Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2010, Article ID 495138, 17 pages
doi:10.1155/2010/495138
Research Article
Existence of Concave Positive Solutions for
Boundary Value Problem of Nonlinear Fractional
Differential Equation with p-Laplacian Operator
Jinhua Wang, Hongjun Xiang, and ZhiGang Liu
Department of Mathematics, Xiangnan University, Chenzhou 423000, China
Correspondence should be addressed to Hongjun Xiang, hunxhjxhj67@126.com
Received 27 July 2009; Accepted 9 March 2010
Academic Editor: Rodica Costin
Copyright q 2010 Jinhua Wang et al. 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 and multiplicity of concave positive solutions for boundary value
γ
α
ut
problem of nonlinear fractional differential equation with p-Laplacian operator D0 φp D0
ρ
α
ft, ut, D0 ut 0, 0 < t < 1, u0 u 1 0, u 0 0, D0 ut|t0 0, where 0 < γ < 1,
α
2 < α < 3, 0 < ρ 1, D0
denotes the Caputo derivative, and f : 0, 1 × 0, ∞ × R → 0, ∞
is continuous function, φp s |s|p−2 s, p > 1, φp −1 φq , 1/p 1/q 1. By using fixed point
theorem, the results for existence and multiplicity of concave positive solutions to the above
boundary value problem are obtained. Finally, an example is given to show the effectiveness of
our works.
1. Introduction
As we know, boundary value problems of integer-order differential equations have been
intensively studied; see 1–5 and therein. Recently, due to the wide development of its theory
of fractional calculus itself as well as its applications, fractional differential equations have
been constantly attracting attention of many scholars; see, for example, 6–15.
In 7, Jafari and Gejji used the adomian decomposition method for solving the
existence of solutions of boundary value problem:
Dα ut μft, ut 0,
u0 0,
0 < t < 1, 1 < α 2,
u1 b.
1.1
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In 9, by using fixed point theorems on cones, Dehghani and Ghanbari considered
triple positive solutions of nonlinear fractional boundary value problem:
Dα xt qtf t, xt, x t 0,
0 < t < 1, 2 α < 3,
x0 x1 0,
1.2
where Dα is the standard Riemann-Liouvill derivative. But we think that Green’s function in
9 is wrong; if α > 2, then, Green’s function cannot be decided by x0 x1 0.
In 11, using fixed point theorems on cones, Zhang investigated the existence and
multiplicity of positive solutions of the following problem:
α
ut ft, ut,
D0
u0 u 0 0,
0 < t < 1, 1 < α 2,
1.3
u1 u 1 0,
α
is the Caputo fractional derivative.
where D0
In 12, by means of Schauder fixed-point theorem, Su and Liu studied the existence
of nonlinear fractional boundary value problem involving Caputo’s derivative:
β
α
ut f t, ut, D0 ut 0,
D0
u0 u 1 0,
or
0 < t < 1, 1 < α 2,
u 0 u1 0,
or
1.4
u0 u1 0.
To the best of our knowledge, the existence of concave positive solutions of fractional
order equation is seldom considered and investigated. Motivated by the above arguments,
the main objective of this paper is to investigate the existence and multiplicity of concave
positive solutions of boundary value problem of fractional differential equation with pLaplacian operator as follows:
α
γ ρ
ut f t, ut, D0 ut 0,
D0 φp D0
u0 u 1 0,
u 0 0,
0 < t < 1,
α
ut|t0
D0
1.5
0,
α
denotes the Caputo derivative, and f : 0, 1 ×
where 0 < γ < 1, 2 < α < 3, 0 < ρ 1, D0
0, ∞×R → 0, ∞ is continuous function, φp s |s|p−2 s, p > 1, φp −1 φq , 1/p1/q 1.
By using fixed point theorem, some results for multiplicity of concave positive
solutions to the above boundary value problems are obtained. Finally, an example is given to
show the effectiveness of our works.
The rest of the paper is organized as follows. In Section 2, we will introduce some
lemmas and definitions which will be used later. In Section 3, the multiplicity of concave
positive solutions for the boundary value problem 1.5 will be discussed.
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2. Basic Definitions and Preliminaries
Firstly we present here some necessary definitions and lemmas.
