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.

Hmm.. OK. I'll try to wrap my head around that. ;-)

> 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...

So let's see if I can work this out. To convert to the weak formulation you'd start with the laplacian and integrate against an arbitrary function. If we just focus on the 'r' part it would go like so:

\int s \frac{1}{r} \frac{d}{dr} (k r \frac{du}{dr}) dV

but the r part of dV is r dr (forgetting about the 2\pi for now) so that's

\int s \frac{1}{r} \frac{d}{dr} (k r \frac{du}{dr}) r dr

which looks like:

k \int s \frac{d}{dr} (r \frac{du}{dr}) dr

But if that gets integrated by parts it's works out to be

k (s r \frac{du}{dr})\big\vert_{\Gamma} - k \int \frac{ds}{dr} \frac{du}{dr} r dr

Which is I think more or less what I had before. Have I got that right?

> 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.