# Positive Solutions for Nonlinearq-Fractional Difference Eigenvalue Problem with Nonlocal Conditions

Authors: Wafa  Shammakh; Maryam Al-Yami
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

The problem of positive solutions for nonlinearq-fractional difference eigenvalue problem with nonlocal boundary conditions is investigated. Based on the fixed point index theory in cones, sufficient existence of positive solutions conditions is derived for the problem.


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## Body

## 1. Introduction

The fractionalq-calculus is the q-extension of ordinary fractional calculus. It has been used by many researchers to adequately describe the evolution of a variety of engineering, economical, physical, and biological processes.We consider a nonlinearq-fractional difference eigenvalue problem with nonlocal boundary conditions given by (1)DqαCut+λgtft,ut=0,0≤t≤1,0<q<1,(2)Dqku0=0,u0=0,2≤k≤n-1,Dqu1=θu,where DqαC denote the fractional q-derivative of the Caputo type, n-1<α≤n,n>2,λ>0 is a parameter, and θu is given by a Riemann-Stieltjes integral θu=∫01utdqAt.This type of BC includes, as particular cases, multipoint problems whenθu=∑i=1m-2αiuζi, (see [1]) and a continuously distributed case when θu=∫01αsusdqs (see [2–4]).More recently, many people pay attention to BVPs involving nonlinearq-difference equations [5–12].In [13], Yuan and Yang dealt with some existence and uniqueness results for nonlinear boundary value problems for delayed q-fractional difference systems based on a contraction mapping principle and Krasnoselskii’s fixed-point theorem.In [14], Yang investigated the sufficient conditions for the existence and nonexistence positive solutions for BVP involving nonlinear q-fractional difference equations.Ferreira [4] studied the existence of positive solutions to the nonlinear q-fractional BVPs by means of Krasnoselskii’s fixed point theorem in cones.In this paper, we obtain the results on the existence of one and two positive solutions by utilizing the results of Webb and Lan [15] involving comparison with the principle characteristic value of a related linear problem to the q-fractional case. We then use the theory worked out by Webb and Infante in [16–19] to study the general nonlocal BCs.


