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The Cauchy problem for heat equation with fractional Laplacian and exponential nonlinearity
A Fino, M Kirane
To cite this version:
A Fino, M Kirane. The Cauchy problem for heat equation with fractional Laplacian and exponential
nonlinearity. 2019. �hal-01889726v2�
The Cauchy problem for heat equation with fractional Laplacian and exponential nonlinearity
A. Z. FINOa, M. KIRANEb,c,d
aLaMA-Liban, Lebanese University, Faculty of Sciences, Department of Mathematics, P.O. Box 826 Tripoli, Lebanon
bLaSIE, Pôle Sciences et Technologies, Université de La Rochelle, Avenue Michel Crépeau, 17031 La Rochelle, France
cNAAM Research Group, Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
dRUDN University, 6 Miklukho-Maklay St, Moscow 117198, Russia
Abstract
We consider the Cauchy problem for heat equation with fractional Laplacian and exponential nonlinearity. We estab- lish local well-posedness result in Orlicz spaces. We derive the existence of global solutions for small initial data. We obtain decay estimates for large time in Lebesgue spaces.
Keywords: Orlicz spaces, fractional Laplacian, Well-posedness, Global existence, Decay estimates 2010 MSC: 35K05, 46E30, 35A01, 35B40, 26A33, 35K55
1. Introduction
This paper concerns the Cauchy problem for the following heat equation
ut+(−∆)β/2u= f(u), t>0,x∈Rn,
u(0,x)=u0(x), x∈Rn, (1.1)
whereuis a real-valued unknown function, 0< β≤2,n≥1, and f :R→Rhaving an exponential growth at infinity (f(u)∼e|u|p,p>1, for largeu) with f(0)=0. Hereafter,k· kq(1≤q≤ ∞) stands for the usualLq(Rn)-norm.
When f(u)=|u|p−1u, the Lebesgue spaces are adapted to study our problem (cf. [3, 14, 15, 16]). By analogy, we consider the Orlicz spaces [5] in order to study heat equations with exponential nonlinearities. The Orlicz space
expLp(Rn)= (
u∈L1loc(Rn);
Z
Rn
exp |u(x)|p λp
!
−1
!
dx<∞,for someλ >0 )
,
endowed with the Luxemburg norm
kukexpLp(Rn) :=inf (
λ >0;
Z
Rn
exp |u(x)|p λp
!
−1
! dx≤1
)
is a Banach space. For the local well-posedness we use the space expL0p(Rn)=
u∈expLp(Rn); there exists{un}∞n=1⊂C∞0(Rn) such that limn→∞kun−ukexpLp(Rn)=0 .
It is also know (see Ioku, Ruf, and Terraneo [9], Majdoub et al. [10, 11]) that expL0p(Rn)=
(
u∈L1loc(Rn);
Z
Rn
exp (α|u(x)|p)−1dx<∞,for everyα >0 )
.
Email addresses:ahmad.fino01@gmail.com; afino@ul.edu.lb(A. Z. FINO),mokhtar.kirane@univ-lr.fr(M. KIRANE)
Whenβ = 2 (i.e. the standard heat equation) and p = 2, Ioku [8] proved the existence of global solutions in expL2(Rn) of (1.1) under the condition (1.4) below withm=1+4n. Later, Ioku et al. [9] studied the local nonexis- tence of solutions of (1.1) for certain data in expL2(R2), and the well-posedness of (1.1) in the subspace expL20(R2) under the condition (1.3) below. In [6], Furioli et al. considered the asymptotic behavior and decay estimates of the global solutions of (1.1) in expL2(Rn) when f(u)=|u|4/nueu2. Next, Majdoub et al. [10] proved the local well- posedness in expL20(Rn) (if f satisfies (1.3) below withm≥1+8n) and the global existence under small initial data in expL2(Rn) (if f satisfies (1.4) below) for the biharmonic heat equation (i.e.ut+ ∆2u= f(u)). Finally, whenβ=2, p>1 andm≥1+2pn, Majdoub and Tayachi [11] proved not only the local well-posedness in expLp0(Rn) but also the global existence of solutions under small initial data in expLp(Rn) of (1.1) and analyzed their decay estimates. We notice that Majdoub and Tayachi [11] considered just the case of when n(p−1)2 > p. In this paper, we generalize the paper of [11] for the fractional laplacian case including the case whenn(p−1)2 ≤pfor the global existence.
