Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 235, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
COMBINED EFFECTS IN NONLINEAR SINGULAR
SECOND-ORDER DIFFERENTIAL EQUATIONS ON THE
HALF-LINE
IMED BACHAR
Abstract. We consider the existence, uniqueness and the asymptotic behavior of positive continuous solutions to the second-order boundary-value problem
1
(Au0 )0 + a1 (t)uσ1 + a2 (t)uσ2 = 0, t ∈ (0, ∞),
A
u(t)
= 0,
lim u(t) = 0,
lim
t→∞ ρ(t)
t→0+
where σ1 , σ2 ∈ (−1, 1), A is a continuous function on [0, ∞), positive and
R
R
1
1
dt < ∞ and 0∞ A(t)
dt = ∞. Here
differentiable on (0, ∞) such that 01 A(t)
Rt 1
ρ(t) = 0 A(s) ds and for i ∈ {1, 2}, ai is a nonnegative continuous function in
(0, ∞) such that there exists c > 0 satisfying for t > 0,
1
hi (m(t))
hi (m(t))
≤ ai (t) ≤ c 2
,
c A2 (t)(1 + ρ(t))µi
A (t)(1 + ρ(t))µi
R z (s)
ρ(t)
where m(t) = 1+ρ(t) and hi (t) = ci t−λi exp( tη is ds), ci > 0, λi ≤ 2,
µi > 2 and zi is continuous on [0, η] for some η > 1 such that zi (0) = 0. The
comparable asymptotic rate of ai (t) determines the asymptotic behavior of the
solution.
1. Introduction
Boundary-value problems on the half-line, have been studied widely in the literature (see, for example, [1, 13, 14, 16, 21, 23] and the references therein). The
motivation for these studies stems from the fact that such problems arise naturally
in the study of radially symmetric solutions of nonlinear elliptic equations (see,
[4, 6, 13, 18, 19, 21, 22] and also many physical models, for example, the model
of gas pressure in a semi-infinite porous medium, the Thomas-Fermi model for determining the electric potential in an isolated neutral atom (see, the Monographs,
[1, 10] and the references therein). Therefore it is very important to investigate the
boundary-value problems for differential equations on half-line.
2010 Mathematics Subject Classification. 34B15, 34B18, 34B27.
Key words and phrases. Green’s function; Karamata regular variation theory;
positive solution; Schauder fixed point theorem.
c 2015 Texas State University - San Marcos.
Submitted May 3, 2015. Published September 11, 2015.
1
2
I. BACHAR
EJDE-2015/235
Zhao [23] considered the problem
u00 + ϕ(., u) = 0,
u > 0,
on (0, ∞),
on (0, ∞),
(1.1)
lim u(t) = 0,
t→0+
where ϕ is a measurable function on (0, ∞)×(0, ∞), dominated by a convex positive
function. Then he showed that there exists b > 0 such that for each µ ∈ (0, b], there
exists a positive continuous solution u of (1.1) satisfying limt→∞ u(t)
t = µ.
On the other hand, in [2], the author studied the singular problem
1
(Au0 )0 + ϕ(·, u) = 0, t ∈ (0, ∞),
A
u > 0, on (0, ∞),
(1.2)
u(t)
= 0,
lim u(t) = 0, lim
t→∞ ρ(t)
t→0+
where A is a continuous function on [0, ∞), positive and differentiable on (0, ∞)
R1 1
R∞ 1
Rt 1
such that 0 A(t)
dt < ∞ and 0 A(t)
dt = ∞. Here ρ(t) = 0 A(s)
ds and the
function ϕ : (0, ∞) × (0, ∞) → [0, ∞) is required to be continuous, non-increasing
with
R ∞ respect to the second variable such that for each c > 0, ϕ(., c) 6= 0 and
A(s) min(1, ρ(s))ϕ(s, c)ds < ∞. The author proved the existence of a unique
0
positive solution u in C([0, ∞)) ∩ C 2 ((0, ∞)) to problem (1.2).
