Uniqueness and monotonicity of solutions for fractional equations with a gradient term

Uniqueness and monotonicity of solutions for fractional equations with a gradient term

Pengyan Wang

School of Mathematics and Statistics, Xinyang Normal University, Xinyang, 464000, P. R. China

Received 4 May 2021, appeared 30 July 2021 Communicated by Patrizia Pucci

Abstract. In this paper, we consider the following fractional equation with a gradient term

(−_∆)^su(x) = f(x,u(x),∇u(x)),

in a bounded domain and the upper half space. Firstly, we prove the monotonicity and uniqueness of solutions to the fractional equation in a bounded domain by the sliding method. In order to obtain maximum principle on unbounded domain, we need to estimate the singular integrals define the fractional Laplacians along a sequence of approximate maximum points by using a generalized average inequality. Then we prove monotonicity and uniqueness of solutions to fractional equation in Rⁿ₊ by the sliding method. In order to solve the difficulties caused by the gradient term, some new techniques are developed. The paper may be considered as an extension of Berestycki and Nirenberg [J. Geom. Phys. 5(1988), 237–275].

Keywords: fractional equation with gradient term, monotonicity, uniqueness, sliding method.

2020 Mathematics Subject Classification: 35R11, 35A09, 35B06, 35B09.

1 Introduction

During the last decades, fractional Laplacian has attracted more and more attention due to its various applications. The methods to study the fractional Laplacian are the extension method [6], moving planes method in integral form [11], the method of moving sphere [26] and direct methods of moving planes [9,22] etc. Recently, to study the monotonicity of the solution, Liu [28], Wu and Chen [35,36] introduced a direct sliding method for fractional Laplacian and fractional p-Laplacian. Berestycki and Nirenberg [3–5] first developed the sliding method, which was used to establish qualitative properties of solutions for nonlinear elliptic equations involving the regular Laplacian such as monotonicity, nonexistence and uniqueness etc. The essential ingredients are different forms of maximum principles. The main idea lies in com- paring values of the solution to the equation at two different points, between which one point is obtained from the other by sliding the domain in a given direction, and then the domain

BCorresponding author. Email: wangpy@xynu.edu.cn

is slide back to a critical position. While in the method of moving planes, one point is the reflection of the other.

Inspired by the above article, in this article, we show the monotonicity, antisymmetry and uniqueness of solutions for the following fractional equation with a gradient term

(−_∆)^su(x) = f(x,u(x),∇u(x)), (1.1) where ∇u denotes the gradient of u, the fractional Laplacian (−_∆)^s with 0 < s < 1 is given by

(−_∆)^su(x) =Cn,sP.V.

Z

Rⁿ

u(x)−u(y)

|x−y|^{n+2s dy}

=C_n,slim

e→0

Z

Rⁿ\Be(x)

u(x)−u(y)

|x−y|^{n+2s dy.}

Define

L_2s =ⁿu:u∈ L¹_loc(_Rⁿ), Z

Rⁿ

|u(x)|

1+|x|^n+2sdx<_∞^o, then it is easy to see that foru∈C_loc^1,1(_Rⁿ)∩ L_2s,(−_∆)^suis well defined.

When s = 1, [3] derived the monotonicity, symmetry and uniqueness of (1.1) in a finite cylinder and a bounded domain which is convex in thex₁ direction by the sliding method. In the cases =1,f(x,u,∇u) = f(u), Gidas, Ni, Nirenberg [21] obtained monotonicity and sym- metry for positive solutions of (1.1), vanishing on the boundary, using the maximum principle and the method of moving planes; in [2,5], Berestycki, Cafferelli and Nirenberg considered the monotonicity and uniqueness of solution for (1.1) by the sliding method. Recently, in the case 0 < s < 1, Chen, Li and Li [9] investigated the semilinear equation in the whole space with f(x,u,∇u) = u^p, 1 < p ≤ ⁿ_n⁺₋^2s_2s, developed a direct method of moving planes for the fractional Laplacian and showed that the nonnegative solution of (1.1) is radially symmetric and monotone decreasing about some point in the critical case p = ⁿ_n⁺₋^2s_2s and nonexistence of positive solutions in the subcritical case 1< p < ⁿ_n⁺₋^2s_2s; Dipierro, Soave and Valdinoci [16]

proved symmetry, monotonicity and rigidity results to (1.1) in an unbounded domain with the epigraph property.

The purpose of the present paper is to extend the results in [3] to the fractional equation.

On the one hand, we extent the cases = 1 in [3] to the fractional case 0< s <1, and extend bounded domain to Rⁿ+. On the other hand, the nonlinear term f(x,u,∇u) has a broader form containing nonlinear term f(u)and f(x,u).

