Hindawi Publishing Corporation
Boundary Value Problems
Volume 2010, Article ID 254928, 16 pages
doi:10.1155/2010/254928
Research Article
Existence and Uniqueness of Positive Solution
for a Singular Nonlinear Second-Order m-Point
Boundary Value Problem
Xuezhe Lv and Minghe Pei
Department of Mathematics, Beihua University, JiLin City 132013, China
Correspondence should be addressed to Minghe Pei, peiminghe@ynu.ac.kr
Received 25 November 2009; Accepted 10 March 2010
Academic Editor: Ivan T. Kiguradze
Copyright q 2010 X. Lv and M. Pei. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
The existence and uniqueness of positive solution is obtained for the singular second-order m
point boundary value problem u t ft, ut 0 for t ∈ 0, 1, u0 0, u1 m−2
i1 αi uηi ,
where m ≥ 3, αi > 0 i 1, 2, . . . , m − 2, 0 < η1 < η2 < · · · < ηm−2 < 1 are constants, and ft, u can
have singularities for t 0 and/or t 1 and for u 0. The main tool is the perturbation technique
and Schauder fixed point theorem.
1. Introduction
In this paper, we investigate the existence and uniqueness of positive solution for the singular
second-order differential equation
u t ft, ut 0,
t ∈ 0, 1
1.1
with the m-point boundary conditions
u0 0,
u1 m−2
αi u ηi ,
1.2
i1
where m ≥ 3, αi > 0 i 1, 2, . . . , m − 2, 0 < η1 < η2 < · · · < ηm−2 < 1 are constants, and ft, u
can have singularities for t 0 and/or t 1 and for u 0.
2
Boundary Value Problems
Multipoint boundary value problems for second-order ordinary differential equations
arise in many areas of applied mathematics and physics; see 1–3 and references therein.
The study of three-point boundary value problems for nonlinear second-order ordinary
differential equations was initiated by Lomtatidze 4, 5. Since then, the nonlinear secondorder multipoint boundary value problems have been studied by many authors; see 1–
3, 6–29 and references therein. Most of all the works in the above mentioned references are
nonsingular multipoint boundary value problems; see 1–3, 10–17, 20–23, 25, 26, 28, 29, but
the works on the singularities have been quite rarely seen; see 4–8, 18, 19, 24, 27.
Recently, Du and Zhao 7, by constructing lower and upper solutions and together
with the maximal principle, proved the existence and uniqueness of positive solutions for the
following singular second-order m-point boundary value problem:
u t ft, ut 0,
u0 0,
u1 t ∈ 0, 1,
m−2
αi u ηi ,
1.3
i1
where m ≥ 3, 0 < αi < 1 i 1, 2, . . . , m − 2, 0 < η1 < η2 < · · · < ηm−2 < 1 are constants,
m−2
i1 αi < 1, ft, u is singular at t 0, t 1 and u 0, under conditions that
H1 ft, u ∈ C0, 1 × 0, ∞, 0, ∞, and ft, u is decreasing in u;
1
H2 ft, λ /
≡ 0, 0 t1 − tft, λt1 − tdt < ∞, for all λ > 0.
The purpose of this paper is to establish existence and uniqueness result of positive
solution to SBVP1.1, 1.2 under conditions that are weaker than conditions in 7 and hence
improve the result in 7 by using perturbation technique and Schauder fixed point theorem
30.
Throughout this paper, we make the following assumptions:
C0 αi > 0, i 1, 2, . . . , m − 2 and m−2
i1 αi ≤ 1;
C1 f : 0, 1 × 0, ∞ → 0, ∞ is continuous and nonincreasing in u for each fixed
t ∈ 0, 1;
1
C2 0 < 0 s1 − sfs, u0 ds < ∞ for each constant u0 ∈ 0, ∞.
2. Preliminary
We consider the perturbation problems that are given by
u t ft, ut 0,
u0 h,
t ∈ 0, 1,
m−2
m−2
αi u ηi 1 −
αi h,
u1 i1
2.1h
i1
where h is any nonnegative constant.
