# Positive Solutions for a Mixed-Order Three-Point Boundary Value Problem forp-Laplacian

Authors: Francisco J. Torres
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The author investigates the existence and multiplicity of positive solutions for boundary value problem of fractional differential equation
withp-Laplacian operator. The main tool is fixed point index theory and
Leggett-Williams fixed point theorem.


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## Body

## 1. Introduction

In this paper, we are interested in the existence of single and multiple positive solutions to nonlinear mixed-order three-point boundary value problem forp-Laplacian. Consider the following:
(1)(φp(D0+αu(t)))′+a(t)f(t,u(t))=0,0<t<1,(2)D0+αu(0)=u(0)=u′′(0)=0,u′(1)=γu′(η),
where η,γ∈(0,1), α∈(2,3], D0+α is the Caputo's derivative, φp(s)=|s|p-2s, and φq=(φp)-1 with (1/p)+(1/q)=1. We assume the following conditions throughout.(H1)
f∈C([0,1]×[0,∞),[0,∞)).(H2)
a∈L1(0,1) is nonnegative and a(t)≢0 on any subinterval of (0,1).The equation with ap-Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics, nonlinear elasticity and glaciology, combustion theory, population biology, nonlinear flow laws, and so on. Liang et al. in [1] used the fixed point theorem of Avery and Henderson to show the existence of at least two positive solutions. Zhao et al. [2] studied the existence of at least three positive solutions by using Leggett-Williams fixed point theorem. Chai [3] obtain results for the existence of at least one nonnegative solution and two positive solutions by using fixed point theorem on cone. Su et al. [4] studied the existence of one and two positive solutions by using the fixed point index theory. Su [5] studied the existence of one and two positive solution by using the method of defining operator by the reverse function of Green function and the fixed point index theory. Tang et al. [6] studied the existence of positive solutions of fractional differential equation with p-laplacian by using the coincidence degree theory.Motivated by the above works, we obtain some sufficient conditions for the existence of at least one and three positive solutions for (1) and (2).The organization of this paper is as follows. In Section2, we present some necessary definitions and preliminary results that will be used to prove our results. In Section 3, we discuss the existence of at least one positive solution for (1) and (2). In Section 4, we discuss the existence of multiple positive solutions for (1) and (2). Finally, we give some examples to illustrate our results in Section 5.


