Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 171, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
TO NONLINEAR RADIAL p-LAPLACIAN EQUATIONS
SYRINE MASMOUDI, SAMIA ZERMANI
Abstract. This article concerns the existence, uniqueness and boundary behavior of positive solutions to the nonlinear problem
1
(AΦp (u0 ))0 + a1 (x)uα1 + a2 (x)uα2 = 0, in (0, 1),
A
lim AΦp (u0 )(x) = 0, u(1) = 0,
x→0
where p > 1, α1 , α2 ∈ (1 − p, p − 1), Φp (t) = t|t|p−2 , t ∈ R, A is a positive
differentiable function and a1 , a2 are two positive measurable functions in (0, 1)
satisfying some assumptions related to Karamata regular variation theory.
1. Introduction
In recent years, the existence of positive solutions for elliptic problems involving
the p-Laplacian has found considerable interest and different approaches have been
developed. This is due to their signification in various areas of pure and applied
mathematics including geological sciences, fluid dynamics, electrostatics, cosmology
(see [1, 2, 6, 10, 12]), as well as in relation to inequalities of Poincaré, Writinger,
Sobolev type and isoperimetric inequalities (see [3, 8, 9, 13, 15]). Motivations for
studying radial solutions can be found in [4, 5, 11, 13, 19] and references therein.
Mâagli et al [15] considered the quasilinear elliptic problem
−∆p u := − div(|∇u|p−2 ∇u) = q(x)uα
u=0
in Ω
on ∂Ω,
(1.1)
where Ω is a C 2 bounded domain of Rn (n ≥ 2), p > 1, the exponent α ∈ (−1, p−1)
and q ∈ C(Ω) is a positive function having singular behavior near the boundary
∂Ω. More precisely, let d(x) be the Euclidean distance of x ∈ Ω to ∂Ω then
q(x) = d(x)−β L(d(x)), with 0 < β < p and L belongs to a functional class K called
Karamata class and defined on (0, η], (η > diam(Ω)) by
Z η z(s) ds : c > 0 and z ∈ C([0, η]), z(0) = 0 .
K := t → L(t) := c exp
s
t
For the convenience of the readers, we briefly describe the result proved in [15].
2010 Mathematics Subject Classification. 34B18, 35J66.
Key words and phrases. p-Laplacian problem; positive solution; boundary behavior;
Schauder fixed point theorem.
c
2015
Texas State University - San Marcos.
Submitted February 2, 2015. Published June 22, 2015.
1
2
S. MASMOUDI, S. ZERMANI
EJDE-2015/171
R η 1−β
1
Theorem 1.1 ([15]). Let L ∈ K ∩ C 2 ((0, η]) such that 0 t p−1 L(t) p−1 dt < ∞.
Then problem (1.1) has a unique positive and continuous solution u satisfying, for
x ∈ (0, 1),

p−1
p−1−α
R d(x) −1
1


p−1 ds
s
L(s)

0



p−β
1

d(x) p−1−α
L(d(x)) p−1−α
u(x) ≈
1
R
p−1−α


η
−1

d(x)
s
L(s)ds


d(x)



