Neumann boundary condition on axis in cylindrical coordinates

14 months ago by
I'm trying to solve the eigenvalue problem from a linear stability analysis about a steady Navier-Stokes solution:

\[ s\hat{\mathbf{u}} + (\mathbf{u}_b . \nabla) \hat{\mathbf{u}} + (\hat{\mathbf{u}}.\nabla)\mathbf{u}_b + \nabla\hat{p} - \nu \nabla^2 \hat{\mathbf{u}} = 0 \]

In particular I'm looking at solutions in cylindrical coordinates where \( \hat{\mathbf{u}}(r,z,\theta) = \tilde{\mathbf{u}}(r,z)e^{mj\theta} \) with \(m=\pm1\).  An asymptotic analysis shows that the boundary conditions for the \(m\pm-1\) modes as \(r\rightarrow 0^+\)  are \(\frac{\partial \tilde{u}_\theta}{\partial r}=\frac{\partial \tilde{u}_r}{\partial r}=0 \) .

I'm using axisymmetric finite elements with \( \int_\Omega d\Omega \rightarrow \int_{0}^L \int_{0}^R \int_{0}^{2\pi} r d\theta dr dz \) and a test function of the form \( \mathbf{v}(r,z,\theta) = \tilde{\mathbf{v}}(r,z)e^{-mj\theta}\). Following recommendation from 'Global mode interaction and pattern selection in the wake of a disk: a weakly nonlinear expansion, Meliga et al. JFM 633 [159-189]' I multiply the governing equations by \(r\) to ensure that there are no \(\frac{1}{r}\) terms left in the variational form. 

Unfortunately I can't work out how I should set the radial derivative boundary conditions on the axis. All the terms involving \(\frac{\partial \tilde{u}_\theta}{\partial r}\) in boundary integrals on the axis look like: 
\[\int_{\Gamma_{\text{axis}}} r^2 \frac{\partial \tilde{u}_r}{\partial r} 2\pi dz\]

Normally I would set Neumann boundary conditions by just setting this boundary integral to zero. Unfortunately, because on the axis \(r=0\),  this term is automatically zero regardless of what the radial derivative is.

Is there something I'm missing? Are there any other common methods of enforcing a Neumann boundary condition? Is there a better way to set up the axisymmetric formulation? 

Thanks for any help.

2 Answers

14 months ago by
Your observation is correct. Since  $r=0$r=0 on the axis, the boundary term goes away naturally. You shouldn't need to set a boundary condition there and it's unphysical to try to set a boundary condition on the axis since you are really solving a problem in a cylinder.
14 months ago by
Thanks for this. I was worrying I'd missed a subtlety somewhere.
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