# Positive Solutions for Higher Order Nonlocal Fractional Differential Equation with Integral Boundary Conditions

Authors: Jiqiang Jiang; Weiwei Liu; Hongchuan Wang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, by using the spectral analysis of the relevant linear operator and Gelfand’s formula, some properties of the first eigenvalue of a fractional differential equation were obtained; combining fixed point index theorem, sufficient conditions for the existence of positive solutions are established. An example is given to demonstrate the application of our main results.


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## Body

## 1. Introduction

Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, control theory, signal and image processing, and aerodynamics (see [1, 2]). In the last decade, boundary value problems of fractional calculus have received a great attention, based on various analytic techniques, and a variety of results concerning the existence of solutions can be found in the literature [3–19]. For example, Zhang [17] studied the following singular fractional boundary value problem (FBVP): (1)D0+αyt+qtft,yt,y′t,…,yn-2t=0,0<t<1,y0=y′0=⋯=yn-20=yn-21=0,where α∈n-1,n,n≥2, D0+α is the standard Riemann-Liouville derivative, the nonlinear term f=g+h, and g and h have different monotone properties. By using a mixed monotone method, a unique positive solution is obtained.Integral boundary conditions arise in thermal conduction problems [20], semiconductor problems [21], and hydrodynamic problems [22]. Recently, the existence and multiplicity of positive solutions for nonlinear ordinary differential equations with integral boundary value problems have received a great deal of attention. To identify a few, see [23–27] and the references therein for integer order integral boundary value problems and [28–33] for fractional order integral boundary value problems. In [28], authors using monotone iterative technique investigated the existence and uniqueness of the positive solutions of higher-order nonlocal fractional differential equations of the type(2)D0+αyt+ft,yt=0,0<t<1,n-1<α≤n,y0=y′0=⋯=yn-20=0,y1=λy,where f∈C((0,1)×R+,R+). By means of Schauders fixed point theorem, FBVP (2) are also studied in [29]. In [30], the authors investigated problems (2) with f(t,y(t)) replaced by q(t)f(t,y(t)), and the existence and multiplicity of positive solutions are obtained by means of the fixed point index theory in cones.Inspired by the work of the above papers, the aim of this paper is to establish the existence of positive solutions for the following nonlinear fractional differential equation involving Stieltjes integrals conditions:(3)-D0+αyt=ft,yt,y′t,…,yn-2t,0<t<1,y0=y′0=⋯=yn-20=0,yn-21=λyn-2,where D0+α is the Riemann-Liouville fractional derivative, n-1<α≤n,n≥2, λ[z]=∫01z(t)dA(t) is a linear functional on C[0,1] given by a Stieltjes integral with A representing a suitable function of bounded variation, and dA can be a signed measure. f:(0,1)×(R0+)n-1→R+ is a continuous function, the nonlinearity f(t,z1,z2,…,zn-1) may be singular at t=0,1, and z1=z2=⋯=zn-1=0. Here R+=[0,+∞) and R0+=(0,+∞).Our work presented in this paper has the following features. First of all, we discuss the boundary value problem with the Stieltjes integral boundary conditionsλ[z]=∫01z(t)dA(t) appearing in the boundary conditions of FBVP (2) as more general which was not considered in the literature [15], covering two-point, multipoint, and nonlocal problems as special cases. The second new feature is that the nonlinearity f is allowed to depend on higher order derivatives of unknown function y(t) up to n-2 order which was not considered in the literature [28–30]. Thirdly, we allow that the nonlinearity f(t,z1,z2,…,zn-1) may be singular at t=0,1 and z1=z2=⋯=zn-1=0. Therefore, our study improves and extends the previous results to some degree in the relevant literature [17, 28–30].The rest of this paper is organized as follows. In Section2, we present some lemmas that are used to prove our main results. In Section 3, the existence of positive solutions is established under some sufficient conditions. In Section 4, an example is given to demonstrate the application of our theoretical results.


## 2. Basic Definitions and Preliminaries

In this paper, our work is based on the theory of fractional calculus, and, for details on definitions of Riemann-Liouville fractional calculus, we refer the reader to [1, 2].LetE=C[0,1], and then E is a Banach space with the norm y=max0≤t≤1⁡yt, for any y∈E. Let P=y∈E:yt≥0fort∈[0,1] be a cone in E and construct a subcone of P as follows: (4)K=y∈P:yt≥tα-n+11-tyfort∈0,1.For any r>0, let Kr={y∈K:y<r}, ∂Kr={y∈K:y=r}, and K¯r={y∈K:y≤r}.Set(5)Gt,s=1Γα-n+2t1-sα-n+1-t-sα-n+1,0≤s≤t≤1,t1-sα-n+1,0≤t≤s≤1.Lemma 1 (see [17]).
Letq∈Cr[0,1](Cr[0,1]={q∈C[0,1],trq∈C[0,1],0≤r<1}). Then the boundary value problem (6)-D0+α-n+2ut=qt,0<t<1,n-1<α≤n,n≥2,u0=0,u1=0has a unique solution (7)ut=∫01Gt,sqsds.According to the theory of fractional calculus, the unique solution of the problem(8)-D0+α-n+2ut=qt,0<t<1,n-1<α≤n,n≥2,u0=0,u1=λuis γ(t)=tα-n+1 with λ[u] replaced by 1 and q(t)≡0. As in [27], Green’s function for boundary value problem (8) is given by(9)Ht,s=γt1-λγGs+Gt,s,where Gs≔∫01G(t,s)dA(t).Lemma 2.
Suppose thatG(s)≥0 for s∈[0,1] and 0≤λ[γ]<1, and then the functions G(t,s) and H(t,s) have the following properties: (1)
G(t,s) and H(t,s) are nonnegative and continuous for (t,s)∈[0,1]×[0,1].(2)
G(t,s) satisfies (10)tα-n+11-ts1-sα-n+1Γα-n+1≤Gt,s≤s1-sα-n+1Γα-n+1,fort,s∈0,1.(3)
H(t,s) satisfies (11)tα-n+11-tΦs≤Ht,s≤Φs,fort,s∈0,1,where (12)Φs=Gs1-λγ+s1-sα-n+1Γα-n+1.Throughout this paper, we adopt the following assumptions:(H0) A is a function of bounded variation, G(s)≥0 for s∈[0,1] and 0≤λ[γ]<1.(H1) f:(0,1)×(R0+)n-1→R+ is a continuous function and, for any 0<r<R<+∞, (13)limm→+∞⁡supz1∈KR/n-2!⋯zn-2∈KRzn-1∈K¯R∖Kr⁡∫HmΦsfs,z1s,z2s,…,zn-1sds=0,whereH(m)=[0,1/m]∪[m-1/m,1].In order to overcome the difficulty due to the dependance off on derivatives, we consider the following modified problem:(14)-D0+α-n+2xt=ft,I0+n-2xt,I0+n-3xt,…,I0+1xt,xt,0<t<1,x0=0,x1=λxwhere n-1<α≤n,n≥2.Lemma 3 (see [15]).
The nonlocal FBVP (3) has a positive solution y(t)=I0+n-2x(t) if and only if the nonlinear fractional integrodifferential equation (14) has a positive solution x(t).Define a nonlinear operatorT:K∖0→P and a linear operator L:E→E as follows:(15)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds,t∈0,1,(16)Lxt=∫01Ht,sxsds,t∈0,1.Observe that problem (14) has solutions if the operator equation x=Tx has fixed points.Lemma 4 (Krein-Rutmann [34]).
LetL:E→E be a continuous linear operator, P be a total cone, and L(P)⊂P. If there exist ψ∈E∖(-P) and a positive constant c such that cL(ψ)≥ψ, then the spectral radius r(L)≠0 and L has a positive eigenfunction corresponding to its first eigenvalue λ=(r(L))-1.Lemma 5 (Gelfand’s formula [34]).
For a bounded linear operatorL and the operator norm ·, the spectral radius of L satisfies (17)rL=limm→+∞⁡Lm1/m.Lemma 6.
Assume that(H0) holds. Then L:K→K defined by (16) is a completely continuous linear operator, and the spectral radius r(L)≠0; moreover L has a positive eigenfunction φ∗ corresponding to its first eigenvalue λ1=(r(L))-1.Proof.
For anyx∈K, by Lemma 2, we can obtain(18)Lx=maxt∈0,1⁡∫01Ht,sxsds≤∫01Φsxsds.
On the other hand, from Lemma2, we also have(19)Lxt=∫01Ht,sxsds≥tα-n+11-t∫01Φsxsds.Then (18) and (19) yield that (20)Lxt≥tα-n+11-tLx.Consequently, L:K→K. And, from the uniform continuity of H(t,s), (t,s)∈[0,1]×[0,1], we have that L:K→K is a completely continuous linear operator.
