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