# Positive Solutions for a System of Fractional Differential Equations with Two Parameters

Authors: Hongyu Li; Junting Zhang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, the existence of positive solutions in terms of different values of two parameters for a system of conformable-type fractional differential equations with the p-Laplacian operator is obtained via Guo-Krasnosel’skii fixed point theorem.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the following system of fractional differential equations:(1)Dα1ϕp1Dα1xt=λgt,xt,yt,0<t<1,Dα2ϕp2Dα2yt=μft,xt,yt,0<t<1,subject to the following boundary condition:(2)x0=x1=0,Dα1x0=Dα1x1=0,y0=y1=0,Dα2y0=Dα2y1=0,where α1,α2∈1,2 are real numbers; Dα1 and Dα2 are the conformable fractional derivative; ϕpi(s)=spi-2s,pi>1,ϕqi=ϕpi-1, 1/pi+1/qi=1,i=1,2; g,f:[0,1]×0,+∞→0,+∞ are continuous; λ and μ are positive parameters.Fractional differential equations have many applications in various fields such as biological science, chemistry, physics, and engineering. Many authors have made large achievements about the study of fractional differential equations boundary value problems. Most results have adopted the Riemann-Liouville and Caputo-type fractional derivatives; we can see [1–28] and the references therein; for example, in [28], by using Guo-Krasnosel’skii fixed point theorem, the authors obtained the various existence results for positive solutions about a system of Riemann-Liouville type fractional boundary value problems with two parameters and the p-Laplacian operator. As we know, there is another kind of fractional derivative which is conformable fractional derivative. Recently, in [29], the authors Khalil R. et al. first introduced a new simple well-behaved definition of the fractional derivative called conformable fractional derivative. They first presented the definition of conformable fractional derivative of order α∈0,1 and generalized the definition to include order α∈n,n+1,n∈N. In [30], Abdeljawad proceeded on to develop the definitions and set the basic concepts in this new simple interesting fractional calculus. Since then, there are a few authors to study the boundary value problems for conformable-type fractional differential equations; for example, we can see [31–33] and the reference therein. In [33], the authors applied approximation method and fixed point theorems on cone to consider the existence and multiplicity of positive solution about the following fractional differential equation with the p-Laplacian operator: (3)DαϕpDαut=λft,ut,0<t<1,u0=u1=0,Dαu0=Dαu1=0,where α∈1,2 is a real number, Dα is the conformable fractional derivative, ϕp(s)=sp-2s,p>1, and f:[0,1]×0,+∞→0,+∞ is continuous.There are few papers about the system of fractional differential equations concerning conformable fractional derivative. System (1), (2) is a new type of conformable fractional differential equations. Motivated by the recent papers [28, 33, 34], we consider the existence of positive solutions of the system for conformable fractional differential equations (1), (2). By using Guo-Krasnosel’skii fixed point theorem, we establish some sufficient conditions on g,f,λ,μ for the existence of at least one positive solutions of system (1), (2) for appropriately chosen parameters.The organization of this paper is as follows. In Section2, we recall some concepts about the conformable fractional derivative and give some lemmas with respect to the corresponding Green’s function. In Section 3, we give some results about the existence of positive solutions of system (1), (2). In Section 4, we summarize the main results of the third section.


## 2. Preliminaries

For the convenience of the reader, we give the following concepts and lemmas of conformable fractional calculus, and some auxiliary results that will be used to prove our main theorems (see [29–33]).Definition 1.
Letα∈n,n+1 and f be a n-differential function at t>0, then the fractional conformable derivative of order α at t>0 is given by (4)Dαft=Dα-nfnt=limϵ→0⁡fnt+ϵtn+1-α-fntϵ,provided the limit of the right hand side exists. If f is α-differentiable in some (0,a), where a>0, and limt→0+Dαf(t) exists, then define Dαf(0)=limt→0+Dαf(t).Definition 2.
Letα∈n,n+1. The fractional integral of order α>0 at t>0 of a function f:(0,+∞)→(0,+∞) is given by (5)Iαft=In+1tα-n-1ft=1n!∫0tt-snsα-n-1fsds,where In+1 denotes the integration operator of order n+1.Lemma 3.