## 2. Preliminaries

In this section, we will present some definitions and lemmas that will be used in the proof of our main results.Letq∈0,1 defined by [20] (3)aq=qa-1q-1=qa-1+⋯+1,a∈R.The q-analogue of the power function a-bn with n∈N is (4)a-b0=1,a-bn=∏k=0n-1a-bqk,a,b∈R,n∈N.More generally, if α∈R, then (5)a-bα=aα∏i=0∞a-bqia-bqα+i.Note that if b=0 then a(α)=aα. The q-gamma function is defined by (6)Γqx=1-qx-11-qx-1,x∈R∖0,-1,-2,…,0<q<1,and satisfies Γq(x+1)=xqΓq(x).Theq-derivative of a function fx is here defined by (7)Dqfx=dqfxdqx=fqx-fxq-1x,and q-derivatives of higher order are defined by(8)Dqnfx=fx,ifn=0,DqDqn-1fxifn∈N.The q-integral of a function f defined in the interval 0,b is given by(9)∫0xftdqt=x1-q∑n=0∞fxqnqn,0≤q<1,x∈0,b.If a∈0,b and f is defined in the interval 0,b, its integral from a to b is defined by (10)∫abftdqt=∫0bftdqt-∫0aftdqt.Similarly as done for derivatives, it can be defined an operator Iqn; namely, (11)Iq0fx=fx,Iqnfx=IqIqn-1fx,n∈N.The fundamental theorem of calculus applies to these operators Iq and Dq; that is, (12)DqIqfx=fx,and if f is continuous at x=0, then (13)IqDqfx=fx-f0.Basic properties of the two operators can be found in the book [20]. We now point out four formulas that will be used later:(14)at-sα=aαt-sα,Dqtt-sα=αqt-sα-1,Dqst-sα=-αqt-qsα-1,Dqx∫0xfx,tdqtx=∫0xDqxfx,tdqt+fqx,x,where Dqi denotes the q-derivative with respect to variable i [21].Remark 1 (see [21]).
We note that ifα>0 and a≤b≤t, then (t-a)(α)≥(t-b)(α).Definition 2 (see [22]).
Letα≥0 and let f be a function defined on 0,1. The fractional q-integral of the Riemann-Liouville type is (Iq0RLf)(x)=f(x) and (15)IqαRLfx=1Γqα∫0xx-qtα-1ftdqt,α∈R+,x∈0,1.Definition 3 (see [22]).
The fractionalq-derivative of the Riemann-Liouville type of order α≥0 is defined by (Dq0RLf)(x)=f(x) and(16)DqαRLfx=DqαIqα-αfx,α>0,where α is the smallest integer greater than or equal to α.Definition 4 (see [22]).
The fractionalq-derivative of the Caputo type of order α≥0 is defined by (17)DqαCfx=Iqα-αDqαfx,α>0.Lemma 5 (see [22]).
Letα,β≥0 and let f be a function defined on 0,1. Then, the next formulas hold:(1)
(IqβIqαf)(x)=(Iqα+βf)(x),(2)
(DqαCIqαf)(x)=f(x).Lemma 6 (see [22]).
Letα∈R+∖N, λ∈-1,∞. Then, the next formulas hold:(18)1Iqαxλ=Γqλ+1Γqλ+1+αxλ+α,2DqαRLxλ=Γqλ+1Γqλ+α+1xλ-α,3DqαCxλ=0ifλ∈N0;α>λ,DqαRLxλifotherwise.Theorem 7 (see [23]).
Letx>0 and α∈R+∖N. Then, the following equality holds:(19)IqαDqαCfx=fx-∑k=0α-1xkΓqk+1Dqkf0.Lemma 8 (see [24]).
SupposeT:K→K is a completely continuous operator and has no fixed points on ∂Kρ∩K. Then the following are true: (i)
IfTu≤u for all u∈∂Kρ∩K, then iT,Kρ∩K,K=1, where i is the fixed point index on K.(ii)
IfTu≥u for all u∈∂Kρ∩K, then iT,Kρ∩K,K=0.Lemma 9 (see [24]).
LetK be a cone in Banach space E. Suppose that T:K-ρ→K is a completely continuous operator. There exists u0∈K∖0 such that u-Tu≠μu0 for any u∈∂Kr and μ≥0,iT,Kρ,K=0.Lemma 10 (see [24]).
LetK be a cone in Banach space E. Suppose that T:K-ρ→K is a completely continuous operator. If Tu≠μu for any u∈∂Kr and μ≥1, then iT,Kρ,K=1.Lemma 11.