In order to solve the difficulty that the nonlinear term at the right side of (1.1) contain the gradient term, in the bounded domain when deriving the contradiction for the minimum point of the function w^τ(x) (see Section 2 below for definition), for the first time, we use the technique of finding the minimum value of the functionw^τ(x)for the variables τ andx at the same time. This is different from the previous sliding process which only finds the minimum value of the variablexfor the fixed τ. In the whole space, we estimate the singular integrals defining the fractional Laplacian along a sequence of approximate maximum, and the estimating is forτand the sequence of approximate maximum at the same time.

In order to apply the sliding method, we give the exterior condition on u. Let u(x) = ϕ(x), x∈_Ω^c, and assume that

(C) for any three pointsx= (x⁰,xn), y= (x⁰,yn)andz= (x⁰,zn)lying on a segment parallel to thex_naxis, y_n <x_n<z_n, withy,z∈_Ω^c, we have

ϕ(y)< u(x)< ϕ(z)_{, if} x∈ _{Ω (1.2)}

and

ϕ(y)≤ ϕ(x)≤ ϕ(z), ifx ∈_Ω^c. (1.3) Remark 1.1. The same monotonicity conditions (1.2) and (1.3) (withΩ^creplaced by∂Ω) were assumed in [4,5,35].

The main result of this paper is

Theorem 1.2. Suppose that u ∈C^1,1_loc(_Ω)∩C(_Ω^¯)satisfies (C) and is a solution of equation ((−_∆)^su(x) = f(x,u,∇u), x∈ _Ω,

u(x) =ϕ(x), x∈ _Ω^c, (1.4)

where Ω is a bounded domain which is convex in x_n direction. Assume that f is continuous in all variables, locally Lipschitz continuous in (u,∇u) and is nondecreasing in xn. Then u is strictly monotone increasing with respect to x_ninΩ, i.e.,for anyτ>0,

u(x⁰,xn+τ)>u(x⁰,xn), for all(x⁰,xn),(x⁰,xn+τ)∈_Ω.

Furthermore, the solution of (1.4)is unique.

Remark 1.3. Theorem (1.2) includes the result of Theorem 2 in [35], and we also prove the uniqueness of solutions in bounded domain. If Ω is the finite cylinder C = {x = (x⁰,x_n) ∈ Rⁿ | |x_n| < l, x⁰ ∈ ω}, where l > 0 and ω is a bounded domain in Rⁿ⁻¹ with smooth boundary, the result of Theorem1.2still holds.

Remark 1.4. The conditions in Theorem 1.2 and Theorem 1 of [14] are different. Neither implies the other. Cheng, Huang and Li [14] studied the positive solutionuand obtained that uis strictly increasing in the half of Ωin x_n-direction withx_n < 0 by the method of moving planes, but the solution we study can be negative and is strictly increasing with respect to xn

in the whole domainΩby the sliding method.

We also have a new antisymmetry result for the equation (1.4) if the bounded domain Ω is symmetric about xn =0.

Corollary 1.5(Antisymmetry). Assume that the conditions of Theorem1.2are satisfied and in addi- tion thatϕis odd in x_nonΩ^c. If f(x,u,∇u)is odd in(x_n,u,∇_x⁰u). Then u is odd, i.e.antisymmetric in x_n:

u(x⁰,−xn) =−u(x⁰,xn), ∀x ∈_Ω.

This follows from the fact that ¯u=−u(x⁰,−xn)is a solution satisfying the same conditions, and so isu.

For the unbounded domain, we give the following result onRⁿ+. Theorem 1.6. Suppose that u ∈C^1,1_loc(_Rⁿ₊)∩ L_2s(_Rⁿ)∩C(_Rⁿ₊)is a solution of











(−_∆)^su(x) = f(u,∇u), x∈ _Rⁿ₊, 0<u(x)≤µ, x∈ _Rⁿ₊, u(x) =_0, x6∈_Rⁿ_+,

(1.5)

and

xnlim→+_∞u(x⁰,x_n) =µ, uniformly for all x⁰ ∈_Rⁿ⁻¹. (1.6)

Assume that f is bounded, continuous in all variables and nonincreasing in u ∈ [µ−δ,µ] for some δ>0. Then u is strictly monotone increasing in xndirection, and moreover it depends on xnonly.

Furthermore, the solution of (1.5)is unique.

Theorem1.6is closely related to the following well-known De Giorgi conjecture [19].

Conjecture(De Giorgi [19]). Ifuis a solution of

−_∆u =u−u³, such that

xnlim→±_∞u(x⁰,x_n) =±1, for all x⁰ ∈_Rⁿ⁻¹, and

|u(x)| ≤1, x∈_Rⁿ, ∂u

∂xn

>0.

