Burgers vortex: finding the angular velocity

Question

I'm trying to solve exercise 5.19 from Elementary Fluid Dynamics by D. J. Acheson. The exercise is as follows:

Seek an exact, steady solution to the Navier-Stokes equations of the form

$$ \mathbf{u} = -\frac{1}{2}\alpha r \mathbf{e}_r + u_\theta (r) \mathbf{e}_{\theta} + \alpha z \mathbf{e}_z $$

where $\alpha$ is a positive constant. Note that $\boldsymbol{\omega} = \omega \mathbf{e}_z$, where

$$ \omega = \frac{1}{r}\frac{d}{dr}(ru_{\theta}) $$

Verify that $\nabla \cdot \mathbf{u} = 0$, and show that the equations of motion imply

$$ -\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r} $$

Deduce that

$$ u_{\theta} = \frac{\Gamma}{2\pi r}\left(1 - e^{-\alpha r^2 / 4\nu}\right) $$

where $\Gamma$ is an arbitrary constant.

So, here's what I've done so far. It's easy enough to verify that $\nabla \cdot \mathbf{u} = 0$. The divergence in cylindrical coordinates is:

$$ \nabla \cdot \mathbf{u} = \frac{1}{r}\frac{\partial}{\partial r}(ru_r) + \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta} + \frac{\partial u_z}{\partial z} $$

Inserting the velocity components I get:

$$ \nabla \cdot \mathbf{u} = \frac{1}{r}(-\alpha r) + \alpha = 0 $$

To show that $-\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r}$ I insert the velocity components into the momentum equation of the $\theta$-component:

$$ \frac{\partial u_{\theta}}{\partial t} + u_r \frac{\partial u_{\theta}}{\partial r} + \frac{u_{\theta}}{r} \frac{\partial u_{\theta}}{\partial \theta} + u_z\frac{\partial u_{\theta}}{\partial z} + \frac{u_r u_{\theta}}{r} = -\frac{1}{\rho r}\frac{\partial P}{\partial \theta} + \nu \left[ \frac{1}{r}\frac{\partial}{\partial r}\left( r \frac{\partial u_{\theta}}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u_{\theta}}{\partial \theta^2} - \frac{u_{\theta}}{r^2} + \frac{2}{r^2}\frac{\partial u_r}{\partial \theta} \right] $$

The pressure term on the right side equals zero since the pressure doesn't vary in the $\theta$-component. Inserting for the velocity components yields:

$$ -\frac{\alpha r}{2} \left[ \frac{\partial u_{\theta}(r)}{\partial r} + \frac{1}{r} u_{\theta}(r) \right] = \nu \left[ \frac{1}{r} \frac{\partial u_{\theta}(r)}{\partial r} + \frac{\partial^2 u_{\theta}(r)}{\partial r^2} - \frac{1}{r^2}u_{\theta}(r) \right] $$

Since

$$ \omega = \frac{1}{r}\frac{d}{dr}(ru_{\theta}) = \frac{u_{\theta}}{r} + \frac{\partial u_{\theta} (r)}{\partial r} $$ $$ \frac{d \omega}{d r} = -\frac{1}{r^2}u_{\theta}(r) + \frac{1}{r}\frac{\partial u_{\theta}(r)}{\partial r} + \frac{\partial^2 u_{\theta}(r)}{\partial r^2} $$

we have that

$$ -\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r} $$

What I need help with figuring out is how to get the following expression for the $\theta$-component velocity:

$$ u_{\theta} = \frac{\Gamma}{2\pi r}\left(1 - e^{-\alpha r^2 / 4\nu}\right) $$

Answer 1

The vorticity is obtained by solving $\,\,\displaystyle \nu \frac{d \omega}{dr} = -\frac{\alpha }{2} r \omega$. This is a first-order linear differential equation that can be solved using seperation of variables

$$\frac{d \omega}{\omega} = -\frac{\alpha }{2\nu} r \, dr,$$

to obtain $\omega = C_1e ^{- \alpha r^2/4 \nu}$ where $C_1$ is a constant of integration.

Next solve for $u_\theta$ using $\displaystyle \frac{1}{r}\frac{d}{dr}(ru_{\theta}) = \omega$.

We have

$$\frac{d}{dr}(ru_{\theta}) = r\omega = C_1 r e ^{- \alpha r^2/4 \nu}$$

Integrating both sides we get (introducing another integration constant $C_2$)

$$r u_\theta = C_2 - \frac{2\nu C_1}{\alpha}e^{- \alpha r^2/4 \nu},$$

which can be rewritten in terms of arbitrary constants $\Gamma$ and $B$ as

$$u_\theta = \frac{\Gamma}{2\pi r}\left(1 - B e^{- \alpha r^2/4 \nu} \right)$$

In order that $u_\theta$ remain finite as $r \to 0$ we require $B= 1$.