Lety∈C0,1 be a given function and n-1<α≤n, then u is a solution of BVP (1)-(2) if and only if u is a solution of the integral equation (20)ut=γtθu+∫01G0t,qsysdqs,where(21)γt=t,G0t,qs=α-1qt1-qsα-2-t-qsα-1Γqα,0≤qs≤t≤1,α-1qt1-qsα-2Γqα,0≤t≤qs≤1.Proof.
Assume thatu is a solution of BVP (1)-(2).
Applying Theorem7, (1) can be reduced to an equivalent integral equation: (22)ut=-1Γqα∫0tt-qsα-1ysdqs+c0+c1t+c2t2+⋯+cn-1tn-1.By (2), we obtain (23)c0=0,c2=⋯=cn-1=0,c1=θu+α-1qΓqα∫011-qsα-2ysdqs.Therefore, we obtain(24)ut=γtθu+tα-1qΓqα∫011-qsα-2ysdqs-1Γqα∫0tt-qsα-1ysdqs=γtθu+∫0tα-1qt1-qsα-2-t-qsα-1Γqαysdqs+∫t1α-1qt1-qsα-2Γqαysdqs=γtθu+∫01G0t,qsysdqs.Conversely, if u is a solution of the integral equation (20), using Lemmas 5 and 6, we have(25)DqαCut=DqαCtθu+DqαCt∫01α-1q1-qsα-2Γqαysdqs-DqαC∫0tt-qsα-1Γqαysdqs=-DqαCIqαyt=-yt.A simple computation shows u0=0,Dqku0=0,2≤k≤n-1,Dqu1=θu.Remark 12.
G
0
t
,
q
s is Green’s function for the local BVP(26)DqαCut+λgtft,ut=0,t∈0,1,n-1<α≤n,0<q<1,Dqku0=0,u0=0,2≤k≤n-1,Dqu1=0.Lemma 13.
FunctionG0t,qs defined in (20) satisfies the following conditions:(H1)
G0(t,qs)≥0 is continuous and G0(t,s)≤Φ0qs for all 0≤t,s≤1;(H2)
G0(t,qs)≥c0tΦ0qs for all 0≤t,s≤1, where(27)Φ0qs=G01,qs=α-1q1-qsα-2-1-qsα-1Γqα,c0t=tα-1.Proof.
It is obvious thatG0(t,qs) is nonnegative and continuous.
(H1) For 0≤qs≤t≤1,(28)G0t,qs=1Γqαα-1qt1-qsα-2-t-qsα-1=1Γqαα-1qt1-qsα-2-tα-11-qstα-1≥1Γqαα-1qtα-11-qsα-2-tα-11-qsα-1=tα-1Γqαα-1q1-qsα-2-1-qsα-1≥0,and for 0≤t≤qs≤1,(29)G0t,qs=α-1qt1-qsα-2Γqα,and it is clear that G0t,qs≥0 and G00,qs=0. Therefore G0t,qs≥0.
For fixeds∈0,1 and t≥qs we have(30)DqtG0t,qs=α-1q1-qsα-2-α-1qt-qsα-2Γqα=1-qsα-2-t-qsα-2Γqα-1≥0;that is, G0t,qs is an increasing function of t. Obviously, G0t,qs,t≤qs, is increasing in t; therefore G0t,qs is an increasing function of t for fixed s∈0,1.
Thus, (H1) holds.
(H2) Suppose now that t≥qs:(31)G0t,qsΦ0qs=α-1qt1-qsα-2-t-qsα-1α-1q1-qsα-2-1-qsα-1≥tα-1α-1q1-qsα-2-1-qsα-1α-1q1-qsα-2-1-qsα-1=tα-1.On the other hand, if t≤qs, then we have(32)G0t,qsΦ0qs=α-1qt1-qsα-2α-1q1-qsα-2-1-qsα-1≥α-1qtα-11-qsα-2-tα-11-qsα-1α-1q1-qsα-2-1-qsα-1=tα-1,and this finished the proof of (H2).DefiningGAqs=∫01G0t,qsdqAt, Green’s function for nonlocal BVP (1)-(2) is given by (33)Gt,qs=γt1-θγGAqs+G0t,qs.Throughout the paper we assume the following:(H3) A is a function of bounded variation, and GAqs=∫01G0t,qsdqAt satisfies GAqs≥0 for almost every s∈0,1. Note that GAqs exists for almost every s by(H1).(H4) The functions g,Φ satisfy g≥0 almost everywhere, gΦ∈L10,1, and (34)∫abΦqsgsdqs>0.(H5)
f
:
0,1
×
0
,
∞
→
0
,
∞ satisfies Caratheodory conditions; that is, f·,u is measurable for each fixed u∈0,∞ and ft,· is continuous for almost every t∈0,1, and for each r>0, there exists ϕr∈L∞0,1 such that 0≤ft,u≤ϕr for all u∈0,r and almost all t∈0,1.(H6) One has the following: γ∈C0,1,γt≥0,0≤θ[γ]<1.Lemma 14.
IfG0 satisfies (H1), (H2), then G satisfies (H1), (H2) for a function Φ, the same interval a,b, and the same constant c, where Φ satisfies (H4) and c=min⁡c0t:t∈a,b.Proof.
We have(35)Gt,qs=γt1-θγGAqs+G0t,qs≤γ1-θγGAqs+Φ0qs≕Φqs,and for t∈a,b(36)Gt,qs≥cγ1-θγGAqs+cΦ0qs=cΦqs.Note that gΦ∈L∞ because A has finite variation and GAqs≤Φqsvar⁡A.
Thus, Green’s functionGt,qs satisfies (H1), (H2) for a function Φ and the constant c.


