# Positive Solutions for Singular Semipositone Fractional Differential Equation Subject to Multipoint Boundary Conditions

Authors: Rui Pu; Xingqiu Zhang; Yujun Cui; Peilong Li; Weiwei Wang
Journal: Journal of Function Spaces
 (2017)

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## Abstract

Existence result together with multiplicity result of positive solutions of higher-order fractional multipoint boundary value problems is given by considering the integrations of height functions on some special bounded sets. The nonlinearity may change its sign and may possess singularities on the time and the space variables at the same time.


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## Body

## 1. Introduction

Very recently, Henderson and Luca [1] obtained the existence result of positive solution for the following singular differential equation with fractional derivative:(1)D0+αut+ft,ut=0,0<t<1,subject to multipoint BCs (2)u0=u′0=⋯=un-20=0,D0+pu1=∑i=1maiD0+quξi.HereD0+α is the traditional Riemann-Liouville derivative with order α, n-1<α≤n  (n≥3), ai≥0, i=1,2,…,m(m∈N+), 0<ξ1<ξ2<⋯<ξm<1, p,q∈R, 1≤p≤n-2, and 0≤q≤p with Δ=Γ(α)/Γ(α-p)-(Γ(α)/Γ(α-q))∑i=1maiξiα-q-1>0. The nonlinear term f permits sign-changing and singularities at t=0,1.Fractional derivative gives a more precise exhibition on long-memory behavior owing to its nonlocal characteristic. Nowadays, researchers have reached a consensus that fractional calculus is one of the effective tools to describe phenomena in almost every field of science and technology. For example, people have realized that electrical conductance of cell membranes of biological organism took on fractional-order style [2]. As a consequence, fractional-order ordinary differential equations can give a more accurate description on spread process of some infectious diseases such as HIV, hand-foot-mouth disease [3], malaria, tuberculosis, and measles. Arafa et al. [4] formulated a model with orders α1,α2,α3>0 to describe the infection of CD4+ T cells.In 2015, Wang et al. [5] concentrated on the positive solutions for a class of modified HIV-1 population dynamics model presented in [6] (3)Dtαut+λft,ut,Dtβut,vt=0,Dtγvt+λgt,ut=0,0<t<1,Dtβu0=Dtβ+1u0=0,Dtβu1=∫01DtβusdAs,v0=v′0=0,v1=∫01vsdBs,where 2<α, γ≤3, 0<β<1, λ>0 is a parameter, α-β>2, and Dtα, Dtβ, and Dtγ are traditional R-L derivatives. ∫01Dtβu(s)dA(s) and ∫01v(s)dB(s) represent R-S integrals and A, B are bounded variations, and the nonlinearities permit sign-changing and singularities at t=0,1. Other relative papers on nonlocal boundary value problems can be found in [7–20].Motivated by the above papers, we devote ourselves to the existence result as well as multiplicity result of positive solutions for BVP (1)-(2). This article admits some new features. First, the nonlinear term f may take negative infinity and change its sign. Compared with [1], f permits singularities on the time and the space variables at the same time. Second, the method exploited in this paper is different from that in [1] in essence. More precisely, height functions with their integrations on some special bounded set are utilized to get the existence result for positive solutions of BVP (1)-(2). Thirdly, a result of multiple positive solutions is also given in this paper. Conditions employed in this paper are easy to be verified.