Recently, in [3], the authors considered problem (1.2) with ϕ(t, u) = a(t)uσ ,
σ < 1, (which include the sublinear case) and a is a nonnegative continuous function on (0, ∞) satisfying some appropriate assumptions related to Karamata regular
variation theory. They have proved the existence, uniqueness and the global asymptotic behavior of positive solutions to problem (1.2).
In this article, we study the boundary-value problem
1
(Au0 )0 + a1 (t)uσ1 + a2 (t)uσ2 = 0, t ∈ (0, ∞),
A
u > 0, on (0, ∞),
(1.3)
u(t)
= 0,
lim u(t) = 0, lim
t→∞ ρ(t)
t→0+
where σ1 , σ2 ∈ (−1, 1), A is a continuous function on [0, ∞), positive and differenR1 1
R∞ 1
Rt 1
tiable on (0, ∞) such that 0 A(t)
dt < ∞, 0 A(t)
dt = ∞ and ρ(t) = 0 A(s)
ds,
t > 0.
Our goal is to study (1.3), especially, when the nonlinearity is the sum of a
singular term and a sublinear term. Under appropriate assumptions on a1 and
a2 related to the Karamata class K (see Definition 1.1), we prove the existence,
uniqueness and the global asymptotic behavior of positive continuous solution to
problem (1.3).
R1 1
dt =
Throughout this paper and without loss of generality, we assume that 0 A(t)
1. To state our result, we need some notation.
Definition 1.1. The class K is the set of all Karamata functions L defined on (0, η]
by
Z η z(s) L(t) := c exp
ds ,
s
t
EJDE-2015/235
NONLINEAR SINGULAR SECOND-ORDER DIFFERENTIAL EQUATIONS
3
for some η > 1, where c > 0 and z ∈ C([0, η]) such that z(0) = 0.
The theory of such functions was initiated by Karamata in the fundamental
paper [15]. On the other hand, we emphasize that the first use of the Karamata
theory in the study of the growth rate of solutions near the boundary was done in
the paper of Cirstea and Rădulescu [8].
Remark 1.2. A function L is in K if and only if L is a positive function in
0
(t)
C 1 ((0, η]), for some η > 1, such that limt→0+ tL
L(t) = 0.
As a typical example of function belonging to the class K (see [5, 17, 20]), we
quote
m
Y
ω
ω
L(t) = 2 + sin(log2 ( )) and L(t) =
(logk ( ))ξk ,
t
t
k=1
where ξk are real numbers, logk x = log ◦ log ◦ . . . log x (k times) and ω is a sufficiently large positive real number such that L is defined and positive on (0, η], for
some η > 1.
In the sequel, we denote by B + ((0, ∞)) the set of nonnegative Borel measurable
functions in (0, ∞) and by C0 ([0, ∞)) the set of continuous functions v on [0, ∞)
such that limt→∞ v(t) = 0. It is easy to see that C0 ([0, ∞)) is a Banach space with
the uniform norm kvk∞ = supt>0 |v(t)|.
For two nonnegative functions f and g defined on a set S, the notation f (t) ≈
g(t), t ∈ S means that there exists c > 0 such that 1c f (t) ≤ g(t) ≤ cf (t), for all
t ∈ S.
Furthermore, let G(t, s) = A(s) min(ρ(t), ρ(s))), be the Green’s function of the
operator u 7→ − A1 (Au0 )0 on (0, ∞) with the Dirichlet conditions limt→0+ u(t) = 0
and limt→∞ u(t)
ρ(t) = 0.
For f ∈ B + ((0, ∞)), we put
Z ∞
V f (t) =
G(t, s)f (s)dt, for t > 0.
0
We point out that if the map s → A(s) min(1, ρ(s))f (s) is continuous and integrable on (0, ∞), then V f is the solution of the boundary-value problem
1
− (Au0 )0 = f, in (0, ∞),
A
lim+ u(t) = 0,
t→0
(1.4)
u(t)
= 0.
lim
t→∞ ρ(t)
For λ ≤ 2, σ ∈ (−1, 1) and L ∈ K defined on
t ∈ (0, η)