Definition 2.1. For each fixed constant h ≥ 0, a function ut is said to be a positive solution of
BVP2.1h if u ∈ C0, 1 ∩ C2 0, 1 with ut > 0 on 0, 1 such that u t ft, ut 0 holds
m−2
for all t ∈ 0, 1 and u0 h, u1 m−2
i1 αi uηi 1 −
i1 αi h.
Boundary Value Problems
3
Lemma 2.2. Assume that conditions C1 and C2 are satisfied. Then, for each fixed constant u0 > 0,
η1
lim t
t → 0
2.2
fs, u0 ds 0,
t
lim− 1 − t
t
t→1
fs, u0 ds 0.
2.3
ηm−2
Proof. We only prove 2.2. And 2.3 can be proved similarly.
For each fixed constant u0 > 0, let
η1
vt t
for t ∈ 0, η1 .
fs, u0 ds
2.4
t
Then from the conditions C1 and C2 , we have
η1
0 ≤ vt ≤
sfs, u0 ds ≤
t
v t η1
η1
for t ∈ 0, η1 ,
sfs, u0 ds < ∞
0
fs, u0 ds − tft, u0 for t ∈ 0, η1 .
2.5
t
Hence from the conditions C1 and C2 , we have
η1
v tdt ≤
η1
η1
dt
0
0
fs, u0 ds η1
t
η1
tft, u0 dt 2 tft, u0 dt < ∞.
0
2.6
0
This implies that v t ∈ L1 0, η1 , and hence for each t ∈ 0, η1 ,
t
0
v τdτ η1
t
dτ
0
fs, u0 ds −
t
η1
τfτ, u0 dτ t
0
τ
fs, u0 ds vt.
2.7
t
Thus, it follows from the absolute continuity of integral that limt → 0 vt 0, that is,
η1
lim t
t → 0
fs, u0 ds 0.
2.8
t
This completes the proof of the lemma.
In the following discussion Gt, s denotes Green’s function for Dirichlet problem:
−u t 0,
t ∈ 0, 1,
u0 u1 0.
2.9
4
Boundary Value Problems
Then Green’s function Gt, s can be expressed as follows:
Gt, s ⎧
⎨1 − ts,
0 ≤ s ≤ t ≤ 1,
⎩1 − st,
0 ≤ t ≤ s ≤ 1.
2.10
It is easy to see that Green’s function Gt, s has the following simple properties:
i 0 ≤ t1 − ts1 − s ≤ Gt, s ≤ s1 − s for t, s ∈ 0, 1 × 0, 1;
ii Gt, s > 0 for t, s ∈ 0, 1 × 0, 1;
iii G0, s G1, s 0 for s ∈ 0, 1.
By direct calculation, we can easily obtain the following result.
Lemma 2.3. Assume that conditions C0 , C1 , and C2 are satisfied. Then, ut is a positive
solution of BVP2.1h h > 0 if and only if u ∈ C0, 1 is a solution of the following integral
equation:
ut 1
Gt, sfs, usds 0
m−2
t
1−
m−2
i1
αi ηi
1
αi
G ηi , s fs, usds h,
0
i1
2.11h
such that ut > h > 0 on 0, 1.
Lemma 2.4. Assume that conditions C0 , C1 , and C2 are satisfied. Suppose also that u ∈ C0, 1
is a solution of the following integral equation:
ut 1
Gt, sfs, usds 0
m−2
t
1−
m−2
i1
αi ηi
1
αi
i1
G ηi , s fs, usds,
2.12
0
such that ut > 0 on 0, 1. Then, ut is a positive solution of SBVP1.1, 1.2.
Proof. Since u ∈ C0, 1 is a solution of 2.12 with ut > 0 on 0, 1, then for each t ∈ 0, 1,
t
1
s1 − tfs, usds < ∞,
0
t1 − sfs, usds < ∞.
2.13
t
So for each t ∈ 0, 1, we have
t
0
sfs, usds < ∞,
1
1 − sfs, usds < ∞.
