Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Colin
Posts:
3
Registered:
10/28/11


Re: PDE toolbox & cylinder coordinates
Posted:
Apr 24, 2013 11:28 AM


"Michael Thomas" <mthomas@mathworks.com> wrote in message <9ugcha$86a$1@news.mathworks.com>... > Taking the generic scalar elliptic PDE, and transforming it to cylindrical > coordinates results in something that looks like > > 1./x*div(x.*grad(u)) + a*u = f >
Michael,
Is the transformation above based on the "yaxis" (vertical axis in PDE Tool window) as the axis of symmetry, or the "xaxis" (horizontal axis)? Would the following be the transformation using the opposite axis?
1./y*div(y.*grad(u)) + a*u = f
Thanks, Michael



