# Weak Comparison Principle for Weighted Fractionalp-Laplacian Equation

Authors: Jin Xie
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

The aim of this paper is to establish a weak comparison principle for a class fractionalp-Laplacian equation with weight. The nonlinear term fx,s>0 is a Carathéodory function which is possibly unbounded both at the origin and at infinity and such that fx,ss1−p decreases with respect to s for a.e. x∈Ω.


---

## Body

## 1. Introduction and Main Results

In this paper, we study a weak comparison principle for the following fractionalp-Laplacian problem
(1)−Δp,βαux=fx,u,x∈Ω,ux>0,x∈Ω,ux=0,x∈ℝN\Ω,whereΩ is a smooth bounded domain of ℝN containing the origin, 0≤β<N−pα/2, 1<p, pα<N, and f:Ω×0,+∞⟶0,+∞ is a general Carathéodory function, which is possibly unbounded both at the origin and at infinity and such that fx,ss1−p decreases with respect to s for a.e. x∈Ω. The weighted fractional p-Laplacian −Δp,βα is the pseudodifferential operator defined as
(2)−Δp,βαux=P.V.∫ℝNux−uyp−2ux−uyx−yn+pαdyxβyβ=limε→0∫ℝN\Bεxux−uyp−2ux−uyx−yn+pαdyxβyβ,here P.V. denotes the principal value of the integral.The simplest model is(3)−Δp,βαux=hxuγx+gxuqx,x∈Ω,ux>0,x∈Ω,ux=0,x∈ℝN\Ω,where γ≥0 and 0≤q≤p, and h and g are nonnegative functions.The interest on the nonlocal operators continues to grow in recent years since such problems arise in various fields. The fractionalp-Laplacian −Δp,βα, on one hand, is an extension of the local operator −divx−β∇up−2∇u. Note that, for this type of operator, the Caffarelli-Kohn-Nirenberg inequality plays an important role, see [1–10]. On the other hand,p=2, which appears in a natural way when dealing with the fractional Laplace problem with Hardy potential. More precisely, let u be a solution to the following problem
(4)−Δαu−λux2α=fx,u,x∈Ω,ux=0,x∈ℝN\Ω,where λ≔ΛN,α+ΦN,αβ, ΛN,α is the Hardy constant, and
(5)ΦN,αβ=22αΓβ+2α/2ΓN−β/2ΓN−β−2α/2Γβ/2−Γ2N+2α/4Γ2N−2α/4.Then, according to ground state representation [11, 12], vx=uxxβ satisfies
(6)Lβv=x−βfx,x−βv,x∈Ω,v=0,x∈ℝN\Ω,where
(7)Lβv=CN,sP.V.∫ℝNvx−vyx−yN+2αdyxβyβ.Fractional Laplace operator−Δα can be defined using Fourier analysis, functional calculus, singular integrals, or Lévy processes. Thus, rich mathematical concepts allow in general rich properties. For some abstract definitions and tools of fractional Laplace operator, see [13]. For more recent results of fractional Laplace elliptic problem, see [14–16] and the reference therein.There are many works on the study of fractionalp-Laplacian equations. Canino et al. [17] investigated the existence and uniqueness of solutions to
(8)−Δpαux=hxuγx,x∈Ω,ux>0,x∈Ω,ux=0,x∈ℝN\Ω.Whenβ=0, Mukherjee and Sreenadh [18] used variational methods to show the existence and multiplicity of positive solution problem (1) with critical growth and singular nonlinearity. Abdellaoui et al. [19] prove the existence of a weak solution to (1) under some hypotheses on fx,u.The main idea of this paper comes from the seminal papers [20, 21]. In [21], Brezis and Oswald have shown the existence and uniqueness of a solution to a Laplace elliptic problem. Recently, Durastanti and Oliva [20] obtained the existence and uniqueness of positive solutions of an elliptic boundary value problem modeled by
(9)−Δpux=hxuγx+gxuqx,x∈Ω,ux>0,x∈Ω,ux=0,x∈∂Ω.The main result of this paper is the following weak comparison principle.Theorem 1.
Assume thatf1,f2 are nonnegative functions such that either f1x,ss1−p or f2x,ss1−p is decreasing with respect to s, and for almost every x∈Ω,
(10)f1x,s≤f2x,s,
for almost everyx∈Ω and for all s∈0,+∞. Suppose that u1 and u2 are weak solutions to the problem
(11)−Δp,βαui=fix,ui,x∈Ω,uix>0,x∈Ω,uix=0,x∈ℝN\Ω,i=1,2.
Then,u1x≤u2x almost everywhere in Ω.Consequently, we obtain the uniqueness of the solution to problem (1).Corollary 2.
Assume thatf is a nonnegative function such that fx,ss1−p is decreasing with respect to s for almost every x∈Ω. Then, there exists at most one weak solution to problem (1).The paper is organized as follows. In Section2, we present some definitions and preliminary tools, which will be used in the proof of Theorem 1. The proof of Theorem 1 and Corollary 2 be given in Section 3.


