I have the following incompressible axisymmetric velocity field. $$u=u_r\hat e_r+u_\theta\hat e_\theta+u_z\hat e_z$$

For the planar analog to this flow (where the swirl velocity $u_\theta=0$) I know a scalar stream function can be defined as $$u_r=\frac{1}{r}\partial_z\psi$$ $$u_z=-\frac{1}{r}\partial_r\psi$$ I am looking for a way to relate $u_\theta$ to $\psi$, and the above equations.

In "Vorticity and Incompressible Flow" by Majda and Bertozzi it states that if there exists a stream function that solves the 2D Euler equations (so that $\Delta\phi=F(\phi)$ where $F(\phi)$ is some unknown smooth function related to the vorticity) then a 3D velocity field can be defined in Cartesian coordinates as $$v=-\partial_2\phi\hat e_1+\partial_1\phi\hat e_2+W(\phi)\hat e_3$$

Here $W(\phi)$ is defined by the following ODE $W'(\phi)W(\phi)=F(\phi)$.

I understand that the axisymmetric planar flow without swirl is not 2D because $u_r$ still has some curvature in it.

For the case of axisymmetric flow with no swirl it turns out that the Laplacian of the stream function is not directly related to vorticity, instead the vorticity is related to the stream function accordingly $$\frac{\omega_\theta}{r}=L\psi$$ where $$L\psi=\frac{1}{r^2}\partial_z^2\psi+\frac{1}{r}\partial_r(\frac{1}{r}\partial_r\psi)$$ and $\omega_\theta/r$ is some constant $\xi$ which is constant with the flow if the swirl is zero. If the swirl is not zero, then $$D_t(\frac{\omega_\theta}{r})\equiv u\cdot\nabla\frac{\omega_\theta}{r}=-\frac{1}{r^4}\partial_z[(ru_\theta)^2]$$ where $ru_\theta$ is constant along the flow in either case.

Is there a way I can do something similar using $L\psi$ in axisymmetric cylindrical coordinates to relate $u_\theta$ to $\psi$ similar to how $W(\phi)$ is related to $\Delta\phi$ in Cartesian coordinates?

  • $\begingroup$ $-\frac{1}{r^4}\partial_z[(ru_\theta)^2]$ should be ${\color{red}{+}}\frac{1}{r^4}\partial_z[(ru_\theta)^2]$. See (2.1a) of ref. "arXiv:1807.00184v1". $\endgroup$
    – mike
    Commented Nov 14, 2022 at 4:03

1 Answer 1


For axisymmetric flow with swirl, the velocity components $u_r(r,z)$, $u_\theta(r,z)$, and $u_z(r,z)$ are independent of the the angular coordinate $\theta.$ Under conditions of incompressible flow, the velocity field is solenoidal and satisfies the continuity equation

$$\nabla \cdot \mathbb{u} = \frac{1}{r}\frac{\partial}{\partial r}(r u_r) + \frac{1}{r}\frac{\partial u_\theta}{\partial \theta} + \frac{\partial u_z}{\partial r} = 0.$$

As in the case of flow without swirl, we can introduce a stream function $\psi$ such that

$$u_r = \frac{1}{r}\frac{\partial \psi}{\partial z}, \,\,\,\, u_z = -\frac{1}{r}\frac{\partial \psi}{\partial r}. $$

This allows the continuity to be satisfied as $u_\theta$ is independent of $\theta$ $ \left( \frac{\partial u_\theta}{\partial \theta}= 0 \right)$ and

$$\nabla \cdot \mathbb{u} = \frac{1}{r}\frac{\partial^2 \psi}{\partial r \partial z} + \frac{1}{r}\frac{\partial u_\theta}{\partial \theta} - \frac{1}{r}\frac{\partial^2 \psi}{\partial z \partial r } = 0.$$

In contrast to axisymmetric flow without swirl there can be three components of vorticity that do not vanish

$$ \omega_r = \frac{1}{r}\frac{\partial u_z}{\partial \theta} - \frac{\partial u_\theta}{\partial z} = - \frac{\partial u_\theta}{\partial z}, \\ \omega_\theta = \frac{\partial u_r}{\partial z} - \frac{\partial u_z}{\partial r}, \\ \omega_z = \frac{1}{r}\frac{\partial}{\partial r}(r u_\theta) - \frac{1}{r} \frac{\partial u_r}{\partial \theta} = \frac{1}{r}\frac{\partial}{\partial r}(r u_\theta). $$

You are specifically interested in how the swirl component of velocity is related to other flow variables.

In inviscid flow, the $\theta$-component of the Euler (momentum) equation is

$$\frac{\partial u_\theta}{\partial t} + u_z \frac{\partial u_\theta}{\partial z} + u_r \frac{\partial u_\theta}{\partial r} + \frac{u_ru_\theta}{r} = 0.$$

Multiplying both sides by $r$ we get

$$\frac{\partial }{\partial t}(r u_\theta) + u_z \frac{\partial }{\partial z}(r u_\theta) + u_r \frac{\partial }{\partial r}(r u_\theta) = 0.$$

This can be written in terms of the convective or material derivative $D/Dt$ as

$$\frac{D}{Dt}(r u_\theta) = 0.$$

Given a circular ring of fluid particles with initial radius $r$ and axial position $z$, the circulation around the ring defined by $C(t) = 2 \pi r u_\theta$ remains invariant as the ring of particles is deformed and convected by the radial and axial components of velocity through time. This also is a consequence of a more general principle known as Kelvin's Theorem.

Streamlines are instantaneously tangential to the velocity field since $\mathbb{u} \cdot \nabla \psi = 0$. Hence, the value of the stream function is fixed along the ring at the initial value $\psi(r,z)$, and is invariant in steady flow.

The implication is that in steady axisymmetric flow with swirl there is an arbitrary function $G$ such that

$$r u_\theta = G(\psi).$$

This, I believe, is the relationship you were looking for. The function $G$ appears along with another arbitrary function in what is called the Bragg-Hawthorne equation which governs the flow dynamics. The functions are determined by solution subject to boundary conditions.

  • $\begingroup$ I'm a little confused as to why you write the $\partial_\theta u_\theta$ term in the continuity equation since it is zero due to axisymmetry of the problem. $\endgroup$
    – WnGatRC456
    Commented Mar 9, 2016 at 18:10
  • $\begingroup$ I wrote the full form of the equation and later used the fact that $\partial_\theta u_\theta = 0$ to show that the continuity condition was satisfied with the other components defined in terms of the stream function. $\endgroup$
    – RRL
    Commented Mar 9, 2016 at 18:11
  • $\begingroup$ Hence, the stream function formulation is also useful for axisymmetric flow with swirl. $\endgroup$
    – RRL
    Commented Mar 9, 2016 at 18:13
  • $\begingroup$ ok that clears it up for me. So the reason that the stream function satisfies continuity is because of the axisymmetry independent of the value of the swirl component. $\endgroup$
    – WnGatRC456
    Commented Mar 9, 2016 at 18:17
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – RRL
    Commented Mar 9, 2016 at 18:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .