# Positive Solutions of a Two-Point Boundary Value Problem for Singular Fractional Differential Equations in Banach Space

Authors: Bo Liu; Yansheng Liu
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

This paper investigates the existence of positive solutions to a two-point boundary value problem (BVP) for singular fractional differential equations in Banach space and presents a number of new results. First, by constructing a novel cone and using the fixed point index theory, a sufficient condition is established for the existence of at least two positive solutions to the approximate problem of the considered singular BVP. Second, using Ascoli-Arzela theorem, a sufficient condition is obtained for the existence of at least two positive solutions to the considered singular BVP from the convergent subsequence of the approximate problem. Finally, an illustrative example is given to support the obtained new results.


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## Body

## 1. Introduction

Fractional differential equations have been widely investigated recently due to its wide applications [1–3] in biology, physics, medicine, control theory, and so forth. As a matter of fact, fractional derivatives provide a more excellent tool for the description of memory and hereditary properties of various materials and processes than integer derivatives. As an important issue for the theory of fractional differential equations, the existence of positive solutions to kinds of boundary value problems (BVPs) has attracted many scholars’ attention, and lots of excellent results have been obtained [4–11] by means of fixed point theorems, upper and lower solutions technique, and so forth.It is noted that as a special class of fractional differential equations, the singular fractional differential equations with kinds of boundary values have been studied in a series of recent works [7, 12, 13]. In [7],  Jiang et al.  studied a singular nonlinear semipositone fractional differential system with coupled boundary conditions and presented some sufficient conditions for the existence of a positive solution by using the fixed point theory in cone and constructing some available integral operators together with approximating technique.  Zhang et al. [13] considered a class of two-point BVP for singular fractional differential equations with a negatively perturbed term and established some results on the multiplicity of positive solutions by using the approximating technique. In [12], Agarwal et al. investigated the existence of positive solutions for a two-point singular fractional boundary value problem and proposed some existence criteria by using sequential techniques. It should be pointed out that the nonlinearities of [7, 13] are singular at t=0,1, while the nonlinearity of [12] is singular at x=0. To our best knowledge, there are fewer results on two-point BVPs for singular fractional differential equations with the nonlinearity being singular at both t=0,1 and x=0. Motivated by this, we consider the following two-point BVP of singular fractional differential equations in Banach space:
(1)D0+βx(t)+f(t,x(t))=θ,t∈J,x(0)=x(1)=θ,
where 1<β≤2 is a real number, J=[0,1], f:J×E→E is continuous, θ denotes the null element in the Banach space E with the norm ∥·∥, D0+β is the standard Riemann-Liouville fractional derivative, and f(t,x) may be singular at t=0,1 and x=θ. Firstly, we establish a sufficient condition for the existence of at least two positive solutions to the approximate problem of BVP (1) by constructing a novel cone and using the fixed point index theory. Secondly, using Ascoli-Arzela theorem, we obtained a sufficient condition for the existence of at least two positive solutions to BVP (1) from the convergent subsequence of the approximate problem. Finally, we give an illustrative example to support the obtained new results.The main features of this paper are as follows. (i) A class of fractional-order two-point boundary value problems with the nonlinearity being singular at botht=0,1 and x=θ is firstly studied in this paper, which generalizes the existing singular fractional differential equations [7, 12, 13] and has wider applications. (ii) A sequential-based method is proposed for singular fractional differential equations with the nonlinearity being singular at both t=0,1 and x=0, which enriches the theory of fractional differential equations.The rest of this paper is organized as follows. Section2 contains the definition of Riemann-Liouville fractional derivative and some notation. The main results are presented in Section 3, which is followed by an illustrative example in Section 4.


## 2. Preliminaries

We first recall some well-known results about Riemann-Liouville derivative. For details, please refer to [14, 15] and the references therein.Definition 1.
The Riemann-Liouville fractional integral of orderβ>0 of a function y:(0,∞)→R is given by
(2)I0+βy(t)=1Γ(β)∫0t‍(t-s)β-1y(s)ds,
provided the right side is pointwise defined on (0,∞).Definition 2.
