# Existence of Concave Positive Solutions for Boundary Value Problem of Nonlinear Fractional Differential Equation withp-Laplacian Operator

Authors: Jinhua Wang; Hongjun Xiang; ZhiGang Liu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

---

## Abstract

We consider the existence and multiplicity of concave positive solutions for boundary value problem of nonlinear fractional differential equation withp-Laplacian operator D0+γ(ϕp(D0+αu(t)))+f(t,u(t),D0+ρu(t))=0, 0<t<1, u(0)=u′(1)=0, u′′(0)=0, D0+αu(t)|t=0=0, where 0<γ<1, 2<α<3, 0<ρ⩽1, D0+α denotes the Caputo derivative, and f:[0,1]×[0,+∞)×R→[0,+∞) is continuous function, ϕp(s)=|s|p-2s, 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.


---

## Body

## 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: (1.1)Dαu(t)+μf(t,u(t))=0,0<t<1,1<α⩽2,u(0)=0,u(1)=b.In [9], by using fixed point theorems on cones,  Dehghani and   Ghanbari considered triple positive solutions of nonlinear fractional boundary value problem: (1.2)Dαx(t)+q(t)f(t,x(t),x′(t))=0,0<t<1,2⩽α<3,x(0)=x(1)=0,
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 x(0)=x(1)=0.In [11], using fixed point theorems on cones, Zhang investigated the existence and multiplicity of positive solutions of the following problem: (1.3)D0+αu(t)=f(t,u(t)),0<t<1,1<α⩽2,u(0)+u′(0)=0,u(1)+u′(1)=0,
where D0+α is the Caputo fractional derivative.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: (1.4)D0+αu(t)=f(t,u(t),D0+βu(t))=0,0<t<1,1<α⩽2,u(0)=u′(1)=0,oru′(0)=u(1)=0,oru(0)=u(1)=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 withp-Laplacian operator as follows: (1.5)D0+γ(ϕp(D0+αu(t)))+f(t,u(t),D0+ρu(t))=0,0<t<1,u(0)=u'(1)=0,u′′(0)=0,D0+αu(t)|t=0=0,
where 0<γ<1, 2<α<3, 0<ρ⩽1, D0+α denotes the Caputo derivative, and f:[0,1]×[0,+∞)×R→[0,+∞) is continuous function, ϕp(s)=|s|p-2s, p>1, (ϕp)-1=ϕq,  1/p+1/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 Section2, 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.


## 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 orderα>0 of a function y:(0,∞)→R is given by
(2.1)Iαy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds,
provided that  the right side is pointwise defined on (0,∞).Definition 2.2 (see [6, 16, 17]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function y:(0,∞)→R is given by
(2.2)Dαy(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α-1y(s)ds,
where n=[α]+1 provided that the right side is pointwise defined on (0,∞).Definition 2.3 (see [17]).
Caputo's derivative of orderα>0 of a function y:(0,∞)→R is defined as
(2.3)Dαy(t)=1Γ(n-α)∫0t(t-s)n-α-1y(n)(s)ds,n-1<α<n,
provided that the right side is pointwise defined on (0,∞).Remark 2.4.
The following properties are well known:(1)
D0+αD0+βy(t)=D0+α+βy(t), α>0,  β>0;(2)
D0+-αy(t)=I0+αy(t), α>0;(3)
I0+α:C[0,1]→C[0,1], α>0.
Ifα is an integer, the derivative for order α is understood in the sense of usual differentiation.Definition 2.5.
LetE 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.Definition 2.6.
LetE 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
(2.4)ψ(λx+(1-λ)y)⩾λψ(x)+(1-λ)ψ(y),
for all x,y∈P and 0⩽λ⩽1.Definition 2.7.
LetE 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
(2.5)φ(λx+(1-λ)y)⩽λφ(x)+(1-λ)φ(y),
for all x,y∈P and 0⩽λ⩽1.
