Hindawi Publishing Corporation
Boundary Value Problems
Volume 2011, Article ID 827510, 15 pages
doi:10.1155/2011/827510
Research Article
Positive Solutions for Integral Boundary Value
Problem with φ-Laplacian Operator
Yonghong Ding
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Yonghong Ding, dyh198510@126.com
Received 20 September 2010; Revised 31 December 2010; Accepted 19 January 2011
Academic Editor: Gary Lieberman
Copyright q 2011 Yonghong Ding. 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.
We consider the existence, multiplicity of positive solutions for the integral boundary value
1
problem with φ-Laplacian φu t ft, ut, u t 0, t ∈ 0, 1, u0 0 urgrdr,
1
u1 0 urhrdr, where φ is an odd, increasing homeomorphism from R onto R. We show
that it has at least one, two, or three positive solutions under some assumptions by applying fixed
point theorems. The interesting point is that the nonlinear term f is involved with the first-order
derivative explicitly.
1. Introduction
We are interested in the existence of positive solutions for the integral boundary value
problem
φ u t f t, ut, u t 0, t ∈ 0, 1,
1
1
urgrdr,
u1 urhrdr,
u0 0
1.1
0
where φ, f, g, and h satisfy the following conditions.
H1 φ is an odd, increasing homeomorphism from R onto R, and there exist two
increasing homeomorphisms ψ1 and ψ2 of 0, ∞ onto 0, ∞ such that
ψ1 uφv ≤ φuv ≤ ψ2 uφv
∀u, v > 0.
Moreover, φ, φ−1 ∈ C1 R, where φ−1 denotes the inverse of φ.
1.2
2
Boundary Value Problems
H2 f : 0, 1 × 0, ∞ × −∞, ∞ → 0, ∞ is continuous. g, h ∈ L1 0, 1 are
1
1
nonnegative, and 0 < 0 gtdt < 1, 0 < 0 htdt < 1.
The assumption H1 on the function φ was first introduced by Wang 1, 2, it covers
two important cases: φu u and φu |u|p−2 u, p > 1. The existence of positive solutions
for two above cases received wide attention see 3–10. For example, Ji and Ge 4 studied
the multiplicity of positive solutions for the multipoint boundary value problem
φp u t qtf t, ut, u t 0,
u0 m
αi uξi ,
u1 i1
t ∈ 0, 1,
m
βi uξi ,
1.3
i1
where φp s |s|p−2 s, p > 1. They provided sufficient conditions for the existence of at
least three positive solutions by using Avery-Peterson fixed point theorem. In 5, Feng et
al. researched the boundary value problem
φp u t qtft, ut 0,
u0 m−2
ai uξi ,
u1 i1
t ∈ 0, 1,
m−2
bi uξi ,
1.4
i1
where the nonlinear term f does not depend on the first-order derivative and φp s |s|p−2 s,
p > 1. They obtained at least one or two positive solutions under some assumptions imposed
on the nonlinearity of f by applying Krasnoselskii fixed point theorem.
As for integral boundary value problem, when φu u is linear, the existence of
positive solutions has been obtained see 8–10. In 8, the author investigated the positive
solutions for the integral boundary value problem
u fu 0,
u0 1
uτdατ,
0
u1 1
1.5
uτdβτ.
0
The main tools are the priori estimate method and the Leray-Schauder fixed point theorem.
However, there are few papers dealing with the existence of positive solutions when φ
satisfies H1 and f depends on both u and u . This paper fills this gap in the literature. The
aim of this paper is to establish some simple criteria for the existence of positive solutions of
BVP1.1. To get rid of the difficulty of f depending on u , we will define a special norm in
Banach space in Section 2.
This paper is organized as follows. In Section 2, we present some lemmas that are
used to prove our main results. In Section 3, the existence of one or two positive solutions for
BVP1.1 is established by applying the Krasnoselskii fixed point theorem. In Section 4, we
give the existence of three positive solutions for BVP1.1 by using a new fixed point theorem
introduced by Avery and Peterson. In Section 5, we give some examples to illustrate our main
results.
Boundary Value Problems
3
2. Preliminaries
The basic space used in this paper is a real Banach space C1 0, 1 with norm · 1 defined by
u1 max{uc , u c }, where uc max0≤t≤1 |ut|. Let
K
u ∈ C 0, 1 | ut ≥ 0, u1 1
1
uthtdt, u is concave on 0, 1 .
