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arXiv:1511.03510v1 [math.AP] 11 Nov 2015

Existence and uniqueness of positive solutions for a class of

logistic type elliptic equations in RNinvolving fractional Laplacian

Alexander Quaas and Aliang Xia

Departamento de Matem´atica, Universidad T´ecnica Federico Santa Mar´ıa

Casilla: V-110, Avda. Espa˜na 1680, Valpara´ıso, Chile.

(alexander.quaas@usm.cl and aliangxia@gmail.com)

Abstract

In this paper, we study the existence and uniqueness of positive

solutions for the following nonlinear fractional elliptic equation:

(−∆)αu=λa(x)u−b(x)upin RN,

where α∈(0,1), N≥2, λ > 0, aand bare positive smooth function

in RNsatisfying

a(x)→a∞>0 and b(x)→b∞>0 as |x| → ∞.

Our proof is based on a comparison principle and existence, uniqueness

and asymptotic behaviors of various boundary blow-up solutions for

a class of elliptic equations involving the fractional Laplacian.

1 Introduction

A celebrated result of Du and Ma [10] asserts that the uniqueness positive

solution of

−∆u=λu −upin RN

for N≥1, λ > 0 and p > 1, is u≡λ1

p−1. Moreover, in [10], the authors also

consider the following logistic type equation:

−∆u=λa(x)u−b(x)upin RN,(1.1)

where p > 1, aand bare positive smooth function in RNsatisfying

a(x)→a∞>0 and b(x)→b∞>0 as |x| → ∞.

AMS Subject Classiﬁcations 2010: 35J60, 47G20.

Key words: Fractional Laplacian, comparison principle, blow-up solution, uniqueness.

1

Then they proved that problem (1.1) has a unique positive solution for each

λ > 0. A similar problem for quasi-linear operator has been studied by Du

and Guo [9].

In the present work, we are interested in understanding whether similar

results hold for equations involving a nonlocal diﬀusion operator, the simplest

of which is perhaps the fractional Laplacian. For α∈(0,1), we study the

following fractional elliptic problem:

(−∆)αu=λu −upin RN,(1.2)

where N≥2. The fractional Laplacian is deﬁned, up to a normalization

constant, by

(−∆)αu(x) = ZRN

2u(x)−u(x+y)−u(x−y)

|y|N+2αdy, ∀x∈RN.

Our ﬁrst main result is

Theorem 1.1 Let λ > 0. Suppose u∈C2α+β

loc (RN)∩L1(RN, ω)for some

β > 0and ω= 1/(1 + |y|N+2α)is a nonnegative solution of (1.2). Then u

must be a constant if pveriﬁes

1 + 2α < p < 1 + α

1−α.(1.3)

Remark 1.1 We notice that

N+ 2α

N−2α≤1 + α

1−α,

if N≥2.

As in [10] and [9], our proof of this result based on a comparison principle

for concave sublinear problems (see Lemma 2.1) and involves boundary blow-

up solutions. We use a rather intuitive squeezing method to proof Theorem

1.1 as follows. Denote BRas a ball centered at the origin with radius R.

Then problem

(−∆)αv=λv −vpin BR,

v= 0 in RN\BR,

has a unique positive solution vRif Ris large enough for any ﬁxed λ > 0.

On the other hand, the following boundary blow-up propblem

(−∆)αw=λw −wpin BR,

limx∈BR,x→∂BRw(x) = +∞,

w=gin RN\¯

BR,

(1.4)

2

for some g∈L1(RN\¯

BR, ω) and λ > 0, has a positive solution wRfor any

R > 0. The comparison principle implies that any entire positive solution

of (1.2) satisﬁes vR≤u≤wRin BR. Moreover, one can show (see Lemmas

2.2 and 2.3 in Section 2) that both vRand wRconverge locally uniformly to

λ1

p−1as R→+∞. Therefore, u≡λ1

p−1in RN.

Next, we make use of Theorem 1.1 to study logistic type fractional elliptic

problems with variable coeﬃcients that are asymptotically positive constants.

More precisely, we study the following problem

(−∆)αu=λa(x)u−b(x)upin RN,(1.5)

where aand bare positive smooth function in RN. Moreover, we suppose

that

a(x)→a∞>0 and b(x)→b∞>0 as |x| → ∞.(1.6)

We can prove that

Theorem 1.2 Let λ > 0. Suppose aand bare positive smooth function in

RNand satisfying (1.6). Then equation (1.5) has a unique positive solution

if pveriﬁes (1.3).

We prove Theorem 1.2 by a similar argument as in the proof of Theorem

1.2, we consider the Dirichlet problem and the boundary blow-up problem

in a ball BR. When Ris large, these problems have positive solutions vR

and wRrespectively. By comparison principle, as R→ ∞,vRincrease to

a minimal positive solution of (1.5) and wRdecrease to a maximal positive

solution of (1.5). Therefore, when (1.5) has a unique positive solution, vR

and wRapproximate this unique solution from below and above, respectively.

We mentioned that, in [10] and [9], the existence and uniqueness results

hold provided p > 1, but in our Theorems 1.1 and 1.2 we require psatisfying

(1.3). This is because we will use Perron’s method (we refer the reader to

User’s guide [6] for the presentation of Perron’s method which extends to the

case of nonlocal equations, see for example [3, 4, 11]) to construct solution

of problem 1.4 by applying Proposition 2.2 and choosing

τ=−2α

p−1∈(−1, τ0(α))

in Vτ(x) (see (2.13)). This implies

p < 1−2α

τ0(α).

