Existence and uniqueness of positive solutions for a class of logistic type elliptic equations in R^N involving fractional Laplacian

Abstract

In this paper, we study the existence and uniqueness of positive solutions
for the following nonlinear fractional elliptic equation: \begin{eqnarray*}
(-\Delta)^\alpha u=\lambda a(x)u-b(x)u^p&{\rm in}\,\,\R^N, \end{eqnarray*}
where α∈(0,1) \alpha\in(0,1) , N≥2 N\ge 2 , λ>0\lambda >0, a and b are positive
smooth function in RN\R^N satisfying \[ a(x)\rightarrow a^\infty>0\quad {\rm
and}\quad b(x)\rightarrow b^\infty>0\quad{\rm as}\,\,|x|\rightarrow\infty. \]
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.

Full text

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 ZΩZRN\Ω
|v1(x)|2
|x−y|N+2αdydx.(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
ZΩZRN\Ω
|v1(x)|2
|x−y|N+2αdydx ≤ZRN\Ω
1
dist(y, ∂D0)N+2αdykv1k2
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+2αdykvεkL∞(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+2αdykwεkL∞(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)
tN1 + 
x+(t−1)x0
t
N+2αdx. (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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