Acta Mathematica Scientia 2012,32B(2):672–694
http://actams.wipm.ac.cn
POSITIVE SOLUTIONS FOR SINGULAR BVPs ON
THE POSITIVE HALF-LINE ARISING FROM
EPIDEMIOLOGY AND COMBUSTION THEORY∗
Smaı̈l Djebali
Department of Mathematics, E.N.S., P.O. Box 92, 16050 Kouba. Algiers, Algeria
E-mail: djebali@ens-kouba.dz
Ouiza Saı̈fi
Department of Economics, Faculty of Economic and Management Sciences, Algiers University, Algeria
E-mail: saifi-kouba@yahoo.fr
)†
Yan Baoqiang (
Department of Mathematics, Shandong Normal University, Jinan 250014, China
E-mail: yanbqcn@yahoo.com.cn
Abstract In this work, we are concerned with the existence and multiplicity of positive
solutions for singular boundary value problems on the half-line. Two problems from epidemiology and combustion theory set on the positive half-line are investigated. We use
upper and lower solution techniques combined with fixed point index on cones in appropriate Banach spaces. The results complement recent ones in the literature.
Key words Fixed points index; positive solution; singular problem; cone; lower and
upper solution; half-line
2000 MR Subject Classification
1
34B15; 34B18; 34B40; 47H10
Introduction
1.1
The mathematical problems
This article is devoted to the study of the existence of multiple positive solutions to the
following boundary value problems set on the positive half-line:

 x′′ (t) − k 2 x(t) + φ(t)f (t, x(t)) = 0, t ∈ I,
(1.1)
 x(0) = 0,
lim x(t) = 0
t→+∞
and

 −y ′′ (t) + cy ′ (t) + λy(t) = φ(t)g(t, y(t)),
∗ Received
 y(0) = 0,
lim y(t) = 0,
t→+∞
November 19, 2009; revised February 14, 2011.
author.
† Corresponding
t ∈ I,
(1.2)
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673
where k, c, and λ are real positive constants and φ ∈ C((0, +∞), (0, +∞)). R+ denotes the
set of nonnegative real numbers and I = (0, +∞). The functions f, g ∈ C(R+ × (0, +∞), R+ )
satisfy lim f (t, x) = +∞ and lim g(t, y) = +∞, that is, f (t, x), g(t, y) may be singular
x→0+
y→0+
at x = 0, y =q
0, respectively. In fact, problem (1.2) can be transformed into problem (1.1)
c
c
c
2
if we set k = λ + c4 , x(t) = y(t)e− 2 t , and f (t, x(t)) = e− 2 t g(t, e 2 t x(t)). In other words,
c
if y is a solution of problem (1.2), then, x = ye− 2 t is a solution of problem (1.1). But if
c
x is a solution of (1.1), then, y = xe 2 t does not necessarily satisfy the boundary condition
at positive infinity in (1.2). Problems of types (1.1) and (1.2) arise in many applications in
physics, combustion theory, and epidemiology. When λ = 0, we recognize Fisher’s Equation
in the equation of problem (1.2), where c is the speed of a traveling wave. In epidemiological
models, λ = 0 stands for a mortality rate (see [3–5, 9, 15] and the references therein). In [9],
problem (1.2) is studied without singularity and the nonlinearity is assumed to satisfy at most
linear growth with respect to y. The Schauder fixed point theorem is used to prove the existence
of positive solutions. However, various assumptions on g including sub-linear and super-linear
conditions were considered in [3]. The existence results of nontrivial solutions obtained in [3]
rely on the Krasnozels’kı̆i fixed point theorem of cone compression and expansion. A recent
fixed point theorem of cone expansion and compression of functional type was used in [4] to
prove the existence of positive solutions with some illustrative examples. Existence of two or
three solutions when φ may be singular at t = 0 was examined in [5]. In [8, 12], B. Yan
et al obtained some existence results of unbounded positive solutions to problem (1.1) with
c=λ=0:

 y ′′ (t) + φ(t)f (t, y(t)) = 0,
(1.3)
 y(a) = 0, lim y ′ (t) = 0.
t→+∞
A similar problem with a Sturm-Liouville operator and mixed boundary conditions was examined in [10]; the existence of a positive solution was obtained when f is positive. In these
three recent works, the fixed point index in cones was employed together with the method of
upper and lower solution. Compactness arguments were obtained via Corduneanu’s criterion
on unbounded intervals [6]. In [11], the authors investigated some questions of existence and
uniqueness for problem (1.1). Motivated by [8, 11], we prove here the existence of solutions for
problems (1.1) and (1.2). The upper and lower solutions techniques are employed. New existence results of multiple positive solutions are also obtained. The proofs rely heavily on detailed
properties of the Green’s function, on suitable choice of a cone in a Bielecki type Banach space
together with the fixed point index on cones [2, 6, 13] and Zima’s compactness criterion [14].
This is developed in Section 2 (Theorems 2.1, 2.2) and in Section 3 (Theorem 3.1) for problem
(1.1) and in Sections 4 (Theorems 4.1, 4.2), and in Section 5 (Theorem 5.1) for problem (1.2).
For each problem, the existence results are illustrated by means of examples of application.
First, some auxiliary results are provided hereafter.
1.2 Preliminaries
Let p : I −→ I be a continuous function. Denote by Y the Banach space consisting of all
weighted functions x, continuous on I, satisfying
sup{|x(t)|p(t)} < ∞,
t∈I
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equipped with a Bielecki’s type norm kxkp = sup{|x(t)|p(t)}.
t∈I
Definition 1.1 A set of functions x ∈ Ω ⊆ Y is said to be almost equicontinuous if it is
equicontinuous on each interval [0, T ].
Lemma 1.1 [14] If the functions x ∈ Ω are almost equi-continuous on I and uniformly
bounded in the sense of the norm kxkq = sup{|x(t)|q(t)}, where the function q is positive,
t∈I
continuous on I, and satisfies
lim
t→+∞
p(t)
= 0,
q(t)
then, Ω is relatively compact in Y .
The following two results will be needed in the sequel [6, 13].
Lemma 1.2 [6, 13] Let Ω be a bounded open set in a real Banach space E, P a cone of
E and A : Ω ∩ P → P a completely continuous map. Suppose λAx 6= x, ∀ x ∈ ∂Ω ∩ P, λ ∈ (0, 1].
Then, i(A, Ω ∩ P, P) = 1.
Lemma 1.3 [6, 13] Let Ω be a bounded open set in a real Banach space E, P a cone of
E, and A : Ω ∩ P → P a completely continuous map. Suppose Ax 6≤ x, ∀ x ∈ ∂Ω ∩ P. Then,
i(A, Ω ∩ P, P) = 0.
2
A Positive Solution for Problem (1.1)
2.1
General setting and assumptions
Given a real parameter θ > k, consider the weighted Banach space
E = {x ∈ C(R+ , R) : sup |x(t)|e−θt < ∞}
t∈R+
endowed with the weighted Bielecki’s sup-norm
kxkθ = sup {|x(t)|e−θt }.
t∈R+
Lemma 2.1 Let z be a function, such that φz ∈ C(R+ ) ∩ L1 (0, +∞). Then, x is a
solution of

 x′′ (t) − k 2 x(t) + φ(t)z(t) = 0, t ∈ I,
(2.1)
 x(0) = 0,
lim x(t) = 0,
t→+∞
if and only if x is a solution of
x(t) =
Z
+∞
G(t, s)φ(s)z(s)ds,
(2.2)
0
where G(t, s) is given by

