# Positive Solutions for a System of Nonlinear Semipositone Boundary Value Problems with Riemann-Liouville Fractional Derivatives

Authors: Xiaowei Qiu; Jiafa Xu; Donal O’Regan; Yujun Cui
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study the existence of positive solutions for the system of nonlinear semipositone boundary value problems with Riemann-Liouville fractional derivativesD0+αD0+αu=f1t,u,u′,v,v′,0<t<1, D0+αD0+αv=f2(t,u,u′,v,v′),0<t<1, u0=u′0=u′(1)=D0+αu(0)=D0+α+1u(0)=D0+α+1u(1)=0, and v(0)=v′(0)=v′(1)=D0+αv(0)=D0+α+1v(0)=D0+α+1v(1)=0, where α∈(2,3] is a real number and D0+α is the standard Riemann-Liouville fractional derivative of order α. Under some appropriate conditions for semipositone nonlinearities, we use the fixed point index to establish two existence theorems. Moreover, nonnegative concave and convex functions are used to depict the coupling behavior of our nonlinearities.


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## Body

## 1. Introduction

In this paper, we investigate the existence of positive solutions for the system of nonlinear semipositone boundary value problems with Riemann-Liouville fractional derivatives(1)D0+αD0+αu=f1t,u,u′,v,v′,0<t<1,D0+αD0+αv=f2t,u,u′,v,v′,0<t<1,u0=u′0=u′1=D0+αu0=D0+α+1u0=D0+α+1u1=0,v0=v′0=v′1=D0+αv0=D0+α+1v0=D0+α+1v1=0,where α∈(2,3] is a real number and D0+α is the standard Riemann-Liouville fractional derivative of order α. The nonlinear terms fi∈C([0,1]×R+4,R)(R+=[0,+∞), R=-∞,+∞) are bounded below; that is, fi(i=1,2) satisfy the following.(H1) there exists a real number M≥0, such that fi(t,x1,x2,x3,x4)+M≥0,∀t∈[0,1], xj∈R+,i=1,2,j=1,2,3,4.Existence and multiplicity of solutions for fractional differential equations are widely studied in the literature; see [1–14] and the references therein. For example, in [1], the authors used the Guo-Krasnosel’skii fixed point theorem to investigate the existence of positive solutions for the singular fractional differential system(2)-D0+αut=λf1t,u,v,t∈0,1,-D0+αvt=λf2t,u,v,t∈0,1,u0=u′0=0,v0=v′0=0,u1=avξ,v1=buη,where fi(i=1,2) satisfy(3)fit,u,vu+v=0,or∞,asu+v⟶+∞,uniformly fort∈0,1or a subinterval.Condition (3) is used to study various types of fractional systems (see [1–12] and the references therein).In this paper we use the fixed point index to study the existence of positive solutions for the system of nonlinear semipositone fractional boundary value problem (1). Under some appropriate conditions for fi(i=1,2), we use the fixed point index to obtain our results. Moreover, nonnegative concave and convex functions are used to depict the coupling behavior of our nonlinearities (see [13–15]), which depend on the unknown functions u,v and their derivatives u′, v′.


## 2. Preliminary

Definition 1 (see [16, 17]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f:(0,+∞)→R is given by(4)D0+αft=1Γn-αddtn∫0tt-sn-α-1fsds,where n=[α]+1 with [α] denoting the integer part of a number α, provided that the right hand side is pointwise defined on (0,+∞).
We first study the Green functions of problem (1). Let (5)G1t,s≔1Γαtα-11-sα-2-t-sα-1,0≤s≤t≤1,tα-11-sα-2,0≤t≤s≤1.
Then we have(6)G2t,s≔∂∂tG1t,s=α-1Γαtα-21-sα-2-t-sα-2,0≤s≤t≤1,tα-21-sα-2,0≤t≤s≤1.Lemma 2.
