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Oct 22, 2022 at 19:40 comment added A. J. Pan-Collantes Sorry, but I don't understand. I have only used differential invariants of the symmetry, in order to get a variable change that let me "rectificate" the vector field of the symmetry, and reduce the PDE. But I don't know what is a differential invariant of a PDE
Oct 19, 2022 at 20:15 history edited Mostafa CC BY-SA 4.0
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Oct 19, 2022 at 10:17 comment added Mustafa Bazghandi @A. J. Pan-Collantes - Differential invariants have many applications including reduction of PDEs. Here equivalence of PDEs is considered.
Oct 14, 2022 at 17:13 history edited Daniele Tampieri CC BY-SA 4.0
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Oct 14, 2022 at 16:34 history edited YCor CC BY-SA 4.0
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Oct 14, 2022 at 15:38 history edited Mostafa CC BY-SA 4.0
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Oct 8, 2022 at 17:00 comment added A. J. Pan-Collantes A PDE has infinitesimal symmetries. In order to "use" one of this symmetries to reduce the PDE I look for differential invariants of this vector field. Trivially, this selected symmetry is a symmetry of the differential invariant. Why is important the relation between other symmetries of the original PDE and other symmetries of the differential invariant? I don't see the point.
Oct 8, 2022 at 8:10 comment added Ben McKay The number 1 is a differential invariant of any geometric structure on any manifold and of every system of differential equations in any variables, and also a conservation law of any system of partial differential equations in any variables. It has every diffeomorphism of every manifold as a symmetry. You can see why theories of differential invariants generally give rise to cohomology theories, quotienting out silly things. So the symmetries of a differential invariant can be a larger pseudogroup than the symmetries of the original geometryic structure or pde.
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S Oct 8, 2022 at 7:09 history asked Mostafa CC BY-SA 4.0