1
R t L(s) 1−σ


ds
,

s
0


1

1−σ

L(t)
,
ΨL,λ,σ (t) = 1
1−σ
R

η L(s)


ds
,

s
t



1,
(0, η] (for some η > 1), we put for
if λ = 2,
if 1 + σ < λ < 2,
if λ = 1 + σ,
if λ < 1 + σ.
Throughout this article we assume the following condition:
(1.5)
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I. BACHAR
EJDE-2015/235
(H1) For i ∈ {1, 2}, ai is a nonnegative continuous function on (0, ∞) such that
1
ai (t) ≈
(ρ(t))−λi (1 + ρ(t))λi −µi Li (m(t)), t > 0,
(1.6)
(A(t))2
ρ(t)
where λi ≤ 2, µi > 2, m(t) := 1+ρ(t)
, for t > 0 and Li ∈ K defined on (0, η]
( for some η > 1) such that
Z η
(1.7)
s1−λi Li (s)ds < ∞.
0
As it will be seen, the numbers
2 − λ1
2 − λ2
β1 = min(1,
) and β2 = min(1,
)
(1.8)
1 − σ1
1 − σ2
will play an important role in the study of asymptotic behavior of solution. Without
loss of generality, we may assume that
2 − λ1
2 − λ2
≤
1 − σ1
1 − σ2
and we define the function θ on (0, ∞) by
(
if β1 < β2
(m(t))β1 ΨL1 ,λ1 ,σ1 (m(t))
θ(t) =
(1.9)
β1
(m(t)) (ΨL1 ,λ1 ,σ1 (m(t)) + ΨL2 ,λ2 ,σ2 (m(t))) if β1 = β2 ,
ρ(t)
where m(t) := 1+ρ(t)
, for t > 0.
For an explicit form of the function θ see (3.1). Now, we are ready to state our
main results.
Theorem 1.3. Let σ1 , σ2 ∈ (−1, 1) and assume that (H1) is fulfilled. Then for
t ∈ (0, ∞),
V (a1 θσ1 + a2 θσ2 )(t) ≈ θ(t).
(1.10)
By applying the above theorem and using the Schauder fixed point theorem, we
prove the following result.
Theorem 1.4. Let σ1 , σ2 ∈ (−1, 1) and assume that (H1) is fulfilled. Then (1.3)
has a unique positive continuous solution u satisfying for t ∈ (0, ∞)
u(t) ≈ θ(t).
(1.11)
The content of this paper is organized as follows. In Section 2, we present some
fundamental properties of the Karamata class K including sharp estimates on some
potential functions. In Section 3, exploiting the results of the previous section,
we first prove Theorem 1.3 which allow us to prove Theorem 1.4 by means of the
Schauder fixed point theorem.
2. Properties of Karamata regular variation theory
We collect in this section some properties of functions belonging to the Karamata
class K.
Proposition 2.1 ([17, 20]). The following hold:
(i) Let L1 , L2 ∈ K and p ∈ R. Then L1 + L2 ∈ K, L1 L2 ∈ K and Lp1 ∈ K.
(ii) Let L ∈ K and ε > 0. Then limt→0+ tε L(t) = 0.
Applying Karamata’s theorem (see [17, 20]), we obtain the following result.
EJDE-2015/235
NONLINEAR SINGULAR SECOND-ORDER DIFFERENTIAL EQUATIONS
5
Lemma 2.2. Let ν ∈ R and L be a function in K defined on (0, η]. We have
Rη
Rη
ν+1
L(t)
(i) If ν < −1, then 0 sν L(s)ds diverges and t sν L(s)ds ∼ − t ν+1
.
t→0+
Rη ν
Rt ν
tν+1 L(t)
(ii) If ν > −1, then 0 s L(s)ds converges and 0 s L(s)ds ∼ + ν+1 .
t→0
The proof of the next lemmas can be found in [7].
Lemma 2.3. Let L be a function in K defined on (0, η] (η > 1). Then
lim+ R η
t→0
t
L(t)
L(s)
s ds
= 0.
Rη
ds ∈ K.
In particular t → t L(s)
R η L(s) s
If further 0 s ds converges, then
lim R t
+
t→0
In particular t →
Rt
0
L(s)
s ds
0
L(t)
L(s)
s ds
= 0.
∈ K.
Lemma 2.4. For i ∈ {1, 2}, let Li ∈ K be defined on (0, η] (η > 1) and put for
t ∈ (0, η),
1
1
Z η L (s) 1−σ
Z η L (s) 1−σ
1
2
1
2
M (t) =
ds
ds
+
.
s
s
t
t
Then for t ∈ (0, η) we have
Z η
(M σ1 L1 + M σ2 L2 )(s)
ds ≈ M (t).
s
t
Lemma 2.5. For i ∈ {1, 2}, let Li ∈ K be defined on (0, η] (η > 1) such that
R η Li (s)
s ds < ∞. Put for t ∈ (0, η),
0
1
1
Z t L (s) 1−σ
Z t L (s) 1−σ
1
2
1
2
ds
ds
N (t) =
+
.
s
s
0
0
Then for t ∈ (0, η) we have
Z t
0
(N σ1 L1 + N σ2 L2 )(s)
ds ≈ N (t).
s
Next, we have the following fundamental sharp estimates on the potential function V b, for
1
e
b(t) =
(ρ(t))−β (1 + ρ(t))β−γ L(m(t)),
(A(t))2
e ∈ K and m(t) =
where β ≤ 2, γ > 2, L
ρ(t)
1+ρ(t)
for t > 0.
e ∈ K be defined on (0, η] (η > 1)
Proposition 2.6 ([3]). Let β ≤ 2, γ > 2 and L
R η 1−β
e
such that 0 s
L(s)ds < ∞. Then for t > 0,
V b(t) ≈ φβ (m(t)),
6
I. BACHAR
EJDE-2015/235
where for r ∈ (0, 1],
R e
r L(s)