## 2. Preliminaries and Several Lemmas

Two traditional Banach spacesE=C[0,1] and L1(0,1) are involved in this article, where E=C[0,1] and L1(0,1) represent the spaces of the continuous functions and Lebesgue integrable functions equipped with the norms u=max0≤t≤1ut and u1=∫01utdt, respectively.Lemma 1 (see [1]).
IfΔ≠0. Given y∈C(0,1)∩L1(0,1), the solution of the following differential equation:(4)D0+αut+yt=0,0<t<1,u0=u′0=⋯=un-20=0,D0+pu1=∑i=1maiD0+quξisatisfies (5)ut=∫01Gt,sysds,t∈0,1,and here (6)Gt,s=G1t,s+tα-1Δ∑i=1maiG2ξi,s,(7)G1t,s=1Γαtα-11-sα-p-1-t-sα-1,0≤s≤t≤1,tα-11-sα-p-1,0≤t≤s≤1,(8)G2t,s=1Γα-qtα-q-11-sα-p-1-t-sα-q-1,0≤s≤t≤1,tα-q-11-sα-p-1,0≤t≤s≤1,∀(t,s)∈[0,1]×[0,1].Lemma 2 (see [1]).
Assume thatai>0(i=1,2,…,m) and Δ>0. Then the Green function G of (4) given by (6) is continuous on [0,1]×[0,1] and meets(a)
G(t,s)≤J(s), ∀t,s∈[0,1], and here J(s)=h1(s)+(1/Δ)∑i=1maiG2(ξi,s), h1(s)=(1-s)α-p-1(1-(1-s)p)/Γ(α), and s∈[0,1];(b)
G(t,s)≥tα-1J(s), ∀t,s∈[0,1];(c)
G(t,s)≤σtα-1  ∀t,s∈[0,1], and here σ=1/Γ(α)+∑i=1maiξiα-q-1/(ΔΓ(α-q)).Lemma 3.
Letu∈C[0,1] satisfy (4), where y∈L1(0,1), y(t)≥0, and 0≤t≤1. Then, u(t)≥tα-1u, and 0≤t≤1.Proof.
It can be easily seen from Lemma5 in [1]; we omit the details.Lemma 4.
Suppose thatw(t)∈C[0,1] be the solution of(9)D0+αut+kt=0,0<t<1,u0=u′0=⋯=un-20=0,D0+pu1=∑i=1maiD0+quξi,where k∈L1(0,1) and k(t)>0. Then, w(t)≤σk1tα-1 and 0≤t≤1.Proof.
For any0≤t≤1, we get by Lemma 2(10)wt=∫01Gt,sksds≤σtα-1∫01ksds=σk1tα-1.Lemma 5 (see [21]).
LetΩ1 and Ω2 be two bounded open sets in Banach space E such that θ∈Ω1 and Ω¯1⊂Ω2 and A:P∩(Ω¯2∖Ω1)→P a completely continuous operator, where θ denotes the zero element of E and P a cone of E. Suppose that one of the two conditions holds:(i)
Au≤u, ∀u∈P∩∂Ω1; Au≥u, ∀u∈P∩∂Ω2.(ii)
Au≥u, ∀u∈P∩∂Ω1; Au≤u, ∀u∈P∩∂Ω2.
ThenA has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

