# Multiple Positive Solutions of a Singular Semipositone  Integral Boundary Value Problem for Fractionalq-Derivatives Equation

Authors: Yulin Zhao; Guobing Ye; Haibo Chen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By using the fixed point index theorem, this paper investigates a class of singular semipositone integral boundary value problem for fractionalq-derivatives equations and obtains sufficient conditions for the existence of at least two and at least three positive solutions. Further, an example is given to illustrate the applications of our main results.


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## Body

## 1. Introduction

Studies onq-difference equations appeared already at the beginning of the 20th century in intensive works especially by Jackson [1], Carmichael [2], and other authors such as Poincare, Picard and, Ramanujan [3]. Up to date, it has evolved into a multidisciplinary subject, for example, see [4–7] and the references therein. For some recent work on q-difference equations, we refer the reader to the papers [8–21], and the basic definitions and properties of q-difference calculus can be found in the book [3, 22]. On the other hand, fractional differential equations have gained importance due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, and probability [23]. Many researchers studied the existence of solutions to fractional boundary value problems, for example, [24–35] and the references therein.The fractionalq-difference calculus had its origin in the works by Al-Salam [36] and Agarwal [37]. More recently, perhaps due to the explosion in research within the fractional differential calculus setting, new developments in this theory of fractional q-difference calculus were made, specifically, q-analogues of the integral and differential fractional operators properties such as the Mittag-Leffler function, the q-Laplace transform, and q-Taylor’s formula [3, 13, 22, 38], just to mention some.However, the theory of boundary value problems for nonlinearq-difference equations is still in the initial stage and many aspects of this theory need to be explored.In [17], Ferreira considered a Dirichlet type nonlinear q-difference boundary value problem as follows:
(1)Dqαu(t)+f(u(t))=0,0<t<1,1<α≤2,u(0)=u(1)=0.
By applying a fixed point theorem in cones, sufficient conditions for the existence of nontrivial solutions were enunciated.In other paper, Ferreira [18] studied the existence of positive solutions to nonlinear q-difference boundary value problem as follows:
(2)Dqαu(t)+f(u(t))=0,0<t<1,2<α≤3,u(0)=Dqu(0)=0,Dqu(1)=β≥0.By using a fixed point theorem in a cone, El-Shahed and Al-Askar [19] were concerned with the existence of positive solutions to nonlinear q-difference equation:
(3)CDqαu(t)+a(t)f(u(t))=0,0<t≤1,2<α≤3,u(0)=Dq2u(0)=0,aDqu(1)+bDq2u(1)=0,
where a,b≥0 and CDqα is the fractional q-derivatives of the Caputo type.Recently, Liang and Zhang [20] discussed the following nonlinear q-fractional three-point boundary value problem:
(4)(Dqαu)(t)+f(t,u(t))=0,0<t<1,2<α≤3,u(0)=(Dqu)(0)=0,Dqu(1)=βDqu(η).
By using a fixed point theorem in partially ordered sets, the authors obtained sufficient conditions for the existence and uniqueness of positive and nondecreasing solutions to the above boundary value problem.In [21], Graef and Kong investigated the following boundary value problem with fractional q-derivatives:
(5)(Dqαu)(t)+f(t,u(t))=0,0<t<1,n-1<α≤n,n∈N,(Dqiu)(0)=0,i=0,…,n-2,bDqu(1)=∑j=1m‍ajDqu(tj)+λ,
where λ≥0 is a parameter, and the uniqueness, existence, and nonexistence of positive solutions are considered in terms of different ranges of λ.Furthermore, Ahmad et al. [11] studied the following nonlinear fractional q-difference equation with nonlocal boundary conditions
(6)(DCqαu)(t)=f(t,u(t)),0≤t≤1,1<α≤2,a1u(0)-b1Dqu(0)=c1u(η1),a2u(1)+b2Dqu(1)=c2u(η2),
where CDqα is the fractional q-derivative of the Caputo type, and ai,bi,ci,ηi∈R. The existence of solutions for the problem is shown by applying some well-known tools of fixed point theory such as Banach’s contraction principle, Krasnoselskii’s fixed point theorem, and the Leray-Schauder nonlinear alternative.It is known that the fractionalq-derivative of Riemann-Liouville type played an important role in the development of the theory of fractional q-derivatives and q-integrals and for its applications in pure mathematics, and is a useful tool in the description of nonconservative models. The details can be found in [22].Sinceq-calculus has a tremendous potential for applications [3, 22], we find it pertinent to investigate problems in this field. Motivated by the papers [20, 21, 30], we will deal with the following integral boundary value problem of nonlinear fractional q-derivatives equation:
(7)(Dqαu)(t)+f(t,u(t))=0,t∈(0,1),2<α≤3,
subject to the boundary conditions
(8)u(0)=(Dqu)(0)=0,u(1)=μ∫01‍u(s)dqs,
where q∈(0,1), μ is parameter with 0<μ<[α]q, Dqα is the q-derivative of Riemann-Liouville type of order α,f:[0,1]×R0+→R is continuous and semipositone and may be singular at u=0, in which R0+=(0,+∞),R=(-∞,+∞). In the present work, we investigate the existence of positive solutions for fractional q-derivatives integral boundary value problem (7) and (8) involving the Riemann-Liouville’s fractional derivative, which is different from [11]. We gave the corresponding Green’s function of the boundary value problem (7) and (8), gave some properties of Green’s function, and constructed a cone by properties of Green’s function. Moreover the existence of at least two and three positive solutions to the boundary value problem (7) and (8) is enunciated.


