Does anyone know a formula of chain rule for fractional laplacian?
say we take the fractional laplacian of order a on function $g(U(x))$ $x\in \mathbb{R}^2$, $U \in \mathbb{R}$, $g \colon \mathbb{R} \to \mathbb{R}$ functional.
Thanks
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$\begingroup$ Is fractional laplacian a fractional derivative or Reisz transorm? $\endgroup$– UdayCommented Apr 13, 2012 at 11:27
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$\begingroup$ Reisz transform( I thought this definition will be consistent with analytical operator theory) $\endgroup$– GrantCommented Apr 13, 2012 at 17:49
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$\begingroup$ I would begin by looking for analogues of the product rule. Once you have that, then the chain rule for the case where $g$ is a polynomial will follow, giving some insight into the general situation. $\endgroup$– Hans EnglerCommented Apr 14, 2012 at 14:56
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2 Answers
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In the fractional case, it turns out that approximate chain rules are more useful than exact formulae (at least for applications to the analysis of PDE). See
http://wiki.math.toronto.edu/TorontoMathWiki/index.php/Fractional_Derivative
In the case $0 \leq a \leq 1/2$, the rule roughly takes the form
$$ (-\Delta)^a g(U) \approx ((-\Delta^a) U) \cdot \nabla g(U) + \ldots$$
where the $\ldots$ error is a paraproduct which is "lower order" than the main term in some sense. One popular way to make this formula precise is the Bony linearisation formula, originally developed in http://www.ams.org/mathscinet-getitem?mr=631751 . This is part of a more general theory known as paradifferential calculus, discussed for instance in Taylor's book http://www.ams.org/mathscinet-getitem?mr=1766415
answered Apr 14, 2012 at 15:56
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$\begingroup$ Is there a good bound for the case $1/2 < a < 1$? Supposing that $g$ is a smooth function. $\endgroup$– KernelCommented Apr 3, 2020 at 13:35
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I showed that the Riesz potential is an integral formulation of general fractional two-side derivative, that is much more general in the sense that is valid for any order greater than -1 for a broad class of funtions. I can give copies by sending a mail to [email protected]