1  e−kt (eks − e−ks ),
G(t, s) =
2k  e−ks (ekt − e−kt ),
0 ≤ s ≤ t < +∞,
0 ≤ t ≤ s < +∞.
(2.3)
Proof (a) The homogeneous problem has only the trivial solution and {ekt , e−kt } is a
fundamental system of solutions. By the method of variation of constants, we deduce that the
unique solution of (2.1) is given by (2.2).
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(b) Conversely, if x satisfies (2.2), then clearly that x(0) = 0 and x′′ (t)−k 2 x(t)+φ(t)z(t) =
0. To show that x(+∞) = 0, we split the integral in (2.2). As φz ∈ L1 (0, +∞), then for any
R +∞
ε > 0, there exists a1 > 0 such that, for all t ≥ a1 , t
φ(s)z(s)ds ≤ 3ε , in particular,
R +∞
ε
−kt
= 0, then, there exists a2 > 0, such that
a1 φ(s)z(s)ds ≤ 3 . As lim e
t→+∞
∀ t ≥ a2 ,
e−kt ≤
ε
3(eka1
−
e−ka1 )
Hence, for any t ≥ max{a1 , a2 }, we have
0
φ(s)z(s)ds
·
Z +∞
(eks − e−ks )φ(s)z(s)ds + (ekt − e−kt )
e−ks φ(s)z(s)ds
0
t
Z a1
Z t
= e−kt
(eks − e−ks )φ(s)z(s)ds + e−kt
(eks − e−ks )φ(s)z(s)ds
2kx(t) = e−kt
Z
R a1
t
0
+(ekt − e−kt )
≤e
≤
−kt
(e
ka1
Z
a1
+∞
e−ks φ(s)z(s)ds
t
Z a1
Z
−ka1
−e
)
φ(s)z(s)ds +
ε ε ε
+ + = ε.
3 3 3
0
t
φ(s)z(s)ds +
a1
Z
+∞
φ(s)z(s)ds
t
Some properties of the Green’s function G are given hereafter. For t ≥ 0, let
γ(t) := (e2kt − 1)e−(θ+3k)t
and
γ
e(t) := γ(t)e−θt .
Lemma 2.2 The Green’s function G satisfies:
(a) G(t, s) ≥ 0, ∀ t, s ∈ R+ .
1
, ∀ t, s ∈ R+ .
(b) G(t, s) ≤ G(s, s) ≤ 2k
(c) G(t, s)e−µt ≤ G(s, s)e−ks , ∀ t, s ∈ I, ∀ µ ≥ k.
(d) G(t, s)e−µt ≥ γ(t)G(τ, s)e−µτ , ∀ t, s, τ ∈ R+ , ∀ µ ≥ k.
Proof It is easy to prove the properties (a), (b), and (c). So, we only prove (d):
 −ks kt
e (e − e−kt ) µτ −µt


 −kτ ks
e e , t ≤ s ≤ τ,


(e − e−ks )

e


−kt ks
−ks

 e (e − e ) µτ −µt

e e , τ ≤ s ≤ t,



e−ks (ekτ − e−kτ )





e−kt µτ −µt


e e ,
s ≤ t ≤ τ,

−µt
G(t, s)e
e−kτ
=

G(τ, s)e−µτ
ekt − e−kt µτ −µt



e e ,
τ ≤ t ≤ s,


ekτ − e−kτ





e−kt


s ≤ τ ≤ t,
 −kτ eµτ e−µt ,


e





ekt − e−kt µτ −µt


e e ,
t ≤ τ ≤ s.
ekτ − e−kτ
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As a consequence,
G(t, s)e−µt
G(τ, s)e−µτ
As µ ≥ k, we have
Consequently,

1


e(µ+k)τ (e2kt − 1)e−(µ+k)t ,

2ks − 1)

(e



 (e2ks − 1)



e(µ+k)τ e−(µ+k)t ,
 2kτ


(e
− 1)



 e(µ+k)τ e−(µ+k)t ,
=
 e(µ+k)τ


(e2kt − 1)e−(µ+k)t ,

2kτ − 1)

(e






e(µ+k)τ e−(µ+k)t




e(µ+k)τ


 2kτ
(e2kt − 1)e−(µ+k)t ,
(e
− 1)


(e2kt − 1)e−(µ+k)t ,






e−(µ+k)t ,




 e−(µ+k)t ,
G(t, s)e−µt
≥

G(τ, s)e−µτ

(e2kt − 1)e−(µ+k)t ,





e−(µ+k)t ,




 (e2kt − 1)e−(µ+k)t ,
G(t, s)e−µt
G(τ, s)e−µτ
2.2
t ≤ s ≤ τ,
τ ≤ s ≤ t,
s ≤ t ≤ τ,
τ ≤ t ≤ s,
s ≤ τ ≤ t,
t ≤ τ ≤ s.
t ≤ s ≤ τ,
τ ≤ s ≤ t,
s ≤ t ≤ τ,
τ ≤ t ≤ s,
s ≤ τ ≤ t,
t ≤ τ ≤ s.


(e2kt − 1)e−(µ+3k)t = γ(t),






(e2kt − 1)e−(µ+3k)t = γ(t),




 (e2kt − 1)e−(µ+3k)t = γ(t),
≥


(e2kt − 1)e−(µ+3k)t = γ(t),





(e2kt − 1)e−(µ+3k)t = γ(t),




 (e2kt − 1)e−(µ+3k)t = γ(t),
t ≤ s ≤ τ,
τ ≤ s ≤ t,
s ≤ t ≤ τ,
τ ≤ t ≤ s,
s ≤ τ ≤ t,
t ≤ τ ≤ s.
An auxiliary problem
For some positive real number h, consider the auxiliary boundary value problem

 x′′ (t) − k 2 x(t) + φ(t)f (t, x(t)) = 0, t > 0,
 x(0) = h,
lim x(t) = 0,
(2.4)
t→+∞
where f ∈ C(I × I, R).
Definition 2.1 A function α ∈ C(R+ , I) ∩ C 2 (I, R) is called a lower solution of (2.4) if
α satisfies

 α′′ (t) − k 2 α(t) + φ(t)f (t, α(t)) ≥ 0,
 α(0) ≤ h,
t→+∞
 β(0) ≥ h,
t→+∞
lim α(t) ≤ 0.
A function β ∈ C(R+ , I) ∩ C 2 (I, R) is called an upper solution of (2.4) if β satisfies

 β ′′ (t) − k 2 β(t) + φ(t)f (t, β(t)) ≤ 0,
lim β(t) ≥ 0.
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677
If there exists an upper solution β and a lower solution α with α(t) ≤ β(t) for all t ∈ R+ ,
then, we can define the set
Dαβ (t) = {x ∈ R : α(t) ≤ x ≤ β(t)}, for each t ≥ 0.
Theorem 2.1 Assume that α, β are lower and upper solutions of problem (2.4) respectively with α(t) ≤ β(t) for all t ∈ R+ . Moreover, suppose that there exists some δ ∈ C(R+ , R+ )
such that
sup e−θt |f (t, x)| ≤ δ(t), ∀ t ∈ R+ ,
β
x∈Dα
(t)
and
Z
+∞
eθs φ(s)δ(s)ds < +∞.
(2.5)
0
Then, problem (2.4) has at least one solution x∗ ∈ E with
α(t) ≤ x∗ (t) ≤ β(t), t ∈ R+ .
Proof We follow the same lines as in [11]. Consider the truncation function
∗
f (t, x) =



 f (t, α(t)),



x < α(t)
f (t, x),
α(t) ≤ x ≤ β(t)
f (t, β(t)),
x > β(t).
As the function f ∗ ∈ C(R+ × R+ , R) has no singularity, define the regular problem

