The Helmholtz decomposition states that "any sufficiently smooth, rapidly decaying 3-D vector field ${\bf F}$ can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field" ref wikipedia : $${\bf F} = -\nabla \phi + \nabla \times {\bf A}$$ where $\nabla \phi$ constitutes the curl-free part, and $\nabla \times {\bf A}$ the divergence-free one.
When $\phi$ happens to be also a solution of the Laplace equation, $$\nabla\cdot(\nabla\phi)=\nabla^2\phi=0$$ then also $\nabla \phi$ is divergence-free.
In this case the solution of the Helmholtz decomposition is not unique: one can find a scalar field $\phi$ and a vector potential $\bf A$ that generate the same vector field $\bf F$, that is, given a scalar function $\phi$ there exist a vector field $\bf A$ such that: $$\nabla \times \bf A=-\nabla \phi.$$
Now, my struggle is exactly this: given an arbitrary scalar field $\phi$ (in my case in form of a 3-D Cartesian mesh), which satisfies the Laplace equation, I would like to find the corresponding vector potential $\bf A$ such that the above equation holds. Only problem, I cannot find how..
My first realisation has been: if one Fourier transforms the Helmholtz equation one finds: $$\tilde{\bf F} = -i{\bf k}\tilde \phi + i{\bf k}\times\tilde{\bf A}.$$ This shows clearly that $\tilde\phi$ can be seen as "something" orthogonal to $\tilde{\bf A}$.
Unfortunately this does not help: something orthogonal to $\tilde{\bf A}$ would not survive the curl operator, so there is not much I can do to "include it" into $\bf A$.
My second realisation was that, yet, such an isomorphism between $\nabla\phi$ and $\nabla\times\bf A$ must exist, and I found a set of examples that beautifully shows it. If one takes the complex expression$$f(x,y) = \left(x + i y\right)^n,$$ then the pair $\phi$ and $\bf A$, $$\begin{align*} \phi & =\Re\left[f(x,y)\right]\\ {\bf A} & =\begin{pmatrix}0\\ 0\\ \Im\left[f(x,y)\right] \end{pmatrix} \end{align*} $$ exactly produce the same identical field $\bf F$, whether one takes the gradient of $\phi$ or the curl of $\bf A$.
This second realisation makes me think that moving to the complex plane could be the way to go. Yet, how to move from this intuition to finding a general relation between an arbitrary (e.g. numerical) $\phi$ and $\bf A$ still escapes me.
Now I wonder: Perhaps the Hodge decomposition, which I am not very familiar with, but that seems to pop up very frequently in this context, could help? If so, can anybody suggest an "easy" reference where I can see how it works on an arbitrary 3-D vector field?
This method, to a given 3-D field ${\bf F}$, always returns a well-determined pair $(\phi, {\bf A})$. And there is no way to get an $\bf A$ our of a $\phi$...
Perhaps I am too biased by this implementation but I cannot see another practical way to implement the Helmholtz decomposition able to do its job and just return the vector potential, and not $\phi$..
– Hans Castrop May 06 '18 at 21:12