Let(11)K=u:ut≥tα-1u.Obviously, K is a cone in E and (E,K) is a partial ordering Banach space.(
A
0
)  f∈C((0,1)×(0,+∞),(-∞,+∞)), and there exists a function k∈L1(0,1) and k(t)>0, such that f(t,u)≥-k(t), ∀t∈(0,1), u>0.(
A
1
) For any positive numbers r1<r2, there exists a nonnegative function γr1,r2∈L1(0,1) such that (12)ft,u≤γr1,r2t,0<t<1,r1tα-1≤u≤r2with (13)∫011-sα-p-1γr1,r2sds<+∞.(
A
2
) There exists a r~1>σk1 such that (14)∫01Jsφs,r~1ds<r~1,where φ(t,r~1)=max{f(t,u):(r~1-σk1)tα-1≤u≤r~1}+k(t).(
A
3
) There exists a r~2>r~1 such that (15)∫01Jsψs,r~2ds>r~2,where ψ(t,r~2)=min{f(t,u):(r~2-σk1)tα-1≤u≤r~2}+k(t).Theorem 6.
Suppose that conditions(A0), (A1), (A2), and (A3) hold. Then BVP (1)-(2) admits at least one positive solution.Proof.
First of all, we concentrate on the following modified approximating BVP (MABVP for short) to overcome difficulties caused by singularities:(16)D0+αut+ft,χnu-wt+kt=0,0<t<1,u0=u′0=⋯=un-20=0,D0+β1=∑i=1maiD0+quξi,and, here,(17)χnu=u,u≥1n,1n,u<1n.The operator Tn is given by (18)Tnut=∫01Gt,sfs,χnu-ws+ksds,0≤t≤1.Next, we will give the proof from three steps.
(I) For Any
σ
k
1
<
r
1
<
r
2, Tn:K∩(Ωr2∖Ωr1)→K, n>1/r1,Is Completely Continuous. For any 0<t<1, considering that u(t)-w(t)≥(r1-σk1)tα-1>0, one has (19)r1-σk1tα-1≤max⁡ut-wt,1n≤r2,n>1r1,which means(20)r1-σk1tα-1≤χnu-wt≤r2,0<t<1.Thus, one gets from (A0), (A1), and Lemma 2 that(21)Tnut=∫01Gt,sfs,χnu-ws+ksds≤∫01Jsfs,χnu-ws+ksds≤∫011-sα-p-11-1-spΓα+1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1fs,χnu-ws+ksds≤∫011-sα-p-11-1-spΓα+1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1γr1-σk1,r2s+ksds<+∞.So, Tn:K∩(Ωr2∖Ωr1)→E is well defined.
At the same time, for givenu∈K∩(Ωr2∖Ωr1), t∈[0,1], one gets from (21) and Lemma 2 that(22)Tnut=∫01Gt,sfs,χnu-ws+ksds≤∫01Jsfs,χnu-ws+ksds.Hence,(23)Tnu≤∫01Jsfs,χnu-ws+ksds.Given u∈K, it follows from Lemma 2 that(24)Tnut=∫01Gt,sfs,χnu-ws+ksds≥tα-1∫01Jsfs,χnu-ws+ksds≥tα-1Tnu,which means that Tn:K∩(Ωr2∖Ωr1)→K.
Given bounded setD⊂K∩(Ωr2∖Ωr1), we can see from (21) that Tm(D) is uniformly bounded. By Ascoli-Arzela theorem, in order to show the complete continuity, we need only to prove that Tn(D) is equicontinuous. It follows from (6), (7), (8), and (A1) that(25)Tnu′t=∫01∂∂tGt,sfs,χnu-ws+ksds≤∫01∂∂tG1t,sfs,χnu-ws+ksds+∫01α-1tα-2Δ∑i=1maiG2ξi,sfs,χnu-ws+ksds≤1Γα∫0tα-1tα-21-sα-p-1-α-1t-sα-2fs,χnu-ws+ksds+∫t1α-1tα-21-sα-p-1fs,χnu-ws+ksds+∫01α-1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1fs,χnu-ws+ksds≤1Γα∫01α-11-sα-p-1+α-11-sα-2γr1-σk1,r2s+ksds+∫01α-1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1γr1-σk1,r2s+ksds<+∞,t∈0,1.Equicontinuity of Tn(D) can be derived from the absolute continuity of Lebesgue integral and (25). Thus, the fact that Tn:K∩(Ωr2∖Ωr1)→K, n>1/r1, is completely continuous has been proved.
(II) For Sufficiently Large
n
, MABVP (16) Admits at Least One Positive Solution. Given u∈K∩∂Ω(r~1), one has u(t)≥r~1tα-1, 0≤t≤1. Thus, for n>1/r~1, similar to (20), we get(26)r~1-σk1tα-1≤χnu-wt≤r~1,0<t<1.From the definition of φ(t,r~1), we have (27)ft,χnu-wt+kt≤φt,r~1.By (A2) and Lemma 2, one gets (28)Tnu=max0≤t≤1⁡∫01Gt,sfs,χnu-ws+ksds≤∫01Jsφs,r~1ds<r~1;that is, (29)Tnu≤u,∀u∈K∩∂Ωr~1.
Givenu∈K∩∂Ω(r~2), one has u(t)≥r~2tα-1, 0≤t≤1. Thus, for n>1/r~2, we get (30)r~2≥χnu-wt≥r~2-σk1tα-1.According to the definition of ψ(t,r~2), one gets(31)ft,χnu-wt+kt≥ψt,r~2.By Lemma 2, (A3), and (31), we have(32)Tnu=max0≤t≤1⁡∫01Gt,sfs,χnu-ws+ksds≥max0≤t≤1⁡tα-1∫01Jsfs,χnu-ws+ksds≥∫01Jsψs,r~2ds≥r~2;that is,(33)Tnu≥u,∀u∈K∩∂Ωr~2.Take m0=max{1/r~1,1/r~2}. Let N={m0,m0+1,…}. Then, for n∈N, both (29) and (33) hold. This together with Lemma 5 shows that Tn(n∈N) has at least one fixed point u~n∈K∩Ω¯r~2∖Ωr~1.
(III) BVP (1)-(2) Admits at Least One Positive Solution. By Lemma 4, one has(34)u~nt≥u~ntα-1≥r~1tα-1>σk1tα-1≥wt,m>N,t∈0,1,(35)u~nt=∫01Gt,sfs,χnu~n-ws+ksds,0<t<1.By (A1), we know {u~n:n>N} are bounded and equicontinuous sets [0,1]. It follows from Arzela-Ascoli theorem that there exist a subsequence N0 of N and a function u~∈C[0,1] such that u~n converges to u~ uniformly on [0,1] as n→∞ through N0. Taking limit as n→∞ on both sides of (35) together with the fact u~n(t)≥tα-1u~n≥r~1tα-1, we get(36)u~t=∫01Gt,sfs,u~s-ws+ksds,0<t<1.Let u(t)=u~(t)-w(t), and then from (36) one has that u(t) is a positive solution for BVP (1)-(2).
A multiplicity result follows if we replace(A2) and (A3) by the following:
(
A
2
′
) There exists r~i>σk1(i=1,2,…,m) such that (37)∫01Jsφs,r~ids<r~i,where φ(t,r~i)=max{f(t,u):(r~i-σk1)tα-1≤u≤r~i}+k(t).
(
A
3
′
) There exists R~i(i=1,2,…,m) with (38)0<r~1<R~1<r~2<R~2<⋯<r~m<R~msuch that(39)∫01Jsψs,R~ids>R~i,where ψ(t,R~i)=min{f(t,u):(R~i-σk1)tα-1≤u≤R~i}+k(t).Theorem 7.
Suppose that conditions(A0), (A1), (A2′), and (A3′) hold. Then BVP (1)-(2) has at least m positive solutions.