Letα∈n,n+1 and f,g be α-differentiable at a point t>0. Then(i)
Dα[af+bg]=aDα(f)+bDα(g), ∀a,b∈R1.(ii)
Dαtk=0, where k=0,1,…,n.(iii)
Dα(C)=0, for all constant functions f(t)=C.(iv)
Dα(fg)=fDα(g)+gDα(f).(v)
Dαf/g=gDαf-fDαg/g2.(vi)
If, in addition,f is differentiable, then Dαf(t)=tn+1-αf(n+1)(t).(vii)
If, in addition,f is differentiable at g(t), then Dα(f∘g)(t)=f(n)(g(t))Dαg(t).Lemma 4.
Givenα∈n,n+1 and f a continuous function defined in the domain of Iα, one has that DαIαf(t)=f(t) for t>0.Lemma 5 (mean value theorem).
Leta>0 and f:[a,b]→R1 be a given function that satisfies(i)
f is continuous on [a,b],(ii)
f is α-differentiable for some α∈(0,1).Then, there existsc∈(a,b) such that (6)Dαfc=fb-fa1/αbα-1/αaα.Lemma 6.
Givenα∈n,n+1 and f:0,+∞→R1 an α-differentiable function, one has that Dαf(t)=0 if and only if f(t)=a0+a1t+⋯+antn, where a1,a2,…,an∈R1.Lemma 7.
Givena∈n,n+1 and x∈C(0,+∞) an α-differentiable function that belongs to C(0,1)∩L(0,1), one has that IαDαx(t)=x(t)+c0+c1t+⋯+cntn, for some ck∈R1, k=0,1,…,n.By Lemma 2.7 in [33], we can obtain the following lemmas.Lemma 8 (see [33]).
Letu∈C[0,1] and α1,α2∈(1,2]. Then the conformable fractional differential equation (7)DαiϕpDαixt=ut,0<t<1,x0=x1=0,Dαix0=Dαix1=0has a unique solution (8)xt=∫01Kit,sϕqi∫01Kis,τuτds,where (9)kit,s=1-tsαi-1,0≤s≤t≤1,tsαi-21-s,0≤t≤s≤1,and i=1,2.Lemma 9 (see [33]).
The functionKi(t,s)(i=1,2) defined by (9) has the following properties:(i)
Ki(t,s)>0, for all t,s∈(0,1);(ii)
min1/4≤t≤3/4Ki(t,s)≥1/4max0≤t≤1Ki(t,s)=1/4Ki(s,s), for s∈(0,1).Lemma 10 (see [35]).
LetE be a Banach space, P⊂E be a cone, Ω1 and Ω2 be bounded open subsets of E, θ∈Ω1, and Ω¯1⊂Ω2. Assume that T:P∩(Ω¯2∖Ω1)→P is a completely continuous operator such that (i)
Tu≤u,∀u∈P∩∂Ω1,Tu≥u,∀u∈P∩∂Ω2or(ii)
Tu≥u,∀u∈P∩∂Ω1,Tu≤u,∀u∈P∩∂Ω2.Then the operatorT has at least one fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

LetX=C[0,1] with supremum norm x=max0≤t≤1xt. Let E=X×X with the norm (x,y)E=x+y. Then E is a Banach space. We define the cone (10)P=x,y∈E∣xt≥0,yt≥0,∀t∈0,1,min1/4≤t≤3/4⁡xt+yt≥14x,yE.In the following, we define the operatorsA,B:E→X and T:E→E: (11)Ax,yt=∫01K1t,sϕq1λ∫01K1s,τgτ,xτ,yτdτds,t∈0,1,Bx,yt=∫01K2t,sϕq2μ∫01K2s,τfτ,xτ,yτdτds,t∈0,1,Tx,y=Ax,y,Bx,y,x,y∈E,where Ki(t,s)(i=1,2) is defined by (9).Obviously, the nontrivial fixed points of the operatorT in P are positive solutions of system (1), (2).Lemma 11.
The operatorT:P→P is a completely continuous operator.Proof.