d(x)
if β = p,
if 1 + α ≤ β < p,
(1.2)
if β = 1 + α,
if β < 1 + α.
Here and throughout this paper, the notation f (x) ≈ g(x), x ∈ S for f and g
nonnegative functions defined on a set S, means that there exists c > 0 such that
1
c g(x) ≤ f (x) ≤ cg(x), for each x ∈ S.
If Ω is the unit ball, similar result was shown in [4] for radial solution of problem
(1.1) which becomes in the radial form
1
(AΦp (u0 ))0 + q(x)uα = 0, in (0, 1),
A
lim AΦp (u0 )(x) = 0, u(1) = 0,
(1.3)
x→0
where Φp (t) = t|t|p−2 , t ∈ R and A(t) = tn−1 . Indeed, by using Karamata variation
theory, the authors in [4] established existence and asymptotic behavior of a unique
positive continuous solution to (1.3) for α < p − 1 and for a large class of functions
A including the example A(t) = tn−1 . More precisely, they proved Theorem 1.3
below under the following assumptions:
(H0) A is a continuous function in [0, 1), positive and differentiable in (0, 1) such
that
A(x) ≈ xλ (1 − x)µ ,
where λ ≥ 0 and µ < p − 1.
(H1) q is a positive measurable function on (0, 1) such that
q(x) ≈ (1 − x)−β L(1 − x)
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
R η 1−β
1
Remark 1.2. We need to verify condition 0 t p−1 L(t) p−1 dt < ∞ in hypothesis
(H1), only for β = p. This is due to Karamata’s theorem which we recall in Lemma
2.2 below.
Theorem 1.3 ([4]). Assume (H0)–(H1) hold. Then problem (1.3) has a unique
positive and continuous solution u satisfying, for x ∈ (0, 1),
p−β
u(x) ≈ (1 − x)min ( p−1−α ,
p−1−µ
p−1 )
ΨL,β,α (x),
(1.4)
EJDE-2015/171
EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
3
where ΨL,β,α is the function defined on (0, 1) by

p−1
p−1−α
R 1−x −1
1


p−1 ds
if β = p,
s
L(s)

