Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 240, pp. 1–10.
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 FOR THE
RADIAL P-LAPLACIAN EQUATION
SONIA BEN OTHMAN, HABIB MÂAGLI
Abstract. We study the existence, uniqueness and asymptotic behavior of
positive solutions to the nonlinear problem
1
(AΦp (u0 ))0 + q(x)uα = 0, in (0, 1),
A
lim AΦp (u0 )(x) = 0, u(1) = 0,
x→0
where α < p − 1, Φp (t) = t|t|p−2 , A is a positive differentiable function and q
is a positive measurable function in (0, 1) such that for some c > 0,
“ Z η z(s) ”
1
≤ q(x)(1 − x)β exp −
ds ≤ c.
c
1−x s
Our arguments combine monotonicity methods with Karamata regular variation theory.
1. Introduction
Let p > 1 and α < p − 1. We consider the boundary-value problem
1
− (AΦp (u0 ))0 + q(x)uα = 0, in (0, 1)
A
AΦp (u0 )(0) := lim AΦp (u0 )(x) = 0, u(1) = 0.
(1.1)
x→0
Here, A is a continuous function in [0, 1), differentiable and positive on (0, 1) and
for all t ∈ R, Φp (t) = t|t|p−2 . Our goal in this paper is to study problem (1.1)
under appropriate conditions on q. We obtain the existence of a unique positive
continuous solution to (1.1) and establish estimates on such solution.
Several articles have been devoted to the study of the differential equation
1
− (AΦp (u0 ))0 + q(x)uα = 0, in (0, 1)
A
with various boundary conditions, especially for the one-dimensional p-Laplacian
equation (see [1, 2, 3, 4, 5, 11, 13, 14, 15]). For α < 0, problem (1.1) has been
studied in [4], where the existence and uniqueness of positive solutions and some
estimates for the solutions have been obtained. Thus, it is interesting to know the
2000 Mathematics Subject Classification. 34B15, 35J65.
Key words and phrases. p-Laplacian; asymptotic behavior; positive solutions;
Schauder’s fixed point theorem.
c
2012
Texas State University - San Marcos.
Submitted September 23, 2012. Published December 28, 2012.
1
2
S. BEN OTHMAN, H. MÂAGLI
EJDE-2012/240
exact asymptotic behavior of such solution as x → 1 and to extend the study of
(1.1) to 0 ≤ α < p − 1.
Asymptotic behavior of solutions of the semilinear elliptic equation
− ∆u = q(x)uα ,
α < 1, x ∈ Ω,
(1.2)
for Ω bounded or an unbounded in Rn (n ≥ 2), with homogeneous Dirichlet
boundary conditions, has been investigated by several authors; see for example
[6, 7, 8, 9, 10, 12, 16, 17, 20] and the references therein. Applying Karamata regular variation theory, Mâagli [16] studied (1.2), when Ω is a bounded C 1,1 -domain.
He showed that (1.2) has a unique positive classical solution that satisfies homogeneous Dirichlet boundary conditions and gave sharp estimates on such solution.
This studied extended the estimates stated in [12, 17, 20]. In this work, we extend
the result established in [16] to the radial case associated to problem (1.1).
To simplify our statements, we need to fix some notation and make some assumptions. Throughout this paper, we shall use K to denote the set of Karamata
functions L defined on (0, η] by
Z η z(s) ds ,
L(t) := c exp
s
t
for some positive constants η, c, and a function z ∈ C([0, η]) such that z(0) = 0.
Recall that L ∈ K if and only if L is a positive function in C 1 ((0, η]), for some
η > 0, such that
tL0 (t)
lim
= 0.
(1.3)
t→0 L(t)
For two nonnegative functions f and g defined on a set S, we write f (x) ≈ g(x),
if there exists a constant c > 0 such that 1c g(x) ≤ f (x) ≤ cg(x), for each x ∈ S.
Furthermore, we refer to Gp f , as the function defined on (0, 1) by
Z 1
Z t
1
p−1
1
Gp f (x) :=
A(s)f (s)ds
dt,
A(t) 0
x
where f is a nonnegative measurable function in (0, 1). We point out that if f is a
nonnegative continuous function such that the mapping x 7→ A(x)f (x) is integrable
in a neighborhood of 0, then Gp f is the solution of the problem
1
− (AΦp (u0 ))0 = f, in (0, 1),
A
AΦp (u0 )(0) = 0, u(1) = 0.
(1.4)
As it is mentioned above, our main purpose in this paper is to establish existence
and global behavior of a positive solution for problem (1.1). Let us introduce our
hypotheses.
The function A is continuous in [0, 1), differentiable and positive in (0, 1) such
that
A(x) ≈ xλ (1 − x)µ
with λ ≥ 0 and µ < p − 1.
The function q is required to satisfy
(H1) q is a positive measurable function on (0, 1) such that
q(x) ≈ (1 − x)−β L(1 − x),
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
3
with β ≤ p and L ∈ K defined on (0, η] (η > 1) such that
Z η
1−β
1
t p−1 (L(t)) p−1 dt < +∞.
0
We need to verify the condition
Z η
1−β
1
t p−1 (L(t)) p−1 dt < +∞
0
in hypothesis (H1), only if β = p (See Lemma 2.2 below).
As a typical example of function q satisfying (H1), we have
q(x) := (1 − x)−β (log
2 −ν
) ,
1−x
x ∈ [0, 1).
Then for β < p and ν ∈ R or β = p and ν > p − 1, the function q satisfies (H1).
Our main result is as follows.
Theorem 1.1. Assume (H1). Then problem (1.1) has a unique positive and continuous solution u satisfying, for x ∈ (0, 1),
u(x) ≈ θβ (x),
where θβ is the function defined on [0, 1) by