s ds
0


r2−β L(r)
e
φβ (r) =
R η L(s)
e
r


s ds
r


r
if β = 2,
if 1 < β < 2,
if β = 1,
if β < 1.
3. Proof of main results
Let σ1 , σ2 ∈ (−1, 1), assume (H1) and for i ∈ {1, 2}, let Li ∈ K defined on (0, η]
(for someη > 1) satisfying (1.6) and (1.7). Let b, L, M and N be the nonnegative
functions defined in (0, η) by
1
Z η L (s) 1−σ
1
1
ds
,
b(t) :=
s
t
1
1
L(t) := (L1 (t)) 1−σ1 + (L2 (t)) 1−σ2 ,
1
1
Z η L (s) 1−σ
Z η L (s) 1−σ
1
2
1
2
M (t) :=
ds
+
ds
,
s
s
t
t
Z η
1
1
Z t L (s) 1−σ
Z t L (s) 1−σ
Li (s)
1
2
1
2
N (t) :=
ds
+
ds
, if
ds < ∞.
s
s
s
0
0
0
First, we give an explicit form of the function θ defined by (1.9). We recall
i
that for i ∈ {1, 2}, λi ≤ 2 and βi = min(1, 2−λ
1−σi ). Since β1 < β2 is equivalent to
2−λ1
2−λ2
1−σ1 < 1−σ2 and 1 + σ1 < λ1 , we deduce that for t ∈ (0, ∞),
 R
1
1−σ
m(t) L1 (s)
1


,
if λ1 = 2 and λ2 < 2,
ds

s
0


2−λ1
1


2−λ2
1
(m(t)) 1−σ1 (L1 (m(t))) 1−σ1 , if 2−λ

1−σ1 < 1−σ2 and 1 + σ1 < λ1 < 2,


2−λ1

2−λ2
(m(t)) 1−σ1 L(m(t)),
1
if 2−λ
1−σ1 = 1−σ2 and 1 + σ1 < λ1 < 2,
(3.1)
θ(t) =
m(t)M (m(t)),
if λ1 = 1 + σ1 and λ2 = 1 + σ2 ,





