Talk:PDE surface

Latest comment: 18 years ago by Lambiam in topic What does the squaring mean?

Is this article promoting the work of a researcher ? I've never heard of Hassan Ugail, and I doubt that most mathematicians have. Can someone clarify this query ? MP (talk) 11:23, 11 April 2006 (UTC)Reply

Yes and no. It did give undue promenance to H. Ugail. I've now fixed that. PDE Surfaces are quite an important methods of generating surfaces, I suspect the topic may be covered elsewhere, although I've not really explored this area of wikipedia. --Salix alba (talk) 14:44, 15 April 2006 (UTC)Reply

What does the squaring mean?

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I don't get the meaning of the squaring of (d^2/du^2+a^2d^2/dv^2). Calling this operator L, is the meaning L(L X(u,v))? Note that L X = 0 is Laplace's equation after a linear rescaling of one of the two ordinates. Given basically any boundary condition on a simple closed boundary, that equation has a unique solution, which is then also a solution of L^2 X = 0. But the latter equation has many solutions. If the meaning is (L X)^2 = 0, that is eqv to the simpler L X = 0. A typo? LambiamTalk 19:31, 15 April 2006 (UTC)Reply

Good point. I found an online paper [1] which indicates this is bassed on the bi-harmonic equation   and is not a typo. However a more important point is that the PDE surface need not necessarily be of this form or even elipitical. I'm week on my differential equations so can't offer much here. --Salix alba (talk) 21:46, 19 April 2006 (UTC)Reply
Ah, I see that I overlooked that the normal derivatives also figure on the boundaries. I think it's useful, next to removing the incorrect "elliptic", to include a link to the online paper, and also to insert the form with nabla^4 as in the C&S paper to make it unambiguous what the intention is. LambiamTalk 22:13, 19 April 2006 (UTC)Reply