# Solutions of Sign-Changing Fractional Differential Equation with the Fractional Derivatives

Authors: Tunhua Wu; Xinguang Zhang; Yinan Lu
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We study the singular fractional-order boundary-value problem with a sign-changing nonlinear term-𝒟αx(t)=p(t)f(t,x(t),𝒟μ1x(t),𝒟μ2x(t),…,𝒟μn-1x(t)),0<t<1,𝒟μix(0)=0,1≤i≤n-1,𝒟μn-1+1x(0)=0,
𝒟μn-1x(1)=∑j=1p-2aj𝒟μn-1x(ξj), where n-1<α≤n, n∈ℕ and n≥3 with 0<μ1<μ2<⋯<μn-2<μn-1 and n-3<μn-1<α-2, aj∈ℝ,0<ξ1<ξ2<⋯<ξp-2<1 satisfying 0<∑j=1p-2ajξjα-μn-1-1<1, 𝒟α is the standard Riemann-Liouville derivative, f:[0,1]×ℝn→ℝ is a sign-changing continuous function and may be unbounded from below with respect to xi, and p:(0,1)→[0,∞) is continuous. Some new results on the existence of nontrivial solutions for the above problem are obtained by computing the topological degree of a completely continuous field.


---

## Body

## 1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines, and particularly in the mathematical modeling of systems and processes in physics, chemistry, aerodynamics, electrodynamics of complex medium, and polymer rheology [1–6]. Fractional-order models have proved to be more accurate than integer-order models, that is, there are more degrees of freedom in the fractional-order models. Hence fractional differential equations have attracted great research interest in recent years, and for more details we refer the reader to [7–16] and the references cited therein.In this paper, we consider the existence of nontrivial solutions for the following singular fractional-order boundary-value problem with a sign-changing nonlinear term  and fractional derivatives:(1.1)-𝒟αx(t)=p(t)f(t,x(t),𝒟μ1x(t),𝒟μ2x(t),…,𝒟μn-1x(t)),0<t<1,𝒟μix(0)=0,1≤i≤n-1,𝒟μn-1+1x(0)=0,𝒟μn-1x(1)=∑j=1p-2aj𝒟μn-1x(ξj),
where n-1<α≤n,n∈ℕ and n≥3 with 0<μ1<μ2<⋯<μn-2<μn-1 and n-3<μn-1<α-2, aj∈ℝ,0<ξ1<ξ2<⋯<ξp-2<1 satisfying 0<∑j=1p-2‍ajξjα-μn-1-1<1, 𝒟α is the standard Riemann-Liouville derivative, f:[0,1]×ℝn→ℝ is a sign-changing continuous function and may be unbounded from below with respect to xi, and p:(0,1)→[0,∞) is continuous.In this paper, we assume thatf:[0,1]×ℝn→ℝ, which implies that the problem (1.1) is changing sign (or semipositone particularly). Differential equations with changing-sign arguments are found to be important mathematical tools for the better understanding of several real-world problems in physics, chemistry, mechanics, engineering, and economics [17–19]. In general, the cone theory is difficult to handle this type of problems since the operator generated by f is not a cone mapping. So to find a new method to solve changing-sign problems is an interesting, important, and difficult work. An effective approach to this problem was recently suggested by Sun [20] based on the topological degree of a completely continuous field. Then, Han and Wu [21, 22] obtained a new Leray-Schauder degree theorem by improving the results of Sun [20]. In [22], Han et al. also investigated a kind of singular two-point boundary-value problems with sign-changing nonlinear terms by applying the new Leray-Schauder degree theorem obtained in [22].To our knowledge, very few results have been established whenf is changing sign [20–24]. In [20, 21, 23], f permits sign changing but required to be bounded from below. In [22, Theorem 1.1], f may be a sign-changing and unbounded function, but the Green function must be symmetric and f is controlled by a special function h(u)=-b-c|u|μ, where b>0,c>0 and μ∈(0,1). Recently, by improving and generalizing the main results of Sun [20] and Han et al. [21, 22], Liu et al. [24] established a generalized Leray-Schauder degree theorem of a completely continuous field for solving  m-point boundary-value problems for singular second-order differential equations.Motivated by [20–24], we established some new results on the existence of nontrivial solutions for the problem (1.1) by computing the topological degree of a completely continuous field. The conditions used in the present paper are weaker than the conditions given in previous works [20–24], and particularly we drop the assumption of even function in [24]. The new features of this paper mainly include the following aspects. Firstly, the nonlinear term f(t,x1,x2,…,xn) in the BVP (1.1) is allowed to be sign changing and unbounded from below with respect to xi. Secondly, the nonlinear term f involves fractional derivatives of unknown functions. Thirdly, the boundary conditions involve fractional derivatives of unknown functions which is a more general case, and include the two-point, three-point, multipoint, and some nonlocal problems as special cases of (1.1).