Suppose thatφ,θ:P→[0,+∞) are two nonnegative continuous convex functionals satisfying
(2.6)∥u∥⩽Mmax{φ(u),θ(u)},foru∈P,
where M is a positive constant and
(2.7)Ω={u∈P∣φ(u)<r,θ(u)<L}≠⌀,forr>0,L>0.
Letr>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:
(2.8)P(φ,r;θ,L)={u∈P∣φ(u)<r,θ(u)<L},P¯(φ,r;θ,L)={u∈P∣φ(u)⩽r,θ(u)⩽L},P(φ,r;θ,L;ψ,a)={u∈P∣φ(u)<r,θ(u)<L,ψ(u)>a},P¯(φ,r;θ,L;ψ,a)={u∈P∣φ(u)⩽r,θ(u)⩽L,ψ(u)⩾a}.Lemma 2.8 (see [2, 5]).
LetE 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 ψ(u)⩽φ(u) for all u∈P¯(φ,r2;θ,L2) and let T:P¯(φ,r2;θ,L2)→P¯(φ,r2;θ,L2) be a completely continuous operator. Suppose the following:(C1)
{u∈P¯(φ,c;θ,L2;ψ,b)∣ψ(u)>b}≠ø,ψ(Tu)>b for all u∈P¯(φ,c;θ,L2;ψ,b);(C2)
φ(Tu)<r1, θ(Tu)<L1,  for all  u∈P¯(φ,r1;θ,L1);(C3)
ψ(Tu)>b,  for all u∈P¯(φ,r2;θ,L2;ψ,b) with  φ(Tu)>c.
ThenT 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)∣ψ(u)>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 thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α that belongs to C(0,1)∩L(0,1). Then
(2.9)I0+αD0+αu(t)=u(t)+C0+C1t+⋯+Cn-1tn-1,
for some Ci∈R,  i=0,1,…,n-1, where n is the smallest integer greater than or equal to α.Lemma 2.10.
Lety∈C[0,1]; then the boundary value problem
(2.10)D0+αu(t)+y(t)=0,0<t<1,2<α<3,(2.11)u(0)=u′(1)=u′′(0)=0
has a unique solution
(2.12)u(t)=∫01G(t,s)y(s)ds,
where
(2.13)G(t,s)={(α-1)t(1-s)α-2-(t-s)α-1Γ(α),0⩽s⩽t⩽1,(α-1)t(1-s)α-2Γ(α),0⩽t⩽s⩽1.Proof.
We may apply Lemma2.9 to reduce (2.10) to an equivalent integral equation:
(2.14)u(t)=-1Γ(α)∫0t(t-s)α-1y(s)ds+C0+C1t+C2t2.
From (2.14), we have
(2.15)u′(t)=-1Γ(α)∫0t(α-1)(t-s)α-2y(s)ds+C1+2C2t,u′′(t)=-1Γ(α)∫0t(α-1)(α-2)(t-s)α-3y(s)ds+2C2,
and by  (2.10) and (2.11), there are C0=C2=0, C1=(1/Γ(α))∫01(α-1)(1-s)α-2y(s)ds.
Therefore, the unique solution of problem (2.10) and (2.11) is
(2.16)u(t)=-1Γ(α)∫0t(t-s)α-1y(s)ds+1Γ(α)∫01(α-1)t(1-s)α-2y(s)ds=∫0t(α-1)t(1-s)α-2-(t-s)α-1Γ(α)y(s)ds+∫t1(α-1)t(1-s)α-2Γ(α)y(s)ds=∫01G(t,s)y(s)ds.Lemma 2.11.
The functionG(t,s) defined by (2.13) satisfies the following conditions:(1)
G(t,s)⩾0, G(t,s)⩽G(1,s)  for  0⩽t, s⩽1,(2)
G(t,s)⩾tα-1G(1,s),   for  0⩽t, s⩽1.Proof.
Since(2.17)g(t,s)=:(α-1)t(1-s)α-2-(t-s)α-1⩾(α-1)tα-1(1-s)α-2-(t-s)α-1=tα-1[(α-1)(1-s)α-2-(1-st)α-1]⩾tα-1[(α-1)(1-s)α-2-(1-s)α-1]⩾0,for2<α<3,0⩽s⩽t⩽1,
observing    (2.13), we have G(t,s)⩾0.