 x′′ (t) − k 2 x(t) + φ(t)f ∗ (t, x(t)) = 0, t > 0
 x(0) = h,
lim x(t) = 0.
(2.6)
t→+∞
To show that problem (2.6) has at least one solution x∗ , let the operator A1 be defined on E
by
Z +∞
A1 x(t) = he−kt +
G(t, s)φ(s)f ∗ (s, x(s))ds.
0
(2.5) implies that A1 is well defined. By Lemma 2.1, x is a solution of (2.6) if and only if x is
a fixed point of A1 .
Step 1 A1 (E) ⊂ E. For x ∈ E and t ∈ R+ , we have the estimates
|A1 x(t)|e
−θt
= he
−(k+θ)t
1
≤ h+
2k
Z
+
Z
+∞
e−θt G(t, s)φ(s)|f ∗ (s, x(s))|ds
0
+∞
0
e−θs φ(s)|f ∗ (s, x(s))|ds
Z +∞
1
≤ h+
φ(s)e−θs |f ∗ (s, x(s))|ds
2k 0
Z +∞
1
≤ h+
φ(s)δ(s)ds < +∞.
2k 0
Then, A1 x ∈ E.
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Vol.32 Ser.B
A1 is continuous. Let some sequence {xn }n≥1 ⊆ E be such that lim xn = x0 ∈
n→+∞
E. By the continuity of the functions f , α, and β, we deduce the continuity of f ∗ too. Then,
for each fixed s ∈ R+ ,
|f ∗ (s, xn (s)) − f ∗ (s, x0 (s))| −→ 0, n → +∞.
In fact, we even have the convergence of f ∗ (·, xn (·)) to f ∗ (·, x0 (·)) on every compact interval of
R+ . Moreover, we have
kA1 xn − A1 x0 kθ = sup |Axn (t) − Ax0 (t)|e−θt
t∈R+
≤ sup
Z
1
≤
2k
0
t∈R+
Z
+∞
0
+∞
e−θt G(t, s)φ(s)|f ∗ (s, xn (s)) − f ∗ (s, x0 (s))|ds
φ(s)e−θs |f ∗ (s, xn (s)) − f ∗ (s, x0 (s))|ds.
As e−θs |f ∗ (s, xn (s)) − f ∗ (s, x0 (s))| ≤ 2δ(s), ∀ s ∈ R+ , the Lebesgue dominated convergence
theorem implies that the right-hand term tends to zero as n → +∞. Our claim follows.
Step 3 A1 (E) is compact. Indeed,
(a) A1 (E) is uniformly bounded. To prove this, let µ ∈ (k, θ). Then, for x ∈ E, we have
Z +∞
|A1 x(t)|e−µt = he−(k+µ)t +
e−µt G(t, s)φ(s)|f ∗ (s, x(s))|ds
0
Z +∞
≤ h+
e−ks G(s, s)φ(s)|f ∗ (s, x(s))|ds
0
Z +∞
1
≤ h+
e−ks φ(s)|f ∗ (s, x(s))|ds
2k 0
Z +∞
1
≤ h+
e(θ−k)s φ(s)δ(s)ds < +∞.
2k 0
Hence, A1 (E) is uniformly bounded with respect to the norm k.kµ .
(b) A1 (E) is equicontinuous. For a given T > 0, x ∈ E, and t, t′ ∈ [0, T ], we have
′
|A1 x(t) − A1 x(t′ )| ≤ h|e−kt − e−kt |
Z +∞
|G(t, s) − G(t′ , s)|φ(s)|f ∗ (s, x(s))|ds
+
0
≤ h|e
+
−kt
Z
−e
+∞
T
−kt′
|+
Z
T
|G(t, s) − G(t′ , s)|φ(s)|f ∗ (s, x(s))|ds
0
|G(t, s) − G(t′ , s)|φ(s)|f ∗ (s, x(s))|ds
′
= h|e−kt − e−kt | +
Z
T
|G(t, s) − G(t′ , s)|φ(s)|f ∗ (s, x(s))|ds
Z +∞
′
′
1
+ [(ekt − e−kt ) − (ekt − e−kt )]
e−ks φ(s)|f ∗ (s, x(s))|ds
2k
T
Z T
′
≤ h|e−kt − e−kt | +
|G(t, s) − G(t′ , s)|φ(s)eθs δ(s)ds
0
Z +∞
′
′
1
kt
−kt
+ [(e − e ) − (ekt − e−kt )]
e(θ−k)s φ(s)δ(s)ds.
2k
T
0
No.2
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679
We have proved that, for any ε > 0 and T > 0, there exists η > 0 such that |A1 x(t)−A1 x(t′ )| < ε
for all t, t′ ∈ [0, T ] and |t − t′ | < η. Then, Zima’s compactness criterion (Lemma 1.1 with
q(t) = e−µt ) guarantees that A1 (E) is relatively compact. Finally, by the Schauder fixed point
theorem, A1 has at least one fixed point x∗ ∈ E, solution of problem (2.6).
Step 4 It only remains to show that α(t) ≤ x∗ (t) ≤ β(t), ∀ t ∈ R+ , in which case x∗ is
also a solution of (2.4). On the contrary, suppose that some point t∗ ∈ R+ exists and satisfies
x∗ (t∗ ) > β(t∗ ) and let z(t) = x∗ (t) − β(t). As z(0) ≤ 0, z(∞) ≤ 0, and z(t∗ ) > 0, then, z must
have a positive maximum at t0 ∈ I. Then, z ′′ (t0 ) ≤ 0, z(t0 ) > 0, and
′′
0 ≥ z ′′ (t0 ) = x∗ (t0 ) − β ′′ (t0 )
≥ [k 2 x∗ (t0 ) − φ(t0 )f ∗ (t0 , x∗ (t0 ))] − [k 2 β(t0 ) − φ(t0 )f ∗ (t0 , β(t0 ))]
= k 2 [x∗ (t0 ) − β(t0 )] + φ(t0 )[f ∗ (t0 , β(t0 )) − f ∗ (t0 , x∗ (t0 ))]
= k 2 [x∗ (t0 ) − β(t0 )] + φ(t0 )[f (t0 , β(t0 )) − f (t0 , β(t0 ))]
= k 2 [x∗ (t0 ) − β(t0 )] > 0,
which is contradictory. Then, x∗ (t) ≤ β(t), ∀ t ∈ R+ . In the same way, we prove α(t) ≤ x∗ (t).
Finally, (2.5) implies that φf ∈ L1 (0, 1); Lemma 2.1 yields that x∗ (+∞) = 0.
2.3 Existence result
Let F (t, z) = f (t, eθt z) and assume
(H1 ) There exist m ∈ C(R+ , I) and p ∈ C(I, I), such that
F (t, z) ≤ m(t)p(z), ∀ t ∈ R+ , ∀ z ∈ I.
There exists a decreasing function q ∈ C(I, I), such that
Z
p(z)
q(z)
(2.7)
is increasing and
+∞
0
φ(s)m(s)q(ce
γ (s))ds < +∞, ∀ c > 0.
(2.8)
(H2 ) For any c > 0, there exists ψc ∈ C(R+ , I), such that
F (t, z) ≥ ψc (t),
with
Z
∀ t ∈ R+ , ∀z ∈ (0, c]
+∞
φ(s)ψc (s)ds < +∞.
(2.9)
0
Using Theorem 2.1, we obtain the following.
Theorem 2.2 Besides Assumptions (H1 ), (H2 ), assume that there exist M > 0 and
h ∈ C(R+ , I) such that
f (t, x) ≤ h(t), ∀ (t, x) ∈ R+ × [M, +∞)
with
Z
(2.10)
+∞
φ(s)h(s)ds < +∞.
0
Then, problem (1.1) has at least one positive solution x∗ ∈ C 2 (R+ , R).
(2.11)
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Choose a decreasing sequence {εn }n≥1 with lim εn = 0 and ε1 < M .
Proof Step 1
n→+∞
Then, consider the sequence of boundary value problems

 x′′ (t) − k 2 x(t) + φ(t)f (t, x(t)) = 0,
 x(0) = εn ,
(2.12)
lim x(t) = 0.
t→+∞
We prove that, for each n ≥ 1, (2.12) has at least one solution xn . Let β be a solution of the
boundary value problem

 x′′ (t) + φ(t)h(t) = 0, t > 0
 x(0) = M,
that is,
β(t) = M +
Z
lim x′ (t) = 0,
t→+∞
+∞
L(t, s)φ(s)h(s)ds,
0
with Green’s function
L(t, s) =

