Electronic Journal of Differential Equations, Vol. 2009(2009), No. 115, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
UPPER AND LOWER SOLUTIONS FOR A SECOND-ORDER
THREE-POINT SINGULAR BOUNDARY-VALUE PROBLEM
QIUMEI ZHANG, DAQING JIANG, SHIYOU WENG, HAIYIN GAO
Abstract. We study the singular boundary-value problem
u00 + q(t)g(t, u) = 0,
u(0) = 0,
t ∈ (0, 1), η ∈ (0, 1), γ > 0
u(1) = γu(η) .
The singularity may appear at t = 0 and the function g may be superlinear
at infinity and may change sign. The existence of solutions is obtained via an
upper and lower solutions method.
1. Introduction
Motivated by the study of multi-point boundary-value problems for linear second order ordinary differential equations, Gupta [7] studied certain three point
boundary-value problems for nonlinear ordinary differential equations. Since then,
more general nonlinear multi-point boundary-value problems have been studied by
several authors using the Leray-Schauder theorem, nonlinear alternative of LeraySchauder or coincidence degree theory. We refer the reader to [3, 4, 5, 9, 12, 13, 14,
15] for some existence results of nonlinear multi-point boundary-value problems.
Recently, Ma [14] proved the existence of positive solutions for the three point
boundary-value problem
u00 + b(t)g(u) = 0,
u(0) = 0,
t ∈ (0, 1)
u(1) = αu(η),
where η ∈ (0, 1), 0 < α < 1/η, b ≥ 0 and g ≥ 0 is either superlinear or sublinear.
He applied a fixed point theorem in cones.
In this paper, we study the singular three-point boundary-value problem
u00 + q(t)g(t, u) = 0,
u(0) = 0,
t ∈ (0, 1), η ∈ (0, 1), γ > 0
u(1) = γu(η).
(1.1)
The singularity may appear at t = 0, and the function g may be superlinear at
u = ∞ and may change sign.
2000 Mathematics Subject Classification. 34B15, 34B16.
Key words and phrases. Singular boundary-value problem; upper and lower solutions;
existence of solutions; superlinear.
c
2009
Texas State University - San Marcos.
Submitted January 27, 2009. Published September 12, 2009.
Supported by grants 10571021 from NSFC of China, and and KLAS from Key Laboratory
for Applied Statistics of MOE.
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Q. ZHANG, D. JIANG, S. WENG, H. GAO
EJDE-2009/115
Some basic results on the singular two point boundary-value problems were obtain in [1, 11, 17], in all these papers the arguments rely on the assumption that
g(t, u) is positive. This implies that the solutions are concave. Recently, some
authors have studied the case when g is allowed to change sign by applying the
modified upper and lower solutions method; see for example [11].
The present work is a direct extension of some results on the singular two-point
boundary-value problems. As in [11], our technique relies essentially on a modified
method of upper and lower solutions method for singular three-point boundaryvalue problems which we believe is well adapted to this type of problems.
2. Upper and lower solutions
Consider the three-point boundary-value problem
u00 + f (t, u) = 0,
t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1/η)
u(0) = A,
u(1) − γu(η) = B.
(2.1)
We use the following assumption:
(A1) f : (0, 1] × R → R is a continuous function, there exist two functions
α, β ∈ C([0, 1], R) and α(t) ≤ β(t), for all t ∈ [0, 1], if there exist a function
h ∈ C( (0, 1], (0, ∞)), such that
|f (t, u)| ≤ h(t)
for α(t) ≤ u ≤ β(t),
Z 1
lim+ t2 h(t) = 0,
th(t)dt < ∞.
t→0
(2.2)
(2.3)
0
We call a function α(t) a lower solution for (2.1), if α ∈ C([0, 1], R)∩C 2 ((0, 1), R),
and
α00 + f (t, α) ≥ 0,
α(0) ≤ A,
for t ∈ (0, 1),
α(1) − γα(η) ≤ B.
Similarly, we call a function β(t) an upper solution for (2.1), if β ∈ C([0, 1], R) ∩
C 2 ((0, 1), R), and
β 00 + f (t, β) ≤ 0,
β(0) ≥ A,
for t ∈ (0, 1),
β(1) − γβ(η) ≥ B.
A function u(t) is said to be a solution to (2.1), if it is both a lower and an upper
solution to (2.1).
Our first result reads as follows.
Theorem 2.1. Assume (A1) and let α, β be, respectively, a lower solution and an
upper solution for (2.1) such that α(t) ≤ β(t) on [0, 1]. Then (2.1) has at least one
solution u(t) such that
α(t) ≤ u(t) ≤ β(t),
for t ∈ [0, 1].
Consider now the modified boundary-value problem
u00 + f1 (t, u) = 0,
u(0) = A,
for t ∈ (0, 1),
u(1) − γu(η) = B,
(2.4)
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UPPER AND LOWER SOLUTIONS
3
where