m(t)(1 + b(m(t))),
if λ1 = 1 + σ1 and λ2 < 1 + σ2 ,




2m(t),
if λ1 < 1 + σ1 ,



N (m(t)),
if λ1 = λ2 = 2,
where m(t) =
ρ(t)
1+ρ(t) .
Proof of Theorem 1.3.
Lemma 3.1. For r, s > 0, we have
2− max(1−σ1 ,1−σ2 ) (r + s) ≤ r1−σ1 (r + s)σ1 + s1−σ2 (r + s)σ2 ≤ 2(r + s).
Proof. Let r, s > 0 and put t =
r
r+s .
(3.2)
Since 0 ≤ t ≤ 1, then we obtain
2− max(1−σ1 ,1−σ2 ) ≤ t1−σ1 + (1 − t)1−σ2 ≤ 2.
Which implies the result.
Now we are ready to prove Theorem 1.3. We recall that for i ∈ {1, 2}, ai is a
nonnegative continuous function on (0, ∞) such that
1
(ρ(t))−λi (1 + ρ(t))λi −µi Li (m(t)), t > 0,
ai (t) ≈
(A(t))2
EJDE-2015/235
NONLINEAR SINGULAR SECOND-ORDER DIFFERENTIAL EQUATIONS
7
where λi ≤ 2 and µi > 2.
Note that throughout the proof, we use Proposition 2.1 and Lemma 2.3 to verify
that some functions are in K. We distinguish the following cases.
Case 1: λ1 = 2 and λ2 < 2. We have
1
Z m(t) L (s) 1−σ
1
1
θ(t) =
.
ds
s
0
Therefore
a1 (t)θσ1 (t) + a2 (t)θσ2 (t)
σ
L1 (s) 1−σ1 1
ds
s
0
Z m(t)
σ
−λ2
λ2 −µ
L1 (s) 1−σ2 1
(ρ(t))
(1 + ρ(t))
L
(m(t))
ds
.
+
2
(A(t))2
s
0
σ
R
e i (t) := Li (t)( t L1 (s) ds) 1−σi 1 ∈ K and
Since for i ∈ {1, 2}, the function t → L
s
0
λ2 < 2, we deduce by Proposition 2.1 that
≈
(ρ(t))−2 (1 + ρ(t))2−µ
L1 (m(t))
2
(A(t))
a1 (t)θσ1 (t) + a2 (t)θσ2 (t) ≈
Z
m(t)
(ρ(t))−2 (1 + ρ(t))2−µ e
L1 (m(t)).
(A(t))2
Moreover, since λ1 = 2, we have
Z η e
1
Z η L (r) 1−σ
L1 (s)
1
1
ds ≤ c
dr
< ∞,
s
r
0
0
it follows by applying Proposition 2.6 with β = λ1 = 2, γ = µ, we obtain
Z
Z m(t)
σ
L1 (s) s L1 (r) 1−σ1 1
σ1
σ2
dr
V (a1 θ + a2 θ )(t) ≈
ds ≈ θ(t).
s
r
0
0
Case 2:
2−λ1
1−σ1
<
2−λ2
1−σ2
and 1 + σ1 < λ1 < 2. Since
2−λ1
1
θ(t) = (m(t)) 1−σ1 (L1 (m(t))) 1−σ1 ,
we obtain
a1 (t)θσ1 (t) + a2 (t)θσ2 (t)
≈
(ρ(t))
+
(2−λ1 )σ1
1−σ1
(ρ(t))
Since in this case
−λ1
(1 + ρ(t))λ1 −
(A(t))2
(2−λ1 )σ2
1−σ1
(2−λ1 )σ2
1−σ1
(2−λ1 )σ1
1−σ1
−λ2
(1 + ρ(t))λ2 −
(A(t))2
− λ2 >
(2−λ1 )σ1
1−σ1
−µ
σ1
(L1 L11−σ1 )(m(t))
(2−λ1 )σ2
1−σ1
−µ
σ2
(L2 L11−σ1 )(m(t)).
− λ1 , we deduce by Proposition 2.1 that
a1 (t)θσ1 (t) + a2 (t)θσ2 (t)
≈
(ρ(t))
(2−λ1 )σ1
1−σ1
−λ1
(1 + ρ(t))λ1 −
(A(t))2
(2−λ1 )σ1
1−σ1
−µ
σ1
(L1 L11−σ1 )(m(t)).
8
I. BACHAR
EJDE-2015/235
Therefore applying Proposition 2.6 with β = λ1 −
σ1
1−σ1
1
1−σ1
= L1
e = L1 L
L
1
(2−λ1 )σ1
1−σ1
∈ (1, 2), γ = µ and
∈ K, we obtain
1
V (a1 θσ1 + a2 θσ2 )(t) ≈ (m(t))2−β (L1 (m(t))) 1−σ1 = θ(t).
Case 3:
2−λ1
1−σ1
=
2−λ2
1−σ2
and 1 + σ1 < λ1 < 2. We have
2−λ1
θ(t) = (m(t)) 1−σ1 L(m(t)).
So
σ1
a1 (t)θ (t) ≈