 s,
 t,
0 ≤ s ≤ t < +∞,
0 ≤ t ≤ s < +∞,
and h is as defined in (2.10). By (2.11), β is well defined. Obviously, β(t) ≥ M, ∀ t ∈ R+ and
by (2.10), we have, for any t > 0, f (t, β(t)) ≤ h(t). Hence,
β ′′ (t) − k 2 β(t) + φ(t)f (t, β(t)) = −φ(t)h(t) − k 2 β(t) + φ(t)f (t, β(t))
= −k 2 β(t) + φ(t)[f (t, β(t)) − h(t)]
≤0
and β(0) = M ≥ εn , β(∞) ≥ 0. Consequently, for any n ≥ 1, β is an upper solution of (2.12).
In addition, if αn (t) = εn e−kt , t > 0, then,
α′′n (t) − k 2 αn (t) + φ(t)f (t, αn (t)) = φ(t)f (t, αn (t)) ≥ 0,
and αn (0) = εn , αn (∞) = 0. Hence, αn is a lower solution of (2.12), ∀ n ≥ 1. Moreover, for all
t ∈ R+ and αn (t) ≤ x ≤ β(t), we have εn e
γ (t) ≤ xe−θt ≤ kβkθ . The monotonicity of q and pq
implies that
sup
e−θt f (t, x) =
sup
e−θt F (t, xe−θt )
αn ≤x≤β
β
x∈Dα
n (t)
≤
≤
sup
e−θt m(t)p(xe−θt )
αn ≤x≤β
sup
e−θt m(t)q(xe−θt )
αn ≤x≤β
≤ e−θt m(t)q(εn γ
e(t))
p(xe−θt )
q(xe−θt )
p(kβk)
:= δ(t).
q(kβk)
R +∞
Using (2.8), we have 0 eθs φ(s)δ(s)ds < +∞. By Theorem 2.1, problem (2.12) has at least
one positive solution xn ∈ E with αn (t) ≤ xn (t) ≤ β(t), ∀ t ∈ R+ .
Step 2 The sequence {xn }n≥1 is relatively compact.
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681
(a) Given µ ∈ (k, θ), we can prove that sup |β(t)|e−µt < +∞. Moreover,
t∈R+
kxn kµ = sup |xn (t)|e−µt ≤ sup |β(t)|e−µt = kβkµ .
t∈R+
t∈R+
Then, {xn }n is uniformly bounded in sense of the norm k.kµ .
(b) From (H2 ), there exists ψkβkµ ∈ C(R+ , I), such that
|F (t, z)| ≥ ψkβkµ (t), for t ∈ R+ and z ∈ (0, kβkµ ]
with
R +∞
0
(2.13)
φ(s)ψkβkµ (s)ds < +∞. As
Z
0
+∞
e−θt G(t, s)φ(s)ψkβkµ (s)ds ≤
1
2k
Z
+∞
0
φ(s)ψkβkµ (s)ds < +∞,
R +∞
then, the map t 7→ 0 e−θt G(t, s)φ(s)ψkβkµ (s)ds belongs to E. From (2.13) and Lemma 2.2,
we find that, for all τ ≥ 0,
xn (t) = εn e−kt +
Z
+∞
G(t, s)φ(s)f (s, xn (s))ds
0
= εn e
−kt
+
Z
+∞
G(t, s)φ(s)F (s, xn (s)e−θs )ds
0
≥ εn e−kt +
≥
Z
Z
+∞
0
G(t, s)φ(s)ψkβkµ (s)ds
+∞
G(t, s)φ(s)ψkβkµ (s)ds
Z +∞
θt
≥ e γ(t)
e−θτ G(τ, s)φ(s)ψkβkµ (s)ds
0
Z +∞
≥ γ(t)
e−θτ G(τ, s)φ(s)ψkβkµ (s)ds.
0
0
Passing to the supremum over τ, we get the lower bound
xn (t) ≥ q ∗ γ(t),
where
q ∗ := sup
τ ∈R+
Z
+∞
e−θτ G(τ, s)φ(s)ψkβkµ (s)ds.
0
Using (H1 ) and the monotonicity of q and pq , we obtain the upper bound
G(t, s)e−θt φ(s)f (s, xn (s)) = G(t, s)e−θt φ(s)F (s, xn (s)e−θs )
≤ G(t, s)e−θt φ(s)m(s)p(xn (s)e−θs )
1
p(xn (s)e−θs )
φ(s)m(s)q(xn (s)e−θs )
2k
q(xn (s)e−θs )
1
p(kβkµ )
≤
φ(s)m(s)q(e
γ (s)q ∗ )
·
2k
q(kβkµ )
≤
(2.14)
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Therefore, for any T > 0 and t, t′ ∈ [0, T ],
′
|xn (t) − xn (t′ )| ≤ εn |e−kt − e−kt |
Z +∞
+
|G(t, s) − G(t′ , s)|φ(s)f (s, xn (s))ds
0
Z +∞
−kt
−kt′
≤ εn |e
−e
|+
|G(t, s) − G(t′ , s)|φ(s)F (s, xn (s)e−θs )ds
0
′
≤ εn |e−kt − e−kt | +
+
Z
+∞
T
≤ εn |e
−kt
Z
T
|G(t, s) − G(t′ , s)|φ(s)m(s)q(e
γ (s)q ∗ )
0
|G(t, s) − G(t′ , s)|φ(s)m(s)q(e
γ (s)q ∗ )
−e
−kt′
|+
Z
0
p(kβkµ )
ds
q(kβkµ )
T
|G(t, s) − G(t′ , s)|m(s)φ(s)q(e
γ (s)q ∗ )
′
′
1
+ [(ekt − e−kt ) − (ekt − e−kt )]
2k
Z
+∞
p(kβkµ )
ds
q(kβkµ )
φ(s)m(s)q(e
γ (s)q ∗ )
T
p(kβkµ )
ds
q(kβkµ )
p(kβkµ )
ds.
q(kβkµ )
It follows that, for any ε > 0 and T > 0, there exists η > 0 such that |xn (t) − xn (t′ )| < ε
for all t, t′ ∈ [0, T ] with |t − t′ | < η. Consequently, {xn }n≥1 is equicontinuous, hence relatively
compact by Lemma 1.1. Therefore, {xn }n≥1 has a subsequence {xnk }k≥1 converging to some
limit x∗ , k → +∞. By continuity of f, we have, for each fixed positive s,
f (s, xnk (s)) −→ f (s, x∗ (s)) as k → +∞.
Using (2.14) and the Lebesgue dominated convergence theorem, we finally deduce that, for any
t ∈ R+ ,
x∗ (t) = lim xnk (t)
k→+∞
Z +∞
= lim [εnk e−kt +
G(t, s)φ(s)f (s, xnk (s))ds]
k→+∞
0
Z +∞
=
G(t, s)φ(s)f (s, x∗ (s))ds.
0
∗
Then, x is a solution of problem (1.1) with x∗ ∈ E, as claimed.
3
Two Positive Solutions for Problem (1.1)
Assume f is superlinear in x, that is,
(H3 ) There exist 0 < a < b < +∞, such that
lim
x→+∞
f (t, x)
= +∞ uniformly in t ∈ [a, b].
x
Given f ∈ C(R+ × I, R+ ), define a sequence of functions {fn }n≥1 by
fn (t, x) = f (t, max{eθt /n, x}),
n ∈ {1, 2, · · ·}.
Let P be the positive cone defined in E by
P = {x ∈ E : x(t) ≥ 0 and x(t) ≥ γ(t)kxkθ , ∀ t ≥ 0}.
No.2
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For x ∈ P, define a sequence of operators by
Z +∞
An x(t) =
G(t, s)φ(s)fn (s, x(s))ds,
0
683
n ∈ {1, 2, · · ·}.
We have
Lemma 3.1 Suppose (H1 ) holds. Then, for each n ≥ 1, the operator An sends P into
P and is completely continuous.
Proof The proof that An is continuous and completely continuous is similar to that of
the operator A in Theorem 2.1; we omit it. Thus, we only show that An P ⊆ P. For x ∈ P, we
have An x(t) ≥ 0, ∀ t ∈ R+ , and from condition (H1 ), we have
|An x(t)|e
−θt
=
Z
+∞
G(t, s)e−θt φ(s)fn (s, x(s))ds
0
=
Z
+∞
0
≤
Z
≤
1
2k
≤
1
2k
+∞
0
Z
G(t, s)e−θt φ(s)F (s, max{1/n, x(s)e−θs})ds
G(t, s)e−θt φ(s)m(s)p(max{1/n, x(s)e−θs })ds
+∞
0
Z
φ(s)m(s)q(max{1/n, x(s)e−θs})
+∞
φ(s)m(s)q(e
γ (s)kxkθ )
0
Z
1 p(max{1/n, kxkθ })
=
2k q(max{1/n, kxkθ })
+∞
p(max{1/n, x(s)e−θs})
ds
q(max{1/n, x(s)e−θs})
p(max{1/n, kxkθ })
ds
q(max{1/n, kxkθ })
φ(s)m(s)q(e
γ (s)kxkθ )ds < +∞.
0
Then, An x ∈ E and, from Lemma 2.2, it holds that
Z +∞
An x(t) =
G(t, s)φ(s)fn (s, x(s))ds
0
Z +∞
e−θt G(t, s)φ(s)fn (s, x(s))ds
= eθt
0
θt
≥ e γ(t)
Z
+∞
e−θτ G(τ, s)φ(s)fn (s, x(s))ds
0