Next we will show thatL has the first eigenvalue λ1>0 by using Krein-Rutmann’s theorem. In fact, by Lemma 2, there exists t0∈(0,1) such that H(t0,t0)>0. Thus there exists [a,b]⊂(0,1) such that t0∈(a,b) and H(t,s)>0 for all t,s∈[a,b]. Choose x∈K such that x(t0)>0 and x(t)=0 for all t∉[a,b]. Then, for any t∈[a,b], we have (21)Lxt=∫01Ht,sxsds≥∫abHt,sxsds>0.So there exists c>0 such that c(Lx)(t)≥x(t) for t∈[0,1]. From Lemma 4, we know that the spectral radius r(L)≠0 and L has a positive eigenfunction φ∗ corresponding to its first eigenvalue λ1=(r(L))-1; i.e., λ1Lφ∗=φ∗. The proof is completed.Lemma 7.
Assume that(H0),(H1) hold; then T:K¯R∖Kr→K is completely continuous.Proof.
First, we proveT(K¯R∖Kr)⊂K. In fact, for any x∈K¯R∖Kr, Lemma 2 implies that (22)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≤∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds,t∈0,1.Hence, (23)Tx≤∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds.
On the other hand, from Lemma2, we also have (24)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≥tα-n+11-t∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≥tα-n+11-tTx.Therefore, T(K¯R∖Kr)⊂K.
Next, for anyr>0, we assert that(25)supx∈∂Kr⁡∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds<+∞,which implies that T:K∖{0}→P is well defined. In fact, it follows from (H1) that there exists a natural number m such that(26)supx∈∂Kr⁡∫HmΦsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds<1.For any x∈∂Kr, as x∈K, we have(27)tα-n+11-tx≤xt≤x=r,t∈0,1,and(28)rtα-n+i+11-tΓα-n+2Γα-n+i+2=xtα-n+i+11-tBi,α-n+2Γi≤xΓi∫0tt-si-1sα-n+11-tds≤I0+ixt=1Γi∫0tt-si-1xsds≤1i!x=ri!,t∈0,1,i=1,2,…,n-2,where B(·,·) is the beta function. So, for any 1/m≤t≤1-1/m, (27) and (28) yield that(29)rmα-n+2≤xt≤r,Γα-n+2rmα-n+i+2Γα-n+i+2≤I0+ixt≤ri!,i=1,2,…,n-2.From (26) and (29), we have(30)supx∈∂Kr⁡∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≤supx∈∂Kr⁡∫HmΦsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds+supx∈∂Kr⁡∫1/m1-1/mΦsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds<1+M1∫01Φsds<+∞,where (31)M1=max⁡ft,z1,z2,…,zn-1:t,z1,z2,…,zn-1∈1m,1-1m×Γα-n+2rmαΓα,rn-2!×Γα-n+2rmα-1Γα-1,rn-3!×⋯×Γα-n+2rmα-n+3Γα-n+3,r×rmα-n+2,r.Thus (25) is true which implies that T is uniformly bounded on any bounded set.
Now we will show thatT:K¯R∖Kr→K is continuous. Let xj,x0∈K¯R∖Kr and xj-x0→0(j→∞). For any ε>0, by (H1), there exists a natural number k>0 such that(32)supx∈K¯R∖Kr⁡∫HkΦsfs,I0+n-2xs,…,xsds<ε4.Since f(t,z1,z2,…,zn-1) is uniformly continuous on [1/k,1-1/k]×Γ(α-n+2)r/kαΓ(α),r/(n-2)!×⋯×[r/kα-n+2,r], we have that (33)limj→+∞⁡ft,I0+n-2xjt,…,xjt-ft,I0+n-2x0t,…,x0t=0holds uniformly on t∈[1/k,1-1/k]. It follows from the Lebesgue control convergence theorem that (34)∫1/k1-1/kΦsfs,I0+n-2xjs,…,xjs-fs,I0+n-2x0s,…,x0sds→0,asj→∞.Thus, for the above ε>0, there exists a natural number J such that, for j>J, we have(35)∫1/k1-1/kΦsfs,I0+n-2xjs,…,xjs-fs,I0+n-2x0s,…,x0sds<ε2.It follows from (32) and (35) that when j>J, (36)Txj-Tx0≤∫HkΦsfs,I0+n-2xjs,…,xjs-fs,I0+n-2x0s,…,x0sds+∫1/k1-1/kΦsfs,I0+n-2xjs,…,xjs-fs,I0+n-2x0s,…,x0sds≤2supx∈K¯R∖Kr⁡∫HkΦsfs,I0+n-2xs,…,xsds+ε2<2×ε4+ε2=ε.Hence, T:K¯R∖Kr→K is continuous.