(ρ(t))
(2−λ1 )σ1
1−σ1
−λ1
(1 + ρ(t))λ1 −
(A(t))2
(2−λ1 )σ1
1−σ1
Hence using again Proposition 2.6 with β = λ1 −
e = L1 Lσ1 ∈ K, we obtain
L
−µ
(L1 Lσ1 )(m(t))
(2−λ1 )σ1
1−σ1
∈ (1, 2), γ = µ and
2−λ1
V (a1 θσ1 )(t) ≈ (m(t)) 1−σ1 (L1 Lσ1 )(m(t)).
On the other hand, since 1 + σ2 < λ2 < 2, we similarly obtain
2−λ1
V (a2 θσ2 )(t) ≈ (m(t)) 1−σ1 (L2 Lσ2 )(m(t)).
Using Lemma 3.1, we deduce that
2−λ1
V (a1 θσ1 + a2 θσ2 )(t) ≈ (m(t)) 1−σ1 L(m(t)) = θ(t).
Case 4: λ1 = 1 + σ1 and λ2 = 1 + σ2 . In this case we have θ(t) = m(t)M (m(t))
By calculations, we obtain
a1 (t)θσ1 (t) + a2 (t)θσ2 (t) ≈
(ρ(t))−1 (1 + ρ(t))1−µ
(M σ1 L1 + M σ2 L2 )(m(t))
(A(t))2
e = M σ1 L1 + M σ2 L2 , we deduce that
Using Proposition 2.6 with β = 1, γ = µ and L
Z η e
L(s)
σ1
σ2
V (a1 θ + a2 θ )(t) ≈ m(t)
ds.
s
m(t)
Hence the results follows from Lemma 2.4.
Case 5: λ1 = 1 + σ1 and λ2 < 1 + σ2 . Since limt→0+ b(t) ∈ (0, ∞], it follows that
θ(t) ≈ m(t)b(m(t)). So, we obtain that
a1 (t)θσ1 (t) + a2 (t)θσ2 (t) ≈
(ρ(t))−1 (1 + ρ(t))1−µ
(L1 bσ1 )(m(t))
(A(t))2
(ρ(t))−λ2 +σ2 (1 + ρ(t))λ2 −σ2 −µ
+
(L2 bσ2 )(m(t)).
(A(t))2
e i (t) := Li (t)bσi (t) ∈ K and
Using the fact that for i ∈ {1, 2}, the function t → L
that λ2 − σ2 < 1, we deduce by Proposition 2.1 that
a1 (t)θσ1 (t) + a2 (t)θσ2 (t) ≈
(ρ(t))−1 (1 + ρ(t))1−µ
(L1 bσ1 )(m(t)).
(A(t))2
EJDE-2015/235
NONLINEAR SINGULAR SECOND-ORDER DIFFERENTIAL EQUATIONS
9
Hence applying Proposition 2.6 with β = 1, γ = µ, we obtain that
Z η e
L1 (s)
V (a1 θσ1 + a2 θσ2 )(t) ≈ m(t)
ds ≈ θ(t).
s
m(t)
Case 6: λ1 < 1 + σ1 . Since θ(t) = 2m(t), we obtain
a1 (t)θσ1 (t) ≈
(ρ(t))−λ1 +σ1 (1 + ρ(t))λ1 −σ1 −µ
L1 (m(t)).
(A(t))2
e = L1 , we obtain
Applying Proposition 2.6 with β = λ1 − σ1 < 1, γ = µ and L
V (a1 θσ1 )(t) ≈ m(t).
On the other hand, since also λ2 < 1 + σ2 , we similarly obtain
V (a2 θσ2 )(t) ≈ m(t).
Hence
V (a1 θσ1 + a2 θσ2 )(t) ≈ m(t) ≈ θ(t).
Case 7: λ1 = λ2 = 2. We have θ(t) = N (m(t)). By calculus, we conclude that
a1 (t)θσ1 (t) + a2 (t)θσ2 (t) ≈
(ρ(t))−2 (1 + ρ(t))2−µ σ1
(N L1 + N σ2 L2 )(m(t))
(A(t))2
On the other hand, since s → (N σ1 L1 + N σ2 L2 )(s) ∈ K and from Lemma 2.5,
Z 1
(N σ1 L1 + N σ2 L2 )(s)
ds ≈ N (1) < ∞,
s
0
e = N σ1 L1 + N σ2 L2 , we deduce
then by Proposition 2.6 with β = 2, γ = µ and L
that
Z m(t)
(N σ1 L1 + N σ2 L2 )(s)
σ1
σ2
V (a1 θ + a2 θ )(t) ≈
ds.
s
0
Hence the results follows from Lemma 2.5.
Proof of Theorem 1.4. The next Lemma will be useful to prove the uniqueness.
Lemma 3.2 ([2]). Let a ≥ 0 and u ∈ C 1 ((a, ∞)) be a function satisfying
