Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 88, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF GROUND STATE SOLUTIONS
FOR SUBLINEAR AND SINGULAR NONLINEAR DIRICHLET
PROBLEMS
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
Abstract. In this article, we are concerned with the asymptotic behavior of
the classical solution to the semilinear boundary-value problem
∆u + a(x)uσ = 0
Rn ,
in
u > 0, lim|x|→∞ u(x) = 0, where σ < 1. The special feature is to
α (Rn ), 0 < α < 1, such that there exists c > 0
consider the function a in Cloc
satisfying
L(|x| + 1)
1 L(|x| + 1)
≤ a(x) ≤ c
,
c (1 + |x|)λ
(1 + |x|)λ
` R t z(s) ´
where L(t) := exp 1 s ds , with z ∈ C([1, ∞)) such that limt→∞ z(t) = 0.
The comparable asymptotic rate of a(x) determines the asymptotic behavior
of the solution.
1. Introduction
In this article, we are interested in estimates for positive solutions of the semilinear problem
−∆u = a(x)g(u), x ∈ Rn , n ≥ 3
in Rn ,
u>0
(1.1)
lim u(x) = 0.
|x|→∞
The existence of such solutions and their asymptotic behavior have been extensively
studied by many authors when (1.1) has a smooth bounded domain Ω with zero
boundary Dirichlet condition. We refer the reader to [1, 3, 6, 8, 9, 14] and the
references therein.
In recent years, the study of ground state solutions of problem (1.1) received a lot
of interest and numerous existence results have been established (see for instance
[2, 4, 5, 7, 10, 12] and the references therein).
More specifically, Lair and Shaker [7] established the existence and the uniqueness of positive classical solution, where g is a positive nonincreasing and continuously differentiable function on (0, ∞) and a is a nontrivial nonnegative function
2000 Mathematics Subject Classification. 31B05, 31C35, 34B27, 60J50.
Key words and phrases. Asymptotic behavior; Dirichlet problem; ground sate solution;
singular equations; sublinear equations.
c
2011
Texas State University - San Marcos.
Submitted April 13, 2011. Published July 5, 2011.
1
2
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
α
in Cloc
(Rn ), satisfying
EJDE-2011/88
∞
Z
t max a(x)dt < ∞.
0
|x|=t
(1.2)
Moreover, they showed that this condition on a is nearly optimal.
Furthermore, Brezis and Kamin [2] proved the existence of a unique positive
solution to the problem
−∆u = a(x)uσ ,
x ∈ Rn , n ≥ 3
u > 0,
lim inf u(x) = 0,
|x|→∞
where 0 < σ < 1 and a isRa nonnegative measurable function potentially bounded,
a(y)
∞
(Rn ).
that is the function x 7→ Rn |x−y|
n−2 dy is in L
Throughout this article, we denote K the set of all functions L defined on [1, ∞),
by
Z t z(s) L(t) := c exp
ds ,
s
1
where c is a positive constant and z ∈ C([1, ∞)) such that limt→∞ z(t) = 0.
Remark 1.1. It is obvious that L ∈ K if and only if L is a positive function in
0
(t)
C 1 ([1, ∞)) such that limt→∞ tL
L(t) = 0.
Example 1.2. Let m ∈ N∗ , (λ1 , λ2 , . . . , λm ) ∈ Rm and ω be a positive real number
sufficiently large such that the function
m
Y
(logk (wt))−λk
L(t) =
k=1
is defined and positive on [1, ∞), where logk x = log ◦ log ◦ · · · ◦ log x (k times).Then
L ∈ K.
In this paper, we give precise asymptotic behavior of the solution to the problem
−∆u = a(x)uσ , x ∈ Rn , n ≥ 3,
u>0
in Rn ,
(1.3)
lim u(x) = 0,
|x|→∞
where σ < 1 and a satisfies the hypothesis
α
(H1) a is a nonnegative function in Cloc
(Rn ), 0 < α < 1, satisfying
L(1 + |x|)
,
(1 + |x|)λ
R∞
where λ ≥ 2 and L ∈ K such that 1 t1−λ L(t)dt < ∞.
Here and throughout the paper, for two nonnegative functions f and g defined on
a set S, the notation f (x) ≈ g(x), for x ∈ S means that there exists c > 0 such
that 1c f (x) ≤ g(x) ≤ cf (x), for all x ∈ S.
R∞
Remark 1.3. (i) Note that we need to verify the condition 1 t1−λ L(t)dt < ∞ in
hypothesis (H1), only for λ = 2 (see Remark 2.2).
(ii) It is obvious to see that if a satisfies hypothesis (H1), then a is potentially
bounded and a verifies (1.2). This implies from [7] and [2], that problem (1.3) has a
a(x) ≈
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ASYMPTOTIC BEHAVIOR
3
unique classical positive solution in C 2,α (Rn ). Thus it becomes interesting to know
the asymptotic behavior of such solution, as t → ∞.
Our main result is the following.
Theorem 1.4. Assume (H1). Then the solution u of problem (1.3) satisfies
u(x) ≈ θλ (x),
n
(1.4)
n
where x ∈ R , and θλ is defined on R by
 R
∞
L(t) 1/(1−σ)