3

Moreover, in [5], the authors proved that τ0(α) has a simplicity formula, that

is, τ0(α) = α−1. Thus, we have

p < 1−2α

τ0(α)=1 + α

1−α.

This article is organized as follows. In Section 2 we present some pre-

liminary lemmas to prove a comparison principle involving the fractional

Laplacian, existence and asymptotic behaviors of boundary blow-up solu-

tions. Section 3 is devoted to prove the existence and uniqueness results of

problems (1.2) and (1.5), i.e., Theorems 1.1 and 1.2.

2 Preliminary lemmas

In this section, we introduce some lemmas which are useful in the proof of

our main results. The ﬁrst important ingredient is the comparison principle

involving the fractional Laplacian which is useful in dealing with boundary

blow-up problems.

Lemma 2.1 (Comparison principle) Suppose that Ωis a bounded domain

in RN,a(x)and b(x)are continuous functions in Ωwith kakL∞(Ω) <∞and

b(x)nonnegative and not identity zero. Suppose u1, u2∈C2α+β(Ω) for some

β > 0are positive in Ωand satisfy

(−∆)αu1−a(x)u1+b(x)up

1≥0≥(−∆)αu2−a(x)u2+b(x)up

2in Ω (2.1)

and lim supx→∂Ω(u2−u1)≤0with u2−u1≤0in RN\¯

Ω, where p > 1. Then

u2≤u1in Ω.

In order to prove Lemma 2.1, we need the following proposition.

Proposition 2.1 For u≥0and v > 0, we have

L(u, v)≥0in RN×RN,

where

L(u, v)(x, y) = (u(x)−u(y))2−(v(y)−v(x)) u(y)2

v(y)−u(x)2

v(x).

Moreover, the equality holds if and only if u=kv a.e. for some contant k.

4

We note that Proposition 2.1 is a special case (p= 2) of Lemma 4.6 in

[13] and we omit the proof here.

Proof of Lemma 2.1. Let φ1and φ2be nonnegative functions in C∞

0(Ω).

By (2.1), we obtain that

ZR2N

(u1(x)−u1(y)) (φ1(x)−φ1(y))

|x−y|N+2α−(u2(x)−u2(y)) (φ2(x)−φ2(y))

|x−y|N+2αdxdy

≥ZΩ

b(x)[up

2φ2−up

1φ1]dx +ZΩ

a(x)(u1φ1−u2φ2)dx.(2.2)

For ε > 0, we denote ε1=εand ε2=ε/2 and let

vi=[(u2+ε2)2−(u1+ε1)2]+

ui+εi

, i = 1,2.

By our our assumption, viis zero near ∂Ω and in RN\¯

Ω. Hence vi∈Xα

0(D0),

where D0⊂⊂ Ω and Xα

0(D0) = {w∈Hα(RN) : w= 0 a.e in RN\D0}. In

fact, it is clear that kv1kL2(RN)=kv1kL2(D0)≤Cand thus it remains to verify

that the Gagliardo norm of v1in RNis bounded by a constant. Using the

symmetry of the integral in the Gagliardo norm with respect to xand yand

the fact that v1= 0 in RN\Ω, we can split as follows

ZRNZRN

|v1(x)−v1(y)|2

|x−y|N+2αdxdy =ZΩZΩ

|v1(x)−v1(y)|2

|x−y|N+2αdxdy

+ 2 ZΩZRN\Ω

|v1(x)|2

|x−y|N+2αdydx.(2.3)

Next, we estimate both integrals in the right hand side of (2.3) is ﬁnite. We

ﬁrst notice that, for any y∈RN\D0,

|v1(x)|2

|x−y|N+2α=χD0(x)|v1(x)|2

|x−y|N+2α≤χD0(x)|v1(x)|2sup

x∈D0

1

|x−y|N+2α.

This implies that

ZΩZRN\Ω

|v1(x)|2

|x−y|N+2αdydx ≤ZRN\Ω

1

dist(y, ∂D0)N+2αdykv1k2

L2(D0)<+∞

since dist(∂Ω, ∂D0)≥γ > 0 and N+2α > N. Hence, the second term in the

right hand side of (2.3) is ﬁnite by the above inequality. In order to show the

5

ﬁrst term in the right hand side of (2.3) is also ﬁnite, we need the following

estimates

(u2(x) + ε2)2−(u1(x) + ε1)2

u1(x) + ε1

−(u2(y) + ε2)2−(u1(y) + ε1)2

u1(y) + ε1

=

(u2(x) + ε2)2

u1(x) + ε1

−(u2(y) + ε2)2

u1(y) + ε1

+ (u1(y)−u1(x))

≤

(u2(x) + ε2)2

u1(x) + ε1

−(u2(y) + ε2)2

u1(y) + ε1

+|u1(y)−u1(x)|(2.4)

and

(u2(x) + ε2)2

u1(x) + ε1

−(u2(y) + ε2)2

u1(y) + ε1

=

(u2(x) + ε2)2−(u2(y) + ε2)2

u1(x) + ε1

+(u2(y) + ε2)2(u1(y)−u1(x))

(u1(x) + ε1)(u1(y) + ε1)

≤u2(x) + u2(y) + 2ε2

u1(x) + ε1

|u2(x)−u2(y)|+(u2(x) + ε2)2

(u1(x) + ε1)(u1(y) + ε1)|u1(y)−u1(x)|.(2.5)

Combining (2.4) and (2.5), we have

(u2(x) + ε2)2−(u1(x) + ε1)2

u1(x) + ε1

−(u2(y) + ε2)2−(u1(y) + ε1)2

u1(y) + ε1

≤C(ε1, ε2,ku1kL∞(Ω),ku2kL∞(Ω) )(|u1(y)−u1(x)| − |u2(y)−u2(x)|)

≤˜

C|x−y|2α+β.