For any bounded setB⊂K¯R∖Kr, we show that T(B) is equicontinuous. In fact, by (H1), for any ε>0, there exists a natural number l such that (37)supx∈K¯R∖Kr⁡∫HlΦsfs,I0+n-2xs,…,xsds<ε4.Let (38)M2=max⁡ft,z1,z2,…,zn-1:t,z1,z2,…,zn-1∈1l,1-1l×Γα-n+2rlαΓα,rn-2!×Γα-n+2rlα-1Γα-1,rn-3!×⋯×Γα-n+2rlα-n+3Γα-n+3,r×rlα-n+2,r.Since H(t,s) is uniformly continuous on [0,1]×[0,1], for the above ε>0 and a fixed s∈[1/l,1-1/l], there exists δ>0 such that (39)Ht,s-Ht′,s<ε2M2+1.for t-t′<δ, t,t′∈[0,1]. Hence, for t-t′<δ, t,t′∈[0,1], and x∈B, we have (40)Txt-Txt′≤∫HkHt,s-Ht′,sfs,I0+n-2xs,…,I0+1xs,xsds+∫1/k1-1/kHt,s-Ht′,sfs,I0+n-2xs,…,I0+1xs,xsds≤2supx∈K¯R∖Kr⁡∫ekΦsfs,I0+n-2xs,…,I0+1xs,xsds+supx∈K¯R∖Kr⁡∫1/k1-1/kHt,s-Ht′,sfs,I0+n-2xs,…,I0+1xs,xsds<2×ε4+ε2=ε,which show that T(B) is equicontinuous. According to the Ascoli-Arzela theorem, T:K¯R∖Kr→K is completely continuous. The proof is completed.Lemma 8 (see [35]).
LetK be a cone in Banach space E. Suppose that T:K¯r→K is a completely continuous operator.(i)
If there existsy0∈K∖{θ} such that y-Ty≠μy0 for any y∈∂Kr and μ≥0, then i(T,Kr,K)=0.(ii)
IfTy≠μy for any y∈∂Kr and μ≥1, then i(T,Kr,K)=1.


## 3. Main Results

Theorem 9.
Suppose that the conditions(H0) and (H1) are satisfied, and(41)liminfz1→0⋯zn-1→0⁡ft,z1,…,zn-1z1+⋯+zn-1>λ1,uniformlyont∈0,1,(42)limsupz1+⋯+zn-1→+∞zn-1→+∞⁡ft,z1,…,zn-1zn-1<λ1,uniformlyont∈0,1,where λ1 is the first eigenvalue of L defined by (16). Then the FBVP (3) has at least one positive solution.Proof.
By (41), there exists r>0 such that(43)ft,z1,z2,…,zn-1≥λ1z1+z2+⋯+zn-1,z1≤rn-2!,…,zn-1≤r,t∈0,1.Let φ∗ be the positive eigenfunction of L corresponding to λ1, and thus φ∗=λ1Lφ∗. For any x∈∂Kr, by virtue of (27) and (28), that is(44)I0+n-2xs≤rn-2!,…,I0+1xs≤r,xs≤r,t∈0,1,from (43) and (44), we have(45)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≥λ1∫01Ht,sI0+n-2xs+I0+n-3xs+⋯+I0+1xs+xsds≥λ1Lxt,t∈0,1.
We may suppose thatT has no fixed point on ∂Kr (otherwise, the proof is finished). Now we show that(46)x-Tx≠μφ∗,x∈∂Kr,μ≥0.
If, otherwise, there existx0∈∂Kr and μ0≥0 such that x0-Tx0=μ0φ∗, obviously, μ0>0 and x0=Tx0+μ0φ∗≥μ0φ∗. Let μ¯=sup⁡μ∣x0≥μφ∗, and then μ¯≥μ0>0 and x0≥μ¯φ∗. Since L(K)⊂K, we have λ1Lx0≥λ1μ¯Lφ∗=μ¯φ∗. Therefore, by (45), we have(47)x0=Tx0+μ0φ∗≥λ1Lx0+μ0φ∗≥μ¯φ∗+μ0φ∗=μ¯+μ0φ∗,which contradicts the definition of μ¯. Hence (46) is true and it follows from Lemma 8 that(48)iT,Kr,K=0.