## 2. Preliminaries and Lemmas

In this section, we give some preliminaries and lemmas.Definition 2.1.
LetE be a real Banach space. A nonempty closed convex set P⊂E is called a cone of E if it satisfies the following two conditions:
(1)x∈P,σ>0 implies σx∈P;
(2)x∈P,-x∈P implies x=θ.Definition 2.2.
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
LetE be a real Banach space, E* the dual space of E, P a total cone in E, that is, E=P-P,¯ and P* the dual cone of P.Lemma 2.3 (Deimling [25]).
LetL:E→E be a continuous linear operator, P 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 has a positive eigenfunction corresponding to its first eigenvalue λ=r(L)-1.Lemma 2.4 (see [25]).
LetP be a cone of the real Banach space E, and Ω a bounded open subsets of E. Suppose that T:Ω¯∩P→P is a completely continuous operator. If there exists x0∈P∖{θ} such that x-Tx≠μx0,∀x∈∂Ω∩P,μ≥0, then the fixed-point index  i(T,Ω∩P,P)=0.
LetL:E→E be a completely continuous linear positive operator with the spectral radius r1≠0. On account of Lemma 2.3, there exist φ1∈P∖{θ} and g1∈P*∖{θ} such that
(2.1)Lφ1=r1φ1,r1L*g1=g1,
where L* is the dual operator of L. Choose a number δ>0 and let
(2.2)P(g1,δ)={u∈P:g1(u)≥δ||u||},
then P(g1,δ) is a cone in E.Lemma 2.5 (see [24]).
Suppose that the following conditions are satisfied.(A1)
T:E→P is a continuous operator satisfying
(2.3)lim∥u∥→+∞∥Tu∥∥u∥=0;(A2)
F:E→E is a bounded continuous operator and there exists u0∈E such that Fu+u0+Tu∈P, for all u∈E;(A3)
r1>0 and there exist v0∈E and η>0 such that
(2.4)LFu≥r1(1+η)Lu-LTu-v0,∀u∈E.Let A=LF, then there exists R>0 such that
(2.5)deg(I-A,BR,θ)=0,
where BR={u∈E:∥u∥<R} is the open ball of radius R in E.Remark 2.6.
If the operatorT which satisfies the conditions of Lemma 2.5 is a null operator, then Lemma 2.5 turns into Theorem  1 in [20]. On the other hand, if the operator T in Lemma 2.5 is such that there exist constants α∈(0,1) and N>0 satisfying ||Tu||≤N||u||α for all u∈E, then Lemma 2.5 turns into Theorem  2.1 in [22] or Theorem  1 in [21]. So Lemma 2.5 is an improvement of the results of paper [20–22].
Now we present the necessary definitions from fractional calculus theory. These definitions can be found in some recent literatures, for example, [26, 27].Definition 2.7 (see [26, 27]).