f (t, α(t)), if u < α(t),
f1 (t, u) = f (t, u),
if α(t) ≤ u ≤ β(t),


f (t, β(t)), if u > β(t).
Lemma 2.2. Assume that (2.3) holds. Then the boundary-value problem
y 00 = −h(t),
y(0) = A,
0 < t < 1,
y(1) − γy(η) = B
(2.5)
has a unique solution y(t) in C([0, 1], [0, ∞)) ∩ C 2 ((0, 1), R), which can be written
as
Z 1
B − A(1 − γ)
t+
G(t, s)h(s)ds, 0 ≤ t ≤ 1,
y(t) = A +
1 − γη
0
where G(t, s) is Green’s function of the boundary-value problem −y 00 = 0, y(0) = 0,
y(1) = γy(η). The function G is explicitly given by: when 0 ≤ s ≤ η,
( s[1−t−γ(η−t)]
, s ≤ t,
1−γη
G(t, s) = t[1−s−γ(η−s)]
, s > t;
1−γη
when η < s ≤ 1,
( s(1−t)+γη(t−s)
G(t, s) =
1−γη
t(1−s)
1−γη ,
,
s ≤ t,
s > t.
Proof. Uniqueness. The proof of the uniqueness of a solution is standard and hence
omitted. Existence. Let
Z 1
B − A(1 − γ)
y(t) := A +
t+
G(t, s)h(s)ds, 0 ≤ t ≤ 1;
1 − γη
0
i.e.,

Rt
A + B−A(1−γ)
t + 0 s[1−t−γ(η−t)]
h(s)ds

1−γη
1−γη


R
R

+ η t[1−s−γ(η−s)] h(s)ds + 1 t(1−s) h(s)ds, 0 ≤ t ≤ η,
1−γη
t
η 1−γη
y(t) =
R η s[1−t−γ(η−t)]
B−A(1−γ)

A + 1−γη t + 0
h(s)ds


1−γη

R 1 t(1−s)
 R t s(1−t)+γη(t−s)
+ η
h(s)ds + t 1−γη h(s)ds, η < t ≤ 1.
1−γη
Then we have

Rt
B−A(1−γ)

+ 0 s(γ−1)

1−γη
1−γη h(s)ds

R

η 1−s−γ(η−s)


+
h(s)ds

1−γη
 Rt1
1−s
0
0 < t ≤ η,
y (t) = + η 1−γη h(s)ds,

R

η
B−A(1−γ)
s(γ−1)