Now we choose a constant0<ρ<1 and define a linear operator L~x=ρλ1Lx, and then L~:E→E is a bounded linear operator and L~(K)⊂K. Moreover L~φ∗=ρλ1Lφ∗=ρφ∗, so the spectral radius of L~ is r(L~)=ρ and L~ also has the first eigenvalue (r(L~))-1=ρ-1>1. By Gelfand’s formula, we know(49)ρ=limm→+∞⁡L~m1/m.
Letε0=1/21-ρ; by (49), there exists a sufficiently large natural number M such that m≥M implies that L~m≤(ρ+ε0)m. For any x∈E, define(50)x∗=∑i=1Mρ+ε0M-iL~i-1x,where L~0=I is the identity operator. Clearly, ·∗ is also the norm of E.
On the other hand, it follows from (42) that there exists R0>r such that(51)ft,z1,z2,…,zn-1≤ρλ1zn-1,z1+⋯+zn-1≥R0,zn-1≥R0.Let (52)M0=supx∈∂KR0⁡∫01Φsfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds,and then, by (25), we know that M0<+∞. For the sake of convenience, let M0∗=M0∗ and choose (53)R>max⁡R0,2M0∗ε0ρ+ε0M-1.
In the following we will prove that(54)Tx≠μx,x∈∂KR,μ≥1.
If, otherwise, there existx1∈∂KR and μ1≥1 such that Tx1=μ1x1. It follows from R>R0 and x1∈C[0,1] that there exists 0<t1≤1 such that x1(t1)=R0. Thus let x~(t)=min⁡{x1(t),R0}, and then we have x~=R0 and x~∈∂KR0. Now set (55)Dx1=t∈0,1:x1t>R0,and then, for any t∈D(x1), we have x1t>R0 and I0+n-2x1t+I0+n-3x1t+⋯+I0+1x1t+x1t>R0. It follows from (51) and Lemma 2 that(56)Tx1t=∫01Ht,sfs,I0+n-2x1s,I0+n-3x1s,…,I0+1x1s,x1sds≤∫Dx1Ht,sfs,I0+n-2x1s,I0+n-3x1s,…,I0+1x1s,x1sds+∫0,1∖Dx1Ht,sfs,I0+n-2x1s,I0+n-3x1s,…,I0+1x1s,x1sds≤ρλ1∫01Ht,sx1sds+∫01Φsfs,I0+n-2x1s,I0+n-3x1s,…,I0+1x1s,x1sds≤L~x1t+M0,t∈0,1.
Noticing thatL~(K)⊂K is a bounded linear operator, and from (56), we have(57)0≤L~jTx1t≤L~jL~x1+M0t,j=0,1,2,…,M-1.Then (57) yields that (58)L~jTx1≤L~jL~x1+M0,j=0,1,2,…,M-1,which leads to(59)Tx1∗=∑i=1Mρ+ε0M-iL~i-1Tx1≤∑i=1Mρ+ε0M-iL~i-1L~x1+M0=L~x1+M0∗.
Sincex1∈∂KR and x1=R, from (50), we have (60)x1∗>ρ+ε0M-1x1=ρ+ε0M-1R>2ε0M0∗,which implies that(61)M0∗<ε02x1∗.
By (50), (59), and (61), we have (62)μ1x1∗=Tx1∗≤L~x1+M0∗=∑i=1Mρ+ε0M-iL~i-1Lx1+M0∗=ρ+ε0∑i=1M-1ρ+ε0M-i-1L~ix1+L~Mx1+M0∗≤ρ+ε0∑i=1M-1ρ+ε0M-i-1L~ix1+ρ+ε0Mx1+M0∗=ρ+ε0∑i=1Mρ+ε0M-iL~i-1x1+M0∗=ρ+ε0x1∗+M0∗≤ρ+ε0x1∗+ε02x1∗=ρ+34x1∗.Notice that μ1≥1, we have ρ+3/4≥1, and then ρ≥1, which is a contradiction with 0<ρ<1. Thus (54) is indeed true, and, by Lemma 8, we have(63)iT,KR,K=1.It follows from (48) and (63) that (64)iT,KR∖K¯r,K=iT,KR,K-iT,Kr,K=1.Then T has at least one fixed point on KR∖K¯r. Consequently, problem (14) has at least one positive solution, which implies that the FBVP (3) has at least one positive solution.Now we consider another case of problem (14). For this, we define a linear operator Lτ for any sufficiently small 0<τ<1 as follows: (65)Lτxt=∫τ1-τHt,sxsds,t∈0,1.From Lemma 6, we know Lτ is also a completely continuous linear operator, and the spectral radius r(Lτ)≠0, and moreover Lτ has a positive eigenfunction φτ corresponding to its first eigenvalue λτ=(r(Lτ))-1.Lemma 10.