Definition 2.1 see 6, 16, 17. The fractional integral of order α > 0 of a function y : 0, ∞ →
R is given by
I α yt 1
Γα
t
t − sα−1 ysds,
2.1
0
provided that the right side is pointwise defined on 0, ∞.
Definition 2.2 see 6, 16, 17. The Riemann-Liouville fractional derivative of order α > 0 of a
continuous function y : 0, ∞ → R is given by
Dα yt n t
d
1
t − sn−α−1 ysds,
Γn − α dt
0
2.2
where n α 1 provided that the right side is pointwise defined on 0, ∞.
Definition 2.3 see 17. Caputo’s derivative of order α > 0 of a function y : 0, ∞ → R is
defined as
Dα yt 1
Γn − α
t
t − sn−α−1 yn sds,
n − 1 < α < n,
2.3
0
provided that the right side is pointwise defined on 0, ∞.
Remark 2.4. The following properties are well known:
β
αβ
α
1 D0
D0 yt D0 yt, α > 0, β > 0;
−α
α
2 D0
yt I0
yt, α > 0;
α
3 I0
: C0, 1 → C0, 1, α > 0.
If α is an integer, the derivative for order α is understood in the sense of usual
differentiation.
Definition 2.5. Let E be a real Banach space over R. A nonempty convex closed set P ⊂ E is
said to be a cone provided that
a au ∈ P , for all u ∈ P , a 0;
b u, −u ∈ P , implies u 0.
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Definition 2.6. Let E be a real Banach space and let P ⊂ E be a cone. A function ψ : P → 0, ∞
is called a nonegative continuous concave functional if ψ is continuous and
ψ λx 1 − λy λψx 1 − λψ y ,
2.4
for all x, y ∈ P and 0 λ 1.
Definition 2.7. Let E be a real Banach space and let P ⊂ E be a cone. A function ϕ : P → 0, ∞
is called a nonegative continuous convex functional if ϕ is continuous and
ϕ λx 1 − λy λϕx 1 − λϕ y ,
2.5
for all x, y ∈ P and 0 λ 1.
Suppose that ϕ, θ : P → 0, ∞ are two nonnegative continuous convex functionals
satisfying
u M max ϕu, θu ,
for u ∈ P,
2.6
where M is a positive constant and
Ω u ∈ P | ϕu < r, θu < L /
Ø,
for r > 0, L > 0.
2.7
Let r > a > 0, L > 0 be given constants, let ϕ, θ : P → 0, ∞ be two nonnegative
continuous convex functionals satisfying 2.6 and 2.7, and let ψ be a nonnegative
continuous concave functional on the cone P . Define the bounded convex sets:
P ϕ, r; θ, L u ∈ P | ϕu < r, θu < L ,
P ϕ, r; θ, L u ∈ P | ϕu r, θu L ,
P ϕ, r; θ, L; ψ, a u ∈ P | ϕu < r, θu < L, ψu > a ,
P ϕ, r; θ, L; ψ, a u ∈ P | ϕu r, θu L, ψu a .
2.8
Lemma 2.8 see 2, 5. Let E be a Banach space, let P ⊂ E be a cone, and let r2 c > b > r1 > 0,
L2 L1 > 0 be given constants. Assume that ϕ, θ are two nonnegative continuous convex functionals
on P , such that 2.6 and 2.7 are satisfied; let ψ be a nonnegative continuous concave functional on
P , such that ψu ϕu for all u ∈ P ϕ, r2 ; θ, L2 and let T : P ϕ, r2 ; θ, L2 → P ϕ, r2 ; θ, L2 be
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5
a completely continuous operator. Suppose the following:
C1 {u ∈ P ϕ, c; θ, L2 ; ψ, b | ψu > b} /
ø, ψT u > b for all u ∈ P ϕ, c; θ, L2 ; ψ, b;
C2 ϕT u < r1 , θT u < L1 , for all u ∈ P ϕ, r1 ; θ, L1 ;
C3 ψT u > b, for all u ∈ P ϕ, r2 ; θ, L2 ; ψ, b with ϕT u > c.
