Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 95, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
OF A SEMILINEAR DIRICHLET PROBLEM
OUTSIDE THE UNIT BALL
HABIB MÂAGLI, SAMEH TURKI, ZAGHARIDE ZINE EL ABIDINE
Abstract. In this article, we are concerned with the existence, uniqueness and
asymptotic behavior of a positive classical solution to the semilinear boundaryvalue problem
−∆u = a(x)uσ
lim u(x) =
|x|→1
in D,
lim u(x) = 0.
|x|→∞
Here D is the complement of the closed unit ball of Rn (n ≥ 3), σ < 1 and the
γ
function a is a nonnegative function in Cloc
(D), 0 < γ < 1, satisfying some
appropriate assumptions related to Karamata regular variation theory.
1. Introduction
The semilinear elliptic equation
−∆u = a(x)uσ ,
σ < 1,
x ∈ Ω ⊂ Rn ,
has been extensively studied for both bounded and unbounded domains Ω in Rn
(n ≥ 2). We refer to [2, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 24, 26]
and the references therein, for various existence and uniqueness results related to
solutions for the above equation with homogeneous Dirichlet boundary conditions.
The asymptotic behavior of such solutions interested many authors who developed
the Karamata regular variation theory (see [3, 7, 9, 15, 21, 28]).
Mâagli [21] considered the problem
−∆u = a(x)uσ in Ω,
u > 0 in Ω,
u=0
(1.1)
on ∂Ω,
1,1
where Ω is a bounded C -domain, σ < 1 and a satisfies an assumption related
to K0 the set of Karamata functions regularly varying at zero (see Definition 1.1
below).
Thanks to the sub-supersolution method and using some potential theory tools,
Mâagli [21] showed that (1.1) has a unique positive classical solution and gave
sharp estimates on the solution. These estimates improve and extend those stated
2000 Mathematics Subject Classification. 31C35, 34B16, 60J50.
Key words and phrases. Asymptotic behavior; Dirichlet problem; subsolution; supersolution.
c
2013
Texas State University - San Marcos.
Submitted February 7, 2013. Published April 11, 2013.
1
2
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
EJDE-2013/95
in [10, 15, 20, 22, 28]. Before stating the result proved in [21], we shall recall the
definition of the Karamata class K0 .
Definition 1.1. A measurable function L is in K0 if there exist η > 0 such that L
is a positive function in C 1 ((0, η]) satisfying
lim+
t→0
tL0 (t)
= 0.
L(t)
As a typical example of function L ∈ K0 , we have
L(t) =
m
Y
(logi (
i=1
2η −µi
)) ,
t
where logi x = log ◦ log ◦ · · · ◦ log x (i times), µi ∈ R. Throughout this paper, for
two nonnegative functions f and g defined on a set S, the notation f (x) ≈ g(x),
x ∈ S, means that there exists a constant c > 0 such that for each x ∈ S, 1c g(x) ≤
f (x) ≤ c g(x). Further, we denote by δ(x) = dist(x,
∂Ω). Also for µ ≤ 2, σ < 1
Rη
and L ∈ K0 defined on (0, η], η > 0 such that 0 t1−µ L(t)dt < ∞, we put ΦL,µ,σ
the function defined on (0, η) by


1,
if µ < 1 + σ,



 R η L(s) ds1/(1−σ) , if µ = 1 + σ,
s
t
ΦL,µ,σ (t) :=
1/(1−σ)