1
− (Au0 )0 ≥ 0, in (a, ∞),
A
u(t)
lim u(t) = 0, lim
= 0.
t→∞ ρ(t)
t→a+
(3.3)
Then u is nondecreasing and nonnegative function on (a, ∞).
Now we are ready to prove Theorem 1.4. Let σ1 , σ2 ∈ (−1, 1), assume (H1) and
put ω
e := a1 θσ1 + a2 θσ2 and v := V (e
ω ). From Theorem 1.3, there exists M > 1
such that for each t > 0,
1
θ(t) ≤ v(t) ≤ M θ(t).
(3.4)
M
On the other hand, using hypothesis (H1), (3.1), Lemma2.2 and Lemma 2.5, we
verify that
Z
∞
A(s) min(1, ρ(s))e
ω (s)ds < ∞.
0
(3.5)
10
I. BACHAR
EJDE-2015/235
1+σ
Put σ = max(|σ1 |, |σ2 |), c = M 1−σ , where the constant M is given in (3.4) and let
Λ = {ω ∈ C0 ([0, ∞)) :
cθ(t)
θ(t)
≤ ω(t) ≤
, t > 0}.
c(1 + ρ(t))
1 + ρ(t)
Using (3.1), Proposition 2.1 and Lemma 2.3, we verify that the function t 7→
θ(t)
1+ρ(t) ∈ C0 ([0, ∞)) and so Λ is not empty.
We define the operator T on Λ by
Z ∞
1
G(t, s) a1 (s)(1 + ρ(s))σ1 ω σ1 (s)
T ω(t) =
1 + ρ(t) 0
(3.6)
σ2 σ2
+ a2 (s)(1 + ρ(s)) ω (s) ds.
We shall prove that T has a fixed point in Λ.
First observe that for this choice of c, by using (3.4), we have for all ω ∈ Λ and
t>0
1
cσ M σ
cθ(t)
T ω(t) ≤
V (a1 cσ M σ θσ1 + a2 cσ M σ θσ2 )(t) =
v(t) ≤
1 + ρ(t)
1 + ρ(t)
1 + ρ(t)
and
θ(t)
c−σ M −σ
v(t) ≥
.
T ω(t) ≥ V (a1 c−σ M −σ θσ1 + a2 c−σ M −σ θσ2 )(t) =
1 + ρ(t)
c(1 + ρ(t))
On the other hand, for all ω ∈ Λ we have
|a1 (t)(1 + ρ(t))σ1 ω σ1 (t) + a2 (t)(1 + ρ(t))σ2 ω σ2 (t)| ≤ cσ M σ ω
e (t),
(3.7)
and for all t, s > 0, we have
G(t, s)
≤ A(s) min(1, ρ(s)).
1 + ρ(t)
(3.8)
G(t,s)
Since for each s > 0, the function t → 1+ρ(t)
is in C0 ([0, ∞)), we deduce by using
(3.7), (3.8) and (3.5) that the family {t → T ω(t), ω ∈ Λ} is relatively compact in
C0 ([0, ∞)). Therefore, T (Λ) ⊂ Λ.
Now, we shall prove the continuity of the operator T in Λ in the supremum norm.
Let (ωk )k∈N be a sequence in Λ which converges uniformly to a function ω in Λ.
Then, for each t > 0, we have
1
|T ωk (t)−T ω(t)| ≤
V [a1 (1+ρ(.))σ1 |ωkσ1 −ω σ1 |+a2 (1+ρ(.))σ2 |ωkσ2 −ω σ2 |](t).
1 + ρ(t)
On the other hand, by similar arguments as above, we have
a1 (1 + ρ(.))σ1 |ωkσ1 − ω σ1 | + a2 (1 + ρ(.))σ2 |ωkσ2 − ω σ2 | ≤ e
cω
e (s).
We conclude by (3.5) and the dominated convergence theorem that for all t > 0,
T ωk (t) → T ω(t)
as k → +∞.
Consequently, as T (Λ) is relatively compact in C0 ([0, ∞)), we deduce that the
pointwise convergence implies the uniform convergence, namely,
kT ωk − T ωk∞ → 0
as k → +∞.
Therefore, T is a continuous mapping from Λ into itself. So the Schauder fixed