Suppose that(H0) holds, and there exists an eigenvalue λ~1 of L such that (66)limτ→0+⁡λτ=λ~1.Proof.
Takeτ1≥τ2≥⋯≥τk≥⋯ and τk→0(k→+∞). So, for any j>k and φ∈E, we have (67)Lτkφt≤Lτjφt≤Lφt,t∈0,1,and (68)Lτkiφt≤Lτjiφt≤Liφt,t∈0,1,i=2,3,…,where Lτki=L(Lτki-1),i=2,3,⋯. Consequently, Lτki≤Lτji≤Li,i=1,2,⋯. From Gelfand’s formula, we get λ1≤λτj≤λτk, where λ1 is the first eigenvalue of L. Since {λτk} is monotonous with lower boundedness λ1, let (69)limk→+∞⁡λτk=λ~1.
Now we shall show thatλ~1 is an eigenvalue of L. Suppose φτk is a positive eigenfunction of Lτk corresponding to λτk with φτk=1,k=1,2,…; i.e.,(70)φτkt=λτk∫τk1-τkHt,sφτksds=λτkLτkφτkt,t∈0,1.
Notice that(71)λτkφτk=max0≤t≤1⁡∫τk1-τkHt,sφτksds≤∫01Φsds,k=1,2,…,and thus {λτkφτk}⊂E is uniformly bounded.
On the other hand, for anyk and t1,t2∈[0,1], we have(72)Lτkφτkt1-Lτkφτkt2≤∫τk1-τkHt1,s-Ht2,sφτksds.It follows from (72) that {Lτkφτk}⊂E is equicontinuous since H(t,s) is uniformly continuous on [0,1]×[0,1]. Without loss of generality, we may suppose that, by Arzela-Ascoli theorem and limk→+∞⁡λτk=λ~1, we get that φτk→φ0 as k→+∞. This leads to φ0=1 and then, by (70), we have (73)φ0t=λ~1∫01Ht,sφ0sds,t∈0,1;that is, φ0=λ~1Lφ0. The proof is completed.Theorem 11.
Suppose that conditions(H0) and (H1) are satisfied, and(74)liminfz1→0⋯zn-1→0⁡ft,z1,…,zn-1zn-1<λ1,uniformlyont∈0,1,(75)limsupz1+⋯+zn-1→+∞⁡ft,z1,…,zn-1z1+⋯+zn-1>λ~1,uniformlyont∈0,1,where λ1,λ~1 are the eigenvalues of L and λ1 is the first eigenvalue of L. Then the FBVP (3) has at least one positive solution.Proof.
It follows from (74) that there exists r>0 such that(76)ft,z1,z2,…,zn-1≤λ1zn-1,z1≤rn-2!,…,zn-1≤r,t∈0,1.Thus for any x∈∂Kr, noticing (77)I0+n-2xs≤rn-2!,…,I0+1xs≤r,xs≤r,t∈0,1,by (76), we have(78)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≤λ1∫01Ht,sxsds=λ1Lxt,t∈0,1.In fact, we may suppose that T has no fixed point on ∂Kr (otherwise, the proof is finished). Now we show that(79)Tx≠μx,x∈∂Kr,μ≥1.
Otherwise, there existx0∈∂Kr and μ0≥1 such that Tx0=μ0x0. We know μ0>1 and from (78) we have(80)μ0x0=Tx0≤λ1Lx0.By using induction for (80), we have (81)μ0mx0≤λ1mLmx0,m=1,2,⋯which implies that (82)Lm≥Lmx0x0≥μ0mx0λ1mx0=μ0mλ1m.By Gelfand’s formula, we know (83)rL=limm→+∞⁡Lm1/m≥μ0λ1>1λ1,which is a contradiction with r(L)=λ1-1. Hence (79) is true and, by Lemma 8, we have(84)iT,Kr,K=1.