## 2. Preliminaries

Definition 1.
LetE be a real Banach space. A nonempty closed convex set K⊂E is called cone if(1)
x∈K, λ≥0 then λx∈K,(2)
x∈K, -x∈K then x=0.Definition 2.
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.Remark 3.
By the positive solution of (1) and (2) we understand a function u(t) which is positive on [0,1] and satisfies the differential equation (1) and the boundary conditions (2).
We will consider the Banach spaceE=C[0,1] equipped with standard norm:
(3)∥u∥=max0≤t≤1|u(t)|
The proof of existence of solution is based upon an application of the following theorem.Theorem 4 (see [7, 8]).
LetE be a Banach space and let K be a cone of E. For r>0, define Kr={u∈K:∥u∥≤r} and assume that T:Kr→K is a completely continuous operator such that Tu≠u for all u∈∂Kr. (1)
If∥Tu∥≤∥u∥ for all u∈∂Kr, then i(T,Kr,K)=1.(2)
If∥Tu∥≥∥u∥ for all u∈∂Kr, then i(T,Kr,K)=0.Lemma 5 (see [9]).
Letn∈ℕ with n≥2, n-1<α≤n. If u∈Cn-1[a,b] and Da+αu∈C(a,b), then
(4)(Ia+αDa+αu)(t)=u(t)-∑k=0n-1‍u(k)(a)k!(t-a)k
holds on (a,b).Lemma 6.
The three-point boundary value problem (1)-(2) has a unique solution
(5)u(t)=∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+γt1-γ∫01‍G2(η,s)×φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds,
where
(6)G1(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,G2(η,s)={(α-1)(1-s)α-2-(α-1)(η-s)α-2Γ(α),0≤s≤η≤1,(α-1)(1-s)α-2Γ(α),0≤η≤s≤1.Proof.
Integrating both sides of (1) on [0,1], we have
(7)φp(Dαu(t))-φp(Dαu(0))=-∫0t‍a(s)f(s,u)ds.
So
(8)Dαu(t)=-φq(∫0t‍a(s)f(s,u)ds).
From Lemma 5, we have
(9)u(t)=-1Γ(α)∫0t‍(t-s)α-1φq(∫0s‍a(τ)f(τ,u)dτ)ds+A+Bt+Ct2,u′(t)=-1Γ(α-1)∫0t‍(t-s)α-2φq(∫0s‍a(τ)f(τ,u)dτ)ds+B+2Ct,u′′(t)=-1Γ(α-2)∫0t‍(t-s)α-3φq×(∫0s‍a(τ)f(τ,u)dτ)ds+2C.
From (2), A=0 and C=0.
Now, consider the following:(10)u′(1)=-1Γ(α-1)∫01‍(1-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds+B,u′(η)=-1Γ(α-1)∫0η‍(η-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds+B,
by the boundary value condition u′(1)=γu′(η), we have
(11)B=1(1-γ)Γ(α-1)∫01‍(1-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds-γ(1-γ)Γ(α-1)∫0η‍(η-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds,
so
(12)u(t)=-1Γ(α)∫0t‍(t-s)α-1φq(∫0s‍a(τ)f(τ,u)dτ)ds+t(1-γ)Γ(α-1)×∫01‍(1-s)α-2φq(∫0s‍a(τ)f(τ,u)dτ)ds-γt(1-γ)Γ(α-1)×∫0η‍(η-s)α-2φq(∫0s‍a(τ)f(τ,u)dτ)ds.
Splitting the second integral in two parts of the form
(13)tΓ(α-1)+k(1-γ)Γ(α-1)=t(1-γ)Γ(α-1),
we have k=γt; thus,
(14)u(t)=-1Γ(α)∫0t‍(t-s)α-1φq×(∫0s‍a(τ)f(τ,u)dτ)ds+tΓ(α-1)∫01‍(1-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds+γt(1-γ)Γ(α-1)∫01‍(1-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds-γt(1-γ)Γ(α-1)∫0η‍(η-s)α-2φq×(∫0s‍a(τ)f(τ,u)dτ)ds;
therefore,
(15)u(t)=∫0t‍(t(1-s)α-2Γ(α-1)-(t-s)α-1Γ(α))φq×(∫0s‍a(τ)f(τ,u)dτ)ds+∫t1‍t(1-s)α-2Γ(α-1)φq(∫0s‍a(τ)f(τ,u)dτ)ds+γt1-γ∫0η‍((1-s)α-2Γ(α-1)-(η-s)α-2Γ(α-1))φq×(∫0s‍a(τ)f(τ,u)dτ)ds+γt1-γ∫η1‍(1-s)α-2Γ(α-1)φq(∫0s‍a(τ)f(τ,u)dτ)ds.
This completes the proof.Lemma 7.