p−1
1
p−1−α
R 1−x (L(s)) p−1


ds
,

s
0


p−β

1
p−1−α
p−1−α ,
θβ (x) := (1 − x) p−1−µ (L(1 − x))


(1 − x) p−1 ,



p−1−µ R η
1

p−1−α ,
(1 − x) p−1 ( 1−x L(s)
s ds)
(1.5)
if β = p
if
if
if
(µ+1)(p−1−α)+αp
<β<
p−1
(µ+1)(p−1−α)+αp
β<
p−1
β = (µ+1)(p−1−α)+αp
.
p−1
p, (1.6)
The article is organized as follows. In Section 2, we prove some basic estimates
and recall some results on functions belonging to K. In Section 3, we prove Theorem
1.1. In the last section, we present some applications.
2. Estimates
In what follows, we give estimates on the functions Gp q and Gp (qθβα ), where q
is a function satisfying (H1) and θβ is the function given by (1.6). To this end,
we recall some fundamental properties of functions belonging to the class K, taken
from [7, 18, 19].
Lemma 2.1 ([18, 19]). Let L1 , L2 ∈ K, m ∈ R and > 0. Then L1 L2 ∈ K,
Lm
1 ∈ K, and limt→0+ t L1 (t) = 0.
Lemma 2.2 ([18, 19]). Let L ∈ K and δ ∈ R. Then we have the following:
Rη
(i) If δ < 2, then 0 t1−δ L(t)dt converges and
Z s
s2−δ L(s)
t1−δ L(t)dt ∼
as s → 0+ .
2
−
δ
0
Rη
(ii) If δ > 2, then 0 t1−δ L(t)dt diverges and
Z η
s2−δ L(s)
t1−δ L(t)dt ∼
as s → 0+ .
δ−2
s
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S. BEN OTHMAN, H. MÂAGLI
EJDE-2012/240
Lemma 2.3 ([7]). Let L ∈ K be defined on (0, η], then we have
Z η
L(s)
ds ∈ K.
t 7→
s
t
Rη
If further 0 L(s)
s ds converges, then
Z t
L(s)
t 7→
ds ∈ K.
s
0
Proposition 2.4. Assume q satisfies (H1). Then for x ∈ (0, 1), we have
Gp q(x) ≈ Ψ(1 − x),
where ψ is the function defined on (0, 1] by
R
1
t (L(s)) p−1

