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by woopwoop
1813 days ago
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Why is the laplacian so ubiquitous? Well, locally any reasonable PDE is well approximated by a linear one. And linear PDEs are of the form Lu = f for a linear differentiable operator L. But why does the particular case of L = laplacian show up so often in physics? Galilean invariance says that the laws of physics should be invariant under translation and rotation. Checking this against Lu = f, you can verify that one requires that L commute with translations and rotations. That is L(u_h) = (Lu)_h, where u_h(x) = u(x+h), and similarly for rotations. What you can show (pretty easily on the Fourier side) is that every linear differential operator with these properties is a polynomial in the Laplacian. |
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I may be tripping over something here but this doesn't sound right. If you mean polynomial in the real number sense, i.e.
Lu = a0 + a1 Delta(u) + a2 Delta(u)^2 + ...
(where Delta = Laplacian, and a0, a1, ... real numbers), then is this true? The famous wave operator doesn't have this form.
And if you mean "polynomial" as in a series over function space, i.e.
Lu ~ a0 + a1 Du + a2 D^2u + ... (infinite terms, not equality but convergence)
where D is the usual differential operator and a0 is a number, a1 is a 1-d vector, a2 is a 2d matrix, so on, then this is standard calculus of variations on any reasonable function space. Taylor series for function spaces if you want. I don't think that's limited to nice Galilean operators.