## 2. Preliminaries onq-Calculus and Lemmas

For the convenience of the reader, below we recall some known facts on fractionalq-calculus. The presentation here can be found in, for example, [1, 3, 12, 19, 21, 22].Letq∈(0,1) and define
(9)[a]q=1-qa1-q,a∈R.
The q-analogue of the power function (a-b)n with n∈N0:={0,1,2,…} is
(10)(a-b)(0)=1,(a-b)(n)=∏k=0n-1‍(a-bqk),n∈N,a,b∈R.
More generally, if γ∈R, then
(11)(a-b)(γ)=aγ∏k=0∞‍a-bqka-bqγ+k,a≠0.
Clearly, if b=0 then a(γ)=aγ. The q-gamma function is defined by
(12)Γq(x)=(1-q)(x-1)(1-q)x-1,x∈R∖{0,-1,-2,…},
and satisfies Γq(x+1)=[x]qΓq(x).Theq-derivative of a function f is defined by
(13)(Dqf)(x)=f(qx)-f(x)(q-1)x,(Dqf)(0)=limx→0(Dqf)(x),
and the q-derivatives of higher order by
(14)(Dq0f)(x)=f(x),(Dqnf)(x)=Dq(Dqn-1f)(x),n∈N.
The q-integral of a function f defined in the interval [0,b] is given by
(15)(Iqf)(x)=∫0x‍f(s)dqs=x(1-q)∑k=0∞‍f(xqk)qk,x∈[0,b].
If a∈[0,b] and f is defined in the interval [0,b], then its integral from a to b is defined by
(16)∫ab‍f(s)dqs=∫0b‍f(s)dqs-∫0a‍f(s)dqs.
Similar to that for derivatives, an operator Iqn is given by
(17)(Iq0f)(x)=f(x),(Iqnf)(x)=Iq(Iqn-1f)(x),n∈N.
The fundamental theorem of calculus applies to these operators Iq and Dq, that is,
(18)(DqIqf)(x)=f(x),
and if f is continuous at x=0, then
(19)(IqDqf)(x)=f(x)-f(0).
The following formulas will be used later, namely, the integration by parts formula:
(20)∫0x‍f(s)(Dqg)(s)dqs=[f(s)g(s)]s=0s=x-∫0x‍(Dqf)(s)g(qs)dqs,[a(t-s)](γ)=aγ(t-s)(γ),Dqt(t-s)(γ)=[γ]q(t-s)(γ-1),(21)Dqs(t-s)(γ)=[γ]q(t-qs)(γ-1),(22)(Dqx∫0x‍f(x,s)dqs)(x)=∫0xDxqf(x,s)dqs+f(qx,x),
where tDq denotes the derivative with respect to the variable t.Definition 1.
Letα≥0 and f be a function defined on [0,1]. The fractional q-integral of Riemann-Liouville type is (Iq0f)(x)=f(x) and
(23)(Iqαf)(x)=1Γq(α)∫0x‍(x-qs)(α-1)f(s)dqs,α>0,x∈[0,1].
The fractionalq-derivative of order α≥0 is defined by (Dq0f)(x)=f(x) and (Dqαf)(x)=(DqkIqk-αf)(x) for α>0, where k is the smallest integer greater than or equal to α.Remark 2.
Letα≥0 and u,v be two functions defined on [0,1], then Dqα(u(t)+v(t))=Dqαu(t)+Dqαv(t).Lemma 3.
Assume thatγ≥0 and a≤b≤t, then (t-a)(γ)≥(t-b)(γ).Lemma 4.
Letα,β≥0 and f be a function defined in [0,1]. Then, the following formulas hold: (1)
(IqβIqαf)(x)=(Iqα+βf)(x),(2)
(DqαIqαf)(x)=f(x).Lemma 5 (see [17]).
Letα>0 and n be a positive integer. Then, the following equality holds: (24)(IqαDqnf)(x)=(DqnIqαf)(x)-∑k=0n-1‍xα-n+kΓq(α+k-n+1)(Dqkf)(0).Lemma 6.