ds,

 0p−β s

1
 p−1
p−1
t
(L(t))
,
Ψ(t) =
p−1−µ

p−1

t
,


1
 p−1−µ R η L(s)
t p−1 ( t s ds) p−1 ,
if β = p,
if µ + 1 < β < p,
(2.1)
if β < µ + 1
if β = µ + 1.
Proof. For x ∈ (0, 1), we have
Z 1
1
p−1
Z t
1
λ
µ−β
s
(1
−
s)
L(1
−
s)ds
Gp q(x) ≈
dt.
µ
λ
0
x t p−1 (1 − t) p−1
Put
Z
1
λ
t p−1 (1 − t)
x
Z
1
h(x) :=
µ
p−1
t
sλ (1 − s)µ−β L(1 − s)ds
1
p−1
dt, x ∈ (0, 1).
0
We shall estimate h(x). Since h is continuous and positive on [0, 1/2], it follows
that h(x) ≈ 1, for x ∈ [0, 1/2]. Now, assume that x ∈ [1/2, 1). Then
Z 1
1
p−1
Z t
1
sλ (1 − s)µ−β L(1 − s)ds
h(x) ≈
dt.
µ
0
x (1 − t) p−1
Moreover, for t ∈ [x, 1), we have
Z t
sλ (1 − s)µ−β L(1 − s)ds
0
Z
1/2
sλ (1 − s)µ−β L(1 − s)ds +
=
Z
t
sλ (1 − s)µ−β L(1 − s)ds
1
2
0
Z
1/2
≈1+
sµ−β L(s)ds.
1−t
Then we distinguish the following cases:
R 1/2
• If β < µ + 1, then by Lemma 2.2, 0 sµ−β L(s)ds < ∞. So, since µ < p − 1, we
obtain
p−1−µ
h(x) ≈ (1 − x) p−1 .
• If p > β > µ + 1, then by Lemma 2.2,
Z 1/2
sµ−β L(s)ds ≈ (1 − t)µ+1−β L(1 − t).
1−t
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
5
So,
Z
1/2
1
sµ−β L(s)ds) p−1 ≈ (1 − t)
(1 +
µ+1−β
p−1
1
L p−1 (1 − t).
1−t
Thus, using the fact that β < p and again Lemma 2.2, we obtain that
p−β
1
h(x) ≈ (1 − x) p−1 L p−1 (1 − x).
• If β = µ + 1, then
Z
1−x
Z
1 h(x) ≈
µ
1
1
L(s) p−1
dt.
ds
s
t p−1
1−t
So, using Lemma 2.3 and the fact that µ < p − 1, by Lemma 2.2 it follows that
Z 1
p−1−µ
1
L(s)
p−1
h(x) ≈ (1 − x)
(
ds) p−1 .
s
1−x
0
• If β = p, we deduce by Lemma 2.2 that
Z 1/2
sµ−β L(s)ds ≈ (1 − t)µ+1−p L(1 − t),
1−t
hence
Z
h(x) ≈
0
1−x
1
(L(s)) p−1
ds.
s
This completes the proof.
The following proposition plays a crucial role in this article.
Proposition 2.5. Let q satisfy (H1) and let θβ be the function given in (1.6).
Then for x ∈ (0, 1), we have
Gp (qθβα )(x) ≈ θβ (x).
Proof. Let β ≤ p and µ < p − 1, a straightforward computation shows that for
x ∈ (0, 1),
q(x)θβα (x) ≈ qe(x),
where