,

 |x|+1 t dt


 (L(1+|x|)1/(1−σ) ,
(λ−2)/(1−σ)
θλ (x) := (1+|x|)
R |x|+2 L(t) 1/(1−σ)
1


,
n−2

(1+|x|)
t dt
1


1

(1+|x|)n−2 ,
for λ = 2,
for 2 < λ < n − (n − 2)σ,
(1.5)
for λ = n − (n − 2)σ,
for λ > n − (n − 2)σ.
To obtain estimates (1.5), we shall adopt a sub-supersolution method. For the
reader’s convenience, we recall the definition.
A positive function v ∈ C 2,α (Rn ) is called a subsolution of problem (1.3) if
−∆v ≤ a(x)v σ
x ∈ Rn ,
lim v(x) = 0.
(1.6)
|x|→∞
If the above inequality is reversed, v is called a supersolution of problem (1.3).
The outline of this article is as follows. In Section 2, we state some already
known results on functions in K, useful for our study and we give estimates on
some potential functions. The proof of Theorem 1.4 is given in Section 3. The last
section is reserved to some applications.
We close this section by giving the following notation. For a nonnegative measurable function a in Rn , we denote by V a the potential of a defined on Rn by
Z
V a(x) =
G(x, y)a(y)dy,
Rn
where G(x, y) =
and cn =
Γ( n
2 −1)
n
4π 2
cn
|x−y|n−2
is the Green function of the Laplacian ∆ in Rn (n ≥ 3),
α
. We point out that for any nonnegative function f in Cloc
(Rn )
2,α
(0 < α < 1) such that V f ∈ L∞ (Rn ), we have V f ∈ Cloc
(Rn ) and satisfies
n
−∆(V f ) = f in R ; see [11, Theorem 6.3].
2. Key estimates
2.1. Technical lemmas. In what follows, we collect some fundamental properties
of functions belonging to the class K. First, we need the following elementary result.
Lemma 2.1 (Karamata’s Theorem). Assume that g ∈ C 1 ([β, ∞), (0, ∞)) and that
limt→∞ tg 0 (t)/g(t) = γ. Then we have the following properties:
R∞
(i) If γ < −1, then β g(s)ds converges and
Z ∞
tg(t)
, as t → ∞.
g(s)ds ∼ −
γ
+1
t
4
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
(ii) If γ > −1, then
R∞
β
g(s)ds diverges and
Z
t
tg(t)
γ+1
g(s)ds ∼
β
EJDE-2011/88
as t → ∞.
Remark 2.2. Let γ ∈ R and L be a function in K. Applying Lemma 2.1 to
g(t) = tγ L(t), we obtain that
R∞
R∞
1+γ
L(t)
• If γ < −1, then 1 sγ L(s)ds diverges and t sγ L(s)ds ∼ − t γ+1
, as
t → ∞;
R∞
Rt
1+γ
L(t)
as
• If γ > −1, then 1 sγ L(s)ds converges and 1 sγ L(s)ds ∼ t γ+1
t → ∞.
Lemma 2.3. (i) Let L1 , L2 ∈ K, p ∈ R. Then L1 L2 ∈ K and Lp1 ∈ K.