(L(t))
,
if 1 + σ < µ < 2,



 R t L(s) ds1/(1−σ) , if µ = 2.
0 s
Now, let us present the result by Mâagli in [21].
γ
Theorem 1.2. Let a ∈ Cloc
(Ω), 0 < γ < 1, satisfying
a(x) ≈ δ(x)−µ L(δ(x))
for x ∈ Ω,
Rη
where µ ≤ 2, L ∈ K0 defined on (0, η], (η > diam(Ω)) such that 0 t1−µ L(t)dt < ∞.
Then problem (1.1) has a unique positive classical solution u satisfying
2−µ
u(x) ≈ δ(x)min(1, 1−σ ) ΦL,µ,σ (δ(x))
for each x ∈ Ω.
Chemmam et al [7] were concerned with K∞ the set of Karamta functions regularly varying at infinity (see Definition 1.4 below). More precisely, by using properties of functions in K∞ , the authors studied the asymptotic behavior of the unique
classical solution of the problem
−∆u = a(x)uσ
u>0
in Rn ,
in Rn ,
(1.2)
lim u(x) = 0,
|x|→∞
where n ≥ 3 and σ < 1. The existence of a unique classical solution of (1.2) has
been proved in [4, 19]. Namely, Chemmam et al in [7] proved the following result.
γ
Theorem 1.3. Let a ∈ Cloc
(Rn ), 0 < γ < 1, satisfying
a(x) ≈ (1 + |x|)−λ L(1 + |x|)
for x ∈ Rn ,
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
3
R∞
where λ ≥ 2, L ∈ K∞ such that 1 t1−λ L(t)dt < ∞. Then the solution u of
problem (1.2) satisfies
1
u(x) ≈
ΨL,λ,σ (1 + |x|) for each x ∈ Rn .
λ−2
,n−2)
min( 1−σ
(1 + |x|)
R∞
In this article, for λ ≥ 2, σ < 1 and L ∈ K∞ such that 1 t1−λ L(t)dt < ∞, the
function ΨL,λ,σ is defined on [1, ∞) by
 R
1/(1−σ)
∞ L(s)

ds
,
if λ = 2,

s
t



1/(1−σ)
(L(t))
,
if 2 < λ < n − σ(n − 2),
ΨL,λ,σ (t) :=
R t+1 L(s) 1/(1−σ)