+ 0 1−γη h(s)ds

1−γη


+ R t γη−s h(s)ds + R 1 1−s h(s)ds, η < t ≤ 1.
η 1−γη
t 1−γη
R1
Rt
and y 00 (t) = −h(t) for all t ∈ (0, 1). Since 0 th(t)dt < ∞, limt→0+ 0 sh(s)ds = 0;
so we have
Z η
1 − s − γ(η − s)
y(0) = A + lim+ t
h(s)ds.
1 − γη
t→0
t
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Q. ZHANG, D. JIANG, S. WENG, H. GAO
EJDE-2009/115
R1
R1
If 0 1−s−γ(η−s)
h(s)ds < ∞, then y(0) = A. If 0 1−s−γ(η−s)
h(s)ds = ∞, then by
1−γη
1−γη
(2.3) we obtain
R η 1−s−γ(η−s)
h(s)ds
1 − γη + t(γ − 1)
1−γη
t
y(0) = A + lim+
= A + lim+ t2 h(t)
= A.
1/t
1 − γη
t→0
t↓0
We have also
y(1) − γy(η)
Z 1
sγ(1 − η)
γη(1 − s)
h(s)ds +
h(s)ds
1
−
γη
1 − γη
0
η
Z η
Z 1
s(1 − η)
η(1 − s)
B − A(1 − γ)
− γ(
η+
h(s)ds +
h(s)ds) = B.
1 − γη
1 − γη
1 − γη
0
η
B − A(1 − γ)
=
+
1 − γη
Z
η
This shows that y(t) is a positive solution of (2.5), and y ∈ C([0, 1], [0, ∞)) ∩
C 2 ((0, 1), R).
Let us define an operator Φ : X → X by
(Φu)(t) = A +
B − A(1 − γ)
t+
1 − γη
Z
1
G(t, s)f1 (s, u(s))ds,
(2.6)
0
where X = {u ∈ C([0, 1], R) with the norm kuk} is a Banach space, with
kuk := sup{|u(t)| : 0 ≤ t ≤ 1}.
Without loss of generality, we assume that A = B = 0.
To prove the existence of a solution to (2.4), we need the following Lemma.
Lemma 2.3. The function Φ is continuous from X to X and Φ(X) is a compact
subset of X.
Proof. As in the proof of Lemma 2.2, from the definition of f1 and from (2.6), we
have
Z 1
Z 1
|(Φu)(t)| ≤
G(t, s)|f1 (s, u(s))|ds ≤
G(t, s)h(s)ds = y(t), t ∈ [0, 1]. (2.7)
0
0
So we have Φu ∈ C([0, 1], R) ∩ C 2 ((0, 1), R), and
kΦuk ≤ kyk.
(2.8)
This shows that Φ(X) is a bounded subset of X.
Noting the facts that y(0) = 0 and the continuity of y(t) on [0, 1], we have from
(2.7) that for any > 0, one can find a δ1 > 0 (independent with u) such that
0 < δ1 < 1/8 and
(2.9)
(Φu)(t) < , t ∈ [0, 2δ1 ].
2
On the other hand, from (2.6), since |f1 (s, u(s))| ≤ h(s), s ∈ (0, 1), we can obtain
|(Φu)0 (t)| ≤ L,
t ∈ [δ1 , 1].
2L ,
then for t1 , t2 ∈ [δ1 , 1], |t2 − t1 | < δ2 , we have
|(Φu)(t1 ) − (Φu)(t2 )| ≤ L|t1 − t2 | < .
2
Define δ = min{δ1 , δ2 }, then using (2.9), (2.10), we obtain
Let δ2 =
|(Φu)(t1 ) − (Φu)(t2 )| < ,
(2.10)
(2.11)
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UPPER AND LOWER SOLUTIONS
5
for t1 , t2 ∈ [0, 1], |t1 − t2 | < δ. This shows that {(Φu)(t) : u ∈ X} is equicontinuous
on [0, 1].
