# Positive Solutions to a Coupled Fractional Differential System withp-Laplacian Operator

Authors: Yuehan Liu; Xiaodi Zhao; Huihui Pang
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

In this paper, we consider a fractional differential system with multistrip and multipoint mixed boundary conditions involvingp-Laplacian operator and fractional derivatives. The existence result of positive solutions is established by the Leggett-Williams fixed point theorem. Also, an example is presented to illustrate our main result.


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## Body

## 1. Introduction

Fractional calculus is a significant branch of mathematical analysis which concerns derivatives and integrals. Fractional differential equations are caused by the intensive development of the theory of fractional calculus and have gained great interest currently, because of their crucial applications in various fields of physics, biology, and engineering such as control, porous media, electromagnetic, and other fields [1–4].Moreover, many mathematical formulations of physical and biological phenomena lead to fractional differential equations withp-Laplacian operator, such as memory, fluid dynamics, predator harvesting, and hereditary properties of various materials, which are difficult to achieve in classical integer-order models. Hence, in recent years, more and more researchers have considered the fractional differential equations with p-Laplacian operator, by applying various mathematical theory and tools; refer to [5–9].For example, authors in [10] discussed the following differential system with p-Laplacian operator (1)Dβ1ϕpDα1μx+ϖ1x,vx=0,x∈0,1,Dβ2ϕpDα2vx+ϖ2x,μx=0,x∈0,1,Dα1μxx=0=0=μxx=0=μ′′xx=0,μ′xx=1=η1IPγ1ψ1μ=η1Γγ1∫0PP-τγ1-1ψ1μτdτ,Dα2vxx=0=0=vxx=0=v′′xx=0,v′xx=1=η2IPγ2ψ2v=η2Γγ2∫0PP-τγ2-1ψ2μτdτ,where 2<αi<3, 0<βi<1, P, ηi,γi>0, ψ1,ψ2∈L[0,1], ϕp(s) is a p-Laplacian operator: ϕp(s)=|s|p-2s, p>1 and ϖi:[0,1]×[0,∞)→(0,∞). Authors established the conclusion of the existence and uniqueness of solution to the above problem by using topological degree theory.By reading the existing literatures [11–24], we note that the fractional differential system involving p-Laplacian operator and multistrip and multipoint boundary conditions has not been studied yet. Thus, in this paper, we first pay close attention to the following fractional differential system, involving p-Laplacian operator and lower fractional derivatives:(2)D0+βφpD0+αut=f1t,ut,vt,D0+αut,D0+αvt,D0+βφpD0+αvt=f2t,ut,vt,D0+αut,D0+αvt,with multistrip and multipoint boundary conditions:(3)u0=D0+αu0=0u1=∑i=1mbi∫0ξiutdAt,φpD0+αu1=∑i=1maiφpD0+αuξi,(4)v0=D0+αv0=0v1=∑i=1mdi∫0ηivtdBt,φpD0+αv1=∑i=1mciφpD0+αvηi,where D0+α and D0+β are the Riemann-Liouville fractional derivatives of order α and β, f:0,1×R+2×R2→R+ is a continuous function. For i=1,2,…,m, 1<α⩽2, 1<β⩽2, 0<ξi≤1, 0<ηi≤1, ai, bi, ci, di are nonnegative constants. φp(s)=|s|p-2s,p>1,φp-1=φq,1/p+1/q=1. And A(t), B(t) are nondecreasing functions of bounded variation in [0,1].We emphasize that, despite the complication of the discrete and integral boundary conditions (3) and (4), operator T can be presented in a concise form based on the piecewise form of Green’s function. And we can accurately estimate the upper and lower bounds of its value, which is fully prepared for the establishment of the main theorem.Accordingly, the conclusion we reached is an extensive result and a meaningful supplement to the theory of the fractional differential systems withp-Laplacian operator.Theorem 1 (the Leggett-Williams fixed point theorem).
LetP be a cone in a real Banach space E, Pc={x∈P∣x⩽c}, Ψ be a nonnegative continuous concave function on P such that Ψ(x)⩽x for all x∈P¯c, and P(Ψ,b,d)={x∈P∣b⩽Ψ(x),x⩽d}. Suppose that T:P¯c→P¯c is completely continuous and there exist constants 0<a<b<d⩽c such that(S1)
{x∈PΨ,b,d∣Ψ(x)>b}≠⌀,andΨ(Tx)>bforx∈P(Ψ,b,d);(S2)
Tx<aforx⩽a;(S3)
Ψ(Tx)>bforx∈P(Ψ,b,c),withTx>d.
