# 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 $\nabla \cdot \vec v =0$?

• I feel like this would reduce $\nabla \Phi$ to be a constant, i.e. be identical to a choice of reference frame (which of course you are free to do, but has no physical significance) – Bort Feb 1 '16 at 9:06
• It sort of depends on what region you're working in and what happens at the boundary. Normally you would want $\phi$ to be bounded at infinity, and that is very restrictive for harmonic functions. – Emilio Pisanty Feb 1 '16 at 21:19
• @EmilioPisanty I was wondering if that might be the solution to my problem---let's say I want the velocity to be bounded at infinity, would that somehow force $\nabla \phi$ to be a constant? – wil3 Feb 1 '16 at 21:34
• It sort of depends on exactly what you're willing to impose. All the partial derivatives of $\phi$ are also harmonic so that's easy. If you want them to be non-singular and harmonic throughout, then they achieve their maximum at the boundary. Whether you can go to something as strong as $\nabla\phi$ being constant isn't quite clear to me at the moment - but that's a good question for Mathematics - must a bounded harmonic function in three dimensions be constant? – Emilio Pisanty Feb 2 '16 at 12:06

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.

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

$$u = \frac{\partial\psi}{\partial y} \ \ \ \ \ \ \ \ v =- \frac{\partial \psi}{\partial x}$$