Then there exists a vectorµ∈_Rⁿ⁻¹ and a functionu₁:R→_Rsuch that u(x⁰,xn) =u₁(µx⁰+xn) inRⁿ.

The other symmetry, uniqueness and monotonicity results on local and nonlocal equations, we also refer readers to [1,18,24,25] for semilinear elliptic equations, [9,13,17,23,30,31] for fractional equations, [34,38] for weighted fractional equation, [14,37] for fractional equations with a gradient term, [27] for integral system with negative exponents, [12] for weighted Hardy-sobolev type system, [8,32,33] for fully nonlinear nonlocal equations with gradient term, [7,15,29] for fractionalp-Laplace equation, and references therein.

The paper is organized as follows. In Section 2 we prove Theorem 1.2 via the sliding method. In Section 3, we first establish a maximum principle in the unbounded domain, then uniqueness and monotonicity for the fractional equation with a gradient term onRⁿ+ are obtained.

2 The proof of Theorem 1.2

For convenience, we list some notations used frequently. For τ ∈ _{R, denote} x = (x⁰,x_n), x⁰ = (x₁,· · · ,x_n−1)∈_Rⁿ⁻¹. Set

u^τ(x) =u(x⁰,xn+τ), w^τ(x) =u^τ(x)−u(x).

Proof of Theorem1.2. For τ > 0, it is defined on the set Ω^τ = _Ω−τe_n which is obtained from Ωby sliding it downward a distance τparallel to the xn axis, whereen = (0, . . . , 0, 1). Set

D^τ :=_Ω^τ∩_Ω, τ˜ =sup{τ|τ>0,D^τ 6= _∅} and

w^τ(x) =u^τ(x)−u(x), x∈ D^τ.

We mainly divide the following two steps to prove thatuis strictly increased in the x_ndirec- tion,i.e.

w^τ(x)>_0, x∈ D^τ, for any 0<τ<τ.˜ (2.1)

Step 1. Forτsufficiently close toτ˜ i.e., D^τ is narrow, we claim that there exists δ >0small enough such that

w^τ(x)≥0, ∀x ∈D^τ, ∀τ∈(τ˜−δ, ˜τ)_{. (2.2)} Otherwise, we set

A₀ = _min

x∈D¯^τ

˜ τ−δ<τ<τ˜

w^τ(x)<_0.

From condition(C),A₀ can be obtained for some(τ⁰,x⁰)∈ {(τ,x)|(τ,x)∈ (τ˜−δ, ˜τ)×D^τ}. Noticing that w^τ0(x)≥ 0,x ∈ ∂D^τ0, we arrive at x⁰ ∈ D^τ0. Sow^τ0(x⁰) = A0. Since (τ⁰,x⁰)is a minimizing point, we have ∇w^τ0(x⁰) = 0, i.e., ∇u^τ0(x⁰) = ∇u(x⁰). Since u^τ0 satisfies the same equation (1.4) inΩ^τ0 asudoes inΩ, and f is nondecreasing inx_n, so we have

(−_∆)^sw^τ0(x⁰) = f((x⁰)⁰,x⁰_n+τ⁰,u^τ0(x⁰),∇u^τ0(x⁰))− f(x⁰,u(x⁰),∇u(x⁰))

≥ f(x⁰,u^τ0(x⁰),∇u^τ0(x⁰))− f(x⁰,u(x⁰),∇u(x⁰))

= f(x⁰,u^τ0(x⁰),∇u(x⁰))− f(x⁰,u(x⁰),∇u(x⁰))

=−c^τ0(x⁰)w^τ0(x⁰)_,

(2.3)

where−c^τ0(x⁰) = ^f(x0,uτ

0(x⁰),∇u(x⁰))−f(x⁰,u(x⁰),∇u(x⁰))

u^τ0(x⁰)−u(x⁰) is aL^∞function satisfying

|c^τ0(x⁰)| ≤C, ∀ x⁰ ∈D^τ0. Hence

(−_∆)^sw^τ0(x⁰) +c^τ0(x⁰)w^τ0(x⁰)≥0.

On the other hand, we obtain

(−_∆)^sw^τ0(x⁰) +c^τ0(x⁰)w^τ0(x⁰)

= C_n,sP.V.

Z

Rⁿ

w^τ0(x⁰)−w^τ0(y)

|x⁰−y|^{n+2s dy}+c^τ0(x⁰)w^τ0(x⁰)

≤ C_n,sw^τ0(x⁰)

Z

(D^τ0)^c

1

|x⁰−y|^n+2sdy+inf

D^τ0

c^τ0(x)w^τ0(x⁰)

≤ w^τ0(x⁰) C₁

d^2s_n −C

<0,

(2.4)

where d_n denotes the width ofD^τ0 in the x_n direction andD^τ0 is narrow. This is a contradic- tion.