2.1
0
It is obvious that K is a cone in C1 0, 1.
Lemma 2.1 see 7. Let u ∈ K, η ∈ 0, 1/2, then ut ≥ η max0≤t≤1 |ut|, t ∈ η, 1 − η.
Lemma 2.2. Let u ∈ K, then there exists a constant M > 0 such that max0≤t≤1 |ut| ≤
Mmax0≤t≤1 |u t|.
Proof. The mean value theorem guarantees that there exists τ ∈ 0, 1, such that
u1 uτ
1
htdt.
2.2
0
Moreover, the mean value theorem of differential guarantees that there exists σ ∈ τ, 1, such
that
1
htdt − 1 uτ u1 − uτ 1 − τu σ.
2.3
0
So we have
⎞
⎛
1
t
2 − 0 htdt
1
−
τ
⎝
⎠
1 max u t ≤
maxu t.
|ut| ≤ |uτ| u sds ≤
1
1
τ
0≤t≤1
0≤t≤1
1 − 0 htdt
1 − 0 htdt
2.4
Denote M 2 −
1
0
htdt/1 −
1
0
htdt; then the proof is complete.
Lemma 2.3. Assume that (H1), (H2) hold. If u is a solution of BVP1.1, there exists a unique δ ∈
0, 1, such that u δ 0 and ut ≥ 0, t ∈ 0, 1.
Proof. From the fact that φu −ft, ut, u t < 0, we know that φu t is strictly
decreasing. It follows that u t is also strictly decreasing. Thus, ut is strictly concave on 0,
1. Without loss of generality, we assume that u0 min{u0, u1}. By the concavity of
1
1
u, we know that ut ≥ u0, t ∈ 0, 1. So we get u0 0 utgtdt ≥ u0 0 gtdt. By
1
0 < 0 gtdt < 1, it is obvious that u0 ≥ 0. Hence, ut ≥ 0, t ∈ 0, 1.
On the other hand, from the concavity of u, we know that there exists a unique δ where
the maximum is attained. By the boundary conditions and ut ≥ 0, we know that δ / 0 or 1,
that is, δ ∈ 0, 1 such that uδ max0≤t≤1 ut and then u δ 0.
4
Boundary Value Problems
Lemma 2.4. Assume that (H1), (H2) hold. Suppose u is a solution of BVP1.1; then
ut 1−
1
0
t
1
1
gr
grdr
δ
φ
0
−1
δ
0
f τ, uτ, u τ dτ ds dr
s
φ−1
0
r
2.5
f τ, uτ, u τ dτ ds
s
or
ut 1
1
1
φ−1
s
f τ, uτ, u τ dτ ds dr
1
r
δ
1 − 0 hrdr 0
1
s
−1
φ
f τ, uτ, u τ dτ ds.
hr
t
2.6
δ
Proof. First, by integrating 1.1 on 0, t, we have
φ u t φ u 0 −
t
f s, us, u s ds,
2.7
0
then
u t φ
−1
φ u 0 −
t
2.8
f s, us, u s ds .
0
Thus
ut u0 s
φ−1 φ u 0 − f τ, uτ, u τ dτ ds
t
0
2.9
0
or
ut u1 −
1
φ
−1
φ u 0 −
s
t
f τ, uτ, u τ dτ ds.
2.10
0
According to the boundary condition, we have
u0 1−
u1 −
1
1−
0
1
grdr
1
0
1
hrdr
1
r
gr
φ
0
0
1
1
hr
0
r
−1
φ u 0 −
s
f τ, uτ, u τ dτ ds dr,
0
φ−1 φ u 0 −
s
0
f τ, uτ, u τ dτ ds dr.
2.11
Boundary Value Problems
5
By a similar argument in 5, φu 0 δ
0
fτ, uτ, u τdτ; then the proof is completed.
Now we define an operator T by
T ut ⎧
δ
1
r
⎪
1
⎪
−1
⎪
gr φ
fτ, uτ, u τdτ ds dr
⎪
1
⎪
⎪
⎪
0
s
⎪ 1 − 0 grdr 0
⎪
⎪
⎪
⎪
δ
t
⎪
⎪
⎪
⎪
−1
⎪
fτ, uτ, u τdτ ds,
⎪
⎨ φ
0 ≤ t ≤ δ,
⎪
1
1
s
⎪
⎪
1
⎪
−1
⎪
⎪
hr φ
fτ, uτ, u τdτ ds dr
1
⎪
⎪
⎪
r
δ
1 − 0 hrdr 0
⎪
⎪
⎪
⎪
1
s
⎪
⎪
⎪
⎪
−1
⎪
φ
fτ,
uτ,
u
ds,
τdτ
⎩
δ ≤ t ≤ 1.
