The First Theorem

Actually, the Helmholtz theorem¹ proves a slightly different thing, from which the statement above follows immediately: a vector field $\vec{V}$ which vanishes at the boundaries can be written as the sum of two terms, one of which is irrotational and the other, solenoidal (that is, divergenceless)[2]. Consider the following well-known identity for an arbitrary vector field $\vec{Z}(\vec{r})$ :

$$-\vec{\nabla}^2\vec{Z} = - \vec{\nabla}(\vec{\nabla}.\vec{Z}) + \vec{\nabla}\times\vec{\nabla}\times\vec{Z}$$ (1)

If we now take our vector field to be

$$\vec{V}=-\vec{\nabla}^2\vec{Z}$$ (2)

then it follows that

$$\vec{V}=-\vec{\nabla}U + \vec{\nabla}\times \vec{W}$$ (3)

with

$$U=\vec{\nabla}.\vec{Z}$$ (4)

and

$$\vec{W}=\vec{\nabla}\times \vec{Z}$$ (5)

Eq.(3) is Helmholtz's theorem, as $\vec{\nabla}U$ is irrotational and $\vec{\nabla}\times \vec{W}$ is solenoidal. But, is it general? It assumes that our vector field can be written as the Laplacian of some other one...
This constitutes, however, no problem if $\vec{V}$ vanishes at infinity fast enough, for, then, the equation

$$\vec{\nabla}^2\vec{Z}=-\vec{V} \;,$$ (6)

which is Poisson's equation, has always the solution

$$\vec{Z}(\vec{r})= \frac{1}{4\pi}\int d^3\vec{r'}\frac{\vec{V}(\vec{r'})} {\vert\vec{r}-\vec{r'}\vert}\;.$$ (7)

It is now a simple matter to prove, from Eq.(3), that $\vec{V}$ is determined from its $div$ and $curl$. Taking, in fact, the divergence of Eq.(3), we have:

$$div \vec{V} = -\vec{\nabla}^2 U$$ (8)

which is, again, Poisson's equation, and, so, determines $U$ as

$$U(\vec{r}) = \frac{1}{4\pi}\int d^3\vec{r'}\frac{\vec{\nabla}'.\vec{V}(\vec{r'})} {\vert\vec{r}-\vec{r'}\vert}$$ (9)

Take now the $curl$ of Eq.(3). We have

$$\vec{\nabla}\times\vec{V} = \vec{\nabla}\times\vec{\nabla}\times\vec{W}$$

$$= \vec{\nabla}(\vec{\nabla}.\vec{W}) - \vec{\nabla}^2\vec{W}$$ (10)

Now, $\vec{\nabla}.\vec{W}=0$, as $\vec{W}=\vec{\nabla}\times\vec{Z}$, so another Poisson equation determines $\vec{W}$. Using $U$ and $\vec{W}$ so determined in Eq.(3) proves our contention.