Letfi(i=1,2) be as in (1). Then (1) is equivalent to(7)D0+αx=-f1t,∫01G1t,sxsds,∫01G2t,sxsds,∫01G1t,sysds,∫01G2t,sysds,D0+αy=-f2t,∫01G1t,sxsds,∫01G2t,sxsds,∫01G1t,sysds,∫01G2t,sysds,x0=x′0=x′1=0,y0=y′0=y′1=0,which takes the form(8)xt=∫01G1t,sf1s,∫01G1s,τxτdτ,∫01G2s,τxτdτ,∫01G1s,τyτdτ,∫01G2s,τyτdτds,yt=∫01G1t,sf2s,∫01G1s,τxτdτ,∫01G2s,τxτdτ,∫01G1s,τyτdτ,∫01G2s,τyτdτds.Let D0+αu=-x, D0+αv=-y. Then an argument similar to that in [18, Lemma 2.7] and [19, Lemma 3] establishes the result (we omit the standard details).Lemma 3 ([19, Lemma 4]).
The functionsGi(t,s)∈C([0,1]×[0,1],R+)(i=1,2). Moreover, the following inequalities are satisfied:(9)tα-1s1-sα-2≤ΓαG1t,s≤s1-sα-2∀t,s∈0,1,(10)α-1α-2tα-21-ts1-sα-2≤ΓαG2t,s≤α-1tα-3s1-sα-2∀t,s∈0,1.Lemma 4 ([19, Lemma 5]).
Letφ(t)=t(1-t)α-2 for all t∈[0,1]. Let(11)k1≔αΓα-1Γ2α≤k2≔1αα-1Γα,k3≔α-1α-2ΓαΓ2α≤k4≔Γα-1Γ2α-2.Then(12)k2i-1φs≤∫01Git,sφtdt≤k2iφs,i=1,2,∀s∈0,1.Lemma 5.
(i) If(x∗(t),y∗(t)) is a positive solution of (7), then (x∗(t)+w(t),y∗(t)+w(t)) is a positive solution of the following differential equation:(13)D0+αx=-F1t,∫01G1t,sxs-wsds,∫01G2t,sxs-wsds,∫01G1t,sys-wsds,∫01G2t,sys-wsds,D0+αy=-F2t,∫01G1t,sxs-wsds,∫01G2t,sxs-wsds,∫01G1t,sys-wsds,∫01G2t,sys-wsds,x0=x′0=x′1=0,y0=y′0=y′1=0,where (14)Fit,x1,x2,x3,x4≔f¯it,x1,x2,x3,x4,t∈0,1,x1,x2,x3,x4≥0,f¯it,0,0,0,0,t∈0,1,x1,x2,x3,x4<0,and f¯i(t,x1,x2,x3,x4)=fi(t,x1,x2,x3,x4)+M,f¯i:[0,1]×R+4→R+ are continuous, and(15)wt≔M∫01G1t,sds=MΓαtα-1α-1-tαα∀t∈0,1.
(ii) Ifxt,yt is a solution of (13) and x(t)≥w(t), y(t)≥w(t),t∈[0,1], then (x∗(t),y∗(t))=(x(t)-w(t),y(t)-w(t)) is a positive solution of (7).Proof.