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,+∞)→ℝ is given by
(2.6)Iαx(t)=1Γ(α)∫0t‍(t-s)α-1x(s)ds
provided that the right-hand side is pointwisely defined on (0,+∞).Definition 2.8 (see [26, 27]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x:(0,+∞)→ℝ is given by
(2.7)𝒟αx(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1x(s)ds,
where n=[α]+1, and [α] denotes the integer part of the number α, provided that the right-hand side is pointwisely defined on (0,+∞).Remark 2.9.
Ifx,y:(0,+∞)→ℝ with order α>0, then
(2.8)𝒟α(x(t)+y(t))=𝒟αx(t)+𝒟αy(t).Lemma 2.10 (see [27]).
(1) Ifx∈L1(0,1),ρ>σ>0, then
(2.9)IρIσx(t)=Iρ+σx(t),𝒟σIρx(t)=Iρ-σx(t),𝒟σIσx(t)=x(t).
(2) Ifρ>0,σ>0, then
(2.10)𝒟ρtσ-1=Γ(σ)Γ(σ-ρ)tσ-ρ-1.Lemma 2.11 (see [27]).
Assume thatx∈C(0,1)∩L1(0,1) with a fractional derivative of order α>0. Then
(2.11)Iα𝒟αx(t)=x(t)+c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈ℝ(i=1,2,…,n), n is the smallest integer greater than or equal to α.Noticing that2<α-μn-1≤n-μn-1<3, let
(2.12)k(t,s)={(t(1-s))α-μn-1-1-(t-s)α-μn-1-1Γ(α-μn-1),0≤s≤t≤1,(t(1-s))α-μn-1-1Γ(α-μn-1),0≤t≤s≤1,
by [28], for t,s∈[0,1], one has
(2.13)tα-μn-1-1(1-t)s(1-s)α-μn-1-1Γ(α-μn-1)≤k(t,s)≤s(1-s)α-μn-1-1Γ(α-μn-1).Lemma 2.12.
If2<α-μn-1<3 and ρ∈L1[0,1], then the boundary-value problem
(2.14)𝒟α-μn-1w(t)+ρ(t)=0,w(0)=w′(0)=0,w(1)=∑j=1p-2‍ajw(ξj),
has the unique solution
(2.15)w(t)=∫01‍K(t,s)ρ(s)ds,
where
(2.16)K(t,s)=k(t,s)+tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1∑j=1p-2‍ajk(ξj,s),
is the Green function of the boundary-value problem (2.14).Proof.
By applying Lemma2.11, we may reduce (2.14) to an equivalent integral equation:
(2.17)w(t)=-Iα-μn-1ρ(t)+c1tα-μn-1-1+c2tα-μn-1-2+c3tα-μn-1-3,c1,c2,c3∈ℝ.
Note that w(0)=w'(0)=0 and (2.17), we have c2=c3=0. Consequently the general solution of (2.14) is
(2.18)w(t)=-Iα-μn-1ρ(t)+c1tα-μn-1-1.
By (2.18) and Lemma 2.10, we have
(2.19)w(t)=-Iα-μn-1ρ(t)+c1tα-μn-1-1=-∫0t‍(t-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds+c1tα-μn-1-1.
So,
(2.20)w(1)=-∫01‍(1-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds+c1,
and for j=1,2,…,p-2,
(2.21)w(ξj)=-∫0ξj‍(ξj-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds+c1ξjα-μn-1-1.
By w(1)=∑j=1p-2‍ajw(ξj), combining with (2.20) and (2.21), we obtain
(2.22)c1=∫01‍(1-s)α-μn-1-1ρ(s)ds-∑j=1p-2aj∫0ξj‍(ξj-s)α-μn-1-1ρ(s)dsΓ(α-μn-1)(1-∑j=1p-2ajξjα-μn-1-1).
