We consider the existence and multiplicity of concave positive solutions for boundary value problem of nonlinear fractional differential equation with p-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.
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: 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: 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: 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: 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 with p-Laplacian operator as follows: 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 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.
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
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 order α>0 of a continuous function y:(0,∞)→R is given by
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 order α>0 of a function y:(0,∞)→R is defined as
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:
D0+αD0+βy(t)=D0+α+βy(t), α>0, β>0;
D0+-αy(t)=I0+αy(t), α>0;
I0+α:C[0,1]→C[0,1], 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
au∈P, for all u∈P, a⩾0;
u,-u∈P, implies u=0.
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-λ)y)⩾λψ(x)+(1-λ)ψ(y),
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-λ)y)⩽λφ(x)+(1-λ)φ(y),
for all x,y∈P and 0⩽λ⩽1.
Suppose that φ,θ:P→[0,+∞) are two nonnegative continuous convex functionals satisfying
∥u∥⩽Mmax{φ(u),θ(u)},foru∈P,
where M is a positive constant and
Ω={u∈P∣φ(u)<r,θ(u)<L}≠⌀,forr>0,L>0.
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∣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]).
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 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:
{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);
φ(Tu)<r1, θ(Tu)<L1, for all u∈P¯(φ,r1;θ,L1);
ψ(Tu)>b, for all u∈P¯(φ,r2;θ,L2;ψ,b) with φ(Tu)>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)∣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 that u∈C(0,1)∩L(0,1) with a fractional derivative of order α that belongs to C(0,1)∩L(0,1). Then
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.
Let y∈C[0,1]; then the boundary value problem
D0+αu(t)+y(t)=0,0<t<1,2<α<3,u(0)=u′(1)=u′′(0)=0
has a unique solution
u(t)=∫01G(t,s)y(s)ds,
where
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 Lemma 2.9 to reduce (2.10) to an equivalent integral equation:
u(t)=-1Γ(α)∫0t(t-s)α-1y(s)ds+C0+C1t+C2t2.
From (2.14), we have
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
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 function G(t,s) defined by (2.13) satisfies the following conditions:
G(t,s)⩾0, G(t,s)⩽G(1,s) for 0⩽t, s⩽1,
G(t,s)⩾tα-1G(1,s), for 0⩽t, s⩽1.
Proof.
Since
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
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, if t⩾s, then,
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).
Let E=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 ∥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 cone P⊂E by P={u∈E∣u(t)⩾0,u(t)isconcaveon[0,1]}.
Let the nonnegative continuous concave functional ψ and the nonnegative continuous convex functionals φ, θ be defined on the cone P by
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]).
Let u∈P, k⩾3; then
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
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
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,
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:
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).
Let T:P→E be the operator defined by
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.
Let u∈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:
(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 Remark 2.4 and noting (3.4) and (3.6), we have
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.
Let Ω∈P be bounded; that is, there exists a positive constant M>0 such that ∥u∥⩽M for all u∈Ω.
Let N=max(t,u,v)∈[0,1]×[0,M]×[-M,M]|f(t,u(t),D0+ρu(t)|; then, for all u∈Ω, we have
|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.
Since G(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,|G(t1,s)-G(t2,s)|<ε(Γ(γ+1)N)q-1.
Therefore,
|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.
Suppose that there exist constants 0<r1<b<kb⩽r2, L2⩾L1>0, such that kb/B⩽min(r2/A,L2/2A,L2/2M), and the following conditions hold:
f(t,u,v)<min{ϕp(r1/A),ϕp(L1/2A),ϕp(L1/2M)}, for (t,u,v)∈[0,1]×[0,r1]×[-L1,L1];
f(t,u,v)>ϕp(kb/B), for (t,u,v)∈[1/k,1]×[b,kb]×[-L2,L2];
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
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 Lemmas 3.3 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 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 φ(u)⩽r2, θ(u)⩽L2.
From (H3), we have
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.
Let u(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.
If u∈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 Lemma 3.2, we have
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.
Let u∈P¯(φ,r1;θ,L1); by (H1), we have
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.
Let u∈P¯(φ,kb;θ,L2;ψ,b), and φ(Tu)>kb; we have
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
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 constants 0<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:
f(t,u,v)<min{ϕp(ri/A),ϕp(Li/2A),ϕp(Li/M)} for (t,u,v)∈[0,1]×[0,ri]×[-Li,Li];
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.
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:
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
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.
Choosing r1=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
By Theorem 3.5, problem (3.21) has at least three concave positive solutions u1, u2, and u3 satisfying
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.
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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