## 3. Main Result

SetE=C0,1 as a Banach space with the norm u=supt∈0,1⁡ut. Let P=u∈E:u≥0 denote the standard cone of nonnegative functions. Define (37)K=u∈P,mina≤t≤but≥cu,where a,b is some subset of 0,1.Note thatγ∈K so K≠0. For any 0<r<R<+∞, let Kr=u∈K:u<r, ∂Kr=u∈K:u=r, K-r=u∈K:u≤r, K-R∖Kr=u∈K:r≤u≤R, and Vr=u∈K:mint∈a,b⁡ut<r and Vr is bounded.Define a nonlinear operatorT:P→K and a linear operator L:P→K by (38)Tut=λ∫01Gt,qsgsfs,usdqs,(39)Lut≔∫01Gt,qsgsusdqs.Lemma 15 (see [18]).
Under hypotheses (H1)–(H6) the maps T:P→E defined in (38) are compact.Theorem 16.
Under hypotheses (H1)–(H6) the maps are T:P→K.Proof.
Foru∈P and t∈0,1 we have(40)Tut≤λ∫01Φqsgsfs,usdqs.Hence,(41)Tu≤λ∫01Φqsgsfs,usdqs.Also, for t∈a,b, we have(42)Tut≥cλ∫01Φqsgsfs,usdqs≥cTu.Similar to the proofs of Lemma 15 and Theorem 16, Lut is compact and maps P into K.We will use the Krein-Rutman theorem. We recall thatλ is an eigenvalue of L with corresponding eigenfunction ϕ if ϕ≠0 and λϕ=Lϕ. The reciprocals of eigenvalues are called characteristic values of L. The radius of the spectrum of L, denoted by rL, is given by the well-known spectral radius formula rL=limn→∞⁡Ln1/n.Theorem 17 (see [15]).
LetK be a total cone in a real Banach space E and let L^:E→E be a compact linear operator with L^K⊆K. If rL^>0 then there is ϕ1∈K∖0 such that L^ϕ1=rL^ϕ1.Thusλ1≔rL^ is an eigenvalue of L^, the largest possible real eigenvalue, and μ1=1/λ1 is the smallest positive characteristic value.Lemma 18 (see [15]).
Assume that (H1)–(H3) hold and let L be as defined in (39). Then rL>0.Theorem 19 (see [15]).
When (H1)–(H3) hold, rL is an eigenvalue of L with eigenfunction ϕ1 in K.Theorem 20 (see [15]).
Letμ1=1/rL and ϕ1t be a corresponding eigenfunction in P of norm 1. Then m≤μ1≤M, where (43)m=supt∈0,1⁡∫01Gt,qsgsdqs-1,M=inft∈a,b⁡∫abGt,qsgsdqs-1.If gt>0 for t∈0,1 and Gt,qs>0 for t,s∈0,1, the first inequality is strict unless ϕ1t is constant for t∈0,1. If gtϕt>0 for t∈a,b, the second inequality is strict unless ϕ1t is constant for t∈a,b.Proof (for the local BVP (1)-(2) if gt≡1).
We now compute the constantm and the optimal value of Ma,b; that is, we determine a,b so that Ma,b is minimal.
Forqs≤t, we have by direct integration(44)∫0tG0t,qsdqs=∫0tα-1qt1-qsα-2-t-qsα-1Γqαdqs=t-t1-tα-1Γqα-tααqΓqα.For qs≥t,(45)∫t1G0t,qsdqs=∫t1α-1qt1-qsα-2Γqαdqs=t1-tα-1Γqα.Then we have(46)∫01G0t,qsdqs=tΓqα-tααqΓqα.And the maximum of this expression occurs when t=1; hence(47)supt∈0,1⁡∫01G0t,qsdqs=1Γqα-1αqΓqα=αq-1αqΓqα.Then m=αqΓqα/(αq-1).
Fora<b, we have by direct integration(48)∫atG0t,qsdqs=-t1-tα-1Γqα+t1-aα-1Γqα-t-aααqΓqα,∫tbG0t,sds=-t1-bα-1Γqα+t1-tα-1Γqα.Then (49)∫abG0t,qsdqs=t1-aα-1Γqα-t-aααqΓqα-t1-bα-1Γqα=t1-aα-1-1-bα-1Γqα-t-aααqΓqα=Rt,a,b,DqtRt,a,b=1-aα-1-1-bα-1Γqα-t-aα-1Γqα.The sign of derivative DtqR shows that this is an increasing function of t so the minimum occurs at t=a. Let(50)Ra,b=aΓqα1-aα-1-1-bα-1.The minimal value of Ma,b corresponds to the maximal value of Ra,b. Consider(51)DqbRa,b=aα-1q1-qbα-2Γqα>0.The quantity Ra,b is an increasing function of b so its maximum occurs when b=1. Let(52)Ra=a1-aα-1Γqα.Then the maximum of Ra occurs when a=1/(1+α-1q). Consider(53)mint∈a,b⁡∫abG0t,qsdqs=R11+α-1q,1.Hence the minimal value of Ma,b is(54)M11+α-1q,1=R11+α-1q,1-1.