(ii) Let L be a function in K then there exists m ≥ 0 such that for every η > 0
and t ≥ 1 ,we have
(1 + η)−m L(t) ≤ L(η + t) ≤ (1 + η)m L(t).
Proof. Assertion (i) is due to Remark 1.1. Let us prove (ii). Let z be the function
Rt
in C([1, ∞)) such that limt→∞ z(t) = 0 and L(t) = exp 1 z(s)
s ds .
Put m = supt∈[1,∞) |z(t)|, then for each η > 0 and t ≥ 1, we have
Z t+η
t
z(s)
t+η
m log
≤
ds ≤ m log
.
t+η
s
t
t
That is,
Z t+η z(s) η
η
(1 + )−m ≤ exp
ds ≤ (1 + )m .
t
s
t
t
So (ii) holds.
Lemma 2.4. Let L ∈ K and ε > 0, then we have
lim t−ε L(t) = 0,
t→∞
lim R t
t→∞
1
If further
R∞
1
L(t)
L(s)/s ds
= 0.
(2.1)
(2.2)
L(s)/s ds converges, then
lim R ∞
t→∞
t
L(t)
= 0.
L(s)/s ds
(2.3)
Proof. Let L ∈ K and ε > 0. It is obvious by Remark 1.1 that the function
ε
t → t− 2 L(t) is non-increasing in [ω, ∞), for ω large enough. Then
ε
ε
t− 2 L(t) ≤ ω − 2 L(ω), for t ≥ ω;
That is,
L(ω)
, for t ≥ ω.
(ωt)ε/2
This proves (2.1). For the rest of the proof, we distinguish two cases.
R∞
Case 1: 1 L(s)
s ds < ∞. Since the function t → tL(t) is nondecreasing in
[ω, ∞), then
Z ∞
Z ∞
ds
L(s)
tL(t)
≤
ds, for t ≥ ω.
2
s
s
t
t
t−ε L(t) ≤
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ASYMPTOTIC BEHAVIOR
Hence
Z
0 < L(t) ≤
t
∞
L(s)
ds,
s
5
for t ≥ ω.
Then limt→∞ L(t) = 0, which implies (2.2).
Moreover, put ϕ(t) = L(t)/t, for t ≥ 1. Since ϕ satisfies limt→∞ tϕ0 (t)/ϕ(t) =
−1, then it follows that
Z ∞
Z ∞
Z ∞
ϕ(s)ds ∼ −
sϕ0 (s)ds = tϕ(t) +
ϕ(s)ds,
t
t
as t → ∞. This implies that
Z ∞
t
L(s)
ds ∼ L(t) +
s
t
Z
∞
t
L(s)
ds,
s
as t → ∞. So we deduce (2.3).
R∞
Case 2: 1 L(s)
s ds = ∞. Put ϕ(t) = L(t)/t, for t ≥ 1. Then for ω sufficiently
large and t ≥ ω, we have
Z t
Z t
ϕ(s)ds ∼ −tϕ(t) + ωϕ(ω) +
ϕ(s)ds,
ω
ω
as t → ∞; that is,
Z
t
ω
L(s)
ds ∼ −L(t) + ωϕ(ω) +
s
Z
t
ω
L(s)
ds,
s
as t → ∞. Which proves (2.2) and completes the proof.
Remark 2.5. Let L ∈ K. Using Remark 1.1 and (2.2), we deduce that
Z t+1
L(s)
t→
ds ∈ K.
s
1
R∞
If further 1 L(s)/s ds converges, we have by (2.3) that
Z ∞
L(s)
ds ∈ K.
t 7→
s
t
2.2. Asymptotic behavior of some potential functions. We are going to give
estimates on the potential functions V a and V (aθλσ ), where θλ is the function given
in (1.5).
Proposition 2.6. Let a be a function satisfying (H1). Then for x ∈ Rn ,
V a(x) ≈ ψ(|x|),
where ψ is the function defined in [0, ∞) by
R ∞