ds
,
if λ = n − σ(n − 2),

s
1


1,
if λ > n − σ(n − 2).
For the convenience of the readers, we recall the definition of the Karamata class
K∞ .
Definition 1.4. A measurable function L defined on [1, ∞) is in K∞ if L is a
positive function in C 1 ([1, ∞)) such that
tL0 (t)
= 0.
t→∞ L(t)
lim
As a typical example of function L ∈ K∞ , we have
m
Y
(logi (ωt))−λi ,
L(t) =
i=1
where ω is a positive real number sufficiently large and λi ∈ R.
In this paper, we are concerned with the existence, uniqueness and estimates of
positive classical solutions to the semilinear Dirichlet problem
−∆u = a(x)uσ in D,
u>0
in D,
(1.3)
lim u(x) = lim u(x) = 0,
|x|→1
|x|→∞
n
where D = {x ∈ R : |x| > 1} is the complementary of the closed unit ball of
Rn (n ≥ 3) and σ < 1. The importance of the sublinear case (0 ≤ σ < 1) in
applications has been widely recognized for many years, see for example Wong
[27] for an extensive bibliography and the significance of the case σ < 0 has been
noticed in studies of boundary layer phenomena for viscous fluids [5, 6]. The main
feature of this paper is the presence of homogeneous Dirichlet boundary conditions
which combines those of [7] and [21]. For this reason the condition imposed on the
function a in problem (1.3) is an appropriate assumption related to K0 and K∞ .
To simplify our statements, we call K the set of functions L defined on (0, ∞) by
t
)N (1 + t),
L(t) := M (
1+t
where M ∈ K0 defined on (0, η] for some η > 1 and N ∈ K∞ . Also, for x ∈ D, we
1
denote by ρ(x) = 1 − |x|
. Let us consider the following hypothesis.
γ
(H1) a is a positive function in Cloc
(D), 0 < γ < 1, satisfying for x ∈ D,
a(x) ≈
L(|x| − 1)
,
|x|λ−µ (|x| − 1)µ
4
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
where µ ≤ 2 ≤ λ, L ∈ K such that
R∞
0
EJDE-2013/95
t1−µ (1 + t)µ−λ L(t)dt < ∞.
To illustrate (H1), we give an example. Let a be the positive function defined on
D by
a(x) =
4|x|
) log(2|x|))−α
(log( |x|−1
|x|λ−µ (|x| − 1)µ
,
where the real numbers µ, λ and α satisfy one of the following conditions
• µ < 2 < λ and α ∈ R;
• µ = 2, λ > 2 and α > 1;
• µ ≤ 2, λ = 2 and α > 1.
Then the function a satisfies (H1). Our main result in this paper is the following.
Theorem 1.5. Assume (H1), then problem (1.3) has a unique classical solution u
satisfying
u(x) ≈ θ(x),
x ∈ D,
(1.4)
where
2−µ
θ(x) :=
(ρ(x))min( 1−σ ,1)
λ−2
|x|min( 1−σ ,n−2)
ΦM,µ,σ (ρ(x)) ΨN,λ,σ (|x|).
(1.5)
The techniques used for proving Theorem 1.5 are based on the sub-supersolution
method. For the convenience of the readers, we shall recall the following definitions.
A positive function v ∈ C 2,γ (D), 0 < γ < 1, is called a subsolution of problem (1.3)
if
−∆v ≤ a(x) v σ
in D,
lim v(x) = lim v(x) = 0.
|x|→1
|x|→∞
As always, a supersolution is defined by reversing the inequality.
Since our approach is based on potential theory tools, we lay out some basic arguments that we are mainly concerned with in this work. An explicit expression for
the Green function G of the Laplace operator ∆ in D with zero Dirichlet boundary
conditions is given in [1] by
G(x, y) =
Γ( n2 − 1)
(|x − y|2−n − [x, y]2−n ),
n
4π 2
x, y ∈ D,
where [x, y]2 = |x − y|2 + (|x|2 − 1)(|y|2 − 1).
We refer in this paper to the potential of a nonnegative measurable function f
defined on D by
Z
V f (x) =
G(x, y) f (y)dy, x ∈ D.
D
γ
Recall that for each nonnegative function f in Cloc
(D), 0 < γ < 1, such that
2,γ
∞
V f ∈ L (D), we have V f ∈ Cloc (D) and satisfies −∆(V f ) = f in D; see [24,
Theorem 6.6].
The rest of the paper is organized as follows. In Section 2, we state and prove
some preliminary lemmas, involving some already known results on functions in K0
and K∞ . In Section 3, we give estimates on some potential functions. Section 4 is
devoted to the proof of Theorem 1.5. The last Section is reserved for an application.
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
5
2. Karamata regular variation theory
2.1. On the Karamata class K0 . In what follows, we recall some fundamental
properties of Karamata functions regularly varying at zero.
Lemma 2.1 (Karamata’s Theorem [25]). Let γ ∈ R and L ∈ K0 defined on (0, η],
η > 1. Then we have the following assertions
Rη
Rt
2−γ
L(t)
(i) If γ < 2, then 0 t1−γ L(t)dt converges and 0 s1−γ L(s)ds ∼ t 2−γ
as
+
t→0 ;
Rη
Rη
2−γ
L(t)
(ii) If γ > 2, then 0 t1−γ L(t)dt diverges and t s1−γ L(s)ds ∼ t γ−2
as
+
t→0 .
Lemma 2.2 ([8, 25]). Let L1 , L2 ∈ K0 defined on (0, η], η > 1, p ∈ R and ε > 0.
Then we have the following assertions:
(i) L1 L2 ∈ K0 and Lp1 ∈ K0 ;
(ii) limt→0+ tε L1 (t) = 0;
Rη
= 0 and t 7→ t L1s(s) ds ∈ K0 ;
(iii) limt→0+ R η LL11(t)
(s)
ds
s
t
Rη
Rt
(iv) If 0 L1s(s) ds converges, then limt→0+ R t LL11(t)
= 0 and t 7→ 0 L1s(s) ds ∈
(s)
0
K0 .
s
ds
Lemma 2.3. If L ∈ K0 defined on (0, η], η > 1, then there exists m ≥ 0 such that
for each t ∈ (0, 1], we have
2−m L(t) ≤ L(
t
) ≤ 2m L(t).
1+t
Proof. Since L ∈ K0 , then by the representation theorem [25], there exist c > 0
and z ∈ C([0, η]) such that z(0) = 0 and satisfying for each r ∈ (0, η], L(r) =
Rη
c exp( r z(s)
s ds).
Put m := sups∈[0,η] |z(s)|. Then for each t ∈ (0, 1], we have
Z t
z(s)
ds ≤ m log(1 + t).
−m log(1 + t) ≤
t
s
1+t
This implies
Z
t
z(s)
ds ≤ m log(2).
s
−m log(2) ≤
t
1+t
That is,
2−m ≤ exp
t
Z
t
1+t
z(s) ds ≤ 2m .
s
It follows that
2−m L(t) ≤ L(
t
) ≤ 2m L(t).
1+t
Lemma 2.4. Let µ ≤ 2, L ∈ K0 defined on (0, η], η > 1, such that
∞ and let
Z
1
s−µ L(s)ds),
I(t) = t(1 +
t
for t ∈ (0, 1].
Rη
0
t1−µ L(t)dt <
6
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
EJDE-2013/95
Then we have


