# Rellich-Necas identity

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.

-
Any chance you want to tell us what the identity is? Maybe we know it but not by that name. –  Deane Yang Mar 11 '10 at 2:08
Note: MathSciNet does not know it by that name. Searching for Rellich and Nicas in the "anywhere" fields yields 0 results. –  Pete L. Clark Mar 11 '10 at 2:58
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". –  Gjergji Zaimi Mar 11 '10 at 3:13
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.) –  Qiaochu Yuan Mar 11 '10 at 3:51

(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.

-
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. –  Deane Yang Mar 11 '10 at 14:52
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. –  Willie Wong Mar 11 '10 at 16:22