0
t
s
2.12
δ
Lemma 2.5. T : K → K is completely continuous.
Proof. Let u ∈ K; then from the definition of T , we have
⎧ δ
⎪
⎪
−1
⎪
fτ, uτ, u τdτ ≥ 0,
φ
⎪
⎪
⎨
t
T u t t
⎪
⎪
⎪
−1
⎪
⎪
fτ, uτ, u τdτ ≤ 0,
⎩−φ
0 ≤ t ≤ δ,
2.13
δ ≤ t ≤ 1.
δ
So T u t is monotone decreasing continuous and T u δ 0. Hence, T ut is
1
nonnegative and concave on 0, 1. By computation, we can get T u1 0 T uthtdt. This
shows that T K ⊂ K. The continuity of T is obvious since φ−1 , f is continuous. Next, we
prove that T is compact on C1 0, 1.
1
Let D be a bounded subset of K and m > 0 is a constant such that 0 fτ,
uτ, u τdτ < m for u ∈ D. From the definition of T , for any u ∈ D, we get
|T ut| <
⎧
φ−1 m
⎪
⎪
⎪
,
⎪
⎪
⎨ 1 − 1 grdr
0 ≤ t ≤ δ,
0
⎪
⎪
⎪
⎪
⎪
⎩
φ−1 m
, δ ≤ t ≤ 1,
1
1 − 0 hrdr
T u t < φ−1 m, 0 ≤ t ≤ 1.
2.14
Hence, T D is uniformly bounded and equicontinuous. So we have that T D is compact on
C0, 1. From 2.13, we know for ∀ε > 0, ∃κ > 0, such that when |t1 − t2 | < κ, we have
6
Boundary Value Problems
|φT u t1 − φT u t2 | < ε. So φT D is compact on C0, 1; it follows that T D is compact
on C0, 1. Therefore, T D is compact on C1 0, 1.
Thus, T : K → K is completely continuous.
It is easy to prove that each fixed point of T is a solution for BVP1.1.
Lemma 2.6 see 1. Assume that (H1) holds. Then for u, v ∈ 0, ∞,
ψ2−1 uv ≤ φ−1 uφv ≤ ψ1−1 uv.
2.15
To obtain positive solution for BVP1.1, the following definitions and fixed point
theorems in a cone are very useful.
Definition 2.7. The map α is said to be a nonnegative continuous concave functional on a cone
of a real Banach space E provided that α : K → 0, ∞ is continuous and
α tx 1 − ty ≥ tαx 1 − tα y
2.16
for all x, y ∈ K and 0 ≤ t ≤ 1. Similarly, we say the map γ is a nonnegative continuous convex
functional on a cone of a real Banach space E provided that γ : K → 0, ∞ is continuous and
γ tx 1 − ty ≤ tγx 1 − tγ y
2.17
for all x, y ∈ K and 0 ≤ t ≤ 1.
Let γ and θ be a nonnegative continuous convex functionals on K, α a nonnegative
continuous concave functional on K, and ψ a nonnegative continuous functional on K. Then
for positive real number a, b, c, and d, we define the following convex sets:
P γ, d u ∈ K | γu < d ,
P γ, α, b, d u ∈ K | αu ≥ b, γu ≤ d ,
P γ, θ, α, b, c, d u ∈ K | αu ≥ b, θu ≤ c, γu ≤ d ,
R γ, ψ, a, d u ∈ K | ψu ≥ a, γu ≤ d .
2.18
Theorem 2.8 see 11. Let E be a real Banach space and K ⊂ E a cone. Assume that Ω1 and Ω2
are two bounded open sets in E with 0 ∈ Ω1 , Ω1 ⊂ Ω2 . Let T : K ∩ Ω2 \ Ω1 → K be completely
continuous. Suppose that one of following two conditions is satisfied:
1 T u ≤ u, u ∈ K ∩ ∂Ω1 , and T u ≥ u, u ∈ K ∩ ∂Ω2 ;
2 T u ≥ u, u ∈ K ∩ ∂Ω1 , and T u ≤ u, u ∈ K ∩ ∂Ω2 .
Then T has at least one fixed point in Ω2 \ Ω1 .