t,R
1
L(s)
I(t) ≈ t t s ds,
1+t

 2−µ
t
t
L( 1+t
),
if µ < 1,
if µ = 1,
if 1 < µ ≤ 2.
Proof. We distinguish three cases.
Rη
Case 1: If µ < 1, then by applying Lemma 2.1 (i), we have 0 s−µ L(s)ds converges.
So we obtain that I(t) ≈ t.
Rη
Case 2: If µ = 1, then since 0 L(s)
s ds < ∞, we have
Z 1
Z 1
L(s)
L(s)
ds ≈
ds,
1+
t
s
s
t
1+t
and hence
Z
1
I(t) ≈ t
t
1+t
L(s)
ds.
s
Rη
Case 3: If 1 < µ ≤ 2, then applying Lemma 2.1 (ii), the integral 0 s−µ L(s)ds
R η −µ
diverges and t s L(s)ds ≈ t1−µ L(t). Combining this with the fact that 1 +
R 1 −µ
Rη
s L(s)ds ≈ t s−µ L(s)ds, we deduce by using Lemma 2.3 that
t
I(t) ≈ t2−µ L(
t
).
1+t
2.2. On the Karamata class K∞ . We quote some properties of functions belonging to the Karamata class K∞ .
Lemma 2.5 ([7]). Let L ∈ K∞ and γ ∈ R. Then we have the following
R∞
R∞
2−γ
L(t)
(i) If γ > 2, then 1 t1−γ L(t)dt converges and t s1−γ L(s)ds ∼ t γ−2
as
t → ∞;
R∞
Rt
2−γ
L(t)
(ii) If γ < 2, then 1 t1−γ L(t)dt diverges and 1 s1−γ L(s)ds ∼ t 2−γ
as
t → ∞.
Lemma 2.6 ([7]). Let L1 , L2 ∈ K∞ , p ∈ R and ε > 0. Then we have the following
assertions:
(i) L1 L2 ∈ K∞ and Lp1 ∈ K∞ ;
(ii) limt→∞ t−ε L1 (t) = 0;
R t+1
(iii) limt→∞ R t LL11(t)
= 0 and t 7→ 1 L1s(s) ds ∈ K∞ ;
(s)
ds
s
1
R∞
R∞
(iv) If 1 L1s(s) ds converges, then limt→∞ R ∞LL11(t)
= 0 and t 7→ t
(s)
t
s
ds
L1 (s)
s ds
K∞ ;
(v) There exists m ≥ 0 such that for every α > 0 and t ≥ 1, we have
(1 + α)−m L1 (t) ≤ L1 (α + t) ≤ (1 + α)m L1 (t).
R∞
Lemma 2.7. Let λ ≥ 2 and L ∈ K∞ be such that 1 t1−λ L(t)dt < ∞. Put
Z t
1
(1 +
sn−λ−1 L(s)ds), for t ∈ [1, ∞).
J(t) =
(1 + t)n−2
1
∈
EJDE-2013/95
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
Then we have
J(t) ≈
 L(t)