≥ γ(t)e−θτ An x(τ ), ∀ τ ∈ R+
≥ γ(t)kAn xkθ .
Therefore, An P ⊆ P.
Theorem 3.1 Assume that Assumptions (H1 ) − (H3 ) hold. Then, there exists k0 > 0
such that, for any k > k0 , problem (1.1) has at least two positive solutions.
Proof Step 1 From Lemma 3.1, An : P −→ P is completely continuous for each n ≥ 1.
Let R > 0 and
R +∞
φ(s)m(s)q(Re
γ (s))p(R)ds
·
k0 := 0
2Rq(R)
Then, for any k > k0 , we have
2kRq(R) >
Z
0
+∞
φ(s)k(s)q(Re
γ (s))p(R)ds.
(3.1)
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Let Ω1 = {x ∈ E : kxkθ < R}. We claim that x 6= λAn x for any x ∈ ∂Ω1 ∩ P, λ ∈ (0, 1] and
n ≥ n0 > 1/R. On the contrary, suppose that there exist n ≥ n0 , x0 ∈ ∂Ω1 ∩ P and λ0 ∈ (0, 1],
such that x0 = λ0 An0 x0 . As x0 ∈ ∂Ω1 ∩ P, one has x0 (t) ≥ γ(t)kx0 kθ = γ(t)R, ∀ t ∈ R+ . Then,
x0 (t)e−θt ≥ γ
e(t)kx0 kθ = γ
e(t)R. Therefore,
R = kx0 kθ = kλ0 An0 x0 kθ
Z +∞
≤ sup
e−θt G(t, s)φ(s)fn (s, x0 (s))ds
t≥0
0
≤ sup
t≥0
1
≤
2k
1
≤
2k
Z
Z
+∞
0
+∞
0
Z
e−θt G(t, s)φ(s)F (s, max{1/n, x0 (s)e−θs })ds
φ(s)m(s)q(max{1/n, x0 (s)e−θs })
+∞
φ(s)m(s)q(e
γ (s)R)
0
which implies
Z
2kRq(R) ≤
p(R)
ds,
q(R)
p(max{1/n, x0 (s)e−θs })
q(max{1/n, x0 (s)e−θs })
+∞
φ(s)m(s)q(Re
γ (s))p(R)ds,
0
contradicting (3.1). By Lemma 1.2, we infer that
i(An , Ω1 ∩ P, P) = 1, for all n ∈ {n0 , n0 + 1, · · ·}.
(3.2)
Hence, there exists an xn ∈ Ω1 ∩ P, such that An xn = xn , ∀ n ≥ n0 . As kxn kθ < R, from (H2 ),
there exists ψR ∈ C(R+ , I) such that, for all t ≥ 0,
Z +∞
fn (t, xn (t)) ≥ ψR (t) with
φ(s)ψR (s)ds < +∞.
0
It is proved that the map t 7→
R +∞
0
G(t, s)φ(s)ψR (s)ds belongs to E. Then,
xn (t) = An xn (t)
Z +∞
=
G(t, s)φ(s)fn (s, xn (s))ds
0
Z +∞
≥
G(t, s)φ(s)ψR (s)ds.
0
Lemma 2.2 implies that
xn (t)e−θt ≥ e−θt
≥ γ(t)
Z
+∞
G(t, s)φ(s)ψR (s)ds
0
Z +∞
0
e−θτ G(τ, s)φ(s)ψR (s)ds, ∀ τ ≥ 0.
Passing to the supremum over τ, we get
xn (t)e−θt ≥ γ
e(t)q ∗ ,
where
q ∗ := sup
τ ∈R+
Z
0
+∞
e−θτ G(τ, s)φ(s)ψR (s)ds.
No.2
Smaı̈l Djebali et al: POSITIVE SOLUTIONS FOR SINGULAR BVPs
685
From (H1 ), we have
Z
+∞
G(t, s)φ(s)fn (s, xn (s))ds ≤
0
Z
+∞
φ(s)m(s)q(e
γ (s)q ∗ )
0
p(R)
< +∞.
q(R)
Step 2 The sequence {xn }n≥n0 is relatively compact.
(a) Given µ ∈ (k, θ), from condition (H1 ), we have
kxn kµ = sup
R+
≤
+∞
G(t, s)e−µt φ(s)fn (s, xn (s))ds
0
+∞
Z
G(t, s)e−µt φ(s)F (s, max{1/n, xn (s)e−θs })ds
0
≤
Z
≤
1
2k
+∞
0
1
≤
2k
=
Z
Z
G(t, s)e−µt φ(s)m(s)p(max{1/n, xn (s)e−θs })ds
+∞
0
Z
+∞
0
1 p(R)
2k q(R)
φ(s)m(s)q(max{1/n, xn (s)e−θs })
φ(s)m(s)q(e
γ (s)q ∗ )
Z
+∞
p(R)
ds
q(R)
p(max{1/n, xn (s)e−θs })
ds
q(max{1/n, xn(s)e−θs })
φ(s)m(s)q(e
γ (s)q ∗ )ds < +∞.
0
(b) For any T > 0 and t, t′ ∈ [0, T ], we have
|xn (t) − xn (t′ )| =
≤
Z
+∞
|G(t, s) − G(t′ , s)|φ(s)fn (s, xn (s))ds
0
Z
0
+∞
|G(t, s) − G(t′ , s)|φ(s)m(s)q(e
γ (s)q ∗ )
p(R)
ds.
q(R)
As in the proof of Theorem 2.2, we verify that, for any ε > 0 and T > 0, there exists η > 0 such
that |xn (t) − xn (t′ )| < ε for all t, t′ ∈ [0, T ] with |t − t′ | < η. Then, {xn }n≥n0 is equicontinuous,
hence relatively compact. Then, there exists a subsequence {xnk }k≥1 with lim xnk = x. As
k→+∞
xnk (t) ≥ γ
e(s)q ∗ , ∀ k ≥ 1, we have x(t) ≥ γ
e(s)q ∗ , ∀ t ∈ R+ . Consequently,
Z
0
+∞
G(t, s)φ(s)f (s, x(s))ds ≤
Z
+∞
0
φ(s)m(s)q(e
γ (s)q ∗ )
p(R)
< +∞.
q(R)
The continuity of f implies that, for all s ∈ R+ ,
lim fnk (s, xnk (s)) = lim f (s, max{eθs /nk , xnk (s)})
k→+∞
k→+∞
= f (s, max{0, x(s)} = f (s, x(s)).
By Lebesgue dominated convergence theorem,
Z +∞
x(t) = lim xnk (t) = lim
G(t, s)φ(s)fnk (s, xnk (s))ds
k→+∞
k→+∞ 0
Z +∞
=
G(t, s)φ(s)f (s, x(s))ds.
0
It is clear that kxkθ ≤ R, and from (3.1), kxkθ < R. Then, x is a positive solution of (1.1).
686
ACTA MATHEMATICA SCIENTIA
Step 3
Let
N∗ = 1 +
r min
Rb
t∈[a,b] a
1
G(t, s)e−θt φ(s)ds
Vol.32 Ser.B
,
where r = min γ(t). By (H3 ), there exists an R′ > R, such that
t∈[a,b]
f (t, x) > N ∗ R′ ,
Let
Ω2 =
∀ t ∈ [a, b], ∀ x ≥ R′ .
R′
x ∈ E : kxkθ <
r
.
Without loss of generality, suppose that R′ > max{1, R}. We show that An x ≤
6 x for all
x ∈ ∂Ω2 ∩ P and n ∈ {1, 2, · · ·}. Suppose, on the contrary, that there exists an n ∈ {1, 2, · · ·}
and x0 ∈ ∂Ω2 ∩ P with An x0 ≤ x0 . As x0 ∈ P, we have
x0 (t) ≥ γ(t)kx0 kθ ≥ min γ(s)
s∈[a,b]
R′
≥ R′ , ∀ t ∈ [a, b].
r
Then, for any t ∈ [a, b], we have
x0 (t)e−θt ≥ An x0 (t)e−θt
Z +∞
=
G(t, s)e−θt φ(s)fn (s, x0 (s))ds
0
≥
≥
Z
b
G(t, s)e−θt φ(s)fn (s, x0 (s))ds
a
Z
b
G(t, s)e−θt φ(s)N ∗ max{
a
≥ min
t∈[a,b]
contradicting kx0 kθ =
R′
r .
Z
eθs
, x0 (s)}ds
n
b
G(t, s)e−θt φ(s)N ∗ R′ >
a
R′
,
r
Finally, Lemma 1.3 yields
i(An , Ω2 ∩ P, P) = 0,
∀ n ∈ N∗ ,
(3.3)
while (3.2) and (3.3) imply
i(An , (Ω2 \ Ω1 ) ∩ P, P) = −1, ∀ n ≥ n0 .
(3.4)
Therefore, An has another fixed point yn ∈ (Ω2 − Ω1 ) ∩ P, ∀ n ≥ n0 . Consider the sequence
′
{yn }n≥n0 . Then, yn (t) ≥ γ(t)R, ∀ t ∈ R+ , and kyn kθ < Rr , ∀ n ≥ n0 . Arguing as above, we
can show that {yn }n≥n0 has a convergent subsequence {ynj }j≥1 with lim ynj = y0 to be
j→+∞
′
a solution of (1.1). Moreover, R < ky0 kθ < Rr · Hence, x0 and y0 are two distinct positive
solutions to problem (1.1).
Remark 3.1 If the following condition holds
(H4 )
2kcq(c)
sup
> 1,
R +∞
c>0 p(c)
φ(τ )m(τ )h(ce
γ (τ ))dτ
0
No.2
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Smaı̈l Djebali et al: POSITIVE SOLUTIONS FOR SINGULAR BVPs
then, Theorem 3.1 holds for each k > 0.
Example 3.1 Consider the singular boundary value problem