Lety∈C[0,1],q∈(0,1),2<α≤3,1<μ<[α]q. Then the unique solution of the equation
(25)(Dqαu)(t)+y(t)=0,t∈(0,1),
subject to BC (8) is given by
(26)u(t)=∫01‍G(t,qs)y(s)dqs,
where
(27)G(t,s)={(tα-1(1-s)(α-1)([α]q-μ+μqα-1s)-([α]q-μ)(t-s)(α-1))×(([α]q-μ)Γq(α))-1,0≤s≤t≤1,(tα-1(1-s)(α-1)([α]q-μ+μqα-1s))×(([α]q-μ)Γq(α))-1,0≤t≤s≤1.Proof.
Let us putn=3. In view of Definition 1 and Lemma 4, we see that
(28)(Dqαu)(t)=-y(t)⟺(IqαDq3Iq3-αu)(t)=-(Iqαy)(t).
Then, it follows from Lemma 5 that the solution u(t) of (25) and BC (8) is given by
(29)u(t)=c1tα-1+c2tα-2+c3tα-3-∫0t‍(t-qs)(α-1)Γq(α)y(s)dqs,
for some constants c1,c2,c3∈R. Since u(0)=0, we have c3=0. Differentiating both sides of (29) and with the help of (20) and (22), we obtain
(30)(Dqu)(t)=[α-1]qc1tα-2+[α-2]qc2tα-3-∫0t‍[α-1]q(t-qs)(α-2)Γq(α)y(s)dqs.
Then by the boundary conditions (Dqu)(0)=0, we get c2=0. Thus, (29) reduces to
(31)u(t)=c1tα-1-∫0t‍(t-qs)(α-1)Γq(α)y(s)dqs.
Using the boundary condition u(1)=μ∫01‍u(s)dqs, we get
(32)c1=μ∫01‍u(s)dqs+∫01‍(1-qs)(α-1)Γq(α)y(s)dqs.
Hence, we have
(33)u(t)=tα-1(μ∫01‍u(s)dqs+∫01‍(1-qs)(α-1)Γq(α)y(s)dqs)-∫0t‍(t-qs)(α-1)Γq(α)y(s)dqs.
Integrate the above equation (33) from 0 to 1, and using (11), (19) and (20), we obtain
(34)∫01‍u(t)dqt=∫01‍tα-1(μ∫01‍u(s)dqs)dqt+∫01‍tα-1(∫01‍(1-qs)(α-1)Γq(α)y(s)dqs)dqt-∫01‍(∫0t‍(t-qs)(α-1)Γq(α)y(s)dqs)dqt=(1-q)μ1-qα∫01‍u(s)dqs+1-q1-qα∫01‍(1-qs)(α-1)Γq(α)y(s)dqs-∫01‍(1-qs)(α)[α]qΓq(α)y(s)dqs=μ[α]q∫01‍u(s)dqs+∫01‍(1-qs)(α-1)qαs[α]qΓq(α)y(s)dqs,
then
(35)∫01‍u(t)dqt=qα[α]q-μ∫01‍(1-qs)(α-1)sΓq(α)y(s)dqs.
Combining this with (29) and (31) yields
(36)u(t)=μqα([α]q-μ)Γq(α)∫01‍tα-1(1-qs)(α-1)sy(s)dqs+1Γq(α)∫01‍tα-1(1-qs)(α-1)y(s)dqs-1Γq(α)∫0t‍(t-qs)(α-1)y(s)dqs=∫01‍[α]q-μ+μqαs([α]q-μ)Γq(α)·tα-1(1-qs)(α-1)y(s)dqs-1Γq(α)∫0t‍(t-qs)(α-1)y(s)dqs=∫01‍G(t,qs)y(s)dqs.
This completes the proof of the lemma.Remark 7.
For the special case whereμ=0, Lemmas 6 has been obtained by Ferreira [18].Lemma 8.
The functionG(t,s) defined by (27) satisfies the following conditions: (i)
G(t,s) is a continuous function on (t,s)∈[0,1]×[0,1], and G(t,qs)≥0, for (t,s)∈[0,1]×[0,1];(ii)
μqαtα-1ρ(s)≤([α]q-μ)Γq(α)G(t,qs)≤λtα-1,([α]q-μ)Γq(α)G(t,qs)≤λρ(s), for (t,s)∈[0,1]×[0,1], where
(37)ρ(s)=(1-qs)(α-1)s,λ=max{[α-1]q([α]q-μ)+μqα,qα-1[α]q}.Proof.