In the last inequality of above estimate, we have used the fact u1, u2∈

C2α+β(Ω). This implies

ZΩZΩ

|v1(x)−v1(y)|2

|x−y|N+2αdxdy < +∞

since the following inequality

w+ (x)−w+(y)≤ |w(x)−w(y)|

for all (x, y)∈RN×RNand function w:RN→R. Therefore, v1∈Xα

0(D0).

Similarly, we can show v2∈Xα

0(D0). On the other hand, by Theorem

6 in [12], we know that vican be approximate arbitrarily closely in the

Xα

0(D0) norm by C∞

0(D0) functions. Hence, we see that (2.2) holds when φi

is replaced by vifor i= 1,2.

Denote

D(ε) = {x∈Ω : u2(x) + ε2> u1(x) + ε1}.

6

We notice that the integrands in the right hand side of (2.2) (with φi=

vi) vanishing outside D(ε). Next, we prove the left hand side of (2.2) in

nonpositive. We ﬁrst divide R2Ninto four disjoint region as:

R2N=RN\D(ε)×RN\D(ε)∪D(ε)×RN\D(ε)

∪RN\D(ε)×D(ε)∪[D(ε)×D(ε)] .

For (x, y)∈RN\D(ε)×RN\D(ε), we know that vi(x) = vi(y) = 0, i= 1,2.

Therefore,

A1:= ZRN\D(ε)ZRN\D(ε)

L(u1, u2)

|x−y|N+2αdxdy = 0,

where

L(u1, u2) = (u1(x)−u1(y)) (v1(x)−v1(y))−(u2(x)−u2(y)) (v2(x)−v2(y)) .

For (x, y)∈D(ε)×RN\D(ε), we notice that v1(y) = v2(y) = 0 and, by the

deﬁnition of D(ε),

u2(x) + ε2> u1(x) + ε1and u2(y) + ε2≤u1(y) + ε1.(2.6)

It follows that

L(u1, u2) = [u1(x)−u1(y)] v1(x)−[u2(x)−u2(y)] v2(x)

= [(u1(x) + ε1)−(u1(y) + ε1)] v1(x)−[(u2(x) + ε2)−(u2(y) + ε2)] v2(x)

=[(u2(x)+ε2)2−(u1(x)+ε1)2]

(u1(x)+ε1)(u2(x)+ε2)·[(u1(x) + ε1)(u2(y) + ε2)−(u1(y) + ε1)(u2(x) + ε2)]

≤0.

Hence,

A2=ZD(ε)ZRN\D(ε)

L(u1, u2)

|x−y|N+2αdydx ≤0.

A similar argument implies that

A3=ZRN\D(ε)ZD(ε)

L(u1, u2)

|x−y|N+2αdydx ≤0.

Finally, if (x, y)∈D(ε)×D(ε), it is easy to check that

L(u1, u2) = (u1(x)−u1(y)) (v1(x)−v1(y)) −(u2(x)−u2(y)) (v2(x)−v2(y))

=−(u1(x)−u1(y))2+ (u1(y)−u1(x)) (u2(y) + ε2)2

u1(y) + ε1

−(u2(x) + ε2)2

u1(x) + ε1

−(u2(x)−u2(y))2+ (u2(y)−u2(x)) (u1(y) + ε1)2

u2(y) + ε2

−(u1(x) + ε1)2

u2(x) + ε2.

7

By Proposition 2.1, we know that L(u1, u2)(x, y)≤0 in D(ε)×D(ε). There-

fore,

A4=ZD(ε)ZD(ε)

L(u1, u2)

|x−y|N+2αdxdy ≤0.

Summing up these estimates from A1to A4, we know that the left hand side

of (2.2) is nonpositive.

On the other hand, as ε→0, the ﬁrst term in the right hand side of (2.2)

converges to ZD(0)

b(x)up−1

2−up−1

1(u2

2−u2

1)dx,

while the last term in the right side of (2.2) converges to 0.

Next, we show that D(0) = ∅. Suppose to the contrary that D(0) 6=∅.

Since the left side of (2.2) is nonpositive by the estimates from A1to A4and

right hand side of (2.2) tends to 0 as ε→0, we easy deduce

ZR2N

L(u1, u2)

|x−y|N+2αdxdy = 0,

where L(u1, u2) = limε→0L(u1, u2) and

ZD(0)

b(x)up−1

2−up−1

1(u2

2−u2

1)dx = 0.

This imply that

b≡0 in D(0)

and

L(u1, u2)≡0 in RN×RN.

Hence, by Proposition 2.1, we know u1=ku2in D(0) for some constant k.

Since b6≡ 0 in Ω, it follows from the above that D(0) 6= Ω. Thus, D(0) ⊂Ω,

∂D(0) ∩Ω6=∅. It follows that the open set D(0) has connected component

Gsuch that ∂G ∩ Ω6=∅. Now on G,u1=ku2. On the other hand, we

have u1|∂G∩Ω=u2|∂G∩Ω>0. Thus, k= 1. So we have u1=u2in G, which

contradicts G ⊂ D(0). Therefore, we must have D(0) = ∅and thus u1≥u2

in Ω. We complete the proof of Lemma 2.1.