ThenT has at least three fixed points x1, x2, and x3 which satisfy (5)x1<a,b<Ψx2,a<x3withΨx1<b.


## 2. Preliminaries

In this section, we will put forward some crucial definitions and theorems, preparing for establishing the existence of solutions.Definition 2.
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,∞)→R is given by (6)I0+αft=1Γα∫0tt-sα-1fsds,provided the right-hand side is pointwise defined on (0,∞), where Γ(α) is the Euler gamma function defined by Γ(α)=∫0∞‍tα-1e-tdt, for α>0.Definition 3.
The Riemann-Liouville fractional derivative of orderα>0 for a function f:(0,∞)→R is given by [0,∞), (7)D0+αft=1Γn-αddtn∫0tt-sn-α-1fsds,where n=[α]+1, [α] stands for the largest integer not greater than α.According to the definitions of Riemann-Liouville’s derivative, the following lemmas can be achieved.Lemma 4.
Forα>0, if we assume that u∈C[0,∞)∩L1(0,1), then we have (8)I0+αD0+αut=ut+m1tα-1+m2tα-2+⋯+mntα-n,for some mi∈R,i=1,2,…,n, while n is the smallest integer greater than or equal to α.Remark 5.
The following properties are useful for our discussion:(i)
As a basic example, we quote forλ>-1, (9)D0+αtλ=Γλ+1Γλ-α+1tλ-α;(ii)
D0+αI0+αu(t)=u(t),foru(t)∈L1(0,1).For convenience, we denote(10)N1=∑i=1mbi∫0ξitα-1dAt,N2=∑i=1mdi∫0ηitα-1dBt,Mk=∑i=1kaiξiβ-1,Ek=∑i=1kciηiβ-1,fork=1,2…m.Besides, it should be noted that M0=E0=0.Lemma 6.
Forh1,h2∈C(0,1), the equations(11)D0+βφpD0+αut=h1t,D0+βφpD0+αvt=h2t,with boundary conditions (3) and (4) have a unique solution as follows: (12)ut=∫01K1t,sφq∫01H¯s,τh1τdτds,vt=∫01K2t,sφq∫01H^s,τh2τdτds,where(13)K1t,s=Gαt,s+tα-11-N1∑i=1mbi∫0ξiGαt,sdAt,K2t,s=Gαt,s+tα-11-N2∑i=1mdi∫0ηiGαt,sdBt,(14)Gαt,s=1Γαtα-11-sα-1-t-sα-1,0⩽s⩽t⩽1,1Γαtα-11-sα-1,0⩽t⩽s⩽1,Gβt,s=1Γβtβ-11-sβ-1-t-sβ-1,0⩽s⩽t⩽1,1Γβtβ-11-sβ-1,0⩽t⩽s⩽1,and H¯(t,s) is a piecewise function; the expression of the kth (k=0,1,…,m) piece is as follows (15)H¯kt,s=Gβt,s+tβ-11-Mm∑i=kmaiGβξi,s+Mk1-MmΓβtβ-11-sβ-1=1Γβtβ-11-sβ-1-t-sβ-1+tβ-11-MmΓβ∑i=k+1maiξiβ-11-sβ-1-ξi-sβ-1+Mk1-MmΓβtβ-11-sβ-1,forξk⩽s⩽t⩽ξk+1,1Γβtβ-11-sβ-1+tβ-11-MmΓβ∑i=k+1maiξiβ-11-sβ-1+Mk1-MmΓβtβ-11-sβ-1,forξk⩽t⩽s⩽ξk+1.Meanwhile, H^(t,s) is also a piecewise function; the expression of kth (k=0,1…,m) piece is as follows (16)H^kt,s=Gβt,s+tβ-11-Em∑i=kmciGβηi,s+Ek1-EmΓβtβ-11-sβ-1=1Γβtβ-11-sβ-1-t-sβ-1+tβ-11-EmΓβ∑i=k+1mciηiβ-11-sβ-1-ηi-sβ-1+Ek1-EmΓβtβ-11-sβ-1,forηk⩽s⩽t⩽ηk+1,1Γβtβ-11-sβ-1+tβ-11-EmΓβ∑i=k+1mciηiβ-11-sβ-1+Ek1-EmΓβtβ-11-sβ-1,forηk⩽t⩽s⩽ηk+1.Proof.