So, the unique solution of problem (2.14) is
(2.23)w(t)=-∫0t‍(t-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds+tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1{∫01‍(1-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds-∑j=1p-2‍aj∫0ξj‍(ξj-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds}=-∫0t‍(t-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds+∫01‍(1-s)α-μn-1-1tα-μn-1-1Γ(α-μn-1)ρ(s)ds+tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1∑j=1p-2‍aj∫01‍(1-s)α-μn-1-1ξjα-μn-1-1Γ(α-μn-1)ρ(s)ds-tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1∑j=1p-2‍aj∫0ξj‍(ξj-s)α-μn-1-1Γ(α-μn-1)ρ(s)ds=∫01‍(k(t,s)+tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1∑j=1p-2‍ajk(ξj,s))ρ(s)ds=∫01‍K(t,s)ρ(s)ds.
The proof is completed.Lemma 2.13.
The functionK(t,s) has the following properties:
(1)K(t,s)>0,fort,s∈(0,1);
(2)K(t,s)≤Ms(1-s)α-μn-1-1,fort,s∈[0,1], where
(2.24)M=1Γ(α-μn-1)(1+∑j=1p-2aj1-∑j=1p-2ajξjα-μn-1-1).Proof.
It is obvious that (1) holds. In the following, we will prove (2). In fact, by (2.13), we have
(2.25)K(t,s)=k(t,s)+tα-μn-1-11-∑j=1p-2ajξjα-μn-1-1∑j=1p-2‍ajk(ξj,s)≤s(1-s)α-μn-1-1Γ(α-μn-1)+∑j=1p-2aj1-∑j=1p-2ajξjα-μn-1-1s(1-s)α-μn-1-1Γ(α-μn-1)≤(1+∑j=1p-2aj1-∑j=1p-2ajξjα-μn-1-1)s(1-s)α-μn-1-1Γ(α-μn-1).
This completes the proof.Now let us consider the following modified problems of the BVP (1.1):
(2.26)-𝒟α-μn-1u(t)=p(t)f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t)),u(0)=0,u′(0)=0,𝒟μ-μn-1u(1)=∑j=1p-2‍aj𝒟μ-μn-1u(ξj).Lemma 2.14.
Letx(t)=Iμn-1u(t),u(t)∈C[0,1], Then (1.1) can be transformed into (2.26). Moreover, if u∈C([0,1],[0,+∞) is a solution of problem (2.26), then, the function x(t)=Iμn-1u(t) is a positive solution of problem (1.1).Proof.
Substitutingx(t)=Iμn-1u(t) into (1.1), by the definition of the Riemann-Liouville fractional derivative and Lemmas 2.10 and 2.11, we obtain that
(2.27)𝒟αx(t)=dndtnIn-αx(t)=dndtnIn-αIμn-1u(t)=dndtnIn-α+μn-1u(t)=𝒟α-μn-1u(t),𝒟μ1x(t)=𝒟μ1Iμn-1u(t)=Iμn-1-μ1u(t),𝒟μ2x(t)=𝒟μ2Iμn-1u(t)=Iμn-1-μ2u(t),⋮⋮𝒟μn-2x(t)=𝒟μn-2Iμn-1u(t)=Iμn-1-μn-2u(t),𝒟μn-1x(t)=𝒟μn-1Iμn-1u(t)=u(t),𝒟μn-1+1x(t)=[𝒟μn-1Iμn-1u(t)]′=u′(t).
Also, we have 𝒟μn-1x(0)=u(0)=0,𝒟μn-1+1x(0)=u'(0)=0, and it follows from 𝒟μn-1x(t)=u(t) that u(1)=∑j=1p-2‍aju(ξj). Hence, by x(t)=Iμn-1u(t),u∈C[0,1], (1.1) is transformed into (2.14).
Now, letu∈C([0,1,0,+∞)) be a solution of problem (2.26). Then, by Lemma 2.10, (2.26), and (2.27), one has
(2.28)-𝒟αx(t)=-dndtnIn-αx(t)=-dndtnIn-αIμn-1u(t)=-dndtnIn-α+μn-1u(t)=-𝒟α-μn-1u(t)=p(t)f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))=p(t)f(t,x(t),𝒟μ1x(t),𝒟μ2x(t),…,𝒟μn-1x(t)),0<t<1.
