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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, [email protected]
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
problem with φ-Laplacian φu
t
ft, ut,u
t 0, t ∈ 0, 1, u0
1
0
urgrdr,
u1
1
0
0, 1
,
u
0
1
0
u
r
g
r
dr, u
1
1
0
u
u
φ
v
∀u, v > 0. 1.2
Moreover, φ, φ
−1
∈ C
1
R, where φ
−1
denotes the inverse of φ.
2 Boundary Value Problems
H2 f : 0, 1 × 0, ∞ × −∞, ∞ → 0, ∞ is continuous. g,h ∈ L
1
0, 1 are
nonnegative, and 0 <
1
0
gtdt<1, 0 <
1
0
htdt<1.
The assumption H1 on the function φ was first introduced by Wang 1, 2, it covers
two important cases: φuu and φu|u|
0,t∈
0, 1
,
u
0
m
i1
α
i
u
ξ
i
,u
1
m
i1
β
f
t, u
t
0,t∈
0, 1
,
u
0
m−2
i1
a
i
u
ξ
i
,u
1
u
0
1
0
u
τ
dα
τ
,u
1
1
0
u
τ
dβ
c
, u
c
}, where u
c
max
0≤t≤1
|ut|.Let
K
u ∈ C
1
0, 1
| u
t
≥ 0,u
1
1
0
u
1
u
τ
1
0
h
t
dt.
2.2
Moreover, the mean value theorem of differential guarantees that there exists σ ∈ τ, 1, such
that
1
0
h
t
dt − 1
u
τ
τ
|
t
τ
u
s
ds
≤
⎛
⎝
1 − τ
1 −
1
1 −
1
0
h
t
dt
max
0≤t≤1
u
t
.
2.4
Denote M 2 −
1
0
htdt/1 −
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
/
0or1,
that is, δ ∈ 0, 1 such that uδmax
0≤t≤1
ut and then u
δ0.
4 Boundary Value Problems
Lemma 2.4. Assume that (H1), ( H2) hold. Suppose u is a solution of BVP1.1;then
u
t
1
1 −
1
0
g
r
dr
1
t
0
φ
−1
δ
s
f
τ,u
τ
,u
τ
dτ
ds
2.5
or
u
t
τ
,u
τ
dτ
ds dr
1
t
φ
−1
s
δ
f
τ,u
τ
,u
s, u
s
,u
s
ds,
2.7
then
u
t
φ
−1
φ
u
0
t
0
φ
−1
φ
u
0
−
s
0
f
τ,u
τ
,u
τ
dτ
0
f
τ,u
τ
,u
τ
dτ
ds.
2.10
According to the boundary condition, we have
u
0
1
1 −
1
0
g
τ,u
τ
,u
τ
dτ
ds dr,
u
1
−
1
1 −
1
0
h
r
dr
1
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
Tu
t
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1
1 −
1
0
g
r
dr
1
0
g
r
r
0
φ
−1
δ
s
,u
τ
dτ
ds, 0 ≤ t ≤ δ,
1
1 −
1
0
h
r
dr
1
0
h
r
1
r
φ
−1
τ,u
τ
,u
τ
dτ
ds, δ ≤ t ≤ 1.
2.12
Lemma 2.5. T : K → K is completely continuous.
Proof. Let u ∈ K; then from the definition of T, we have
Tu
t
⎧
⎪
⎪
⎪
⎪
⎪
⎨
t
δ
f
τ,u
τ
,u
τ
dτ
≤ 0,δ≤ t ≤ 1.
2.13
So Tu
t is monotone decreasing continuous and Tu
δ0. Hence, Tut is
nonnegative and concave on 0, 1. By computation, we can get Tu1
1
0
Tuthtdt.This
shows that TK ⊂ K. The continuity of T is obvious since φ
−1
,f is continuous. Next, we
⎩
φ
−1
m
1 −
1
0
g
r
dr
, 0 ≤ t ≤ δ,
φ
−1
m
1 −
1
0
h
r
dr
− φTu
t
2
| <ε.SoφTD
is compact on C0, 1; it follows that TD
is compact
on C0, 1. Therefore, TD is compact on C
1
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, ∞,
ψ
−1
2
u
v ≤ φ
−1
uφ
v
≤ ψ
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 t hat γ : K → 0, ∞ is continuous and
γ
tx
1 − t
y
≤ tγ
x
1 − t
γ
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,andψ a nonnegative continuous functional on K. Then
for positive real number a, b, c,andd, we define the following convex sets:
P
γ,θ,α,b,c,d
u ∈ K | α
u
≥ b, θ
u
≤
c, γ
u
≤ d
,
R
γ,ψ,a,d
u ∈ K | ψ
u
2 Tu≥u, u ∈ K ∩ ∂Ω
1
, and Tu≤u, u ∈ K ∩ ∂Ω
2
.