The continuity ofG is easily checked. On the other hand, when 0≤qs≤t≤1, in view of Lemma 3, we have
(38)([α]q-μ)Γq(α)G(t,qs)=tα-1(1-qs)(α-1)([α]q-μ+μqαs)-([α]q-μ)(t-qs)(α-1)=tα-1(1-qs)(α-1)([α]q-μ+μqαs)-([α]q-μ)tα-1(1-qst)(α-1)≥tα-1(1-qs)(α-1)([α]q-μ+μqαs)-([α]q-μ)tα-1(1-qs)(α-1)=μqαtα-1(1-qs)(α-1)s=μqαtα-1ρ(s)≥0.
Further, since we have(39)G(t,qs)=(([α]q-μ)[tα-1(1-qs)(α-1)-(t-qs)(α-1)]+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1=(([α]q-μ)[(t-tqs)(α-1)-(t-qs)(α-1)]+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1=(([α]q-μ)∫t-qst-tqs‍Dqx(α-1)dqx+μqαtα-1(1-qs)(α-1)s{([α]q-μ)∫t-qst-tqs‍})×(([α]q-μ)Γq(α))-1≤([α-1]q([α]q-μ)(t-tqs)(α-2)×[(t-tqs)-(t-qs)]+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1=(q[α-1]q([α]q-μ)tα-2(1-qs)(α-2)s(1-t)+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1≤(q[α-1]q([α]q-μ)tα-2(1-qs)(α-2)s(1-qs)+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1≤(q[α-1]q([α]q-μ)tα-2(1-qs)(α-2)s(1-qαs)+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1=(q[α-1]q([α]q-μ)tα-2(1-qs)(α-1)s+μqαtα-1(1-qs)(α-1)s)×(([α]q-μ)Γq(α))-1,
we get
(40)([α]q-μ)Γq(α)G(t,qs)≤(q[α-1]q([α]q-μ)+μqα)(1-qs)(α-1)s≤λρ(s),([α]q-μ)Γq(α)G(t,qs)≤[α-1]q([α]q-μ)tα-1(1-qs)(α-1)+μqαtα-1(1-qs)(α-1)s≤([α-1]q([α]q-μ)(1-qs)(α-1)+μqα(1-qs)(α-1)s)tα-1≤([α-1]q([α]q-μ)+μqα)tα-1≤λtα-1.
When 0≤t≤qs≤1, since μ<[α]q, we have
(41)([α]q-μ)Γq(α)G(t,qs)=tα-1(1-qs)(α-1)([α]q-μ+μqαs)≥μqαtα-1(1-qs)(α-1)s=μqαtα-1ρ(s),([α]q-μ)Γq(α)G(t,qs)=tα-1(1-qs)(α-1)([α]q-μ+μqαs)≤qα-1[α]qsα-1(1-qs)(α-1)≤λρ(s).
This completes the proof of the lemma.Remark 9.
If we let0<τ<1, then
(42)mint∈[τ,1]G(t,qs)≥μqατα-1([α]q-μ)Γq(α)ρ(s),fors∈[0,1].
According to [16], we may take τ=qn,n∈N.Lemma 10.
Letp∈C[0,1]. Then the boundary value problem
(43)(Dqαu)(t)+p(t)=0,t∈(0,1),2<α≤3,u(0)=(Dqu)(0)=0,u(1)=μ∫01‍u(s)dqs,
has a unique solution w(t)=∫01‍G(t,qs)p(s)dqs with
(44)w(t)≤λ([α]q-μ)Γq(α)tα-1∫01‍p(s)dqs=λ0tα-1∫01‍p(s)dqs,
where λ0=λ/([α]q-μ)Γq(α).
From [39, Theorem  2.3.1], one has the following definition. Let Q be a retract of real Banach space E, Ω be a relatively bounded open subset of Q,T:Ω→Q be completely continuous operator. The integer i(T,Ω,Q) be defined by
(45)i(T,Ω,Q)=deg(I-T·χ,B(θ,r)∩χ-1(Ω),θ),
where χ:E→Q is an arbitrary retraction and r>0 such that B(θ,r)⊃Ω. Then the integer i(T,Ω,Q) is called the fixed point index of T on Ω with respect to Q.Lemma 11 (see [39]).
LetP be a cone in a Banach space E. Let Ω be an open bounded subset of E with Ωr=Ω∩P≠ϕ and Ω-r≠P. Assume that T:Ω-r→P is a compact map such that u≠Tu for u∈∂Ωr. Then (i)
ifT(Ω-r)⊂Ωr for u∈Ωr, then i(T,Ωr,P)=1;(ii)
if there existsu0∈P∖{θ} such that u≠Tu+λu0 for all u∈∂Ωr and λ>0, then i(T,Ωr,P)=0.