 (1+t)λ−2 ,R
7
if 2 ≤ λ < n,
t+1 L(s)
1
(1+t)n−2 1
s ds,


1
(1+t)n−2 ,
if λ = n,
if λ > n.
Proof. We split the proof into three cases.
Case 1: If 2 ≤ λ < n, then it follows by Lemma 2.5 (ii) that
1
(1 + tn−λ L(t)).
J(t) ≈
(1 + t)n−2
Which implies from Lemma 2.6 (i) and (ii) that
J(t) ≈
L(t)
.
(1 + t)λ−2
Case 2: If λ = n, then we get
J(t) ≈
1
(1 + t)n−2
t+1
Z
1
L(s)
ds.
s
Case 3: If λ > n, then it follows from Lemma 2.5 (i) that
converges. So we reach
1
.
J(t) ≈
(1 + t)n−2
R∞
1
sn−λ−1 L(s)ds
3. Asymptotic behavior of some potential functions
In what follows, we are going to give estimates on the potential functions V a
and V (a θσ ), where a is a function satisfying (H1) and θ is the function given in
(1.5). These estimates will be useful in the proof of our main result.
Remark 3.1. Let L ∈ K. Then there exist M ∈ K0 defined on (0, η] for some
η > 1 and N ∈ K∞ such that
t
L(t) = M (
)N (1 + t).
1+t
Since M ∈ C 1 ((0, η] and N ∈ C 1 ([1, ∞)), we obtain by virtue of Lemmas 2.3 and
2.6 that
L(t) ≈ M (t), for 0 < t ≤ 1
Which implies that
Z ∞
and L(t) ≈ N (t), for t ≥ 1.
t1−µ (1 + t)µ−λ L(t)dt < ∞
0
⇔
Z
1
t1−µ M (t)dt < ∞ and
0
(3.1)
Z
∞
(3.2)
t1−λ N (t)dt < ∞ .
1
According to Lemmas 2.1 and 2.5, we need to verify that
∞ in hypothesis (H1), only if λ = 2 or µ = 2.
R∞
0
t1−µ (1+t)µ−λ L(t)dt <
Proposition 3.2. Let a be a function satisfying (H1). Then for x ∈ D, we have
V a(x) ≈
(ρ(x))min(2−µ,1)
ΦM,µ,0 (ρ(x)) ΨN,λ,0 (|x|).
|x|min(λ−2,n−2)
8
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
EJDE-2013/95
Proof. Let a be a function satisfying (H1). Let µ ≤ R2 ≤ λ, M ∈ K0 and N ∈ K∞
∞
t
such that L(t) = M ( 1+t
)N (1 + t), for t ∈ (0, ∞) and 0 t1−µ (1 + t)µ−λ L(t)dt < ∞
satisfying
L(|x| − 1)
, x ∈ D.
a(x) ≈
λ−µ
|x|
(|x| − 1)µ
For x ∈ D, we have
Z
L(|y| − 1)
V a(x) ≈
dy.
G(x, y) λ−µ
|y|
(|y| − 1)µ
D
Since the function
L(|y| − 1)
dy
λ−µ (|y| − 1)µ
|y|
D
is radial, then by elementary calculus, we obtain that
Z
Z ∞ n+µ−λ−1
L(|y| − 1)
r
(1 − (min(|x|, r))2−n )L(r − 1)
G(x, y) λ−µ
dy
=
c
dr,
n
µ
|y|
(|y| − 1)
(max(|x|, r))n−2 (r − 1)µ
D
1
Z
x 7→
G(x, y)
where cn > 0. That is,
Z
Z |x| rn+µ−λ−1 (1 − r2−n )L(r − 1)
L(|y| − 1)
G(x, y) λ−µ
dy = cn
dr
|y|
(|y| − 1)µ
|x|n−2 (r − 1)µ
D
1
Z ∞ µ−λ+1
r
(1 − |x|2−n )L(r − 1) +
dr .
(r − 1)µ
|x|
In what follows, we distinguish two cases.
Case 1: 1 < |x| ≤ 2. We have
Z |x| n+µ−λ−1
1
r
(1 − r2−n )L(r − 1)
V a(x) ≈
dr
|x|n−2 1
(r − 1)µ
Z 2 µ−λ+1
Z ∞ µ−λ+1
r
L(r − 1)
r
L(r − 1)
+ (1 − |x|2−n )(
dr +
dr).
µ
(r
−
1)
(r
−
1)µ
|x|
2
Since for |x| ∈ (1, 2],
V a(x) ≈
Z
|x|
Z
|x|
1
|x|n−2
≈ 1 and 1 − |x|2−n ≈ |x| − 1, it follows that
rn+µ−λ−1 (1 − r2−n )L(r − 1)
dr
(r − 1)µ
1
Z ∞ µ−λ+1
Z 2 rµ−λ+1 L(r − 1)
r
L(r − 1) + (|x| − 1)
dr
+
dr .
(r − 1)µ
(r − 1)µ
|x|
2
This implies
V a(x) ≈
(r − 1)
1−µ
L(r − 1)dr + (|x| − 1)
Z
2
(r − 1)−µ L(r − 1)dr
|x|
1
∞
Z
+
(r − 1)1−λ L(r − 1)dr .
2
That is,
Z
V a(x) ≈
0
|x|−1
Z
s1−µ L(s)ds + (|x| − 1)
1
|x|−1
s−µ L(s)ds +
Z
1
∞
s1−λ L(s)ds .
EJDE-2013/95
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
Using (3.1), we obtain that
Z |x|−1
Z
s1−µ M (s)ds + (|x| − 1)
V a(x) ≈
1
s−µ M (s)ds +
|x|−1
0
Z
∞
9
s1−λ N (s)ds .
1
R∞
Taking into account of the fact that 0 t1−µ (1 + t)µ−λ L(t)dt < ∞ and using (3.2),
we reach
Z |x|−1
Z 1
V a(x) ≈
s1−µ M (s)ds + (|x| − 1) 1 +
s−µ M (s)ds
|x|−1
0
|x|−1
Z
s1−µ M (s)ds + I(|x| − 1),
=
0
where I is the function given in Lemma 2.4 by replacing L by M . Using Lemmas
2.1 and 2.2, we have
(
Z |x|−1
(|x| − 1)2−µ M (|x| − 1), if µ < 2,
1−µ
s
M (s)ds ≈ R |x|−1 M (s)
if µ = 2.
0
s ds,
0
Hence, we deduce from Lemma 2.4 that