0



1

L(1 − x) p−1−α
if (µ+1)(p−1−α)+αp
< β < p,
p−1
ΨL,β,α (x) := (1.5)
1
p−1−α

R
η

(µ+1)(p−1−α)+αp
−1

if
β
=
,
s
L(s)ds

p−1
1−x




(µ+1)(p−1−α)+αp
1
if β <
.
p−1
For the special case when A(t) = tn−1 the estimates (1.2) and (1.4) are the same.
In this article, we study the boundary-value problem
1
(AΦp (u0 ))0 + a1 (x)uα1 + a2 (x)uα2 = 0, in (0, 1),
A
(1.6)
lim AΦp (u0 )(x) = 0, u(1) = 0,
x→0
where α1 , α2 ∈ (1 − p, p − 1) and A satisfies (H0). Our purpose is to establish an
existence and a uniqueness of a continuous solution to (1.6) and to give estimates
on such solution, where appear the combined effects of singular and sublinear terms
in the nonlinearity.
The pure elliptic semilinear problem corresponding to (1.6) is
∆u + a1 (x)uα1 + a2 (x)uα2 = 0,
u = 0,
in Ω,
on ∂Ω,
(1.7)
which has been studied by several authors on smooth domains, see for example
[7, 14, 16, 17, 20] and references therein.
Let us introduce our conditions on the functions ai :
(H2) For i ∈ {1, 2}, ai is a positive measurable function and satisfies for each
x ∈ (0, 1)
ai (x) ≈ (1 − x)−βi Li (1 − x),
where βi ≤ p and Li ∈ K defined on (0, η], (η > 1) such that
Z η
1−βi
1
t p−1 Li (t) p−1 dt < ∞.
0
p−β
As it turns out, estimates (1.4) depend closely on min ( p−1−α
, p−1−µ
p−1 ). Also, as
it will be seen, the numbers
p − β1
p−1−µ
δ1 = min (
,
),
p − 1 − α1
p−1
p − β2
p−1−µ
δ2 = min (
,
)
p − 1 − α2
p−1
play a crucial role in the combined effects of singular and sublinear nonlinearities
in problem (1.6) and lead to a competition. However, without loss of generality,
p−β1
p−β2
we can suppose that p−1−α
≤ p−1−α
and we introduce the function θ defined on
1
2
(0, 1) by
(
(1 − x)δ1 ΨL1 ,β1 ,α1 (x)
if δ1 < δ2 ,
θ(x) :=
(1.8)
δ1
(1 − x) (ΨL1 ,β1 ,α1 (x) + ΨL2 ,β2 ,α2 (x)) if δ1 = δ2 ,
where for i ∈ {1, 2}, ΨLi ,βi ,αi is the function defined by (1.5).
Now, we are ready to state our main result.
4
S. MASMOUDI, S. ZERMANI
EJDE-2015/171
Theorem 1.4. Assume (H0) and (H2) hold and suppose that α1 , α2 ∈ (1−p, p−1).
Then problem (1.6) has a unique positive and continuous solution u satisfying, for
each x ∈ (0, 1),
u(x) ≈ θ(x),
(1.9)
where θ is the function defined by (1.8).
The outline of this article is as follows. In Section 2, we give some already
known results on functions in K useful for our study and we give estimates of some
potential functions. In Section 3, we prove our main result.
2. Preliminary results
Our arguments combine a method of fixed point theorem with Karamata regular
variation theory. So, we are quoting some properties of functions in K useful for
our study.
2.1. The Karamata class K. It is obvious to see that a function L is in K if and
only if L is a positive function in C 1 ((0, η]) such that
tL0 (t)
= 0.
t→0 L(t)
lim
A standard function belonging to the class K is given by
L(t) :=
p
Y
ω
(logk ( ))λk ,
t
k=1
∗
p
where p ∈ N , (λ1 , λ2 , . . . , λp ) ∈ R , ω is a positive real number sufficiently large
and logk (x) = log o log o . . . .o log(x) (k times).
Lemma 2.1 ([18]).
(i) Let L ∈ K and ε > 0. Then limt→0 tε L(t) = 0.
(ii) Let L1 , L2 ∈ K, p ∈ R. Then L1 + L2 ∈ K, L1 L2 ∈ K and Lp1 ∈ K.
Lemma 2.2 (Karamata’s Theorem). Let L ∈ K be defined on (0, η] and σ ∈ R.
Rη
(i) If σ > −1, then 0 tσ L(t)dt converges and
Z t
t1+σ L(t)
sσ L(s)ds ∼
as t → 0+ .
σ
+
1
0
Rη
(ii) If σ < −1, then 0 tσ L(t)dt diverges and
Z η
t1+σ L(t)
sσ L(s)ds ∼ −
as t → 0+ .
σ+1
t
Lemma 2.3 ([7]). Let L ∈ K be defined on (0, η]. Then
Z η
L(s)
t→
ds ∈ K.
s
t
Rη
If further 0 L(t)
t dt converges, then
Z t
L(s)
t→
ds ∈ K.
s
0
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EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
5
Lemma 2.4 ([7]). For i ∈ {1, 2}, let ηi < 1 and Li ∈ K. For t ∈ (0, η), put
1
1
Z η L (s) 1−η
Z η L (s) 1−η
1
2
1
2
J(t) =
ds
ds
+
.
s
s
t
t
Then, for t ∈ (0, η), we have
Z η η1
(J L1 + J η2 L2 )(s)
ds ≈ J(t).
s
t
Lemma 2.5 ([7]). For i ∈ {1, 2}, let ηi < 1 and Li ∈ K such that
For t ∈ (0, η), put
1
1
Z t L (s) 1−η
Z t L (s) 1−η
1
2
1
2
I(t) =
+
.
ds
ds
s
s
0
0
Then, for t ∈ (0, η), we have
Z t
0
Rη
0
Li (s)
s ds
< ∞.
(I η1 L1 + I η2 L2 )(s)
ds ≈ I(t).
s
2.2. Potential estimates. For a nonnegative measurable function f in (0, 1), let
Z 1
Z t
1
1
Gp f (x) :=
(
A(s)f (s)ds) p−1 dt.
x A(t) 0
We point out that if f is a nonnegative measurable function such that the mapping
x → A(x)f (x) is integrable in [0, 1], then Gp f is the solution of the problem
1
(AΦp (u0 ))0 + f = 0, in (0, 1),
A
(2.1)
lim AΦp (u0 )(x) = 0, u(1) = 0.
x→0
In what follows, we aim to prove Proposition 2.8. To this end, we need the following
two lemmas which are proved in [4] and [7].
Lemma 2.6 ([4]). Assume (H0) and (H1) hold. Then for

