Can a vector field $\mathbf{A}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ always be expressed as the sum of a solenoidal (divergenceless) part $\mathbf{A}_S$, where $\nabla \cdot \mathbf{A}_S = 0$, and an irrotational part $\mathbf{A}_R$, where $\nabla\times\mathbf{A}_R = 0$?
$$
“\,\forall{\mathbf{A}}\;
\exists{\mathbf{A}_S}\;
\exists{\mathbf{A}_R} :
\mathbf{A} = \mathbf{A}_S + \mathbf{A}_R \;\wedge\;
\nabla \cdot \mathbf{A}_S = 0 \;\wedge\;
\nabla \times \mathbf{A}_R = 0
\,”\,?
$$
Since irrotational vector fields are gradients of some scalar field ($\mathbf{A}_R = \nabla R$), this is equivalent to asking:
$$
“\,
\forall\mathbf{A}\;
\exists\mathbf{A}_S\;
\exists R \in \left(\mathbb{R}^3\rightarrow\mathbb{R}\right) :
\mathbf{A} = \mathbf{A}_S + \nabla R
\;\wedge\;
\nabla\cdot\mathbf{A}_S = 0
\,”\,?
$$
What’s a proof of the result? If true, are $\mathbf{A}_R$ and $\mathbf{A}_S$ unique?
If it’s not always true, then what kind of fields can/cannot be decomposed in this way?
If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.
Best Answer
This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $\mathbf{F}$ on $\mathbb{R}^3$ can be written as $$ \mathbf{F} = \underbrace{-\nabla\Phi}_\text{irrotational} + \underbrace{\nabla\times\mathbf{A}}_\text{solenoidal} $$ provided 1) that $\mathbf{F}$ is twice continuously differentiable and 2) that $\mathbf{F}$ vanishes faster than $1/r$ as $r \rightarrow \infty$. If $\mathbf{F}$ is on a bounded domain $V \subset \mathbb{R}^3$, then the condition 2) that $\mathbf{F}$ vanishes can be relaxed.
Helmholtz’s theorem even gives $\Phi$ and $\mathbf{A}$ in terms of $\mathbf{F}$. More details can be found here: https://en.wikipedia.org/wiki/Helmholtz_decomposition