The Riemann-Liouville fractional derivative of orderβ>0 of a continuous function y:(0,∞)→R is given by
(3)D0+βy(t)=1Γ(n-β)(ddt)n∫0t‍y(s)(t-s)β-n+1ds,
where n is the smallest integer greater than or equal to β, provided that the right side is pointwise defined on (0,∞).
One can easily obtain the following properties from the definition of Riemann-Liouville derivative.Proposition 3 (see [15]).
Letβ>0; if one assumes that u∈C(0,1)∩L(0,1), then, the fractional differential equation D0+βu(t)=0 has u(t)=C1tα-1+C2tβ-2+⋯+Cntβ-n, Ci∈R,i=1,2,…,n as unique solutions, where n is the smallest integer greater than or equal to β.Proposition 4 (see [15]).
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order β>0 that belongs to C(0,1)∩L(0,1). Then,
(4)I0+βD0+βu(t)=u(t)+C1tβ-1+C2tβ-2+⋯+Cntβ-n,
for some Ci∈R, i=1,2,…,n, where n is the smallest integer greater than or equal to β.The following lemmas will be used in the proof of the main results.Lemma 5 (see [16]).
IfS⊂C[J,E] is bounded and equicontinuous, then
(5)αc(S)=supt∈J(αS(t)),
where α(·) and αc(·) denote the Kuratowski noncompactness measure of bounded sets in E and C[J,E], respectively, S(t)={x(t):x∈S}(t∈J), and C[J,E] is the Banach space of all continuous functions x:J→E with the norm ∥x∥c=maxt∈J∥x(t)∥.Lemma 6 (see [16]).
LetP be a cone in E, and  let  Pr={x∈P:∥x∥<r}. Let F:Pr→P be a strict set contraction. Assume that there  exist  a u0∈P and u0≠θ such that x-Fx≠λu0 for any x∈∂Pr and λ≥0. Then i(F,Pr,P)=0.Lemma 7 (see [16]).
LetV={xn}∈L[J,E], and there exists a g∈L[J,R+] such that ∥xn(t)∥≤g(t),a.e.t∈J for all xn∈V; then α({∫at‍xn(s)ds:n∈N})≤2∫at‍α(V(s))ds, t∈J.Lemma 8 (Ascoli-Arzela theorem [16]).
H
⊂
C
[
J
,
E
] is relative compact if and only if H is equicontinuous, and for any t∈J, H(t) is a relatively compact set in E.Lemma 9 (see [17]).
LetB⊆E be bounded open set. A:P∩B¯→B is condensing. If there exists 1>μ>0, such that Ax=μx, where x∈P⋂‍∂B, then i(A,P⋂‍B,P)=1.


## 3. Main Results

LetP={x∈E:x(t)≥θ,∀t∈J}. Then, one can see that P is a normal solid cone of E. Define Pd={x∈P:∥x∥<d} and Pd¯={x∈P:∥x∥≤d}. Let P* denote the dual cone of P. We consider BVP (1) in C[J,E]. x is called a solution to BVP (1), if x satisfies (1). In addition, we call x(t) a positive solution to BVP (1), if x(t)>θ,∀t∈J.For convenience, let us list the following assumptions.(H1)
f∈C[(0,1)×P∖{θ},P] and
(6)∥f(t,tβ-2x)∥≤k(t)∥q(x)∥,t∈(0,1),x∈P∖{θ},
where k:(0,1)→(0,+∞), ∫01‍s(1-s)β-1k(s)ds<+∞, and q:P∖{θ}→P.(H2)
1/Γ(β)∫01‍s(1-s)β-1k(s)q[(β-1)s(1-s)β-1r1,R1]ds<+∞ for all R1>r1>0, and there exists R>0 such that 1/Γ(β)∫01‍s(1-s)β-1k(s)q[(β-1)s(1-s)β-1R,R+1]ds<R, where
