Example of conservative vector field that is not irrotational Can anyone suggest an example of a vector field $F: A \subset \mathbb{R}^3 \to \mathbb{R}^3$ that satisfies all the following conditions?


*

*$F$ does not belong to $C^1(A)$

*$F$ is conservative in $A$

*It is not true that $\mathrm{curl} F(x)=\bar{0}$ $\,\,\,\,\forall x \in A$


(Also an example in dimensions other than $\mathbb{R}^3$ would be good)
 A: Here we give an example in $\mathbb{R}^2$. 
Generalizing to $\mathbb{R}^3$ is straightforward. 
Let 
$$\varphi = \begin{cases}
\frac{x y(x^2-y^2)}{x^2+y^2}, & (x,y)\ne (0,0) \\
0, & \textrm{else}
\end{cases}$$
and consider the gradient field ${\bf F}=\nabla\varphi$ on $\mathbb{R}^2$. 
It is a standard exercise to show that $\mathrm{curl}\,{\bf F}$ is undefined at the origin.  
A: The field $F$ is conservative, so it has antiderivative: we have $F=\mathrm{grad}\,\varphi$ with some scalar field $\varphi$. If $F$ is differentiable at some point $a$, then $\mathrm{curl}\,F(a)=0$ by Young's theorem. Therefore, curl is either zero or undefined.
An example for undefined curl is:
$$
F(x,y,z,\ldots) = \big(|x|,0,0,\ldots\big).
$$
This field has the antiderivative $\frac{x|x|}{2}$ so it is conservative; but 
$\mathrm{curl}\,F$ is undefined along the plane $x=0$.
If you want to define curl using only partial derivatives (although curl should be independent from the co-ordinate system), a modified construction is
$$
F = \mathrm{grad} \frac{(x+y)|x+y|}2 = \big( |x+y|,|x+y|, 0 \big).
$$
