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Research article

A coupled system of p-Laplacian implicit fractional differential equations depending on boundary conditions of integral type

  • Received: 11 March 2023 Revised: 18 April 2023 Accepted: 22 April 2023 Published: 09 May 2023
  • MSC : 26A33, 34A08, 34B27

  • The objective of this article is to investigate a coupled implicit Caputo fractional p-Laplacian system, depending on boundary conditions of integral type, by the substitution method. The Avery-Peterson fixed point theorem is utilized for finding at least three solutions of the proposed coupled system. Furthermore, different types of Ulam stability, i.e., Hyers-Ulam stability, generalized Hyers-Ulam stability, Hyers-Ulam-Rassias stability and generalized Hyers-Ulam-Rassias stability, are achieved. Finally, an example is provided to authenticate the theoretical result.

    Citation: Dongming Nie, Usman Riaz, Sumbel Begum, Akbar Zada. A coupled system of p-Laplacian implicit fractional differential equations depending on boundary conditions of integral type[J]. AIMS Mathematics, 2023, 8(7): 16417-16445. doi: 10.3934/math.2023839

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  • The objective of this article is to investigate a coupled implicit Caputo fractional p-Laplacian system, depending on boundary conditions of integral type, by the substitution method. The Avery-Peterson fixed point theorem is utilized for finding at least three solutions of the proposed coupled system. Furthermore, different types of Ulam stability, i.e., Hyers-Ulam stability, generalized Hyers-Ulam stability, Hyers-Ulam-Rassias stability and generalized Hyers-Ulam-Rassias stability, are achieved. Finally, an example is provided to authenticate the theoretical result.



    Derivatives of integer orders are the particular form of fractional order derivatives. When the notation dndtn was presented by Leibniz at the end of 17th century for nth order derivative, L'Hospital asked from him can we take n=12? Leibniz replied "this is an apparent Paradox, from which one day useful consequences will be drawn", and this was the origin of fractional derivatives Riemann-Liouville derivative was originated after the great contributions of mathematicians Fourier and Laplace, and due to this notion of fractional derivative the fractional calculus was developed. After that, researchers showed interest in fractional calculus [4,5,6,7,8,9,13,17,18,19,20,28]. Fractional derivatives are universal operators, which cover different physical phenomena and areas of mathematical modeling such as: control theory [26], dynamical process [24], electro-chemistry [16], image and signal processing [22], mathematical biology [21], etc.

    The existence of solutions is the basic subject for the investigation of the fractional differential equations. Numerous mathematicians worked on the existence of solutions for different Boundary Value Problems, Using different fixed point theorems (see [23,29,32]). In [2] Ahmad et al. studied the existence of coupled fractional differential equations involving a p-Laplacian operator by applying fixed point theorems. In [11,12], the authors got theoretical results related to fractional differential equations with p-Laplacian operator (p-LO).

    It can be easily observed that obtaining the exact solution of non linear models is a tough and challenging task, to overcome the difficulty, a lot of approximation methods were established. Hyers-Ulam Stability (HUS) can predict the gap between exactness and approximations of the solutions. This notion was initiated (in 1940) by Ulam [25] and further extended by Hyers to abstract spaces, after one year. Recent results shows that different models with different boundary conditions are investigated for HUS [3,15,30,31].

    Mohammed et al. [14], explored the existence and uniqueness of solution of a model which involves the Caputo-Katugampola fractional derivative:

    {cDβ(ϕp[cDα,ρ1ϖ(ϱ)])=A(ϕp[cDα,ρ1ϖ(ϱ)])+f(ϱ,ϖ(ϱ),cDα,ρ1ϖ(ϱ))=0, ϱ[ϱ0,T],ϖ(ϱ0)+B1ϖ(T)=C1Tϱ0g1(s,ϖ(s),cDγ,ρ1ϖ(s))ds,ϖ(ϱ0)+B2ϖ(T)=C2Tϱ0g1(s,ϖ(s),cDγ,ρ1ϖ(s))ds,ϕp(cDα,ρ1ϖ(ϱ0))=ϖ0,   (ϕp(cDα,ρ1ϖ(ϱ0)))=ϖ1,

    where cDβ is a Caputo fractional derivative of order β(1,2), cDγ,ρ1, γ(0,1) and cDα,ρ, α(1,2), ρ>0 are Caputo-Katugampola fractional derivative, ϕp, p>1 is a p-LO.