1
R 1−x L(s) p−1


ds

s
0



1
L(1 − x) p−1
p−β p−1−µ
Gp (q)(x) ≈ (1 − x)min ( p−1 , p−1 ) R
1
η


p−1
( 1−x L(s)

s ds)




1
x ∈ (0, 1), we have
if β = p,
if µ + 1 < β < p,
if β = µ + 1,
if β < µ + 1.
Lemma 2.7 ([7]). For s, t > 0, η1 < 1 and η2 < 1, we have
2−max(1−η1 ,1−η2 ) (t + s) ≤ t1−η1 (t + s)η1 + s1−η2 (t + s)η2 ≤ 2(t + s).
Proposition 2.8. Assume (H0) and (H2) hold. Let θ be the function given by
(1.8). Then for x ∈ (0, 1),
Gp (a1 θα1 + a2 θα2 )(x) ≈ θ(x).
Proof. For t ∈ (0, 1), let
1
1
K(t) := L1p−1−α1 + L2p−1−α2 (t),
1
1
Z η L (s) p−1−α
Z η L (s) p−1−α
1
2
1
2
N (t) :=
ds
ds
+
,
s
s
t
t
6
S. MASMOUDI, S. ZERMANI
M (t) :=
t
Z
0
EJDE-2015/171
1
1
Z
p−1
p−1
(L1 (s)) p−1 p−1−α1 t (L2 (s)) p−1 p−1−α2
ds
+
ds
.
s
s
0
p−β1
p−β2
1 )+pα1
Since δ1 < δ2 is equivalent to p−1−α
< p−1−α
and (µ+1)(p−1−α
< β1 < p,
p−1
1
2
we can write

1
p−1
R

 1−x L1 (s) p−1 ds p−1−α1 if β1 = p, β2 < p,

s

0


p−β1

1


(1 − x) p−1−α1 L1 (1 − x) p−1−α1



p−β2
p−β1
1 )+pα1
 if p−1−α
< p−1−α
, (µ+1)(p−1−α
< β1 < p,

p−1
1
2



p−β1



(1 − x) p−1−α1 K(1 − x)



p−β1
p−β2
1 )+pα1

 if p−1−α
= p−1−α
, (µ+1)(p−1−α
< β1 < p,
p−1
1
2
θ(x) =
p−1−µ

(1 − x) p−1 N (1 − x)