In order to state our main results, we note that the linear semigroupe−t(−∆)β/2is continuous att=0 in expLp0(Rn) (see Proposition 2) which is not the case in expLp(Rn) (cf. [9] in the case ofβ=2), therefore, we have to define two kinds of mild solutions, the standard one where the space expL0p(Rn) is used, and the weak-mild solution where we use the space expLp(Rn).
Definition 1. (Mild solution)
Given u0∈expLp0(Rn)and T>0. We say that u is a mild solution for the Cauchy problem (1.1) if u∈C([0,T]; expL0p(Rn)) satisfying
u(t)=e−t(−∆)β/2u0+Z t 0
e−(t−s)(−∆)β/2f(u(s))ds, (1.2) where e−t(−∆)β/2is defined in (2.7) below.
Definition 2. (Weak-mild solution)
Given u0 ∈ expLp(Rn)and T > 0. We say that u is a weak-mild solution for the Cauchy problem (1.1) if u ∈ L∞((0,T); expLp(Rn))satisfying the associated integral equation (1.2) inexpLp(Rn)for almost all t ∈ (0,T)and u(t)→u0in the weak∗topology as t→0.
We recall thatu(t)→u0in weak∗sense if and only if lim
t→0
Z
Rn
[u(t,x)ϕ(x)−u0(x)ϕ(x)]dx=0, for everyϕ∈L1(lnL)1/p(Rn), where
L1(lnL)1/p(Rn) := (
f ∈L1loc(Rn);
Z
Rn
|f(x)|ln1/p(2+|f(x)|)dx<∞ )
is a predual space of expLp(Rn) (see [2, 12]).
First, we interest in the local well-posedness. We assume that f satisfies
f(0)=0, |f(u)−f(3)| ≤C|u−3|(eλ|u|p+eλ|3|p), ∀u,3∈R, (1.3) for some constantsC>0,p>1, andλ >0. Typical example satisfying (1.3) is: f(u)=±ue|u|p.
Theorem 1. (Local well-posedness)
Suppose that f satisfies (1.3). Given u0 ∈ expL0p(Rn), there exist a time T =T(u0)>0and a unique mild solution u∈C([0,T]; expL0p(Rn))to (1.1).
Next, our second interest is the global existence and the decay estimate. In this case, the behaviour of f(u) near u=0 plays a crucial role, therefore the following behaviour near zero will be allowed
|f(u)| ∼ |u|m,
wheren(m−1)β ≥p. More precisely, we suppose that
f(0)=0, |f(u)−f(3)| ≤C|u−3|(|u|m−1eλ|u|p+|3|m−1eλ|3|p), ∀u,3∈R, (1.4) wheren(m−1)β ≥p>1,C>0, andλ >0 are constants. Typical example satisfying (1.4) is: f(u)=±|u|m−1ue|u|pwhere m≥1+βpn.
Theorem 2. (Global existence)
Let n ≥ 1, p > 1, and0 < β ≤ 2. Suppose that f satisfies (1.4) for m ≥ p. Then there exists a positive constant ε >0such that every initial data u0∈expLp(Rn)withku0kexpLp(Rn) ≤ε, there exists a global weak-mild solution u∈L∞((0,∞); expLp(Rn))to (1.1) satisfying
limt→0
u(t)−e−t(−∆)β/2u0
expLp(Rn)=0. (1.5)
Moreover, there exists a constant C>0such that
ku(t)kLq(Rn)≤Ct−σ, for all t>0, (1.6) where
σ= 1
m−1 − n βq >0, and n(m−1)
β <q<∞ if β= n(p−1)
p , and n(m−1)
β <q<n(m−1) β
1
(2−m)+ if β, n(p−1)
p ,
with(·)+stands for the positive part.
Remark 1. In Theorem 2, we have to distinguish 3 cases:β < n(p−1)p ,β >n(p−1)p , andβ=n(p−1)p . We note that in the case ofβ > n(p−1)p we have to takem> p. Indeed, ifm= p, it follows thatβ > n(m−1)m , butn(m−1)/β ≥ p, which implies thatβ≤ n(m−1)m , therefore n(p−1)p < n(p−1)p ; contradiction.
This paper is organized as follows: in Section 2, we present several preliminaries. Section 3 contains the proof of the local well-posedness theorem (Theorem 1). Finally, we prove the global existence theorem (Theorem 2) in Section 4.