α(p−1)
1
p−1−α

L(1−x) R 1−x (L(s)) p−1


ds
if β = p
p

(1−x)
s
0


p−1


p−1−α

 (L(1−x))α(p−β)
if (µ+1)(p−1−α)+αp
< β < p,
p−1
)
(β−
p−1−α
qe(x) := (1−x)
L(1−x)


if β < (µ+1)(p−1−α)+αp

p−1
 (1−x)(β− α(p−1−µ)
)

p−1

α

 L(1−x)
R η L(s)
p−1−α


ds
.
if β = (µ+1)(p−1−α)+αp
p−1
1−x s
(1−x)(µ+1)
So, we deduce that
e − x),
qe(x) = (1 − x)−δ L(1
e ∈ K and
where δ ≤ p. Then, using Lemmas 2.1 and 2.3, we verify that L
R η 1−δ
1
e
t p−1 (L(t)) p−1 dt < +∞. Hence, by Proposition 2.4,
0
e − x),
Gp (qθβα )(x) ≈ Gp qe(x) ≈ ψ(1
x ∈ (0, 1),
e and β by δ. This
where ψe is the function defined in (2.1) by replacing L by L
completes the proof.
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S. BEN OTHMAN, H. MÂAGLI
EJDE-2012/240
3. Proof of Theorem 1.1
3.1. Existence and asymptotic behavior. Let q satisfy (H1) and let θβ be the
function given by (1.6). By Proposition 2.5, there exists a constant m ≥ 1 such
that for each x ∈ (0, 1),
1
θβ (x) ≤ Gp (qθβα )(x) ≤ mθβ (x).
(3.1)
m
Now we look at the existence of positive solution of problem (1.1) satisfying (1.5).
For the case α < 0, we refer to [4]. So prove the existence result only for the case
0 ≤ α < p − 1, and then give the precise asymptotic behavior of such solution for
α < p − 1. We will split the proof into two cases.
Case 1: α < 0. Let u be a positive continuous solution of (1.1). To obtain
estimates (1.5) on the function u, we need the following comparison result.
Lemma 3.1. Let α < 0 and u1 , u2 ∈ C 1 ((0, 1)) ∩ C([0, 1]) be two positive functions
such that
1
− (AΦp (u01 ))0 ≤ q(x)uα
in (0, 1),
1,
A
(3.2)
AΦp (u01 )(0) = 0, u1 (1) = 0
and
1
in (0, 1),
− (AΦp (u02 ))0 ≥ q(x)uα
2,
A
(3.3)
AΦp (u02 )(0) = 0, u2 (1) = 0.
Then u1 ≤ u2 .
Proof. Suppose that u1 (x0 ) > u2 (x0 ) for some x0 ∈ (0, 1). Then there exists
x1 , x2 ∈ [0, 1], such that 0 ≤ x1 < x0 < x2 ≤ 1 and for x1 < x < x2 , u1 (x) > u2 (x)
with u1 (x2 ) = u2 (x2 ), u1 (x1 ) = u2 (x1 ) or x1 = 0.
We deduce that
AΦp (u02 )(x1 ) ≤ AΦp (u01 )(x1 ).
(3.4)
α
α
On the other hand, since α < 0, we have u1 (x) < u2 (x), for each x ∈ (x1 , x2 ). This
yields
1
1
α
(AΦp (u01 ))0 − (AΦp (u02 ))0 ≥ q(uα
2 − u1 ) ≥ 0 on (x1 , x2 ).
A
A
Using further (3.4), we deduce that the function ω(x) := (AΦp (u01 ) − AΦp (u02 ))(x)
is nondecreasing on (x1 , x2 ) with ω(x1 ) ≥ 0. Hence, from the monotonicity of Φp ,
we obtain that the function x 7→ (u1 − u2 )(x) is nondecreasing on (x1 , x2 ) with
(u1 − u2 )(x1 ) ≥ 0 and (u1 − u2 )(x2 ) = 0. This yields to a contradiction. The proof
is complete.
α
Now, we are ready to prove (1.5). Put c = m− p−1−α and v := Gp (qθβα ). It
follows from (1.4) that the function v satisfies
1
− (AΦp (v 0 ))0 = qθβα , in (0, 1).
A
According to (3.1), we obtain by simple calculation that 1c v and cv satisfy respectively (3.2) and (3.3). Thus, we deduce by Lemma 3.1 that
1
v(x) ≤ u(x) ≤ cv(x), x ∈ (0, 1).
c
This proves the result.
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ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
7
p−1
Case 2: 0 ≤ α < p − 1. Put c0 = m p−1−α and let
1
θβ ≤ u ≤ c0 θβ .
Λ := u ∈ C([0, 1]);
c0
Obviously, the function θβ belongs to C([0, 1]) and so Λ is not empty. We consider
the integral operator T on Λ defined by
T u(x) := Gp (quα )(x),
x ∈ [0, 1].
We prove that T has a fixed point in Λ, in order to construct a solution of problem
(1.1). For this aim, first we observe that T Λ ⊂ Λ. Let u ∈ Λ, then for each x ∈ [0, 1)
1
α
(qθβα )(x) ≤ q(x)uα (x) ≤ cα
0 (qθβ )(x).