    Hira et al. [27], investigated the existence, uniqueness and HU stability of solutions to nonlinear coupled FDEs:

    {cDβ10+(Lp(cDα10+ϖ(ϱ)))=A1(ϱ)ϖ(ϱ)+Φ(ϱ,ϖ(ϱ),cDβ10+(Lp(cDα10+ω(ϱ)))),cDβ20+(Lp(cDα20+y(ϱ)))=A2(ϱ)ω(ϱ)+Ψ(ϱ,cDβ20(Lp(cDα20+ϖ(ϱ))),ω(ϱ)), ϱT,cDα10+ϖ(0))=ϖ(0)=ϖ(0)=0,ϖ(T)=C1T0g1(s,ϖ(s),cDβ10+(Lp(cDα10+ω(s)))),cDα20+ω(0))=ω(0)=ω(0)=0,ω(T)=C2T0g2(s,cDβ10+(Lp(cDα10+ϖ(s))),ω(s)), (1.1)

    where 2<αi3, 0<β11, ηi, γi>0, ψiL[0,T], and cDαi0+ and cDβi0+ are the Caputo derivatives of order αi and βi, i=1,2, respectively. Lp(s)=|s|p2s is a p-LO, where 1p+1q=1, and Lq denotes inverse of p-Laplacian. Ai:TR are closed bounded linear operators for any ρT=[0,T], and Φ, Ψ, gk:T×R×RR, (k=1,2) are continuous functions i=1,2. CkRn×n, (k=1,2).

    In [12], Lu et al. investigated the nonlinear fractional BVP with p-Laplacian operator

    {DB0+[ϑpDZ0+X(α)]=ϝ(α,X(α)),0<α<1,X(0)=X(0)=X(1)=0,DZ0+X(0)=DZ0+X(1))=0,

    where Z(2,3], and B(1,2]. DZ, DB represents the standard Riemann-Liouville fractional derivatives.

    In [2], Ahmad et al. investigated the system:

    {cDZ1[ϑpDBX(α)]+K1(α)ϝ1(α,X(α),cDZ2[ϑp(DBY(α)])=0,α]0,1[,cDZ2[ϑpDBY(α)]+K2(α)ϝ2(α,cDZ1[ϑpDBX(α)],Y(α))=0,α]0,1[,([ϑpDBX(α)])i=([ϑp(DBY(α)])i=0,i=¯0,m1,IkBX(0)=IkBY(0)=0,k=¯2,m,Dδ(X(1))=Dδ(Y(1))=0,

    where cDZ and DB respectively denotes the Caputo and Riemann-Liouville FD of order Z and B, m1<Z1,Z2,Bm, m{4,5,}, and 1<δ2, K1(), K2() are linear and bounded operators on R.

    Zhang et al. [33], studied the existence of

    {cDBϑp(cDZX(α))+ϝ(α,cDBX(α))=0,α(0,1),(ϑp(cDZX(0)))i=ϑp(cDZX(1))=0,i=1,2,,m1=:¯1,m1,X(0)+X(0)=10S1(ϖ)X(ϖ)dϖ+a,X(1)+X(1)=10S2(ϖ)X(ϖ)dϖ+b,Xj(0)=0,j=¯2,n1,

    where Z and B are the orders of Caputo fractional derivatives cDZ and cDB respectively.

    Following [33], in this article, we present existence and stability analysis of the model of the form

    {cDB1ϑp(cDZ1X(α))ϝ1(α,cDγ1X(α),cDγ2Y(α))=0,α(0,1),cDB2ϑp(cDZ2)Y(α)ϝ2(α,cDγ1X(α),cDγ2Y(α))=0,α(0,1),(ϑp(cDZ1X(0)))i=ϑp(cDZ1X(1))=0,i=¯1,m1,(ϑp(cDZ2Y(0)))i=ϑp(cDZ2Y(1))=0,i=¯1,m1,X(0)+X(0)=10S1(ϖ)X(ϖ)dϖ+10H1(ϖ)dϖ,X(1)+X(1)=10S2(ϖ)X(ϖ)dϖ+10H2(ϖ)dϖ,Y(0)+Y(0)=10S3(ϖ)Y(ϖ)dϖ+10H3(ϖ)dϖ,Y(1)+Y(1)=10S4(ϖ)Y(ϖ)dϖ+10H4(ϖ)dϖ,Xj(0)=0,Yj(0)=0,j=¯2,n1, (1.2)

    where 1<m1<B1;B2<m; 1<n1<Z1;Z2<n; Z1B1;Z2B2>1, Z1;Z2;B1, and B2 be the orders of Caputo fractional derivatives cDZ1, cDZ2, cDB1 and cDB2 respectively, and 0<γ1,γ21. The p-Laplacian operator is represented by ϑp and is defined as ϑp(ϖ)=|ϖ|p2ϖ,p>1,ϑ1p=ϑq,1p+1q = 1. S1,S2,S3,S4C([0,1],R+), ϝ1, and ϝ2 are appropriate functions. H1,H2,H3,H4C([0,1],R+) are perturbation functions.

