When is a vector field the curl of another vector field?

Question

Under what conditions does a given vector field $\bf X$ on some open subset $U \subseteq \mathbb{R}^3$ satisfy ${\bf X} = \text{curl } {\bf Y}$ for some vector field $\bf Y$ on $U$?

Answer 1

If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$\text{div } {\bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $\text{curl}$ and $\text{div}$ operators and the exterior derivative operator.

If $H^2(U)$ is not trivial, then $\text{div } {\bf X} = 0$ is a necessary but not a sufficient condition.

Answer 2

Given $\mathbf{X}=(X_1,X_2,X_3)$, a vector field $\mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $\mathbf{X}=\text{curl }\mathbf{Y}$ will satisfy:

$$\begin{cases} X_1&=\dfrac{\partial Y_3}{\partial y}-\dfrac{\partial Y_2}{\partial z}\\ X_2&=\dfrac{\partial Y_1}{\partial z}-\dfrac{\partial Y_3}{\partial x}\\ X_3&=\dfrac{\partial Y_2}{\partial x}-\dfrac{\partial Y_1}{\partial y} \end {cases}$$

Answer 3

The condition $\text{div }{\bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin: $${\bf Y}({\bf r}) = \int_0^1{\bf X}(t{\bf r})\times t{\bf r}\,dt.$$ See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps