle applies. $\endgroup$
    – Carlo Beenakker
    Commented May 8, 2022 at 16:39
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    $\begingroup$ The scalar field $\Phi$ solves after applying the divergence using the fact that the curl of $A$ is divergence-free the equation $\Delta \Phi = \nabla \cdot \mathbf{F}$. I expect that its solution $\Phi$ for generic not divergence-free $\mathbf{F}$ is supported on all of $\mathbb{R}^3$, since it is the convolution of $\nabla \cdot \mathbf{F}$ with the fundamental solution $1/|x|$ for the Laplacian in $\mathbb{R}^3$. $\endgroup$
    – André Schlichting
    Commented May 8, 2022 at 20:20
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    $\begingroup$ See this post with $f=\nabla \cdot F$ mathoverflow.net/questions/417344/… $\endgroup$
    – Giorgio Metafune
    Commented May 10, 2022 at 18:42

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