(Note $\vec{F}$ and $\vec{G}$ are arbitrary 3D vector fields)
So I have been messing with some PDE recently. Some expressions came to mind include
$$\nabla \cdot \vec{F}=0 \hspace{12mm}[1]$$ $$\nabla \times \vec{F}=\vec{0}\hspace{12mm}[2]$$ $$\nabla \phi=0\hspace{12mm}[3]$$
I often seen in some workings in physics textbooks that $[2]$ usually implies
$$\vec{F}=\nabla\phi\hspace{12mm}[4]$$
In physics we are taught about scalar potentials and vector potentials which satisfy the following because of vector calculus identities
$$\nabla\phi=\vec{F},\nabla \times \vec{F}=\vec{0}\hspace{12mm}[5]$$
$$\nabla\times\vec{G}=\vec{F}, \nabla \cdot \vec{F}=0\hspace{12mm}[6]$$
However I noticed for $[1]$ it does not necessary means $[6]$ because there exist irrotational vector fields e.g. $\vec{K}$ where the divergence vanishes everywhere as shown in this link because of how the net inflow/outflow of vectors in an infinitesimal volume is zero
$$\vec{K}=\begin{pmatrix} \frac{x}{r^3} \\\frac{y}{r^3}\\\frac{z}{r^3}\end{pmatrix}$$
where $r^2=x^2+y^2+z^2$
Q1 But can we also have an analogous case for $[2]$, where there exists a vector field $\vec{S}$ without singularities such that it satisfy $[2]$ but not $[4]$ i.e. there is no $\phi$ such that
$$\vec{S}=\nabla\phi$$
is true?
Q2
If no such $\vec{S}$ exists, how to explain why we cannot have a notion of net microscopic circulation analogous to net inflow/outflow of vectors in divergence?
Attempt at imagining what it might look like based on the intuition that $\vec{S}$ has no net microscopic circulation seemed to give me singularities...
No. The underlying reason is topological: There's an object called de Rham cohomology that detects whether such functions exist. Unfortunately, it's hard to go into any details without introducing a lot of topological machinery. For $\mathbb{R}^3$, we can explicitly construct $\phi$ with $F = \nabla \phi$ by integrating: $\phi(x) = \int_0^x F.ds$, where the integration is taken along a path from $0$ to $x$. By Stokes' theorem, this integral is path-independent, and so defines a nicely behaved function $\phi$.
On the other hand, if $\phi$ has a singularity, then there's no guarantee that the path won't go through it, and the resulting integral may not be path-independent even when the path does avoid it. The classic example is the function $F$ you give in your post. It's easier to consider the $2$-dimensional version. There, $\int_C F.ds = 2\pi$, where $C$ is the unit circle. If $F$ were a gradient, that integral would have to vanish by path-independence. (Think of the circle as two paths from $1$ to $-1$ joined at their endpoints.) It turns out that in many situations (specifying the exact conditions requires a bit of topology), that's the only obstruction to being the gradient of a field. Specifically, there are integrals over finitely many spaces that vanish iff a irrotational field $F$ has $F = \nabla \phi$ for some $\phi$.
I realize that the above is vague and possibly not very useful, but the answer to your question is 'no' in general: Irrotational fields on $\mathbb{R}^3$ with singularities, or more generally fields on other spaces, are not necessarily of the form $F = \nabla \phi$. To go further into the matter, the next place to look is de Rham cohomology.