L(r)/r dr,

t+1



 L(1+t)
,
(1+t)λ−2
ψ(t) =
R t+2
1

L(r)/r dr,

n−2
(1+t)
1




1
(1+t)n−2 ,
for λ = 2,
for 2 < λ < n,
for λ = n,
for λ > n.
(2.4)
6
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
EJDE-2011/88
Proof. First, we recall the following well known result. Let ϕ be a nonnegative
radial measurable function and x ∈ Rn , then we have
Z
Z ∞
rn−1
ϕ(y)
dy
=
c
ϕ(r)dr.
n−2
max(|x|, r)n−2
Rn |x − y|
0
R∞
Now, let λ ≥ 2 and L ∈ K satisfying 1 t1−λ L(t)dt < ∞ and such that
a(x) ≈
L(1 + |x|)
.
(1 + |x|)λ
Thus
1
L(1 + |y|)
dy = cn I(|x|),
λ
n−2
Rn (1 + |y|) |x − y|
where I is the function defined on [0, ∞) by
Z ∞
rn−1 L(1 + r)
dr.
I(t) =
max(t, r)n−2 (1 + r)λ
0
Z
V a(x) ≈
So to prove the result, it is sufficient to show that I(t) ≈ ψ(t) for t ∈ [0, ∞). We
have
Z ∞
Z 1 n−1
Z t n−1
r
L(1 + r)
1
r
L(1 + r)
rL(1 + r)
1
dr + n−2
dr +
dr
I(t) = n−2
λ
λ
t
(1
+
r)
t
(1
+
r)
(1 + r)λ
0
1
t
:= I1 (t) + I2 (t) + I3 (t).
It is clear that for t ≥ 2,
I1 (t) ≈
1
tn−2
.
(2.5)
To estimate I2 and I3 , we distinguish two cases.
Case 1: λ > 2. Using Lemma 2.3 (ii) and Remark 2.2, for t ≥ 2 we have
Z ∞
L(t)
(2.6)
I3 (t) ≈
r1−λ L(r)dr ≈ λ−2 .
t
t
R∞
• If 2 < λ < n, then applying again Remark 2.2, we have 1 rn−1−λ L(r)dr = ∞
Rt
and 1 rn−1−λ L(r)dr ∼ t2−λ L(t), as t → ∞. So by Lemma 2.3 (ii), for t ≥ 2 we
obtain
Z t
L(t)
1
rn−1−λ L(r)dr ≈ λ−2 .
I2 (t) ≈ n−2
t
t
1
Then by (2.5), (2.6) and (2.1), for t ≥ 2 we have
I(t) ≈
1
L(t)
L(t)
+ λ−2 ≈ λ−2 .
tn−2
t
t
L(1+t)
Now, since the functions t → I(t) and t → (1+t)
λ−2 are positive and continuous in
[0, ∞), for t ≥ 0 we obtain
L(1 + t)
I(t) ≈
.
(1 + t)λ−2
Rt
• If λ > n, then applying Remark 2.2, we have 1 rn−1−λ L(r)dr < ∞. So by
Lemma 2.3 (ii), for t ≥ 2, we obtain
Z t
1
1
I2 (t) ≈ n−2
rn−1−λ L(r)dr ≈ n−2 .
t
t
1
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ASYMPTOTIC BEHAVIOR
7
This together with (2.5), (2.6) and (2.1) implies that for t ≥ 2,
1
I(t) ≈ n−2 .
t
Then by the same argument as above, we deduce that for t ≥ 0,
1
I(t) ≈
.
(1 + t)n−2
• If λ = n, then using (2.5), (2.6) and (2.2), for t ≥ 2, we have
Z t
Z t
L(r)
1
L(r)
1
I(t) ≈ n−2 (1 +
dr + L(t)) ≈ n−2
dr.
t
r
t
r
1
1
So for t ≥ 0, we obtain
Z t+2
L(r)
1
dr.
I(t) ≈
n−2
(1 + t)
r
1
Case 2: λ = 2. By Remark 2.2, for t ≥ 2, we have I2 (t) ≈ L(t). So for t ≥ 2,
we have
Z ∞
1
L(r)
I(t) ≈ n−2 + L(t) +
dr.
t
r
t
Hence using (2.1) and (2.3), for t ≥ 2, we have
Z ∞
L(r)
I(t) ≈
dr.
r
t
So for t ≥ 0, we obtain
Z ∞
L(r)
I(t) ≈
dr.
r
t+1
This completes the proof.
The following Proposition plays a key role in this paper.
Proposition 2.7. Let a be a function satisfying (H1) and let θλ be the function
given by (1.5). Then for x ∈ Rn ,
V (aθλσ )(x) ≈ θλ (x).
R∞
Proof. Let λ ≥ 2 and L ∈ K satisfying 1 t1−λ L(t)dt < ∞ and such that
a(x) ≈
L(1 + |x|)
.
(1 + |x|)λ
Then for every x ∈ Rn , we have

L(1+|x|) R ∞
L(t) σ/(1−σ)

, λ = 2,

2
(1+|x|)
t dt
|x|+1



1/(1−σ)