We can obtain the continuity of Φ in a similar way as above. In fact, if un , u ∈ X
and kun − uk → 0 as n → ∞, then we have
Z 1
G(t, s)h(s)ds = 2y(t), t ∈ [0, 1],
(2.12)
|(Φun )(t) − (Φu)(t)| ≤ 2
0
Noting the facts that y(0) = 0 and the continuity of y(t) on [0, 1], then for any
> 0, one can find a δ1 > 0 (independent of un ) such that 0 < δ1 < 1/8 and
|(Φun )(t) − (Φu)(t)| < ,
t ∈ [0, δ1 ].
(2.13)
On the other hand, from the continuity of f1 , one has
|(Φun )(t) − (Φu)(t)| → 0,
t ∈ [δ1 , 1],
(2.14)
as n → ∞. This together with (2.13) implies that kΦun − Φuk → 0 as n → ∞.
Therefore, Φ : X → X is completely continuous. The proof is complete.
Lemma 2.4. Let u(t) be a solution to (2.4). Then α(t) ≤ u(t) ≤ β(t) for all
t ∈ [0, 1]; i.e., u(t) is a solution to (2.1).
Proof. We first prove that u(t) ≤ β(t) on [0, 1]. Let x(t) := u(t) − β(t). Assume
that u(t) > β(t) for some t ∈ [0, 1]. Since u(0) = 0 ≤ β(0), it follows that
x(0) ≤ 0,
x(1) = u(1) − β(1) ≤ γu(η) − γβ(η) = γx(η).
Let σ ∈ (0, 1] be such that x(σ) = maxt∈[0,1] x(t). Then x(σ) > 0.
Case(i): σ ∈ (0, 1). So there exists an interval (a, σ] ⊂ (0, 1) such that x(t) > 0
in (a, σ], and
x(a) = 0, x(σ) = max x(t) > 0, x0 (σ) = 0.
t∈[0,1]
For t ∈ (a, σ] we have that f1 (t, u(t)) = f (t, β(t)) and therefore
u00 (t) + f1 (t, u(t)) = u00 (t) + f (t, β(t)) = 0
for all t ∈ (a, σ].
On the other hand, as β is an upper solution for (2.1), we have
β 00 (t) + f (t, β(t)) ≤ 0
00
for all t ∈ (a, σ].
00
Thus, we obtain u (t) ≥ β (t) for all t ∈ (a, σ], and hence, x00 (t) ≥ 0. Then
x0 (t) ≤ 0 on (a, 1] which is a contradiction.
Case(ii): σ = 1. So there exists (a, 1] ⊂ (0, 1] such that
x(a) = 0,
x(1) = max x(t), x(1) − γx(η) ≤ 0.
t∈[0,1]
In the same way as in Case(i), we can obtain that x(t) > 0, x00 (t) ≥ 0, t ∈ (a, 1].
Since x(η) ≥ γ1 x(1) > 0, then η > a.
Consider the three-point boundary-value problem
x00 = h(t) > 0,
x(a) = 0,
a < t < 1,
x(1) − γx(η) = b1 ≤ 0.
(2.15)
Then this equation has a unique solution x(t) ∈ C([a, σ], [0, ∞)) ∩ C 2 ((a, 1), R),
which can be represented as
Z 1
b1 (t − a)
x(t) =
−
G[a,1] (t, s)h(s)ds, a ≤ t ≤ 1,
1 − a − γ(η − a)
a
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Q. ZHANG, D. JIANG, S. WENG, H. GAO
EJDE-2009/115
where G[a,1] (t, s) is the Green’s function of the boundary-value problem −y 00 = 0,
y(a) = 0, y(1) = γy(η), which is explicitly given by: when a ≤ s ≤ η,