2. Preliminaries
2.1. Orlicz spaces: basic properties
In this section we present the definition of the so-called Orlicz spaces onRnand some related properties. More details and complete presentations can be found in [1, 12, 13].
Definition 3. (Orlicz space)
Letφ:R+→R+be a convex increasing function such that φ(0)=0= lim
s→0+φ(s), lim
s→∞φ(s)=∞.
The Orlicz space Lφ(Rn)is defined by Lφ(Rn)=
(
u∈L1loc(Rn);
Z
Rn
φ |u(x)| λ
!
dx<∞, for someλ >0 )
,
endowed with the Luxemburg norm
kukLφ :=inf (
λ >0;
Z
Rn
φ |u(x)|
λ
! dx≤1
) .
On the other hand, we denote by Lφ0(Rn)=
(
u∈L1loc(Rn);
Z
Rn
φ |u(x)|
λ
!
dx<∞, for everyλ >0 )
.
It can be shown (as in Ioku et al. [9]) that
Lφ0(Rn)=C∞0(Rn)k·kLφ = the closure ofC0∞(Rn) inLφ(Rn).
It is known that (Lφ(Rn),k· kLφ) and (Lφ0(Rn),k· kLφ) are Banach spaces. Note that, ifφ(s) = sp, 1 ≤ p < ∞, then Lφ(Rn)=Lφ0(Rn)=Lp(Rn), and ifφ(s)=esp−1, 1≤p<∞, thenLφ(Rn) is the space expLp(Rn), whileLφ0(Rn) is the space expL0p(Rn). Moreover, foru∈LφandK:=kukLφ >0, we can easy check by the definition of the infimum that
(
λ >0, Z
Rn
φ |u(x)| λ
! dx≤1
)
=[K;∞[, in particular
Z
Rn
φ |u(x)|
kukLφ
!
dx≤1. (2.1)
The following Lemmas summarize the embedding between Orlicz and Lebesgue spaces.
Lemma 1. [11, Lemma 2.3]
For every1≤q≤p, we have Lq(Rn)∩L∞(Rn),→expL0p(Rn),→expLp(Rn), more precisely kukexpLp(Rn)≤ 1
(ln 2)1/p(kukq+kuk∞). (2.2)
Similarly, we have Lemma 2.
Letφ(s)=esp−1−sp, p>1. For every q≤2p, we have Lq(Rn)∩L∞(Rn),→Lφ0(Rn),→Lφ(Rn), more precisely kukLφ(Rn)≤C(p)(kukq+kuk∞). (2.3) Proof. Letg(s)=esp−sp;gis a strictly increasing. Letα≥C(p)(kukq+kuk∞) whereC(p) :=1/g−1(2), then
Z
Rn
exp |u(x)|p αp
!
−1− |u(x)|p αp
!!
dx =
∞
X
k=2
1 k!αpkkukpk
pk
≤
∞
X
k=2
1
k!αpk(kukq+kuk∞)pk
= exp kukq+kuk∞ α
!p
−1− kukq+kuk∞ α
!p
= g kukq+kuk∞
α
!
−1
≤ 1,
where we have used the interpolation inequalitykukr ≤ kukq/rq kuk1−q/r∞ ≤ kukq +kuk∞ for allq ≤ r ≤ ∞and all u∈Lq∩L∞. Therefore
[C(p)(kukq+kuk∞);∞[⊆
( α >0;
Z
Rn
φ |u(x)|
α
! dx≤1
) ,
which implies that kukLφ(Rn)=inf
( α >0;
Z
Rn
φ |u(x)| α
! dx≤1
)
≤infn
α >0;α∈[C(p)(kukq+kuk∞);∞[o
=C(p)(kukq+kuk∞).
Lemma 3. [11, Lemma 2.4]
For every1≤p≤q<∞, we haveexpLp(Rn),→Lq(Rn), more precisely kukq≤ Γ q
p+1
!!1/q
kukexpLp(Rn), (2.4)
whereΓis the gamma function.