It is obvious thatA(x,y)(t)≥0,B(x,y)(t)≥0 for (x,y)∈P,t∈[0,1]. By Lemma 8, we have (12)Ax,yt=∫01K1t,sϕq1λ∫01K1s,τgτ,xτ,yτdτds≤ϕq1λ∫01s1-sα1-1ϕq1∫01K1s,τgτ,xτ,yτdτds,and (13)Bx,yt=∫01K2t,sϕq2μ∫01K2s,τfτ,xτ,yτdτds≤ϕq2μ∫01s1-sα2-1ϕq2∫01K2s,τfτ,xτ,yτdτds.Then (14)Tx,yE=Ax,y+Bx,y≤ϕq1λ∫01s1-sα1-1ϕq1∫01K1s,τgτ,xτ,yτdτds+ϕq2μ∫01s1-sα2-1ϕq2∫01K2s,τfτ,xτ,yτdτds.
On the other hand, by Lemma9, we have (15)min1/4≤t≤3/4⁡Ax,yt≥14ϕq1λ∫01K1s,sϕq1∫01K1s,τgτ,xτ,yτdτds≥14Ax,y,and (16)min1/4≤t≤3/4⁡Bx,yt≥14ϕq2μ∫01K2s,sϕq2∫01K2s,τfτ,xτ,yτdτds≥14Bx,y.
So we have(17)min1/4≤t≤3/4⁡Ax,yt+Bx,yt≥min1/4≤t≤3/4⁡Ax,yt+min1/4≤t≤3/4⁡Bx,yt≥14x,yE,i.e., T(P)⊂P.
By the paper [33], we know that A and B are completely continuous operator. It is obvious that T is completely continuous. The proof is completed.Denote(18)g0=limx+y→0+⁡maxt∈0,1⁡gt,x,yϕp1x+y.f0=limx+y→0+⁡maxt∈0,1⁡ft,x,yϕp2x+y.g∞=limx+y→+∞⁡min1/4≤t≤3/4⁡gt,x,yϕp1x+y.f∞=limx+y→+∞⁡min1/4≤t≤3/4⁡ft,x,yϕp2x+y.C1=∫01s1-sα1-1ϕq1∫01K1s,τdτds,C2=∫01s1-sα2-1ϕq2∫01K2s,τdτds.C3=∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τdτds,C4=∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τdτds.Theorem 12.
Assume thatg0,f0,g∞,f∞∈(0,+∞), M1<M2 and M3<M4, then for each λ∈(M1,M2) and μ∈(M3,M4), system (1), (2) has a positive solution (x(t),y(t)), t∈[0,1], where (19)M1=ϕp18C31g∞,M2=ϕp112C11g0,M3=ϕp28C41f∞,M4=ϕp212C21f0.Proof.
Letλ∈(M1,M2) and μ∈(M3,M4). Then there exists a number ϵ>0 such that ϵ<min⁡g∞,f∞, and (20)ϕp18C31g∞-ϵ≤λ≤ϕp112C11g0+ϵ,ϕp28C41f∞-ϵ≤μ≤ϕp212C21f0+ϵ.
For the aboveϵ>0, we know that there exists R1>0 such that (21)gt,x,y<g0+ϵϕp1x+y,0≤x+y≤R1,t∈0,1.ft,x,y<f0+ϵϕp2x+y,0≤x+y≤R1,t∈0,1.
LetΩ1=x,y∈E∣x,yE<R1. For any (x,y)∈P∩∂Ω1, we have (22)Ax,yt=∫01K1t,sϕq1λ∫01K1s,τgτ,xτ,yτdτds≤ϕq1λ∫01K1t,sϕq1∫01K1s,τg0+ϵϕp1xτ+yτdτds≤ϕq1λϕq1g0+ϵ∫01s1-sα1-1ϕq1∫01K1s,τϕp1xτ+yτdτds≤ϕq1λϕq1g0+ϵ∫01s1-sα1-1ϕq1∫01K1s,τϕp1x+ydτds=ϕq1λg0+ϵC1x+y≤12x,yE.So (23)Ax,y≤12x,yE,∀x,y∈P∩∂Ω1.
