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2024 | OriginalPaper | Buchkapitel

1. The Square Root of the Laplacian

verfasst von : Xavier Fernández-Real, Xavier Ros-Oton

Erschienen in: Integro-Differential Elliptic Equations

Verlag: Springer Nature Switzerland

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Abstract

In this chapter, we study the simplest integro-differential elliptic operator: the square-root of the Laplacian, \(\sqrt {-\varDelta }\). We start by establishing its basic properties, including the harmonic extension representation and the corresponding heat kernel and fundamental solution. We then prove the comparison principle, compute its Poisson kernel in a ball and find the corresponding mean value property, deduce the Harnack inequality, and establish interior regularity estimates. Finally, we construct some explicit solutions and develop the analogous results for the fractional Laplacian \((-\Delta )^s\), with \(s \in (0, 1)\).

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Fußnoten
1
Integrability at infinity is not a problem at this point; we can assume u to be compactly supported.
 
2
We refer to [29] for a direct proof of Proposition 1.6.3, not based on Kelvin transforms nor the Poisson kernel of the half-space.
 
3
Alternatively, a function that satisfies \({\sqrt {-\Delta }\,} u = f\) in \(B_1\), as seen in the extension, corresponds to a harmonic function \(\tilde u\) in \(\mathbb {R}^{n+1}_+\) such that \(\partial _{x_{n+1}} \tilde u = f\) on \(\{x_{n+1} = 0\}\). Hence, we expect \(\tilde u\) (and, therefore, u) to have one more “derivative” of regularity than f.
 
4
This is the unique harmonic extension given by the Poisson kernel, because it is sublinear at infinity. A priori, if we just required harmonicity of the extension, we could have subtracted \(\lambda y\) for any \(\lambda \in \mathbb {R}\), but only \(\lambda = 1\) gives the sublinear growth at infinity.
 
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Metadaten
Titel
The Square Root of the Laplacian
verfasst von
Xavier Fernández-Real
Xavier Ros-Oton
Copyright-Jahr
2024
DOI
https://doi.org/10.1007/978-3-031-54242-8_1

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