Rellich-Necas identity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:45:50Z http://mathoverflow.net/feeds/question/17794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17794/rellich-necas-identity Rellich-Necas identity PDE 2010-03-11T01:31:18Z 2010-03-19T00:26:46Z <p>I am looking for a book/paper which has the proof of the Rellich-Nicas identity.</p> <p><b>[EDIT by Yemon Choi]</b> It seems that what was meant is "the Rellich-Necas identity", although the original poster hasn't really clarified or expanded on the request.</p> http://mathoverflow.net/questions/17794/rellich-necas-identity/17838#17838 Answer by Ady for Rellich-Necas identity Ady 2010-03-11T08:37:26Z 2010-03-11T08:37:26Z <p>My guessing is that the question is about the so-called <strong>Rellich-Necas</strong> identity, named after the late Czech mathematician Jindrich NECAS. See e.g. [C24], [C14], and [C11] in <a href="http://dml.cz/bitstream/handle/10338.dmlcz/134050/MathBohem_129-2004-4_8.pdf" rel="nofollow">http://dml.cz/bitstream/handle/10338.dmlcz/134050/MathBohem_129-2004-4_8.pdf</a>.</p> http://mathoverflow.net/questions/17794/rellich-necas-identity/17863#17863 Answer by Willie Wong for Rellich-Necas identity Willie Wong 2010-03-11T14:05:23Z 2010-03-11T14:05:23Z <p>(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.)</p> <p>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 R<sup>N</sup>, $v$ a vector field, $u$ a function, and $\delta$ denoting the Euclidean divergence, we have</p> <p>$\delta( A(\nabla u,\nabla u) v) = 2 \delta( v(u) A(\nabla u)) + \delta(v) A(\nabla u,\nabla u)$ $- 2A(\nabla u) \cdot \nabla v \cdot \nabla u - 2 v(u) \delta(A(\nabla u)) + v(A)(\nabla u,\nabla u)$</p> <p>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 $A$. </p> <p>Verifying that the identity is true should just be a basic application of multivariable calculus. </p>