We shall reduce BVP (11), (3), and (4) to two fractional differential problems. To this goal, first, by means of transformation, we convert problem (11) into(17)D0+βy1t=h1t,D0+βy2t=h2t,y10=0,y11=∑i=1maiy1ξi,y20=0,y21=∑i=1mciy2ηi,(18)φpD0+αut=y1t,φpD0+αvt=y2t,u0=0,u1=∑i=1mbi∫0ξiutdAt,v0=0,v1=∑i=1mdi∫0ηivtdBt.
In view of Lemma4, the solution of problem (17) can be written as(19)y1t=I0+βh1t=1Γβ∫0tt-sβ-1h1sds+c1tβ-1+c2tβ-2,where c1, c2 are arbitrary real constants.
Byy1(0)=0, we find c2=0. Setting t=1 in (19), together with the boundary condition y1(1)=∑i=1m‍aiy1(ξi), we have (20)1Γβ∫011-sβ-1h1sds-1Γβ∑i=1mai∫0ξiξi-sβ-1h1sds=c1∑i=1maiξiβ-1-1.Accordingly, we can easily find out(21)c1=1Γβ1-Mm∑i=1mai∫0ξiξi-sβ-1h1sds-∫011-sβ-1h1sds.Combining (21) with (19), we can obtain (22)y1t=1Γβ∫0tt-sβ-1h1sds+tβ-11-Mm∑i=1mai∫0ξiξi-sβ-1h1sds-∫011-sβ-1h1sds.
Next, considering the multipoint boundary conditions, we divide the domain of integration intom+1 pieces. For ξk⩽t⩽ξk+1, the unique solution of (17) can be expressed as (23)y1t=1Γβ∫0tt-sβ-1h1sds+tβ-11-Mm∑i=1mai∫0ξiξi-sβ-1h1s-∫011-sβ-1h1sds=1Γβ∫0ξ1t-sβ-1+tβ-11-Mm∑i=1maiξi-sβ-1-tβ-11-sβ-11-Mmh1sds+1Γβ∫ξ1ξ2t-sβ-1+tβ-11-Mm∑i=2maiξi-sβ-1-tβ-11-sβ-11-Mmh1sds+⋯+1Γβ∫ξktt-sβ-1+tβ-11-Mm∑i=k+1maiξi-sβ-1-tβ-11-sβ-11-Mmh1sds+1Γβ∫tξk+1tβ-11-Mm∑i=k+1maiξi-sβ-1-tβ-11-sβ-11-Mmh1sds+⋯+1Γβ∫ξm1-tβ-11-sβ-11-Mmh1sds.When ξk⩽s⩽t, we observe that the kth integrand in (23) can be taken as follows: (24)1Γβt-sβ-1+tβ-11-Mm∑i=k+1maiξi-sβ-1-tβ-11-sβ-11-Mm=1Γβt-sβ-1-tβ-11-sβ-1+tβ-11-Mm∑i=k+1maiξi-sβ-1-ξiβ-11-sβ-1-tβ-11-sβ-11-Mm+tβ-11-sβ-1+tβ-11-Mm∑i=k+1maiξiβ-11-sβ-1=1Γβt-sβ-1-tβ-11-sβ-1+tβ-11-Mm∑i=k+1maiξi-sβ-1-ξiβ-11-sβ-1-Mk1-Mmtβ-11-sβ-1=-Gβt,s-tβ-11-Mm∑i=k+1maiGβξi,s-Mk1-MmΓβtβ-11-sβ-1=-H¯kt,s,where ∑i=m+1m‍aiGβ(ξi,s) is also noted by 0.
As a result, the Green functionH¯k(t,s) is always valid and can exactly express each integrand in the above formula, for k=0,1,⋯m. Thus, we can take (23) as a unified form(25)y1t=-∫01H¯t,sh1sds.H¯(t,s) is a piecewise function such that the form of the kth piece is given by (15).
We take (25) into (18); then (26)φpD0+αut=-∫01H¯t,sh1sds.From Lemma 4, we have(27)ut=-1Γα∫0tt-sα-1φq∫01H¯s,τh1τdτds+c1tα-1+c2tα-2,where c1, c2 are arbitrary real constants.