1 )+pα1
2 )+pα2

if β1 = (µ+1)(p−1−α
, β2 = (µ+1)(p−1−α
,

p−1
p−1



p−1−µ R η
1


(1 − x) p−1 ( 1−x L1s(s) ds) p−1−α1




1 )+pα1
2 )+pα2

if β1 = (µ+1)(p−1−α
, β2 < (µ+1)(p−1−α
,

p−1
p−1


p−1−µ

(µ+1)(p−1−α
)+pα
1
1

(1 − x) p−1
if β1 <
,


p−1


M (1 − x) if β1 = β2 = p.
The main idea is to prove that the function a1 θα1 + a2 θα2 satisfies (H1) and then
to apply Lemma 2.6.
Throughout the proof, we use Lemma 2.1 and Lemma 2.3 to verify that some
functions are in K. We distinguish the following cases.
Case 1. β1 = p and β2 < p. Using (H2) we have
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)−p L1 (1 − x)θα1 (x) + (1 − x)−β2 L2 (1 − x)θα2 (x).
p−1
R 1−x
1
Since θ(x) = ( 0 s−1 L1 (s) p−1 ds) p−1−α1 < ∞, by Lemmas 2.1 and 2.3, the function x → Li (1 − x)θαi (x) ∈ K, for i ∈ {1, 2}.
Now, using β2 < p we deduce by Lemma 2.1(i) that
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)−p L1 (1 − x)θα1 (x).
Moreover, by calculus we have
1
1
1
Z η
Z η
Z t
α1
(L1 (t)θα1 (1 − t)) p−1
L1 (t) p−1
L1 (s) p−1
dt =
(
ds) p−1−α1 dt
t
t
s
0
0
0
1
Z η
p−1
p − 1 − α1
L1 (t) p−1
=
(
dt) p−1−α1 < ∞ .
p−1
t
0
So applying Lemma 2.6, for β = p and L(t) = L1 (t)θα1 (1 − t), we obtain
1
p−1
Z 1−x L (s) p−1
p−1−α
1
1
α1
α2
Gp (a1 θ + a2 θ )(x) ≈
ds
= θ(x).
s
0
Case 2.
p−β1
p−1−α1
<
p−β2
p−1−α2
and
(µ+1)(p−1−α1 )+pα1
p−1
a1 (x)θα1 (x) ≈ (1 − x)
pα1 +β1 (1−p)
p−1−α1
< β1 < p. Using (H2) we have
p−1
L1 (1 − x) p−1−α1
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EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
and
7
α2
p−β1
a2 (x)θα2 (x) ≈ (1 − x)α2 p−1−α1 −β2 (L2 L1p−1−α1 )(1 − x).
Now, since
p−β1
p−1−α1
<
p−β2
p−1−α2 ,
we have
(p − β1 )(1 +
α1 − α2
) < p − β2 ,
p − 1 − α1
and so
pα1 + β1 (1 − p)
p − β1
< α2
− β2 .
p − 1 − α1
p − 1 − α1
We deduce by Lemma 2.1(i) that
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)
(µ+1)(p−1−α1 )+pα1
p−1
Moreover, since
pα1 +β1 (1−p)
p−1−α1
p−1
L1 (1 − x) p−1−α1 .
< β1 < p, we deduce by a simple calculus that
−pα1 + β1 (p − 1)
∈ (µ + 1, p).
p − 1 − α1
p−1
1 +β1 (p−1)
and L(t) = (L1 (t)) p−1−α1 we obtain
So applying Lemma 2.6, for β = −pαp−1−α
1
that
p−β1
1
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x) p−1−α1 L1 (1 − x) p−1−α1 = θ(x).
Case 3.
p−β1
p−1−α1
=
p−β2
p−1−α2
and
(µ+1)(p−1−α1 )+pα1
p−1
a1 (x)θα1 (x) ≈ (1 − x)
pα1 +β1 (1−p)
p−1−α1
and
< β1 < p. Using (H2) we have
(L1 K α1 )(1 − x)
p−β1
a2 (x)θα2 (x) ≈ (1 − x)α2 p−1−α1 −β2 (L2 K α2 )(1 − x).
Now,
p−β1
p−1−α1
=
p−β2
p−1−α2
is equivalent to
(p − β1 )(1 +
α1 − α2
) = p − β2 ;
p − 1 − α1
that is,
pα1 + β1 (1 − p)
p − β1
= α2
− β2 .
p − 1 − α1
p − 1 − α1
Then
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)
pα1 +β1 (1−p)
p−1−α1
(L1 K α1 + L2 K α2 )(1 − x).
Furthermore, the function x → (L1 K α1 + L2 K α2 )(1 − x) ∈ K. Now using that