By applying this comparison principle together with the Perron’s method

for the nonlocal equation, we can obtain the following two lemmas.

Lemma 2.2 Let Ωbe a bounded domain in RNwith smooth boundary and

p > 1. Suppose aand bare smooth positive functions in ¯

Ω, and let µ1denote

8

the ﬁrst eigenvalue of (−∆)αu=µa(x)uin Ωwith u= 0 in RN\Ω. Then

equation

(−∆)αu=µu[a(x)−b(x)up−1] in Ω,

u= 0 in RN\Ω

has a unique positive solution for every µ > µ1. Furthermore, the unique

solution uµsatisﬁes uµ→[a(x)/b(x)]1/p−1uniformly in amy compact subset

of Ωas µ→+∞.

Proof. (Existence) The existence follows from a simple sub- and super-

solution argument. In fact, any constant great than or equal to M=

max¯

Ω[a(x)/b(x)]1/(p−1) is a super-solution. Let φbe a positive eigenfunction

corresponding to µ1(for the existence of the ﬁrst eigenvalue and correspond-

ing eigenfunction has been obtained in [13] and [15]), then for each ﬁxed

µ > µ1and small positive ε,εφ < M and is a sub-solution. Therefore, by

the sub- and super-solution method (see [14]), there exist at least one positive

solution.

(Uniqueness) If u1and u2are two positive solutions, by Lemma 2.1, we

have u1≤u2and u2≤u1both hold in Ω. Hence, u1=u2. This proves the

uniqueness.

(Asymptotic behaviour) Given any compact subset Kof Ω and any small

ε > 0 such that ε < v0(x) = [a(x)/b(x)]1/(p−1) in Ω. Let

vε(x) =

v0(x) + εin K,

l(x) in Ω \K,

0,in RN\Ω,

where l(x) is nonnegative function such that vεis smooth in Ω and satisfying

D0:= supp(vε)⊂⊂ Ω. Thus, for any x∈Ω,

|(−∆)αvε(x)| ≤ ZRN

|vε(x)−vε(y)|

|x−y|N+2αdy

=ZΩ

|vε(x)−vε(y)|

|x−y|N+2αdy +ZRN\Ω

|vε(x)|

|x−y|N+2αdy

≤ZΩ

|vε(x)−vε(y)|

|x−y|N+2αdy +ZRN\Ω

1

dist(y, ∂D0)N+2αdykvεkL∞(RN)

≤C

for some positive constant C=C(ε) since vεis smooth and dist(∂Ω, ∂D0)≥

γ > 0. On the other hand, we notice that vε(a(x)−b(x)vp−1

ε)≤ −δin Ω for

9

some positive constant δ=δ(ε). Hence, for all large µ,vεis a super-solution

of our problem.

On the other hand, let φbe a positive eigenfunction corresponding to µ1.

Then we can ﬁnd a small neighborhood of ∂Ω in Ω, say U, such that φis

very small in U. Therefore, for all µ > µ1+ 1, we have

(−∆)αφ=µ1a(x)φ≤µφ[a(x)−b(x)φp−1] in U. (2.7)

By shrinking Ufurther if necessary, we can assume that ¯

U∩K=∅and

φ < v0−εin U. Next, we choose smooth function wεas

wε(x) =

v0(x)−εin K,

φ(x) in U,

l(x) in the rest of Ω,

0,in RN\Ω,

where lis a positive function such that wεis smooth in Ω and satisfying

l≤v0−ε/2. Moreover, we let

φ(x)≤wε(x) in Ω (2.8)

otherwise we choose ˜

φ=φ/C for some constant C > 0 large replace φ. Then

we can see that, for x∈Ω\U,

|(−∆)αwε(x)| ≤ ZRN

|wε(x)−wε(y)|

|x−y|N+2αdy

=ZΩ

|wε(x)−wε(y)|

|x−y|N+2αdy +ZRN\Ω

|wε(x)−wε(y)|

|x−y|N+2αdy

=ZΩ

|wε(x)−wε(y)|

|x−y|N+2αdy +ZRN\Ω

|wε(x)|

|x−y|N+2αdy

=ZΩ

|wε(x)−wε(y)|

|x−y|N+2αdy +ZRN\Ω

1

dist(y, ∂U ∩Ω)N+2αdykwεkL∞(RN)

≤C,

for some positive constant C=C(ε) since dist(∂Ω, ∂U ∩Ω) ≥γ > 0. More-

over, we know wε(a(x)−b(x)wp−1

ε)≥δin Ω \Ufor some positive constant

δ=δ(ε). Therefore,

(−∆)αwε≤µwε(a(x)−b(x)wp−1

ε) in Ω \U(2.9)

for all large µ. For x∈U, by (2.7) and (3.7), we have

(−∆)αwε(x) = ZΩ

wε(x)−wε(y)

|x−y|N+2αdy +ZRN\Ω

wε(x)

|x−y|N+2αdy

10

=ZΩ

φ(x)−wε(y)

|x−y|N+2αdy +ZRN\Ω

φ(x)

|x−y|N+2αdy

≤ZΩ

φ(x)−φ(y)

|x−y|N+2αdy +ZRN\Ω

φ(x)

|x−y|N+2αdy

= (−∆)αφ(x)

≤µφ[a(x)−b(x)φp−1]

=µwε[a(x)−b(x)wp−1

ε],(2.10)

for µ > µ1+ 1. Finally, combining (2.9) and (2.10), we know wεis a sub-

solution of our problem for all large µ.