If(x∗(t),y∗(t)) is a positive solution of (7) then (note w(t)=M∫01G1t,sds) we obtain x∗(0)+w(0)=x∗′(0)+w′0=x∗′1+w′1=0 and (16)D0+αx∗t+wt+F1t,∫01G1t,sx∗sds,∫01G2t,sx∗sds,∫01G1t,sy∗sds,∫01G2t,sy∗sds=D0+αx∗t+D0+αwt+f1t,∫01G1t,sx∗sds,∫01G2t,sx∗sds,∫01G1t,sy∗sds,∫01G2t,s×y∗sds+M=D0+αwt+M=D0+αM∫01G1t,sds+M=-M+M=0.Similarly, we have (17)D0+αy∗t+wt+F2t,∫01G1t,sx∗sds,∫01G2t,sx∗sds,∫01G1t,sy∗sds,∫01G2t,sy∗sds=-M+M=0;that is, x∗t+wt,y∗t+wt satisfies (13). Therefore, (i) holds. Similarly, it is easy to prove (ii). This completes the proof.From Lemma5, to obtain a positive solution of (7), we only need to find solutions xt, yt of (13) satisfying (t)≥w(t), y(t)≥w(t), t∈[0,1]. If x(t), y(t) are solutions of (13), then x(t), y(t)satisfy(18)xt=∫01G1t,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τyτ-wτdτ,∫01G2s,τyτ-wτdτds,yt=∫01G1t,sF2s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τyτ-wτdτ,∫01G2s,τyτ-wτdτds.LetE≔C[0,1], x≔maxt∈0,1⁡xt, P≔{x∈E:xt≥0,t∈[0,1]}. Then (E,·) is a real Banach space, and P is a cone on E. We denote Bρ≔x∈E:x<ρ for ρ>0. Now, note that u, v solve (1) if and only if x≔-D0+αu, y≔-D0+αv are fixed points of operator (19)Aix,yt≔∫01G1t,sFis,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds,Ax,yt=A1,A2x,ytforx,y∈E.Therefore, if (x,y) is a positive fixed for A with x(t)≥w(t), y(t)≥w(t) for t∈[0,1], then (x∗,y∗)=(x-w,y-w) is a positive solution for (1). Moreover, from the continuity of Gi and Fi(i=1,2), we know that Ai:P×P→P, A:P×P→P×P are continuous and completely continuous operators.Lemma 6.
LetP0≔{x∈P:x(t)≥tα-1x, t∈[0,1]}. Then P0 is a cone in E and A(P×P)⊂P02.Proof.
From (9) for t∈[0,1] we have (20)A1x,yt=∫01G1t,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≤1Γα∫01s1-sα-2F1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds.Also from (9) and the above inequality, for every (x,y)∈P×P, we obtain (21)A1x,yt=∫01G1t,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≥tα-1Γα∫01s1-sα-2F1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≥tα-1A1x,yfor all t∈[0,1]. Similarly, A2(x,y)(t)≥tα-1A2(x,y). Therefore A(P×P)⊂P02. This completes the proof.To obtain a positive solution of (1), we seek a positive fixed point (x∗,y∗) of A with x∗≥w, y∗≥w (note mean that x∗t=A1x∗,y∗t, y∗(t)=A2(x∗,y∗)(t) for t∈[0,1]). From Lemma 6, we have x∗,y∗∈P0. For x∗∈P0 we have (22)x∗t-wt=x∗t-M∫01G1t,sds=x∗t-MΓαtα-1α-1-tαα=x∗t-Mtα-1Γα1α-1-tα≥x∗t-x∗tx∗Mα-1Γα.As a result, x∗(t)≥w(t) for t∈[0,1] if x∗≥M/α-1Γα≔k5. Similarly, if y∗≥k5, we have y∗(t)≥w(t), for t∈[0,1].Lemma 7 (see [20]).
LetΩ⊂E be a bounded open set and A:Ω¯∩P→P a continuous or completely continuous operator. If there exists u0∈P∖{0} such that u-Au≠μu0 for all μ≥0 and u∈∂Ω∩P, then i(A,Ω∩P,P)=0, where i denotes the fixed point index on P.Lemma 8 (see [20]).
LetΩ⊂E be a bounded open set with 0∈Ω. Suppose A:Ω¯∩P→P is a continuous or completely continuous operator. If u≠μAu for all u∈∂Ω∩P and 0≤μ≤1, then i(A,Ω∩P,P)=1.