Therefore we derive (2.2) is true forτsufficiently close to ˜τ.

Step 2. The inequality(2.2)provides a starting point, from which we can carry out the sliding. Now we decreaseτas long as(2.2)holds to its limiting position. Define

τ₀=inf{τ|w^τ(x)≥0, x ∈D^τ, 0< τ<τ˜}. We will prove

τ₀ =_0.

Otherwise, assume τ₀ > 0, we show that the domain Ωcan be slided upward a little bit more and we still have

w^τ(x)≥0, x∈ D^τ, for anyτ₀−ε<τ≤τ₀, (2.5) which contradicts the definition ofτ₀.

Sincew^τ0(x)>0, x∈_Ω∩∂D^τ0 by condition (C)andw^τ0(x)≥0, x∈ D^τ0, then w^τ0(x)6≡0, x∈D^τ0.

If there exists a point ˜x∈ D^τ0 such thatw^τ0(x˜) =0, then ˜xis the minimum point. So we have

∇w^τ0(x˜) =0 and

(−_∆)^sw^τ0(x˜) =C_n,sP.V.

Z

Rⁿ

w^τ0(x˜)−w^τ0(y)

|x˜−y|^{n+2s dy}<0, which contradicts to

(−_∆)^sw^τ0(x˜) = f(x˜⁰, ˜xn+τ₀,u^τ0(x˜),∇u^τ0(x˜))− f(x,˜ u(x˜),∇u(x˜))

≥ f(x,˜ u^τ0(x˜)_,∇u^τ0(x˜))− f(x,˜ u(x˜)_,∇u(x˜))

=0.

Hence,

w^τ0(x)>0, x ∈D^τ0. (2.6)

Next we will prove (2.5). Suppose (2.5) is not true, one has A₁= min

x∈D^τ τ0−ε<τ<τ0

w^τ(x)<0.

The minimum A1 can be obtained for some µ ∈ (τ₀−ε,τ₀)_{, ¯x} ∈ D^µ where w^µ(x¯) = _{A1 by} condition(C). We carve out ofD^τ0 a closed setK⊂ D^τ0 such thatD^τ0\Kis narrow. According to (2.6),

w^τ0(x)≥C0>0, x ∈K.

From the continuity ofw^τ inτ, we have for smallε>0,

w^µ(x)≥0, x ∈K. (2.7)

From(C), it follows

w^µ(x)≥0, x ∈(D^µ)^c.

So ¯x∈ D^µ\Kand∇w^µ(x¯) =0. SinceD^τ0 ⊂D^µand smallε, we obtain thatD^µ\Kis a narrow domain. Similar to (2.3), we have

(−_∆)^sw^µ(x¯) +c(x¯)w^µ(x¯)≥0.

Similar to (2.4) and narrow domainD^µ\K, we have

(−_∆)^sw^µ(x¯) +c(x¯)w^µ(x¯)<0.

This is a contradiction. Hence we derive (2.5), which contradicts to the definition of τ₀. So τ₀ =0. Therefore, we have shown that

w^τ(x)≥_0, x∈ D^τ, for any 0<τ<τ.˜ (2.8)

Next we prove (2.1). Since

w^τ(x)6≡0, x∈ D^τ, for any 0<τ<τ,˜

if there exists a point x⁰ for someτ₁ ∈ (0, ˜τ) such that w^τ1(x⁰) = 0, then x⁰ is the minimum point and

(−_∆)^sw^τ1(x⁰) =Cn,sP.V.

Z

Rⁿ

w^τ1(x⁰)−w^τ1(y)

|x⁰−y|^{n+2s dy}<0.

This contradicts to

(−_∆)^sw^τ1(x⁰) = f((x⁰)⁰,x⁰_n+τ₁,u^τ1(x⁰),∇u^τ1(x⁰))− f(x⁰,u(x⁰),∇u(x⁰))≥0.

Therefore, we arrive at (2.1).

Now we prove uniqueness. If u is another solution satisfying the same conditions, the same argument as before but replace w^τ = u^τ−u with w^τ = u^τ−u. Similarly to (2.8), we haveu^τ(x)≥ u in D^τ for any 0 < τ < τ. Hence,˜ u ≥ u. Interchanging the roles ofu andu, we find the opposite inequality. Therefore,u= u.

This completes the proof of Theorem1.2.

3 The uniqueness and monotonicity of solution on R

In the section, we will prove Theorem 1.6. We first establish a maximum principle in the unbounded domain for the fractional equation with a gradient term.

