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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 $\frac{\delta}{\delta m}$? Thanks. |
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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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