Simple solutions in cylindrical coordinates

Writing the Navier-Stokes equations

Vectors and derivatives

We work in cylindrical coordinates \((r,\theta,z)\). There is a natural local basis of orthonormal vectors \({\hat e}_r(\theta)\), \({\hat e}_{\theta}(\theta)\), \({\hat e}_z\). It is important that the directions of the first two of these depend on theta.

We can expand a vector such as velocity in these coordinates as:

\[ {\vec u} = u_r {\hat e}_r + u_{\theta} {\hat e}_{\theta} + u_z {\hat e}_z \]

In taking derivatives of such a vctor, we must take into account that \(\partial_{\theta} {\hat e}_r = {\hat e}_{\theta}\), and \(\partial_{\theta} {\hat e}_{\theta} = - {\hat e}_r\) (\({\hat e}_z\) is constant). Thus,

\[ \partial_{\theta} {\vec u} = \left(\partial_{\theta} u_r - u_{\theta}\right){\hat e}_r + \left(\partial_{\theta} u_{\theta} + u_r\right) {\hat e}_{\theta} + \partial_{\theta} u_z {\hat e}_z \]

This is the basis for the formulae in (A.35) of Acheson.

Steady flow between rotating cylinders.

Consider a fluid between concentric cylindrs with radii \(r_1 < r_2\), rotating with angular velocities \(\Omega_{1,2}\) about \(r = 0\). We are interested in a static solution with constant pressure throughout. We expect that the velocity will be entirely in the \(heta\) direction and that it will be independent of \(\theta, z\). The \(\theta\) component of the Navier-Stokes equation for \(u_{\theta}(r,t)\) is:

\[ \nu \left( \partial_r^2 u_{\theta} + \frac{1}{r} \partial_r u_{\theta} - \frac{1}{r^2} u_{\theta}\right) = 0 \]

These equations are homogenous in \(r\). Each term, actng on \(r^k\), outputs the same power of \(r\). We thus take the ansätz \(u_{\theta} = c r^{\alpha}\) and get the indicial equation \(\alpha (\alpha - 1) + \alpha - 1 = \alpha^2 -1 = 0\). The solutions ae \(\alpha = \pm 1\), so that

\[ u = \frac{B}{r} + A r \]

From here we can apply the “no-slip” boundary conditions \(u(r_k) = r_k \Omega_k\) and solve for \(A,B\):

\[ A = \frac{\Omega_2 r_2^2 - \Omega_1 r_1^2}{r_2^2 - r_1^2}\ ; B = \frac{(\Omega_1 - \Omega_2) r_1^2 r_2^2}{r_2^2 - r_1^2} \]