## 2. Preparations

First of all, we give the definition of truncation function.Definition 3.
For everyk≥0 and σ∈ℝN, define
(12)Tkσ=max−k,mink,σ,Gkσ=σ−Tkσ.
LetΩ⊂ℝN, the weighted fractional Sobolev space Wβα,pΩ is defined by
(13)Wβα,pΩ=φ∈LpΩ,dμ:∫Ω∫Ωφx−φypdν<∞,endowed with the norm
(14)φWβα,pΩ=∫Ωφxpdμ1/p+∫Ω∫Ωφx−φypdν1/p,where
(15)dμ≔dxx2β,dν≔dxdyx−yn+pαxβyβ.
The spaceW0,βα,pΩ is the completion of C0∞Ω with respect to the above norm.Definition 4.
A positive functionu∈W0,βα,pΩ∩Lp−1Ω is a weak solution to problem (1) if fx,u∈Llocp−1Ω and for any ϕ∈C01Ω,
(16)−Δp,βαu,ϕ=∫Ωfx,uϕxdx,where
(17)−Δp,βαu,ϕ=∬DΩux−uyp−2ux−uyϕx−ϕydν,DΩ=ℝN×ℝN\Ωc×Ωc.
It is worth pointing out that the formulation (16) can be extended for W0α,pΩ-test functions by standard argument.The following fractional Picone inequality appears in Proposition 2.15 of [22] with β=0. A slight change in this proof actually shows that the fractional Picone inequality also holds for fractional p-Laplacian −Δp,βα.Lemma 5.
Consideru,v∈W0,βα,pΩ with u≥0. Assume that −Δp,βαu is a positive bounded Radon measure in Ω. Then
(18)2∫Ω−Δp,βαuvpup−1≤vW0,βα,pΩp.Proof.
Takevk=Tkv and u~=u+η, where η>0 is a constant. It is easy to show that vkp/u~p−1∈W0,βα,pΩ. By similar arguments as in the proof of Proposition 2.10 of [22], we can easily prove the following equation
(19)2∫Ω−Δp,βαuvkpu~p−1=−Δp,βαu,vkpu~p−1=∬DΩux−uyp−2ux−uyvkpxu~p−1x−vkpyu~p−1ydν≤∬DΩvkx−vkypdν≤vW0,βα,pΩp,
where we use discrete Picone inequality. For more details, see Proposition 4.2 of [23]. Letting k⟶+∞ and η⟶0 in (19) leads to (18).