(7)q[r1,R1]:=supx∈P-R1∖Pr1∥q(x)∥<+∞,R1>r1>0.(H3)
f(t,tβ-2x) is uniformly continuous with respect to t on [γ,1-γ]×P¯R1∖Pr1, where γ∈(0,1/2) and R1>r1>0.(H4)
There exists a constantL≥0, such that
(8)α(f(t,tβ-2D))≤Lα(D),∀t∈(0,1),D⊆P-R1∖Pr1,
where L<min{Γ(β)ββ/2(β-1)β-1,Γ(β)β(β+1)/2}, and R1>r1>0.(H5)
There existsφ*∈P* with ∥φ*∥=1 and φ∈L[0,1], such that
(9)lim∥x∥→0infx∈Pφ*(f(t,tβ-2x))≥φ(t),
uniformly in t∈(0,1), where 0<∫01‍s(1-s)β-1φ(s)ds<+∞.(H6)
There existψ*∈P* with ∥ψ*∥=1 and [a,b]⊆(0,1), such that
(10)lim∥x∥→+∞infx∈Pψ*(f(t,tβ-2x))∥x∥=+∞,
uniformly on t∈[a,b].According to [18], BVP (1) is equivalent to
(11)x(t)=∫01‍G(t,s)f(s,x(s))ds,
where
(12)G(t,s)={[t(1-s)]β-1-(t-s)β-1Γ(β),0≤s≤t≤1,[t(1-s)]β-1Γ(β),0≤t≤s≤1.Consider the operatorA associated with the singular boundary value problem (1), which is defined by
(13)(Ax)(t)=∫01‍G(t,s)f(s,x(s))ds.
Set G*(t,s)=t2-βG(t,s). As in [19], we have
(14)G*(t,s)≥(β-1)t(1-t)G*(τ,s),G*(t,s)≤1Γ(β)s(1-s)β-1.Define(15)(A*x)(t)=∫01‍G*(t,s)f(s,tβ-2x(s)).
Then, one can see that x(t) is a fixed point of the operator A* if and only if tβ-2x(t) is a solution of BVP (1).Choosee∈Po with ∥e∥=1. We consider the following approximate problem of (15):
(16)(An*x)(t)=∫01‍G*(t,s)f(s,tβ-2(x(s)+en))ds,n∈N.Denote(17)Q={x∈C[J,P]:x(t)≥(β-1)t(1-t)x(s),∀t,s∈J}.
It is easy to check that Q is a cone in C[J,E]. Let Qr={x∈Q:∥x∥<r}. By (H1) and (H2), we can conclude that
(18)∥A*x∥≤∫01‍s(1-s)β-1Γ(α)k(s)×q[(β-1)s(1-s)β-1r,r]ds<+∞,∀x∈Qr,
which implies that the operator A* is well defined. In addition, from the definition of Q, we can prove that
(19)∥x(t)∥≥(β-1)t(1-t)∥x∥c,∀x∈Q,t∈J.
By (16), we have (An*x)(t)≥(β-1)t(1-t)(An*x)(s),∀t,s∈J. Thus An*Q⊂Q.In the following, we first investigate the existence of two positive solutions to the approximate problem (16) and then establish the existence criterion for the existence of two positive solutions to BVP (1) by using the sequential-based technique and Ascoli-Arzela theorem.Lemma 10.
Let (H1) and (H2) be satisfied. Then for anyr>0, the operator An* is a continuous bounded operator from Q¯r into Q.Proof.
By (H1), we can easily see thatAn* is bounded from Qr to Q.
Next, we prove thatAn* is continuous.
Let∥xm-x∥c→0 as m→+∞(xm,x∈Q¯r). From (H1), we have
(20)∥f(s,(xm(s)+en)sβ-2)∥≤k(s)q(xm(s)+en)≤k(s)q[1n,r+1n].
Hence
(21)∥(An*xm)(t)∥=∥∫01‍G*(t,s)f(s,(xm(s)+en)sβ-2)ds∥≤1Γ(β)∫01‍s(1-s)β-1k(s)q[1n,r+1n]ds<+∞,
which together with the dominated convergence theorem imply that
(22)limm→+∞(An*xm)(t)=(An*x)(t).
We now show that(23)limm→+∞∥An*xm-An*x∥c=0.