Byu(0)=0, we easily find c2=0. Setting t=1 in (27), together with the boundary condition u(1)=∑i=1m‍bi∫0ξi‍u(t)dA(t), we have (28)∑i=1mbi∫0ξiutdAt=-1Γα∫011-sα-1φq∫01H¯s,τh1τdτds+c1.Set c1 and c2 into (27); we find out(29)ut=∫01Gαt,sφq∫01H¯s,τh1τdτds+tα-1∑i=1mbi∫0ξiutdAt.Furthermore, by integrating and summing on both sides of (29), it is easy to see that (30)∑i=1mbi∫0ξiutdAt=∑i=1mbi∫0ξi∫01Gαt,sφq∫01H¯s,τh1τdτdsdAt+N1∑i=1mbi∫0ξiutdAt.Then, we get (31)∑i=1mbi∫0ξiutdAt=11-N1∑i=1mbi∫0ξi∫01Gαt,sφq∫01H¯s,τh1τdτdsdAt.
Take (31) into (29); we conclude that the unique solution u(t) of BVP (11), (3), and (4) is (32)ut=∫01Gαt,s+tα-11-N1∑i=1mbi∫0ξiGαt,sdAtφq∫01H¯s,τh1τdτds.Analogically, the unique solution v(t) of BVP (11), (3), and (4) is (33)vt=∫01Gαt,s+tα-11-N2∑i=1mdi∫0ηiGαt,sdBtφq∫01H^s,τh2τdτds,where H^(t,s) is also a piecewise function, the kth piece H^k(t,s) is defined by (16), and the calculation process is similar.
This completes the proof.Denote(34)Φαt,s=tα-11-sα-1-t-tα-1sα-1,0⩽s⩽t⩽1,tα-11-sα-1,0⩽t⩽s⩽1.Φβt,s=1Γβtβ-11-sβ-1-t-tβ-1sβ-1,0⩽s⩽t⩽1,tβ-11-sβ-1,0⩽t⩽s⩽1.Lemma 7.
Fort,s∈[0,1] and 1<γ⩽2, it holds that (35)tγ-11-sγ-1-t-sγ-1⩾tγ-11-sγ-1-t-tγ-1sγ-1⩾0.Proof.
For1<γ⩽2 and 0⩽t⩽1, we have t⩽tγ-1⩽1, while tγ-1 is a monotone increasing function. Therefore, we can infer that t-s⩽t-tγ-1s⩽t-ts. For (t-s)γ-1⩽(t-tγ-1s)γ-1⩽(t-ts)γ-1, we finally find out (t-ts)γ-1-(t-s)γ-1⩾(t-ts)γ-1-(t-tγ-1s)γ-1⩾(t-ts)γ-1-(t-ts)γ-1=0, which implies Φα(t,s)⩾0 and Φβ(t,s)⩾0, for 1<α,β⩽2, 0⩽t,s⩽1.
This completes the proof.Lemma 8.
Fort,s∈[0,1], the functions Gα(t,s),Gβ(t,s) defined by (14) have the following properties:(1)
Gα(t,s)>0, Gβ(t,s)>0, for t,s∈(0,1);(2)
Φα(t,s)(1-s)α-1⩽Γ(α)Gα(t,s)⩽(1-s)α-1,fort,s∈(0,1);Φβ(t,s)1-sβ-1⩽Gβ(t,s)⩽1-sβ-1/Γ(β),fort,s∈(0,1).Proof.
(1) The positive ofGα(t,s) and Gβ(t,s) is easily checked.
(2) For0⩽s⩽t⩽1, we have (36)ΓαGαt,s=tα-11-sα-1-t-sα-1⩽tα-11-sα-1⩽1-sα-1,ΓαGαt,s=tα-11-sα-1-t-sα-1⩾tα-11-sα-1-t-tα-1sα-1⩾tα-11-sα-1-t-tα-1sα-11-sα-1=Φαt,s1-sα-1.For 0⩽t⩽s⩽1, we have (37)ΓαGαt,s=tα-11-sα-1⩽1-sα-1,ΓαGαt,s=tα-11-sα-1⩾tα-11-sα-11-sα-1=Φαt,s1-sα-1.The other inequality is achieved similarly. All these complete the proof of the lemma.Fors∈[0,1], denote (38)ϱ1s=11-N1Γα∑i=1mbi∫0ξiΦαt,sdAt,ϱ2s=11-N2Γα∑i=1mdi∫0ηiΦαt,sdBt,ρ1=1Γα1+11-N1∑i=1mbi∫0ξidAt,ρ2=1Γβ11-Mm∑i=1mai+11-Mm,ρ3=1Γα1+11-N2∑i=1mdi∫0ηidBt,ρ4=1Γβ11-Em∑i=1mci+11-Em,and the constant ρ is given by the maximum value of ρ1, ρ2, ρ3, and ρ4.Lemma 9.