Next, we present some definitions and results concerning the fractional Laplacian that will be used hereafter. The fundamental solutionSβof the usual linear fractional diffusion equation
ut+(−∆)β/2u=0, β∈(0,2], x∈Rn, t>0, (2.5) can be represented via the Fourier transform by
Sβ(t)(x) :=Sβ(x,t)= 1 (2π)n/2
Z
Rn
eix.ξ−t|ξ|βdξ. (2.6)
This mean that the solution of (2.5) with any initial datau(0)=u0can be written as
u(x,t)=Sβ(x,t)∗u0(x)=:e−t(−∆)β/2u0, (2.7) wheree−t(−∆)β/2 is a strongly continuous semigroup on Lp(Rn),p > 1, generated by the fractional power−(−∆)β/2. Moreover,Sβsatisfies
Sβ(1)∈L∞(Rn)∩L1(Rn), Sβ(x,t)≥0, Z
Rn
Sβ(x,t)dx=1, (2.8)
for all x ∈ Rn andt > 0.Hence, using Young’s inequality for the convolution and the following self-similar form Sβ(x,t)=t−n/βSβ(xt−1/β,1),we get theLr−Lqestimate
e−t(−∆)β/23
q ≤ Ct−nβ(1r−1q)k3kr, (2.9) for all3∈Lr(Rn) and all 1≤r≤q≤ ∞,t>0.In particular, using Young’s inequality for the convolution and (2.8), we have
e−t(−∆)β/23 q=
Sβ(x,t)∗3 q ≤
Sβ(t)
1k3kq=k3kq, (2.10)
for all3∈Lq(Rn) and all 1≤q≤ ∞,t>0.
The following proposition is a generalization of Proposition 3.2 in [11] and it is presented (without proof) by Furioli et al. [6, Lemma 3.1].
Proposition 1.
Let1≤q≤p,1≤r≤ ∞, and0< β≤2. Then the following estimates hold.
(i)
e−t(−∆)β/2ϕ
expLp(Rn)≤ kϕkexpLp(Rn), for all t>0, ϕ∈expLp(Rn).
(ii)
e−t(−∆)β/2ϕ
expLp(Rn)≤C t−βqn
ln(t−nβ+1)−1/p
kϕkq, for all t>0, ϕ∈Lq(Rn).
(iii)
e−t(−∆)β/2ϕ
expLp(Rn)≤ 1
(ln 2)1/p
hC t−βrn kϕkr+kϕkqi
, for all t>0, ϕ∈Lr(Rn)∩Lq(Rn).
Proof. We start by proving (i). For anyλ >0, using (2.10) and Taylor expansion, we have Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx=
∞
X
k=1
e−t(−∆)β/2ϕ
pk pk
k!λpk ≤
∞
X
k=1
kϕkpkpk k!λpk =Z
Rn
exp |ϕ|p λp
!
−1
! dx.
Then
( λ >0;
Z
Rn
exp |ϕ|p λp
!
−1
! dx≤1
)
⊆
λ >0;
Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx≤1
,
and therefore
e−t(−∆)β/2ϕ
expLp(Rn) = inf
λ >0;
Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx≤1
≤ inf (
λ >0;
Z
Rn
exp |ϕ|p λp
!
−1
! dx≤1
)
= kϕkexpLp(Rn).
This proves (i). Similarly, to prove (ii), we use again (2.9) and Taylor expansion. For anyλ >0, we have Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx=
∞
X
k=1
e−t(−∆)β/2ϕ
pk pk
k!λpk ≤
∞
X
k=1
Cpkt−nβ(1q−pk1)pkkϕkqpk k!λpk =tnβ
exp
Ct−βqnkϕkq
λ
p
−1
. As
tnβ
exp
Ct−βqnkϕkq λ
p
−1
≤1⇐⇒λ≥C t−βqn
ln(t−nβ +1)−1/p
kϕkq,
we conclude that
λ >0, λ∈[C t−βqn
ln(t−nβ+1)−1/p
kϕkq;∞[
⊆
λ >0, Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx≤1
; whereupon
e−t(−∆)β/2ϕ
expLp(Rn) = inf
λ >0, Z
Rn
exp
|e−t(−∆)β/2ϕ|p λp
−1
dx≤1
≤ inf
λ >0, λ∈[C t−βqn
ln(t−nβ+1)−1/p
kϕkq;∞[
= C t−βqn
ln(t−nβ+1)−1/p
kϕkq.
This proves (ii). Finally, to prove (iii), we use the embeddingLq(Rn)∩L∞(Rn),→expL0p(Rn) (2.2); we get
e−t(−∆)β/2ϕ
expLp(Rn)≤ 1 (ln 2)1/p
e−t(−∆)β/2ϕ q+
e−t(−∆)β/2ϕ ∞
.