(|x| − 1)+ (|x| − 1)2−µ M (|x| − 1), 


M (s)
(|x| − 1) M (|x| − 1) + R 1
s ds ,
|x|−1
V a(x) ≈
2−µ

− 1)
M (|x| − 1),

(|x|

R |x|−1 M (s)
s ds + M (|x| − 1),
0
if µ < 1,
if µ = 1,
if 1 < µ < 2,
if µ = 2.
Now, applying Lemma 2.2, we obtain for 1 < |x| ≤ 2 that


|x| − 1,
if µ < 1,



M (s)
(|x| − 1) R η
ds,
if
µ = 1,
s
|x|−1
V a(x) ≈
2−µ

(|x| − 1)
M (|x| − 1), if 1 < µ < 2,



R |x|−1 M (s) ds,
if µ = 2.
s
0
Thus, we obtain that for 1 < |x| ≤ 2,
V a(x) ≈ (|x| − 1)min(2−µ,1) ΦM,µ,0 (|x| − 1)
So, since |x| − 1 ≈ ρ(x), it follows from Lemmas 2.2 and 2.3 that
V a(x) ≈ (ρ(x))min(2−µ,1) ΦM,µ,0 (ρ(x)),
for 1 < |x| ≤ 2.
(3.3)
Case 2: |x| > 2. We have
V a(x) ≈
Z
1 2 rn+µ−λ−1 (1 − r2−n )L(r − 1)
dr
|x|n−2 1
(r − 1)µ
Z |x| n+µ−λ−1
r
(1 − r2−n )L(r − 1) dr
+
(r − 1)µ
2
Z ∞ µ−λ+1
r
L(r − 1)
2−n
dr.
+ (1 − |x|
)
(r
−
1)µ
|x|
Which implies
1 V a(x) ≈
|x|n−2
Z
2
(r − 1)
1
1−µ
Z
L(r − 1)dr +
2
|x|
(r − 1)n−λ−1 L(r − 1)dr
10
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
Z
EJDE-2013/95
∞
+
(r − 1)1−λ L(r − 1)dr.
|x|
That is
1 V a(x) ≈
|x|n−2
Z
1
1−µ
s
Z
|x|−1
n−λ−1
s
L(s)ds +
Z
L(s)ds +
s1−λ L(s)ds.
|x|−1
1
0
∞
Using (3.1), we reach
Z
Z |x|−1
Z ∞
1 1 1−µ
n−λ−1
V a(x) ≈
s
M (s)ds +
s
N (s)ds +
s1−λ N (s)ds.
|x|n−2 0
1
|x|−1
R∞
We deduce from the fact that 0 t1−µ (1 + t)µ−λ L(t)dt < ∞ and (3.2) that
Z |x|−1
Z ∞
1 n−λ−1
V a(x) ≈
1+
s
N (s)ds +
s1−λ N (s)ds
|x|n−2
1
|x|−1
Z ∞
= J(|x| − 1) +
s1−λ N (s)ds,
|x|−1
where J is the function given in Lemma 2.7 by replacing L by N . By applying
Lemma 2.5, we have
(R ∞ N (s)
Z ∞
s ds, if λ = 2,
1−λ
s
N (s)ds ≈ N|x|−1
(|x|−1)
,
if λ > 2.
|x|−1
|x|λ−2
Then, we deduce from Lemma 2.7 that

R∞
N (|x| − 1) + |x|−1 N s(s) ds,




 N (|x|−1)
,
|x|λ−2
R |x| N (s)
V a(x) ≈
1

 |x|n−2 (N (|x| − 1) + 1
s ds),


 1 + N (|x|−1) ,
|x|n−2
|x|λ−2
if λ = 2,
if 2 < λ < n,
if λ = n,
if λ > n.
Therefore, using Lemma 2.6, we obtain that for |x| > 2,
R ∞ N (s)

ds,
if λ = 2,

|x|−1 s

 N (|x|−1)