For(t,s)∈[0,1]×[0,1], the functions Kk(t,s)(k=1,2), H¯(t,s), and H^(t,s) defined by Lemma 6 satisfy the following results:(1)
Kk(t,s), H¯(t,s), H^(t,s) are continuous, Kk(t,s)⩾0, H¯(t,s)⩾0, H^(t,s)⩾0;(2)
ϱ1(s)tα-1(1-s)α-1⩽K1(t,s)⩽ρ(1-s)α-1,ϱ2(s)tα-1(1-s)α-1⩽K2(t,s)⩽ρ(1-s)α-1;(3)
Φβ(t,s)tβ-1(1-s)β-1⩽H¯(t,s)⩽ρ(1-s)β-1,Φβ(t,s)tβ-1(1-s)β-1⩽H^(t,s)⩽ρ(1-s)β-1.Proof.
The nonnegativeness ofKk(t,s), H¯(t,s), and H^(t,s) is easily checked.
(2) According to the property (2) in Lemma8, we have (39)K1t,s=Gαt,s+tα-11-N1∑i=1mbi∫0ξiGαt,sdAt⩾Φαt,s1-sα-1Γα+tα-11-N1∑i=1mbi∫0ξiΦαt,s1-sα-1ΓαdAt⩾tα-11-N1∑i=1mbi∫0ξiΦαt,s1-sα-1ΓαdAt=ϱ1stα-11-sα-1,K1t,s=Gαt,s+tα-11-N1∑i=1mbi∫0ξiGαt,sdAt⩽1-sα-1Γα+tα-11-N1∑i=1mbi∫0ξi1-sα-1ΓαdAt⩽1Γα1+11-N1∑i=1mbi∫0ξidAt1-sα-1=ρ11-sα-1⩽ρ1-sα-1.
Fork=0,1…m, we have (40)Hkt,s=Gβt,s+tβ-11-Mm∑i=k+1maiGβξi,s+Mm1-MmΓβtβ-11-sβ-1⩽1Γβ11-Mm∑i=k+1mai+11-Mm1-sβ-1⩽1Γβ11-Mm∑i=1mai+11-Mm1-sβ-1=ρ21-sβ-1⩽ρ1-sβ-1.Hkt,s=Gβt,s+tβ-11-Mm∑i=k+1maiGβξi,s+Mm1-MmΓβtβ-11-sβ-1⩾tβ-11-sβ-1Φβt,s+∑i=k+1maiΦβξi,s1-Mm+Mm1-MmΓβ⩾tβ-11-sβ-1Φβt,s+∑i=k+1maiΦβξi,s1-Mm⩾tβ-11-sβ-1Φβt,s.Because of the randomicity of k, we have (41)Φβt,stβ-11-sβ-1⩽H¯t,s⩽ρ1-sβ-1
According to the above, from the property (2) of Lemma8, we easily get (42)ϱ2stα-11-sα-1⩽K2t,s⩽ρ31-sα-1,Φβt,stβ-11-sβ-1⩽H^t,s⩽ρ41-sβ-1.
This completes the proof of the lemma.Define the spaceX={ut∣u(t)∈C[0,1] and D0+αu(t)∈C[0,1]} is endowed with the norm u=max{u0,D0+αu0}, where u0=maxt∈[0,1]u(t), D0+αu0=maxt∈[0,1]D0+αut.Then we introduce the product space(X×X,(u,v)) endowed with the norm (u,v)=max{u,v} and define a partial order u1v1⩾u2v2 where u1(t)⩾u2(t), v1(t)⩾v2(t), for t∈[0,1].Lemma 10.
( X × X , ( u , v ) ) is a Banach space.Proof.
Let{un}n=1∞ be a Cauchy sequence in the space (X,·). Then clearly {un}n=1∞ and {D0+αu(t)}n=1∞ are Cauchy sequences in the space C[0,1]. Therefore, {un}n=1∞ and {D0+αu(t)}n=1∞ converge to some v and w on [0,1] uniformly and v,w∈C[0,1]. Then, we need to prove w=D0+αv.