Then T has at least one fixed point in
Ω
2
\ Ω
1
.
Theorem 2.9 see 12. Let K be a cone in a real Banach space E.Letγ 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,
α
u
≤ ψ
u
, u≤Mγ
u
2.19
for all u ∈
L max
⎧
⎨
⎩
1
0
ψ
−1
1
1 − s
ds
1 −
1
0
g
s
ds
, 1
⎫
⎬
⎭
,N min
v
c
→ μ
max
t∈0,1
f
t, u
t
,v
t
φ
u
c
v
c
,f
μ
lim inf
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
∞
<ψ
1
1/2L,f
0
>ψ
2
1/N;
iii f
0
<ψ
1
1/2L,f
∞
>ψ
2
1/N.
Then BVP1.1 has at least one positive solution.
8 Boundary Value Problems
Proof. i Let Ω
1
{u ∈ K |u
g
r
dr
1
0
g
r
r
0
φ
−1
δ
s
f
τ,u
τ
,u
ds
1
1 −
1
0
h
r
dr
1
0
h
r
1
r
φ
−1
s
δ
f
τ
dτ
ds
≥ min
⎧
⎨
⎩
1
1 −
1
0
g
r
dr
1
0
g
r
1/2
s
f
τ,u
τ
,u
τ
dτ
ds,
1
1 −
1
0
h
r
dr
1
1
1/2
φ
−1
s
1/2
f
τ,u
τ
,u
τ
dτ
ds
≥ min
1/2
0
f
τ,u
τ
,u
τ
dτ
ds
≥ min
1/2
0
φ
−1
φ
r
N
0
ψ
−1
2
1
2
− s
ds,
1
1/2
ψ
−1
2
s −
1
2
ds
r
u
1
.
3.2
1
1 −
1
0
g
r
dr
1
0
g
r
1
0
φ
−1
1
s
f
τ,u
τ
dτ
ds
≤
1
1 −
1
0
g
r
dr
1
0
φ
−1
1 − s
φ
R
L
3.4
From 2.13, we have
Tu
c
max
φ
−1
δ
0
f
τ,u
τ
,u
τ
τ,u
τ
,u
τ
dτ
≤ φ
−1
φ
R
L
≤ R
u
1
.
3.5
This implies that
≤ ψ
1
1
2L
φ
u
c
v
c
for t ∈
0, 1
,
u
c
1
1
2L
φ
M
, 3.8
10 Boundary Value Problems
then for all ρ>
M,letΩ
3
{u ∈ K |u
1
<ρ}. For every u ∈ ∂Ω
3
, we have u
c
u
c
≤ 2ρ.
In the following, we consider two cases.
Case 1 u
c
u
. 3.9
Case 2 2ρ
0
≤u
c
u
c
≤ 2ρ. In this case,
f
t, u, u
≤ ψ
1
1
2L
φ
u
c
0
>ψ
2
1/N, there exists 0 <ξ<ρsuch that
f
t, u, v
≥ ψ
2
1
N
φ
u
c
v
c
for t ∈
0, 1
2
1
N
φ
u
c
u
c
≥ ψ
2
1
N
φ
u
∞
<ψ
1
1/2L, and there exists m
1
> 0 such that
f
t, u, v
≥ φ
m
1
N
for t ∈
0, 1
,m
1
≤
u
c
v
,
u
c
v
c
≤ 2m
2
. 3.14
Then BVP1.1 has at least two positive solutions.
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 t he nonnegative continuous concave functional α, the nonnegative continuous
convex functionals γ, θ, and nonnegative continuous functional ψ be defined on cone K by
γ
u
max
0≤t≤1
u
t
|
.
4.1
By Lemmas 2.1 and 2.2, the functionals defined above satisfy
ηθ
u
≤ α
u
≤ ψ
u
θ
u
,
u
1
max
γ
1
0
ht1 −
tdt × −d, d.