Noticing that
(2.29)Iαu(t)=1Γ(α)∫0t‍(t-s)α-1u(s)ds,
which implies that Iαu(0)=0, from (2.27), for i=1,2,…,n-1, we have
(2.30)𝒟μix(0)=0,𝒟μn-1+1x(0)=0,𝒟μn-1x(1)=∑j=1p-2‍aj𝒟μn-1x(ξj).
Moreover, it follows from the monotonicity and property ofIμn-1 that
(2.31)Iμn-1u∈C([0,1],[0,+∞)).
Consequently,x(t)=Iμn-1u(t) is a positive solution of problem (1.1).In the following let us list some assumptions to be used in the rest of this paper.(H1)p:(0,1)→[0,+∞) is continuous, p(t)≢0 on any subinterval of (0,1), and
(2.32)∫01‍p(s)ds<+∞.(H2)f:[0,1]×ℝn→(-∞,+∞) is continuous.In order to use Lemma2.5, let E=C[0,1] be our real Banach space with the norm ∥u∥=maxt∈[0,1]|u(t)| and P={u∈C[0,1]:u(t)≥0,forallt∈[0,1]}, then P is a total cone in E.Define two linear operatorsL,J:C[0,1]→C[0,1] by
(2.33)(Lu)(t)=∫01‍K(t,s)p(s)u(s)ds,t∈[0,1],u∈C[0,1],(2.34)(Ju)(t)=∫01‍K(s,t)p(s)u(s)ds,t∈[0,1],u∈C[0,1],
and define a nonlinear operator A:C[0,1]→C[0,1] by
(2.35)(Au)(t)=∫01‍K(t,s)p(s)f(t,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds,t∈[0,1].Lemma 2.15.
Assume (H1) holds. Then(i)
L,J:C[0,1]→C[0,1] are completely continuous positive linear operators with the first eigenvalue r>0 and λ>0, respectively.(ii)
L satisfies L(P)⊂P(g1,δ).Proof.
(i)  By using the similar method of paper [22], it is easy to know that L,J:C[0,1]→C[0,1] are completely continuous positive linear operators. In the following, by using the Krein-Rutmann’s theorem, we prove that L,J have the first eigenvalue r>0 and λ>0, respectively.
In fact, it is obvious that there ist1∈(0,1) such that K(t1,t1)p(t1)>0. Thus there exists [a,b]⊂(0,1) such that t1∈(a,b) and K(t,s)p(s)>0 for all t,s∈[a,b]. Choose ψ∈P such that ψ(t1)>0 and ψ(t)=0 for all t∉[a,b]. Then for t∈[a,b],
(2.36)(Lψ)(t)=∫01‍K(t,s)p(s)ψ(s)ds≥∫ab‍K(t,s)p(s)ψ(s)ds>0.
So there existsν>0 such that ν(Lψ)(t)≥ψ(t) for t∈[0,1]. It follows from Lemma 2.3 that the spectral radius r1≠0. Thus corresponding to r=r1-1, the first eigenvalue of L, and L has a positive eigenvector φ1 such that
(2.37)rLφ1=φ1.
In the same way,J has a positive first eigenvalue λ and a positive eigenvector φ2 corresponding to the first eigenvalue λ, which satisfy
(2.38)λJφ2=φ2.
(ii) Notice thatK(t,0)=K(t,1)≡0 for t∈[0,1], by λJφ2=φ2 and (2.12)–(2.16), we have φ2(0)=φ2(1)=0. This implies that φ2′(0)>0 and φ2′(1)<0 (see [29]). Define a function χ on [0,1] by
(2.39)χ(s)={φ2′(0),s=0,φ2(s)(1-s)s,0<s<1,-φ2′(1),s=1.