2.14
t
m−2 1
For convenience, let c : 1/1 − m−2
i1 αi ηi i1 αi 0 Gηi , sfs, usds. Take t ∈ 0, 1 and
Δt such that t Δt ∈ 0, 1, then from the definition of derivative, the mean value theorem of
Boundary Value Problems
5
integral, and the absolute continuity of integral, we have
lim
Δt → 0
ut Δt − ut
Δt
tΔt
1
1
lim
s1 − t − Δtfs, usds 1 − st Δtfs, usds
Δt → 0 Δt
0
tΔt
1
t
− s1 − tfs, usds − t1 − sfs, usds c
0
t
t
tΔt
1
− sΔtfs, usds s1 − t − Δtfs, usds
lim
Δt → 0 Δt
0
t
1
tΔt
t1 − sfs, usds c
1 − sΔtfs, usds −
−
t
tΔt
sfs, usds t1 − tft, ut 0
−
t
sfs, usds 0
1
1
t
1 − sfs, usds − t1 − tft, ut c
t
1 − sfs, usds c.
t
2.15
Hence
u t −
t
sfs, usds 0
1
1 − sfs, usds c
for t ∈ 0, 1.
2.16
t
Consequently u ∈ C0, 1.
Again, from the definition of derivative and the mean value theorem of integrals, we
have
u t Δt − u t
Δt → 0
Δt
tΔt
1
1
lim
−
sfs, usds 1 − sfs, usds
Δt → 0 Δt
0
tΔt
1
t
sfs, usds − 1 − sfs, usds
lim
0
t
t1
tΔt
1
−
sfs, usds −
lim
1 − sfs, usds
Δt → 0 Δt
t
t
tΔt
1
−
fs, usds
lim
Δt → 0 Δt
t
−ft, ut for t ∈ 0, 1.
Hence u t −ft, ut for t ∈ 0, 1. In particular, u ∈ C0, 1.
2.17
6
Boundary Value Problems
On the other hand, from 2.12, we have u0 0 and
m−2
m−2
αi u ηi αi
i1
1
G ηi , s fs, usds 0
i1
m−2
1 αi G ηi , s fs, usds 0
i1
m−2
1
1−
m−2
i1
αi ηi
1
αi
i1
ηi
m−2
1−
m−2
i1
m−2
αi ηi
i1 αi ηi
m−2
1 − i1 αi ηi
1
αi
G ηi , s fs, usds
0
i1
m−2
1 αi G ηi , s fs, usds
i1
0
G ηi , s fs, usds
0
u1.
2.18
In summary, ut is a positive solution of SBVP1.1, 1.2. This completes the proof of the
lemma.
Remark 2.5. Assume that all conditions in Lemma 2.4 hold. Then
1 if f ∈ C0, 1 × 0, ∞, 0, ∞, we have
2.19
u ∈ C0, 1 ∩ C1 0, 1 ∩ C2 0, 1;
2 if f ∈ C0, 1 × 0, ∞, 0, ∞, we get
2.20
u ∈ C0, 1 ∩ C1 0, 1 ∩ C2 0, 1.
Lemma 2.6. Assume that conditions C0 , C1 , and C2 are satisfied. Then, for each constant h > 0,
BVP2.1h has a unique solution ut; h with ut; h ≥ h on 0, 1.
Proof. We begin by defining an operator T in Dh by
T ut 1
Gt, sfs, usds 0
m−2
t
1−
m−2
i1
αi ηi
1
αi
i1
G ηi , s fs, usds h,
2.21
0
where Dh : {u ∈ C0, 1 : ut ≥ h on 0, 1} is a convex closed set. Then from Lemma 2.2
and the condition C2 , we have T u ∈ C0, 1 and T u satisfies
T u t ft, ut 0,
T u0 h,
t ∈ 0, 1,
m−2
m−2
αi T u ηi 1 −
αi h.
T u1 i1
2.22
i1
We now apply Schauder fixed point theorem 30 to obtain the existence of a fixed
point for T . To do this, it suffices to verify that T is continuous in Dh and T Dh is a compact
set.
Boundary Value Problems
7
Take u0 ∈ Dh , and let {uk }∞
k1 ⊂ Dh such that
uk − u0 C0,1 −→ 0 as k −→ ∞.
2.23
ft, uk t −→ ft, u0 t as k −→ ∞.
2.24
Then for each t ∈ 0, 1,
From the definition of T , we have
T uk t 1
Gt, sfs, uk sds 0
m−2
t
1−
m−2
i1
αi ηi
1
αi
i1
G ηi , s fs, uk sds h.