cα
0
This together with (3.1) implies that
α
1
p−1
θβ .
α θβ ≤ T u ≤ mc0
mc0p−1
α
Since mc0p−1 = c0 and T Λ ⊂ C([0, 1]), then T leaves invariant the convex Λ.
Moreover, since α ≥ 0, then the operator T is nondecreasing on Λ. Now, let {uk }k
be a sequence of functions in C([0, 1]) defined by
1
u0 = θβ , uk+1 = T uk , for k ∈ N.
c0
Since T Λ ⊂ Λ, we deduce from the monotonicity of T that for k ∈ N, we have
u0 ≤ u1 ≤ · · · ≤ uk ≤ uk+1 ≤ c0 θβ .
Applying the monotone convergence theorem, we deduce that the sequence {uk }k
converges to a function u ∈ Λ which satisfies
u(x) = Gp (quα )(x), x ∈ [0, 1].
We conclude that u is a positive continuous solution of problem (1.1) which satisfies
(1.5).
3.2. Uniqueness. Assume that q satisfies (H1). For α < 0, the uniqueness of
solution to problem (1.1) follows from Lemma 3.1. Thus, we look at the case
0 ≤ α < p − 1. Let
Γ = {u ∈ C([0, 1]) : u(x) ≈ θβ (x)}.
Let u and v be two positives solutions of problem (1.1) in Γ. Then there exists a
constant k ≥ 1 such that
1
v
≤ ≤ k.
k
u
This implies that the set
1
J = {t ∈ (1, +∞) : u ≤ v ≤ tu}
t
is not empty. Now, put c := inf J, then we aim to show that c = 1. Suppose that
c > 1, then
−α
1
1
− (AΦp (v 0 ))0 + (AΦp (c p−1 u0 ))0 = q(x)(v α − c−α uα ), in (0, 1),
A
A
−α
lim+ (AΦp (v 0 ) − AΦp (c p−1 u0 ))(x) = 0,
x→0
8
S. BEN OTHMAN, H. MÂAGLI
EJDE-2012/240
−α
(v − c p−1 u)(1) = 0.
So, we have
−α
1
1
− (AΦp (v 0 ))0 + (AΦp (c p−1 u0 ))0 ≥ 0
A
A
in (0, 1),
−α
which implies that the function θ(x) := (AΦp (c p−1 u0 ) − AΦp (v 0 ))(x) is nondecreasing on (0, 1) with limx→0+ θ(x) = 0. Hence from the monotonicity of Φp ,
−α
we obtain that the function x 7→ (c p−1 u − v)(x) is nondecreasing on [0, 1) with
−α
α
(c− p−1 u − v)(1) = 0. This implies that c p−1 u ≤ v. On the other hand, we deduce
α
α
by symmetry that v ≤ c p−1 u. Hence c p−1 ∈ J. Now, since α < p − 1 and c > 1, we
α
have c p−1 < c. This yields to a contradiction with the fact that c := inf J. Hence,
c = 1 and then u = v.
4. Applications
First application. Let q be a positive measurable function in [0, 1) satisfying for
x ∈ [0, 1)
3 −σ
q(x) ≈ (1 − x)−β log
,
1−x
where the real numbers β and σ satisfy one of the following two conditions:
• β < p and σ ∈ R,
• β = p and σ > p − 1.
Using Theorem 1.1, we deduce that problem (1.1) has a positive continuous solution
u in [0, 1] satisfying
(i) If β <
(µ+1)(p−1−α)+αp
,
p−1
then for x ∈ (0, 1),
u(x) ≈ (1 − x)
(ii) If β =
(µ+1)(p−1−α)+αp
p−1
(µ+1)(p−1−α)+αp
p−1
p−1−µ
p−1
(µ+1)(p−1−α)+αp
p−1
log log
p−1−µ
p−1
1−σ
log
(v) If
3 p−1−α
.
1−x
and σ > 1, then for x ∈ (0, 1),
u(x) ≈ (1 − x)
(µ+1)(p−1−α)+αp
p−1
1
3 p−1−α
.
1−x
and σ < 1, then for x ∈ (0, 1),
u(x) ≈ (1 − x)
(iv) If β =
.
and σ = 1, then for x ∈ (0, 1),
u(x) ≈ (1 − x)
(iii) If β =
p−1−µ
p−1
p−1−µ
p−1
.
< β < p, then for x ∈ (0, 1),
p−β
u(x) ≈ (1 − x) p−1−α log
−σ
3 p−1−α
.
1−x
(vi) If β = p and σ > p − 1, then for x ∈ (0, 1),
p−1−σ
3 p−1−α
u(x) ≈ log
1−x
EJDE-2012/240
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS
9
Second application. Let q be a function satisfying (H1) and let α, γ < p − 1. We
are interested in the nonlinear problem
γ
1
− (AΦp (u0 ))0 + Φp (u0 )u0 = q(x)uα , in (0, 1),
A
u
(4.1)
AΦp (u0 )(0) = 0, u(1) = 0.
γ
Put v = u1− p−1 ; then v satisfies
(α−γ)(p−1)
1
p − 1 − γ p−1
− (AΦp (v 0 ))0 = (
) q(x)v p−1−γ ,
A
p−1
AΦp (v 0 )(0) = 0, v(1) = 0.
in (0, 1),
(4.2)
Using Theorem 1.1, we deduce that (4.2) has a unique solution v such that v(x) ≈
θeβ (x), where