## 3. Proof of Main Theorem

Proof.
Suppose thatf1x,ss1−p is decreasing with respect to s for almost every x∈Ω. A slight change will be needed provided f2x,ss1−p is decreasing with respect to s for almost every x∈Ω but no essential difference.
For fixedε>0 and k∈ℕ, define
(20)u1,εx=u1x+ε,u2,εx=u2x+ε,wεx=u1,εpx−u2,εpx,ϕ1x=Tkwε+xu1,εp−1x,ϕ2x=Tkwε+xu2,εp−1x.
whereTk is defined by (12) and wε+≔maxwε,0 is the positive part of the function wε.
In the following proof, I show thatwε+=0, which leads to u1x≤u2x almost everywhere in Ω.
Choosingϕ1 and ϕ2 as test functions in equations of u1 and u2, respectively, we find
(21)−Δp,βαu1,ϕ1=∫Ωf1x,u1ϕ1xdx,(22)−Δp,βαu2,ϕ2=∫Ωf2x,u2ϕ2xdx.
Subtracting the two equations (21) and (22), we obtain
(23)−Δp,βαu1,ϕ1−−Δp,βαu2,ϕ2=∫Ωf1x,u1ϕ1x−f2x,u2ϕ2xdx=∫Ωf1x,u1u1,εp−1x−f2x,u2u2,εp−1xTkwε+dx.
DecomposeℝN as ℝN=D1∪D2∪D3, where
(24)D1=x∈ℝN:wε+x=0,D2=x∈ℝN:0<wε+x<k,D3=x∈ℝN:wε+x≥k.
Therefore,(25)Tkwε+x=0,x∈D1,wε+,x∈D2,k,x∈D3,ℝN×ℝN=⋃i,j=1i,j=3Dij,
hereDij=x,y∈ℝN×ℝN:x∈Di,y∈Dj.
By the definition ofϕi and Dij, we get
(26)ϕix−ϕiy=0,x∈D11,−wε+yui,εp−1y,x∈D12,−kui,εp−1y,x∈D13,wε+xui,εp−1x,x∈D21,wε+xui,εp−1x−wε+yui,εp−1y,x∈D22,wε+xui,εp−1x−kui,εp−1y,x∈D23,kui,εp−1x,x∈D31,kui,εp−1x−wε+yui,εp−1y,x∈D32,kui,εp−1x−kui,εp−1y,x∈D33.
Now, rewrite−Δp,βαu1,ϕ1−−Δp,βαu2,ϕ2 as
(27)−Δp,βαu1,ϕ1−−Δp,βαu2,ϕ2=∬⋃i,j=1i,j=3Diju1x−u1yp−2u1x−u1yϕ1x−ϕ1ydν−∬⋃i,j=1i,j=3Diju2x−u2yp−2u2x−u2yϕ2x−ϕ2ydν=∑i,j=1i,j=3Iij−Jij,where
(28)Iij=∬Diju1x−u1yp−2u1x−u1yϕ1x−ϕ1ydν,Jij=∬Diju2x−u2yp−2u2x−u2yϕ2x−ϕ2ydν.
For simplicity of notation, denote(29)Aiu=uix−uiyp−2uix−uiy,i=1,2.
Obviously,(30)I11=J11=0.
Now considerI12−J12. By (26), we get
(31)I12−J12=−∬D12A1uu1,εpy−u2,εpyu1,εp−1ydν+∬D12A2uu1,εpy−u2,εpyu2,εp−1ydν=∬D12u1,εpy−u2,εpyM12dν,where
(32)M12=A2uu2,εp−1y−A1uu1,εp−1y.
Note thatu1x≤u2x and u1y≥u2y for x,y∈D12. Thus, u1x−u1y≤u2x−u2y. By the monotonicity of the function ht=tp−2t, we find M12≥0, which implies that
(33)I12−J12≥0.
ForI13−J13, we have
(34)I13−J13=k∬D13M12dν≥0,
sinceM12≥0 for x,y∈D13.
ForI21−J21, we find
(35)I21−J21=∬D21A1uu1,εpx−u2,εpxu1,εp−1xdν−∬D21A2uu1,εpx−u2,εpxu2,εp−1xdν=∬D21u1,εpx−u2,εpxM21dν,where
(36)M21=A1uu1,εp−1x−A2uu2,εp−1x.
Recalling thatu1x−u1y≥u2x−u2y since u1x≥u2x and u1y≤u2y for x,y∈D21. This fact, together with Aiux/ui,εp−1x<1 for i=1,2, yields M21≥0. Consequently
(37)I21−J21≥0.
Now, we considerI22−J22.