Then T has at least three fixed points u1 , u2 , and u3 in P ϕ, r2 ; θ, L2 . Further u1 ∈ P ϕ,
r1 ; θ, L1 , u2 ∈ {P ϕ, r2 ; θ, L2 ; ψ, b | ψu > b}, and u3 ∈ P ϕ, r2 ; θ, L2 \ P ϕ, r2 ; θ, L2 ;
ψ, b P ϕ, r1 ; θ, L1 .
Lemma 2.9 see 12. Assume that u ∈ C0, 1 ∩ L0, 1 with a fractional derivative of order α that
belongs to C0, 1 ∩ L0, 1. Then
α
α
I0
D0
ut ut C0 C1 t · · · Cn−1 tn−1 ,
2.9
for some Ci ∈ R, i 0, 1, . . . , n − 1, where n is the smallest integer greater than or equal to α.
Lemma 2.10. Let y ∈ C0, 1; then the boundary value problem
α
ut yt 0,
D0
0 < t < 1, 2 < α < 3,
u0 u 1 u 0 0
2.10
2.11
has a unique solution
ut 1
2.12
Gt, sysds,
0
where
Gt, s ⎧
α − 1t1 − sα−2 − t − sα−1
⎪
⎪
⎪
,
⎪
⎨
Γα
0 s t 1,
⎪
⎪
⎪
α − 1t1 − sα−2
⎪
⎩
,
Γα
0 t s 1.
2.13
Proof. We may apply Lemma 2.9 to reduce 2.10 to an equivalent integral equation:
1
ut −
Γα
t
0
t − sα−1 ysds C0 C1 t C2 t2 .
2.14
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From 2.14, we have
u t −
1
Γα
1
u t −
Γα
t
α − 1t − sα−2 ysds C1 2C2 t,
0
t
2.15
α − 1α − 2t − sα−3 ysds 2C2 ,
0
1
and by 2.10 and 2.11, there are C0 C2 0, C1 1/Γα 0 α − 11 − sα−2 ysds.
Therefore, the unique solution of problem 2.10 and 2.11 is
ut −
1
Γα
t
0
t
t − sα−1 ysds 0
1
Γα
1
α − 1t1 − sα−2 ysds
0
α − 1t1 − sα−2 − t − sα−1
ysds Γα
1
t
α − 1t1 − sα−2
ysds
Γα
2.16
1
Gt, sysds.
0
Lemma 2.11. The function Gt, s defined by 2.13 satisfies the following conditions:
1 Gt, s 0, Gt, s G1, s for 0 t, s 1,
2 Gt, s tα−1 G1, s, for 0 t, s 1.
Proof. Since
gt, s : α − 1t1 − sα−2 − t − sα−1
α − 1tα−1 1 − sα−2 − t − sα−1
s α−1
tα−1 α − 11 − sα−2 − 1 −
t
tα−1 α − 11 − sα−2 − 1 − sα−1
0,
observing 2.13, we have Gt, s 0.
for 2 < α < 3, 0 s t 1,
2.17
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Form 2.13, we obtain
Gt t, s ⎧
α − 11 − sα−2 − α − 1t − sα−2
⎪
⎪
⎪
, 0 s t 1,
⎪
⎨
Γα
⎪
⎪
⎪
α − 11 − sα−2
⎪
⎩
,
Γα
2.18
0 t s 1.
Clearly, Gt t, s 0, for 0 t, s 1, we have that Gt, s is increasing with respect to t ∈ 0, 1,
and therefore, Gt, s G1, s, for 0 t, s 1. 1 of Lemma 2.11 holds.
On the other hand, if t s, then,
Gt, s α − 1t1 − sα−2 − t − sα−1
G1, s α − 11 − sα−2 − 1 − sα−1
tα−1 α − 11 − sα−2 − 1 − sα−1
α − 11 − s
α−2
− 1 − s
2.19
α−1
tα−1 .
If t s, then Gt, s/G1, s tα−1 ; therefore, Gt, s tα−1 G1, s, for 0 s, t 1. 2 of
Lemma 2.11 holds.
3. Existence of Three Concave Positive Solutions
In this section, we study the existence of concave positive solution for problem 1.5.