Letβ∈(0,1) be fixed. The kernel, G1(t,s), satisfies the following properties.(1)
0≤G1(t,s)≤G1(1,s) for all s∈(0,1),(2)
minβ≤t≤1G1(t,s)≥βG1(1,s) for all s∈[0,1].Proof.
(
1
) As 2<α≤3 and 0≤s≤t≤1, we have
(16)(α-1)t(1-s)α-2>t(1-s)α-2≥(t-s)(t-s)α-2=(t-s)α-1;
thus, G1(t,s)>0. Note ∂G1(t,s)/∂t≥0 then, G1(t,s) is increasing as a function of t; therefore,
(17)G1(t,s)≤G1(1,s)∀s∈[0,1].
(
2
) For β≤t≤1, we have
(18)minβ≤t≤1G1(t,s)=G1(β,s),
where
(19)G1(β,s)={(α-1)β(1-s)α-2-(β-s)α-1Γ(α)0≤s≤β,(α-1)β(1-s)α-2Γ(α)β≤s≤1.
(a) If 0<s≤β,
(20)minβ≤t≤1G1(t,s)=β(α-1)(1-s)α-2Γ(α)-(β-s)α-1Γ(α);
on the other hand,
(21)βG1(1,s)=β(α-1)(1-s)α-2Γ(α)-β(1-s)α-1Γ(α).
Since 2<α≤3 and(i)
α-1>1, β∈(0,1)⇒βα-1<β,(ii)
s≤β⇒s/β≤1⇒1-(s/β)≥0,(iii)
β<1⇒1<1/β⇒-s(1/β)<-s⇒1-s(1/β)<1-s;
thus, we have(22)(1-sβ)α-1<(1-s)α-1.
From (20), we obtain
(23)(β-s)α-1=(β(1-sβ))α-1=βα-1(1-sβ)α-1≤β(1-sβ)α-1<β(1-s)α-1.
It follows from (20), (21), and (23), that item 2 in the proof hold.
(b) Ifβ≤s<1,
(24)minβ≤t≤1G1(t,s)=β(α-1)(1-s)α-2Γ(α),βG1(1,s)=β(α-1)(1-s)α-2Γ(α)-β(1-s)α-1Γ(α).
It follows from (24) that item 2 in the proof holds.Lemma 8 (see [10]).
The unique solutionu(t) of (1), (2) is nonnegative and satisfies
(25)minβ≤t≤1u(t)≥β∥u∥.Define the coneK by
(26)K={u∈C[0,1]:u(t)≥0,minβ≤t≤1u(t)≥β∥u∥}
and the operator T:K→E by
(27)Tu(t)=∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+γt1-γ∫01‍G2(η,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds.Remark 9.
By Lemma6, the problem (1)-(2) has a positive solution u(t) if and only if u is a fixed point of T.Lemma 10.
T is completely continuous and T(K)⊆K.Proof.
By Lemma8, T(K)⊆K. In view of the assumption of nonnegativeness and continuity of functions Gi(x,y) with i=1,2 and a(t)f(t,u(t)), we conclude that T:K→K is continuous.
LetΩ⊆K be bounded; that is, there exists M>0 such that ∥u∥≤M for all u∈Ω.
Let(28)L=max0≤t≤1,0≤u≤M|f(t,u)|.
Then fromu∈Ω and from Lemmas 6 and 7, we have
(29)|Tu(t)|=|∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u)dτ)ds+γt1-γ∫01‍G2(η,s)×φq(∫0s‍a(τ)f(τ,u)dτ)ds∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u)dτ)|≤∫01‍G1(1,s)φq(∫01‍a(τ)Ldτ)ds+γ1-γ∫01‍G2(η,s)×φq(∫01‍a(τ)Ldτ)ds≤Lq-1φq(∫01‍a(τ)dτ)×[∫01‍(α-1)Γ(α)(1-s)α-2ds+γ1-γ×∫01‍(α-1)Γ(α)(1-s)α-2ds]=(α-1)Lq-1(1-γ)Γ(α)φq(∫01‍a(τ)dτ)∫01‍(1-s)α-2ds=Lq-1(1-γ)Γ(α)φq(∫01‍a(τ)dτ)≐l.
Hence, T(Ω) is bounded.
On the other hand, letu∈Ω, t1,t2∈[0,1] with t1<t2; Then
(30)|Tu(t2)-Tu(t1)|≤Lq-1[∫01‍|G1(t2,s)-G1(t2,s)|φq×(∫0s‍a(τ)f(τ,u)dτ)ds]+Lq-1γ|t2-t1|1-γ×∫01‍G2(η,s)φq(∫0s‍a(τ)f(τ,u)dτ)ds.
The continuity of G1 implies that the right-side of the above inequality tends to zero if t2→t1. Therefore, T is completely continuous by Arzela-Ascoli theorem.
We introduce the notation(31)fa:=liminfu→aminβ≤t≤1f(t,u)up-1,fb:=limsupu→bmax0≤t≤1f(t,u)up-1,
where a,b=0+,∞. Consider the following:
(32)Λ1=[β(∫β1‍G1(1,s)φq(∫0s‍a(τ)dτ)ds+γ1-γ×∫β1‍G2(η,s)φq(∫0s‍a(τ)dτ)ds)]-1,Λ2=[1(1-γ)Γ(α)φq(∫01‍a(τ)dτ)]-1.