Lemma 3.1(Maximum principle). Let D be an open set inRⁿ, possibly unbounded and disconnected, suppose that

lim

k→_∞

|D^c∩(B₂k+1(q)\B₂k(q))|

|(B₂k+1(q)\B₂k(q))| >0,

where q is any point in D. Let w∈C^1,1_loc(D)∩ L_2sbe bounded from above and satisfy







(−_∆)^s_w(_x) +_c(_x)_w(_x) +

∑

b_j(_x)_wj(_x)≤0, x∈ D,

w(x)≤0, x∈_Rⁿ\D,

(3.1)

for some nonnegation function c(x). Then

w(x)≤0, x∈ D.

Furthermore, we have

either w(x)<0in D or w(x)≡0inRⁿ. (3.2) Remark 3.2. The proof of Lemma3.1is different from Theorem 3 in [35]. Here we mainly use the following generalized average inequality.

Lemma 3.3([35] A generalized average inequality). Suppose that w∈ C^1,1_loc(_Rⁿ)∩ L_2s andx is a¯ maximum point of w inRⁿ. Then for any r>0, we have

C₀

C_n,sr^2s(−_∆)^sw(x¯) +C₀ Z

B_r^c(x¯)

r^2s

|x¯−y|^n+2sw(y)dy≥ w(x¯), where C₀ satisfies

C0

Z

B^c_r(x¯)

r^2s

|x¯−y|^n+2sdy=1.

Proof of Lemma3.1. Suppose on the contrary, there is some point x such thatw(x) > 0 in D, then

0< A:= sup

x∈_Rⁿ

w(x)<_∞. (3.3)

There exists a sequence{x^k} ⊂Dsuch that

w(x^k)→ A>0, ask →_∞. (3.4)

Let

η(x) =





 ce

1

|x|²−1, |x|<1,

0, |x| ≥1, (3.5)

wherec>0 is a constant, taking c=e such thatη(0) =max_Rⁿη(x) =1.

Set

ψ_k(x) =η(x−x^k). (3.6)

From (3.4), there exists a sequence{ε_k}with ε_k >0 such that w(x^k) +ε_kψ_k(x^k)≥ A.

Since w(x) ≤ 0, x ∈ _Rⁿ\D, it follows from (3.4) that x^k is away from ∂D. Without loss of generality, we may assume that dist(x^k,∂D) = 2. So B₁(x^k) ⊂ D. Since for any x ∈ D\B₁(x^k), w(x)≤ Aandψ_k(x) =0, hence

w(x^k) +ε_kψ_k(x^k)≥w(x) +ε_kψ_k(x), for any x∈_Rⁿ\B₁(x^k). It follows that there exists a point ¯x^k ∈ B₁(x^k)such that

w(x¯^k) +ε_kψ_k(x¯^k) =max

Rⁿ (w(x) +ε_kψ_k(x))> A. (3.7) So(w(x¯^k) +ε_kψ_k(x¯^k))_j =0 and

w_j(x¯^k)→0, ask→_∞. (3.8)

Forw+ε_kψ_k, using Lemma3.3, we obtain

(w+ε_kψ_k)(x¯^k)≤C₁(−_∆)^s(w+ε_kψ_k)(x¯^k) +C₂ Z

B^c₂(x¯^k)

(w+ε_kψ_k)(y)

|x¯^k−y|^{n+2s dy.}

Letε_k →0, by the first inequality of (3.1), it implies that w(x¯^k)≤C₁(−_∆)^sw(x¯^k) +C₂

Z

B^c₂(x¯^k)

w(y)

|x¯^k−y|^n+2sdy

≤ −c(x¯^k)w(x¯^k)−

∑

b_j(x)w_j(x¯^k) +C₂ Z

B^c₂(x¯^k)

w(y)

|x¯^k−y|^n+2sdy.

(3.9)

Lettingk →_∞, combining (3.4), (3.8), (3.9) and nonnegative functionc(x), we arrive at 0<(c(x) +1)A←(c(x¯^k) +1)w(x¯^k)≤ C₂

Z

B^c₂(x¯^k)

w(y)

|x¯^k−y|^n+2sdy, this is impossible because of (3.3) and the second inequality of (3.1).

Based on above result, ifw=0 at some pointx₀ ∈ D, thenx₀ is a maximum point ofwin D. And we still havew_j =0 in the maximum point. Ifw6≡0 inRⁿ, then we have

(−_∆)^sw(x₀) +c(x₀)w(x₀) +

∑

b_j(x₀)w_j(x₀) =C_n,sP.V.

Z

Rⁿ

−w(y)

|x₀−y|^n+2sdy>0.

This is a contradiction with (3.1). So we have eitherw<0 in Dor w≡0 inRⁿ. This completes the proof Lemma3.1.

We also need the following lemma.

Lemma 3.4 ([9], Maximum principle). Let Γ be a bounded domain in Rⁿ. Assume that u ∈ C_loc^1,1(_Γ)∩ L_2sand u be lower semi-continuous onΓ, and satisfy¯

((−_∆)^su(x)≥0, x ∈_Γ, u(x)≥0, x ∈_Rⁿ\_Γ.