Using theLr−L∞andLq−Lqestimates (2.9), we conclude that
e−t(−∆)β/2ϕ
expLp(Rn)≤ 1 (ln 2)1/p
kϕkq+C t−βrn kϕkr .
We will also need the following smoothing results.
Proposition 2.
Ifϕ∈expL0p(Rn), then e−t(−∆)β/2ϕ∈C([0,∞); expLp0(Rn)).
Proof. Letϕ∈expL0p(Rn). By (i) of Proposition 1 and the definition of expL0p(Rn), we havee−t(−∆)β/2ϕ∈expLp0(Rn) for everyt>0. Thus, by the linearity of the semigroupe−t(−∆)β/2, it remains to prove the continuity att=0,
lim
t→0
e−t(−∆)β/2ϕ−ϕ
expLp(Rn)=0.
Sinceϕ∈expL0p(Rn), there exists a sequence (ϕn)n ⊆C0∞(Rn) such that limn→∞kϕn−ϕkexpLp =0. By (2.2), and estimation (i) of Proposition 1, we obtain
e−t(−∆)β/2ϕ−ϕ
expLp(Rn) ≤
e−t(−∆)β/2(ϕ−ϕn)
expLp+
e−t(−∆)β/2ϕn−ϕn
expLp+kϕn−ϕkexpLp
≤ 1
(ln 2)1/p
ke−t(−∆)β/2ϕn−ϕnkp+ke−t(−∆)β/2ϕn−ϕnk∞
+2kϕn−ϕkexpLp. Sinceϕn∈C0∞(Rn), using the fact thate−t(−∆)β/2 is a strongly continuous semigroup onLr(Rn) (1<r≤ ∞), we have limt→0
ke−t(−∆)β/2ϕn−ϕnkp+ke−t(−∆)β/2ϕn−ϕnk∞
=0. Hence lim sup
t→0
e−t(−∆)β/2ϕ−ϕ
expLp(Rn)≤2kϕn−ϕkexpLp,
for everyn∈N. This finishes the proof of the proposition.
It is known thate−t(−∆)β/2is aC0-semigroup onLp(Rn). By Proposition 2, it is aC0-semigroup on expLp0(Rn).
Lemma 4. [4, Lemma 4.1.5]
Let X be a Banach space and g∈L1(0,T;X), thenRt
0 e−(t−s)(−∆)β/2g(s)ds∈C([0,T];X). Moreover
Z t 0
e−(t−s)(−∆)β/2g(s)ds L∞(0,T;X)
≤ kgkL1(0,T;X). The following lemmas are essential for the proof of the global existence (Theorem 2).
Lemma 5.
Letλ >0,1≤q<∞and K >0be such thatλqKp≤1. Assume that u∈expLp(Rn)satisfies kukexpLp(Rn)≤K,
thenexp|u|p
λp
−1∈Lq(Rn)and
eλ|u|p−1
Lq(Rn)≤(λqKp)1/q. Proof. Using the elementary inequality (z−1)q≤zq−1,z≥1, we have
Z
Rn
eλ|u|p−1q
dx≤ Z
Rn
eλq|u|p−1 dx≤
Z
Rn
e
λqKp |u|p
kukp
expLp(Rn) −1
dx≤λqKpZ
Rn
e
|u|p kukp
expLp(Rn) −1
dx≤λqKp, where we have used (2.1) and the fact thateθs−1≤θ(es−1), 0≤θ≤1,s≥0. This completes the proof.
Lemma 6.
Let p>1,0< β≤2be such thatβ < n(p−1)p . Then, for every r>nβ, there exists C=C(n,p, β,r)such that
Z t 0
e−(t−s)(−∆)β/2g(s)ds
L∞(0,∞;expLp(Rn))
≤CkgkL∞(0,∞;L1(Rn)∩Lr(Rn)), for every g∈L∞(0,∞;L1(Rn)∩Lr(Rn)).
Proof. By Proposition 1 (ii) withq=1, we have
e−t(−∆)β/2ϕ
expLp(Rn)≤C t−nβ
ln(t−nβ +1)−1/p
kϕk1, (2.11)
for allt>0,ϕ∈L1(Rn)∩Lr(Rn) (kϕkL1∩Lr=kϕkL1+kϕkLr), while by Proposition 1 (iii) withq=1, we obtain
e−t(−∆)β/2ϕ
expLp(Rn)≤C(t−βrn +1)kϕkr+kϕk1. (2.12) Combining (2.11) and (2.12), we get
e−t(−∆)β/2ϕ
expLp(Rn)≤κ(t)kϕkr+kgk1, where
κ(t)=min
C(t−βrn +1),C t−nβ
ln(t−nβ +1)−1/p .