ρ
1−ρ
Let E C1 0, 1. From Definitions 2.1 and 2.3, we obtain D0 ut I0 u t, 0 < ρ < 1,
ρ
ρ
and D0 ut u t, ρ 1. So, by 3 of Remark 2.4, we know that D0 ut is continuous for
all ut ∈ E. Hence, for all ut ∈ E, we can define
u ⎧
ρ
⎪
max |ut| max D0 ut max |u t|,
⎪
⎨0t1
0t1
0t1
0 < ρ < 1,
⎪
⎪
⎩ max |ut| max |u t|,
ρ 1.
0t1
0t1
3.1
Lemma 3.1 see 12. E, · is a Banach space.
Define the cone P ⊂ E by P {u ∈ E | ut 0, ut is concave on 0, 1}.
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Let the nonnegative continuous concave functional ψ and the nonnegative continuous convex
functionals ϕ, θ be defined on the cone P by
ψu θu min
1/ktk−1/k
ϕu max |ut|,
|ut|,
0t1
⎧
ρ
⎪
⎪
max D0 ut max |u t|,
⎨0t1
0t1
0 < ρ < 1,
⎪
⎪
⎩ max |u t|,
ρ 1.
0t1
3.2
Lemma 3.2 see 1. Let u ∈ P , k 3; then
min
1/ktk−1/k
|ut| 1
max |ut|.
k 0t1
3.3
Lemma 3.3. BVP 1.5 is equivalent to the integral equation
ut 1
Gt, sφq
0
1
Γ γ
s
s − τ
γ−1
f
ρ
τ, uτ, D0 uτ
0
dτ ds.
3.4
Proof. From BVP 1.5 and Lemma 2.9, we have
α
γ
ρ
φp D0
ut −I0 f t, ut, D0 ut C,
1
− Γ γ
t
ρ
t − τγ−1 f τ, uτ, D0 uτ dτ C.
0
3.5
α
By D0
ut|t0 0, there is C 0, and then,
α
D0
ut
−φq
1
Γ γ
t
t − τ
γ−1
0
f
ρ
τ, uτ, D0 uτ
dτ .
3.6
Therefore, BVP1.5 is equivalent to following problem:
α
D0
ut
φq
1
Γ γ
t
0
γ−1
t − τ
f
ρ
τ, uτ, D0 uτ
dτ
0,
0 < t < 1, 2 < α < 3,
u0 u 1 u 0 0.
By Lemma 2.10, BVP 1.5 is equivalent to the integral equation 3.4.
3.7
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Let T : P → E be the operator defined by
T ut 1
Gt, sφq
0
1
Γ γ
s
γ−1
s − τ
f
ρ
τ, uτ, D0 uτ
0
dτ ds : Ft.
3.8
Lemma 3.4. T : P → P is completely continuous.
Proof. Let u ∈ P ; in view of nonnegativeness and continuity of Gt, s and ft, u, v, we have
T u 0, and t ∈ 0, 1 is continuous:
T u t 1
0
t
0
Gt t, sφq
1
Γ γ
s
γ−1
s − τ
f
ρ
τ, uτ, D0 uτ
0
α−11−sα−2 −α−1t−sα−2
φq
Γα
1
t
α − 11 − sα−2
φq
Γα
s
1
Γ γ
1
Γ γ
s − τ
s
γ−1
0
s−τ
dτ ds
γ−1
f
0
f
ρ
τ, uτ, D0 uτ
ρ
τ, uτ, D0 uτ
dτ ds
dτ ds.
3.9
Clearly, T u t is continuous for α < 2.
By Remark 2.4 and noting 3.4 and 3.6, we have
2
2−α α
D0 T ut
D0
T ut D0
α
2−α
D0
D0 Ft
2−α
−D0
φq
α−2
−I0
φq
1
−
Γα − 2
1
Γ γ
1
Γ γ
t
0
t
t − τ
γ−1
f
0
ρ
τ, uτ, D0 uτ
t
t − τ
γ−1
0
t − s
α−3
φq
f
ρ
τ, uτ, D0 uτ
1
Γ γ
t
0
t − τ
γ−1
f
dτ
dτ
ρ
τ, uτ, D0 uτ
dτ ds
0.
3.10
So, T u is concave on 0, 1 and T u ∈ C1 0, 1; we obtain T P ⊂ P .
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Let Ω ∈ P be bounded; that is, there exists a positive constant M > 0 such that u M
for all u ∈ Ω.