(38)I22−J22=∬D22A1uwεxu1,εp−1x−wεyu1,εp−1ydν−∬D22A2uwεxu1,εp−1x−wεyu1,εp−1ydν=∬D22A1uu1,εx−u2,εpxu1,εp−1x−u1,εy+u2,εpyu1,εp−1ydν+∬D22A2uu2,εx−u1,εpxu2,εp−1x−u2,εy+u1,εpyu2,εp−1ydν=u1,εWβs,pp−u2,εpxu1,εp−1x,−Δp,βαu1,ε+u2,εWβs,pp−u1,εpxu2,εp−1x,−Δp,βαu2,ε≥0,
here we use the fractional Picone inequality; see Lemma5.
ForI23−J23, we have
(39)I23−J23=∬D23A1uu1,εpx−u2,εpxu1,εp−1x−ku1,εp−1ydν−∬D23A2uu1,εpx−u2,εpxu2,εp−1x−ku2,εp−1ydν=∬D23A1uu1,εpx−u2,εpxu1,εp−1x−ku1,εp−1ydν+∬D23A2uu2,εpx−u1,εpxu2,εp−1x+ku2,εp−1ydν.
Now, we consider the first term of the right-hand side of (39). Suppose that u1x−u1y≥0,
(40)∬D23A1uu1,εpx−u2,εpxu1,εp−1x−ku1,εp−1ydν=∬D23A1uu1,εpx−u1,εpyu1,εp−1x+u1,εpy−u2,εpxu1,εp−1x−ku1,εp−1ydν≥∬D23A1uu1,εpy−u2,εpxu1,εp−1x−ku1,εp−1ydν≥u1,εWβs,pD23≥0.
Suppose thatu1x−u1y≤0, the above inequation holds also since k≤u1,εp−1y for x,y∈D23.
For the second term of the right-hand side of (39), by a similar argument, we get
(41)∬D23A2uu2,εpx−u1,εpxu2,εp−1x+ku2,εp−1ydν=∬D23A2uu2,εpx−u2,εpyu2,εp−1x+u2,εpy−u1,εpxu2,εp−1x+ku2,εp−1ydν≥∬D23A2uu2,εpy−u1,εpxu2,εp−1x+ku2,εp−1ydν≥u2,εWβs,pD23≥0.
Thus, taking into account (40) and (41), we derive that
(42)I23−J23≥0.
ForI31−J31, it is easily seen that
(43)I31−J31=k∬D21u1x−u1yp−2u1x−u1yu1,εp−1xdν−k∬D21u2x−u2yp−2u2x−u2yu1,εp−1xdν≥0.
ForI32−J32, repeating the previous argument of I23−J23 leads to
(44)I32−J32=∬D32A1uku1,εp−1x−u1,εpy−u2,εpyu1,εp−1ydν−∬D32A2uku2,εp−1x−u1,εpy−u2,εpyu2,εp−1ydν≥0.
ForI33−J33, we have
(45)I33−J33=∬D33A1uku1,εp−1x−ku1,εp−1ydν−∬D33A2uku2,εp−1x−ku2,εp−1ydν=−Δp,βαu1,ε,ku1,εp−1D33−−Δp,βαu2,ε,ku2,εp−1D33=f1x,u1,ku1,εp−1D−f2x,u2,ku2,εp−1D=k∫Df1x,u1u1,εp−1−f2x,u2u2,εp−1,
whereD satisfies D33=ℝN\Dc×Dc.
For the right-hand side of (23), according to (10), we have
(46)∫Ωf1x,u1u1,εp−1−f2x,u2u2,εp−1Tkwε+dx≤∫Ωf1x,u1u1,εp−1−f1x,u2u2,εp−1Tkwε+dx.
Using the monotonicity off1x,ss1−p, we know that
(47)f1x,u1u1p−1≤f1x,u2u2p−1,
sinceu1x≥u2x for x∈D, where D appears in (45). Thus, taking into account (45)–(47), we obtain, for small enough ε>0,
(48)∫Ωf1x,u1u1,εp−1−f2x,u2u2,εp−1Tkwε+dx−I33−J33=∫Ω\Df1x,u1u1,εp−1−f2x,u2u2,εp−1wε+dx≤0.
This fact, together with (23), (30), (33), (34), (37), (38), (42)–(44), leads to
(49)0≤∫Ω\Df1x,u1u1,εp−1−f2x,u2u2,εp−1wε+dx≤0,
which implies thatwε+=0 for x∈Ω\D, that is u1x≥u2x for any x∈Ω. This completes the proof of Theorem 1.The proof of Corollary2 is immediate, which is omitted.



---

*Source: 6675031-2020-12-21.xml*