In fact, if (23) is not true, then there exist a positive number ε0 and a subsequence {xmi}⊂{xm} such that
(24)∥An*xmi-An*x∥c≥ε0.
Since {An*xm} is relatively compact, there is a subsequence of {An*xmi} which converges to some y∈Q. Without loss of generality, we assume that {An*xmi} itself converges to y, that is,
(25)limi→∞∥An*xmi-y∥c=0.
By virtue of (22) and (25), we have y=An*x, which contradicts with (24). Hence (23) holds, and the continuity of An* is proved.Lemma 11.
Let (H1)–(H4) be satisfied. Then for anyr>0, the operator An* is a strict set contraction from Qr to Q.Proof.
By virtue of Lemma10, we know that An*S is bounded and equicontinuous on J. Thus, from Lemma 5, one can see that
(26)αc(An*S)=supt∈Jα((An*S)(t)),r>0,S⊆Qr,
where An*S(t)={(An*x)(t):x∈S},t∈J.
Set(27)Dδ={∫δ1-δ‍G*(t,s)f(s,(x(s)+en)sβ-2)ds:x∈S},δ∈(0,12).
Based on (H1) and (H2), we have
(28)∥∫δ1-δ‍G*(t,s)f(s,(x(s)+en)sβ-2)ds-∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds∥≤c(∫0δ‍s(1-s)β-1k(s)ds+∫11-δ‍s(1-s)β-1k(s)ds),x∈S,δ∈(0,12),
where c=q[1/n,r+(1/n)]. Hence
(29)dH(Dδ,An*S)⟶0,δ⟶0+,
where dH(·,·) denotes the Hausdorff metrics, which implies that
(30)α(An*S)=limδ→0+α(Dδ).
Since
(31)∫δ1-δ‍G*(t,s)f(s,sβ-2(x(s)+en))ds∈(1-2δ)×co¯{G*(t,s)f(s,sβ-2(x(s)+en)):s∈[δ,1-δ]},
by Lemmas 5 and 7, we have
(32)α(Dδ)=α({∫δ1-δ‍G*(t,s)f(s,sβ-2(x(s)+en))ds,x∈S})≤(1-2δ)×α(co¯{f(s,(x(s)+en)sβ-2):s∈[δ,1-δ],x∈S}G*(t,s)×f(s,(x(s)+en)sβ-2):s∈[δ,1-δ],x∈S})≤α({(s,(x(s)+en)sβ-2)G*(t,s)×f(s,(x(s)+en)sβ-2):s∈[δ,1-δ],x∈S})≤1Γ(β)(β-1)β-1ββ×α({f(s,(x(s)+en)sβ-2):s∈[δ,1-δ],x∈S})≤1Γ(β)L(β-1)β-1ββα(S(Iδ))≤1Γ(β)2L(β-1)β-1ββαc(S),
where Iδ=[δ,1-δ]. By (H4), we can obtain
(33)α(AnS)=limδ→0+α(Dδ)≤1Γ(β)(β-1)β-1ββ2Lαc(S)<αc(S).
Consequently, the operator An is a strict set contraction from Qr into Q.Theorem 12.
Let (H1)–(H6) hold. Then, there existsr∈(0,R) such that the operator An* has two fixed points in QR∖Q¯r and Q∖Q¯R, respectively, for arbitrary sufficiently large positive integer n, where R is given in (H2).Proof.
For any givent∈(0,1), we have ∫01‍G*(t,s)φ(s)ds≤∫01‍s(1-s)β-1φ(s)ds. Since ∫01‍G*(t,s)φ(s)ds>0, there exists a ε′>0 such that
(34)r′=∫01‍∫01‍G*(t,s)(φ(s)-ε′)dsdt>0.
Then by (H5), there exists r′′∈(0,R) such that
(35)φ*(f(t,tβ-2x(t)))≥φ(t)-ε′,∀t∈(0,1)
holds for x∈Q and ∥x∥c≤r′′. Letting 0<r<l:=min{r′,r′′}, we prove that x-An*x≠λe for any x∈∂Qr, λ≥0 as n>1/(l-r). In fact, suppose that this is false. Then there exist λ>0, x∈∂Qr such that x-An*x=λe. It is obvious that
(36)x(t)=(An*x)(t)+λe≥(An*x)(t)=∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds.
