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Topic: PDE toolbox & cylinder coordinates
Replies: 2   Last Post: Apr 24, 2013 11:28 AM

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Posts: 3
Registered: 10/28/11
Re: PDE toolbox & cylinder coordinates
Posted: Apr 24, 2013 11:28 AM
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"Michael Thomas" <> wrote in message <9ugcha$86a$>...
> 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


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