p−1−γ
1
p−1−α
R 1−x (L(s)) p−1


ds
if β = p,


s
0


(p−β)(p−1−γ)
p−1−γ


(1 − x) (p−1)(p−1−α) (L(1 − x)) (p−1)(p−1−α) , if (µ+1)(p−1−α)+(α−γ)p
p−1−γ
θeβ (x) =
<
β
<
p,


p−1−µ



if β < (µ+1)(p−1−α)+(α−γ)p
(1 − x) p−1
,

p−1−γ

p−1−γ

p−1−µ R η
(1 − x) p−1 (
(µ+1)(p−1−α)+(α−γ)p
L(s)
ds) (p−1)(p−1−α)
if β =
.
p−1−γ
1−x s
Consequently, (4.1) has a unique solution u satisfying

p−1
1
p−1−α
R 1−x (L(s)) p−1


ds
,
if β = p

s
 0

p−β

1
p−1−α
p−1−α ,
if (µ+1)(p−1−α)+(α−γ)p
< β < p,
p−1−γ
u(x) ≈ (1 − x) p−1−µ (L(1 − x))

(µ+1)(p−1−α)+(α−γ)p

(1 − x) p−1−γ ,
if β <

p−1−γ


p−1−µ R η
1

L(s)
p−1−γ
p−1−α
.
(1 − x)
( 1−x s ds)
, if β = (µ+1)(p−1−α)+(α−γ)p
p−1−γ
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Sonia Ben Othman
Département de Mathématiques, Faculté des Sciences de Tunis, Campus universitaire,
2092 Tunis, Tunisia
E-mail address: Sonia.benothman@fsb.rnu.tn
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