## 3. Single Solutions

In what follows, the numberβ∈(0,1).Theorem 11.
Suppose that conditions (H1) and (H2) hold. Assume thatf also satisfy (A1)f(t,u)≥(mr)p-1, βr≤u≤r, β≤t≤1,(A2)f(t,u)≤(MR)p-1, 0≤u≤R, 0≤t≤1,
wherem∈(Λ1,∞) and M∈(0,Λ2). Then (1)-(2) has at least one positive solution u such that r≤∥u∥≤R.Proof.
Without loss of generality, we suppose thatr<R. For any u∈K, we have
(33)u(t)≥β∥u∥,β≤t≤1.
We define two open subsets of E,
(34)Ω1={u∈K:∥u∥<r},Ω2={u∈K:∥u∥<R}.
For u∈∂Ω1, by (33), we have
(35)r=∥u∥≥u(t)≥β∥u∥=βr,t∈[β,1].
For t∈[β,1], by (A1), (27), and Lemma 7, we have
(36)Tu(t)=∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+γt1-γ∫01‍G2(η,s)φq×(∫0s‍a(τ)f(τ,u(τ))dτ)ds≥∫β1‍βG1(1,s)φq(∫0s‍a(τ)(mr)p-1dτ)ds+γβ1-γ∫β1‍G2(η,s)φq(∫0s‍a(τ)(mr)p-1dτ)ds=rmβ[∫β1‍G1(1,s)φq(∫0s‍a(τ)dτ)ds+γ1-γ∫β1‍G2(η,s)φq(∫0s‍a(τ)dτ)ds]>r=∥u∥.
Therefore,
(37)∥Tu∥>∥u∥,∀u∈∂Ω1∩K.
Then by Theorem 4,
(38)i(T,Ω1,K)=0.
On the other hand, asu∈∂Ω2, we have
(39)0≤u(t)≤∥u∥=R,
thus, by (A2), (27), Lemma 7, and (29), we have
(40)Tu(t)=∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+γt1-γ∫01‍G2(η,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds≤MR[∫01‍G1(t,s)φq(∫0s‍a(τ)dτ)ds+γt1-γ∫01‍G2(η,s)φq(∫0s‍a(τ)dτ)ds]≤MR[∫01‍G1(t,s)φq(∫01‍a(τ)dτ)ds+γt1-γ∫01‍G2(η,s)φq(∫01‍a(τ)dτ)ds]≤MR(1-γ)Γ(α)φq(∫01‍a(τ)dτ)ds<R=∥u∥.
Therefore,
(41)∥Tu∥<∥u∥,∀u∈∂Ω2∩K.
Then by Theorem 4,
(42)i(T,Ω1,K)=1.
By (38) and (42):
(43)i(T,Ω2∖Ω1¯,K)=1.
Then, T has a fixed point u∈Ω2∖Ω1¯, u is positive solution of problem (1)-(2), and r<∥u∥<R.Corollary 12.
Suppose that conditions (H1) and (H2) hold. Assume thatf also satisfy (A3)  f∞=λ∈((2Λ1/β)p-1,∞),(A4)  f0=ψ∈[0,(Λ2/2)p-1).
Then (1)-(2) has at least one positive solution u such that r≤∥u∥≤R.Proof.
By(A4), for ɛ=(Λ2/2)p-1-ψ, there exists a suitably small positive number H1, as 0<u≤H1 and 0≤t≤1, such that
(44)f(t,u)≤(ψ+ɛ)up-1≤(Λ22)p-1up-1≤(Λ22H1)p-1.
Let R=H1 and M=(Λ2/2)∈(0,Λ2), then, by (44), condition (A2) holds.
By(A3), for ɛ=λ-(2Λ1/β)p-1, there exists a sufficiently large r≠R such that
(45)f(t,u)up-1≥λ-ɛ=(2Λ1β)p-1,βr≤u<∞,β≤t≤1.
Thus, when βr≤u≤r, one has
(46)f(t,u)≥(2Λ1β)p-1up-1≥(2Λ1r)p-1.
Let m=2Λ1∈(Λ1,∞). Then, by (46), condition (A1) holds.
Hence, from Theorem11 the desired result hold.Corollary 13.
Suppose that conditions (H1) and (H2) hold. Assume thatf also satisfy (A5)  f0=λ∈((2Λ1/β)p-1,∞),(A6)  f∞=ψ∈[0,(Λ2/2)p-1).
Then (1)-(2) has at least one positive solution u such that r≤∥u∥≤R.Proof.
By(A5), for ɛ=λ-(2Λ1/β)p-1, there exists a sufficiently small r>0, such that
(47)f(t,u)up-1≥λ-ɛ=(2Λ1β)p-1,0<u<r,β≤t≤1.
Thus, when βr≤u≤r, we have
(48)f(t,u)≥(2Λ1β)p-1up-1≥(2Λ1r)p-1.
Let m=2Λ1∈(Λ1,∞). Then by (48), condition (A1) holds.
By(A6), for ɛ=(Λ2/2)p-1-ψ, there exist a sufficiently large H2≠r such that
(49)f(t,u)up-1≤ψ+ɛ=(Λ22)p-1,H2≤u<∞,0≤t≤1.
We consider the following two cases.
(a) Suppose thatf(t,u) is unbounded, then we know from (H1) that there is a R≠r(>H2) such that
(50)f(t,u)≤f(t,R),0≤u≤R,0≤t≤1.
Since R>H2, then from (49) and (50) we have
(51)f(t,u)≤f(t,R)≤(Λ22R)p-1,H2≤u≤R,0≤t≤1.
Letting M=(Λ2/2)∈(0,Λ2), we have
(52)f(t,u)≤(MR)p-1,0≤u≤R,0≤t≤1.
Thus, (A2) holds.
(b) Suppose thatf(t,u) is bounded, say
(53)f(t,u)≤Lp-1,0≤u<∞,0≤t≤1.
In this case, taking sufficiently large R≥(2/Λ2)L, then letting M=Λ2/2∈(0,Λ2), we have
(54)f(t,u)≤Lp-1≤(Λ22R)p-1≤(MR)p-1,0≤u≤R,0≤t≤1.
Thus, (A2) holds.
Hence, from Theorem11 the desired result holds.