ρ
Let N maxt,u,v∈0,1×0,M×−M,M |ft, ut, D0 ut|; then, for all u ∈ Ω, we have
s
1
1
ρ
γ−1
s − τ f τ, uτ, D0 uτ dτ ds
|T ut| Gt, sφq
0
Γ γ 0
N
Γ γ 1
N
Γ γ 1
q−1 1
G1, sds
0
q−1 N
Γ γ 1
q−1
1
1
−
Γα Γα 1
1
,
Γα
s
1 1
ρ
γ−1
T u t G t, sφq
s − τ f τ, uτ, D0 uτ dτ ds
t
0
Γ γ 0
N
Γ γ 1
N
Γ γ 1
q−1 1
0
q−1
Gt t, sds
1
,
Γα
t
1
ρ
−ρ
t − s T u sds
D0 T ut Γ 1 − ρ 0
1
Γ 1−ρ
t
t−s−ρ
0
v
1
1
ρ
γ−1
× Gs s, vφq v−τ f τ, uτ, D0 uτ dτ ds
0
Γ γ 0
1
Γ 1−ρ
N
Γ γ 1
N
Γ γ 1
q−1
N
Γ γ 1
q−1
1
Γα
t
t − s−ρ ds
0
t1−ρ
1 − ρ Γ 1 − ρ Γα
q−1
1
.
Γ 2 − ρ Γα
3.11
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So, for all u ∈ Ω, T ut N/Γγ 1q−1 2/Γα1/Γ2−ρΓα. Hence, T Ω is uniformly
bounded.
Since Gt, s is continuous on 0, 1 × 0, 1, it is uniformly continuous on 0, 1 × 0, 1.
Thus for fixed s ∈ 0, 1 and for any ε > 0, there exists a constant δ > 0, such that any
t1 , t2 ∈ 0, 1 and |t1 − t2 | < δ,
q−1
Γ γ 1
.
|Gt1 , s − Gt2 , s| < ε
N
3.12
Therefore,
|T ut2 − T ut1 | 1
|Gt2 , s − Gt1 , s|φq
0
N
Γ γ 1
q−1 1
1
Γ γ
s
s − τ
γ−1
f
0
ρ
τ, uτ, D0 uτ
dτ ds
|Gt2 , s − Gt1 , s|ds < ε.
0
3.13
That is to say, T Ω is equicontinuous. By the means of the Arzela-Ascoli Theorem, we have
that T : P → P is completely continuous. The proof is completed.
Let
A
1
Γ γ 1
q−1
1
,
Γα
q−1 1
γq−1 1
1
B
G1, sds,
k
Γ γ 1
1/k
M
1
Γ γ 1
q−1
3.14
1
.
ΓαΓ 2 − ρ
Theorem 3.5. Suppose that there exist constants 0 < r1 < b < kb r2 , L2 L1 > 0, such that
kb/B minr2 /A, L2 /2A, L2 /2M, and the following conditions hold:
H1 ft, u, v < min{φp r1 /A, φp L1 /2A, φp L1 /2M}, for t, u, v ∈ 0, 1 × 0, r1 ×
−L1 , L1 ;
H2 ft, u, v > φp kb/B, for t, u, v ∈ 1/k, 1 × b, kb × −L2 , L2 ;
H3 ft, u, v min{φp r2 /A, φp L2 /2A, φp L2 /2M}, for t, u, v ∈ 0, 1 × 0, r2 ×
−L2 , L2 .
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Then, the BVP 1.5 has at least three concave positive solutions u1 , u2 , and u3 , such that
ρ
max D0 u1 t max u1 t < L1 ,
max |u1 t| < r1 ,
0t1
b<
0t1
min
1/ktk−1/k
0t1
|u2 t| < max |u2 t| r2
0t1
ρ
max D0 u2 t max u2 t < L2 ,
0t1
0t1
r1 < max |u3 t| kb,
0t1
min
1/ktk−1/k
3.15
|u3 t| < b,
ρ
L1 < max D0 u3 t max u3 t < L2 .
0t1
0t1
Proof. By Lemmas 3.3 and 3.4, we have that T : P → P is completely continuous and problem
1.5 has a solution u ut if and only if ut satisfies the operator equation u T u.