## 3. Main Results

LetK≔α/Γα≥maxt,s∈[0,1]⁡G1t,s+G2t,s, In the sequel, we use c1,c2,… and d1,d2,… to stand for different positive constants. Now, we list our assumptions on Fi(i=1,2).(H2) There exist h,g∈C(R+,R+) such that(i)
h is concave and strictly increasing on R+ (and limx→+∞⁡h(x)=+∞);(ii)
there existc1>0, d1>1/k12k1+k32, for all t,x1,x2,y1,y2∈[0,1]×R+4 such that (23)F1t,x1,x2,y1,y2≥d1hy1+y2-c1,F2t,x1,x2,y1,y2≥gx1+x2-c1;(iii)
h(K2g(x))≥K2x-c1 for x∈R+.(H3) For all (t,x1,x2,y1,y2)∈[0,1]×[0,k5]4, there is a constant M1∈(0,k5k2-1) such that (24)Fit,x1,x2,y1,y2≤M1,i=1,2.(H4) There exist β, γ∈C(R+,R+) such that(i)
β is convex and strictly increasing on R+ (and limx→+∞⁡β(x)=+∞);(ii)
for all(t,x1,x2,y1,y2)∈[0,1]×R+4, (25)F1t,x1,x2,y1,y2≤βy1+y2,F2t,x1,x2,y1,y2≤γx1+x2;(iii)
there existd2>0 such that β(K2γ(x))≤K2x+d2, for x∈R+.(H5) There exist Q:[0,1]→R, θ∈(0,1], t0∈[θ,1], for all (t,x1,x2,y1,y2)∈[θ,1]×0,k54, such that (26)fit,x1,x2,y1,y2+M≥Qt,i=1,2,where (27)∫θ1G1t0,sQsds>Mα-1Γα.Theorem 9.
Suppose that (H1)–(H3) hold. Then (1) has at least one positive solution.Proof.
We first prove that there existsR>k5 such that(28)x,y≠Ax,y+λϕ,ϕ,∀x,y∈∂BR∩P×P,λ≥0,where ϕ∈P0 is a given function. Suppose there exist (x,y)∈∂BR∩(P×P), λ≥0 with (x,y)=A(x,y)+λ(ϕ,ϕ), then x(t)≥A1(x,y)(t), y(t)≥A2(x,y)(t) for t∈[0,1]. From (i), (ii) of (H2) we have(29)
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.From (ii) of (H2) we have(30)Kyt≥K∫01G1t,sF2s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≥K∫01G1t,sg∫01G1s,τ+G2s,τxτ-wτdτds-Kc1∫01G1t,sds=K∫01G1t,sg∫01G1s,τ+G2s,τxτ-wτdτds-c3.From (30) and (i) of (H2) we obtain(31)hKyt+c3≥hK∫01G1t,sg∫01G1s,τ+G2s,τxτ-wτdτds.This together with (iii) of (H2) yields(32)hKyt≥hKyt+c3-hc3≥hK∫01G1t,sg∫01G1s,τ+G2s,τxτ-wτdτds-hc3≥∫01hG1t,sKK2g∫01G1s,τ+G2s,τxτ-wτdτds-hc3≥∫01G1t,sKhK2g∫01G1s,τ+G2s,τxτ-wτdτds-hc3≥∫01KG1t,s∫01G1s,τ+G2s,τxτ-wτdτds-c4.Then (32) is substituted into (29) and we