 (s−a)[1−t−γ(η−t)] , s ≤ t,
1−a−γ(η−a)
G[a,1] (t, s) = (t−a)[1−s−γ(η−s)]

, s > t;
1−a−γ(η−a)
when η < s ≤ 1,
G[a,1] (t, s) =

 (s−a)(1−t)+γ(t−s)(η−a) ,
1−a−γ(η−a)
 (t−a)(1−s) ; s > t.
1−a−γ(η−a)
s ≤ t,
1−a
Since 0 < γ < η1 < η−a
, then G[a,1] (t, s) ≥ 0, and hence x(t) ≤ 0 on [a, 1], which
is a contradiction. In very much the same way, we can prove that u(t) ≥ α(t) on
[0, 1].
3. Main results
Let g : [0, 1] × (0, ∞) → R be a continuous function and q ∈ C((0, 1], R+
0 ).
Consider the three-point boundary-value problem
u00 + q(t)g(t, u) = 0,
t ∈ (0, 1), η ∈ (0, 1), γ ∈ (0, 1]
u(0) = 0,
u(1) = γu(η).
(3.1)
Theorem 3.1. Assume that
(H1) |g(t, x)| ≤ F (x) + Q(x) on [0, 1] × (0, ∞) with F > 0 continuous and nonincreasing on (0, ∞), Q ≥ 0 continuous on [0, ∞), and Q
F nondecreasing on
(0, ∞);
(H2) there exist constants L > 0 and ε > 0 such that g(t, x) > L for all (t, x) ∈
[0, 1] × (0, ε], and F (x) > L, x ∈ (0, ε];
(H3)
Z 1
tq(t)dt < ∞,
(3.2)
lim t2 q(t) = 0,
t→0+
0
Z c
1
du sup
> b0 ,
(3.3)
Q(c)
c∈(0,∞) 1 +
0 F (u)
F (c)
where b0 =
R1
0
rq(r)dr.
Then (3.1) has at least one solution u ∈ C([0, 1], [0, ∞))∩C 2 ((0, 1), R) with u(t) > 0
on (0, 1].
From Lemma 2.2, we obtain the following result.
Lemma 3.2. There exists an unique solution W ∈ C([0, 1], [0, ∞)) ∩ C 2 ((0, 1), R),
with W (t) > 0 on (0, 1] to the problem
W 00 + q(t) = 0,
W (0) = 0,
0 < t < 1,
W (1) = γW (η).
Choose M > 0, δ > 0 (δ < M ) such that
Z M
du
1
> b0 .
Q(M )
F
(u)
1+
δ
F (M )
(3.4)
(3.5)
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UPPER AND LOWER SOLUTIONS
7
Let n0 ∈ {1, 2, . . . } be chosen so that 1/n0 < min{ε − mkW k, δ}, where W is
the solution of (3.4), and 0 < m < min{L, ε/kW k, 1} is chosen and fixed. Let
N + = {n0 , n0 + 1, . . . }.
We first show that the boundary-value problem
u00 + q(t)g(t, u) = 0, 0 < t < 1,
(3.6)
1
1−γ
u(0) = , u(1) − γu(η) =
, n ∈ N+
n
n
has a solution un for each n ∈ N + with un (t) ≥ n1 for t ∈ [0, 1] and kun k < M .
We have the following Claim
Claim: Let αn (t) = mW (t) + n1 , t ∈ [0, 1], then αn (t) is a (strict) lower solution
for problem (3.6).
Proof. For the choice of m and n, we have mW (t) + n1 ≤ mkW k + n10 < ε, then
from (H2),
1
g(t, mW (t) + ) > L > m for all t ∈ [0, 1].
n
Then we obtain
1
1
αn00 (t) + q(t)g(t, αn (t)) = (mW (t) + )00 + q(t)g(t, mW (t) + )
n
n
1
= mW 00 (t) + q(t)g(t, mW (t) + )
n
1
= q(t)(g(t, mW (t) + ) − m) > 0, 0 < t < 1.
n
We obtain αn (0) = mW (0) + n1 = n1 , and
1
1
− γ(mW (η) + )
n
n
1−γ
1−γ
= m(W (1) − γW (η)) +
=
.
n
n
Thus the proof of Claim is complete.
αn (1) − γαn (η) = mW (1) +
To find the upper solution of (3.6), we consider the problem
u00 + q(t)F (u)(1 +
Q(M )
) = 0,
F (M )
0 < t < 1,
(3.7)
1
1−γ
, u(1) − γu(η) =
.
n
n
To show that this problem has a solution we study
u(0) =
u00 + q(t)F ∗ (u)(1 +
Q(M )
) = 0,
F (M )
0 < t < 1,
(3.8)
1
1−γ
, u(1) − γu(η) =
,
n
n
(
F (u), u ≥ 1/n,
∗
F (u) =