Now, we show that all the conditions of Lemma 2.8 hold.
Step 1. We will show that T : P ϕ, r2 ; θ, L2 → P ϕ, r2 ; θ, L2 .
If u ∈ P ϕ, r2 ; θ, L2 , then ϕu r2 , θu L2 .
From H3 , we have
ϕT ut max|T ut|
0≤t≤1
s
1
1
ρ
r−1
max Gt, sφq
s − τ f τ, uτ, D0 uτ dτ ds
0≤t≤1 0
Γr 0
r2
A
1
G1, sφq
0
1
Γr
s
r−1
s − τ
0
q−1 r2
1
1
1
−
A rΓr
Γα Γα 1
r2
A
r2 ,
1
Γr 1
q−1
1
Γα
dτ ds
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s
1
1
ρ
maxT u t max Gt t, sφq
s − τr−1 f τ, uτ, D0 uτ dτ ds
0≤t≤1
0≤t≤1 0
Γr 0
13
q−1 1
L2
1
Gt t, sds
2A Γ γ 1
0
q−1
1
1
L2
2A Γr 1
Γα
L2
,
2
t
1
ρ
−ρ
maxD0 T ut max t − s T u sds
0≤t≤1
0≤t≤1 Γ 1 − ρ
0
1
Γ 1−ρ
t
t − s−ρ
0
v
1
1
ρ
γ−1
× Gs s, vφq
v−τ f τ, uτ, D0 uτ dτ ds
0
Γ γ 0
L2
1
2M Γ 1 − ρ
1
Γ γ 1
q−1
1
Γα
t
t − s−ρ ds
0
⎞
⎛
q−1
1
1
L2 ⎝
⎠
2M
Γ γ 1
ΓαΓ 2 − ρ
L2
.
2
3.16
Then, θT ut L2 .
So, T : P ϕ, r2 ; θ, L2 → P ϕ, r2 ; θ, L2 .
Step 2. Let ut kb/2, 0 t 1. It is easy to see that ut kb/2 ∈ P ϕ, kb; θ, L2 , ψ, b, and
ψu ψkb/2 > b. Consequently, {u ∈ P ϕ, kb; θ, L2 , ψ, b | ψu > b} /
ø.
If u ∈ P ϕ, kb; θ, L2 , ψ, b, then for all 1/k t 1, b ut kb, θut L2 . By
ρ
H2 , we obtain ft, ut, D0 ut > φp kb/B, for 1/k t 1.
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From Lemma 3.2, we have
ψT u min
1/ktk−1/k
|T ut|
1
max |T ut|
k 0t1
s
1
1
1
ρ
r−1
max Gt, sφq
s − τ f τ, uτ, D0 uτ dτ ds
0t1
k
Γr 0
0
1
kb 1
sγ
α−1
>
max t
G1, sφq
ds
B k 0t1
γΓ γ
0
b
B
b
>
B
1
Γ γ 1
q−1 1
3.17
G1, sφq sγ ds
0
q−1 1
γq−1 1
1
G1, sds b;
k
Γ γ 1
1/k
that is, ψT u > b, for all u ∈ P ϕ, kb; θ, L2 , ψ, b. This shows that condition C1 of Lemma 2.8
holds.
Step 3. Let u ∈ P ϕ, r1 ; θ, L1 ; by H1 , we have
r1
L1
L1
ρ
, φp
, φp
.
f t, ut, D0 ut < min φp
A
2A
2M
3.18
Similarly, we can prove that T : P ϕ, r1 ; θ, L1 → P ϕ, r1 ; θ, L1 . C2 of Lemma 2.8 holds.
Step 4. Let u ∈ P ϕ, kb; θ, L2 ; ψ, b, and ϕT u > kb; we have
ψT u min
1/ktk−1/k
|T ut| 1
1
max |T ut| ϕT u > b.
k 0t1
k
3.19
C3 of Lemma 2.8 holds. Therefore, the BVP 1.5 has at least three positive solutions
u1 , u2 , and u3 satisfying
ρ
max D0 u1 t max u1 t L1 ,
max |u1 t| r1 ,
0t1
0t1
b<
min
1/ktk−1/k
0t1
|u2 t| < max |u2 t| r2 ,
0t1
ρ
max D0 u2 t max u2 t L2 ,
0t1
min
1/ktk−1/k
The proof is completed.
