It's often possible to prove that the solutions retain the symmetry of the problem; it may be that you could prove this for the class of problems that is interesting you, and automate the transform to 2D for this class only. On Tue, May 31, 2011 at 11:35 AM, Henry Baker <hbaker1@pipeline.com> wrote:

Hmmmm... I guess that's both good news & bad news: bad news because it involves more work, but good news because it may produce more interesting results.

At 08:29 AM 5/31/2011, Allan Wechsler wrote:

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The p-orbitals of electrons, and some of the vibration modes of a circular drumhead, are examples that occur to me. I'm sure there are simpler "textbook" examples, but I don't know them offhand.

On Tue, May 31, 2011 at 11:07 AM, Henry Baker <hbaker1@pipeline.com> wrote: Interesting! Can you think of any specific examples?

At 08:05 AM 5/31/2011, Allan Wechsler wrote:

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I think this is a forlorn hope in the case of minimization problems in general. Although this particular problem has a solution with circular symmetry, such problems in general won't. A minimization problem with a completely symmetrical formulation might have a solution which breaks the symmetry.

On Tue, May 31, 2011 at 10:56 AM, Henry Baker <hbaker1@pipeline.com> wrote: I was hoping that there was some sort of generic way to interconvert 2D <-> 3D problems like this, rather than having to solve each one individually. I noticed from MathWorld & Wikipedia that there is a generic way to compute surfaces & volumes of revolution.

At 07:49 AM 5/31/2011, Allan Wechsler wrote:

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Oh, OK, that's much clearer, thank you.

I think the problem is still not well-formed, though for a much more subtle reason. In the 2-dimensional version of this problem, we are to preserve the length of the chain. What are we to preserve in the 3-dimensional case? Area? There is no corresponding physical model, since no physical fabric can shunt curvature around in such a way as to preserve area. And if we pre-guess the curvature at each point, we have essentially pre-guessed the solution.

I suppose the area formulation is well-defined: what surface of a given area, bounded by a horizontal surface, has minimum gravitational potential? We need to clarify that the surface has a constant density in mass per unit area.

If we are completely satisfied that the solution will be rotationally symmetric (I am, but don't know how to prove it), then certainly we can come up with an r-z formulation to replace the x-y-z one. It's a classic Calculus of Variations problem (so is the full 3D version, for that matter). In the r-z formulation we have to be careful to calculate the potential correctly, since length-elements near the periphery will contribute more area than ones near the center. If we have z = f(r), then I think the potential element is r dr sqrt( 1 + (dz/dr)^2 ), but I don't remember enough Calculus of Variations to know what to do next.

On Tue, May 31, 2011 at 10:30 AM, Henry Baker <hbaker1@pipeline.com> wrote: You are right; I didn't do a very good job of explaining.

Here's an example: in 2D, a catenary models a hanging chain. What is the equivalent 3D structure (presumably some sort of hanging soap bubble) ? Is it a revolved catenary or something else?

At 07:24 AM 5/31/2011, Allan Wechsler wrote:

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Somehow this problem statement is confusing me. Is the solution of your system a function f(x,y,z), and then the "surface of revolution" of which you speak is the surface f(x,y,z)=0? If I have gotten this wrong, please clarify with the correct form of the solution.

Intuitively, it would certainly seem as if the answer to your question should be "yes". Suppose the axis of rotational symmetry to be the z axis. Then new coordinates would be z and r = sqrt(x^2 + y^2).

On Tue, May 31, 2011 at 9:24 AM, Henry Baker <hbaker1@pipeline.com> wrote:

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Suppose I have a system of partial differential equations in 3D with rotational symmetry about an axis. Is it trivial to convert this system into an ordinary differential equation in 2D, such that the solution, considered as a surface of revolution, is the 3D solution?

Ditto in the reverse direction -- from 2D to 3D.