Note that(43)I0+αD0+αunt-I0+αwt⩽1Γα∫0tt-sα-1D0+αuns-wsds⩽1Γα+1maxs∈0,1D0+αuns-ws.
By the convergence of{D0+αu(t)}n=1∞, we get(44)limn→∞I0+αD0+αunt=I0+αwtuniformly for t∈[0,1]. On the other hand, by Remark 5, one has I0+αD0+αun(t)=un(t)+c1tα-1+c2tα-2, for t∈[0,1] and some c1∈R. Furthermore, we can get(45)limn→∞I0+αD0+αunt=limn→∞unt+c1tα-1+c2tα-2=vt+c1tα-1+c2tα-2.
From (44) and (45),we obtain (46)I0+αwt=vt+c1tα-1+c2tα-2,fort∈0,1.Taking the α-order derivative on both sides of yields, we have (47)I0+αD0+αwt=D0+αvt+c1tα-1+c2tα-2,fort∈0,1.In view of Remark 5 and Lemma 4, we find that (48)wt=D0+αvt,fort∈0,1.
Above all, we can achieve that(X×X,(u,v)) is a Banach space. This completes the proof.Define the coneP⊂X×X by(49)P=u,v∈X×X∣ut⩾0,vt⩾0.Let the nonnegative continuous concave functional Ψ on the cone P be defined by (50)Ψu,v=minminθ⩽t⩽1-θut,minθ⩽t⩽1-θvt,where θ is a positive constant, satisfying 0<θ<1-θ<1.For convenience, for0⩽s⩽1, we denote (51)cs=11-N1∑i=1mbi∫0ξiGαt,sdAt,eks=11-Mm∑i=kmaiGβξi,s+Mk1-sβ-11-MmΓβ,where k=0,1⋯m.For any(u,v)∈X×X, we introduce the integral operator T by (52)Tu,vt=T1u,vtT2u,vt,where(53)T1u,vt=∫01K1t,sφq∫01H¯s,τf1τ,uτ,vτ,D0+αuτ,D0+αvτdτds,T2u,vt=∫01K2t,sφq∫01H^s,τf2τ,uτ,vτ,D0+αuτ,D0+αvτdτds.Consequently, by calculation (54)D0+αTu,vt=D0+αT1u,vtD0+αT2u,vt,where (55)D0+αT1u,vt=-φq∫01H¯t,sf1s,us,vs,D0+αus,D0+αvsds,D0+αT2u,vt=-φq∫01H^t,sf2s,us,vs,D0+αus,D0+αvsds.Lemma 11.
T : P → P is completely continuous.Proof.
Firstly, we prove thatT1:P→P. In view of the nonnegativeness and continuity of K1(t,s), H¯(t,s) and f1(t,u(t),v(t),D0+αu(t),D0+αv(t)), T1 is continuous and T1(u,v)(t)⩾0 for (u,v)∈P. Hence, T1P⊂P.
Next, we showT1 is uniformly bounded. Let Ω⊂P be bounded; that is, there exists a positive constant L>0 such that 0⩽u⩽L, 0⩽v⩽L, -L⩽D0+αu(t)⩽L, -L⩽D0+αv(t)⩽L for all (u,v)∈Ω. Let A1=maxt∈[0,1]|f1(t,u(t),v(t),D0+αu(t),D0+αv(t))|. Then for (u,v)∈Ω, from the Lemma 9, we have (56)T1u,vt=∫01K1t,sφq∫01H¯s,τf1τ,uτ,vτ,D0+αuτ,D0+αvτdτds⩽A1q-1∫01K1t,sφq∫01H¯s,τdτds⩽A1q-1∫01ρ1-sα-1φq∫01ρ1-τβ-1dτds=A1q-1ρβq-1ρα,D0+αT1u,vt=-φq∫01H¯t,sf1s,us,vs,D0+αus,D0+αvsds⩽A1q-1φq∫01H¯t,sds⩽A1q-1φq∫01ρ1-sβ-1ds=A1q-1ρβq-1.Hence, T1(Ω) is uniformly bounded. Then, we show T1 is equicontinuous. Indeed, for any (u,v)∈Ω, t1,t2∈[0,1], t1<t2, we have (57)T1ut2,vt2-T1ut1,vt1⩽∫01K1t2,s-K1t1,sφq∫01H¯s,τf1τ,uτ,vτ,D0+αuτ,D0+αvτdτds⩽A1q-1ρβq-1∫01K1t2,s-K1t1,sds⩽A1q-1ρβq-1∫01Gαt2,s-Gαt1,s+cst2α-1-t1α-1ds.Moreover, (58)D0+αT1ut2,vt2-D0+αT1ut1,vt1⩽φq∫01H¯t2,s-H¯t1,sf1s,us,vs,D0+αus,D0+αvsds⩽A1q-1φq∫01H¯t2,s-H¯t1,sds⩽A1q-1φq∫01Gβt2,s-Gβt1,s+est2β-1-t1β-1ds.Therefore, the above process implies that T1:P→P is equicontinuous for all (u,v)∈Ω.
