TITLE: Is there a vector field that is equal to its own curl?
QUESTION [47 upvotes]: I was wondering if there is a vector field that satisfies the following condition: 
$$\vec F=\nabla \times \vec F$$

REPLY [8 votes]: Jesse Madnick's answer notwithstanding one might be interested in nontrivial explicit solutions from which more complicated solutions can be generated by transportation and superposition.
I'm looking for solutions that do not depend on the third variable – "plane waves", so to speak. When $F=(F_1,F_2,F_3)$ is such a solution then
$${\rm curl}\,F=(F_{3.2},-F_{3.1}, F_{2.1}-F_{1.2})\ .$$
We write $F_3=:u$. The condition ${\rm curl}\,F=F$ then leads to
$F_1=u_y$, $F_2=-u_x$, and above all to 
$$u+\Delta u=0\ .\tag{1}$$
The well known equation $(1)$ can be treated using separation of variables. From a geometrical standpoint it would be natural to try $u=R(r)\>\Phi(\phi)$, but this leads to Bessel's differential equation $r^2 R''+r R'+(r^2-k^2)R=0$, $\>k\in{\mathbb N}$, whose solutions are non elementary.
When we write $u=X(x)\>Y(y)$ instead we are lead to
$$X''+\lambda X=0,\qquad Y''+(1-\lambda) Y=0\ ;$$
here $\lambda\in{\mathbb R}$ is arbitrary. When $0<\lambda<1$ we obtain solutions $u$ which are periodic in $x$ and in $y$. For other values of $\lambda$ the resulting $u$ is periodic in one variable, and a hyperbolic function in the other. The values $\lambda=0$ and $\lambda=1$ are special; here one of $X(\cdot)$ and $Y(\cdot)$ is linear, the other periodic with period $2\pi$.