2.25
0
Also, from the conditions C1 and C2 , we have
ft, u0 t ft, uk t ≤ 2ft, h for t ∈ 0, 1,
1
s1 − sfs, hds < ∞.
2.26
0
Thus by Lebesgue-dominated convergence theorem, we have
max |T uk t − T u0 t| ≤
1
t∈0,1
Gs, sfs, uk s − fs, u0 sds
0
1−
m−2
1
−→ 0
i1 αi
m−2
i1 αi ηi
1
m−2
1−
Gs, sfs, uk s − fs, u0 sds
0
i1 αi
m−2
i1 αi ηi
1
s1 − sfs, uk s − fs, u0 sds
0
as k −→ ∞.
2.27
Therefore, T : Dh → Dh is continuous.
Next we need to show that T Dh is a relatively compact subset of C0, 1.
1 From the definition of T and the conditions C1 and C2 , for each u ∈ Dh we have
0 < h ≤ T ut ≤ T ht for t ∈ 0, 1.
This implies that T Dh is uniformly bounded.
2.28
8
Boundary Value Problems
2 For each u ∈ Dh , since
T u t −
t
1
sfs, usds 0
1 − sfs, usds
t
m−2
1−
i1
1
m−2
1
αi
αi ηi
G ηi , s fs, usds
2.29
for t ∈ 0, 1,
0
i1
then
T u t ≤
t
sfs, hds 1
0
m−2
: Mt
i1
1
m−2
1
1−
1 − sfs, hds
t
αi ηi
αi
G ηi , s fs, hds
2.30
0
i1
for t ∈ 0, 1.
Obviously Mt ≥ 0 on 0, 1, and
1
1
Mtdt 2 s1 − sfs, hds 0
0
1
≤ 2 s1 − sfs, hds 0
2
m−2
1−
i1 αi
m−2
i1 αi ηi
m−2
1
1−
1−
1
m−2
i1
1
m−2
i1
αi ηi
αi ηi
1
αi
i1
G ηi , s fs, hds
0
m−2
1
αi s1 − sfs, hds
i1
2.31
0
s1 − sfs, hds < ∞.
0
Thus M ∈ L1 0, 1. From the absolute continuity of integral, we have that for each number
ε > 0, there is a positive number δ > 0 such that for all t1 , t2 ∈ 0, 1, if |t1 − t2 | < δ, then
t
| t21 Mtdt| < ε. It follows that for all t1 , t2 ∈ 0, 1 with |t1 − t2 | < δ, we have
t2
t2 t2
T u t dt ≤ Mtdt < ε.
|T ut2 − T ut1 | T u tdt ≤ t1
t1
t1
2.32
Therefore T Dh is equicontinuous on 0, 1. It follows from Ascoli-Arzela theorem that T Dh is a relatively compact subset of C0, 1. Consequently, by Schauder fixed point theorem 30,
T has a fixed point ut; h ∈ Dh . Obviously, ut; h > h > 0 on 0, 1. Hence from Lemma 2.3,
ut; h is a solution of BVP 2.1h .
Next, we will show the uniqueness of solution. Let us suppose that u1 t; h, u2 t; h are
two different solutions of BVP2.1h . Then there exists t0 ∈ 0, 1 such that u1 t0 ; h / u2 t0 ; h.
Without loss of generality, assume that u1 t0 ; h > u2 t0 ; h. Let wt : u1 t; h − u2 t; h, then
w0 0, wt0 > 0, and hence there exists t1 ∈ 0, t0 such that
wt1 0,
wt > 0
for t ∈ t1 , t0 .
2.33
Boundary Value Problems
9
Further we have wt > 0 on t1 , 1. In fact, assume to the contrary that the conclusion is false.
Then there exists t2 ∈ t0 , 1 such that wt2 ≤ 0. Thus there exists t3 ∈ t0 , t2 such that
wt3 0,
wt > 0
for t ∈ t0 , t3 .
2.34
Since wt1 0, wt > 0 on t1 , t0 , then
w t −ft, u1 t; h ft, u2 t; h ≥ 0 for t ∈ t1 , t3 .