−4t 2
−17
x +1

 x′′ (t) − 9 x(t) + (e3t − 1)e 2 t e √
= 0,
4
x

 x(0) = 0,
lim x(t) = 0.
(3.5)
t→+∞
If θ = 2, then,
f (t, x) = e
−17
2 t
2
−19 z + 1
e−4t x2 + 1
√
and F (t, z) = e 2 t √ ·
x
z
Let
m(t) = e
−19
2 t
, p(z) =
z2 + 1
1
√ , q(z) = , and φ(t) = (e3t − 1).
z
z
Then, Assumptions (H1 )–(H3 ) are satisfied. Indeed
(H1 ) The function q ∈ C(R+
0 , I) is decreasing and
F (t, z) ≤ m(t)p(z),
The function
p(z)
q(z)
=
∀ t ∈ R+ , ∀ z > 0.
√ 2
z(z + 1) is increasing, and for any c > 0,
Z
+∞
φ(s)m(s)q(ce
γ (s))ds =
0
(H2 ) For any c > 0, there exists ψc =
F (t, z) ≥ ψc (t),
−19
√1 e 2 t
c
1
< +∞.
c
∈ C(R+ , I), such that
∀ t ∈ R+ , ∀ z ∈ (0, c] with
Z
+∞
φ(s)ψc (s)ds < +∞.
0
(H3 ) There exist 0 < a < b < +∞, such that
lim
x→+∞
f (t, x)
= +∞ uniformly for t ∈ [a, b].
x
Therefore, Theorem 3.1 implies that problem (3.5) has at least two positive solutions.
4
A Solution for Problem (1.2)
4.1
General setting
Notice that every solution of problem (1.2) is also a fixed point of the following integral
operator:
Z +∞
By(t) =
K(t, s)φ(s)g(s, y(s))ds,
0
where K(t, s) is the Green’s function to the corresponding homogenous Dirichlet boundary
value problem (1.2). It is seen that K is given by

 er2 t (e−r2 s − e−r1 s ), 0 ≤ s ≤ t < +∞,
1
K(t, s) =
(4.1)
r1 − r2  e−r1 s (er1 t − er2 t ),
0 ≤ t ≤ s < +∞,
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where
√
√
c2 + 4λ
c + c2 + 4λ
r2 =
< 0 < r1 =
·
2
2
For some θ > r1 , consider the weighted Banach space
c−
E ′ = {y ∈ C(R+ , R) : sup |y(t)|e−θt < ∞}
t∈R+
endowed with the Bielecki’s sup-norm
kykθ = sup {|y(t)|e−θt }.
t∈R+
′
Let P be the positive cone defined in E by
P = {y ∈ E ′ : y(t) ≥ 0 and y(t) ≥ γ1 (t)kykθ , ∀ t ∈ R+ },
where
γ1 (t) = e(r1 −r2 )t − 1)e−(θ+r1 −2r2 )t ,
∀ t ∈ R+ .
Let z = ye−θt , ge(t, z) = g(t, eθt z), and γe1 (t) = γ1 (t)e−θt .
Lemma 4.1 The Green’s function K satisfies the estimates:
(a) K(t, s) ≥ 0, ∀ t, s ∈ R+ ,
1
(b) K(s, s) ≤ r1 −r
, ∀ t, s ∈ R+ ,
2
−µt
(c) K(t, s)e
≤ K(s, s)e−r1 s , ∀ t, s ∈ I, ∀ µ ≥ r1 ,
(d) K(t, s)e−µt ≥ γ1 (t)K(τ, s)e−µτ , ∀ t, s, τ ∈ R+ , ∀ µ ≥ r1 .
Proof As Properties (a)-(c) are easy to prove, we only check (d):
 −r1 s r1 t
e
(e − er2 t ) µτ −µt


e e , t ≤ s ≤ τ,

r
τ
−r

2
e (e 2 s − e−r1 s )




r2 t −r2 s

− e−r1 s ) µτ −µt

 e (e
e e , τ ≤ s ≤ t,


−r
s
r
τ

e 1 (e 1 − er2 τ )





er2 t µτ −µt


e e ,
s ≤ t ≤ τ,

−µt
K(t, s)e
e r2 τ
=

K(τ, s)e−µτ
er1 t − er2 t µτ −µt


e e ,
τ ≤ t ≤ s,



e r1 τ − e r2 τ



r2 t


 e eµτ e−µt ,

s ≤ τ ≤ t,


 e r2 τ



r1 t
r2 t


 e − e eµτ e−µt ,
t ≤ τ ≤ s,
e r1 τ − e r2 τ
which implies

1


e(µ−r2 )τ (e(r1 −r2 )t − 1)e−(µ−r2 )t , t ≤ s ≤ τ,

 (e(r1 −r2 )s − 1)





(e(r1 −r2 )s − 1) (µ−r2 )τ −(µ−r2 )t


e
e
,
τ ≤ s ≤ t,



(e(r1 −r2 )τ − 1)



 e(µ−r2 )τ e−(µ−r2 )t ,
s ≤ t ≤ τ,
K(t, s)e−µt
=
(µ−r
)τ
−µτ
2

K(τ, s)e
e


(e(r1 −r2 )t − 1)e−(µ−r2 )t ,
τ ≤ t ≤ s,

(r1 −r2 )τ − 1)