## 3. The Main Results

In order to abbreviate our discussion, we give the following assumptions.(H0)
There existsp(t)∈C[0,1], such that
(46)ϕ(t)h0(u)≤f(t,u)+p(t)≤φ(t)(g(u)+h(u))for all (t,u)∈[0,1]×R0+, where ϕ, φ∈C([0,1],R0+), g∈C(R0+,R0+), and g(u) is nonincreasing with respect to u;h0, h∈(R+,R+) and h0(u), h(u) are nondecreasing with respect to u, where R+=[0,+∞).(H1)
0<∫01‍φ(s)ρ(s)dqs<+∞.LetX=C[0,1] be the Banach space endowed with morm ∥u∥=maxt∈[0,1]|u(t)|, and define the cone Ω∈X by
(47)Ω={u∈X:u(t)≥μqαtα-1λ∥u∥X,q∈(0,1),2<α≤3}.
By a positive solution of BVP (7) and (8), we mean a function u∈C[0,1] such that u(t) satisfies (7) and (8) and u(t)>0 on (0,1).SettingF(t,u)=f(t,u)+p(t) and for any u∈Ω, m∈N, we consider the following singular nonlinear boundary value problem:
(48)(Dqαu)(t)+F(t,[u(t)]*+1m)=0,t∈(0,1),u(0)=(Dqu)(0)=0,u(1)=μ∫01‍u(s)dqs,
where [u(t)]*=max{u(t)-w(t),0}.According to Remark2, we can see that if u(t)≥w(t) for t∈[0,1] is a positive solution of BVP (48), then u-w is a positive solution of BVP (7) and (8).Lemma 12.
For anym∈N, let Tm:Ω→Ω be the operator defined by
(49)(Tmu)(t)=∫01‍G(t,qs)F(s,[u(s)]*+1m)dqs.
Then Tm:Ω→Ω is completely continuous.Proof.
For anyu∈Ω,m∈N, Lemma 8 implies that (Tmu)(t)≥0 on [0,1], and (50)(Tmu)(t)≥μqαtα-1([α]q-μ)Γq(α)∫01‍ρ(s)F(s,[u(s)]*+1m)dqs.
On the other hand,
(51)∥Tmu∥X=max0≤t≤1(Tmu)(t)≤λ([α]q-μ)Γq(α)∫01‍ρ(s)F(s,[u(s)]*+1m)dqs.
Then (Tmu)(t)≥(μqαtα-1/λ)∥Tmu∥X, which leads to Tm(Ω)⊂Ω. Thus Tm:Ω→Ω.
It follows from the nonnegativeness and continuity ofG(t,s) and F that the operator Tm is continuous. Suppose B⊂Ω is any bounded set; then, for any u∈B, there is a constant number M>0 such that ∥u∥X≤M. Let
(52)L(m-1,M,λ0)=λ0[g(1m)+h(M+1)].
for all u∈B, by Lemma 8, we have
(53)|(Tmu)(t)|≤∫01‍|G(t,qs)F(s,[u(s)]*+1m)|dqs≤λ0∫01‍ρ(s)|F(s,[u(s)]*+1m)|dqs≤λ0∫01‍ρ(s)φ(s)(g([u(s)]*+1m)λ0∫01‍ρ(s)φ(s)+h([u(s)]*+1m))dqs≤L(m-1,M,λ0)∫01‍ρ(s)φ(s)dqs<+∞.
Hence, Tm(B) is bounded.
On the other hand, for anyε>0, according to (H1), there is a constant δ>0 such that
(54)∫0δ‍φ(s)ρ(s)dqs<ε6L(m-1,M,λ0),∫1-δ1‍φ(s)ρ(s)dqs<ε6L(m-1,M,λ0).
From the property of continuity of G, there exists η with 0<η<δ such that for any t1,t2∈[0,1], and s∈[0,1], when t2-t1<η we have
(55)|G(t2,qs)-G(t1,qs)|<ε3φ0L(m-1,M,λ0),|(Tmu)(t2)-(Tmu)(t1)|≤∫01‍|G(t2,qs)-G(t1,qs)|F(s,[u(s)]*+1m)dqs≤L(m-1,M,λ0)∫01‍|G(t2,qs)-G(t1,qs)|φ(s)dqs=2L(m-1,M,λ0)(∫0δ‍φ(s)G(s,qs)dqs2L(m-1,M,λ0)+∫1-δ1‍φ(s)G(s,qs)dqs)+L(m-1,M,λ0)∫δ1-δ‍|G(t2,qs)-G(t1,qs)|φ(s)dqs<2ε3+L(m-1,M,λ0)φ0×∫δ1-δ‍|G(t2,qs)-G(t1,qs)|dqs<ε,
where φ0=max{φ(t):δ≤t≤1-δ}. By means of the Arzela-Ascoli Theorem, Tm:Ω→Ω is completely continuous.Theorem 13.