Then

u(x)≥0, x∈_Γ.

If u(x) =0at some point x∈ _Γ, then

u(x) =0 almost everywhere inRⁿ. Proof of Theorem1.6. DefineRⁿ+={x= (x₁, . . . ,x_n)|x_n>0}. Let

u^τ(x) =u(x⁰,x_n+τ) and U^τ(x) =u(x)−u^τ(x).

Outline of the proof: We will use the sliding method to prove the monotonicity and unique- ness of uand divide the proof into three steps.

In Step 1, we will show that forτsufficiently large, we haveU^τ(x)≤0, x ∈_Rⁿ. Especially, since u → µuniformly asx_n → +_{∞, for}δ >0, there exists a M₀ > 0 such that forx_n ≥ M₀, u∈[µ−_δ,µ]_and f is nondecreasing inu∈ [µ−_δ,µ]. Hence we will show that

U^τ(x)≤_0, x∈_Rⁿ_, ∀ τ≥ M₀. (3.10) This provides the starting point for the sliding method. Then in Step 2, we decreaseτcontin- uously as long as (3.10) holds to its limiting position. Define

τ0 :=inf{τ|U^τ(x)≤0, x ∈_Rⁿ, 0<τ< M0}. (3.11) We first will show that τ0 =0. Then we deduce that the solutionumust be strictly monotone increasing inx_n. In Step 3, we obtain that the solutionudepends onx_n. Finally we will prove the uniqueness.

Now we show the details in the three steps.

Step 1. Sinceu(x) =0,x ∈_Rⁿ\_Rⁿ₊, it yields that

U^τ(x)≤0, ∀x ∈_Rⁿ\_Rⁿ₊.

For τ≥ M₀, suppose (3.10) is violated, there exists a constant A>0 such that sup

x∈Rⁿ+

U^τ(x) = A, (3.12)

hence for someτ₁ ≥ M₀ there exists a sequence{x^k} ⊂_Rⁿ₊such that

U^τ1(x^k)→ A, ask→_∞. (3.13)

We will apply Lemma3.1to functionU^τ1(x)− ^A₂. Sinceτ₁ ≥ M₀, we haveu^τ1(x)∈ [µ−δ,µ]. Let

D=

x∈_Rⁿ|U^τ1(x)− ^A 2 >₀

.

Forx ∈D, we haveu(x)≥ u^τ1 ≥µ−δ. From equation (1.5),U^τ1(x)satisfied (−_∆)^sU^τ1(x) = f(u,∇u)− f(u^τ1,∇u^τ1)

:= −b_j(x)(U^τ1)_j(x)−c(x)U^τ1(x), wherec(x) =−^f(u,∇u_u⁾⁻_−u^f(τ1^uτ1,∇u) ≤0 by the monotonicity of f.

HenceU^τ1(x)− ^A₂ satisfies

((−_∆)^sU^τ1(x) +b_j(x)(U^τ1)_j(x) +c(x)(U^τ1(x)− ^A₂) =_0, x∈ D,

U^τ1(x)− ^A₂ ≤0, x∈ _Rⁿ\D.

By Lemma3.1, we derive

U^τ1(x)− ^A

2 ≤0, x ∈_Rⁿ,

which contradicts (3.13). Hence we obtain (3.10) and finish the proof of Step 1.

We also give an alternative proof which is an application of the general average inequality (Lemma3.3), and this idea can be applied to other problems.

For τ ≥ M₀, if (3.10) is violated, we have (3.13). Obviously,U^τ1(x) ≤ 0,x ∈ _∂Rⁿ₊. So by (3.13) we have x^k is away from ∂Rⁿ+, without loss of generality, assume dist(x^k,∂Rⁿ+) > 2.

Thus there exists 0< ε_k →_{0, ¯}x^k ∈B₁(x^k)_{such that} U^τ1(x¯^k) +ε_kψ_k(x¯^k) =max

Rⁿ (U^τ1(x) +ε_kψ_k(x))≥ A, whereψ_k(x¯^k)is as stated in (3.6). So∇(U^τ1(x¯^k) +ε_kψ_k(x¯^k)) =0 and

∇U^τ1(x¯^k)→0, ask→_∞. (3.14) Since

[U^τ1+ε_kψ_k](x¯^k)≥[U^τ1 +ε_kψ_k](x^k) andψ_k(x¯^k)≤ ψ_k(x^k), we obtain

U^τ1(x¯^k)≥U^τ1(x^k). (3.15) Hence forτ₁≥ M₀,

u(x¯^k)≥u^τ1(x¯^k)≥µ−_δ.