## 4. The Existence of at Least One Positive Solution

For convenience, we introduce the following notations:(55)f-u≔supt∈0,1ft,u,f_u≔inft∈0,1ft,u;f0≔limsupu→0+⁡f-uu,f0≔liminfu→0+⁡f_uu;f∞≔limsupu→∞⁡f-uu,f∞≔liminfu→∞⁡f_uu,f0,r≔sup0≤t≤1,0≤u≤r⁡ft,ur,fr,r/c≔infa≤t≤b,r≤u≤r/c⁡ft,ur.Under hypotheses (H1)–(H4) let L~ be defined by(56)L~ut=∫abGt,qsgsusdqs.Then L~ is a compact linear operator and L~P⊆K.HencerL~ is an eigenvalue of L~ with an eigenfunction ϕ~1 in K. Let μ~1≔1/rL~. Note that μ~1≥μ1; hence the condition in the following theorem is more stringent compared with the case if rL could be used.Theorem 21.
Assume that(A1)
0≤λf0<μ1,(A2)
μ~1<λf∞≤∞.Then (1)-(2) had at least one positive solution.Proof.
Letε>0 be such that f0≤(1/λ)μ1-ε. Then there exists ρ0>0 such that(57)ft,u≤1λμ1-εu,∀u∈0,ρ0andalmostallt∈0,1.Let ρ∈0,ρ0. We prove that (58)Tu≠βuforu∈∂Kρ,β≥1,which implies the result. In fact, if (58) does not hold, then there exist u∈∂Kρ and β≥1 such that Tu=βu.
This implies(59)βut=λ∫01Gt,qsgsfs,usdqs≤μ1-ε∫01Gt,qsgsusdqs=μ1-εLut.Thus, we have shown ut≤μ1-εLut. This gives (60)ut≤μ1-εLμ1-εLut=μ1-ε2L2ut.And by iterating (61)ut≤μ1-εnLnutforn∈N.Therefore(62)u≤μ1-εnLnu,1≤μ1-εnLn,and we have (63)1≤μ1-εlimn→+∞⁡Ln1/n=μ1-ε1μ1<1,a contradiction. It follows that (64)ikT,Kρ=1,foreachρ∈0,ρ0.Let ρ1>0, ρ1>ρ be chosen so that ft,u>(μ~1/λ)u for all u≥cρ1, c as in (H2), and almost all t∈0,1.
We claim thatu≠Tu+βϕ~1 for all β>0 and u∈∂Kρ∗ when ρ∗>ρ1. Note that u∈K with u=ρ∗≥ρ1.
We haveut≥cρ1 for all t∈a,b.
Now, if our claim is false, then we have(65)ut=λ∫01Gt,qsgsfs,usdqs+βϕ~1t.Therefore,(66)ut≥μ~1∫abGt,qsgsusdqs+βϕ~1t=μ~1L~ut+βϕ~1t.From (66) we firstly deduce that ut≥βϕ~1t on 0,1.
Then we have(67)μ~1L~ut≥μ~1L~βϕ~1t=βϕ~1t.Inserting this into (66) we obtain ut≥2βϕ~1t for t∈0,1.
Repeating this process gives(68)ut≥nβϕ~1tfort∈0,1,n∈N.Since ϕ~1t is strictly positive on 0,1 this is a contradiction; then (69)iKT,Kρ∗=0,foru∈∂Kρ∗.By (64) and (69), one has(70)iKT,Kρ∗∖K-ρ=iKT,Kρ∗-iKT,Kρ=-1.Therefore, T has at least one fixed point u0∈Kρ∗∖K-ρ, and u0 is a positive solution of BVP (1)-(2).Theorem 22.
