Rotational and irrotational flow 
*

*Can a rotational flow take place if there is no viscosity? 

*If yes then will it always be rotational or will it become irrotational after some time?
 A: If by rotational flow you mean
$$
\vec{\omega} := \nabla \times \vec{u} \ne 0
$$
everywhere, then the answer to your question is


*

*Yes, vorticity can be introduced through the fluid boundaries (see Klein's Kafeeloffel experiment)

*No, conservation of total vorticity means any vorticity in the fluid stays with the fluid (unless there is a vorticity sink somewhere)


Here are more details:
Consider the vorticity form of the Navier-Stoke's equation (with a barotropic fluid):
   $$
   \frac{D \vec{\omega}}{D t} = -\vec{\omega} \cdot \nabla \vec{u} + \nu \nabla^2 \vec{\omega}.
   $$
Notice that if $\vec{\omega}$ is zero everywhere to begin with, the governing equation reduces to
$$
   \frac{D \vec{\omega}}{D t} = 0.
$$
So an irrotational fluid will not develop vorticity on its own.
However, it is possible to introduce rotational fluid elements into the fluid through its boundaries.  A body moving in a fluid will have a bound vortex sheet on its surface that maintains the no-flow-through condition.  If $\nu \ne 0$, then that bound vortex sheet is constantly being diffused into the fluid.  But even for an inviscid fluid where $\nu = 0$, the bound vortex sheet can be released into the fluid if the body is dissolved or instantaneously removed from the fluid (Klein's Kafeeloffel experiment).
We can determine what happens to the rotational elements in an inviscid fluid by tracking the flux of vorticity through the fluid surface $\mathcal{S}(t)$.  We have
$$
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathcal{S}(t)} \vec{\omega}\cdot\vec{n} \,\mathrm{d}\mathcal{S} 
& = \int_{\mathcal{S}(t)} \frac{D\vec{\omega}}{Dt} \cdot\vec{n} \,\mathrm{d}\mathcal{S} 
  + \int_{\mathcal{S}(t)} \vec{\omega}\cdot\frac{D\vec{n}}{Dt} \,\mathrm{d}\mathcal{S} \\
& = \int_{\mathcal{S}(t)} \left[ {D\vec{\omega}}{Dt} - \vec{\omega}\cdot\nabla \vec{u} \right] \cdot \vec{n} \,\mathrm{d}\mathcal{S}.
\end{align*}
$$
From the vorticity transport equation, we see that the term inside the brackets is 
$$
\frac{D\vec{\omega}}{Dt} - \vec{\omega} \cdot \nabla \vec{u} = \nu \nabla^2 \vec{\omega}
$$
which is zero for inviscid flows!  That means that 
$$
\frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathcal{S}(t)} \vec{\omega}\cdot\vec{n} \,\mathrm{d}\mathcal{S} = 0,
$$
so vorticity never leaves a material volume of an inviscid fluid.
A: 1) Yes. The rotational flow of a rigidly rotating fluid (in a rotating cylinder, for example) is a perfectly good solution of the Euler equation.
2) No. For one, the rigidly rotating flow described in 1) is perfectly stable, even if viscosity is taken into account. Furthermore, the main phenomenon described by the Navier-Stokes equation is vorticity diffusion. This will tend to do the opposite of what you ask about. For example, a vortex line (which is a flow that is almost everywhere irrotational, with voricity concentrated in a line) will decay by vorticity diffusion to rotational flow.  