Similarly, we have(24)Bx,yt=∫01K2t,sϕq2μ∫01K2s,τfτ,xτ,yτdτds≤ϕq2μ∫01K2t,sϕq2∫01K2s,τf0+ϵϕp2xτ+yτdτds≤ϕq2μϕq2f0+ϵ∫01s1-sα2-1ϕq2∫01K2s,τϕp2xτ+yτdτds≤ϕq2μϕq2f0+ϵ∫01s1-sα2-1ϕq2∫01K2s,τϕp2x+ydτds=ϕq2λf0+ϵC2x+y≤12x,yE.So (25)Bx,y≤12x,yE,∀x,y∈P∩∂Ω1.
Hence(26)Tx,yE=Ax,y+bx,y≤x,yE,∀x,y∈P∩∂Ω1.
On the other hand, for the aboveϵ>0, there exists R~2>0 such that (27)gt,x,y≥g∞-ϵϕp1x+y,x+y≥R~2>0,t∈14,34,ft,x,y≥f∞-ϵϕp2x+y,x+y≥R~2>0,t∈14,34.
LetR2=max⁡2R1,4R~2. Let Ω2=x,y∈E∣x,yE<R2. For any (x,y)∈P∩∂Ω2, we have min1/4≤t≤3/4xt+yt≥1/4(x+y)=1/4(x,y)E=R2/4≥R~2. So by Lemma 9, we have (28)Ax,y14=∫01K114,sϕq1λ∫01K1s,τgτ,xτ,yτdτds≥14ϕq1λ∫01s1-sα1-1ϕq1∫1/43/4K1s,τgτ,xτ,yτdτds≥14ϕq1λ∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τg∞-ϵϕp1xτ+yτdτds≥14ϕq1λϕq1g∞-ϵ∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τϕp114x+ydτds=116ϕq1λg∞-ϵC3x,yE≥12x,yE.So (29)Ax,y≥12x,yE,∀x,y∈P∩∂Ω2.
Similarly, we have(30)Bx,y34=∫01K234,sϕq2μ∫01K2s,τfτ,xτ,yτdτds≥14ϕq2μ∫01s1-sα2-1ϕq2∫1/43/4K2s,τfτ,xτ,yτdτds≥14ϕq2μ∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τf∞-ϵϕp2xτ+yτdτds≥14ϕq2μϕq2f∞-ϵ∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τϕp214x+ydτds=116ϕq2λf∞-ϵC4x,yE≥12x,yE.So (31)Bx,y≥12x,yE,∀x,y∈P∩∂Ω2.
Hence(32)Tx,yE=Ax,y+Bx,y≥x,yE,∀x,y∈P∩∂Ω2.
By Lemma10 and (26) (32), the operator T has one fixed point (x,y)∈P∩(Ω¯2∖Ω1). That is, (x,y) is a positive solution of system (1), (2).Similar to the proof of Theorem12, we can easily obtain the following results.Theorem 13.
Assume thatg0=0,g∞,f0,f∞∈(0,+∞) and M3<M4, then for each λ∈(M1,+∞) and μ∈(M3,M4), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 14.
Assume thatf0=0,g∞,g0,f∞∈(0,+∞) and M1<M2, then for each λ∈(M1,M2) and μ∈(M3,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 15.
Assume thatg0=0,f0=0,g∞,f∞∈(0,+∞), then for each λ∈(M1,+∞) and μ∈(M3,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 16.
Assume thatg0,f0∈(0,+∞),g∞=+∞ or g0,f0∈(0,+∞),f∞=+∞, then for each λ∈(0,M2) and μ∈(0,M4), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 17.
Assume thatg0∈(0,+∞),f0=0,f∞=+∞ or g0∈(0,+∞),f0=0,g∞=+∞, then for each λ∈(0,M2) and μ∈(0,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 18.
Assume thatg0=f0=0,f∞=+∞ or g0=f0=0,g∞=+∞, then for each λ∈(0,+∞) and μ∈(0,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].