This meansu(x¯^k),u^τ1(x¯^k)are all in the nondecreasing interval of f. So f(u(x¯^k),∇u(x¯^k))− f(u^τ1(x¯^k),∇u^τ1(x¯^k))

= f(u,∇u(x¯^k))− f(u,∇u^τ1(x¯^k)) + f(u,∇u^τ1(x¯^k))− f(u^τ1,∇u^τ1(x¯^k))

≤ f(u(x¯^k),∇u(x¯^k))− f(u(x¯^k),∇u^τ1(x¯^k)).

(3.16)

Using Lemma3.3to the functionU^τ1+ε_kψ_k at ¯x^k, we obtain (U^τ1 +ε_kψ_k)(x¯^k)≤ C₁(−_∆)^s(U^τ1+ε_kψ_k)(x¯^k) +C₂

Z

B^c₂(x¯^k)

(U^τ1+ε_kψ_k)(y)

|x¯^k−y|^{n+2s dy.}

Letε_k →0, by the equation (1.5), it implies that U^τ1(x¯^k)≤C₁(−_∆)^sU^τ1(x¯^k) +C₂

Z

B^c₂(x¯^k)

U^τ1(y)

|x¯^k−y|^n+2sdy

=C₁[f(u,∇u(x¯^k))− f(u,∇u^τ1(x¯^k))] +C₂ Z

B^c₂(x¯^k)

U^τ1(y)

|x¯^k−y|^n+2sdy.

(3.17)

From (3.13) and (3.15), we have

U^τ1(x¯^k)→ A>0, ask →_∞. (3.18) Lettingk→∞, combining (3.14), (3.17) and (3.18), we arrive at

0< A←U^τ1(x¯^k)≤C₂ Z

B₂^c(x¯^k)

U^τ1(y)

|x¯^k−y|^n+2sdy, this is impossible because of (3.12) andU^τ1(y)≤0,y∈_Rⁿ\_Rⁿ₊.

Hence (3.10) is correct and we have finished the proof of Step 1.

Step 2. Firstly, we will check that

τ₀ =0, (3.19)

whereτ₀as defined in (3.11). In fact, suppose on the contraryτ₀>0, thenτ₀can be decreased a little bit. To be more rigorously, there exists a e > 0 such that for any τ ∈ (τ₀−e,τ₀], one has

U^τ(x)≤0, for any x∈_Rⁿ₊. (3.20) This is a contradiction with the definition of τ₀. Hence (3.19) is correct. In the sequel, we will prove (3.20).

To do so, we just need to prove sup

Rⁿ⁻¹×(0,M0+1]

U^τ(x)<0, ∀ τ∈(τ₀−e,τ₀] (3.21) and

sup

Rⁿ⁻¹×(M0+1,+_∞)

U^τ(x)≤0, ∀τ∈ (τ0−e,τ0]. (3.22) In order to prove (3.21) we need to show that

sup

Rⁿ⁻¹×(0,M0+1]

U^τ0(x)<0. (3.23)

If not, then

sup

Rⁿ⁻¹×(0,M0+1]

U^τ0(x) =_0.

So there exists a sequence{x^k} ⊂_Rⁿ⁻¹×(0,M₀+1]such that

U^τ0(x^k)→_{0, as}k →_{∞. (3.24)}

We first show that x^k is away from the boundary ∂Rⁿ+. Suppose thatz be a point on∂Rⁿ+. Denoterz :=dist(z+τ₀en,∂Rⁿ+), en= (0, . . . , 0, 1). For each fixedτ₀ >0, we have

x∈inf_∂Rⁿ₊dist(z+τ₀e_n,∂Rⁿ+):=r₀ >0.

For every pointzon ∂Rⁿ+, there exists a ballB_rz(z+τ₀e_n)⊂_Rⁿ₊with radius ofr_z centered at z+τ0en. For simplicity of notation, we useBinstead ofBrz(z+τ0en).

Let

E= {x∈_Rⁿ₊|dist(x,∂Rⁿ+)≥2}. We construct a subsolution

¯

u(x) =u_E(x) +εΦ(x), x ∈B, whereΦ(x) = (1− |x|²)^s₊,u_E :=u·χ_E andχ_E is define as

χ_E(x) =

(1, x∈ E, 0, x∈_Rⁿ\E.

By(−_∆)^s_Φ(x) =C[20], for x∈ Bit yields

(−_∆)^su¯(x) = (−_∆)^s(u_E+_εΦ)(x)

=ε(−_∆)^s_Φ(x) + (−_∆)^suE(x)

≤εC−ε₁C_n,s Z

E

1

|x−y|^n+2sdy

≤εC−ε₁CC_n,s.