(µ+1)(p−1−α1 )+pα1
< β1 < p, we deduce that
p−1
−pα1 + β1 (p − 1)
∈ (µ + 1, p).
p − 1 − α1
So applying Lemma 2.6, for β =
that
−pα1 +β1 (p−1)
p−1−α1
and L = L1 K α1 + L2 K α2 , we obtain
p−β1
1
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x) p−1−α1 (L1 K α1 + L2 K α2 ) p−1 (1 − x).
Hence, since
αi
p−1
< 1 for i ∈ {1, 2} and
1
1
1
(s + t) p−1 ≈ s p−1 + t p−1 ,
for s, t > 0,
(2.2)
8
S. MASMOUDI, S. ZERMANI
EJDE-2015/171
1
1
applying Lemma 2.7 for t = L1 (x) p−1−α1 , s = L2 (x) p−1−α2 and ηi =
{1, 2}), we obtain that
αi
p−1 ,
(i ∈
p−β1
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x) p−1−α1 K(1 − x) = θ(x).
Case 4. β1 =
(µ+1)(p−1−α1 )+pα1
p−1
and β2 =
(µ+1)(p−1−α2 )+pα2
.
p−1
Using (H2) we have
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)−(1+µ) (L1 N α1 + L2 N α2 )(1 − x).
Furthermore, the function x → (L1 N α1 + L2 N α2 )(1 − x) ∈ K. So applying Lemma
2.6 for β = 1 + µ and L = L1 N α1 + L2 N α2 , we obtain that
1
Z η (L N α1 + L N α2 )(t) p−1
p−1−µ
1
2
α1
α2
p−1
.
Gp (a1 θ + a2 θ )(x) ≈ (1 − x)
dt
t
1−x
Since
η
Z
p−1
p−1
L1 (s) p−1−α1 η L2 (s) p−1−α2
N
(t) ≈
ds
+
ds
,
s
s
t
t
it follows that N p−1 ≈ J, where J is the function given in Lemma 2.4, for ηi =
αi
p−1 , i ∈ {1, 2}. So, we have
1
Z η (L J η1 + L J η2 )(t) p−1
p−1−µ
1
2
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x) p−1
dt
.
t
1−x
p−1
Z
Hence, since ηi < 1 for i ∈ {1, 2}, by Lemma 2.4, it follows that
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x)
≈ (1 − x)
1 )+pα1
Case 5. β1 = (µ+1)(p−1−α
and β2 <
p−1
1
p−1−α
R
η
1
L1 (s)
. Using (H2), we have
s ds
1−x
p−1−µ
p−1
p−1−µ
p−1
1
J p−1 (1 − x)
N (1 − x) = θ(x).
(µ+1)(p−1−α2 )+pα2
.
p−1
Let b(1 − x) =
a1 (x)θα1 (x) + a2 (x)θα2 (x)
≈ (1 − x)−(1+µ) (L1 bα1 )(1 − x) + (1 − x)−β2 +α2
p−1−µ
p−1
(L2 bα2 )(1 − x).
Now, since for i ∈ {1, 2}, the function x → (Li bαi )(1 − x) ∈ K and −(1 + µ) <
−β2 + α2 p−1−µ
p−1 , it follows from Lemma 2.1(i) that
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)−(1+µ) (L1 bα1 )(1 − x).
Applying again Lemma 2.6, for β = 1 + µ and L = L1 bα1 , we obtain
1
Z η (L bα1 )(s) p−1−α
p−1−µ
1
1
α1
α2
p−1
Gp (a1 θ + a2 θ )(x) ≈ (1 − x)
ds
≈ θ(x).
s
1−x
Case 6. β1 <
(µ+1)(p−1−α1 )+pα1
.
p−1
Using (H2), we have
a1 (x)θα1 (x) ≈ (1 − x)−β1 +α1
Applying Lemma 2.6 for β = β1 −
α1 p−1−µ
p−1
p−1−µ
p−1
L1 (1 − x).
< 1 + µ and L = L1 , we deduce that
Gp (a1 θα1 )(x) ≈ (1 − x)
p−1−µ
p−1
.
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EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
p−β1
p−β2
Moreover, since p−1−µ
p−1 < p−1−α1 ≤ p−1−α2 , it follows that β2 <
So, in the same manner we obtain
Gp (a2 θα2 )(x) ≈ (1 − x)
p−1−µ
p−1
9
(µ+1)(p−1−α2 )+pα2
.
p−1
.
Then, we conclude that
Gp (a1 θα1 + a2 θα2 )(x) ≈ (1 − x)
p−1−µ
p−1
= θ(x).
Case 7. β1 = β2 = p. Using (H2), we have