obtain(33)xt≥d1∫01G1t,s∫01G1s,τ+G2s,τK∫01KG1τ,r∫01G1r,l+G2r,l×xl-wldldr-c4dτds-c2≥d1∫01G1t,s∫01G1s,τ+G2s,τ∫01G1τ,r∫01G1r,l+G2r,l×xl-wldldrdτds-c5≥d1∫01G1t,s∫01G1s,τ+G2s,τ∫01G1τ,r∫01G1r,l+G2r,l×xldldrdτds-c6.Multiplying by φ(t) for (33) and integrating over [0,1], we have(34)∫01φtxtdt≥d1k12k1+k32∫01φtxtdt-c6k2Γα,using the fact that(35)k1φs≤∫01G1t,sφtdt≤k2φs,k1+k3φs≤∫01G1t,s+G2t,sφtdt≤k2+k4φs,which can be derived from (12) in Lemma 4. From (34) we obtain(36)∫01φtxtdt≤c6k2Γαd1k12k1+k32-1.Note that x∈P0 (note that x=A1(x,y)+λϕ and A1(x,y)∈P0 from Lemma 6 and ϕ∈P0) and we have(37)∫01φttα-1xdt≤∫01φtxtdt≤c6k2Γαd1k12k1+k32-1,x≤c6k2d1k13k1+k32-k1.From (29) we have(38)d1∫01G1t,s∫01G1s,τ+G2s,τKhKyτdτds≤xt+c2≤xt+c2≤c6k2d1k13k1+k32-k1+c2.Multiplying by φ(t) and integrating over [0,1] we obtain (39)∫01xt+c2φtdt≥∫01φtd1∫01G1t,s∫G1s,τ+G2s,τKhKyτdτdsdt≥d1∫01k1φs∫01G1s,τ+G2s,τKhKyτdsdt≥d1k1k1+k3K∫01φthKytdt.Consequently, we have (40)∫01φthKytdt≤Kd1k1k1+k3∫01c6k2d1k13k1+k32-k1+c2φtdt=KΓαk2d1k1k1+k3c6k2d1k13k1+k32-k1+c2≔N1.Note that we may assume y(t)≢0 for t∈[0,1]. Then y>0 and h(Ky)>0. For y∈P0, we have (41)k1ΓαKy≤∫01Kφtytdt=KyhKy∫01φtKytKyhKydt≤KyhKy∫01φthKytdt≤KyhKyN1.Hence, h(Ky)≤N1/k1Γα. Note limz→+∞⁡h(z)=+∞, and thus there exists N2≥0 such that Ky≤N2. Therefore if (x,y)∈∂BR∩(P×P), λ≥0 with (x,y)=A(x,y)+λ(ϕ,ϕ) then x≤c6k2/d1k13k1+k32-k1 and y≤N2/K. Thus if we take R>max⁡k5,c6k2/d1k13k1+k32-k1,N2/K then (28) is true. Lemma 7 implies(42)iA,BR∩P×P,P×P=0.Let x,y∈∂Bk5∩P. From (H3) we have (43)A1x,yt=∫01G1t,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≤∫01G1t,sM1ds≤∫01s1-sα-2ΓαM1ds<Mα-1Γα=k5=x,so A1(x,y)<x. Similarly A2(x,y)<y. Hence A(x,y)<(x,y) for x,y∈∂Bk5∩P. Thus (44)x,y≠λAx,y,∀x,y∈∂Bk5∩P×P,λ∈0,1.It follows from Lemma 8 that(45)iA,Bk5∩P×P,P×P=1.From (42) and (45) we have (46)iA,BR∖B¯k5∩P×P,P×P=0-1=-1.Therefore the operator A has at least one fixed point in (BR∖B¯k5)∩(P×P) and so (1) has at least a positive solution. This completes the proof.Theorem 10.
Suppose that (H1), (H4), and (H5) hold. Then (1) has at least one positive solution.Proof.