We can choose ε ≤ ε₁Cn,sCC⁻¹ := ε₀ such that (−_∆)^su(x) ≤ 0, x ∈ B. Then fixing ε = ^ε₂⁰, combiningu(x)≥u(x), x∈ B^c and Lemma3.4, we derive

u^τ0(z) =u(z+τ₀en)≥u(z+τ₀en)≥ ^ε0

2Φ(z+τ₀en)≥Cτ₀ >0, ∀ z∈_∂Rⁿ₊. Then, we infer that

U^τ0(z) =u^τ0(z)>C_τ0 >0, ∀ z∈∂Rⁿ+. (3.25) By (3.24) and (3.25), we obtain that x^k is away from the boundary ∂Rⁿ+. Without loss of generality, we may assumeB₁(x^k)⊂ _Rⁿ₊. Similar to the argument as Lemma 3.1, let ψ(x) = η(x−x^k), whereηis as stated in (3.5),x^k satisfies dist(x^k,∂Rⁿ+)≥ 2 and B₁(x^k)⊂ _Rⁿ₊. Then there exists a sequenceε_k →0 such that

U^τ0(x^k) +ε_kψ(x^k)>_0.

Since forx ∈_Rⁿ₊\B₁(x^k), noting thatU^τ0(x)≤0 andψ(x) =0, we have U^τ0(x^k) +ε_kψ(x^k)>U^τ0(x) +ε_kψ(x), for any x∈_Rⁿ\B₁(x^k). Then there exists ¯x^k ∈ B₁(x^k)such that

U^τ0(x¯^k) +ε_kψ(x¯^k) =max

Rⁿ (U^τ0(x) +ε_kψ(x))>0. (3.26)

It can be seen from

U^τ0(x¯^k) +ε_kψ(x¯^k)≥U^τ0(x^k) +ε_kψ(x^k), andψ(x¯^k)≤ψ(x^k)that

0>_U^τ0(x¯^k)≥U^τ0(_x^k) +ε_kψ(_x^k)−ε_kψ(x¯^k)≥U^τ0(_x^k)→0, ask →_∞.

Hence

U^τ0(x¯^k)→0, ask →_∞. Since f is continuous, we have

f(u(x¯^k),∇u(x¯^k))− f(u^τ0(x¯^k),∇u^τ0(x¯^k))→0, ask →_∞. (3.27) On one hand, we have

(−_∆)^s(U^τ0 +ε_kψ)(x¯^k) = (−_∆)^sU^τ0(x¯^k) + (−_∆)^s(ε_kψ)(x¯^k)

= f(u(x¯^k)_,∇u(x¯^k))− f(u^τ0(x¯^k)_,∇u^τ0(x¯^k)) +ε_k(−_∆)^sψ(x¯^k)_. ^(3.28) On the other hand,

(−_∆)^s(U^τ0+ε_kψ)(x¯^k) =Cn,sP.V.

Z

Rⁿ

U^τ0(x¯^k) +ε_kψ(x¯^k)−U^τ0(y)−ε_kψ(y)

|x¯^k−y|^{n+2s dy}

≥C Z

B^c₂(x^k)

|U^τ0(y)|

|x^k−y|^n+2sdy

=C Z

B^c₂(0)

|U^τ0(z+x^k)|

|z|^{n+2s dz.}

(3.29)

Denote

u_k(x) =u(x+x^k) _and U_k^τ0(x) =U^τ0(x+x^k)_.

Since f is bounded, one can derive (see [10]) that u(x) is at least uniformly Hölder contin- uous, so u(x) is uniformly continuous, by the Arzelà–Ascoli theorem, up to extraction of a subsequence, one has

u_k(x)→u_∞(x), x∈_Rⁿ₊, ask →_∞. Combining (3.27), (3.28) and (3.29), lettingk→∞, we obtain

U_k^τ0(x)→0, x∈ B^c₂(0), uniformly, ask →_∞.

Therefore,

U_k^τ0(x)→u_∞(x)−u^τ_∞⁰(x)≡0, x∈ B^c₂(0). (3.30) Recall that u > 0 in Rⁿ+ while u(x) ≡ 0, x ∈ _Rⁿ\_Rⁿ₊. Since x^k ∈ _Rⁿ⁻¹×(0,M₀], there exists x⁰ such thatu_∞(x⁰) =0, then by (3.30),

0=u_∞(x⁰) =u^τ_∞⁰(x⁰) =u_∞((x⁰)⁰,x⁰_n+τ₀) =u^τ_∞⁰((x⁰)⁰,x⁰_n+τ₀)

=u_∞((x⁰)⁰_,x⁰_n+_2τ0) =· · · =u_∞((x⁰)⁰_,x⁰_n+kτ₀)_. ^(3.31) We obtain from (1.6) that

xnlim→+_∞u_∞(x) =µ>0, uniformly inx⁰ = (x₁, . . . ,x_n−1),