a1 (x)θα1 (x) + a2 (x)θα2 (x) ≈ (1 − x)−p (L1 M α1 + L2 M α2 )(1 − x).
αi
Furthermore, the function x → (L1 M α1 + L2 M α2 )(1 − x) ∈ K. Now, since p−1
<1
1
R η (Li (s)) p−1
αi
ds < ∞, for i ∈ {1, 2}, applying (2.2) and Lemma 2.5 for ηi = p−1
,
and 0
s
we obtain
1
1
α1
α2
1
Z η
Z η
(L1 M α1 + L2 M α2 ) p−1 (t)
(L1p−1 M p−1 + L2p−1 M p−1 )(t)
dt ≈
dt
t
t
0
0
≈ M (η) < ∞.
So applying Lemma 2.6 for β = p and L = L1 M α1 + L2 M α2 , by (2.2) we deduce
that
1
1
Z 1−x
((L1 M α1 ) p−1 + (L2 M α2 ) p−1 )(t)
α1
α2
Gp (a1 θ + a2 θ )(x) ≈
dt.
t
0
Then, using again Lemma 2.5 we conclude that
Gp (a1 θα1 + a2 θα2 )(x) ≈ M (1 − x) = θ(x).
Proposition 2.9. Assume (H0)–(H1) hold. Then the family of functions
Fq = {x → Gp f (x); f ∈ B((0, 1)), |f | ≤ q}
is uniformly bounded and equicontinuous in [0, 1]. Consequently Fq is relatively
compact in C([0, 1]).
Proof. Let f be a measurable function such that |f (x)| ≤ q(x), x ∈ (0, 1). By
Proposition 2.6, we have
p−β
|Gp f (x)| ≤ Gp q(x) ≈ (1 − x)min ( p−1 ,
p−1−µ
p−1 )
ΨL,β,0 (x).
p−β
p−1−µ
From Lemma 2.1 and Lemma 2.3, the function x → (1 − x)min ( p−1 , p−1 ) ΨL,β,0 (x)
is continuous on [0, 1) and tends to zero as x → 1. Then, we prove that Fq is
uniformly bounded and limx→1 Gp (f )(x) = 0, uniformly on f .
Moreover, let x, x0 ∈ [0, 1] such that x < x0 . Then
Z t
Z x0 1
p−1
1
0
|Gp f (x) − Gp f (x )| ≤
A(s)q(s)ds
dt.
A(t) 0
x
Since Gp q(0) < ∞, it follows by the dominated convergence theorem, the equicontinuity of Fq in [0, 1]. Hence, by Ascoli’s theorem, we deduce that Fq is relatively
compact in C([0, 1]).
10
S. MASMOUDI, S. ZERMANI
EJDE-2015/171
3. Proof of main result
3.1. Existence and boundary behavior. Assume (H0) and (H2) hold. Let θ be
the function given by (1.8). By Proposition 2.8, there exists a constant m ≥ 1 such
that for each x ∈ [0, 1),
1
θ(x) ≤ Gp (a1 θα1 + a2 θα2 )(x) ≤ mθ(x).
(3.1)
m
Now, we look at the existence of positive solution of problem (1.6) satisfying u(x) ≈
θ(x). We will use a fixed-point argument. We consider the set
Γ := {u ∈ C([0, 1]) :
1
θ ≤ u ≤ c0 θ},
c0
p−1
where c0 = m p−1−max(|α1 |,|α2 |) . Obviously, the set Γ is not empty. We consider the
integral operator T on Γ defined by
T u(x) := Gp (a1 uα1 + a2 uα2 )(x),
x ∈ [0, 1].
First, we observe that T Γ ⊂ Γ. Indeed, let u ∈ Γ, then for i ∈ {1, 2}, we have
−|αi |
c0
|α |
ai (x)θαi (x) ≤ ai (x)uαi (x) ≤ c0 i ai (x)θαi (x),
x ∈ [0, 1).
This and (3.1) imply
max(|α1 |,|α2 |)
p−1
1
max(|α1 |,|α2 |)
p−1
θ(x) ≤ T u(x) ≤ mco
θ(x).
mco
max(|α1 |,|α2 |)
p−1
= c0 and T Γ ⊂ C([0, 1]), it follows that T leaves invariant the
Since mco
convex Γ.
|α |
|α |
Next, let q = c0 1 a1 θα1 +c0 2 a2 θα2 . By the proof of Proposition 2.8, the function
q satisfies (H1). Since T Γ is a closed set of Fq , it follows by Proposition 2.9, that