By means of the Arzela-Ascoli theorem,T1:P→P is completely continuous. Likewise, the proof of T2:P→P follows the same method as T1. Thus, T:P→P is completely continuous.


## 3. Existence Result

For convenience, we denote(59)Q1=∫01ϱ1s1-sα-1φq∫01Φβs,τsβ-11-τβ-1dτds,Q2=∫01ϱ2s1-sα-1φq∫01Φβs,τsβ-11-τβ-1dτds,M=minαρq-1βρ,βρ,N=maxk=1,2Qkθα-1-1/q-1.Theorem 12.
There exist nonnegative numbers0⩽a⩽b⩽θ2/Γ(α-1)c, and fk(t,u,v,w,z)(k=1,2) satisfies the following conditions:(H1)
fk(t,u,v,w,z)⩽Mc1/(q-1), for (t,u,v,w,z)∈[0,1]×[0,c]×[0,c]×[-c,c]×[-c,c];(H2)
fk(t,u,v,w,z)<Ma1/(q-1), for (t,u,v,w,z)∈[0,1]×[0,a]×[0,a]×[-a,a]×[-a,a];(H3)
fk(t,u,v,w,z)>Nb1/(q-1), for (t,u,v,w,z)∈[θ,1-θ]×[b,c]×[b,c]×[-c,c]×[-c,c].
Then BVP (2)-(4) has at least three positive solutions (u1,v1), (u2,v2), and (u3,v3) such that (60)u1,v1<a,b<Ψu2,v2<u2,v2⩽c,a<u3,v3⩽cwithΨu3,v3<b.Proof.
We will verify that conditions(S1)-(S3) of Theorem 1 are satisfied.
LetP¯c={(u,v)∈X×X∣u(t)⩾0,v(t)⩾0and(u,v)⩽c}. We first prove that T:P¯c→P¯c. For any (u,v)∈P¯c, t∈[0,1], we have 0⩽u(t)⩽c, 0⩽v(t)⩽c, -c⩽D0+αu(t)⩽c, -c⩽D0+αv(t)⩽c. Consequently, combining with properties (2) and (3) of Lemma 9, (61)T1u,vt=∫01K1t,sφq∫01H¯s,τf1τ,uτ,vτ,D0+αuτ,D0+αvτdτds⩽∫01ρ1-sα-1φq∫01ρ1-τα-1Mc1/q-1dτds=∫01ρ1-sα-1Mq-1cφq∫01ρ1-τα-1dτds=Mq-1cρβq-1ρα⩽c,D0+αT1u,vt=-φq∫01H¯t,sf1s,us,vs,D0+αus,D0+αvsds⩽φq∫01ρ1-sβ-1Mc1/q-1ds=Mq-1cφqρβ=Mq-1cρβq-1⩽c.
Thus,T1(u,v)⩽c and further to get T1P¯c⊂P¯c. Therefore T1:P¯c→P¯c is completely continuous. The proof of T2 is the same as T1. Thus, T:P¯c→P¯c is completely continuous.
By the above, condition(S2) of Theorem 1 can be satisfied, using assumption (H2).
Finally, in order to verify{u,v∈PΨ,b,c∣Ψ(u,v)>b}≠⌀, we set (u,v)=(btα/θ2,btα/θ2), for 0⩽t⩽1. It is easy to find that (62)Ψu,v=minθ⩽t⩽1-θmin⁡btαθ2,min⁡btαθ2=minθ⩽t⩽1-θbtαθ2⩾bθαθ2=bθα-2>b.Hence, (63)u,v=maxmax0⩽t⩽1ut,max0⩽t⩽1vt,max0⩽t⩽1D0+αut,max0⩽t⩽1D0+αvt=max⁡bθ2,bθ2,bθ2Γα+1,bθ2Γα+1=bθ2Γα+1<c.
