Implication of divergence of a vector field is zero

Question

I know that if the curl of a vector field is zero, then the vector field must be the gradient of a scalar field. But if the divergence of a vector field is zero, is it a must that the vector field be the curl of some vector field? If so please provide mathematical proof.

Answer 1

Without further assumptions, neither of the statements you made are true. It is possible to have a vector field with $0$ curl, yet it not be the gradient of some function, and it is also possible to have a divergence-free vector field yet it not be the curl of some vector field. You need to impose certain topological restrictions on the domain of the vector fields.

If you assume the vector field is defined on a contractible subset then both statements are true (for example if it's defined on an open ball, or for example a star-shaped open set or for example on the whole of $\Bbb{R}^3$). In this case, the theorems you're looking for are special cases of Poincare's lemma.

For the first statement, if you have a vector field $\mathbf{F}:\Bbb{R}^n \to \Bbb{R}^n$ of class $\mathcal{C}^1$ such that for all $i,j\in\{1,\dots, n\}$, $\frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i}$ (in the case $n=3$, this is equivalent to saying the curl is zero), then the function $f:\Bbb{R}^n\to \Bbb{R}$ given by \begin{align} f(x) &= \int_0^1 \sum_{i=1}^n F_i(tx)\cdot x^i \, dt. \end{align} is a scalar potential: $\text{grad}(f) = \mathbf{F}$ (proof is a direct calculation).

For simplicity, let's say your vector field $\mathbf{F}:\Bbb{R}^3\to \Bbb{R}^3$ is defined everywhere, is of class $\mathcal{C}^1$, and is divergence free. Then, the vector field $\mathbf{A}:\Bbb{R}^3\to \Bbb{R}^3$ defined as \begin{align} \mathbf{A}(\mathbf{x}) &:= \int_0^1 t\cdot [\mathbf{F}(t\mathbf{x}) \times \mathbf{x}]\, dt \end{align}, where $\times$ is the cross product in $\Bbb{R}^3$, will satisfy $\text{curl}(\mathbf{A}) = \mathbf{F}$. The proof is a tedious, but direct calculation. Note that we need $\mathbf{F}$ to be defined everywhere so that the integral above makes sense (or at the very least, if we want this specific proof to work, we need the domain to be star-shaped with respect to the origin).