Theorem 2.9 see 12. Let K be a cone in a real Banach space E. Let γ and θ be a nonnegative
continuous convex functionals on K, α a nonnegative continuous concave functional on K, and ψ
Boundary Value Problems
7
a nonnegative continuous functional on K satisfying ψλu ≤ λψu for 0 ≤ λ ≤ 1, such that for
positive number M and d,
αu ≤ ψu,
u ≤ Mγu
2.19
for all u ∈ P γ, d. Suppose T : P γ, d → P γ, d is completely continuous and there exist positive
numbers a, b, and c with a < b such that
S1 {u ∈ P γ, θ, α, b, c, d | αu > b} /
∅ and αT u > b for u ∈ P γ, θ, α, b, c, d;
S2 αT u > b for u ∈ P γ, α, b, d with θT u > c;
S3 0 ∈
/ Rγ, ψ, a, d and ψT u < a for u ∈ Rγ, ψ, a, d with ψu a.
Then T has at least three fixed points u1 , u2 , u3 ∈ P γ, d, such that
γui ≤ d for i 1, 2, 3,
αu1 > b,
ψu2 > a with αu2 < b,
ψu3 < a.
3. The Existence of One or Two Positive Solutions
For convenience, we denote
⎫
⎧
⎨ 1 ψ −1 1 − sds ⎬
0 1
,1 ,
L max
⎩ 1 − 1 gsds ⎭
0
N min
ft, ut, vt
,
uc vc → μ t∈0,1 φuc vc f μ lim sup max
1/2
0
fμ ψ2−1
1
1
1
−1
− s ds,
ds ,
ψ2 s −
2
2
1/2
ft, ut, vt
,
uc vc → μ t∈0,1 φuc vc lim inf
min
3.1
where μ denotes 0 or ∞.
Theorem 3.1. Assume that (H1) and (H2) hold. In addition, suppose that one of following conditions
is satisfied.
i There exist two constants r, R with 0 < r < N/LR such that
a ft, u, v ≥ φr/N for t, u, v ∈ 0, 1 × 0, r × −r, r and
b ft, u, v ≤ φR/L for t, u, v ∈ 0, 1 × 0, R × −R, R;
ii f ∞ < ψ1 1/2L, f0 > ψ2 1/N;
iii f 0 < ψ1 1/2L, f∞ > ψ2 1/N.
Then BVP1.1 has at least one positive solution.
8
Boundary Value Problems
Proof. i Let Ω1 {u ∈ K | u1 < r}, Ω2 {u ∈ K | u1 < R}.
For u ∈ ∂Ω1 , we obtain u ∈ 0, r and u ∈ −r, r, which implies ft, u, u ≥ φr/N.
Hence, by 2.12 and Lemma 2.6,
T uc max|T ut|
0≤t≤1
1−
1
0
δ
1
1
φ
1
gr
grdr
δ
φ
0
−1
δ
0
f τ, uτ, u τ dτ ds dr
s
f τ, uτ, u τ dτ ds
−1
0
r
s
1
1
1
hr
φ
−1
s
f τ, uτ, u τ dτ ds dr
r
δ
1 − 0 hrdr 0
1
s
−1
φ
f τ, uτ, u τ dτ ds
δ
≥ min
δ
⎧
⎨
1
1
⎩ 1 − 1 grdr
0
1/2
φ
−1
gr
0
1
0
f τ, uτ, u τ dτ ds dr
s
f τ, uτ, u τ dτ ds,
1
1
φ
−1
1/2
φ
−1
1
hr
hrdr
1
0
δ
s
0
φ−1
s
r
f τ, uτ, u τ dτ ds dr
δ
f τ, uτ, u τ dτ ds
s
1/2
≥ min
φ
−1
0
1/2
0
1−
r
1/2
1/2
f τ, uτ, u τ dτ ds,
s
1
φ
−1
1/2
s
f τ, uτ, u τ dτ ds
1/2
1
1
1
r
r
−1
−1
−s
ds,
s−
ds
φ
φ
φ
φ
≥ min
N
2
N
2
0
1/2
1/2
1
1
r
−1 1
−1
min
− s ds,
ψ2
ψ2 s −
≥
ds
N
2
2
0
1/2
1/2
r u1 .