T Γ is relatively compact in C([0, 1]).
Now, we shall prove the continuity of operator T in Γ. Let (un ) be a sequence
in Γ converging uniformly to u in Γ. Then, by applying the dominated convergence
theorem, we conclude that for each x ∈ [0, 1], T un (x) → T u(x) as n → +∞.
Consequently, as T Γ is relatively compact in C([0, 1]), we deduce that the pointwise
convergence implies the uniform convergence. Namely, kT un − T uk∞ → 0 as n →
+∞.
Therefore, T is a compact operator from Γ into itself. So the Schauder fixed-point
theorem implies the existence of u ∈ Γ such that
u = Gp (a1 uα1 + a2 uα2 )(x), x ∈ [0, 1].
We conclude that u is a positive continuous solution of problem (1.6) which satisfies
(1.9).
3.2. Uniqueness. Let u and v be two solutions of (1.6) in Γ. Then, there exists a
constant M > 1 such that
1
u
≤ ≤ M.
M
v
This implies that the set
J = {t ∈ (1, ∞) :
1
u ≤ v ≤ tu}
t
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EXISTENCE AND ASYMPTOTIC BEHAVIOR OF SOLUTIONS
11
is not empty. Now, put c := inf J. We aim to show that c = 1. Suppose that c > 1
and let α = max(|α1 |, |α2 |), then we have
−α
1
1
(AΦp (v 0 ))0 + (AΦp (c p−1 u0 ))0
A
A
= a1 (x)(v α1 − c−α uα1 ) + a2 (x)(v α2 − c−α uα2 ) ≥ 0,
−
lim (AΦp (v 0 ) − AΦp (c
−α
p−1
x→0
in (0, 1),
u0 ))(x) = 0,
u(1) = v(1) = 0.
This implies that the function
−α
ψ(x) := AΦp (c p−1 u0 ) − AΦp (v 0 )(x)
is nondecreasing on (0, 1) with limx→0 ψ(x) = 0. Hence from the monotonicity
−α
of Φp , we obtain that the function c p−1 u − v is nondecreasing on (0, 1) satisfying
−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. Then, we have c p−1 ∈ J. Now, since α ∈ (0, p − 1),
α
it follows that c p−1 < c and this yields to a contradiction with the definition of c.
Hence, c = 1 and so u = v.
Example 3.1. Let 1 − p < α1 < 0 < α2 < p − 1 and β1 , β2 < p such that
p−β2
p−1−α2 .
p−β1
p−1−α1 ≤
−βi
Let a1 , a2 be continuous functions on (0, 1) such that ai (x) ≈ (1 − x) ,
for i ∈ {1, 2}. Then problem (1.6) has a unique continuous solution u satisfying for
x ∈ (0, 1),

p−β1
(µ+1)(p−1−α1 )+pα1

< β1 < p,
(1 − x) p−1−α1 if
p−1



p−1−µ
1

η


(1 − x) p−1 (log( 1−x
)) p−1−α2



 if β1 = (µ+1)(p−1−α1 )+pα1 , β2 = (µ+1)(p−1−α2 )+pα2 ,
p−1
p−1
u(x) ≈
1
(1 − x) p−1−µ
p−1 (log( η )) p−1−α1


1−x


(µ+1)(p−1−α1 )+pα1
2 )+pα2


, β2 < (µ+1)(p−1−α
,
if
β
=
1

p−1
p−1



p−1−µ

1 )+pα1
(1 − x) p−1
if β1 < (µ+1)(p−1−α
.
p−1
Acknowledgements. The authors gratefully acknowledge the referees for valuable
comments.
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Syrine Masmoudi
Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire,
2092 Tunis, Tunisia
E-mail address: syrine.sassi@fst.rnu.tn
Samia Zermani
Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire,
2092 Tunis, Tunisia
E-mail address: zermani.samia@yahoo.fr