## 4. Multiple Solutions

To show the existence of multiple solutions we will use the Leggett-Williams fixed point theorem [11]. To this end define the following subsets of a cone K as
(55)Kc={u∈K:∥u∥<c},K(ψ,b,d)={u∈K:b≤ψ(u),∥u∥≤d}.Definition 14.
A mapα:K→[0,+∞) is said to be a nonnegative continuous concave functional on a cone K of a real Banach space E if α is continuous and
(56)α(tx+(1-t)y)≥tα(x)+(1-t)α(y)
for allx,y∈K and t∈[0,1].Theorem 15 (see [11]).
SupposeT:Kc¯→Kc¯ is completely continuous and suppose that there exists a concave positive functional ψ on K such that ψ(u)≤∥u∥ for u∈Kc¯. Suppose that there exist constants 0<a<b<d≤c such that(B1)
{u∈K(ψ,b,d):ψ(u)>b}≠∅ and ψ(Tu)>b if u∈K(ψ,b,d),(B2)
∥Tu∥<a if u∈Ka,(B3)
ψ(Tu)>b for u∈K(ψ,b,c) with ∥Tu∥>d.
Then,T has at least three fixed points u1, u2, and u3 such that ∥u1∥<a, b<ψ(u2) and ∥u3∥>a with ψ(u3)<b.Theorem 16.
Suppose that there exista, b, c with 0<a<βb<b≤c such that(C1)
f(t,u)<(aΛ2)p-1, (t,u)∈[0,1]×[0,a],(C2)
f(t,u)>(bΛ1)p-1, (t,u)∈[β,1]×[βb,b],(C3)
f(t,u)<(cΛ2)p-1, (t,u)∈[0,1]×[0,c].
Then (1)-(2) has at least three positive solutions.Proof.
By Lemma10, T:K→K is completely continuous.
Let(57)ψ(u)=minβ≤t≤1u(t).
It is obvious that ψ is a nonnegative continuous concave functional on K with ψ(u)≤∥u∥, for u∈Kc¯. Now we will show that the conditions of Theorem 15 are satisfied. For u∈Kc¯, then ∥u∥≤c. For t∈[0,1] by (27), (29), and (C3), one has
(58)Tu(t)=∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u)dτ)ds+γt1-γ×∫01‍G2(η,s)φq(∫0s‍a(τ)f(τ,u)dτ)ds<∫01‍G1(t,s)φq(∫01‍a(τ)(cΛ2)p-1dτ)ds+γ1-γ×∫01‍G2(η,s)φq(∫01‍a(τ)(cΛ2)p-1dτ)ds<cΛ2(1-γ)Γ(α)φq(∫01‍a(τ)dτ)=c.
This implies T:Kc¯→Kc. By the same method, if u∈Ka¯, then we can get ∥Tu∥<a and therefore (B2) is satisfied. Next, we assert that {u∈K(ψ,βb,b):ψ(u)>βb}≠ϕ and ψ(Tu)>βb for all u∈K(ψ,βb,b). In fact, the constant function
(59)βb+b2∈{u∈K(ψ,βb,b):ψ(u)>βb}.
On the other hand, for u∈K(ψ,βb,b), we have
(60)βb≤ψ(u)=minβ≤t≤1u(t)≤u(t)≤∥u∥≤b,∀t∈[β,1].
Thus, in view of (27), Lemma 7, and (C2), one has
(61)ψ(Tu)=minβ≤t≤1[∫01‍G1(t,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+γt1-γ×∫01‍G2(η,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds]≥∫β1‍βG1(1,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds+βγ1-γ∫β1‍G2(η,s)φq(∫0s‍a(τ)f(τ,u(τ))dτ)ds>bΛ1β[∫β1‍G1(1,s)φq(∫0s‍a(τ)dτ)ds+γ1-γ∫β1‍G2(η,s)φq(∫0s‍a(τ)dτ)ds]=b>βb
as required and therefore (B1) is satisfied.
Finally, we assert that ifu∈K(ψ,βb,c) with ∥Tu∥>b then ψ(u)>βb. To see this, suppose that u∈K(ψ,βb,c) and ∥Tu∥>b, then it follows from Lemma 10 that
(62)ψ(Tu)=minβ≤t≤1(Tu)(t)≥β∥Tu∥>βb.
Thus, (B3) is satisfied.
Therefore, by the conclusion of Theorem15, the operator T has at least three fixed points. This implies that (1)-(2) has at least three solutions.