We first show that there existsR>k5 such that(47)x,y≠λAx,y,∀x,y∈∂BR∩P×P,λ∈0,1.Suppose there exist (x,y)∈∂BR∩(P×P), λ∈[0,1] with (x,y)=λA(x,y), then x(t)≤A1(x,y)(t), y(t)≤A2(x,y)(t) for t∈[0,1]. From (i), (ii) of (H4), we have(48)
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.From (ii) of (H4), we get(49)Kyt≤K∫01G1t,sF2s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τ×yτ-wτdτ,∫01G2s,τyτ-wτdτds≤K∫01G1t,sγ∫01G1s,τ+G2s,τxτ-wτdτds.From (49) and (i), (iii) of (H4) we have(50)βKyt≤βK∫01G1t,sγ∫01G1s,τ+G2s,τxτ-wτdτds≤∫01βG1t,sKK2γ∫01G1s,τ+G2s,τxτ-wτdτds≤∫01G1t,sKβK2γ∫01G1s,τ+G2s,τxτ-wτdτds≤K∫01G1t,s∫01G1s,τ+G2s,τxτ-wτdτ+d2ds≤K∫01G1t,s∫01G1s,τ+G2s,τxτ-wτdτds+d3.Then substitute (50) into (48) and we obtain(51)xt≤∫01G1t,s∫01G1s,τ+G2s,τKβKyτdτds≤∫01G1t,s∫01G1s,τ+G2s,τKK∫01G1τ,r∫01G1r,l+G2r,l×xl-wldldr+d3dτds≤∫01G1t,s∫01G1s,τ+G2s,τ∫01G1τ,r∫01G1r,l+G2r,l×xl-wldldrdτds+d4.Multiplying by φ(t) for (51) and integrating over [0,1], from (35), we obtain(52)∫01φtxtdt≤k22k2+k42∫01φtxtdt+Γαk2d4.Consequently, we have (53)∫01φtxtdt≤Γαk2d41-k22k2+k42.Note that x∈P0 (note x=λA1(x,y) and A1(x,y)∈P0) and we have (54)∫01φttα-1xtdt≤∫01φtxtdt≤Γαk2d41-k22k2+k42,x≤k2d4k1-k1k22k2+k42.From (50) and Lemma 3 we have (55)βKyt≤K∫01G1t,s∫01G1s,τ+G2s,τxτ-wτdτds+d3≤Kx∫01G1t,s∫01G1s,τ+G2s,τdτds+d3≤Kx∫01s1-sα-2∫011Γατ1-τα-2+α-1sα-3τ1-τα-2dτds+d3≤Kxα2α-12Γα2+π23-2αΓα-1Kxα-1Γα-1/2α2Γα2+d3,so β(Ky(t)) is bounded. Note limz→+∞⁡β(z)=+∞, and thus there exists N3≥0 such that Ky≤N3. Therefore if (x,y)∈∂BR∩(P×P),λ∈[0,1] with (x,y)=λA(x,y) then x≤k2d4/k1-k1k22k2+k42 and y≤N3/K. Thus if we take R>max⁡k5,k2d4/k1-k1k22k2+k42,N3/K then (47) is true. Lemma 8 implies(56)iA,BR∩P×P,P×P=1.Let Bk5≔{x∈P:x<k5} and consider x∈∂Bk5∩P. It follows from (H5) that (57)A1x,yt0=∫01G1t0,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τyτ-wτdτ,∫01G2s,τyτ-wτdτds≥∫θ1G1t0,sF1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τyτ-wτdτ,∫01G2s,τyτ-wτdτds=∫θ1G1t0,sf1s,∫01G1s,τxτ-wτdτ,∫01G2s,τxτ-wτdτ,∫01G1s,τyτ-wτdτ,∫01G2s,τyτ-wτdτ+Mds≥∫θ1G1t0,sQsds>Mα-1Γα,and, hence, A1(x,y)≥A1x,yt0>x. Similarly A2(x,y)≥A2(x,y)(t0)>y for y∈∂Bk5∩P. Therefore A(x,y)>(x,y) for all x,y∈∂Bk5∩(P×P). Thus (58)x,y≠Ax,y+λϕ,ϕ,∀x,y∈∂Bk5∩P×P,ϕ∈P,λ∈0,1.It follows from Lemma 7 that(59)iA,Bk5∩P×P,P×P=0.From (56) and (59), we have (60)iA,BR∖B¯k5∩P×P,P×P=1-0=1.
Therefore the operatorA has at least one fixed point on (BR∖B¯k5)∩(P×P) and so (1) has at least one positive solution, which completes the proof.



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*Source: 7351653-2018-05-06.xml*