Suppose(H0) and (H1) hold. In addition, assume that the following conditions are satisfied. (C1)
There exists a constantr>2λ2k0/μqα such that
(56)λ0∫01‍ρ(s)φ(s)(g(μqαsα-12λr)+h(r+1))dqs<r,
where k0=∫01‍p(s)dqs/([α]q-μ)Γq(α).(C2)
There exist constantsξ1,ξ2 with ξ2>ξ1>r such that
(57)μ*h0(12ξi)∫τ1‍ρ(s)ϕ(s)dqs>ξi,i=1,2,
where μ*=μqατα-1/([α]q-μ)Γq(α),τ=qn,n∈N.(C3)
(58)limu→+∞h(u)u=0.
Then BVP (7) and (8) has at least two positive solutions u*,u** with r≤∥u*∥X≤ξ1<ξ2≤∥u**∥X.Proof.
First, we prove that(59)i(Tm,Ω0,Ω)=1,
where Ω0={u∈Ω:∥u∥X<r}.
To see this, letu∈Ω∩∂Ω0. Then ∥u∥X=r and u(t)≥(μqα/λ)tα-1r for t∈[0,1]. Now for t∈(0,1), we get
(60)u(t)-w(t)≥u(t)-λ([α]q-μ)Γq(α)tα-1∫01‍p(s)dqs≥u(t)-μqα2λtα-1r≥12u(t)≥μqαtα-12λ∥u∥X=μqαtα-12λr.
So, for any u∈Ω∩∂Ω0,t∈(0,1), we get
(61)μqαtα-12λr≤u(t)-w(t)≤r.
It follows from (C1), (61) and Lemma 8 that, for any u∈Ω∩∂Ω0,
(62)(Tmu)(t)=∫01‍G(t,qs)F(s,[u(s)]*+1m)dqs≤λ0∫01‍ρ(s)φ(s)(g([u(s)]*+1m)λ0∫01‍ρ(s)φ(s)+h([u(s)]*+1m))dqs≤λ0∫01‍ρ(s)φ(s)(g(μqαsα-12λr)+h(r+1))dqs<r=∥u∥X.
This together with (56) yields Tm(Ω-0)⊂Ω0. From the (i) of Lemma 11, (59) is satisfied.
Let us chooseϵ>0 such that
(63)ϵλ0∫01‍ρ(s)φ(s)dqs<1.
Then for the above ϵ, according to (C2) and (C3), there exists R>ξ2>0 such that, for any u≥R,
(64)h(u)≤ϵu.
Take(65)R1=R+(λ0∫01‍ρ(s)φ(s)×(g(μqαsα-12λR)+h(R+1)+ϵ)dqs)×(1-ϵλ0∫01‍ρ(s)φ(s)dqs)-1,
then R1>R>ξ2.
Now letΩ1={u∈Ω:∥u∥X<R1} and ∂Ω1={u∈Ω:∥u∥X=R1}. Then, for any u∈Ω∩∂Ω1, we have
(66)(Tmu)(t)=∫01‍G(t,qs)F(s,[u(s)]*+1m)dqs≤λ0∫01‍ρ(s)φ(s)×(g([u(s)]*+1m)+h([u(s)]*+1m))dqs≤λ0∫01‍ρ(s)φ(s)×({maxu∈[0,R]h([u(s)]*+1m)}g([u(s)]*+1m)+maxu∈[0,R]h([u(s)]*+1m))dqs+λ0∫01‍ρ(s)φ(s)ϵ(∥u∥X+1)dqs≤λ0∫01‍ρ(s)φ(s)×(g(μqαsα-12λR1)+h(R+1))dqs+λ0∫01‍ρ(s)φ(s)ϵ(R1+1)dqs≤λ0∫01‍ρ(s)φ(s)×(g(μqαsα-12λR)+h(R+1)+ϵ)dqs+ϵλ0∫01‍ρ(s)φ(s)dqs·R1<R1=∥u∥X.
This together with (56) yields Tm(Ω-1)⊂Ω1, which implies that
(67)i(Tm,Ω1,Ω)=1.
In the following, let(68)Ω10={u∈Ω:∥u∥<R1,mint∈[τ,1]u(t)>ξ1},Ω11={u∈Ω:∥u∥<R1,mint∈[τ,1]u(t)>ξ2}.
It is easy to see that Ωi,Ω1i,(i=0,1) are bounded sets and satisfy
(69)Ω0⊂Ω1,Ω10⊂Ω1,Ω11⊂Ω10,Ω10∩Ω0=∅.
For any u∈Ω∩∂Ω10, by Remark 9 and (61), we have
(70)u(t)-w(t)≥12u(t)≥12ξ1,for  anyt∈[τ,1].