Thenχ is continuous on [0,1] and χ(s)>0 for all s∈[0,1]. So, there exist δ1,δ2>0 such that δ1≤χ(s)≤δ2 for all s∈[0,1]. Thus
(2.40)δ1(1-s)s≤φ2(s)≤δ2(1-s)s,
for all s∈[0,1].
LetL* be the dual operator of L, we will show that there exists g1∈P*∖{θ} such that
(2.41)λL*g1=g1.
In fact, let
(2.42)g1(u)=∫01‍p(t)φ2(t)u(t)dt∀u∈E.
Then by (H1) and (2.40), we have
(2.43)∫01‍p(t)φ2(t)u(t)dt≤δ2∥u∥∫01‍t(1-t)p(t)dt≤δ2∥u∥∫01‍p(t)dt<+∞,
which implies that g1 is well defined. We state that g1 of (2.42) satisfies (2.41). In fact, by (2.40), (2.41), and interchanging the order of integration, for any s,t∈[0,1], we have
(2.44)λ-1g1(u)=∫01‍p(t)(λ-1φ2(t))u(t)dt=∫01‍p(t)(Jφ2)(t)u(t)dt=∫01‍p(t)u(t)∫01‍K(s,t)p(s)φ2(s)dsdt=∫01‍p(s)φ2(s)∫01‍K(s,t)p(t)u(t)dtds=∫01‍p(s)φ2(s)(Lu)(s)ds=g1(Lu)=(L*g1)(u)∀u∈E.
So (2.41) holds.
In the following we prove thatL(P)⊂P(g1,δ). In fact, by (2.41) and α-μn-1-1>1, we have
(2.45)φ2(s)≥δ1(1-s)s≥δ1(1-s)α-μn-1-1s≥δ1M-1K(t,s),t,s∈[0,1].
Take δ=δ1M-1λ-1>0 in (2.2). For any u∈P, by (2.44), (2.45), we have
(2.46)g1(Lu)=λ-1g1(u)=λ-1∫01‍p(s)φ2(s)u(s)ds≥δ1M-1λ-1∫01‍K(t,s)p(s)u(s)ds=δ(Lu)(t)∀t∈[0,1].
Hence, g1(Lu)≥δ||Lu||, that is, L(P)⊂P(g1,δ). The proof is completed.


## 3. Main Result

Theorem 3.1.
Assume that (H1)(H2) hold, and the following conditions are satisfied.
(H3) There exist nonnegative continuous functionsb,c:[0,1]→(0,+∞) and a nondecreasing continuous function h:ℝn→[0,+∞) satisfying
(3.1)lim∑i=1n|xi|→+∞h(x1,x2,…,xn)∑i=1n|xi|=0,f(t,x1,x2,…,xn)≥-b(t)-c(t)h(x1,x2,…,xn),∀xi∈ℝ.
(H4)f also satisfies
(3.2)liminf∑i=1n-1xi≥0∑i=1nxi→+∞f(t,x1,x2,…,xn)∑i=1nxi>λlimsup∑i=1n|xi|→0|f(t,x1,x2,…,xn)|∑i=1n|xi|<λ,
uniformly on t∈[0,1], where λ is the first eigenvalue of the operator J defined by (2.34).
Then the singular fractional-order boundary-value problem (1.1) has at least one nontrivial solution.Proof.
According to Lemma2.15, L satisfies L(P)⊂P(g1,δ). Let
(3.3)(Tu)(t)=ch(Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))foru∈E,
where c=maxt∈[0,1]c(t). It follows from (H3) that T:E→P is a continuous operator. Note that
(3.4)Iμn-1u(t)=∫0t‍(t-s)μn-1-1u(s)Γ(μn-1)ds≤∥u∥Γ(μn-1),Iμn-1-μiu(t)=∫0t‍(t-s)μn-1-μi-1u(s)Γ(μn-1-μi)ds≤∥u∥Γ(μn-1-μi),i=1,2,…,n-2.