F ( n1 ), u < 1/n.
u(0) =
where
Then F ∗ (u) ≤ F (u) for u > 0.
In the same way as in the Claim, we can easily prove αn (t) =
a (strict) lower solution of (3.8).
1
n
+ mW (t) is also
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Q. ZHANG, D. JIANG, S. WENG, H. GAO
EJDE-2009/115
By Lemma 2.2, let βn0 ∈ C([0, 1], R) ∩ C 2 ((0, 1), R) be the unique solution of the
boundary-value problem
Q(M )
) = 0, 0 < t < 1,
F (M )
1−γ
u(1) − γu(η) =
.
n
u00 + q(t)F (αn (t))(1 +
1
u(0) = ,
n
(3.9)
Since βn0 is a solution of this equation,
00
βn0 + q(t)F (αn (t))(1 +
βn0 (0) =
1
,
n
Q(M )
) = 0,
F (M )
βn0 (1) − γβn0 (η) =
0 < t < 1,
1−γ
.
n
On the other hand, as αn is a lower solution of (3.8), and αn ≥ 1/n, we have
αn 00 + q(t)F (αn (t))(1 +
αn (0) =
1
,
n
Q(M )
) ≥ 0,
F (M )
αn (1) − γαn (η) =
0 < t < 1,
1−γ
.
n
So we obtain αn (t) ≤ βn0 (t) for t ∈ [0, 1]. Thus
Q(M )
)
F (M )
Q(M )
Q(M )
) + q(t)F (βn0 )(1 +
)
= −q(t)F (αn )(1 +
F (M )
F (M )
Q(M )
= q(t)(1 +
)(F (βn0 ) − F (αn )) ≤ 0,
F (M )
00
βn0 + q(t)F ∗ (βn0 )(1 +
so that βn0 is an upper solution for problem (3.8).
If we now take αn0 ≡ αn , we have that αn0 and βn0 are, respectively, a lower and
an upper solution of (3.8) with αn0 (t) ≤ βn0 (t), for all t ∈ [0, 1]. So by the Lemma
2.4, we know that there exists a solution βn ∈ C([0, 1], R) ∩ C 2 ((0, 1), R) of (3.8)
such that
αn (t) = αn0 (t) ≤ βn (t) ≤ βn0 (t), ∀t ∈ [0, 1].
Now we claim that kβn k < M . Suppose this is false; i.e., suppose kβn k ≥ M . Since
βn (1)− n1 = γ(βn (η)− n1 ) ≤ βn (η)− n1 , βn00 (t) ≤ 0 on (0,1) and βn ≥ n1 on [0, 1], there
exists σ ∈ (0, 1) with βn0 (t) ≥ 0 on (0, σ), βn0 (t) ≤ 0 on (σ, 1) and βn (σ) = kβn k.
Then for z ∈ (0, 1), we have
−βn00 (z) ≤ F (βn (z))(1 +
Q(M )
)q(z).
F (M )
Integrate from t(0 < t ≤ σ) to σ to obtain
Z σ
Q(M )
)
F (βn (z))q(z)dz;
βn0 (t) ≤ (1 +
F (M ) t
so we have
βn0 (t)
Q(M )
≤ (1 +
)
F (βn (t))
F (M )
Z
σ
q(z)dz .
t
(3.10)
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UPPER AND LOWER SOLUTIONS
9
Then integrate from 0 to σ to obtain
Z βn (σ)
Z
Z
Z
Q(M ) σ
dy
Q(M ) σ σ
q(z)dz dt = 1 +
tq(t)dt.
≤ 1+
1
F (y)
F (M ) 0
F (M ) 0
t
n
Consequently
M
Z
dy
Q(M ) 1
tq(t)dt.
(3.11)
≤ 1+
F (y)
F (M ) 0
δ
This contradicts (3.5) and consequently kβn k < M .
It follows from the fact βn ≥ 1/n, we can obtain βn is a solution of (3.7) also.
Since F is nonincreasing on (0, ∞), similar to the proof of Lemma 2.4, we can
obtain the uniqueness of solutions to (3.7).
Next we show that βn is an upper solution of (3.6). Observe that
Z
|g(t, x)| ≤ F (x) + Q(x)
on [0, 1] × (0, ∞).
We have
Q(M ) + q(t)|g(t, βn (t))|
βn00 (t) + q(t)g(t, βn (t)) ≤ −q(t)F (βn (t)) 1 +
F (M )
Q(β (t)) Q(M ) n
≤ q(t)F (βn (t))
−
≤ 0, t ∈ (0, 1).
F (βn (t)) F (M )
Thus βn is an upper solution for problem (3.6).
This together with the Claim yields that αn and βn are, respectively, a lower
and an upper solution for (3.6) with αn ≤ βn for all t ∈ [0, 1]. So we conclude (3.6)