If(u,v)∈P(Ψ,b,c), we have b⩽u(t)⩽c, b⩽v(t)⩽c, -c⩽D0+αu(t)⩽c, -c⩽D0+αv(t)⩽c, for all t∈[θ,1-θ]. Then (64)ΨTu,vt=minminθ⩽t⩽1-θT1u,vt,minθ⩽t⩽1-θT2u,vt>min⁡Nq-1b∫01K1t,sφq∫01Φβs,τsβ-11-τβ-1dτds,Nq-1b∫01K2t,sφq∫01Φβs,τsβ-11-τβ-1dτds⩾min⁡Nq-1bθα-1∫01ϱ1s1-sα-1φq∫01Φβs,τsβ-11-τβ-1dτds,Nq-1bθα-1∫01ϱ2s1-sα-1φq∫01Φβs,τsβ-11-τβ-1dτds=mink=1,2Nq-1bθα-1Qk=b.
That is,Ψ(T(u,v)(t))>b for all (u,v)∈P(Ψ,b,c). This shows that condition (S1) of Theorem 1 is also satisfied.
From the above, BVP (2)-(4) has at least three positive solutions (u1,v1), (u2,v2), and (u3,v3) such that (65)u1,v1<a,b<Ψu2,v2<u2,v2⩽c,a<u3,v3⩽cwithΨu3,v3<b.
The proof is completed.Example 13.
To verify the feasibility of our conclusion, here we consider the boundary value problem of nonlinear fractional differential system as follows:(66)D0+1.7φpD0+1.2ut=f1t,ut,vt,D0+1.2ut,D0+1.2vt,D0+1.7φpD0+1.2vt=f2t,ut,vt,D0+1.2ut,D0+1.2vt,with boundary conditions (67)u0=D0+1.2u0=0u1=∑i=12bi∫0ξiutdAt,φpD0+1.2u1=∑i=12aiφpD0+1.2uξi,v0=D0+1.2v0=0v1=∑i=12di∫0ηivtdBt,φpD0+1.2v1=∑i=12ciφpD0+1.2vηi,In this model, we set p=2, and (68)a1=19,a2=49,b1=14,b2=34,c1=25,c2=45,d1=13,d2=12,η1=12,η2=35,ξ1=13,ξ2=23.
LetA(t)=0.5t, B(t)=0.8t for t∈[0,1], (69)fkt,u,v,w,z=t200+u+v+w+z350,0⩽u⩽5,0⩽v⩽5,w∈R,z∈R,43.007u+v-430.010,5<u⩽10,5<v⩽10,w∈R,z∈R,400cos⁡t+2v4000+100,else.It is obvious that fk(t,u,v,w,z):[0,1]×R+2×R2→[0,+∞) (k=1,2) is continuous. Take θ=0.4; by calculation, we obtain (70)q=2,ρ=12.463,N1=0.220,N2=0.277,Q1=0.034,Q2=0.044,M=0.013,N=35.327.
Fork=1,2, set a=5, b=10, and c=40000, which satisfy 0⩽a⩽b⩽θ2/Γ(α-1)c, such that (F1)
fk(t,u,v,w,z)⩽Mc, for (t,u,v,w,z)∈[0,1]×[0,40000]2×[-40000,40000]2;(F2)
fk(t,u,v,w,z)<Ma, for (t,u,v,w,z)∈[0,1]×[0,5]2×[-5,5]2;(F3)
fk(t,u,v,w,z)⩾Nb, for (t,u,v,w,z)∈[0.4,0.6]×[10,40000]2×[-40000,40000]2,
Thus, all the conditions in Theorem12 are achieved. Consequently, BVP (66) has at least three positive solutions (u1,v1), (u2,v2), and (u3,v3) such that (u1,v1)<5, 10<mint∈[0.4,0.6](u2,v2)(t)<(u2,v2)⩽40000, and 5<(u3,v3)⩽40000 with mint∈[0.4,0.6](u3,v3)(t)<10.



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*Source: 3543670-2019-06-17.xml*