2.35
It follows from wt1 wt3 0 that wt ≤ 0 on t1 , t3 . This is a contradiction to wt > 0
on t1 , t3 .
Now we prove that wt ≥ 0 on 0, t1 . In fact, assume to the contrary that the
conclusion is false. Then there exists t4 ∈ 0, t1 such that wt4 < 0. Since w0 wt1 0,
then there exist t5 , t6 with 0 ≤ t5 < t4 < t6 ≤ t1 such that
wt5 wt6 0,
wt < 0
for t ∈ t5 , t6 .
2.36
Thus,
w t −ft, u1 t; h ft, u2 t; h ≤ 0 for t ∈ t5 , t6 .
2.37
It follows from wt5 wt6 that wt ≥ 0 on t5 , t6 . This is a contradiction to wt < 0 on
t5 , t6 .
In summary, we have wt ≥ 0 on 0, t1 and wt > 0 on t1 , 1. Thus
wt 1
Gt, s fs, u1 s; h − fs, u2 s; h ds
0
1−
t
m−2
i1
m−2
αi ηi
1
αi
i1
G ηi , s fs, u1 s; h − fs, u2 s; h ds
2.38
0
≤ 0 for t ∈ 0, 1.
This is a contradiction to wt > 0 on t1 , 1. This completes the proof of the lemma.
Lemma 2.7. Assume that conditions C0 , C1 , and C2 are satisfied. Then, the unique solution
ut; h of BVP2.1h is nondecreasing in h.
Proof. Let 0 < h2 < h1 , and let ut; h1 , ut; h2 be the solutions of BVP2.1h1 and BVP2.1h2 ,
respectively. We will show
ut; h1 ≥ ut; h2 for t ∈ 0, 1.
2.39
10
Boundary Value Problems
Assume to the contrary that the above inequality is false. Then there exists t0 ∈ 0, 1 such that
ut0 ; h1 < ut0 ; h2 . Since u0; h1 h1 > h2 u0; h2 , we have that there exists t1 ∈ 0, t0 such that
ut1 ; h1 ut1 ; h2 ,
ut; h1 < ut; h2 for t ∈ t1 , t0 .
2.40
Next we prove ut; h1 < ut; h2 on t0 , 1. In fact, assume to the contrary that the
conclusion is false. Then there exists t2 ∈ t0 , 1 such that
ut2 ; h1 ut2 ; h2 ,
ut; h1 < ut; h2 for t ∈ t0 , t2 .
2.41
Hence
u t; h1 − u t; h2 −ft, ut; h1 ft, ut; h2 ≤ 0
for t ∈ t1 , t2 .
2.42
It follows from uti ; h1 uti ; h2 , i 1, 2 that ut; h1 ≥ ut; h2 on t1 , t2 . This is a
contradiction to ut; h1 < ut; h2 on t1 , t2 . Thus ut; h1 < ut; h2 on t1 , 1. This implies
that
u t; h1 − u t; h2 −ft, ut; h1 ft, ut; h2 ≤ 0 for t ∈ t1 , 1.
2.43
It follows from u t1 ; h1 − u t1 ; h2 ≤ 0 that u t; h1 − u t; h2 ≤ 0 on t1 , 1. Hence, from
ut; h1 < ut; h2 on t1 , 1, we have u 1; h1 − u 1; h2 < 0. Thus
u1; h1 − u1; h2 < u ηm−2 ; h1 − u ηm−2 ; h2 .
2.44
There are two cases to consider.
Case 1 see t1 ≥ ηm−2 . In this case, we have
u ηi ; h1 − u ηi ; h2 ≥ 0,
i 1, 2, . . . , m − 2.
2.45
Hence from the boundary conditions of BVP2.1h , we have
u1; h1 − u1; h2 m−2
αi u ηi ; h1 i1
−
m−2
αi u ηi ; h2 −
i1
≥
m−2
1−
m−2
αi h1
i1
1−
m−2
αi h2
i1
αi u ηi ; h1 − u ηi ; h2 ≥ 0.
i1
This is a contradiction to u1; h1 − u1; h2 < 0.