Assume that(A3)
μ1<λf0≤∞,(A4)
0≤λf∞<μ1.Then (1)-(2) had at least one positive solution.Proof.
Letε>0 satisfy f0>(1/λ)μ1+ε. Then there exists R1>0 such that (71)ft,u≥1λμ1+εu,∀t∈0,1,u∈0,R1.For any u∈∂KR1 we have by (71) that(72)Tut=λ∫01Gt,qsgsfs,usdqs≥μ1+ε∫01Gt,qsgsusdqs≥μ1Lut,∀t∈0,1.Let u~1 be the positive eigenfunction of L corresponding to μ1; that is, u~1=μ1Lu~1. We may suppose that T has no fixed point on ∂KR1; otherwise, the proof is finished. In the following we will show that (73)u-Tu≠βu~1,∀u∈∂KR1,β≥0.If (73) is not true, then there is u~0∈∂KR1and β0≥0 such that u~0-Tu~0=β0u~1. It is clear that β0>0 and u~0=Tu~0+β0u~1≥β0u~1.
Set(74)β∗=sup⁡β:u~0≥βu~1.Obviously, β∗≥β0>0. It follows from LP⊂P that (75)μ1Lu~0≥μ1Lβ∗u~1=β∗μ1Lu~1=β∗u~1,and using this and (72), we have (76)u~0=Tu~0+β0u~1≥μ1Lu~0+β0u~1≥β∗u~1+β0u~1,which contradicts (74). Thus, (73) holds.
By Lemma9, we have(77)iKT,KR1=0.On the other hand, let ε>0 satisfy f∞<(1/λ)μ1-ε. Then there exists R2>R1 such that(78)ft,u≤1λμ1-εu,∀t∈0,1,u≥R2.By (H5) there exists an L∞ function φ1 such that(79)ft,u≤1λφ1t,∀u∈0,R2,t∈0,1.Hence, we have (80)ft,u≤1λμ1-εu+φ1t,∀u∈R+,t∈0,1.Since 1/μ1 is the radius of the spectrum of L, I/μ1-ε-L-1 exists.
Let(81)C=∫01φ1sΦsgsdqs,R0=Iμ1-ε-L-1cμ1-ε.We prove that, for each R>R0, (82)Tu≠βu,∀u∈∂KR,β≥1.In fact, if not, there exist u∈∂KR and β≥1 such that Tu=βu.
This together with (80) implies (83)ut≤∫01Gt,qsgsμ1-εus+φ1sdqs=μ1-ε∫01Gt,qsgsusdqs+∫01Gt,qsgsφ1sdqs≤μ1-εLut+C.This implies (84)Iμ1-ε-Lut≤Cμ1-ε,ut≤Iμ1-ε-L-1Cμ1-ε=R0.Therefore, we have u≤R0<R, a contradiction. Taking R>R2, it follows from (74) and properties of index that (85)iKT,KR=1,∀R>R0.Now (77) and (85) combined imply(86)iKT,KR∖K-R1=iKT,KR-iKT,K-R1=1.Therefore, T has at least one fixed point u0∈KR/K-R1, and u0 is a positive solution of BVP (1)-(2).