Thus
(37)φ*(x(t))≥∫01‍G*(t,s)φ*(f(s,(x(s)+en)sβ-2))ds≥∫01‍G*(t,s)(φ(s)-ε′)ds.
This implies that
(38)∫01‍φ*(x(t))dt≥∫01‍∫01‍G*(t,s)(φ(s)-ε′)dsdt=r′>r,
which is in contradiction with
(39)|φ*(x(t))|≤∥x(t)∥≤∥x∥c=r,t∈J.
Combining with Lemma 6, we obtain that
(40)i(An*,Qr,Q)=0.
Now, we show thati(An*,QR,Q)=1. By Lemma 9, we need only to prove that An*x≠λx for x∈∂QR and λ≥1. In fact, if there exist x∈∂QR and some λ≥1 such that An*x=λx, then
(41)x(t)=1λ∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds≤∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds≤1Γ(β)∫01‍s(1-s)β-1f(s,(x(s)+en)sβ-2)ds.
Therefore, by (H1) and (H2), we have
(42)R=∥x∥c<1Γ(β)∫01‍s(1-s)β-1k(s)∥q(x(s)+en)∥ds<1Γ(β)∫01‍s(1-s)β-1k(s)×q[(β-1)s(1-s)β-1R,R+1]ds<R,
which is a contradiction. Consequently,
(43)i(An*,QR,Q)=1.
In the following, we choose(44)R′>((β-1)a(1-b)∫ab‍maxt∈JG*(t,s)ds)-1.
By (H6), there exists M>R such that φ*(f(t,tβ-2x))≥R′∥x∥ holds for x>M.
Let(45)R¯=max{1(β-1)R′a(1-b)maxt∈J∫ab‍G*(t,s)ds-1M+1(β-1)a(1-b),1+1(β-1)R′a(1-b)maxt∈J∫ab‍G*(t,s)ds-1}.
We now claim that x-An*x≠λe for x∈∂QR¯ and λ≥0. As a matter of fact, if this is not true, then there exists λ>0 such that x-An*x=λe, that is, x(t)>An*x(t), which implies that
(46)R¯≥φ*(x(t))≥φ*((An*x)(t))≥∫01‍G*(t,s)φ*(f(s,(x(s)+en)sβ-2))ds≥∫01‍G*(t,s)R′((β-1)a(1-b)(∥x∥-1))ds≥(β-1)a(1-b)R′(R¯-1)∫ab‍G*(t,s)ds>R¯,
which is a contradiction. Therefore, by Lemma 6, we can obtain that
(47)i(An*,QR¯,Q)=0.
This together with (40), (43), and (47) implies that
(48)i(An*,QR∖Q¯r,Q)=i(An*,QR,Q)-i(An*,Qr,Q)=1,i(An*,QR¯∖Q¯R,Q)=i(An*,QR¯,Q)-i(An*,QR,Q)=-1.
Thus, there exist x∈QR∖Q¯r and y∈QR¯∖Q¯R such that An*x=x and An*y=y. This completes the proof.Theorem 13.
Let (H1)–(H6) be satisfied. Then BVP (1) has at least two positive solutions in C[J,E].Proof.
From Theorem12, there exists an integer n0>0, such that the operator An*, n≥n0 has two fixed points in xn∈QR∖Q¯r and yn∈Q∖Q¯R. Denote D:={xn:n≥n0}. Obviously D is uniformly bounded. Next, we show that D is equicontinuous.
Firstly, we prove that(49)limt→0+∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds=0,(50)limt→1-∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds=0.
On one hand, by the absolute continuity of integration [19], for any ε>0, there exists δ1>0 such that ∫0t‍(1/Γ(β))s(1-s)β-1f(s,(x(s)+e/n)sβ-2)ds<ε/2,∀0<t≤δ1.