Denote(33)g¯0=limx+y→0+⁡mint∈1/4,3/4⁡gt,x,yϕp1x+y.f¯0=limx+y→0+⁡mint∈1/4,3/4⁡ft,x,yϕp2x+y.g¯∞=limx+y→+∞⁡maxt∈0,1⁡gt,x,yϕp1x+y.f¯∞=limx+y→+∞⁡maxt∈0,1⁡ft,x,yϕp2x+y.Theorem 19.
Assume thatg¯0,f¯0,g¯∞,f¯∞∈(0,+∞), M¯1<M¯2 and M¯3<M¯4, then for each λ∈(M¯1,M¯2) and μ∈(M¯3,M¯4), system (1), (2) has a positive solution (x(t),y(t)), t∈[0,1], where (34)M¯1=ϕp18C31g¯0,M¯2=ϕp112C11g¯∞,M¯3=ϕp28C41f¯0,M¯4=ϕp212C21f¯∞.Proof.
Letλ∈(M¯1,M¯2) and μ∈(M¯3,M¯4). Then there exists a number ϵ>0 such that ϵ<min⁡{g¯0,f¯0}, and (35)ϕp18C31g¯0-ϵ≤λ≤ϕp112C11g¯∞+ϵ,ϕp28C41f¯0-ϵ≤μ≤ϕp212C21f¯∞+ϵ.
For the aboveϵ>0, we know that there exists R3>0 such that (36)gt,x,y>g¯0-ϵϕp1x+y,x≥0,y≥0,x+y≤R3,t∈14,34,ft,x,y>f¯0-ϵϕp2x+y,x≥0,y≥0,x+y≤R3,t∈14,34.
LetΩ3=x,y∈E∣x,yE<R3. For any (x,y)∈P∩∂Ω3, we have min1/4≤t≤3/4xt+yt≥1/4(x+y)=1/4(x,y)E. So by Lemma 9, we obtain (37)Ax,y14=∫01K114,sϕq1λ∫01K1s,τgτ,xτ,yτdτds≥14ϕq1λ∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τgτ,xτ,yτdτds≥14ϕq1λ∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τg¯0-ϵϕp1xτ+yτdτds≥14ϕq1λϕq1g¯0-ϵ∫1/43/4s1-sα1-1ϕq1∫1/43/4K1s,τϕp114x+ydτds=116ϕq1λg¯0-ϵC3x,yE≥12x,yE.So (38)Ax,y≥12x,yE,∀x,y∈P∩∂Ω3.
Similarly, we get(39)Bx,y34=∫01K234,sϕq2μ∫01K2s,τfτ,xτ,yτdτds≥14ϕq2μ∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τfτ,xτ,yτdτds≥14ϕq2μ∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τf¯0-ϵϕp2xτ+yτdτds≥14ϕq2μϕq2f¯0-ϵ∫1/43/4s1-sα2-1ϕq2∫1/43/4K2s,τϕp214x+ydτds=116ϕq2μf¯0-ϵC3x,yE≥12x,yE.So (40)Bx,y≥12x,yE,∀x,y∈P∩∂Ω3.
Hence(41)Tx,yE=Ax,y+Bx,y≥x,yE,∀x,y∈P∩∂Ω3.
We define the functionsg~,f~:[0,1]×0,+∞→0,+∞, g~(t,u)=max0≤x+y≤ug(t,x,y), and f~(t,u)=max0≤x+y≤uf(t,x,y). So it is obvious that g~(t,u) and f~(t,u) are nondecreasing on u for every t∈[0,1]; and g(t,x,y)≤g~(t,u),f(t,x,y)≤f~(t,u),x≥0,y≥0,x+y≤u,t∈[0,1]; and they satisfy the conditions (42)limsupu→+∞⁡maxt∈0,1⁡g~t,uϕp1u≤g¯∞,limsupu→+∞⁡maxt∈0,1⁡f~t,uϕp2u≤f¯∞.
For the aboveϵ>0, there exists R~4>0 such that (43)g~t,uϕp1u≤limsupu→+∞⁡maxt∈0,1⁡g~t,uϕp1u+ϵ≤g¯∞+ϵ,t∈0,1,u≥R~4,f~t,uϕp2u≤limsupu→+∞⁡maxt∈0,1⁡f~t,uϕp2u+ϵ≤f¯∞+ϵ,t∈0,1,u≥R~4.