3.2
This implies that
T u1 ≥ u1
for u ∈ ∂Ω1 .
3.3
Boundary Value Problems
9
Next, for u ∈ ∂Ω2 , we have ft, u, v ≤ φR/L. Thus, by 2.12 and Lemma 2.6,
T uc max|T ut|
0≤t≤1
≤
1−
1
0
1
1
1
grdr
1
φ
−1
0
1
gr
φ
0
−1
1
0
f τ, uτ, u τ dτ ds dr
s
f τ, uτ, u τ dτ ds
s
1
1
φ−1
1
1 − 0 grdr 0
1 −1
R 0 ψ1 1 − sds
≤
L 1 − 1 grdr
≤
3.4
R
ds
1 − sφ
L
0
≤ R u1 .
From 2.13, we have
1
δ
−1
T u max φ−1
f τ, uτ, u τ dτ , φ
f τ, uτ, u τ dτ
c
0
≤ φ−1
1
δ
f τ, uτ, u τ dτ
3.5
0
R
−1
φ
≤φ
L
≤ R u1 .
This implies that
T u1 ≤ u1
for u ∈ ∂Ω2 .
3.6
Therefore, by Theorem 2.8, it follows that T has a fixed point in Ω2 \ Ω1 . That is BVP1.1 has
at least one positive solution such that 0 < r ≤ u1 ≤ R.
ii Considering f ∞ < ψ1 1/2L, there exists ρ0 > 0 such that
ft, u, v ≤ ψ1
1
φuc vc 2L
for t ∈ 0, 1, uc vc ≥ 2ρ0 .
3.7
Choosing M > ρ0 such that
max ft, u, v | uc vc ≤ 2ρ0 ≤ ψ1
1
φ M ,
2L
3.8
10
Boundary Value Problems
then for all ρ > M, let Ω3 {u ∈ K | u1 < ρ}. For every u ∈ ∂Ω3 , we have uc u c ≤ 2ρ.
In the following, we consider two cases.
Case 1 uc u c ≤ 2ρ0 . In this case,
ρ
M
1
φ M ≤φ
≤φ
.
2L
2L
L
f t, u, u ≤ ψ1
3.9
Case 2 2ρ0 ≤ uc u c ≤ 2ρ. In this case,
f t, u, u ≤ ψ1
ρ
1
1
φ uc u c ≤ ψ1
φ 2ρ ≤ φ
.
2L
2L
L
3.10
Then it is similar to the proof of 3.6; we have T u1 ≤ u1 for u ∈ ∂Ω3 .
Next, turning to f0 > ψ2 1/N, there exists 0 < ξ < ρ such that
ft, u, v ≥ ψ2
1
φuc vc N
for t ∈ 0, 1, uc vc ≤ 2ξ.
3.11
Let Ω4 {u ∈ K | u1 < ξ}. For every u ∈ ∂Ω4 , we have uc u c ≤ 2ξ. So
f t, u, u ≥ ψ2
1
1
ξ
φ uc u c ≥ ψ2
φu1 ≥ φ
.
N
N
N
3.12
Then like in the proof of 3.3, we have T u1 ≥ u1 for u ∈ ∂Ω4 . Hence, BVP1.1 has at least
one positive solution such that 0 < ξ ≤ u1 ≤ ρ.
iii The proof is similar to the i and ii; here we omit it.
In the following, we present a result for the existence of at least two positive solutions
of BVP1.1.
Theorem 3.2. Assume that (H1) and (H2) hold. In addition, suppose that one of following conditions
is satisfied.
I f 0 < ψ1 1/2L, f ∞ < ψ1 1/2L, and there exists m1 > 0 such that
ft, u, v ≥ φ
m 1
N
for t ∈ 0, 1, m1 ≤ uc vc ≤ 2m1 ;
3.13
II f0 > ψ2 1/N, f∞ > ψ2 1/N, and there exists m2 > 0 such that
ft, u, v ≤ φ
m 2
L
for t ∈ 0, 1, uc vc ≤ 2m2 .
Then BVP1.1 has at least two positive solutions.
3.14
Boundary Value Problems
11
4. The Existence of Three Positive Solutions
In this section, we impose growth conditions on f which allow us to apply Theorem 2.9 of
BVP1.1.