## 5. The Existence of Two Positive Solutions

Theorem 23.
Suppose (A2), (A3), and(A5)
λf0,ρ′≤m for some ρ′>0.
Then (1)-(2) had at least two positive solutions.Proof.
By (A5), we have(87)Tut=λ∫01Gt,qsgsfs,usdqs≤∫01Gt,qsgsρ′mdqs,so that Tu≤ρ′=u, for all u∈∂Vρ′. Now Lemma 8 yields (88)ikT,Vρ′=1.On the other hand, in view of (A2), we may take ρ∗>ρ′ so that (69) holds (see the proof of Theorem 21). From (A3), we may take R1∈0,ρ′ so that (77) holds (see the proof of Theorem 22).
Combining (88), (69), and (77), we arrive at (89)ikT,Kρ∗∖V-ρ′=0-1=-1,ikT,Vρ′∖K-R1=1-0=1.Consequently, T has at least two fixed points, with one on Kρ∗∖V-ρ′ and the other on Vρ′∖K-R1. Therefore, (1)-(2) had at least two positive solutions.Theorem 24.
Suppose (A1), (A4), and(A6)
λfρ′,ρ′/c≥M for some ρ′>0.Then (1)-(2) had at least two positive solutions.Proof.
By (A6), we have(90)Tut=λ∫01Gt,qsgsfs,usdqs≥λ∫abGt,qsgsfs,usdqs≥∫abGt,qsgsMρ′dqs,so that Tu≥ρ′=u, for all u∈∂Vρ′, and by Lemma 8 this yields (91)ikT,Vρ′=0.On the other hand, in view of (A1), we may take ρ∈0,ρ′ so that (64) holds (see the proof of Theorem 21). In addition, from (A4), we may take R>ρ′ so that (85) holds (see the proof of Theorem 22).
Combining (91), (64), and (85), we arrive at (92)ikT,KR∖V-ρ′=1-0=1,ikT,Vρ′∖K-ρ=0-1=-1.Hence, T has at least two fixed points, with one on Vρ′∖K-ρ and the other on KR∖V-ρ′. Therefore, (1)-(2) had at least two positive solutions.We illustrate the applicability of these results with some examples.Example 25.
Consider the problem(93)D0.52.5ut+λ5t+37u2+uu+12+cos⁡u=0,t∈0,1,D0.52u0=0,u0=0,D0.5u1=0.Here we have gt=5t+3, fu=2+cos⁡u((7u2+u)/(u+1)), and 2<α≤3.
It is readily shown thatf0=f0=3, f∞=21, f∞=7.
Also,3u≤fu≤21u for u≥0. By calculation, we find m=0.19722, and the smallest M calculated is Ma,b≈M0.484405,1≈0.74665. We find μ1≈0.30366. Hence, by Theorem 21, there is at least one positive solution if 3λ<μ1 and 7λ>μ1; that is, there is a positive solution if λ∈0.47047,1.09773.
By Theorem22, there does not exist a positive solution if either 3λ>μ1 or 21λ<μ1; that is, if λ<1.09773 or λ>0.15682 no positive solution exists.



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*Source: 759378-2015-12-21.xml*