The equations of electrostatics,$$\nabla \cdot \vec{E} = \rho, \quad \nabla \times \vec{E} = 0$$make it possible to introduce the scalar potential $\vec{E} = - \nabla \phi$, which then satisfies the scalar Poisson equation$$\nabla^2 \phi = -\rho.$$Similarly, equations of magnetostatics,$$\nabla \cdot \vec{B} = 0, \quad \nabla \times \vec{B} = \vec{J}$$make it possible to introduce the divergenceless vector potential$$\vec{B} = \nabla \times \vec{A}, \quad \nabla \cdot \vec{A} = 0$$which is found to satisfy the vector Poisson equation$$\nabla^2 \vec{A} = -\vec{J}.$$All this is well known.

I am searching for a missing part in this trilogy. Is these a physical theory that uses the following formalism: a scalar function $u$ satisfies $$\nabla u = \vec{F},$$where $\vec{F}$ is a vector source for $u$. Then one may introduce the longitudinal vector potential$$u = - \nabla \cdot \vec{W}, \quad \nabla \times \vec{W} = 0$$so that it again satisfies the vector Poisson equation$$\nabla^2 \vec{W} = -\vec{F}.$$

I am aware that it usually isn't practical to introduce a vector field in order to describe a simpler scalar field. However, there may be some circumstances in which it is advantageous to do so which I am not aware of.

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    $\begingroup$ I cannot answer the original question since I do not recall an example where such a representation of a scalar field by a "vector potential" would be useful. But I would point out that the situation you describe is mathematically equivalent to the example of electrostatics. For $\nabla u=F$ to hold, the field $F$ must be conservative and thus has a potential, $F=\nabla\phi$. The field $W$ is curl-free and thus also conservative, so it has a potential too, $W=\nabla\chi$. Your equation for $u$ is then equivalent to $\nabla\nabla^2\chi=-\nabla\phi$, so $\nabla^2\chi=-\phi$ up to constant. $\endgroup$ – Tomáš Brauner Jun 16 at 20:49

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