Date: Apr 24, 2013 11:28 AM
Subject: Re: PDE toolbox & cylinder coordinates
"Michael Thomas" <firstname.lastname@example.org> wrote in message <email@example.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
Is the transformation above based on the "y-axis" (vertical axis in PDE Tool window) as the axis of symmetry, or the "x-axis" (horizontal axis)? Would the following be the transformation using the opposite axis?
-1./y*div(y.*grad(u)) + a*u = f