Since wε< vε, we deduce that wε≤uµ< vεin Ω. In particular,

[a(x)/b(x)]1/(p−1) −ε≤uµ≤[a(x)/b(x)]1/(p−1) +ε

in Kfor all large µ. Hence, uµ→[a(x)/b(x)]1/p−1as µ→+∞in K, as

required.

Lemma 2.3 Let Ω,aand bbe as in Lemma 2.2. Suppose pveriﬁes (1.3),

then equation

(−∆)αu=µu[a(x)−b(x)up−1] in Ω,

limx∈Ω,x→∂Ωu= +∞,

u=gµin RN\Ω

(2.11)

has at least one positive solution for each µ > 0if the measurable function

gµsatisfying

ZRN\Ω

gµ(y)

1 + |y|N+2αdy ≤C, (2.12)

where positive constant Cis independent of µ. Furthermore, suppose uµis a

positive solution of (2.11), then uµsatisﬁes uµ→[a(x)/b(x)]1/(p−1) uniformly

in amny compact subset of Ωas µ→+∞.

We ﬁrst recall the following result in [4]. Assume that δ > 0 such that

the distance function d(x) = dist(x, ∂Ω) is of C2in Aδ={x∈Ω : d(x)< δ}

and deﬁne

Vτ(x) =

l(x), x ∈Ω\Aδ,

d(x)τ, x ∈Aδ,

0, x ∈RN\Ω,

(2.13)

where τis a parameter in (−1,0) and the function lis positive such that Vτ

is C2in Ω.

11

Proposition 2.2 ([4], Proposition 3.2) Assume that Ωis a bounded, open

subset of RNwith a C2boundary. Then there exists δ1∈(0, δ)and s constant

C > 1shch that if τ∈(−1, τ0(α)) where τ0(α)is the unique solution of

C(τ) = Z+∞

0

χ(0,1)|1−t|τ+ (1 + t)τ−2

t1+2αdt

for τ∈(−1,0) and χ(0,1) is the characteristic function of the interval (0,1),

then

1

Cd(x)τ−2α≤ −(−∆)αVτ(x)≤Cd(x)τ−2α, f or all x ∈Aδ1.

Next, we will the existence result in Lemma 2.3 by applying Perrod’s

method and thus we need to ﬁnd ordered sub and super-solution of (2.11).

As in [4], we begin with a simple lemma that reduce the problem to ﬁnd

them only in Aδ.

Lemma 2.4 Let Ω,aand bbe as in Lemma 2.2. Suppose Uand Ware

order super and sub-solution of (2.11) in the sub-domain Aδ. Then there

exists λlarge such that Uλ=U+λη and Wλ=W−λη are ordered super

and sub-solution of (2.11), where η∈C∞

0(RN)satisfying 0≤η≤1and

supp(η)⊂Ω\Aδ.

Proof. The proof is similar as Lemma 4.1in [4] and we just need replace

¯

Vin Lemma 4.1in [4] to ηfor our lemma. So we omit the proof here.

Now we in position to prove Lemma 2.3.

Proof of Lemma 2.3. (Existence) We deﬁne

Gµ(x) = 1

2ZRN

˜gµ(x+y)

|y|N+2αdy for x∈Ω,

where

˜gµ(x) = 0, x ∈Ω,

gµ(x), x ∈RN\Ω.

We observe that

G(x) = −(−∆)α˜gµ(x) for x∈Ω.

Moreover, we know that Gµis continuous (see Lemma 2.1 in [4]) and nonneg-

ative in Ω. Therefore, if uis a solution of (2.11), then u−˜gµis the solution

of

(−∆)αu=µu[a(x)−b(x)up−1] + Gµ(x) in Ω,

limx∈Ω,x→∂Ωu= +∞,

u= 0 in RN\Ω

(2.14)

12

and vice versa, if uis a solution of (2.14), then u+ ˜gµis a solution of (2.11).

Next, we will look for solution of (2.14) instead of (2.11).

Deﬁne

Uλ(x) = λVτ(x) and Wλ(x) = λWτ(x),

where τ=−2α/(p−1). Notice that τ=−2α/(p−1) ∈(−1, α −1),

τp =τ−2αand τp < τ < 0.

By Proposition 2.2, we ﬁnd that for x∈Aδand δ > 0 small

(−∆)αUλ+µb(x)Up

λ−µa(x)Uλ−Gµ(x)

≥ −Cλd(x)τ−2α+µb(x)λpd(x)τp −µa(x)λd(x)τ−Gµ(x)

≥ −Cλd(x)τ−2α+µb(x)λpd(x)τp −µa(x)λd(x)τ p −Gµ(x),

for some C > 0. Then there exists a large λ > 0 such that Uλis a super-

solution of (2.14) with the ﬁrst equation in Aδsince Gµis continuous in Ω.

Similarly, by Proposition 2.2, we have that for x∈Aδand δ > 0 small

(−∆)αWλ+µb(x)Wp

λ−µa(x)Wλ−Gµ(x)

≤ − λ

Cd(x)τ−2α+µb(x)λpd(x)τp −µa(x)λd(x)τ−Gµ(x)

≤ − λ

Cd(x)τ−2α+µb(x)λpd(x)τp

≤0,

if λ > 0 small. Here we have used the fact Gµis nonnegative.