For a fixed sufficiently small0<τ<1, let (85)χτ=ταΓα-n+2Γα+τα-1Γα-n+2Γα-1+⋯+τα-n+3Γα-n+2Γα-n+3+τα-n+2.So it follows from (74) and limτ→0+⁡λτ=λ~1 that there exist a sufficiently small τ>0 and R>r such that(86)ft,z1,z2,…,zn-1≥λτz1+z2+⋯+zn-1,z1+z2+⋯+zn-1≥χτR,t∈0,1,where λτ is the first eigenvalue of Lτ.
Letφτ be the positive eigenfunction of Lτ corresponding to λτ, and then φτ=λτLτφτ. For any x∈∂KR, s∈[τ,1-τ], we have(87)I0+n-2xs+I0+n-3xs+⋯,I0+1xs+xs=1Γn-2∫0ss-τn-3xτdτ+1Γn-3∫0ss-τn-4xτdτ+⋯+1Γ1∫0ss-τ1-1xτdτ+xs≥1Γn-2∫0ss-τn-3τα-n+11-τdτx+1Γn-3∫0ss-τn-4τα-n+11-τdτx+⋯+1Γ1∫0ss-τ1-1τα-n+11-τdτx+sα-n+11-sx≥ταΓα-n+2Γα+τα-1Γα-n+2Γα-1+⋯+τα-n+3Γα-n+2Γα-n+3+τα-n+2R=χτR.
By (86) and (87), we have (88)Txt=∫01Ht,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≥∫τ1-τHt,sfs,I0+n-2xs,I0+n-3xs,…,I0+1xs,xsds≥λτ∫τ1-τHt,sI0+n-2xs+I0+n-3xs+⋯,I0+1xs+xsds≥λτ∫τ1-τHt,sxsds=λτLτxt,t∈0,1.
Proceeding as for the proof of Theorem9, we get (89)x-Tx≠μφτ,x∈∂KR,μ≥0.By Lemma 8, we have(90)iT,KR,K=0.It follows from (84) and (90) that (91)iT,KR∖K¯r,K=iT,KR,K-iT,Kr,K=-1.Then T has at least one fixed point on KR∖K¯r. Consequently, problem (14) has at least one positive solution, which implies that the FBVP (3) has at least one positive solution.Remark 12.
In this work, the nonlinearityf(t,z1,z2,…,zn-1) may be singular at t=0,1 and z1=z2=⋯=zn-1=0. To the best of our knowledge, very little work has been done for the case where f can be singular at z1=z2=⋯=zn-1=0.


## 4. An Example

Example 1.
Consider the following problem:(92)D0+7/2yt+y+y′+y″-1/4+ln⁡y″1-t=0,0<t<1,y0=y′0=y″0=0,y″1=λy″,where α=7/2, f(t,z1,z2,z3)=z1+z2+z3-1/4+ln⁡z3/1-t, and (t,z1,z2,z3)∈(0,1)×(R0+)3. Obviously, f is singular at t=1 and z1=z2=z3=0. Clearly, (93)Gt,s=G1t,s=t1-s1/2Γ3/2,0≤t≤s≤1,G2t,s=t1-s1/2-t-s1/2Γ3/2,0≤s≤t≤1.In the following we discuss that condition (H0) holds when λ[·] take different cases.(1)
Letλ[y″]=0. In this case, we have (94)λγ=0,Gs=0,Φs=s1-s1/2Γ1/2≤131-s1/2Γ1/2.(2)
Now, letλ[y″]=∫01y″(t)(4-5t)dt. Note that the function h(t)=4-5t changes the sign on the interval [0,1]. In this case, we have (95)λγ=∫01t1/2dAt=∫01t1/24-5tdt=23<1,Gs=∫01Gt,s4-5tdt=1Γ5/21-s1/2s3-2s≥0,s∈0,1,Φs=Gs1-λγ+s1-sα-n+1Γα-n+1=3s1-s1/23-2sΓ5/2+s1-s1/2Γ1/2≤131-s1/2Γ1/2.(3)
Letλy″=1/2y″1/2. In this case, we have (96)λγ=∫01t1/2dAt=24≈0.353553<1,0≤Gs=12G12,s=1/21-s1/2Γ3/2,12≤s≤1,1/21-s1/2-1/2-s1/2Γ3/2,0≤s≤12.≤2/41-s1/2Γ3/2,s∈0,1,Φs=Gs1-λγ+s1-sα-n+1Γα-n+1≤21-s1/24-2Γ3/2+s1-s1/2Γ1/2≤131-s1/2Γ1/2.(4)
Letλ[y″]=2y″1/2-y″3/4. Note that the coefficients b1=2, b2=-1; i.e., not all of the coefficients must be positive, some coefficients of bi can be negative. In this case, we have (97)
λ
γ
=
∫
0
1
t
1
/
2
d
A
t
=
2
-
3
2
≈
0.548188
<
1
,
G
s
=
2
G
2
1
2
,
s
-
G
2
3
4
,
s
,
0
≤
s
<
1
2
,
2
G
1
1
2
,
s
-
G
2
3
4
,
s
,
1
2
≤
s
≤
3
4
,
2
G
1
1
2
,
s
-
G
1
3
4
,
s
,
3
4
<
s
≤
1
.Then 0≤G(s)<1,s∈[0,1], and (98)Φs=Gs1-λγ+s1-sα-n+1Γα-n+1≤221-s1/22-22+3Γ3/2+s1-s1/2Γ1/2≤131-s1/2Γ1/2.