,
if 2 < λ < n,
|x|λ−2
V a(x) ≈
R |x| N (s)
1

 |x|n−2 1
s ds, if λ = n,


 1 ,
if λ > n.
|x|n−2
Hence, we get that for |x| > 2,
V a(x) ≈
ΨN,λ,0 (|x| − 1)
,
|x|min(λ−2,n−2)
which implies by Lemma 2.6 that
V a(x) ≈
ΨN,λ,0 (|x|)
.
|x|min(λ−2,n−2)
(3.4)
Combining (3.3) and (3.4), we deduce that for x ∈ D,
V a(x) ≈
This completes the proof.
(ρ(x))min(2−µ,1)
ΦM,µ,0 (ρ(x))ΨN,λ,0 (|x|).
|x|min(λ−2,n−2)
EJDE-2013/95
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
11
The following proposition plays a crucial role in the proof of Theorem 1.5.
Proposition 3.3. Let a be a function satisfying (H1) and let θ be the function
given by (1.5). Then for x ∈ D, we have
V (a θσ )(x) ≈ θ(x).
Proof. LetR a be a function satisfying (H1). Let µ ≤ 2 ≤ λ and L ∈ K satisfying the
∞
condition 0 t1−µ (1 + t)µ−λ L(t)dt < ∞ and such that for x ∈ D, we have
M (ρ(x))N (|x|)
L(|x| − 1)
:=
,
(3.5)
|x|λ−µ (|x| − 1)µ
|x|λ−µ (|x| − 1)µ
where M ∈ K0 and N ∈ K∞ . Put
λ−2
2−µ
λ1 = λ + σ min(
, n − 2) and µ1 = µ − σ min(
, 1).
1−σ
1−σ
We verify that µ1 ≤ 2 ≤ λ1 and by using (1.5) and (3.5), we obtain
e
(M ΦσM,µ,σ )(ρ(x))(N ΨσN,λ,σ )(|x|)
L(|x|
− 1)
:=
.
a(x)θσ (x) ≈
λ
−µ
µ
λ
−µ
|x| 1 1 (|x| − 1) 1
|x| 1 1 (|x| − 1)µ1
a(x) ≈
σ
e
f := M Φσ
Since M
M,µ,σ ∈ K0 and N := N ΨN,λ,σ ∈ K∞ , we have that the function
t
e defined on (0, ∞) by L(t)
e
f(
e (1 + t) is in K and by Lemmas 2.1 and
L
=M
)N
R ∞ 1+t
1−µ1
e
2.5, we obtain that the integral 0 t
(1 + t)µ1 −λ1 L(t)dt
< ∞. Hence, it follows
from Proposition 3.2 that
V (aθσ )(x) ≈
(ρ(x))min(2−µ1 ,1)
Φf
(ρ(x))ΨNe ,λ1 ,0 (|x|),
|x|min(λ1 −2,n−2) M ,µ1 ,0
x ∈ D.
Now, using
2−µ
λ−2
, 1), min(λ1 − 2, n − 2) = min(
, n − 2),
1−σ
1−σ
we obtain by elementary calculus that for x ∈ D,
min(2 − µ1 , 1) = min(
ΦM
f,µ1 ,0 (ρ(x)) = ΦM,µ,σ (ρ(x)),
ΨNe ,λ1 ,0 (|x|) = ΨN,λ,σ (|x|).
This completes the proof.
4. Proof of Theorem 1.5
4.1. Existence and asymptotic behavior. Let a be a function satisfying (H1).
We look now at the existence of positive solution of problem (1.3) satisfying (1.4).
The main idea is to find a subsolution and a supersolution to problem (1.3) of the
L0 (|x|−1)
form cV (aω σ ), where c > 0 and ω(x) = |x|β−α
, which will satisfy
(|x|−1)α
V (aω σ ) ≈ ω.
(4.1)
So the choice of the real numbers α, β and the function L0 in K is such that (4.1)
is satisfied. Setting ω(x) = θ(x), where θ is the function given by (1.5), we have by
Proposition 3.3, that the function θ satisfies (4.1). Hence, let v = V (aθσ ) and let
m ≥ 1 be such that
1
θ ≤ v ≤ mθ.
(4.2)
m
This implies that for σ < 1, we have
1 σ
θ ≤ v σ ≤ m|σ| θσ .
m|σ|
12
H. MÂAGLI, S. TURKI, Z. ZINE EL ABIDINE
EJDE-2013/95
Put c = m|σ|/(1−σ) , then it is easy to show that u = 1c v and u = cv are respectively
a subsolution and a supersolution of problem (1.3).
Now, since c ≥ 1, we get u ≤ u on D and thanks to the method of subsupersolution (see [23]), it follows that problem (1.3) has a classical solution u
such that u ≤ u ≤ u in D. Using (4.2), we deduce that u satisfies (1.4). This
