On 06/04/2012 05:33 PM, steve wrote:
Hi Robert,
Where "BT" is some kind of boundary term, that I haven't sorted out yet.
Since it's a Neumann BC I'm not sure how it gets incorporated into the
weak
form statement.
Yes, those are simple and standard. But I meant the weak form of the equations in the cylindrical (and other) coordinates :). (With all those 1/r and other stuff.)
Ah... good point. So assuming u has no theta dependence (which I realize as I'm typing this is not necessarily true in the acoustics case.. but more on that later) the volume integrals reduce to:
2\pi \int_\Omega (u_{,r} s_{,r} + u_{,z} s_{,z}) r,dr,dz
and
2\pi k^2 \int_\Omega (s u ) r,dr,dz (for the acoustics case only)
OK. So it seems that for the above, you could use directly dw_laplace and dw_volume_dot terms, with material arguments equal to 2\pi r and 2\pi k^2 r, respectively. Or just r, and put the known constants directly into the equations string.
Where \Omega is now the 2-D interior of the 'r-z' plane. \Gamma is reduced to a 1D boundary, but d\Gamma is going to get a factor of 2\pi r as well to take into account the 'distance' around in the theta direction.
Since the weak form only depends on the gradient, and '1/r' only shows up in the theta part of the gradient, it doesn't appear at this point. The only 'r' factor is in the volume element itself.
I thought (googled) that the 'r' part of the cylindrical Laplacian was (d = \partial):
\frac{1}{r} \frac{d}{dr} (k r \frac{du}{dr}), which is
k \frac{d^2 u}{dr^2} + \frac{k}{r} \frac{du}{dr}.
so in weak form it comes to a laplace-like term with different coefficients by the 'u' and 'z' parts, and a gradient-like term w.r.t. 'r' (the second one in the line above). Just curious...
Now.. about the theta dependence of u. In the acoustics case there will be modes that have theta dependence, but they will have a continuity condition that will require the theta part go like exp(+mj theta) where m is an integer, so this will add a term to the laplacian. But I can worry about that after I get the basic concept working I think.
Good.
Cheers, r.