Seen from above, condition(H0) holds in different cases 1-4. Now we define a cone(99)K=y∈C0,1:yt≥t1/21-tyfort∈0,1.For any 0<r<R<+∞ and z3∈K¯R∖Kr, we have(100)0≤rt5/21-t≤t1/21-tr=t1/21-tz3≤z3t≤R,t∈0,1,0≤23rt5/21-t≤23rt3/21-t=rt3/21-tΓ3/2Γ5/2≤z2t=I0+1z3t≤R,t∈0,1,0≤415rt5/21-t=rt5/21-tΓ3/2Γ7/2≤z1t=I0+2z3t≤R2,t∈0,1.Noticing that ln⁡z3 is decreasing on (0,1) and is increasing on [1,+∞), so we have (101)ln⁡z3t≤ln⁡rt5/21-t+ln⁡R≤ln⁡r+ln⁡R+ln⁡t5/21-t,z1t+z2t+z3t-1/4≤2915r-1/4t-5/81-t-1/4.Consequently(102)∫01ln⁡t1/21-t+2915r-1/4t-5/81-t-1/4dt≤12∫01ln⁡tdt+∫01ln⁡1-tdt+2915r-1/4∫01t-5/81-t-1/4dt≤32+2915r-1/4B38,34<+∞.The absolute continuity of integral leads to (103)limm→+∞⁡∫Hmln⁡t1/21-t+2915r-1/4t-5/81-t-1/4dt=0.Hence(104)limm→+∞⁡supz1∈KR/2z2∈KRz3∈K¯R∖Kr⁡∫HmΦsfs,z1s,z2s,z3sds≤limm→+∞⁡supz1∈KR/2z2∈KRz3∈K¯R∖Kr⁡∫Hm131-s1/2Γ1/2×z1s+z2s+z3s-1/4+ln⁡z3s1-sds≤limm→+∞⁡supz1∈KR/2z2∈KRz3∈K¯R∖Kr⁡∫Hm13Γ1/2ln⁡R+ln⁡r+ln⁡s1/21-s+2915r-1/4s-5/81-s-1/4ds=26Γ1/2ln⁡R+ln⁡rlimm→+∞⁡1m+13Γ1/2limm→+∞⁡∫Hmln⁡s1/21-s+2915r-1/4s-5/81-s-1/4ds=0,
On the other hand, it is obvious that(105)liminfz1→0z2→0z3→0⁡ft,z1,z2,z3z1+z2+z3=liminfz1→0z2→0z3→0⁡z1+z2+z3-1/2+ln⁡z3t1-tz1+z2+z3=+∞,limsupz1+z2+z3→+∞z3→+∞⁡ft,z1,z2,z3z3=limsupz1+z2+z3→+∞z3→+∞⁡z1+z2+z3-1/2+ln⁡z3z3t1-t=0,uniformly on t∈[0,1], which implies that (106)limsupz1+z2+z3→+∞z3→+∞⁡ft,z1,z2,z3z3<λ1<liminfz1→0z2→0z3→0⁡ft,z1,z2,z3z1+z2+z3.
Therefore, all conditions of Theorem9 are satisfied. Thus Theorem 9 ensures that FBVP (92) has at least one positive solution.



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*Source: 6598351-2018-10-01.xml*