completes the proof.
4.2. Uniqueness. Let a be a function satisfying (H1) and θ be the function defined
in (1.5). We aim to show that problem (1.3) has a unique positive solution in the
cone
Γ = {u ∈ C 2,γ (D) : u(x) ≈ θ(x)}.
To this end, we consider the following cases.
Case σ < 0 Let u and v be two solutions of (1.3) in Γ and put w = u − v. Then
the function w satisfies
w + V (hw) = 0 in D,
where h is the nonnegative measurable function defined in D by
(
σ
−(u(x))σ
if u(x) =
6 v(x),
a(x) (v(x))
u(x)−v(x)
h(x) =
0
if u(x) = v(x).
Furthermore, it is clear to see that V (h|w|) < ∞. Then, we get by [2, Lemma 4.1]
that w = 0. This proves the uniqueness.
Case 0 ≤ σ < 1 Let us now assume that u and v are arbitrary solutions of problem
(1.3) in Γ. Since u, v ∈ Γ, then there exists a constant m ≥ 1 such that
1
u
≤ ≤ m in D.
m
v
This implies that the set J := {t ∈ (0, 1] : tu ≤ v} is not empty. Now put c := sup J.
It is easy to see that 0 < c ≤ 1. On the other hand, we have
−∆(v − cσ u) = a(x)(v σ − cσ uσ ) ≥ 0
σ
in D,
σ
lim (v − c u)(x) = lim (v − c u)(x) = 0.
|x|→1
|x|→∞
Then, by the maximum principle, we deduce that cσ u ≤ v. Which implies that
cσ ≤ c. Using the fact that σ < 1, we get that c ≥ 1. Hence, we arrive at u ≤ v
and by symmetry, we obtain that u = v. This completes the proof.
5. Applications
Let σ, β < 1 and let Ra be a function satisfying (H1). Let µ ≤ 2 ≤ λ and L ∈ K
∞
satisfying the condition 0 t1−µ (1 + t)µ−λ L(t)dt < ∞ and such that for x ∈ D, we
have
L(|x| − 1)
M (ρ(x))N (|x|)
a(x) ≈
:=
,
|x|λ−µ (|x| − 1)µ
|x|λ−µ (|x| − 1)µ
where M ∈ K0 and N ∈ K∞ . We are interested in the problem
−∆u +
β
|∇u|2 = a(x)uσ
u
u > 0 in D,
in D,
lim u(x) = lim u(x) = 0.
|x|→1
|x|→∞
(5.1)
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
13
Put v = u1−β , then by calculus, we obtain that v satisfies
σ−β
−∆v = (1 − β)a(x)v 1−β
v>0
in D,
in D,
(5.2)
lim v(x) = lim v(x) = 0.
|x|→1
|x|→∞
Since σ1 := σ−β
1−β < 1, we obtain by applying Theorem 1.5 that problem (5.2) has a
unique solution v such that, for each x ∈ D,
v(x) ≈
(ρ(x))min(
|x|min(
(2−µ)(1−β)
,1)
1−σ
(λ−2)(1−β)
,n−2)
1−σ
ΦM,µ,σ1 (ρ(x)) ΨN,λ,σ1 (|x|).
Consequently, we deduce that problem (5.1) has a unique positive solution u satisfying for each x ∈ D,
2−µ
u(x) ≈
1
(ρ(x))min( 1−σ , 1−β )
λ−2 n−2
, 1−β )
min( 1−σ
|x|
1
1
1−β
1−β
(ρ(x)) ΨN,λ,σ
(|x|).
ΦM,µ,σ
1
1
Acknowledgements. We thank the anonymous referee for the careful reading of
our manuscript.
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Habib Mâagli
King Abdulaziz University, College of Sciences and Arts, Rabigh Campus, Department
of Mathematics, P. O. Box 344, Rabigh 21911, Saudi Arabia
E-mail address: habib.maagli@fst.rnu.tn
Sameh Turki
Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire,
2092 Tunis, Tunisia
E-mail address: sameh.turki@ipein.rnu.tn
Zagharide Zine El Abidine
Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire,
2092 Tunis, Tunisia
E-mail address: Zagharide.Zinelabidine@ipeib.rnu.tn