Thus from the monotone assumption of h on xi, we have
(3.5)h(Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))≤h(∥u∥Γ(μn-1),∥u∥Γ(μn-1-μ1),…,∥u∥Γ(μn-1-μn-2),∥u∥),for  anyu∈E,t∈[0,1],
which implies that
(3.6)∥Tu∥=maxt∈[0,1]{ch(Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))}≤ch(∥u∥Γ(μn-1),∥u∥Γ(μn-1-μ1),…,∥u∥Γ(μn-1-μn-2),∥u∥)for  anyu∈E.
Let
(3.7)τ=1Γ(μn-1)+1Γ(μn-1-μ1)+⋯+1Γ(μn-1-μn-2)+1,
then
(3.8)lim∥u∥→+∞∥Tu∥∥u∥≤lim∥u∥→+∞ch(∥u∥/Γ(μn-1),∥u∥/Γ(μn-1-μ1),…,∥u∥/Γ(μn-1-μn-2),∥u∥)∥u∥=lim∥u∥→+∞cτh(∥u∥/Γ(μn-1),∥u∥/Γ(μn-1-μ1),…,∥u∥/Γ(μn-1-μn-2),∥u∥)τ∥u∥≤lim∥u∥→+∞cτh(∥u∥/Γ(μn-1),∥u∥/Γ(μn-1-μ1),…,∥u∥/Γ(μn-1-μn-2),∥u∥)∥u∥/Γ(μn-1)+∥u∥/Γ(μn-1-μ1)+⋯+∥u∥/Γ(μn-1-μn-2)+∥u∥=cτlim∑i=1n|xi|→+∞h(x1,x2,…,xn)∑i=1n|xi|=0,
that is,
(3.9)lim∥u∥→+∞∥Tu∥∥u∥=0.
Hence T satisfies condition (A1) in Lemma 2.5.
Next takeu0(t)≡b(t), and (Fu)(t)=f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t)) for u∈E. Then it follows from (H3) that
(3.10)Fu+u0+Tu=f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))+b(t)+ch(Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))≥0,
that yields
(3.11)Fu+u0+Tu∈P,∀u∈E,
namely, condition (A2) in Lemma 2.5 holds.
From (H4), there existsε>0 and a sufficiently large l1>0 such that, for any t∈[0,1],
(3.12)f(t,x1,x2,…,xn)≥λ(1+ε)(x1+x2+⋯+xn)≥λ(1+ε)xn,x1+x2+⋯+xn>l1.
Combining (H3) with (3.12), there exists b1≥0 such that
(3.13)Fu=f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))≥λ(1+ε)u(t)-b1-ch(Iμn-1u(t),…,Iμn-1-μn-2u(t),u(t)),∀u∈E,
that is,
(3.14)Fu≥λ(1+ε)u-b1-Tu∀u∈E.
As L is a positive linear operator, it follows from (3.14) that
(3.15)(LFu)(t)≥λ(1+ε)(Lu)(t)-Lb1-(LTu)(t),∀t∈[0,1].
So condition (A3) in Lemma 2.5 holds. According to Lemma 2.5, there exists a sufficiently large number R>0 such that
(3.16)deg(I-A,BR,θ)=0.
On the other hand, it follows from (H4) that there exist0<ε<1 and 0<r<R, for any t∈[0,1], such that
(3.17)|f(t,x1,x2,…,xn)|≤1-εMτ∫01‍p(s)ds(|x1|+|x2|+⋯+|xn|),|x1|+|x2|+⋯+|xn|<r.
Thus for any u∈E with ||u||≤r/τ≤r≤R, we have
(3.18)|Iμn-1u(t)|+|Iμn-1-μ1u(t)|+⋯+|Iμn-1-μn-2u(t)|+|u(t)|≤∥u∥Γ(μn-1)+∥u∥Γ(μn-1-μ1)+⋯+∥u∥Γ(μn-1-μn-2)+∥u∥≤τ∥u∥≤r.