has a solution un ∈ C([0, 1], R) ∩ C 2 ((0, 1), R) such that
1
mW (t) + = αn (t) ≤ un (t) ≤ βn (t) ≤ M, ∀t ∈ [0, 1].
n
Consider now the pointwise limit
z(t) := lim un (t),
∀t ∈ [0, 1].
(3.12)
Let e = [a, 1] ⊂ (0, 1], Let tn ∈ (a, 1) such that u0n (tn ) = un (1) − un (a) /(1 − a).
We obtain
Z tn
un (1) − un (a)
0
un (t) =
+
q(s)g(s, un (s)ds, t ∈ e.
1−a
t
n→+∞
Since mW (t) ≤ un (t) ≤ M , then we have
Z
2M
Q(M ) 1
+ 1+
q(t)F (mW (t))dt := C(a, 1),
|u0n (t)| ≤
1−a
F (M ) a
t ∈ e.
(3.13)
Set vn = maxt∈e |u0n (t)|, which implies vn is bounded. That means u0n (t) is bounded
on e.
Then, by the Ascoli-Arzela theorem, it is standard to conclude that z(t) is a
solution of (3.1) on the interval e = [a, 1]. Since e is arbitrary, we find that
z ∈ C((0, 1], [0, ∞)) ∩ C 2 ((0, 1), R),
and z 00 (t) + q(t)g(t, z(t)) = 0,
t ∈ (0, 1).
Also, we have
1
1−γ
= 0, z(1) − γz(η) = lim
= 0.
n→+∞
n
n
The same argument as in [11] works, we can prove the continuity of z(t) at t = 0
and t = 1. The proof is complete.
z(0) = lim
n→+∞
10
Q. ZHANG, D. JIANG, S. WENG, H. GAO
EJDE-2009/115
By essentially the same argument as in Theorem 3.1 and [2, Theorem 4.2], we
have the following result.
Theorem 3.3. Assume that
(H1*) for any r > 0 there is hr ∈ C((0, 1], (0, ∞)): |q(t)g(t, x)| ≤ hr (t) for all
(t, x) ∈ (0, 1] × [r, ∞), such that
Z 1
thr (t)dt < +∞;
lim+ t2 hr (t) = 0,
t→0
0
(H2*) there exist constants L > 0 and ε > 0 such that g(t, x) > L for all (t, x) ∈
[0, 1] × (0, ε].
Then (3.1) has at least one solution u ∈ C([0, 1], [0, ∞) ∩ C 2 ((0, 1), R). Moreover,
if g(t, x) is non-increasing in x > 0, then the solution is unique.
4. An example
Consider the singular boundary-value problem
u00 + σt−m (u−α + uβ − T sin(8πt)) = 0,
u(0) = 0,
zu(1) = γu(η),
t ∈ (0, 1)
η ∈ (0, 1), γ ∈ (0, 1]
(4.1)
with 0 ≤ m < 2, σ > 0, α > 0, β ≥ 0. Set
F (u) = u−α ,
Q(u) = uβ + 1, q(t) = σt−m ,
Z 1
σ
b0 =
rq(r)dr =
.
2−m
0
Applying Theorem 3.1, we find that (4.1) has a positive solutions if
xα+1
.
α
α+β )
x∈(0,∞) (α + 1)(1 + x + x
σ < (2 − m) sup
(4.2)
Obviously, (H1)-(H3) in Theorem 3.1 are satisfied. Thus, (4.1) has a solution
u ∈ C([0, 1], [0, ∞) ∩ C 2 ((0, 1), R) with u > 0 on (0, 1].
We remark that if 0 ≤ β < 1, then (4.1) has at least one positive solution for all
σ > 0, since the right-hand side of (4.2) is infinity.
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Qiumei Zhang
School of Science, Changchun University, Changchun 130022, China.
Department of Mathematics, Northeast Normal University, Changchun 130024, China
E-mail address: zhangqm1110@yahoo.com.cn
Daqing Jiang
Department of Mathematics, Northeast Normal University, Changchun 130024, China
E-mail address: jiangdq067@nenu.edu.cn
School of Science, Changchun University, Changchun 130022, China
E-mail address, Haiyin Gao: gaohaiyinhealthy@yahoo.com.cn
E-mail address, Shiyou Weng: wengshiyou2001@yahoo.com.cn