(e






e(µ−r2 )τ e−(µ−r2 )t ,
s ≤ τ ≤ t,



(µ−r2 )τ

e


 (r −r )τ
(e(r1 −r2 )t − 1)e−(µ−r2 )t ,
t ≤ τ ≤ s.
(e 1 2 − 1)
No.2
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As µ ≥ r1 , we derive the bounds
K(t, s)e−µt
K(τ, s)e−µτ
Consequently,
K(t, s)e−µt
K(τ, s)e−µτ
4.2


(e(r1 −r2 )t − 1)e−(µ−r2 )t ,





 e−(µ−r2 )t ,



 −(µ−r2 )t

e
,
≥

 (e(r1 −r2 )t − 1)e−(µ−r2 )t ,



 −(µ−r2 )t

e
,




 (e(r1 −r2 )t − 1)e−(µ−r2 )t ,
689
t ≤ s ≤ τ,
τ ≤ s ≤ t,
s ≤ t ≤ τ,
τ ≤ t ≤ s,
s ≤ τ ≤ t,
t ≤ τ ≤ s.


(e(r1 −r2 )t − 1)e−(µ+r1 −2r2 )t = γ1 (t),






(e(r1 −r2 )t − 1)e−(µ+r1 −2r2 )t = γ1 (t),




 (e(r1 −r2 )t − 1)e−(µ+r1 −2r2 )t = γ (t),
1
≥
(r
−(µ+r

1 −r2 )t
1 −2r2 )t

(e
− 1)e
= γ1 (t),




(r
−r
)t
−(µ+r
−2r
)t
1
2

(e 1 2 − 1)e
= γ1 (t),




 (e(r1 −r2 )t − 1)e−(µ+r1 −2r2 )t = γ (t),
1
t ≤ s ≤ τ,
τ ≤ s ≤ t,
s ≤ t ≤ τ,
τ ≤ t ≤ s,
s ≤ τ ≤ t,
t ≤ τ ≤ s.
Assumptions
In this section, consider the hypotheses:
(A1 ) There exist some k ∈ C(R+ , I), p ∈ C(I, I), and a decreasing function q ∈ C(I, I),
such that
g(t, z) ≤ k(t)p(z), ∀ t ∈ R+ , ∀ z ∈ I,
e
R +∞
φ(s)k(s)q(cγe1 (s))ds < +∞ for each c > 0.
where p(z)
q(z) is an increasing function and 0
(A2 ) For any c > 0, there exists ψc ∈ C(R+ , I), such that
Z +∞
g(t, z) ≥ ψc (t), ∀ t ∈ R+ , ∀ z ∈ (0, c] with
e
φ(s)ψc (s)ds < +∞.
0
Consider the auxiliary boundary value problem

 −y ′′ (t) + cy ′ (t) + λy(t) = φ(t)g(t, y(t)),
 y(0) = h > 0,
lim y(t) = 0,
(4.2)
t→+∞
where g ∈ C(R+ × I, R) and φ ∈ C(I, I).
Definition 4.1 A function α ∈ C(R+ , I) ∩ C 2 (I, R) is called a lower solution of (4.2) if
it satisfies

 −α′′ (t) + cα′ (t) + λα(t) ≤ φ(t)g(t, α(t)),
 α(0) ≤ h,
t→+∞
 β(0) ≥ h,
t→+∞
lim α(t) ≤ 0.
A function β ∈ C(R+ , I) ∩ C 2 (I, R) is called an upper solution of (4.2) if it satisfies

 −β ′′ (t) + cβ ′ (t) + λβ(t) ≥ φ(t)g(t, β(t)),
lim β(t) ≥ 0.
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If there exists an upper solution β and a lower solution α of (4.2) with α(t) ≤ β(t) for all
t ∈ R+ , then, we can define
Dαβ (t) = {y ∈ R : α(t) ≤ y ≤ β(t)},
t ∈ R+ .
Theorem 4.1 Assume that α, β are respectively lower and upper solutions of problem
(4.2) with α(t) ≤ β(t) for all t ∈ R+ . Moreover, suppose that there exists δ ∈ C(R+ , R+ ) such
that
sup e−r1 t |g(t, y)| ≤ δ(t),
β
y∈Dα
(t)
∀ t ∈ R+ ,
R +∞
and 0 er1 s φ(s)δ(s)ds < +∞. Then, problem (4.2) has at least one solution y ∗ ∈ E ′ with
α(t) ≤ y ∗ (t) ≤ β(t), t ∈ R+ .
Proof Let



 g(t, α(t)),
∗
g (t, y) = g(t, y),



g(t, β(t)),
y < α(t),
α(t) ≤ y ≤ β(t),
y > β(t).
Note that g ∗ is continuous which allows us to define the regular problem