 (L(1+|x|)
2 < λ < n − (n − 2)σ,
(λ−2σ)/(1−σ) ,
a(x)θλσ (x) ≈ h(x) := (1+|x|) R |x|+2
σ/(1−σ)
L(t)
L(1+|x|)


, λ = n − (n − 2)σ,

(1+|x|)n
t dt
1



 L(1+|x|) ,
λ > n − (n − 2)σ.
(1+|x|)λ+(n−2)σ
e
L(1+|x|)
(1+|x|)µ ,
where µ ≥ 2. Moreover, due to Lemma 2.3
R∞
e
e
and Remark 2.5, we deduce that L ∈ K and satisfies 1 t1−µ L(t)dt
< ∞. Hence,
it follows by Proposition 2.6, that
e
V (aθσ )(x) ≈ V (h)(x) ≈ ψ(|x|),
in Rn ,
We point out that h(x) =
λ
8
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
EJDE-2011/88
e and λ by µ. This
where ψe is the function defined by (2.4) by replacing L by L
completes the proof by a simple calculus.
3. Proof of Theorem 1.4
Let a be a function satisfying (H1). The main idea is to find a subsolution and
a supersolution of problem (1.3) of the form cV (aφσ ), where c > 0 and φ(x) =
L0 (1+|x|)
, which will satisfy necessarily
(1+|x|)β
V (aφσ ) ≈ φ.
(3.1)
So, the choice of the real β and the function L0 in K is such that (3.1) is satisfied. Setting φ(x) = θλ (x), where θλ is the function given by (1.5), we have by
Proposition 2.7 that the function θλ satisfies (3.1).
Let v := V (aθλσ ) and let M > 1 be such that
1
v ≤ θλ ≤ M v.
M
Which implies that for σ < 1,
vσ
≤ θλσ ≤ M |σ| v σ .
M |σ|
Put c := M |σ|/(1−σ) , then it is easy to verify 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 Rn and thanks to the method of sub and
supersolution, it follows that problem (1.3) has a solution u satisfying u ≤ u ≤ u,
in Rn .
Finally, by using Remark 1.3 (ii) and Proposition 2.7, we deduce that the unique
classical positive solution of problem (1.3) satisfies (1.4). This completes the proof.
4. Applications
α
4.1. First application. Let σ < 1 and a be a positive function in Cloc
(Rn ) satisfying for x ∈ Rn
m
Y
1
(logk (w(1 + |x|)))−λk ,
a(x) ≈
(1 + |x|)λ
k=1
where m ∈ N∗ and w is a positive constant large enough. The real numbers λ and
λk , 1 ≤ k ≤ m, satisfy one of the following two conditions
• λ > 2 and λk ∈ R for 1 ≤ k ≤ m.
• λ = 2 and λ1 = λ2 = · · · = λl−1 = 1, λl > 1, λk ∈ R for l + 1 ≤ k ≤ m.
Using Theorem 1.4, we deduce that problem (1.3) has a unique classical positive
solution u in Rn satisfying
(i) If λ = 2, then for x ∈ Rn
m
Y
u(x) ≈ (logl w(1 + |x|))(1−λl )/(1−σ)
(logk w(1 + |x|))−λk /(1−σ) .
k=l+1
(ii) If 2 < λ < n − σ(n − 2), then for x ∈ Rn
u(x) ≈
1
(1 + |x|)(λ−2)/(1−σ)
m
Y
k=1
(logk w(1 + |x|))−λk /(1−σ) .
EJDE-2011/88
ASYMPTOTIC BEHAVIOR
9
(iii) If λ = n − σ(n − 2) and λ1 = λ2 = · · · = λm = 1, then for x ∈ Rn
1
u(x) ≈
(logm+1 w(1 + |x|))1/(1−σ) .
(1 + |x|)n−2
(iv) If λ = n − σ(n − 2) and λ1 = λ2 = · · · = λl−1 = 1, λl < 1, λk ∈ R, for
l + 1 ≤ k ≤ m, then for x ∈ Rn
m
Y
1
(1−λl )/(1−σ)
u(x) ≈
(log
w(1
+
|x|))
(logk w(1 + |x|))−λk /(1−σ) .
l
(1 + |x|)n−2
k=l+1
(v) If λ = n − σ(n − 2) and λ1 = λ2 = · · · = λl−1 = 1, λl > 1, λk ∈ R, for
l + 1 ≤ k ≤ m, then for x ∈ Rn
1
u(x) ≈
.
(1 + |x|)n−2
(vi) If λ > n + σ(n − 2), then for x ∈ Rn
1
u(x) ≈
.
(1 + |x|)n−2
4.2. Second application. Let a be a function satisfying (H1) and let σ, β < 1.
We are interested in the problem
β
−∆u + |∇u|2 = a(x)uσ in Rn ,
u
(4.1)
u > 0, in Rn ,
lim u(x) = 0.
|x|→∞
Put v = u1−β , then by a simple calculus, we obtain that v satisfies
σ−β
−∆v = (1 − β)a(x)v 1−β
v>0
in Rn ,
in Rn ,
(4.2)
lim v(x) = 0.
|x|→∞
Applying Theorem 1.4 to problem (4.2), we obtain
tion v such that
 R
(1−β)/(1−σ)
∞
L(s)