By (3.17), for any t∈[0,1], we have
(3.19)|f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t))|≤1-εMτ∫01‍p(s)ds(|Iμn-1u(t)|+|Iμn-1-μ1u(t)|+⋯+|Iμn-1-μn-2u(t)|+|u(t)|).
Thus if there exist u1∈∂Br/τ and μ1∈[0,1] such that u1=μ1Au1, then by (3.19), we have
(3.20)g1(∥u1∥)=g1(μ1∥Au1∥)=μ1g1(∥Au1∥)=μ1g1(max0≤t≤1|∫01‍K(t,s)p(s)f(s,Iμn-1u1(s),…,Iμn-1-μn-2u1(s),u1(s))ds|)≤1-εMτ∫01‍p(s)dsg1(∫01‍Mp(s)(|Iμn-1u1(s)|+⋯+|Iμn-1-μn-2u1(s)|+|u1(s)|)ds)≤1-εMτ∫01‍p(s)dsMτ∫01‍p(s)dsg1(∥u1∥)=(1-ε)g1(∥u1∥).
Therefore, g1(∥u1∥)≤0.
Butφ2(t)>0 for all t∈(0,1) by the maximum principle, and u1(t) attains zero on isolated points by the Sturm theorem. Hence, from (2.42),
(3.21)g1(∥u1∥)=∫01‍p(t)φ2(t)∥u1∥dt>0.
This is a contradiction. Thus
(3.22)u≠μAu,∀u∈∂Br,μ∈[0,1].
It follows from the homotopy invariance of the Leray-Shauder degree that
(3.23)deg(I-A,Br/τ,θ)=1.
By (3.16), (3.23), and the additivity of Leray-Shauder degree, we obtain
(3.24)deg(I-A,BR∖Br/τ,θ)=deg(I-A,BR,θ)-deg(I-A,Br/τ,θ)=-1.
As a result, A has at least one fixed point u on BR∖Br/τ, namely, the BVP (1.1) has at least one nontrivial solution x(t)=Iμn-1u(t).Corollary 3.2.
Assume that (H1), (H2), and (H4) hold. If the following condition is satisfied.
(H*3) There exists a nonnegative continuous functionsb:[0,1]→(0,+∞) such that
(3.25)f(t,x1,x2,…,xn)≥-b(t),for  anyt∈[0,1],∀xi∈ℝ,
then the singular higher multipoint boundary-value problems (1.1) have at least one nontrivial solution.Corollary 3.3.
Assume that (H1), (H2), and (H4) hold. If the following condition is satisfied.
(H3**) There exist three constantsb>0,c>0, and αi∈(0,1) such that
(3.26)f(t,x1,x2,…,xn)≥-b-c∑i=1n‍|xi|αi,for  anyt∈[0,1],∀xi∈ℝ,
then the singular higher multipoint boundary-value problems (1.1) have at least one nontrivial solution.Remark 3.4.
Noticing that the Green function of the BVP (1.1) is not symmetrical, which implies that the existence of nontrivial solutions of the BVP (1.1) cannot be obtained by Theorem  2.1 in [22] and Theorem  1 in [20]. It is interesting that we construct a new linear operator J instead of K in paper [22] and use its first eigenvalue and its corresponding eigenfunction to seek a linear continuous functional g of P. As a result, we overcome the difficulty caused by the nonsymmetry of the Green function. In [24], the nonlinearity does not contain derivatives and a stronger condition is required, that is, h must be an even function; here we omit this stronger assumption.Remark 3.5.
The results of [20–22] is a special case of the Corollary 3.2 and Corollary 3.3 when α1(i=1,2,…,n) are integer and the nonlinear term does not involve derivatives of unknown functions.



---

*Source: 797398-2012-09-05.xml*