I am looking for a book/paper which has the proof of the Rellich-Nicas identity.
[EDIT by Yemon Choi] It seems that what was meant is "the Rellich-Necas identity", although the original poster hasn't really clarified or expanded on the request.
$\begingroup$
$\endgroup$
4
-
6$\begingroup$ Any chance you want to tell us what the identity is? Maybe we know it but not by that name. $\endgroup$– Deane YangCommented Mar 11, 2010 at 2:08
-
3$\begingroup$ Note: MathSciNet does not know it by that name. Searching for Rellich and Nicas in the "anywhere" fields yields 0 results. $\endgroup$– Pete L. ClarkCommented Mar 11, 2010 at 2:58
-
$\begingroup$ This identity doesn't seem to be mentioned anywhere. Unless the question is edited and you point to more information, the thread might get closed as "not a real question". $\endgroup$– Gjergji ZaimiCommented Mar 11, 2010 at 3:13
-
1$\begingroup$ This question is literally the only Google result for "Rellich-Nicas." (On the other hand, there appears to be a notion of "Rellich-type identity" but it's not clear which one your name refers to.) $\endgroup$– Qiaochu YuanCommented Mar 11, 2010 at 3:51
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
10
(Boo! I tried to post this in a comment to Ady, but the HTML Math won't parse right. So here goes. Sorry about the really long equation being broken up not very neatly.)
Googling Rellich-Necas turns up a bunch of recent papers by LUIS ESCAURIAZA in which the identities are used. But as far as I can tell the identity is just a simple differential equality obtained from symbolic manipulation of terms. The following seems to be a straight-forward version of the identity: let $A = (A_{ij})$ be a symmetric bilinear form (with variable coefficients) on RN, $v$ a vector field, $u$ a function, and $\delta$ denoting the Euclidean divergence, we have
$ \delta( A(\nabla u,\nabla u) v) = 2 \delta( v(u) A(\nabla u)) + \delta(v) A(\nabla u,\nabla u)$ $- 2A(\nabla u) \cdot \nabla v \cdot \nabla u - 2 v(u) \delta(A(\nabla u)) + v(A)(\nabla u,\nabla u)$
Where $v(u)$ is the partial derivative of $u$ in the direction of $v$, and $A(\nabla u)\cdot\nabla v \cdot \nabla u$ is, in coordinates, $\partial_i u A_{ij} \partial_j v_k \partial_k u$ with implied summation, and $v(A)$ is the symmetric bilinear form obtained by taking the $v$ partial derivative of the coefficients of $A$.
Verifying that the identity is true should just be a basic application of multivariable calculus.
answered Mar 11, 2010 at 14:05
-
2$\begingroup$ 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. $\endgroup$ Commented Mar 11, 2010 at 14:52
-
$\begingroup$ 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. $\endgroup$ Commented Mar 11, 2010 at 16:22
-
1$\begingroup$ @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. $\endgroup$ Commented Jan 3, 2023 at 14:31
-
1$\begingroup$ 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 $\endgroup$ Commented May 6 at 1:41
-
1$\begingroup$ @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) $\endgroup$ Commented May 6 at 12:42
$\begingroup$
$\endgroup$
0
My guessing is that the question is about the so-called Rellich-Necas identity, named after the late Czech mathematician Jindrich NECAS. See e.g. [C24], [C14], and [C11] in http://dml.cz/bitstream/handle/10338.dmlcz/134050/MathBohem_129-2004-4_8.pdf.