 −y ′′ (t) + cy ′ (t) + λy(t) = φ(t)g ∗ (t, y(t)),
 y(0) = h > 0,
lim y(t) = 0.
(4.3)
t→+∞
To show (4.3) has at least one solution y ∗ , consider for y ∈ E ′ the operator
By(t) = her2 t +
Z
+∞
K(t, s)φ(s)g ∗ (s, y(s))ds.
0
Step 1
For y ∈ E ′ and t ∈ R+ , we have
|By(t)|e
−θt
= he
−(−r2 +θ)t
Z
+
Z
+∞
e−θt K(t, s)φ(s)|g ∗ (s, y(s))|ds
0
+∞
e−r1 s K(s, s)φ(s)|g ∗ (s, y(s))|ds
Z +∞
1
≤ h+
e−r1 s φ(s)|g ∗ (s, y(s))|ds
r1 − r2 0
Z +∞
1
≤ h+
φ(s)e−r1 s |g ∗ (s, y(s))|ds
r1 − r2 0
Z +∞
1
≤ h+
φ(s)δ(s)ds < +∞,
r1 − r2 0
≤ h+
0
proving that By ∈ E ′ for all y ∈ E ′ .
Step 2 Assume that {yn }n≥1 ⊆ E ′ and y0 ∈ E ′ with
+
∗
lim yn = y0 . Then, yn (t) −→
n→+∞
y0 (t) as n → +∞, t ∈ R . Thus, the continuity of g implies that, for t ≥ 0,
|g ∗ (s, yn (s)) − g ∗ (s, y0 (s))| −→ 0, n → +∞.
No.2
Smaı̈l Djebali et al: POSITIVE SOLUTIONS FOR SINGULAR BVPs
691
Moreover, as e−r1 s |g ∗ (s, yn (s)) − g ∗ (s, y0 (s))| ≤ 2δ(s), ∀ s ≥ 0, the Lebesgue dominated convergence theorem guarantees that
kByn − By0 kθ = sup |Byn (t) − By0 (t)|e−θt
R+
≤ sup
Z
+∞
R+ 0
Z +∞
e−θt K(t, s)φ(s)|g ∗ (s, yn (s)) − g ∗ (s, y0 (s))|ds
K(s, s)e−r1 s φ(s)|g ∗ (s, yn (s)) − g ∗ (s, y0 (s))|ds
0
Z +∞
1
≤
φ(s)e−r1 s |g ∗ (s, yn (s)) − g ∗ (s, y0 (s))|ds.
r1 − r2 0
≤
As the right-hand term tends to 0 as n → ∞, B : E ′ −→ E ′ is continuous.
Step 3 The set B(E ′ ) is relatively compact.
(a) Let r1 < µ < θ and q(t) = e−µt in Lemma 1.1. Then, for y ∈ E ′ ,
|By(t)|e
−µt
= he
−(−r2 +µ)t
Z
≤ h+
+∞
e−µt K(t, s)φ(s)|g ∗ (s, y(s))|ds
0
+∞
e−r1 s K(s, s)φ(s)|g ∗ (s, y(s))|ds
0
1
≤ h+
r1 − r2
1
r1 − r2
≤ h+
+
Z
Z
+∞
e−r1 s φ(s)|g ∗ (s, y(s))|ds
0
Z
+∞
φ(s)δ(s)ds < +∞.
0
So, B(E ′ ) is uniformly bounded in the sense of the norm k.kµ .
(b) For a given T > 0, y ∈ E ′ , and t, t′ ∈ [0, T ], we have
′
|By(t) − By(t′ )| ≤ h|er2 t − er2 t |
Z +∞
+
|K(t, s) − K(t′ , s)|φ(s)|g ∗ (s, y(s))|ds
0
Z T
′
≤ h|er2 t − er2 t | +
|K(t, s) − K(t′ , s)|φ(s)|g ∗ (s, y(s))|ds
0
Z +∞
+
|K(t, s) − K(t′ , s)|φ(s)|g ∗ (s, y(s))|ds
T
′
+
1
r1 − r2
Z
T
|K(t, s) − K(t′ , s)|φ(s)|g ∗ (s, y(s))|ds
0
Z +∞
′
′
r1 t
[(e − er2 t ) − (er1 t − er2 t )]
e−r1 s φ(s)|g ∗ (s, y(s))|ds
= h|er2 t − er2 t | +
′
≤ h|e−kt − e−kt | +
Z
0
T
T
|K(t, s) − K(t′ , s)|φ(s)g ∗ (s, y(s))|ds
′
′
1
+
[(er1 t − er2 t ) − (er1 t − er2 t )]
r1 − r2
Z
+∞
φ(s)δ(s)ds.
T
Then, for any ε > 0 and T > 0, there exists η > 0 such that |By(t) − By(t′ )| < ε for all
t, t′ ∈ [0, T ] and |t − t′ | < η. Hence, Lemma 1.1 implies that B(E ′ ) is relatively compact. The
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Schauder fixed point theorem guarantees that B has at least one fixed point y ∗ ∈ E ′ . Also, y ∗
is a solution of (4.3).
Step 4 Next, we show that y ∗ satisfies α(t) ≤ y ∗ (t) ≤ β(t), ∀ t ∈ R+ , which implies
that y ∗ is a solution of (4.2). Suppose there is some t∗ ∈ R+ with y ∗ (t∗ ) > β(t∗ ) and let
z(t) = y ∗ (t) − β(t). As z(0) ≤ 0, z(∞) ≤ 0, and z(t∗ ) > 0, z must have positive maximum at
some point t0 ∈ I and then z ′′ (t0 ) ≤ 0, z ′ (t0 ) = 0, and z(t0 ) > 0. Therefore,
′′
0 ≥ z ′′ (t0 ) = y ∗ (t0 ) − β ′′ (t0 )
′
≥ [cy ∗ (t0 ) + λy ∗ (t0 ) − φ(t0 )g ∗ (t0 , y ∗ (t0 ))] − [cβ ′ (t0 ) + λβ(t0 ) − φ(t0 )g ∗ (t0 , β(t0 ))]
′
= c[y ∗ (t0 ) − β ′ (t0 )] + λ[y ∗ (t0 ) − β(t0 )] + φ(t0 )[g ∗ (t0 , β(t0 )) − g ∗ (t0 , y ∗ (t0 ))]
= λ[y ∗ (t0 ) − β(t0 )] + φ(t0 )[g(t0 , β(t0 )) − g(t0 , β(t0 ))]
= λ[y ∗ (t0 ) − β(t0 )] > 0,
which is a contradiction. Then, y ∗ (t) ≤ β(t), ∀ t ∈ R+ . Similarly, we can prove that α(t) ≤ y ∗ (t).
4.3 An existence result
Using Theorem 4.1, we obtain the following.
Theorem 4.2 Let g ∈ C(R+ × I, R+ ) and conditions (A1 ), (A2 ) hold. Also, assume that
there exist M > 0 and h ∈ C(R+ , I), such that
g(t, y) ≤ h(t), ∀ (t, y) ∈ R+ × [M, +∞),
with
Z
(4.4)
+∞
φ(s)h(s)ds < +∞.
(4.5)
0
Then, problem (1.2) has at least one positive solution y ∗ ∈ C(R+ , R+ ) ∩ C 2 (R+ , R).
Proof Choose a decreasing sequence {εn }n≥1 with lim εn = 0 and ε1 < M , and
n→+∞
consider the sequence of boundary value problems

 −y ′′ (t) + cy ′ (t) + λy(t) = φ(t)g(t, y(t)),
 y(0) = εn ,
lim y(t) = 0.
(4.6)
t→+∞
To show that (4.6) has at least one solution yn , for n ≥ 1, let β be a solution of the boundary
value problem

 y ′′ (t) + φ(t)h(t) = 0, t > 0
 y(0) = M,
lim y ′ (t) = 0.
t→+∞
From [8], we know that β(t) ≥ M and β ′ (t) ≥ 0 for any t ∈ R+ , which implies that β is an
upper solution of (4.6), ∀ n ≥ 1. Let αn (t) = εn er2 t , t ≥ 0. Then, for all n ≥ 1, αn is a lower
solution of (4.6). By Theorem 4.1, we infer that problem (4.6) has at least one positive solution
yn ∈ E ′ , ∀ n ≥ 1, with
αn (t) ≤ yn (t) ≤ β(t), ∀ t ∈ R+ .
Using similar arguments as in the proof of Theorem 2.2, we can prove that {yn }n≥1 is relatively
compact and then we can extract a convergent subsequence {ynk }k≥1 such that ynk → y ∗ as
k → +∞ and y ∗ is a solution of problem (1.2).
No.2
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Smaı̈l Djebali et al: POSITIVE SOLUTIONS FOR SINGULAR BVPs
693
Two Positive Solutions for Problem (1.2)
Theorem 5.1 Besides (A1 ) − (A2 ), assume that
(A3 ) there exist 0 < a < b < +∞, such that
g(t, y)
= +∞ uniformly for t ∈ [a, b].
y
√
Then, there exists ξ0 > 0 such that, for any c and λ with c2 + 4λ > ξ0 , problem (1.2)
has at least two positive solutions.
Proof The proof is similar to that of the proof of Theorem 3.1. We replace 2k by
√
c2 + 4λ = r1 − r2 .
Remark 5.1 If the following condition holds:
(A4 )
(r1 − r2 )cq(c)
sup
> 1,
R +∞
c>0 p(c)
φ(τ )m(τ )h(ce
γ (τ ))dτ
0
lim
y→+∞
then, Theorem 5.1 holds for each c, λ > 0.
Example 5.1 Consider the singular boundary value problem

2
e−6t
 −y ′′ (t) + y ′ (t) + 2y(t) = (e3t − 1)e− 17
2 t
√y +2 ,
y
 y(0) = 0,
lim y(t) = 0.
(5.1)
t→+∞
Here, r1 = 2, r2 = −1, and θ = 2. Then,
z2 + 2
ge(t, z) = e−10t √ , φ(t) = (e3t − 1), and γ1 (t) = (e3t − 1)e−7t .
z
Let
k(t) = e−10t , p(z) =
1
z2 + 1
√ , and q(z) = ·
z
z
Then, we check the validity of the assumptions:
(A1 ) The function q ∈ C(I, I) is decreasing and
where
p(z)
q(z)
=
g(t, z) ≤ k(t)p(z),
e
∀ t ∈ R+ , ∀ z ∈ I,
√ 2
z(z + 1) is an increasing function and for any c > 0, it holds that
Z
+∞
0
(A2 ) There exists ψc (t) =
ge(t, z) ≥ ψc (t),
φ(s)k(s)q(cγe1 (s))ds =
√2 e−10t
c
1
< +∞.
c
∈ C(R+ , I), such that
∀ t ∈ R+ , ∀z ∈ (0, c] with
Z
+∞
φ(s)ψc (s)ds < +∞.
0
(A3 ) For any 0 < a < b < +∞, we have
g(t, x)
= +∞, uniformly in t ∈ [a, b].
x→+∞
x
lim
Therefore, all conditions of Theorem 5.1 are fulfilled and so problem (5.1) has at least two
positive solutions.
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