Thus
(71)mint∈[τ,1](Tmu)(t)=mint∈[τ,1]∫01‍G(t,qs)F(s,[u(s)]*+1m)dqs≥mint∈[τ,1]∫01‍G(t,qs)ϕ(s)h0([u(s)]*+1m)dqs≥μ*h0(12ξ1)∫τ1‍ρ(s)ϕ(s)dqs>ξ1.
This means that Tm(Ω-10)⊂Ω10, for u∈Ω-10. Thus, it follows from the (i) of Lemma 11 that
(72)i(Tm,Ω10,Ω)=1.
Similarly, we can prove that, for any u∈Ω-11,
(73)i(Tm,Ω11,Ω)=1.
Thus, using (59), (67), and (72), we obtain
(74)i(Tm,Ω1∖(Ω-0∩Ω-10),Ω)=i(Tm,Ω1,Ω)-i(Tm,Ω0,Ω)-i(Tm,Ω10,Ω)=-1
which implies that Tm has at least one fixed point um1∈Ω1∖(Ω-0∩Ω-10) and satisfies
(75)Tmum1(t)=∫01‍G(t,qs)F(s,[um1(s)]*+1m)dqs,t∈[0,1],
and r≤∥um1∥X≤ξ1.
Obviously,F(t,[um1(t)]*+1/m) is continuous for any um1∈C([0,1],R+),t∈(0,1). Also, it can be seen that um1 has uniform lower and upper bounds. This directly comes from um1∈Ω1∖(Ω-0∩Ω-10). Hence, in order to pass the solution um1 of the problem (48) to that of the original problem (7) and (8), we need the following fact:
(76){um1}m=1∞is  an  equicontinuous  family  on[0,1].
As in the proof of Lemma12, we can prove that the sequence {um1}m=1∞ is equicontinuous on [0,1]. Now the Arzela-Ascoli Theorem guarantees that the sequence {um1}m=1∞ has a subsequence {umk,1}k=1∞, converging uniformly on [0,1] to u1∈X. As for (61) and the fact r≤∥um1∥X≤ξ1, we obtain that
(77)μqαtα-12λr≤u1(t)-w(t)≤ξ1,∀t∈[0,1].
Moreover, umk,1 satisfies the following integral equation:
(78)umk,1=∫01‍G(t,qs)F(s,[umk,1(s)]*+1mk)dqs.
Letting k→+∞, we have
(79)u1=∫01‍G(t,qs)F(s,[u1(s)]*)dqs.
Let u*(t)=u1(t)-w(t), then u* is a positive solution of BVP (7) and (8).
From (73), Tm has at least one fixed point um2∈Ω-11. Similar to (76), there exists a subsequence {umk,2}k=1∞⊆{um2}m=1∞ such that limk→+∞umk,2(t)=u2(t)∈X, and ξ2≤∥u2∥X≤R1. Let u**(t)=u2(t)-w(t), then u** is also a positive solution of BVP (7) and (8).
Sinceξ1<ξ2, we have u*≠u**. This implies that u*,u** are two different positive solutions of BVP (7) and (8).Theorem 14.
Suppose(H0), (H1), (C1), and (C2) hold. In addition, assume that the following conditions are satisfied. (C4)
There existsR1 with R1>ξ2(ξ2 as given in (C2)) such that
(80)λ0∫01‍ρ(s)φ(s)(g(μqαsα-12λR1)+h(R1+1))dqs<R1.(C5)
(81)limu→+∞h0(u)u=+∞.Then BVP (7) and (8) has at least three different positive solutions.Proof.
It can be seen that condition(C4) is equivalent to condition (C3). As a consequence we obtain that the BVP (7) and (8) has at least two different positive solutions u1, u2 with r≤∥u1∥X<∥u2∥X<R1.
On the other hand, choose a real numberM*>0 such that
(82)M*≥2λμ*μqατα-1∫τ1‍ρ(s)ϕ(s)dqs.
By(C5), there exists R1*>R1 such that
(83)h0(u)≥M*u,u≥R1*.
Choose
(84)R2>max{R1,2λR1*μqατα-1},
and let Ω2={u∈Ω:∥u∥X<R2}.
In the following, we will prove that(85)u≠Tmu+νℏ,ℏ∈Ω∖{θ},∀u∈∂Ω2,ν∈[0,1],m∈N.
Suppose that (85) is false; then there exists ν0∈[0,1],u0∈∂Ω2, such that
(86)u0=Tmu0+ν0ℏ.
Foru0∈∂Ω2 and for any t∈[τ,1], we have
(87)u0(t)-w(t)≥u0(t)-λ([α]q-μ)Γq(α)tα-1∫01‍p(s)dqs≥(1-λ2∫01‍ρ(s)dqsμqα([α]q-μ)Γq(α)R2)u0(t)≥12u0(t)≥μqαtα-12λ∥u0∥X≥μqατα-12λR2>R1*>0.