ds
,

|x|+1 s



(1−β)/(1−σ)

(L(1+|x|))

,
(1+|x|)(λ−2)/(1−σ)
v(x) ≈ θeλ (x) :=
1−β
1−σ
R
|x|+2 L(s)

1

,

(1+|x|)n−2
s ds
1




1
,
(1+|x|)n−2
that there exists a unique soluif λ = 2,
if 2 < λ < n − (n − 2) σ−β
1−β ,
if λ = n − (n − 2) σ−β
1−β ,
if λ > n − (n − 2) σ−β
1−β .
Consequently, we deduce that (4.1) has a unique positive solution u satisfying
1/(1−β)
u(x) ≈ θeλ (x)
 R
1/(1−σ)
∞
L(s)

ds
,
if λ = 2,


s
|x|+1


2−λ

(1 + |x|) (1−σ)(1−β) (L(1 + |x|))1/(1−σ) ,
if 2 < λ < n − (n − 2) σ−β
1−β ,
=
1/(1−σ)
R

|x|+2 L(s)


(1 + |x|)(2−n)/(1−β) 1
, if λ = n − (n − 2) σ−β

s ds
1−β ,



(1 + |x|)(2−n)/(1−β) ,
if λ > n − (n − 2) σ−β
1−β .
10
RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
EJDE-2011/88
4.3. Third application. Let a be a function satisfying (H1) and L be a function
in K such that
L(1 + |x|)
a(x) ≈
.
(1 + |x|)λ
γ
Let b ∈ Cloc
(Rn ), 0 < γ < 1 satisfying for x ∈ Rn ,
b(x) ≈
L1 (1 + |x|)
,
(1 + |x|)µ
where µ ∈ R and L1 ∈ K. Let σ, β < 1 and p ∈ R. We are interested in the system
−∆u = a(x)uσ
−∆v = b(x)up v β
u, v > 0
n
in R ,
in Rn ,
in Rn ,
(4.3)
lim u(x) = lim v(x) = 0.
|x|→∞
|x|→∞
By Theorem 1.4, it follows that there exists a unique classical solution u to (1.3)
satisfying (1.4). So, we distinguish the following cases.
R∞
Case 1: λ = 2. By hypothesis (H1), we have 1 L(t)
t dt < ∞ and using estimates
(1.5), we deduce that
Z
L1 (1 + |x|) ∞ L(s) p/(1−σ)
L2 (1 + |x|)
b(x)up (x) ≈
.
ds
:=
(1 + |x|)µ
s
(1 + |x|)µ
|x|+1
It is obviousR to see by Lemma 2.3 and Remark 2.5 that L2 ∈ K. Now suppose that
∞
µ ≥ 2 and 1 t1−µ L2 (t)dt < ∞. Then applying Theorem 1.4, we conclude that
(4.3) has a unique classical solution (u, v) such that u(x) ≈ θλ (x) and
 R
1/(1−β)
∞
L2 (s)

ds
,
if µ = 2,


s
|x|+1

 (L (1+|x|))1/(1−β)

 2
if 2 < µ < n − (n − 2)β,
(µ−2)/(1−β) ,
v(x) ≈ (1+|x|)
R
1/(1−β)
|x|+2
L
(s)

1
2

, if µ = n − (n − 2)β,
n−2

s ds
1

 (1+|x|)