## 4. An Example

Consider the fractional differential equation(40)D0+8/3ut+116t21-t3u6+1u8-112t3=0,0<t<1,with multipoint BCs(41)u0=u′0=0,D0+1/2u1=12D0+1/3u14+23D0+1/3u12+14D0+1/4u45.It is clear that (40)-(41) has the form of (1)-(2), where α=8/3, n=3, p=1/2, q=1/3, m=3, ξ1=1/2, ξ2=1/4, ξ3=4/5, a1=1/2, a2=2/3, and a3=1/4. Obviously, (A0) holds for k(t)=1/12t3. After direct calculation, we have α-p-1=7/6, α-q-1=4/3, Γ(α)=1.5046, Γ(α-p)=1.0823, Γ(α-q)=1.1906, ∑i=13aiξiα-q-1=0.5290, Δ=0.7217, σ=1.2803, k1=0.1250, and σk1=0.1600. It is clear that (A1) is valid for γr1,r2=(1/16t2(1-t)3)[r26+(r1t5/3)-1/8]+1/12t3. Next, we check (A2) and (A3).Taker~1=1>0.1600=σk1, and then by Lemma 2 we have(42)∫01Jsφs,1ds≤∫011-sα-p-11-1-spΓα+1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1116s21-s3max⁡u6+1u8:0.8400s5/3≤u≤1ds=∫011-s7/61-1-s1/2Γα+∑i=13aiξi4/3ΔΓα-q1-s7/6116s21-s3max⁡u6+1u8:0.8400s5/3≤u≤1ds<116∫010.66461-s7/6-0.66461-s5/3+0.6156×1-s7/6s-2/31-s-1/31+10.8400s5/38ds<116∫011.28021-s7/6-0.66461-s5/3s-2/31-s-1/31+1.0220s-5/24ds=1161.2802×2.3283-0.6646×2.1200+1.2802×1.0220×7.2087-0.6646×1.0220×6.9455=116×6.2858=0.3929<1.Thus, (A2) is verified.Taker~2=20, and, then, we have(43)∫01Jsψs,20ds≥∫011-sα-p-11-1-spΓα+1ΔΓα-q∑i=1maiξiα-q-11-sα-p-1-1-sα-q-1116s21-s3min⁡u6+1u8:9.8400s5/3≤u≤20ds≥∫011-s7/61-1-s1/2Γα+∑i=13aiξi4/3ΔΓα-q1-s7/6-1ΔΓα-q1-s4/3116s21-s39.84006s10+1208ds≥9.8400616∫011Γα+∑i=13aiξi4/3ΔΓα-qs28/31-s5/6-1Γαs28/31-s4/3-1ΔΓα-qs28/31-sds=56735∫010.6646+0.6156s28/31-s5/6-0.6646s28/31-s4/3-1.1637s28/31-sds=56735×1.2802×0.0121-0.6646×0.0044-1.1637×0.0085=56735×0.0027=153.1845>20.Thus, (A3) is valid. According to Theorem 6, BVP (40)-(41) admits at least one positive solution.



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