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If $m=u-u_{xx}$, I would like to ask if the following formula is true?

$\frac{\delta}{\delta u}=(1-\partial_{xx})\frac{\delta}{\delta m}$, where $\frac{\delta}{\delta u}$ and $\frac{\delta}{\delta m}$ are the variational derivatives;

If not, is there some other relation between $\frac{\delta}{\delta u}$ and $\frac{\delta}{\delta m}$?


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Can you give some motivations? – Anand Aug 17 '12 at 10:05

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 $F$: $dF(u)v=F(v)$.

See: Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997(pdf)

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I don't think this is correct; on the one hand,

$\frac{\delta}{\delta m(y)}F[m(x)]=F'[m(x)]\delta(x-y)$

on the other hand,

$\frac{\delta}{\delta u(y)}F[m(x)]=F'[m(x)] (1-\partial_{xx})\delta(x-y)$

which is not equal to $(1-\partial_{xx})F'[m(x)]\delta(x-y)$

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