Let the nonnegative continuous concave functional α, the nonnegative continuous
convex functionals γ, θ, and nonnegative continuous functional ψ be defined on cone K by
γu maxu t,
ψu θu max|ut|,
0≤t≤1
αu min |ut|.
4.1
η≤t≤1−η
0≤t≤1
By Lemmas 2.1 and 2.2, the functionals defined above satisfy
ηθu ≤ αu ≤ ψu θu,
u1 max γu, θu ≤ Mγu,
4.2
for all u ∈ K. Therefore, the condition 2.19 of Theorem 2.9 is satisfied.
Theorem 4.1. Assume that (H1) and (H2) hold. Let 0 < a < b ≤ dη/11−
tdt and suppose that f satisfies the following conditions:
P 1 ft, u, v ≤ φd for t, u, v ∈ 0, 1 × 0, Md × −d, d;
P 2 ft, u, v > φb/ηK for t, u, v ∈ η, 1 − η × b, b/η1 1 −
tdt × −d, d.
1
0
1
0
htdt/
htdt/
1
0
1
0
ht1−
ht1 −
P 3 ft, u, v < φa/L for t, u, v ∈ 0, 1 × 0, a × −d, d;
Then BVP1.1 has at least three positive solutions u1 , u2 , and u3 satisfying
maxui t ≤ d
0≤t≤1
for i 1, 2, 3,
min |u1 t| > b,
η≤t≤1−η
max0≤t≤1 |u2 t| > a with minη≤t≤1−η |u2 t| < b,
4.3
max0≤t≤1 |u3 t| < a,
1/2
1−η
where L defined as 3.1, K min{ η ψ2−1 1/2 − sds, 1/2 ψ2−1 s − 1/2ds}.
Proof. We will show that all the conditions of Theorem 2.9 are satisfied.
If u ∈ P γ, d, then γu max0≤t≤1 |u t| ≤ d. With Lemma 2.2 implying
max0≤t≤1 |ut| ≤ Md, so by P 1, we have ft, ut, u t ≤ φd when 0 ≤ t ≤ 1. Thus
γT u maxT u t
0≤t≤1
max φ
−1
δ
f τ, uτ, u τ dτ , φ
0
≤ φ−1
1
f τ, uτ, u τ dτ
0
≤φ
−1
φd d.
This proves that T : P γ, d → P γ, d.
−1
1
δ
f τ, uτ, u τ dτ
4.4
12
Boundary Value Problems
To check condition S1 of Theorem 2.9, we choose
1
b 1 − 0 htdt
b
u0 t 1 − t,
η η 1 ht1 − tdt
0
0 ≤ t ≤ 1.
4.5
Let
⎛
c
1−
1
⎞
htdt
b⎝
0
⎠.
1 1
η
ht1
−
tdt
0
4.6
Then u0 t ∈ P γ, θ, α, b, c, d and αu0 > b, so {u ∈ P γ, θ, α, b, c, d | αu > b} / ∅. Hence,
for u ∈ P γ, θ, α, b, c, d, there is b ≤ ut ≤ c, |u t| ≤ d when η ≤ t ≤ 1 − η. From assumption
P 2, we have
b
f t, ut, u t > φ
ηK
for t ∈ η, 1 − η .
4.7
It is similar to the proof of assumption i of Theorem 3.1; we can easily get that
αT u min |T ut| ≥ η max|T ut| > b
η≤t≤1−η
0≤t≤1
for u ∈ P γ, θ, α, b, c, d .
4.8
This shows that condition S1 of Theorem 2.9 is satisfied.
Secondly, for u ∈ P γ, α, b, d with θT u > c, we have
αT u ≥ ηθT u ≥ ηc > b.
4.9
Thus condition S2 of Theorem 2.9 holds.
Finally, as ψ0 0 < a, there holds 0 ∈
/ Rγ, ψ, a, d. Suppose that u ∈ Rγ, ψ, a, d with
ψu a; then by the assumption P 3,
a
f t, ut, u t < φ
L
for t ∈ 0, 1.
4.10
So like in the proof of assumption i of Theorem 3.1, we can get
ψT u max|T ut| < a.
0≤t≤1
Hence condition S3 of Theorem 2.9 is also satisfied.
