3
$\begingroup$

Let $\Omega$ be a bounded smooth domain of $\mathbb{R}^n$, $0<s<1$ and $(-\Delta)^s$ denotes the restricted fractional Laplacian. Let consider the following fractional Heat equation:

$‎‎$‎ ‎\begin{cases}‎ ‎u_t = (-\Delta)^s u + ‎f(x,t) & \quad \mathrm{in} ‎\Omega \times (0,T),\\ u(x,0)=u_0 & \quad \mathrm{in} ‎\mathbb{R}^N‎, ‎\\ u(x,t)=0 & \quad \mathrm{in} ‎(\mathbb{R}^N \setminus \Omega )\times (0,T)‎. ‎\end{cases}‎ ‎$‎‎$‎ I want to know that, does $(-\Delta)^s$ generate a semigroup and is it analytic? Is the regularity results for this problem, depending on the regularities of $f(x,t)$ and $u_0$, well-known?

Can someone give a survey of reference for these questions?

$\endgroup$
  • $\begingroup$ I am not sure what is the meaning of "restricted fractional Laplacian". Does this simply refers to the restriction of the fractional Laplacian to the class of functions which vanish in the complement of the domain? Or rather to the operator defined as the usual fractional Laplacian, but with the integration region restricted to the domain? $\endgroup$ – Mateusz Kwaśnicki Sep 8 '19 at 19:52
  • $\begingroup$ May I ask on which space you would be interested in a semigroup? $L^2(\Omega)$? $\endgroup$ – Jochen Glueck Sep 8 '19 at 20:40
1
$\begingroup$

It depends on $s$, if $1/2\leq s \leq 1$, then you get analyticity. If $0<s<1/2$, then you get Gevrey class only. See Section 8.3, arXiv:1606.00873

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The semigroup being analytic (as asked by the OP) has nothing to do with it mapping the initial condition into some space of analytic functions, which seems to be what you're mentioning... $\endgroup$ – Martin Hairer Sep 28 at 20:41
1
$\begingroup$

Pablo Raúl Stinga's User’s guide to the fractional Laplacian and the method of semigroups (2018) may provide a helpful entry point to the literature. The semigroup connection is expressed by:

The fractional Laplacian $L^s=(-\Delta)^s$, $0<s<1$ can be expressed in terms of the heat diffusion semigroup $v=e^{-tL}u$ generated by $L$ acting on $u$ through the integral formula $$L^s > u=\frac{1}{\Gamma(-s)}\int_0^\infty\left(e^{-tL}u-u\right)\frac{dt}{t^{1+s}}.$$ The solution to $L^s u=f$ can then be written as $$u=\frac{1}{\Gamma(s)}\int_0^\infty e^{-tL}f\frac{dt}{t^{1-s}}.$$

This connection forms the starting point of the regularity study reviewed by Stinga, see in particular theorems 13-15 (Schauder–Hölder–Zygmund estimates).

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Regarding analyticity, there is the following nice result by Gomilko and Tomilov (On Subordination of Holomorphic Semigroups, Theorem 1.1):

Gomilko, Alexander; Tomilov, Yuri, On subordination of holomorphic semigroups, Adv. Math. 283, 155-194 (2015). ZBL1319.47034.

A smooth function $\psi\colon (0,\infty)\to (0,\infty)$ is called Bernstein function if $(-1)^n f^{(n+1)}\geq 0$ for all $n\in\mathbb{N}$. If $\psi$ is a Bernstein function and $A$ generates a bounded holomorphic semigroup of angle $\theta$ on the Banach space $X$, then $\psi(A)$ also generates a bounded holomorphic semigroup of angle $\theta$.

Clearly, $\lambda\mapsto \lambda^s$ is a Bernstein function for $s\in (0,1)$. Thus the fractional Laplacian generates a bounded holomorphic semigroup whenever the Laplacian does. In particular, this is the case on $L^p(\Omega)$ for $p\in[1,\infty)$. As a consequence you get $W^{\alpha,p}$ regularity of $u(t,\cdot)$ in terms of $u_0$ and $f$ by the usual semigroup methods. Of course, if you are only interested in the case $p=2$, all of this can also be established via the spectral theorem.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.