Finally, by using Lemma 2.4, there exists a solution ˜uµof problem (2.14)

and thus a solution uµ= ˜uµ+ ˜gµis a solution of (2.11). Moreover, uµ>0 in

Ω. Indeed, since 0 is a sub-solution of (2.11), by Lemma 2.1, we have uµ≥0

in Ω. If uµ(x′) = 0 for some points x′∈Ω and uµ6≡ 0 in RN, then by the

deﬁnition of fractional Laplacian (−∆)αuµ(x′)<0 which is a contradiction.

Therefore, uµ>0 in Ω.

(Asymptotic behaviour) Let Kbe an arbitrary compact subset of Ω,

v0(x) = [a(x)/b(x)]1/(p−1) in Ω and ε > 0 any small positive number satisﬁes

v0> ε in ¯

Ω. Deﬁne

˜wε(x) =

v0(x)−ε+λη(x) in K,

µ−1d(x)τin Aδ,

l(x) in the rest of Ω,

0 in RN\Ω,

where τis a parameter in (−1,0), λand ηdeﬁned as in Lemma 2.4 and

the function lis positive such that wεis C2in Ω. By a similar argument as

13

Proposition 3.2 in [4], there exists δ1∈(0, δ) and constants c > 0 and C > 0

such that

c(1 + µ−1d(x)τ−2α)≤ −(−∆)α˜wε(x)≤C(1 + µ−1d(x)τ−2α)

for all x∈Aδ1and τ∈(−1, α −1). Hence, for x∈Aδand δ > 0 small,

(−∆)α˜wε+µb(x) ˜wp

ε−µa(x) ˜wε−Gµ(x)

≤ −cµ−1d(x)τ−2α+µ1−pb(x)d(x)τp −a(x)λd(x)τ−Gµ(x)−c

≤ −cµ−1d(x)τ−2α+µ1−pb(x)d(x)τp

≤0,

if µis large enough. Hence, ˜wεis sub-solution in Aδ. By applying Lemma

2.4, we know that wε= ˜wε−λη + ˜gµis a sub-solution of problem (2.11) for

all large µ > 0.

On the other hand, we deﬁne choose a function

˜vε(x) =

v0(x) + ε−λη in K,

µd(x)τin Aδ,

l(x) in the rest of Ω,

0,in RN\Ω,

where τis a parameter in (−1,0), λand ηdeﬁned as in Lemma 2.4 and

the function lis positive such that vεis C2in Ω. By a similar argument as

Proposition 3.2 in [4], there exists δ1∈(0, δ) and constants c > 0 and C > 0

such that

c(1 + µd(x)τ−2α)≤ −(−∆)α˜vε(x)≤C(1 + µd(x)τ−2α)

for all x∈Aδ1and τ∈(−1, α −1). Hence, for x∈Aδand δ > 0 small,

(−∆)α˜vε+µb(x)˜vp

ε−µa(x)˜vε−Gµ(x)

≥ −Cµd(x)τ−2α+µp+1b(x)d(x)τp −µ2a(x)λd(x)τ−Gµ(x)−C

≥ −Cµd(x)τ−2α+µp+1b(x)d(x)τp −µ2a(x)λd(x)τp −Gµ(x)−C

≥0,

if µis large enough since kGµkL∞(¯

Ω) ≤Cfor some constant C > 0 indepen-

dent of µby (2.14). Hence, ˜vεis sub-solution in Aδ. By applying Lemma

2.4, we know that vε= ˜vε+λη + ˜gµis a super-solution of problem (2.11) for

all large µ > 0.

As wε< vεin Ω, we must have wε≤uµ≤vεin Ω. This implies that

uµ→v0in Kas µ→ ∞, as required. We complete the proof of this Lemma.

14

3 Proofs

The main purpose of this section is to prove Theorems 1.1 and 1.2.

Proof of Theorem 1.1. Let us ﬁrst observe that a nonnegative entire

solution of (1.2) is either identically zero or positive everywhere. Indeed, if

u(x′) = 0 for some points x′∈RNand u6≡ 0 in RN, then by the deﬁnition

of fractional Laplacian (−∆)αu(x′)<0 which is a contradiction. Therefore,

we only consider positive solution.

Suppose λ > 0 and let ube an arbitrary positive entire solution of (1.2).

We will show that u(x0) = λ1/(p−1) for any point x0∈RNby using pointwise

convergence of Lemmas 2.2 and 2.3.

For any t > 0, deﬁne

ut(x) = u[x0+t(x−x0)].

Then utsatisﬁes

(−∆)αu=t2α(λu −up) in RN.

Let Bdenote the unit ball with center x0. By Lemma 2.2, for all large t,

the problem

(−∆)αv=t2αu(λ−up−1) in B,

u= 0 in RN\B

has a unique positive solution vtand vt→λ1/(p−1) as t→ ∞ at x=x0∈B.

By Lemma 2.1, we have that ut≥vtin Band thus

u(x0) = ut(x0)≥vt(x0).

Letting t→ ∞ in the above inequality we conclude that u(x0)≥λ1/(p−1).

Let wtbe a positive solution of

(−∆)αw=t2αw(λ−wp−1) in B,

limx∈B,x→∂B w= +∞,

w=utin RN\¯

B.