0t1
|u3 t| < b,
r1 < max |u3 t| kb,
0t1
ρ
L1 < max D0 u3 t max u3 t L2 .
0t1
0t1
3.20
International Journal of Mathematics and Mathematical Sciences
15
Corollary 3.6. If there exist constants 0 < r1 < b1 < kb1 r2 < b2 < kb2 · · · rn , and
0 < L1 L2 · · · Ln−1 , n ∈ N, such that kbi /B min{ri1 /A, Li1 /2A, Li1 /2M}, for
1 i n − 1 and the following conditions are satisfied:
I1 ft, u, v < min{φp ri /A, φp Li /2A, φp Li /M} for t, u, v ∈ 0, 1 × 0, ri ×
−Li , Li ;
I2 ft, u, v > φp kbi /B, for t, u, v ∈ 1/k, 1 × bi , kbi × −Li1 , Li1 ;
then the problem 1.5 has at least 2n − 1 concave positive solutions.
Proof. If n 1, by Condition I1 and Step 1 of the proof of Theorem 3.5, we can obtain that
T : P ϕ, r1 ; θ, L1 → P ϕ, r1 ; θ, L1 ⊂ P ϕ, r1 ; θ, L1 . From the Schauder fixed-point theorem,
the problem 1.5 has at least one fixed-point u1 ∈ P ϕ, r1 ; θ, L1 .
If n 2, by Theorem 3.5, there exist at least three concave positive solutions u2 , u3 , and
u4 . By the induction method, we finish the proof.
Finally, we present an example to check our results.
Example 3.7. Consider the boundary value problem:
1/2
5/2
1/2
D0
φ3/2 D0
ut f t, ut, D0
ut 0,
u0 u 1 u 0 0,
0 < t < 1,
3.21
α
utt0 0,
D0
where
⎧
4
t
|v|
⎪
2
⎪
,
⎨ 6u 4
ft, u, v 20
4 × 10 4
⎪
|v|
⎪
⎩ t 96 ,
20
4 × 104
for u 4,
3.22
for u > 4.
Let k 4; note that p 3/2, q 3, α 5/2, γ 1/2, ρ 1/2; we have A ≈ 0.9585, B ≈ 0.1089,
M ≈ 1.0819.
Choosing r1 1/4, b 1, r2 104 , L1 2 × 104 , L2 4 × 104 . It is easy to see that
kb/B minr2 /A, L2 /2A, L2 /2M, and ft, u, v satisfying
1 ft, u, v 0.4875 < min{φp r1 /A, φp L1 /2A, φp L1 /2M} ≈ 0.5107, t, u, v ∈
0, 1 × 0, 1/4 × −2 × 104 , 2 × 104 ,
2 ft, u, v 7.0125 > φp kb/B ≈ 6.060, t, u, v ∈ 1/4, 1 × 1, 4 × −4 × 104 , 4 × 104 ,
3 ft, u, v 97.05 < min{φp r2 /A, φp L2 /2A, φp L2 /2M} ≈ 102.1419, t, u, v ∈
0, 1 × 0, 104 × −4 × 104 , 4 × 104 .
16
International Journal of Mathematics and Mathematical Sciences
By Theorem 3.5, problem 3.21 has at least three concave positive solutions u1 , u2 , and
u3 satisfying
max |u1 t| <
0t1
1<
ρ
max D0 u1 t max u1 t < 2 × 104 ,
1
,
4
0t1
min
1/ktk−1/k
0t1
|u2 t| < max |u2 t| 104 ,
0t1
ρ
max D0 u2 t max u2 t < 4 × 104 ,
0t1
0t1
3.23
1
< max |u3 t| 4,
min
|u3 t| < 1,
1/ktk−1/k
4 0t1
ρ
2 × 104 < max D0 u3 t max u3 t < 4 × 104 .
0t1
0t1
Acknowledgments
This work was jointly supported by the Natural Science Foundation of Hunan Provincial
under Grants 09JJ3005 and 2009JT3042, the Construct Program of the Key Discipline in Hunan
Province, and Aid Program for Science and Technology Innovative Research Team in Higher
Educational Institutions of Hunan Province.
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