point theorem implies the existence of ω ∈ Λ such that
1
V (a1 (1 + ρ(.))σ1 ω σ1 + a2 (1 + ρ(.))σ2 ω σ2 )(t).
ω(t) =
1 + ρ(t)
EJDE-2015/235
NONLINEAR SINGULAR SECOND-ORDER DIFFERENTIAL EQUATIONS 11
Put u(t) = 1 + ρ(t)ω(t). Then u is continuous and satisfies
u(t) = V (a1 uσ1 + a2 uσ2 )(t).
Since the function s → A(s) min(1, ρ(s))[a1 (s)uσ1 (s) + a2 (s)uσ2 (s)] is continuous
and integrable on (0, ∞), then it follows that u is a solution of problem (1.3).
Finally, it remains to prove that u is the unique positive continuous solution satisfying (1.11). To this end, assume that problem (1.3) has two positive continuous
solutions u, v satisfying (1.11). Then there exists a constant m > 1 such that
1
u
≤ ≤ m.
m
v
This implies that the set
J = {m ≥ 1 :
u
1
≤ ≤ m}
m
v
is not empty. Let σ := max(|σ1 |, |σ2 |) and put c0 := inf J. Then c0 ≥ 1 and we
have c10 v ≤ u ≤ c0 v. It follows that for i ∈ {1, 2}, uσi ≤ cσ0 v σi and that the function
w := cσ0 v − u satisfies
1
− (Aw0 )0 = a1 (cσ0 v σ1 − uσ1 ) + a2 (cσ0 v σ2 − uσ2 ) ≥ 0,
A
lim+ w(t) = 0,
t→0
lim
t→∞
w(t)
= 0.
ρ(t)
By Lemma 3.2, this implies that the function w = cσ0 v − u is nonnegative. By
symmetry, we also have v ≤ cσ0 u. Hence cσ0 ∈ J and c0 ≤ cσ0 . Since 0 ≤ σ < 1, then
c0 = 1 and therefore u = v.
2−λ2
1
Example 3.3. Let σ1 ∈ (−1, 0), σ2 ∈ (0, 1) and λ1 , λ2 < 2, such that 2−λ
1−σ1 ≤ 1−σ2 .
Let µ1 , µ2 > 2 and a1 , a2 be a positive continuous function on (0, ∞) such that
ai (t) ≈
1
(ρ(t))−λi (1 + ρ(t))λi −µi ,
(A(t))2
Then by Theorem 1.4, problem (1.3) has
satisfying for t > 0,
 ρ(t) 2−λ1

( 1+ρ(t) ) 1−σ1 ,



1
 ρ(t) (log( 2+2ρ(t) )) 1−σ
2 ,
ρ(t)
u(t) ≈ 1+ρ(t)
1
ρ(t)
2+2ρ(t) 1−σ


 1+ρ(t) (log( ρ(t) )) 1 ,

 ρ(t)
1+ρ(t) ,
for i ∈ {1, 2}.
a unique positive continuous solution u
if 1 + σ1 < λ1 < 2,
if λ1 = 1 + σ1 and λ2 = 1 + σ2 ,
if λ1 = 1 + σ1 and λ2 < 1 + σ2 ,
if λ1 < 1 + σ1 .
Acknowledgments. This Project was funded by the National Plan for Science,
Technology and Innovation (MAARIFAH), King Abdulaziz City for Science and
Technology, Kingdom of Saudi Arabia, Award Number (13-MAT1813-02).
The author wants to thank the anonymous referee for the careful reading of the
manuscript, and the useful suggestions.
12
I. BACHAR
EJDE-2015/235
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Imed Bachar
King Saud University College of Science, Mathematics Department, P.O. Box 2455,
Riyadh 11451, Saudi Arabia
E-mail address: abachar@ksu.edu.sa