Timeline for Rellich-Necas identity
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7 at 19:35 | comment | added | Adi | What I'm looking for is in the case v =Ax, can one get an identity without any derivatives of A? | |
| May 7 at 13:18 | comment | added | Willie Wong | @Adi: I don't think I understand what you are looking for. I don't even understand the first step of "rephrasing the rellich identity". What exactly is the LHS here? If you start with a calculus identity and take all the terms involving derivatives of A and put them on one side, then tautologically they would all cancel to zero. So I don't think LHS=Err can be a rephrasing of any calculus identities. | |
| May 7 at 13:04 | comment | added | Adi | What I was looking for is the following (though my question might be a bit vague). Let me phrase the Rellich identity as LHS = Err, where Err are all the terms that contain a derivative of A. After Cancellation, Err becomes "smaller" (as in less terms). With this reduced expression of Err, can one "go back" to get an expression "similar" to the LHS, as in containing quantities without derivatives of A nor Hessian of u? | |
| May 6 at 12:42 | comment | added | Willie Wong | @Adi: "nice" is in the eye of the beholder. As this is really a calculus identity, the question is (using my best sleazy used-car salesperson voice) "what kinda identity do you have in mind?" :-) (depending on what you are looking for you can consider asking a new MO question) | |
| May 6 at 8:02 | comment | added | Adi | Thanks for the reply. I was hoping there would be a nice identity that took all the cancellations into consideration. | |
| May 6 at 1:41 | comment | added | Willie Wong | Even without setting $v = Ax$ there are a lot of cancellations on the RHS. For example, there's a bit of cancellations between $2\delta(v(u) A(\nabla u))$ and $2 v(u) \delta(A(\nabla u))$. In fact the terms involving derivatives of $A$ cancel completely between those two. // I am not an expert on how this identity is used, so don't ask me why the identity itself has the duplication built-in. My naive guess is that it is helpful to keep all second derivatives of $u$ in divergence form somehow. @Adi | |
| May 5 at 10:30 | comment | added | Adi | @WillieWong if one takes $v =Ax$, then if one collected all the terms that contain derivatives of $A$, a lot of cancellations occur. Is there a way to see this clearly? | |
| Jan 3, 2023 at 14:31 | comment | added | Willie Wong | @Adi: regard $A$ as a matrix-valued function, and $v$ a vector field. Then you can take componentwise the partial derivative of $A$ in the direction of $v$. As long as $A$ is symmetric, this resulting function (which we can call $v(A)$), is a also symmetric. And so we can use it to create a symmetric bilinear form. | |
| Mar 19, 2010 at 9:06 | vote | accept | PDE | ||
| Mar 18, 2010 at 23:44 | vote | accept | PDE | ||
| Mar 19, 2010 at 9:06 | |||||
| Mar 11, 2010 at 16:22 | comment | added | Willie Wong | I guess it is more useful in the integral form then, since you mentioned geometric analysis. I was under the impression that Rellich-types usually don't involve commuting derivatives, while Bochner-types are the ones that give you curvature terms. | |
| Mar 11, 2010 at 14:52 | comment | added | Deane Yang | Something like this is often called a "Rellich-type identity" by geometric analysts. If this is done on a Riemannian manifold using the Laplace-Beltrami operator, then the process of commuting covariant derivatives produces curvature terms and these are often quite important. The calculation can sometimes be quite involved and requires some care, but uses only the standard rules of calculus and, on a Riemannian manifold, covariant differentiation. | |
| Mar 11, 2010 at 14:05 | history | answered | Willie Wong | CC BY-SA 2.5 |