Sog~(t,u)≤(g¯∞+ϵ)ϕp1(u), f~(t,u)≤(f¯∞+ϵ)ϕp2(u), and t∈[0,1],u≥R~4.
By the definition ofg~,f~, we get g(t,x,y)≤g~(t,(x,y)E), f(t,x,y)≤f~(t,(x,y)E). Let R4=max⁡2R3,R~4. Let Ω4=x,y∈E∣x,yE<R4. For any (x,y)∈P∩∂Ω4, we have (44)Ax,yt=∫01K1t,sϕq1λ∫01K1s,τgτ,xτ,yτdτds≤ϕq1λ∫01s1-sα1-1ϕq1∫01K1s,τg~τ,x,yEdτds≤ϕq1λϕq1g¯∞+ϵ∫01s1-sα1-1ϕq1∫01K1s,τϕp1x,yEdτds=ϕq1λg¯∞+ϵC1x,yE≤12x,yE.So (45)Ax,y≤12x,yE,∀x,y∈P∩∂Ω4.
Similarly, we have(46)Bx,yt=∫01K2t,sϕq2μ∫01K2s,τfτ,xτ,yτdτds≤ϕq2μ∫01s1-sα2-1ϕq2∫01K2s,τf~τ,x,yEdτds≤ϕq2μϕq2f¯∞+ϵ∫01s1-sα2-1ϕq2∫01K2s,τϕp2x,yEdτds=ϕq2μf¯∞+ϵC2x,yE≤12x,yE.So (47)Bx,y≤12x,yE,∀x,y∈P∩∂Ω4.
Hence(48)Tx,yE=Ax,y+Bx,y≤x,yE,∀x,y∈P∩∂Ω4.
By Lemma10 and (41) (48), the operator T has one fixed point (x,y)∈P∩(Ω¯4∖Ω3). So (x,y) is a positive solution of system (1), (2).Similar to the proof of Theorem19, we can easily obtain the following results.Theorem 20.
Assume thatg¯0,g¯∞,f¯0∈(0,+∞),f¯∞=0 and M¯1<M¯2, then for each λ∈(M¯1,M¯2) and μ∈(M¯3,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 21.
Assume thatg¯0,f¯∞,f¯0∈(0,+∞),g¯∞=0 and M¯3<M¯4, then for each λ∈(M¯1,+∞) and μ∈(M¯3,M¯4), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 22.
Assume thatg¯0,f¯0∈(0,+∞),g¯∞=0,f¯∞=0, then for each λ∈(M¯1,+∞) and μ∈(M¯3,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 23.
Assume thatg¯∞,f¯∞∈(0,+∞),g¯0=+∞ or f¯0=+∞,g¯∞,f¯∞∈(0,+∞), then for each λ∈(0,M¯2) and μ∈(0,M¯4), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 24.
Assume thatg¯0=+∞,g¯∞∈(0,+∞),f¯∞=0 or f¯0=+∞,g¯∞∈(0,+∞),f¯∞=0, then for each λ∈(0,M¯2) and μ∈(0,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 25.
Assume thatf¯∞∈(0,+∞),g¯∞=0,g¯0=+∞ or f¯∞∈(0,+∞),f¯0=+∞,g¯∞=0, then for each λ∈(0,+∞) and μ∈(0,M¯4), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].Theorem 26.
Assume thatg¯0=+∞,g¯∞=0,f¯∞=0 or g¯∞=0,f¯∞=0,f¯0=+∞, then for each λ∈(0,+∞) and μ∈(0,+∞), system (1), (2) has a positive solution (x(t),y(t)),t∈[0,1].


## 4. Conclusion

In this paper, we investigate the existence of positive solutions for a system of conformable-type fractional differential equations with two parameters and the p-Laplacian operator. By employing Guo-Krasnosel’skii fixed point theorem, under different combinations of sublinearity and superlinearity of the nonlinearitiesf,g, various sufficient conditions for the existence of at least one positive solutions of system (1), (2) are derived in terms of appropriately chosen parameters λ,μ.



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