1

if µ > n − (n − 2)β.
(1+|x|)n−2 ,
λ−2
Case 2: 2 < λ < n − (n − 2)σ. Put γ = µ + 1−σ
p. From the estimates (1.5), we
deduce that
L1 (1 + |x|)(L(1 + |x|))p/(1−σ)
L2 (1 + |x|)
:=
.
b(x)up (x) ≈
(1 + |x|)γ
(1 + |x|)γ
by Lemma 2.3 we have that L2 ∈ K. Now suppose that γ ≥ 2 and
RObviously
∞ 1−γ
t
L
(t)dt
< ∞. Then applying Theorem 1.4, we conclude that system (4.3)
2
1
has a unique classical solution (u, v) such that u(x) ≈ θλ (x) and
 R
1/(1−β)
∞
L2 (s)

ds
,
if γ = 2,


s
|x|+1


1/(1−β)

L
((1+|x|))
 2
if 2 < γ < n − (n − 2)β,
(γ−2)/(1−β) ,
v(x) ≈ (1+|x|)
R |x|+2 L2 (s) 1/(1−β)

1

, if γ = n − (n − 2)β,

(1+|x|)n−2
s ds
1



1

if γ > n − (n − 2)β.
(1+|x|)n−2 ,
Case 3: λ = n − (n − 2)σ. We have
Z
L1 (1 + |x|) |x|+2 L(s) p/(1−σ)
L2 (1 + |x|)
p
b(x)u (x) ≈
ds
:=
.
s
(1 + |x|)µ+(n−2)p 1
(1 + |x|)µ+(n−2)p
EJDE-2011/88
ASYMPTOTIC BEHAVIOR
11
By Lemma 2.3 and Remark
R ∞ 2.5, obviously we have that L2 ∈ K. Now suppose that
µ + (n − 2)p ≥ 2 and 1 t1−µ−(n−2)p L2 (t)dt < ∞. Then applying Theorem 1.4,
we conclude that (4.3) has a unique classical solution (u, v) such that u(x) ≈ θλ (x)
and
 R ∞ L (s) 1/(1−β)
2

ds
,
if µ + (n − 2)p = 2,

 |x|+1 s

1/(1−β)


 (L2 (1+|x|))
if 2 < µ + (n − 2)p < n − (n − 2)β,
µ+(n−2)p ,
1−β
(1+|x|)
v(x) ≈
R
1/(1−β)

|x|+2 L2 (s)
1

, if µ + (n − 2)p = n − (n − 2)β,


(1+|x|)n−2
s ds
1



1
if µ + (n − 2)p > n − (n − 2)β.
(1+|x|)n−2 ,
Case 4: λ > n − (n − 2)σ. We have
b(x)up (x) ≈
L1 (1 + |x|)
.
(1 + |x|)n−2+µ
R∞
Suppose that n − 2 + µ ≥ 2 and 1 t1−(n−2+µ) L1 (t)dt < ∞. Then applying
Theorem 1.4, we conclude that (4.3) has a unique classical solution (u, v) such that
u(x) ≈ θλ (x) and
 R ∞ L (s) 1/(1−β)
1

,
if n − 2 + µ = 2,

s ds
|x|+1


1/(1−β)

(L
(1+|x|))
 1
if 2 < n − 2 + µ < n − (n − 2)β,
(µ+n−4)/(1−β) ,
v(x) ≈ (1+|x|)
R |x|+2 L1 (s) 1/(1−β)

1

, if n − 2 + µ = n − (n − 2)β,

(1+|x|)n−2
s ds
1



1
if n − 2 + µ > n − (n − 2)β.
(1+|x|)n−2 ,
Acknowledgments. We want thank the anonymous referee for a careful reading
of the original manuscript.
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RYM CHEMMAM, ABDELWAHEB DHIFLI, HABIB MÂAGLI
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Rym Chemmam
Département de Mathématiques, Faculté des Sciences de Tunis, Université Tunis El
Manar, Campus universitaire, 2092 Tunis, Tunisia
E-mail address: Rym.Chemmam@fst.rnu.tn
Abdelwaheb Dhifli
Département de Mathématiques, Faculté des Sciences de Tunis, Université Tunis El
Manar, Campus universitaire, 2092 Tunis, Tunisia
E-mail address: dhifli waheb@yahoo.fr
Habib Mâagli
Département de Mathématiques, Faculté des Sciences de Tunis, Université Tunis El
Manar, Campus universitaire, 2092 Tunis, Tunisia
E-mail address: habib.maagli@fst.rnu.tn