On the other hand, from (H2), we set(51)M¯:=∫δ11‍s(1-s)β-1Γ(β)k(s)q[(β-1)s(1-s)β-1r,R]ds,
and δ=min{δ1,δ1ε/M¯}. For 0<t<δ, we have
(52)∫01‍G*(t,s)f(s,(x(s)+en)sβ-2)ds=∫0δ1‍G*(t,s)f(s,(x(s)+en)sβ-2)ds+∫δ11‍G*(t,s)f(s,(x(s)+en)sβ-2)ds<ε2+∫δ11‍t(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds≤ε2+tδ1∫δ11‍s(1-s)β-1Γ(β)k(s)q(x(s)+en)ds≤ε2+tδ1∫δ11‍s(1-s)β-1Γ(β)k(s)q[(β-1)s(1-s)β-1r,R]ds≤ε,
which implies that (49) holds. Similarly, one can prove that (50) holds.
Next, we show thatD is equicontinuous for t∈[γ,1-γ], 0<γ<1/2.
From (H3), we define(53)M^=maxt∈[γ,1-γ]f(t,(x(t)+en)tβ-2).
By the absolute continuity of integration, forγ∈(0,1/2) and the previous ε>0, there exists 0<δ2<γ such that
(54)|∫0δ2‍s(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|<ε6.
Since((t2(1-s)β-1-(t2-s)β-1t22-β)/Γ(β))f(s,(x(s)+e/n)sβ-2) is bounded for s∈[t1,t2], there exists δ3>0, such that
(55)|∫t1t2‍t2(1-s)β-1-(t2-s)β-1t22-βΓ(β)f(s,(x(s)+en)sβ-2)ds|<ε6,|t2-t1|<δ3.
For|t2-t1|<δ4:=γε/(6∫01‍(s(1-s)β-1/Γ(β))k(s)q[(β-1)s(1-s)β-1r,R]ds), one can see that
(56)|∫t21‍(t2-t1)(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|<ε6.
Similarly, for |t2-t1|<δ5:=δ2ε/(6∫01‍(s(1-s)β-1/Γ(β))k(s)q[(β-1)s(1-s)β-1r,R]ds), we have
(57)|∫δ2t1‍(t2-t1)(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|<ε6.
Since(58)|∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)×f(s,(x(s)+en)sβ-2)ds∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)|≤|1δ2(1-t1)β-1∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)×s(1-s)β-1f(s,(x(s)+en)sβ-2)ds1δ2(1-t1)β-1∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)|,f(t,(x(t)+e/n)tβ-2) is bounded on t∈[δ2,t1], and (t22-β(t2-s)β-1-t12-β(t1-s)β-1)/Γ(β)→0 as t1→t2, there exists δ6>0 such that
(59)|∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)×f(s,(x(s)+en)sβ-2)ds∫δ2t1‍t22-β(t2-s)β-1-t12-β(t1-s)β-1Γ(β)|<ε6,|t2-t1|<δ6.
We chooseδ′=min{δi:i=2,3,4,5,6}. For |t2-t1|<δ′, we have
(60)|x(t2)-x(t1)|=|∫01‍G*(t2,s)f(s,(x(s)+en)sβ-2)ds-∫01‍G*(t1,s)f(s,(x(s)+en)sβ-2)ds|≤|∫0δ2‍G*(t2,s)f(s,(x(s)+en)sβ-2)ds|+|∫0δ2‍G*(t1,s)f(s,(x(s)+en)sβ-2)ds|+|∫δ2t2‍G*(t1,s)f(s,(x(s)+en)sβ-2)ds+∫t21‍G*(t2,s)f(s,(x(s)+en)sβ-2)ds-∫δ2t1‍G*(t1,s)f(s,(x(s)+en)sβ-2)ds-∫t11‍G*(t1,s)f(s,(x(s)+en)sβ-2)ds|≤2|∫0δ2‍s(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|+|∫δ2t2‍t2(1-s)β-1-(t2-s)β-1t22-βΓ(β)×f(s,(x(s)+en)sβ-2)ds∫δ2t2‍t2(1-s)β-1-(t2-s)β-1t22-βΓ(β)|+|∫t2t1‍t1(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|+|∫t2t1‍(t2-t1)(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|+|∫δ2t1‍(t2-t1)(1-s)β-1Γ(β)f(s,(x(s)+en)sβ-2)ds|+|∫δ2t1‍(t22-β(t2-s)β-1-t12-β(t1-s)β-1)Γ(β)×f(s,(x(s)+en)sβ-2)ds∫δ2t1‍(t22-β(t2-s)β-1-t12-β(t1-s)β-1)Γ(β)|<ε.