Due to the assumptionsβ < n(p−1)p andr>nβ, we see thatκ∈L1(0,∞). Thus, forg∈L∞(0,∞;L1(Rn)∩Lr(Rn)), we have
Z t 0
e−(t−s)(−∆)β/2g(s)ds
expLp(Rn)
≤ Z t
0
e−(t−s)(−∆)β/2g(s)
expLp(Rn)ds
≤ Z t
0
κ(t−s)
kg(s)kL1(Rn)+kg(s)kLr(Rn)
ds
≤ kgkL∞(0,∞;L1(Rn)∩Lr(Rn))
Z ∞
0
κ(s)ds,
for everyt>0. This proves Lemma 6.
We remark that n(p−1)p may not included in (0,2]. So if n(p−1)p >2, we have n(p−1)p > β, and this case is recovered by Lemma 6. If n(p−1)p ≤ 2, we have three case to study: the case ofβ < n(p−1)p is done by Lemma 6, and the case β > n(p−1)p can be done separately without using any kind of an a priori estimate, so it remains to study the case of β= n(p−1)p where we have a similar result as in Lemma 6. For this, we need to introduce an appropriate Orlicz space.
LetLφ(Rn) this space, withφ(u)=e|u|p−1− |u|p, associated with its Luxemburg norm. From the definition ofk· kLφ, (2.4), and the standard inequalityeθs−1≤θ(es−1), 0≤θ≤1,s≥0, we can easily get
C1kukexpLp(Rn)≤ kukLp(Rn)+kukLφ(Rn)≤C2kukexpLp(Rn), (2.13) for someC1,C2>0.
Lemma 7.
Let p>1,0< β≤2be such thatβ= n(p−1)p . Then, there exists C=C(n,p)such that
Z t 0
e−(t−s)(−∆)β/2g(s)ds
L∞(0,∞;Lφ(Rn))
≤Ckgk
L∞(0,∞;L1(Rn)∩L2p(Rn)∩L
2p p−1(Rn)),
for every g∈L∞(0,∞;L1(Rn)∩L2p(Rn)∩Lp−12p(Rn)).
Proof. On the one hand, by (2.9), we have Z
Rn
φ
|e−t(−∆)β/2ϕ|
λ
dx=
∞
X
k=2
e−t(−∆)β/2ϕ
pk pk
k!λpk ≤
∞
X
k=2
Cpkt−nβ(1−pk1)pkkϕk1pk k!λpk =tnβφ
Ct−nβkϕk1
λ
≤tnβ
exp
Ct−nβkϕk1
λ
2p
−1
,
for allt>0,ϕ∈L1(Rn), where we have used the fact thate|x|p−1− |x|p≤e|x|2p−1, for allx∈R. As tnβ
exp
Ct−βnkϕk1
λ
2p
−1
≤1⇐⇒λ≥C t−nβ
ln(t−nβ +1)−1/2p
kϕk1;
hence,
λ >0, λ∈[C t−nβ
ln(t−nβ +1)−1/2p
kϕk1;∞[
⊆
λ >0, Z
Rn
φ
|e−t(−∆)β/2ϕ| λ
dx≤1
; whereupon
e−t(−∆)β/2ϕ
Lφ(Rn) ≤C t−nβ
ln(t−nβ+1)−1/2p
kϕk1, (2.14)
for allt>0, ϕ∈L1(Rn). On the other hand, from (2.9) and the embeddingL2p(Rn)∩L∞(Rn),→Lφ0(Rn) (see Lemma 2), we have
e−t(−∆)β/2ϕ Lφ(Rn)
≤
e−t(−∆)β/2ϕ
L∞(Rn)+
e−t(−∆)β/2ϕ L2p(Rn)
≤ Ct−nβ(p−12p−0)kϕk
L
2p
p−1(Rn)+kϕkL2p(Rn)
= C t−12kϕk
L
2p
p−1(Rn)+kϕkL2p(Rn)
≤ C(t−12+1) kϕk
L
2p
p−1(Rn)+kϕkL2p(Rn)
, (2.15)
for allt>0,ϕ∈L2p(Rn)∩Lp−12p(Rn), where we have used the fact thatβ= n(p−1)p . Now, letg ∈L∞(0,∞;L1(Rn)∩ L2p(Rn)∩Lp−12p(Rn)), we conclude from (2.14) and (4.2) that
e−t(−∆)β/2g(t)
Lφ(Rn)≤κ(t)kg(t)k
L1(Rn)∩L2p(Rn)∩L
2p p−1(Rn), for allt>0, where
κ(t) :=min
C(t−12 +1);C t−nβ
ln(t−nβ +1)−1/2p . We can easily check thatκ∈L1(0,∞). Therefore
Z t 0
e−(t−s)(−∆)β/2g(s)ds Lφ(Rn)
≤ Z t
0
e−(t−s)(−∆)β/2g(s) Lφ(Rn) ds
≤ Z t
0
κ(t−s)kg(s)k
L1(Rn)∩L2p(Rn)∩L
2p p−1(Rn)
ds
≤ kgk
L∞(0,∞;L1(Rn)∩L2p(Rn)∩L
2p p−1(Rn))
Z ∞
0
κ(s)ds,
for everyt>0. This proves Lemma 7.