By our assumption, we know that

ZRN

ut(x)

1 + |x|N+2αdx ≤C, (3.1)

where constant C > 0 independent of tfor tlarge enough. In fact

ZRN

ut(x)

1 + |x|N+2αdx =ZRN

u(x0+t(x−x0))

1 + |x|N+2αdx

=ZRN

u(x)

tN1 +

x+(t−1)x0

t

N+2αdx. (3.2)

15

Deﬁne function

f(t) = tN 1 +

x+ (t−1)x0

t

N+2α!,

we know f(1) = 1 + |x|N+2αand f(t)→+∞as t→+∞. Then, we can

choose tlarge enough such that f(t)≥f(1). So by (3.2), for tlarge enough,

we have ZRN

ut(x)

1 + |x|N+2αdx ≤ZRN

u(x)

1 + |x|N+2αdx ≤C,

since u∈L1(RN, ω).

Then, applying Lemma 2.3, we see that wt→λ1/(p−1) as t→ ∞ at

x=x0∈B. Applying Lemma 2.1, we have that ut≤wtin Band thus

u(x0) = ut(x0)≤wt(x0).

Letting t→ ∞ in the above inequality we conclude that u(x0)≤λ1/(p−1) .

Therefore, u(x0) = λ1/(p−1). Since x0is arbitrary, we conclude that u≡

λ1/(p−1) in RNfor λ > 0, the unique constant solution of (1.2).

Next, we will extend Theorem 1.1 to similar problem with variable coef-

ﬁcients, that is, Theorem 1.2. We ﬁrst consider the following equation which

is more general than (1.5):

(−∆)αu=a(x)u−b(x)up, x ∈RN,(3.3)

where a(x) and b(x) are continuous functions in RNand satisfying

lim

|x|→∞ a(x) = a∞>0,lim

|x|→∞ b(x) = b∞>0.(3.4)

Here we allow aand bcan be change sign which is more general than (1.5).

Theorem 3.1 Under the above assumptions, if u∈C2α+β

loc (RN)∩L1(RN, ω)

for some β > 0is a positive solution of (3.3) with pveriﬁes (1.3), then

lim

|x|→∞ u(x) = a∞

b∞1

p−1

.

We postpone the proof of Theorem 3.1 and ﬁrst we use it to prove the

following result.

Corollary 3.1 Under the assumptions in Theorem 3.1, if we further assume

that bis a nonnegative, then problem (3.3) has at most one positive solution.

16

Proof. Suppose u1and u2are two positive solutions of (3.3). By Theorem

3.1, we have

lim

|x|→∞[(1 + ε)u1−u2] = ε(a∞/b∞)1/(p−1) >0

for any positive constant ε.

Since bis nonnegative, then (1 + ε)u1is a super solution of (3.3). There-

fore, applying Lemma 2.1 in a large ball to conclude that (1 + ε)u1≥u2in

a large ball. It follows that this is true in all of RN. Hence, u1≥u2in RN

since εis arbitrary. Similarly, we also can deduce u2≥u1in RN. Finally, we

must have u1=u2in RN, that is, (3.3) has at most one positive solution.

Now we are in the position to prove Theorem 3.1.

Proof of Theorem 3.1. We prove it by a contradiction argument.

Assume that there exists a sequence points xn∈RNsatisfying |xn| → ∞

such that |u(xn)−(a∞/b∞)1/(p−1)| ≥ ε0for some constant ε0>0.

We deﬁne

an(x) = a(xn+x), bn(x) = b(xn+x) and un(x) = u(xn+x).

Then unsatisﬁes

(−∆)αun=an(x)un−bn(x)up

nin RN.(3.5)

If we let

Ln

αu=−(−∆)αu−(a∞−an)u,

then we can rewrite (3.5) as

−Ln

αun=a∞un−bn(x)up

nin RN.

Next, we ﬁx a ball Br={x∈RN| |x|< r}and consider the following

problem

−Ln

αw=a∞w−bnwpin Br,

w= 0 in RN\Br.(3.6)

By using the variational characterization of the ﬁrst eigenvalue and (3.4),

we see that λ1(−Ln

α, Br)→λ1((−∆)α, Br) as n→ ∞, where λ1(−Ln

α, Br),

λ1((−∆)α, Br) denote the ﬁrst eigenvalues of −Ln

αand (−∆)αin Brwith

Dirichlet boundary conditions in RN\Br, respectively. Since we can choose

rlarge enough such that λ1((−∆)α, Br)< a∞, then we may assume that

λ1(−Ln

α, Br)< a∞for all n. On the other hand, we know that bn→b∞

17

uniformly in Brand thus we may also assume that bn≥b∞/2 in Brfor all

n.

Let φn∈Xα

0(Br) be the ﬁrst eigenfunction corresponding to λ1(−Ln

α, Br),

that is,

(−∆)αφn+ (a∞−an)φn=λ1(−Ln

α, Br)φnin Br,

φn= 0 in RN\Br,(3.7)

with kφnkL∞(Br)= 1. By Theorems 1 and 2 in [16] and using (3.4), we know

φnis also a viscosity solution of (3.7). Then by Theorem 2.6 in [8], we have

φn∈Cβ

loc(Br). Then, by Corollary 4.6 in [7], φnconverges uniformly to a φ∞

and φ∞satisﬁes

(−∆)αφ∞=λ1((−∆)α, Br)φ∞in Br,

φ∞= 0 in RN\Br.

in viscosity sense. Next, by a similar argument as Theorem 2.1 in [4], we

know φ∞∈C2α+β

loc (Br) and is a classical solution. Then φ∞is the normalized

positive eigenfunction corresponding to λ1((−∆)α, Br).