This together with (49) and (50) implies that D is equicontinuous for t∈[0,1].
Now, we show thatD(t) is relatively compact. By Lemma 7, we have
(61)α(D(t))≤α({∫01‍G*(t,s)f(s,(xn(s)+en)sβ-2ds):n≥n0})≤2∫01‍G*(t,s)×α({f(s,(xn(s)+en)sβ-2):n≥n0})ds≤2L∫01‍G*(t,s)α(D(s))ds,
which together with Lemma 5 and (H4) implies that
(62)αc(D)≤2L1Γ(β)(∫01‍s(1-s)β-1ds)αc(D).
Thus, αc(D)=0. It follows from Lemma 8 that there is a convergent subsequence of {xn}. Without loss of generality, we assume that {xn} itself converges to some x∈Q. Then, the dominated convergence theorem and (16) imply that
(63)x(t)=∫01‍G*(t,s)f(s,sβ-2x(s))ds.
Thus, the singular BVP (1) has a positive solution tβ-2x(t). Similarly, {yn} has a convergent subsequence, which converges to y(t). Then, tβ-2y(t) is also a positive solution to BVP (1).
Finally, we showx≠y. We only need to prove that the operator A* has no fixed point in ∂QR.
In fact, if it is not true, then we assume thatz is a fixed point of the operator A* in ∂QR. Then
(64)z(t)=∫01‍G*(t,s)f(s,sβ-2z(s))ds,
with (β-1)t(1-t)β-1R≤∥z(t)∥≤R+1.
By (H1) and (H2), one can see that(65)R=maxt∈J∥z(t)∥≤∫01‍s(1-s)β-1k(s)q[(β-1)s(1-s)β-1R,R]ds≤∫01‍s(1-s)β-1k(s)q[(β-1)s(1-s)β-1R,R+1]ds<R,
which is a contradiction. The proof of this theorem is completed.Remark 14.
It is noted that the singularity off(t,x) at t=0,1 is overcome by (H2), while the singularity at x=θ is handled by solving the approximate problem and using the sequential-based technique.


## 4. Example

Consider the following BVP:(66)-D0+3/2xn(t)=12πt(1-t)×(1|∑i=1mxi(t)|t1/2+txn+1(t)+txn+22(t)),t∈(0,1),xn(0)=xn(1)=0,n=1,2,…,m,
where xm+n=xn. Then, BVP (66) has at least two positive solutions.Proof.
We consider the problem (66) in m-dimensional Euclidean space Rm={x=(x1,x2,…,xm):xn∈R,n=1,2,…,m} with the norm ∥x∥=∑n=1m‍|xn|. Then, BVP (66) has the form of (1) with
(67)x(t)=(x1(t),x2(t),…xm(t)),fn(t,x)=12πt(1-t)(1|∑i=1mxi|t1/2+txn+1+txn+22).
It is easy to see thatf(t,x) is singular at t=0,1 and x=θ. Set k(t)=1/(2πt(1-t)) and q(x)=(1/|∑i=1m‍xi|)+xn+1+xn+22. One can easily see that (H1) holds.
We chooseR=1 and φ*=ϕ*=(1,1,…,1). Considering that
(68)∫01‍s(1-s)ds=π8,
we can easily check that (H2)–(H6) hold. From Theorem 13, BVP (66) has at least two positive solutions x=(x1,x2,…xm) and (y1,y2,…ym) which satisfy 0<∥x∥<1<∥y∥<+∞.



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*Source: 585639-2013-08-01.xml*