Helmholtz decomposition allows incompressible flow with an irrotational component? A vector field can be written in terms of irrotational and a divergence-free components. Using a 2D velocity field as an example,
$ \vec v = -\nabla \phi + \nabla \times \vec \Psi$
Where $\vec \Psi$ is a vector potential, which in fluid mechanics is only guaranteed to exist if we're working in two dimensions so that $\vec \Psi = (0,0,\psi)$, where $\psi$ is called the stream function.
There are many sources I can find that say that an incompressible flow ($\nabla \cdot \vec v = 0$) simplifies to $ \vec v = \nabla \times \vec \Psi$  Here is one such example, although the Wikipedia article on stream functions implies the same.
This seems incorrect to me, since taking the divergence of both sides of this equation
$ \nabla \cdot \vec v = -\nabla \cdot \nabla \phi + \nabla \cdot \nabla \times \vec \Psi$
simply yields the Laplace equation, $\nabla^2 \phi = 0$. This means that as long as $\phi$ is a nonzero harmonic function, I can have a velocity field in an incompressible fluid that has an irrotational component. Is there an additional constraint that forces $\nabla \phi = 0$ in order for $\nabla \cdot \vec v =0$?
 A: The key concept needed here is that the Hemholtz decomposition is not necessarily unique.
Non-uniqueness can occur because there exist nontrivial vector fields which are both irrotational and divergence-free. For example, the constant 2D velocity field $\vec v = (1,0)$ can be expressed as either $\vec v=-\nabla \phi$ with $\phi(x,y)=-x$, or as $\vec v=\nabla \times \vec \Psi$ with $\vec \Psi (x,y) = (0,0,y)$.
Because of this non-uniqueness, showing that it’s possible to express a particular velocity field with an incompressible flow as a Hemholtz decomposition with a nonzero $\phi$ doesn’t mean that it isn’t also possible to express the same velocity field as a different Hemholtz decomposition in which $\phi = 0$.
An arbitrary 2-D velocity field with $\nabla \cdot \vec v = 0$ can be written purely in terms of a vector potential in which the stream function is
$$\psi(x,y)=\int^{y}_{0}v_{x}(0,y')dy'-\int^{x}_{0}v_{y}(x',y)dx' .$$
I'll leave verification of that equation as an exercise to the reader.
A: You can use a trick to solve that: Instead to take the divergence, take the curl as follows:
$\nabla \times  \vec v = \nabla \times  (-\nabla \phi + \nabla \times \vec \Psi)$
but  $ \nabla \times \nabla \phi=0 $, so
$\nabla \times  \vec v = \nabla \times  (\nabla \times \vec \Psi) = \nabla(\nabla \cdot \vec \Psi)-\nabla^{2} \vec \Psi$, (just a vector identity)
but  $ \nabla(\nabla \cdot \vec \Psi) = 0$, because the left hand side only has a component perpendicular to the plane, so this gradient must be a gradient along the axes orthogonal to the plane. But we know that the third component of the field $\vec \Psi$ has dependency in $x$ and $y$ only, that is, $\vec \Psi = (0,0,\psi(x,y))$ and $\frac{\partial \psi}{\partial z} =0$, therefore
$$\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -\frac{\partial^{2} \psi}{\partial x^{2}} - \frac{\partial^{2} \psi}{\partial y^{2}} $$ Integrate,
$$ \int (\frac{\partial v}{\partial x} + \frac{\partial^{2} \psi}{\partial x^{2}})dx  = \int(- \frac{\partial^{2} \psi}{\partial y^{2}} +\frac{\partial u}{\partial y})dy, \ \ \ then $$
$$ v + \frac{\partial \psi}{\partial x} + f(y) = u -\frac{\partial\psi}{\partial y} + g(x), \ \ \ where \ f(y) \ and \ g(x) \ are \ some \ function. $$
A trivial solution is then when $f(y)=g(x)=0$ and both sides of the equation are set to zero, from where we get our stream functions
$$ u = \frac{\partial\psi}{\partial y}   \ \ \ \ \ \ \ \ v =- \frac{\partial \psi}{\partial x} $$