It is easily to check that εφnis a subsolution of (3.6) for every nif we

choose εsmall enough. Furthermore, (2a∞/b∞)1/(p−1) is a supersolution of

(3.6) for all n. Then (3.6) has a positive solution wnsatisﬁes εφn≤wn≤

(2a∞/b∞)1/(p−1). Then, using the regularity results again, we know wncon-

verges in C2α+β

loc (Br) to some function wsatisfying εφ∞≤v≤(2a∞/b∞)1/(p−1)

and

(−∆)αw=a∞w−b∞wpin Br,

w= 0 in RN\Br.

Applying Lemma 2.2, we know the above problem has a unique positive

solution. Therefore, w=wris uniquely determined and the whole sequence

wnconverges to wr.

By the comparison principle (see Lemma 2.1), we know that

un≥wn→wrin Br.(3.8)

Next, we show unhas a uniformly bounded in RNfor all nlarge enough, that

is, there exists a positive constant Cindependent of nsuch that un(x0)≤C

for any x0∈RN. We deﬁne, for any t > 0,

ut,n(x) = un[x0+t(x−x0)].

Then ut,n satisfying

(−∆)αu=t2α(˜anu−˜

bnup) in RN,

18

where ˜an(x) = an(x0+t(x−x0)) and ˜

bn(x) = bn(x0+t(x−x0)). On the

other hand, since ˜an→a∞and ˜

bn→b∞uniformly in Bwhere Bdenote the

unit ball with center x0, we may assume ˜an≤2a∞and ˜

bn≥b∞/2 in Bfor

all n.

We consider the following problem

(−∆)αv=t2α(2a∞v−(b∞/2)vp) in B,

limx∈B,x→∂B v= +∞,

v=ut,n in RN\B.

(3.9)

As a argument before, we know ut,n ∈L1(RN, ω) for tand nlarge enough.

Thus, by applying Lemma 2.3, we know this problem has at least one positive

solution. Let vtis a solution of (3.9), then vt→(4a∞/b∞)1/(p−1) as t→ ∞

at x=x0∈B. Then the comparison principle deduce that ut,n ≤vtin B

and thus

un(x0) = ut,n(x0)≤vt(x0).

Letting t→ ∞ in the above inequality we conclude that un(x0)≤(4a∞/b∞)1/(p−1)

as we required.

Hence, kunkL∞(RN)≤Cfor all nlarge enough, where constant C > 0

independent of n. On the other hand, un∈C2α+β

loc (RN) implies that un

converges uniformly to some function u∞and

(−∆)αun→(−∆)αu∞in Br

is strongly as n→+∞. Hence, u∞is nonnegative and satisﬁes

(−∆)αu=a∞u−b∞upin Br.

Furthermore, u∞≥wr>0. Thus u∞is a positive solution and |u∞(0) −

(a∞/b∞)1/(p−1)| ≥ ε0due to the choice of xn.

Choose a sequence r=r1≤r2≤ · · · ≤ rm→ ∞ as m→ ∞. We can

apply the above argument to each rmand then use a diagonal process to

obtain a positive solution Uof

(−∆)αu=a∞u−b∞upin RN,(3.10)

which satisﬁes U(0) ≥wrm(0) and |U(0)−(a∞/b∞)1/(p−1)| ≥ ε0. By changing

of variables of the form x=θy,θ∈R, then (3.10) can write as

(−∆)αv= (a∞/b∞)v−vpin RN,

where v(y) = u(x) = u(θy). In fact, we can choose θ= (b∞)−1/(2α). Then

applying Theorem 1.1 to the above equation, we have v≡(a∞/b∞)1/(p−1).

19

Hence, u≡(a∞/b∞)1/(p−1). This a contradiction. We complete the proof.

Proof of Theorem 1.2. First, we let λ > 0. We consider the following

eigenvalue problem with weight function:

(−∆)αu=λa(x)uin Br,

u= 0 in RN\Br.

We denote µ1be the ﬁrst eigenvalue of this problem. Since µ1→0 as r→ ∞,

we can choose r1>0 large enough such that µ1≤λwhen r≥r1. So we

can choose an increasing sequence r1< r2<···< rn→ ∞ and consider the

following problem

(−∆)αu=λa(x)u−b(x)upin Bn,

u= 0 in RN\Bn,(3.11)

where Bn=Brn. By Lemma 2.2, problem (3.11) has a unique positive

solution unfor each n. Furthermore, by the comparison principle (see Lemma

2.1), we know un≤un+1. On the other hand, any positive constant M

satisfying Mp−1≥Mp−1

0=λsupRNa(x)/infRNb(x) is a supersolution of

(3.11). It follows that un≤M0for all n. Therefore, unis increasing in n

and u∞(x) = limn→∞ un(x) is well deﬁned in RN. Then, u∞satisfying (1.5).

Since u∞≥un>0 in Bnfor each n, we know that u∞is a positive solution

of (1.5). Moreover, by Corollary 3.1, u∞is the unique solution of (1.5). We

complete the proof.

4 Acknowledgements

A. Quaas was partially supported by Fondecyt Grant No. 1151180 Programa

Basal, CMM. U. de Chile and Millennium Nucleus Center for Analysis of

PDE NC130017.

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