P3 ft, u, v <φa/L for t, u, v ∈ 0, 1 × 0,a × −d, d;
Then BVP1.1 has at least three positive solutions u
1
,u
2
, and u
3
satisfying
max
0≤t≤1
u
i
t
≤ d for i 1, 2, 3, min
η≤t≤1−η
|
u
1
|
<a,
4.3
where L defined as 3.1, K min{
1/2
η
ψ
−1
2
1/2 − sds,
1−η
1/2
ψ
−1
2
s − 1/2ds}.
Proof. We will show that all the conditions of Theorem 2.9 are satisfied.
If u ∈
Pγ,d, then γumax
0≤t≤1
|u
t|≤d.WithLemma 2.2 implying
max
0≤t≤1
|ut|≤Md,sobyP1, we have ft, ut,u
τ
,u
τ
dτ
,φ
−1
1
δ
f
τ,u
τ
,u
τ
dτ
This proves that T :
Pγ,d → Pγ,d.
12 Boundary Value Problems
To check condition S1 of Theorem 2.9, we choose
u
0
t
b
η
b
1 −
1
0
h
t
dt
η
1
0
0
h
t
1 − t
dt
⎞
⎠
. 4.6
Then u
0
t ∈ Pγ,θ,α,b,c,d and αu
0
>b,so{u ∈ Pγ,θ,α,b,c,d | αu >b}
/
∅. Hence,
for u ∈ Pγ,θ,α,b,c,d, there is b ≤ ut ≤ c, |u
t|≤d when η ≤ t ≤ 1 − η. From assumption
P2, we have
f
t, u
t
,u
0≤t≤1
|
Tu
t
|
>b 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 ∈ Pγ,α,b,d with θTu >c, we have
α
Tu
≥ ηθ
Tu
≥ ηc > b. 4.9
Thus condition S2 of Theorem 2.9 holds.
Finally, as ψ00 <a, there holds 0 /∈ Rγ,ψ,a,d. Suppose that u ∈ Rγ,ψ,a,d with
ψua; then by the assumption P3,
f
Tu
t
|
<a. 4.11
Hence condition S3 of Theorem 2.9 is also satisfied.
Boundary Value Problems 13
Thus BVP1.1 has at least three positive solutions u
1
,u
2
,andu
3
satisfying
max
0≤t≤1
u
i
t
≤ d for i 1, 2, 3, min
η≤t≤1−η
|
t
|
<a.
4.12
5. Examples
In this section, we give three examples as applications.
Example 5.1. Let φu|u|u, gtht1/2. Now we consider the BVP
φ
u
f
t, u
t
,u
t
0,t∈
dt,
5.1
where ft, u, v1 t18 u4 cos v for t, u, v ∈ 0, 1 × 0, ∞ × −∞, ∞.
Let ψ
1
uψ
2
uu
2
, u>0. Choosing r 1,R 100. By calculations we obtain
L
4
3
,N
2
3
1
2
3/2
,φ
r
N
18,φ
R
18 u
4 cos v
≤
2 × 118 × 5 < 75
2
.
5.4
Hence, by Theorem 3.1,BVP5.1 has at least one positive solution.
Example 5.2. Let φuu, gtht1/2. Consider the BVP
φ
u
f
t, u
t
,u
t
t
dt,
5.5
where ft, u, v1 t1/10 u1/100 v
2
1 u
c
v
c
2
for t, u, v ∈ 0, 1 × 0, ∞×
−∞, ∞.
14 Boundary Value Problems
Let ψ
1
uψ
2
uu, u>0. Then L 1,N 1/8. It easy to see
f
0
f
∞
∞ >ψ
2
1
N
1
u
c
v
c
2
≤ 2
1
10
1
5
1
100
1
25
1
1
f
t, u
t
,u
t
0,t∈
0, 1
,
u
0
u
1
1
2
1
4
v
10
5
3
,u≤ 12,
sin t
10
4
2500 · 12
6
1
10
4
v
10
5
3
,u>12.
5.9
Choosing a 1/10, b 1, η 1/4, d 10
5
, then by calculations we obtain that
L
4
d
10
10
for 0 ≤ t ≤ 1, 0 ≤ u ≤ 3 · 10
5
, −10
5
≤ v ≤ 10
5
,
f
t, u, v
> 2304 for
1
4
≤ t ≤
3
4
, 1 ≤ u ≤ 12, −10
5
≤ v ≤ 10
5
,
f
t, u, v
≤ 10
5
for i 1, 2, 3, min
1/4≤t≤3/4
|
u
1
t
|
> 1,
max
0≤t≤1
|
u
2
t
|
>
1
10
with min
1/4≤t≤3/4
|
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