A question on variational derivatives? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:39:10Z http://mathoverflow.net/feeds/question/104802 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104802/a-question-on-variational-derivatives A question on variational derivatives? utpp 2012-08-16T00:04:56Z 2012-12-14T06:14:12Z <p>If $m=u-u_{xx}$, I would like to ask if the following formula is true?</p> <p>$\frac{\delta}{\delta u}=(1-\partial_{xx})\frac{\delta}{\delta m}$, where $\frac{\delta}{\delta u}$ and $\frac{\delta}{\delta m}$ are the variational derivatives;</p> <p>If not, is there some other relation between $\frac{\delta}{\delta u}$ and $\frac{\delta}{\delta m}$?</p> <p>Thanks.</p> http://mathoverflow.net/questions/104802/a-question-on-variational-derivatives/105574#105574 Answer by Carlo Beenakker for A question on variational derivatives? Carlo Beenakker 2012-08-26T20:27:06Z 2012-08-26T20:27:06Z <p>I don't think this is correct; on the one hand,</p> <p>$\frac{\delta}{\delta m(y)}F[m(x)]=F'[m(x)]\delta(x-y)$</p> <p>on the other hand,</p> <p>$\frac{\delta}{\delta u(y)}F[m(x)]=F'[m(x)] (1-\partial_{xx})\delta(x-y)$</p> <p>which is not equal to $(1-\partial_{xx})F'[m(x)]\delta(x-y)$</p> http://mathoverflow.net/questions/104802/a-question-on-variational-derivatives/116350#116350 Answer by Peter Michor for A question on variational derivatives? Peter Michor 2012-12-14T06:14:12Z 2012-12-14T06:14:12Z <p>This is the trap of notation. A clean way to treat this is to consider the mapping $F(u) = u-\partial_{xx}u = (1-\partial_{xx})u$ which is linear and smooth if applied to the right space of $u$'s. Thus its derivative is again $F$: $dF(u)v=F(v)$.</p> <p>See: Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997<a href="http://www.mat.univie.ac.at/~michor/apbookh-ams.pdf" rel="nofollow">(pdf)</a></p>