Form (2.13), we obtain
(2.18)Gt′(t,s)={(α-1)(1-s)α-2-(α-1)(t-s)α-2Γ(α),0⩽s⩽t⩽1,(α-1)(1-s)α-2Γ(α),0⩽t⩽s⩽1.
Clearly, Gt′(t,s)⩾0, for 0⩽t, s⩽1, we have that G(t,s) is increasing with respect to t∈[0,1], and  therefore, G(t,s)⩽G(1,s), for 0⩽t, s⩽1. (1) of Lemma 2.11 holds.
On the other hand, ift⩾s, then,
(2.19)G(t,s)G(1,s)=(α-1)t(1-s)α-2-(t-s)α-1(α-1)(1-s)α-2-(1-s)α-1⩾tα-1[(α-1)(1-s)α-2-(1-s)α-1](α-1)(1-s)α-2-(1-s)α-1=tα-1.
If t⩽s, then G(t,s)/G(1,s)=tα-1; therefore, G(t,s)⩾tα-1G(1,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).LetE=C1[0,1]. From    Definitions 2.1 and 2.3, we obtain D0+ρu(t)=I0+1-ρu′(t), 0<ρ<1, and D0+ρu(t)=u′(t), ρ=1. So, by (3) of Remark 2.4, we know that D0+ρu(t) is continuous for all u(t)∈E. Hence, for all u(t)∈E, we can  define (3.1)∥u∥={max0⩽t⩽1|u(t)|+max0⩽t⩽1|D0+ρu(t)|+max0⩽t⩽1|u′(t)|,0<ρ<1,max0⩽t⩽1|u(t)|+max0⩽t⩽1|u′(t)|,ρ=1.Lemma 3.1 (see [12]).
(E,∥·∥) is a Banach  space.
Define the coneP⊂E by  P={u∈E∣u(t)⩾0,u(t)isconcaveon[0,1]}.
Let the nonnegative continuous concave functionalψ and the nonnegative continuous convex functionals φ, θ be defined on the cone P by
(3.2)ψ(u)=min1/k⩽t⩽(k-1)/k|u(t)|,φ(u)=max0⩽t⩽1|u(t)|,θ(u)={max0⩽t⩽1|D0+ρu(t)|+max0⩽t⩽1|u′(t)|,0<ρ<1,max0⩽t⩽1|u′(t)|,ρ=1.Lemma 3.2 (see [1]).
Letu∈P, k⩾3; then
(3.3)min1/k⩽t⩽(k-1)/k|u(t)|⩾1kmax0⩽t⩽1|u(t)|.Lemma 3.3.
BVP (1.5) is equivalent to the integral equation
(3.4)u(t)=∫01G(t,s)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds.Proof.
From BVP (1.5) and Lemma 2.9, we have
(3.5)ϕp(D0+αu(t))=-I0+γf(t,u(t),D0+ρu(t))+C,=-1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ+C.
By D0+αu(t)|t=0=0, there is C=0, and then,
(3.6)D0+αu(t)=-ϕq(1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ).
Therefore, BVP(1.5) is equivalent to following problem:
(3.7)D0+αu(t)+ϕq(1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)=0,0<t<1,2<α<3,u(0)=u′(1)=u′′(0)=0.
By Lemma 2.10, BVP (1.5) is equivalent to the integral equation (3.4).
LetT:P→E be the operator defined by
(3.8)Tu(t)=∫01G(t,s)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds=:F(t).Lemma 3.4.
T:P→P is completely continuous.Proof.
Letu∈P; in view of nonnegativeness and continuity of G(t,s) and f(t,u,v), we have Tu⩾0, and t∈[0,1] is continuous:
(3.9)(Tu)′(t)=∫01Gt′(t,s)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds=∫0t(α-1)(1-s)α-2-(α-1)(t-s)α-2Γ(α)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds+∫t1(α-1)(1-s)α-2Γ(α)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds.
Clearly,(Tu)′(t) is continuous for α<2.
