fluid dynamics - Does irrotational imply inviscid?

Let us consider a 2D irrotational flow, such that $\nabla\times\boldsymbol u =\boldsymbol 0$. Defining the stream function such that $\boldsymbol u =\nabla\times\psi \boldsymbol n$ where $\boldsymbol n$ is the unit vector perpendicular to the plane of the motion. The incompressibility condition implies $\nabla^2\psi =0$, so $\nabla^2 \boldsymbol u= \nabla^2 \nabla\times\psi \boldsymbol n=\nabla\times\nabla^2\psi \boldsymbol n=\boldsymbol 0$. From this follows that in the Navier-Stokes equations the term $\nu\nabla^2\boldsymbol u=0$ vanishes, no matter the viscosity.

Now, does this allow us to say that the fluid is inviscid? Or, after considering the nature of the inertia, we can say that:

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  if inertia can be neglected, the equation of the motion is $\nabla p=\boldsymbol 0$.

  
  


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  if inertia is important, the flow is at high Reynolds number.

  
  
  

My problem is: Is every irrotational flow inviscid? This is kind of counter-intuitive. But I think my error is in the "no matter the viscosity"...

Answer

(I try to answer to my own question, after some reflections made with the help of Luboš.)

For an incompressible and irrotational flow, the conditions $\nabla\times \boldsymbol u=\boldsymbol 0$ and $\nabla\cdot \boldsymbol u = 0$ imply, $\nabla^2\boldsymbol u =\boldsymbol 0$. Indeed:

$$\nabla^2\boldsymbol u = \nabla(\nabla\cdot\boldsymbol u)-\nabla\times(\nabla\times \boldsymbol u) = \boldsymbol 0$$

This forces us to write down the Navier-Stokes equation for the motion of the fluid without the viscous term $\mu\nabla^2\boldsymbol u$, no matter the viscosity:

$$ \rho(\partial_t \boldsymbol u + u\cdot\nabla \boldsymbol u) = -\nabla p \ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$

Now, it could seem that this implies the flow is automatically a high-Reynolds number flow (for which we could have written down the same equation, but for a different reason: $\mu=\rho\nu\simeq 0$, and this would have been an approximation). But, even if the viscosity is far from neglectable, we can make another kind of approximation, saying that the inertia, represented by the left-hand-side terms in N-S equation, can be neglected because of $Re=UL\rho/\mu\ll 1$ (this can happen in a lot of situations: microobjects, extra-slow flows, and - of course - high viscosity. In this case, the equations of motion become: $$-\nabla p = \boldsymbol 0\ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$ which are, in fact, the equations we would arrive at if we started by the Stokes equation (for inertia-less flows) for irrotational flows.

Then, an irrotational flow is not necessarily governed by the Euler equation, i.e., it's not necessarily inviscid.