It follows from (83) and (85) that we have
(88)R2=∥u0∥X=∥Tmu0+ν0ℏ∥X≥∫01‍G(t,qs)F(s,[u0(s)]*+1m)dqs>μ*∫τ1‍ρ(s)ϕ(s)h0([u0(s)]*+1m)dqs≥μ*∫τ1‍ρ(s)ϕ(s)dqs·M*([u0(s)]*+1m)≥M*μ*μqατα-1∫τ1‍ρ(s)ϕ(s)dqs2λ·R2≥R2,
which is a contradiction. Hence, (85) is ture; from the (ii) of Lemma 11, we get
(89)i(Tm,Ω2,Ω)=0.
Combining this with (67) yields
(90)i(Tm,Ω2∖Ω-1,Ω)=i(Tm,Ω2,Ω)-i(Tm,Ω1,Ω)=-1.
This implies that Tm has at least one fixed point um3∈Ω2∖Ω-1 with R1<∥um3∥X≤R2. Similar to (76), there exists a subsequence {umk,3}k=1∞⊆{um3}m=1∞ such that limk→+∞umk,3(t)=u3*(t)∈X, and R1<∥u3*∥X≤R2. Let u3(t)=u3*(t)-w(t), then u3 is also a positive solution of BVP (7) and (8). So the proof is complete.By the induction method, we can obtain the following multiplicity results for BVP (7) and (8).Corollary 15.
Suppose(H0), (H1), (C1), and (C5) hold. In addition, there exist constants ri, ξik, (i=1,…,m;k=1,2) with r<ξ11<ξ12<r1<ξ21<ξ22<r2<⋯<ξm1<ξm2<rm such that  (91)λ∫01‍ρ(s)φ(s)(g(μqαsα-12λri)+h(ri+1))dqs<ri,i=1,2,…,m,μ*h0(12ξik)∫τ1‍ρ(s)ϕ(s)dqs>ξik,i=1,2,…,m;k=1,2,
where μ*=μqατα-1/([α]q-μ)Γq(α).
Then BVP (7) and (8) has at least 2m+1 different positive solutions.Now we present an example to illustrate our main results.Example 16.
Consider the following problem:(92)D0.52.5u(t)+t10(120u+h(u))-40(1+t)=0,0<t<1,u(0)=(D0.5u)(0)=0,u(1)=1.5∫01‍u(s)d0.5s,
where
(93)h(u)={8u,0≤u≤60,1490u3,60≤u≤900,30u2,900≤u≤1600,3.2×105u,u≥1600.
Then BVP (91) has at least two positive solutions.Proof.
In this case,α=2.5, q=0.5, μ=1.5, and
(94)f(t,u)=t10(120u+h(u))-40(1+t).
Let
(95)p(t)=40(1+t),ϕ(t)=t10,φ(t)=t10,g(u)=120u,h0(u)=78h(u).
It is easy to see that the assumptions (H0), (H1) hold.
Taken=2, then τ=qn=0.25. By calculation, we get
(96)λ≈0.7335,λ0≈1.3913,μ*≈0.0524,k0=∫01‍p(s)dqs([α]q-μ)Γq(α)≈∫01‍40(1+s)d0.5s0.5272≈12.6454.
Set r=52, then
(97)r=52>2λ2k0μqα≈51.3086,Δ:=λ0∫01‍ρ(s)φ(s)(g(μqαsα-12λr)+h(r+1))dqs=λ010∫01‍(1-12s)(3/2)s3/2λ010∫01‍×(λ10μqαs3/2r+h(r+1))dqs≤λ010(λ520μqα+h(r+1))×∫01‍(1-12s)(3/2)dqs≈35.8280<r=52,
which implies that the assumption (C1) holds.
On the other hand, takeξ1=1×103, ξ2=1.1×103, then
(98)μ*h0(12ξ1)∫τ1‍ρ(s)ϕ(s)dqs=μ*10h0(12ξ1)∫1/41‍(1-12s)(3/2)s3/2dqs≥μ*160h0(12ξ1)∫1/41‍(1-12s)(3/2)dqs≈1.1681×103>1×103=ξ1,μ*h0(12ξ2)∫τ1‍ρ(s)ϕ(s)dqs≥1.4135×103>1.1×103=ξ2
which implies that the assumption (C2) holds.
Finally, we have(99)limu→+∞h(u)u=limu→+∞3.2×105uu=limu→+∞3.2×105u=0.
Thus (C3) also holds. It follows from Theorem 13 that the BVP (91) has at least two positive solutions u*,u** with 52≤∥u*∥≤1×103<1.1×103≤∥u**∥.



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*Source: 643571-2013-02-11.xml*