By Remark2.4 and noting (3.4) and (3.6), we  have
(3.10)D0+2(Tu)(t)=D0+2-αD0+α(Tu)(t)=D0+2-α(D0+αF(t))=-D0+2-α(ϕq(1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ))=-I0+α-2(ϕq(1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ))=-1Γ(α-2)∫0t(t-s)α-3(ϕq(1Γ(γ)∫0t(t-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds)⩽0.
So,Tu is concave on [0,1] and Tu∈C1[0,1]; we obtain T(P)⊂P.
LetΩ∈P be bounded; that is, there exists a positive constant M>0 such that ∥u∥⩽M for all u∈Ω.
LetN=max(t,u,v)∈[0,1]×[0,M]×[-M,M]|f(t,u(t),D0+ρu(t)|;  then, for  all u∈Ω, we have
(3.11)|Tu(t)|=|∫01G(t,s)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds|⩽(NΓ(γ+1))q-1∫01G(1,s)ds=(NΓ(γ+1))q-1[1Γ(α)-1Γ(α+1)]⩽(NΓ(γ+1))q-11Γ(α),|Tu'(t)|=|∫01Gt′(t,s)ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds|⩽(NΓ(γ+1))q-1∫01Gt′(t,s)ds⩽(NΓ(γ+1))q-11Γ(α),|D0+ρ(Tu)(t)|=|1Γ(1-ρ)∫0t(t-s)-ρ(Tu)′(s)ds|⩽1Γ(1-ρ)∫0t(t-s)-ρ×|∫01Gs′(s,v)ϕq(1Γ(γ)∫0v(v-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)|ds⩽1Γ(1-ρ)(NΓ(γ+1))q-11Γ(α)∫0t(t-s)-ρds=(NΓ(γ+1))q-1t1-ρ(1-ρ)Γ(1-ρ)Γ(α)⩽(NΓ(γ+1))q-11Γ(2-ρ)Γ(α).
So, for  all u∈Ω,   ∥Tu(t)∥⩽(N/Γ(γ+1))q-1[2/Γ(α)+1/Γ(2-ρ)Γ(α)].  Hence, T(Ω) is uniformly  bounded.
SinceG(t,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,(3.12)|G(t1,s)-G(t2,s)|<ε(Γ(γ+1)N)q-1.
Therefore,
(3.13)|Tu(t2)-Tu(t1)|⩽∫01|G(t2,s)-G(t1,s)|ϕq(1Γ(γ)∫0s(s-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)ds⩽(NΓ(γ+1))q-1∫01|G(t2,s)-G(t1,s)|ds<ε.
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(3.14)A=(1Γ(γ+1))q-11Γ(α),B=(1k)γ(q-1)(1Γ(γ+1))q-1∫1/k1G(1,s)ds,M=(1Γ(γ+1))q-11Γ(α)Γ(2-ρ).Theorem 3.5.
Suppose that there exist constants0<r1<b<kb⩽r2,  L2⩾L1>0, such that kb/B⩽min(r2/A,L2/2A,L2/2M), and the following conditions hold:(H1)
f(t,u,v)<min{ϕp(r1/A),ϕp(L1/2A),ϕp(L1/2M)},  for (t,u,v)∈[0,1]×[0,r1]×[-L1,L1];(H2)
f(t,u,v)>ϕp(kb/B),  for  (t,u,v)∈[1/k,1]×[b,kb]×[-L2,L2];(H3)
f(t,u,v)⩽min{ϕp(r2/A),ϕp(L2/2A),ϕp(L2/2M)}, for (t,u,v)∈[0,1]×[0,r2]×[-L2,L2].
Then, the BVP (1.5) has at least three concave positive solutions u1, u2, and u3, such that
(3.15)max0⩽t⩽1|u1(t)|<r1,max0⩽t⩽1|D0+ρu1(t)|+max0⩽t⩽1|u1′(t)|<L1,b<min1/k⩽t⩽(k-1)/k|u2(t)|<max0⩽t⩽1|u2(t)|⩽r2max0⩽t⩽1|D0+ρu2(t)|+max0⩽t⩽1|u2′(t)|<L2,r1<max0⩽t⩽1|u3(t)|⩽kb,min1/k⩽t⩽(k-1)/k|u3(t)|<b,L1<max0⩽t⩽1|D0+ρu3(t)|+max0⩽t⩽1|u3′(t)|<L2.Proof.