2.46
Boundary Value Problems
11
Case 2 see t1 < ηm−2 . In this case, we have
u1; h1 − u1; h2 < u ηm−2 ; h1 − u ηm−2 ; h2 < 0,
u ηm−2 ; h1 − u ηm−2 ; h2 ≤ u ηi ; h1 − u ηi ; h2 , i 1, 2, . . . , m − 3.
2.47
It follows from C0 that
u1; h1 − u1; h2 <
m−2
m−2
αi u ηm−2 ; h1 − u ηm−2 ; h2 ≤
αi u ηi ; h1 − u ηi ; h2 .
i1
2.48
i1
This is a contradiction to the boundary conditions of BVP2.1h .
In summary, we have ut; h1 ≥ ut; h2 on 0, 1. This completes the proof of the
lemma.
3. Main Results
We now state and prove our main results for singular second-order m-point boundary value
problem 1.1, 1.2.
Theorem 3.1. Assume that conditions C0 , C1 , and C2 are satisfied. Then, SBVP1.1, 1.2 has
at most one positive solution.
Proof. Suppose that u1 t and u2 t are any two positive solutions of SBVP1.1, 1.2. We now
prove that u1 t ≡ u2 t on 0, 1. To do this, let vt u1 t − u2 t on 0, 1. We will show that
vt ≡ 0 on 0, 1. There are three cases to consider.
Case 1 see v1 > 0. In this case, we have that vt ≥ 0 on 0, 1. In fact, assume to the
contrary that the conclusion is false. Then, there exists t0 ∈ 0, 1 such that vt0 < 0. Since
v0 0 and v1 > 0, then there exist t1 , t2 ∈ 0, 1 with t1 < t0 < t2 such that
vt < 0 on t1 , t2 ,
vt1 vt2 0.
3.1
Thus
v t u1 t − u2 t −ft, u1 t ft, u2 t ≤ 0
for t ∈ t1 , t2 .
3.2
Hence vt ≥ 0 on t1 , t2 , which is a contradiction to vt < 0 on t1 , t2 . Therefore vt ≥ 0 on
0, 1. Consequently
v t −ft, u1 t ft, u2 t ≥ 0 for t ∈ 0, 1.
3.3
Thus vt is convex on 0, 1. Since v1 > 0 and
v1 u1 1 − u2 1 m−2
m−2
m−2
αi u1 ηi −
αi u2 ηi αi v ηi ,
i1
i1
i1
3.4
12
Boundary Value Problems
then there exists i0 ∈ {1, 2, . . . , m − 2} such that
v ηi0 max v ηi : i 1, 2, . . . , m − 2 > 0,
3.5
and hence from C0 and 0 < ηi0 < 1, we have
v1 ≤
m−2
1 αi v ηi0 ≤ v ηi0 <
v ηi0 ,
η
i0
i1
3.6
which is a contradiction to that vt is convex on 0, 1.
Case 2 see v1 0. In this case, we have that vt ≡ 0 on 0, 1. In fact, assume to the
0. We may
contrary that the conclusion is false. Then, there exists t0 ∈ 0, 1 such that vt0 /
assume without loss of generality that vt0 > 0. Then from v0 v1 0, there exist
t1 , t2 ∈ 0, 1 with t1 < t0 < t2 such that
vt > 0
on t1 , t2 ,
vt1 vt2 0.
3.7
Thus
v t −ft, u1 t ft, u2 t ≥ 0 for t ∈ t1 , t2 .
3.8
Since vt1 vt2 0, then
vt ≤ 0 for t ∈ t1 , t2 ,
3.9
which is a contradiction to that vt > 0 on t1 , t2 .
Case 3 see v1 < 0. In this case, similar to the proof of Case 1 we can easily show that
vt ≤ 0 on 0, 1. Consequently
v t −ft, u1 t ft, u2 t ≤ 0 for t ∈ 0, 1.
3.10
Thus vt is concave on 0, 1. Since v1 m−2
i1 αi vηi < 0, then there exists i1 ∈ {1, 2, . . . , m−
2} such that vηi1 min{vηi : i 1, 2, . . . , m − 2} < 0, and hence from 0 < ηi1 < 1, we have
v1 ≥
m−2
1 αi v ηi1 ≥ v ηi1 >
v ηi1 ,
η
i1
i1
3.11
which is a contradiction to that vt is concave on 0, 1.