4.11
Boundary Value Problems
13
Thus BVP1.1 has at least three positive solutions u1 , u2 , and u3 satisfying
maxui t ≤ d
0≤t≤1
for i 1, 2, 3,
min |u1 t| > b,
η≤t≤1−η
max|u2 t| > a with min |u2 t| < b,
η≤t≤1−η
0≤t≤1
max|u3 t| < a.
4.12
0≤t≤1
5. Examples
In this section, we give three examples as applications.
Example 5.1. Let φu |u|u, gt ht 1/2. Now we consider the BVP
f t, ut, u t 0, t ∈ 0, 1,
φ u
1 1
1 1
utdt,
u1 utdt,
u0 2 0
2 0
5.1
where ft, u, v 1 t18 u4 cos v for t, u, v ∈ 0, 1 × 0, ∞ × −∞, ∞.
Let ψ1 u ψ2 u u2 , u > 0. Choosing r 1, R 100. By calculations we obtain
L
4
,
3
N
2 1 3/2
,
3 2
φ
r 18,
N
φ
R
752 .
L
5.2
For t, u, v ∈ 0, 1 × 0, 1 × −1, 1,
ft, u, v 1 t18 u4 cos v
≥ 18 × 4 > 18,
5.3
for t, u, v ∈ 0, 1 × 0, 100 × −100, 100,
ft, u, v 1 t18 u4 cos v
≤ 2 × 118 × 5 < 752 .
5.4
Hence, by Theorem 3.1, BVP5.1 has at least one positive solution.
Example 5.2. Let φu u, gt ht 1/2. Consider the BVP
f t, ut, u t 0, t ∈ 0, 1,
φ u
1 1
1 1
utdt,
u1 utdt,
u0 2 0
2 0
5.5
where ft, u, v 1 t1/10 u1/100 v2 1 uc vc 2 for t, u, v ∈ 0, 1 × 0, ∞ ×
−∞, ∞.
14
Boundary Value Problems
Let ψ1 u ψ2 u u, u > 0. Then L 1, N 1/8. It easy to see
f0 f∞ ∞ > ψ2
1
N
8.
5.6
Choosing m2 1/10, for t ∈ 0, 1, uc vc ≤ 2m2 .
!
1
1
ft, u, v 1 t
u
v2 1 uc vc 2
10
100
1
1
1
1
1
≤2
1
.
10 5
100 25
25
m 39
1
2
<
φ
.
1250 10
L
5.7
Hence, by Theorem 3.2, BVP5.5 has at least two positive solutions.
Example 5.3. Let φu |u|u, gt ht 1/2; consider the boundary value problem
u u f t, ut, u t 0, t ∈ 0, 1,
1 1
utdt,
u0 u1 2 0
5.8
where
⎧
sin t
1
v 3
⎪
6
⎪
2500u
,
⎪
⎨ 104
104 105
ft, u, v ⎪
⎪
v 3
⎪
⎩ sin t 2500 · 126 1
,
104
104 105
u ≤ 12,
5.9
u > 12.
Choosing a 1/10, b 1, η 1/4, d 105 , then by calculations we obtain that
4
L ,
3
2 1 3/2
K
,
3 4
b
φ
ηK
2304,
φ
a
L
9
.
1600
5.10
It is easy to check that
ft, u, v < φd 1010
ft, u, v > 2304
ft, u, v <
9
1600
for 0 ≤ t ≤ 1, 0 ≤ u ≤ 3 · 105 , −105 ≤ v ≤ 105 ,
for
3
1
≤ t ≤ , 1 ≤ u ≤ 12, −105 ≤ v ≤ 105 ,
4
4
for 0 ≤ t ≤ 1, 0 ≤ u ≤
1
, −105 ≤ v ≤ 105 .
10
5.11
Boundary Value Problems
15
Thus, according to Theorem 4.1, BVP5.8 has at least three positive solutions u1 , u2 , and u3
satisfying
maxui t ≤ 105
0≤t≤1
1
max|u2 t| >
0≤t≤1
10
with
for i 1, 2, 3,
min |u2 t| < 1,
1/4≤t≤3/4
min |u1 t| > 1,
1/4≤t≤3/4
1
.
max|u3 t| <
0≤t≤1
10
5.12
Acknowledgments
The research was supported by NNSF of China 10871160, the NSF of Gansu Province
0710RJZA103, and Project of NWNU-KJCXGC-3-47.
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