By Lemmas3.3 and 3.4, we have that  T:P→P is completely continuous and problem (1.5) has a solution u=u(t) if and only if u(t) satisfies the operator equation u=Tu.
Now, we show that all the conditions of Lemma2.8 hold.
Step 1.
We will show thatT:P¯(φ,r2;θ,L2)→P¯(φ,r2;θ,L2).
Ifu∈P¯(φ,r2;θ,L2), then φ(u)⩽r2, θ(u)⩽L2.
From(H3), we have
(3.16)φ(Tu(t))=max0≤t≤1|(Tu)(t)|=max0≤t≤1|∫01G(t,s)ϕq(1Γ(r)∫0s(s-τ)r-1f(τ,u(τ),D0+ρu(τ))dτ)ds|⩽r2A∫01G(1,s)ϕq(1Γ(r)∫0s(s-τ)r-1dτ)ds⩽r2A[1rΓ(r)]q-1(1Γ(α)-1Γ(α+1))⩽r2A(1Γ(r+1))q-11Γ(α)=r2,max0≤t≤1|(Tu)'(t)|=max0≤t≤1|∫01Gt′(t,s)ϕq(1Γ(r)∫0s(s-τ)r-1f(τ,u(τ),D0+ρu(τ))dτ)ds|⩽L22A[1Γ(γ+1)]q-1∫01Gt′(t,s)ds=L22A(1Γ(r+1))q-11Γ(α)⩽L22,max0≤t≤1|D0+ρ(Tu)(t)|=max0≤t≤1|1Γ(1-ρ)∫0t(t-s)-ρ(Tu)'(s)ds|⩽1Γ(1-ρ)∫0t(t-s)-ρ×|∫01Gs′(s,v)ϕq(1Γ(γ)∫0v(v-τ)γ-1f(τ,u(τ),D0+ρu(τ))dτ)|ds⩽L22M1Γ(1-ρ)(1Γ(γ+1))q-11Γ(α)∫0t(t-s)-ρds=L22M((1Γ(γ+1))q-11Γ(α)Γ(2-ρ))⩽L22.
Then, θ((Tu)(t))⩽L2.
So,T:P¯(φ,r2;θ,L2)→P¯(φ,r2;θ,L2).Step 2.
Letu(t)=kb/2, 0⩽t⩽1. It is easy to see that u(t)=kb/2∈P¯(φ,kb;θ,L2,ψ,b), and ψ(u)=ψ(kb/2)>b. Consequently, {u∈P¯(φ,kb;θ,L2,ψ,b)∣ψ(u)>b}≠ø.
Ifu∈P¯(φ,kb;θ,L2,ψ,b), then for all 1/k⩽t⩽1, b⩽u(t)⩽kb, θ(u(t))⩽L2. By (H2), we obtain f(t,u(t),D0+ρu(t))>ϕp(kb/B), for 1/k⩽t⩽1.
From Lemma3.2, we have
(3.17)ψ(Tu)=min1/k⩽t⩽(k-1)/k|Tu(t)|⩾1kmax0⩽t⩽1|(Tu)(t)|=1kmax0⩽t⩽1|∫01G(t,s)ϕq(1Γ(r)∫0s(s-τ)r-1f(τ,u(τ),D0+ρu(τ))dτ)ds|>kbB1kmax0⩽t⩽1tα-1∫01G(1,s)ϕq(sγγΓ(γ))ds=bB(1Γ(γ+1))q-1∫01G(1,s)ϕq(sγ)ds>bB(1k)γ(q-1)(1Γ(γ+1))q-1∫1/k1G(1,s)ds=b;
that is, ψ(Tu)>b, for all u∈P¯(φ,kb;θ,L2,ψ,b). This shows that condition (C1) of Lemma 2.8 holds.Step 3.