In summary, vt ≡ 0 on 0, 1, that is, u1 t ≡ u2 t on 0, 1. This completes the proof
of the theorem.
Boundary Value Problems
13
Theorem 3.2. Assume that conditions C0 , C1 , and C2 are satisfied. Then SBVP1.1, 1.2 has
exactly one positive solution.
Proof. The uniqueness of positive solution to SBVP1.1, 1.2 follows from Theorem 3.1
immediately. Thus we only need to show the existence.
Let {hj }∞
j1 be a decreasing sequence that converges to the number 0. Then from
Lemma 2.6, BVP2.1hj has a unique solution ut; hj : uj t. From Lemma 2.7 and 2.11h ,
we have that for each j < k,
0 ≤ uj t − uk t ≤ hj − hk
for t ∈ 0, 1.
3.12
Thus there exists u ∈ C0, 1 such that
lim uj t ut ≥ 0,
uniformly on 0, 1.
j →∞
3.13
It is easy to see that ut satisfies boundary conditions 1.2.
Now we prove that
ut > 0
for t ∈ 0, 1.
3.14
At first, we prove that
u ηi0 max u ηi : i 1, 2, . . . , m − 2 > 0,
3.15
where i0 ∈ {1, 2, . . . , m − 2}. In fact, assume to the contrary that the conclusion is false. Then
u1 m−2
αi u ηi 0.
3.16
i1
From the fact that each function in the sequence {uj }∞
j1 is concave, we have that ut is
concave. It follows from u0 uηi0 u1 0 that ut ≡ 0 on 0, 1. Thus when j is
large enough, uj t is small enough such that uj t ≤ h1 on 0, 1. Hence from condition C1 ,
we have
uj ηi0 1
G ηi0 , s f s, uj s ds
0
1−
1
>
0
ηi0
m−2
i1
m−2
αi ηi
1
αi
i1
G ηi , s f s, uj s ds hj
0
G ηi0 , s fs, h1 ds > 0.
3.17
14
Boundary Value Problems
Let j → ∞, we have
1
u ηi0 ≥
G ηi0 , s fs, h1 ds > 0.
3.18
0
This is a contradiction to uηi0 0. Thus uηi0 > 0, and hence u1 > 0. Since ut is concave,
then ut > 0 on 0, 1. Since
uj t 1
Gt, sf s, uj s ds 0
1−
m−2
i1
1
m−2
t
αi ηi
αi
G ηi , s f s, uj s ds hj ,
3.19
0
i1
then passing to the limit, by Monotone convergence theorem 31, we have
ut 1
Gt, sfs, usds 0
1−
m−2
i1
1
m−2
t
αi ηi
αi
G ηi , s fs, usds.
3.20
0
i1
Therefore by Lemma 2.4, ut is a positive solution of SBVP1.1, 1.2. This completes the
proof of the theorem.
Finally, we give an example to which our results can be applicable.
Example 3.3. Consider the singular nonlinear second-order m-point boundary value problem:
u 1
tβ1 1
− tβ2 u2−β1
u0 0,
0,
u1 t ∈ 0, 1,
3.21
m−2
αi u ηi ,
i1
where m ≥ 3, 0 < η1 < η2 < · · · < ηm−2 < 1, αi > 0 i 1, 2, . . . , m − 2,
β1 , β2 ∈ 0, 2.
Let
ft, u 1
tβ1 1
− tβ2 u2−β1
for t, u ∈ 0, 1 × 0, ∞.
m−2
i1
αi ≤ 1, and
3.22
Obviously, the function ft, u is singular at t 0, 1 and u 0. It is easy to verify that ft, u
satisfies conditions C1 and C2 . So from Theorem 3.2, SBVP3.21 has exactly one positive
solution. However, we note that Theorem 2 in 7 cannot guarantee that SBVP3.21 has a
unique positive solution, since
1
0
t1 − tft, λt1 − tdt ∞ for λ > 0.
3.23
Boundary Value Problems
15
Acknowledgment
The authors thank the referee for valuable suggestions which led to improvement of the
original manuscript.
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