Finally, the following proposition is needed for the local well-posedness result in the space expL0p(Rn).
Proposition 3. [11, Proposition 2.9]
Let1≤p<∞and u∈C([0,T]; expL0p(Rn))for some T >0. Then, for everyα >0, it holds eα|u|p−1
∈C([0,T];Lr(Rn)), 1≤r<∞.
Corollary 1. [11, Corollary 2.13]
Let1≤p<∞and u∈C([0,T]; expL0p(Rn))for some T >0. Assume that f satisfies (1.3). Then, for every p≤r<∞, it holds
f(u)∈C([0,T];Lr(Rn)).
3. Proof of Theorem 1
In this section, we prove Theorem 1 i.e. the local existence and the uniqueness of a mild solution to (1.1) in C([0,T]; expL0p(Rn)) for someT >0. Throughout this section, we assume that the nonlinearity f satisfies (1.3). In order to find the required solution, we will apply the Banach fixed-point theorem to the integral formulation (1.2), using a decomposition argument developed in [7] and used in [9, 10, 11]. The idea is to split the initial datau0 ∈ expL0p(Rn), using the density ofC∞0Rn), into a small part in expLp(Rn) and a smooth one. Letu0∈expL0p(Rn). Then, for everyε >0 there exists30 ∈C∞0Rn) such that
k40kexpLp(Rn)≤ε,
where40:=u0−30. Now, we split our problem (1.1) into the following two problems. The first one is the fractional semilinear heat equation with smooth initial data:
3t+(−∆)β/23= f(3), t>0,x∈Rn, 3(0)=30∈C∞0(Rn), x∈Rn,
(3.1) and the second one is a fractional semilinear heat equation with small initial data in expLp(Rn):
4t+(−∆)β/24= f(4+3)−f(3), t>0,x∈Rn, 4(0)=40, k40kexpLp ≤ε, x∈Rn.
(3.2) We notice that if3is a mild solution of (3.1) and4 is a mild solution of (3.2), thenu =3+4is a solution of our problem (1.2), where the definition of the mild solutions for problems (3.1)- (3.2) are defined similarly as in definition 1. We now prove the local existence result concerning (3.1) and (3.2).
Lemma 8.
Let0 < β ≤ 2, p > 1 and 30 ∈ Lp(Rn)∩L∞(Rn). Then, there exist a time T = T(30) > 0 and a mild solution 3∈C([0,T]; expL0p(Rn))∩L∞(0,T;L∞(Rn))of (3.1).
Lemma 9.
Let0 < β ≤ 2, p > 1, and40 ∈ expL0p(Rn). Let T > 0and 3 ∈ L∞(0,T;L∞(Rn))be given by Lemma 8. Then, for k40kexpLp ≤ ε, with ε 1 small enough, there exist a time Te = Te(40, ε,3) > 0 and a mild solution4 ∈ C([0,Te]; expL0p(Rn))to problem (3.2).