Letu∈P¯(φ,r1;θ,L1); by (H1), we have
(3.18)f(t,u(t),D0+ρu(t))<min{ϕp(r1A),ϕp(L12A),ϕp(L12M)}.
Similarly, we can prove that T:P¯(φ,r1;θ,L1)→P¯(φ,r1;θ,L1). (C2) of Lemma 2.8 holds.Step 4.
Letu∈P¯(φ,kb;θ,L2;ψ,b), and φ(Tu)>kb; we have
(3.19)ψ(Tu)=min1/k⩽t⩽(k-1)/k|Tu(t)|⩾1kmax0⩽t⩽1|Tu(t)|=1kφ(Tu)>b.
(C3) of Lemma 2.8 holds. Therefore, the BVP (1.5) has at least three positive solutions u1, u2,  and u3 satisfying
(3.20)max0⩽t⩽1|u1(t)|⩽r1,max0⩽t⩽1|D0+ρu1(t)|+max0⩽t⩽1|u1′(t)|⩽L1,b<min1/k⩽t⩽(k-1)/k|u2(t)|<max0⩽t⩽1|u2(t)|⩽r2,max0⩽t⩽1|D0+ρu2(t)|+max0⩽t⩽1|u2′(t)|⩽L2,r1<max0⩽t⩽1|u3(t)|⩽kb,min1/k⩽t⩽(k-1)/k|u3(t)|<b,L1<max0⩽t⩽1|D0+ρu3(t)|+max0⩽t⩽1|u3′(t)|⩽L2.
The proof is completed.Corollary 3.6.
If there exist constants0<r1<b1<kb1⩽r2<b2<kb2⩽⋯⩽rn, and 0<L1⩽L2⩽⋯⩽Ln-1, n∈N, such that kbi/B⩽min{ri+1/A,Li+1/2A,Li+1/2M}, for 1⩽i⩽n-1 and the following conditions are satisfied:(I1)
f(t,u,v)<min{ϕp(ri/A),ϕp(Li/2A),ϕp(Li/M)} for (t,u,v)∈[0,1]×[0,ri]×[-Li,Li];(I2)
f(t,u,v)>ϕp(kbi/B),  for  (t,u,v)∈[1/k,1]×[bi,kbi]×[-Li+1,Li+1]; then the problem (1.5) has at least 2n-1 concave positive solutions.Proof.
Ifn=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).
Ifn=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:(3.21)D0+1/2(ϕ3/2(D0+5/2u(t)))+f(t,u(t),D0+1/2u(t))=0,0<t<1,u(0)=u′(1)=u′′(0)=0,D0+αu(t)t=0=0,
where
(3.22)f(t,u,v)={t20+6u2+(|v|4×104)4,foru⩽4,t20+96+(|v|4×104)4,foru>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.
Choosingr1=1/4, b=1, r2=104, L1=2×104, L2=4×104. It is easy to see that kb/B⩽min(r2/A,L2/2A,L2/2M), and f(t,u,v) satisfying(1)
f(t,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)
f(t,u,v)⩾7.0125>ϕp(kb/B)≈6.060,(t,u,v)∈[1/4,1]×[1,4]×[-4×104,4×104],(3)
f(t,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].
By Theorem3.5, problem (3.21)  has at least three concave positive solutions u1, u2, and u3 satisfying
(3.23)max0⩽t⩽1|u1(t)|<14,max0⩽t⩽1|D0+ρu1(t)|+max0⩽t⩽1|u1′(t)|<2×104,1<min1/k⩽t⩽(k-1)/k|u2(t)|<max0⩽t⩽1|u2(t)|⩽104,max0⩽t⩽1|D0+ρu2(t)|+max0⩽t⩽1|u2′(t)|<4×104,14<max0⩽t⩽1|u3(t)|⩽4,min1/k⩽t⩽(k-1)/k|u3(t)|<1,2×104<max0⩽t⩽1|D0+ρu3(t)|+max0⩽t⩽1|u3′(t)|<4×104.



---

*Source: 495138-2010-06-03.xml*