## 5. Examples

Example 1.
Consider the boundary value problem withp-Laplacian:
(63)(φp(D0+5/2u(t)))′+14t-1/2au1/2e2ub+eu+e2u=0,0<t<1,D0+5/2u(0)=u(0)=u′′(0)=0,u′(1)=89u′(12),
where α=5/2, η=1/2, γ=8/9, β=1/2, p=3/2, q=3, f(t,u)=au1/2e2u/(b+eu+e2u), a(t)=(1/4)t-1/2, and φ3(∫01‍a(t)dt)=1/4, φ3(∫0s‍a(t)dt)=(1/4)s.
Then(64)f0=ψ=ab+2,f∞=λ=a.
Next,
(65)Λ1=[12(∫1/21‍G1(1,s)14sds+8∫1/21‍G2(12,s)14sds)]-1=[9352π]-1≈4.873999426.
Then, for a=5 we have f∞∈((2Λ1/β)p-1,∞), so condition (A3) holds.
Now, Consider(66)Λ2=π3≈0.5908179503.
Then, for a=5 and b=8 we have f0∈[0,(Λ2/2)p-1), so condition (A4) holds. Therefore, by Corollary 12, (63) has at least one positive solution.Example 2.
Consider the boundary value problem withp-Laplacian:
(67)(φp(D0+5/2u(t)))′+14t-1/2f(t,u)=0,0<t<1,D0+5/2u(0)=u(0)=u′′(0)=0,u′(1)=89u′(12),
where α=5/2, η=1/2, γ=8/9, β=1/2, p=3/2, q=3, a(t)=(1/4)t-1/2, φ3(∫01‍a(t)dt)=1/4, and φ3(∫0s‍a(t)dt)=(1/4)s.
Let(68)f(t,u)={t40+14u2,u≤113+t40+u1/4,u>1.
By Example 1, we have Λ1≈4.873999426 and Λ2≈0.5908179503.
Choosinga=1/14, b=18, and c=1296, then
(69)f(t,u)=t40+14u2≈0.096…<(aΛ2)1/2≈0.20…,(t,u)∈[0,1]×[0,114],f(t,u)=t40+13+u1/4≈14.7…>(9Λ1)1/2≈9.3…,(t,u)∈[12,1]×[9,18],f(t,u)=t40+13+u1/4≈19…<(1296Λ2)1/2≈27…,(t,u)∈[0,1]×[0,1296].
Then the conditions (C1)–(C3) are satisfied. Therefore, it follows from Theorem 16 that (67) has at least three positive solutions u1, u2, and u3 such that
(70)∥u1∥<114,18<ψ(u2),∥u3∥>114withψ(u3)<18.



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*Source: 912576-2013-11-26.xml*