Gevrey semigroup generated by \(- (\Lambda^\alpha + b \cdot \nabla)\) in \(L^p(\mathbb{R}^n)\) (Q2326012)
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| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Gevrey semigroup generated by \(- (\Lambda^\alpha + b \cdot \nabla)\) in \(L^p(\mathbb{R}^n)\) |
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Gevrey semigroup generated by \(- (\Lambda^\alpha + b \cdot \nabla)\) in \(L^p(\mathbb{R}^n)\) (English)
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4 October 2019
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In this paper semigroup generated by the fractional Laplacian \( \Lambda ^\alpha \) perturbed by \(b\nabla\) (with \(n-\)dimensional vector \(b\)) is considered. While \( - (\Lambda ^\alpha + b\nabla) \) generates an analytic semigroup in \( L^p ( \mathbb{R}^d) \), \( 1 \leq p< \infty\) when \( 1 \leq \alpha \leq 2\), the perturbation argument does not work in the case \( 0< \alpha < 1.\) The main result of the paper states that for \( 0< \alpha < 1\) the semigroup \( e^{- t(\Lambda ^\alpha + b\nabla)}\) is of Gevrey class \( \delta \) in \( L^p ( \mathbb{R}^d) \), \( 1 \leq p< \infty\) for any \( \delta > \frac{1}{\alpha},\) \( t > 0.\) Furthermore, the authors prove that the result is sharp in the following sense: If \( 0< \alpha < 1\), \( 1 \leq p< \infty\), \( b = (1,0,\dots,0) \in \mathbb{R}^d,\) then \( e^{- t(\Lambda ^\alpha + b\nabla)}\) is not of Gevrey class \( \delta \) with \( \delta \leq \frac{1}{\alpha},\) \( t > 0.\) See [\textit{J. Hao} et al., J. Differ. Equations 259, No. 9, 4763--4798 (2015; Zbl 1326.35059)] for the definition of the Gevrey class of strongly continuous semigroup on a Banach space. As an independent auxiliary result, a Bernstein type estimate for Fourier multipliers is proved.
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Gevrey semigroup
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fractional Laplacian operator
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resolvent estimates
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Bernstein type estimate
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