solenoid and irrotational vector

Let $V$ be a vector point function. $V$ is solenoid if $\operatorname{div} V =0$ and irrotational if $\operatorname{curl} V =0$.

How can one visualize examples of solenoid or irrotational functions? Are there any functions such that they are both irrotational and solenoid? (Except constant functions)

An example of a solenoid field is the vector field $V(x,y)= (y,-x)$. This vector field is ''swirly" in that when you plot a bunch of its vectors, it looks like a vortex. It is solenoid since $$\text{div} V = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(-x) = 0.$$ The divergence being zero means that locally no field is being "created" at each point, much as is the case with this vector field. For real world examples of this, think of the magnetic field, $\vec{B}$. One of Maxwell's Equations says that the magnetic field must be solenoid.

An irrotational vector field is, intuitively, irrotational. Take for example $W(x,y) = (x,y)$. At each point, $W$ is just a vector pointing away from the origin. When you plot a few of these vectors, you don't see swirly-ness, as is the case for $V$.

As for a vector field which has both properties, see this post.

A good example of an irrotational vector field, that is always worth pointing out as warning case, is the irrotational vortex:

$$\vec{V}(x,y,z) = \left ( \frac{-y}{x^2 + y^2} , \frac{x}{x^2 + y^2}, 0 \right )$$

It has zero curl, even though it looks like a vortex that is clearly "rotating" around the origin:

So it is important to take that intuition for "rotation" carefully. This is one of the reasons why Maxwell called it "curl" in the first place, as he did not want to attach the idea of rotation to it.

Consider an arbitrary analytic function $$f(x,y)=u(x,y)+iv(x,y)$$. Then its complex conjugate is $$\overline{f(x,y)}=u(x,y)-iv(x,y)$$ (which is not analytic unless $$f$$ is constant). According to Cauchy-Riemann equations for $$f$$, we have:

$$u_x=v_y$$ and $$v_x=-u_y$$.

The first equation is equivalent to $$∇.\overline{f}=0$$ (i.e.$$\overline{f}$$ is incompressible or solenoidal) and the second one is equivalent to $$∇×\overline{f}=0$$ (i.e.$$\overline{f}$$ is irrotational).

Just to add to the answer above, under fairly mild conditions, you can decompose a vector field (in $R^3$) into its solenoidal and irrotational parts (Helmholtz Decomposition). So you can think of general vector fields as having "constituents", one solenoidal and the other irrotational.

Draw its field lines and local eqipotential surfaces, which are alays perpendicular to eacheck other. The field lines are continuous for an incompressible (solenoid) field, while the eqipotentials are continuous for irrotational (conservative) fields. The terms in parentheses indicate the existence of a scalar or vector potential, respectively, which is always the case for these fields in a space without any holes or gaps (multiply connected), by potential theory (Poincare or de Rham).

By the Helmholtz or Hodge decomposition theorem, there is only one field per winding number in a given space satisfying external boundary conditions which is both incompressible and irrotational. It has both continuous field lines and eqipotentials that break only at the boundary. This 'harmonic field' is as uniformly spread out and even as possible, and its scalar and vector potentials satisfies the Laplace equation.