Proof of Lemma 8.In order to use the Banach fixed-point theorem, we introduce the following Banach space YT :=n
3∈C([0,T]; expL0p(Rn))∩L∞(0,T;L∞(Rn)); k3kYT ≤2k30kLp∩L∞
o,
wherek3kYT :=k3kL∞(0,T;Lp)+k3kL∞(0,T;L∞)andk30kLp∩L∞ :=k30kLp+k30kL∞. For3∈YT, we defineΦ(3) by Φ(3) :=e−t(−∆)β/230+Z t
0
e−(t−s)(−∆)β/2f(3(s))ds.
We will prove that ifT >0 is small enough, then,Φis a contraction fromYTinto itself.
•Φ:YT →YT.Let3∈YT. As30∈ Lp(Rn)∩L∞(Rn), then, by Lemma 1, we conclude that30 ∈expL0p(Rn). Then, using Proposition 2,e−t(−∆)β/230 ∈C([0,T]; expL0p(Rn)). Next, forq=porq=∞, we have
kf(3)kLq≤Ceλk3k∞pk3kq≤Ceλk3kp∞(2k30kLp∩L∞), (3.3) which implies, using again Lemma 1, that f(3) ∈ expL0p(Rn) and more precisely f(3) ∈ L1(0,T; expL0p(Rn)) . It follows, by density and smoothing effect of the fractional semigroupe−t(−∆)β/2(Lemma 4), that
Z t 0
e−(t−s)(−∆)β/2f(3(s))ds∈C([0,T]; expL0p(Rn)).
SoΦ(3)∈C([0,T]; expLp0(Rn)). Moreover, using (2.10) and (3.3), we have
kΦ(3)kYT ≤ k30kLp∩L∞+2T C(2k30kLp∩L∞)eλ(2k30kLp∩L∞)p≤2k30kLp∩L∞, forT >0 small enough, namely 4T Ceλ(2k30kLp∩L∞)p≤1. This proves thatΦ(3)∈YT.
•Φis a contraction.Let31,32∈YT. Forq=porq=∞, we have
kf(31)−f(32)kLq ≤Ck31−32kq(eλk31k∞p +eλk32kp∞)≤2Ck31−32kqeλ(2k30kLp∩L∞)p ≤2Ck31−32kYTeλ(2k30kLp∩L∞)p. By (2.10), it holds
kΦ(31)−Φ(32)kYT ≤2T Ck31−32kYTeλ(2k30kLp∩L∞)p≤1
2k31−32kYT.
This finishes the proof of Lemma 8.
Proof of Lemma 9.To prove Lemma 9, we need the following result.
Lemma 10. [11, Lemma 4.4]
Let3∈L∞(0,T;L∞(Rn))for some T >0. Let1<p≤q<∞, and41,42 ∈expLp(Rn)withk41kexpLp,k42kexpLp ≤ M for sufficiently small M > 0 (namely 2pλqMp ≤ 1, whereλ is given in (1.3)). Then, there exists a constant C=C(q)>0such that
kf(41+3)−f(42+3)kq≤Ce2p−1λk3k∞pk41−42kexpLp. ForTe>0, we define the following Banach space
WeT :=
4∈C([0,Te]; expL0p(Rn)); k4kL∞(0,eT;expL0p) ≤2ε , and we consider the mapΦdefined, for4∈W
Te, by Φ(4) :=e−t(−∆)β/240+Z t
0
e−(t−s)(−∆)β/2(f(4(s)+3(s))−f(3(s)))ds.
We will prove that ifεandTe>0 are small enough, then,Φis a contraction fromW
Teinto itself.
•Φis a contraction. Let41,42 ∈ WTe. Using Lemma 1, i.e. the embeddingLp(Rn)∩L∞(Rn) ,→ expL0p(Rn), we have
kΦ(41)−Φ(42)kexpLp≤ 1 (ln 2)1/p
kΦ(41)−Φ(42)kp+kΦ(41)−Φ(42)k∞
. (3.4)
Letr>0 be an auxiliary constant such thatr>max{p,nβ}. Then kΦ(41)−Φ(42)k∞≤C
Z t 0
(t−s)−βrnkf(41(s)+3(s))−f(42(s)+3(s))krds,
thanks to theLr−L∞ estimate (2.9). Applying Lemma 10 withq = rand under the condition 2pλr(2ε)p ≤1, we obtain
kΦ(41)−Φ(42)k∞ ≤ Ce2p−1λk3k∞p Z t
0
(t−s)−βrn ds
!
k41−42kL∞(0,eT;expLp)
≤ Ce2p−1λk3k∞pTe1−βrnk41−42kL∞(0,eT;expLp). (3.5)
