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

  • $\begingroup$ 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) $\endgroup$ – Bort Feb 1 '16 at 9:06
  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Feb 1 '16 at 21:19
  • $\begingroup$ @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? $\endgroup$ – wil3 Feb 1 '16 at 21:34
  • $\begingroup$ 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? $\endgroup$ – 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.