    (C1) 0<10H1(ϖ)dϖ<10H2(ϖ)dϖ<210H1(ϖ)dϖ<+, 0S1(α)S2(α)2S1(α), 010S1(ϖ)dϖ, 10S2(ϖ)dϖ<1, 0S3(α)S4(α)2S3(α), 010S3(ϖ)dϖ, 10S4(ϖ)dϖ<1.

    The remaining manuscript is as follows: Section 2 contains basic definitions, auxiliary lemmas and related theorems. In Section 3, we present the existence theory for the problem (1.2) and gives some related properties of Green function. In Section 4, we obtain at least three positive solutions of coupled system (1.2) by using the Avery-Peterson fixed point theorem. Section 5 contains HU type stability results, and Section 6 provide an example to authenticate the theoretical result.

    Here, we are presenting an important literature concerning the Caputo fractional derivatives and integral, which gives us help throughout this article, for the details, reader should study [1,10].

    Definition 2.1. Let XL1([0,T],R+) be a function. Then the Z order fractional integral is defined by

    IZX(ϖ)=1Γ(Z)α0(αϖ)Z1X(ϖ)dϖ  α>0,  Z>0,

    on the condition that the integral on right side exists, where Γ is the Euler Gamma function, defined as

    Γ(Z)=0αZ1eαdα.

    Definition 2.2. Let X:[0,T]R be a function. Then the Z> order fractional derivative is defined as

    cDZX(α)=1Γ(nZ)α0(αϖ)nZ1Xn(ϖ)dϖ,

    on the condition that the integral on the right side exists, and [Z] denote the integer part of a real number Z, where n=1+[Z].

    Definition 2.3. Let X:[0,T]R be a function. Then the Z order sequential fractional derivative is defined as:

    DZX(α)=DZ1DZ2DZ3DZmX(α),

    where Z=(Z1,Z2,Z3,,Zm) is any multi-index, and the operator DZ can either be Riemann-Liouville or Caputo or any other kind of integro-differential operator.

    Lemma 2.1. For any Z>0, the Caputo FDE cDZX(α)=0 has a solution of the form

    X(α)=e0+e1α+e2α2++en1αn1,

    where eiR,i=¯0,n1, and n=1+[Z].

    Lemma 2.2. For any Z>0, we have

    IZ(cDZX(α))=X(α)+e0+e1α+e2α2++en1αn1,

    where eiR,i=¯0,n1, and n=1+[Z].

    Lemma 2.3. [33] Let ϝ1C([0,1],R), and 1<m1<B1<m. Then

    {DB1ω(α)=ϝ1(α),0<α<1,ω(1)=ωi(0)=0,i=¯1,m1, (2.1)

    has a solution (unique)

    ω(α)=10KB1(α,ϖ)ϝ1(ϖ)dϖ,

    where

    KB1(α,ϖ)=1Γ(B1){(1ϖ)B11+(αϖ)B11,0ϖα1,(1ϖ)B11,0αϖ1. (2.2)

    Let

    MS1=10S1(ϖ)dϖ,MS1=10ϖS1(ϖ)dϖ,MS2=10S2(ϖ)dϖ,MS2=10ϖS2(ϖ)dϖ,MS3=10S3(ϖ)dϖ,MS3=10ϖS3(ϖ)dϖ,MS4=10S4(ϖ)dϖ,MS4=10ϖS4(ϖ)dϖ,η1(α)=(2α)δS1+(α1)δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1), (2.3)
    η2(α)=(2α)δS1δS2(MS1MS1)+(α1)δS21δS1δS2(2MS2MS2)(MS1MS1), (2.4)
    ψ1(α)=(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ+10[η1(α)S1(ϖ)+η2(α)S2(ϖ)]×[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]dϖ, (2.5)
    ψ2(α)=(2α)10H3(ϖ)dϖ+(α1)10H4(ϖ)dϖ+10[η3(α)S3(ϖ)+η4(α)S4(ϖ)]×[(2ϖ)10H3(θ)dθ+(ϖ1)10H4(θ)dθ]dϖ. (2.6)

    From (C1), we know for α(0,1),

    2S1(α)>S2(α)>αS2(α)>αS1(α),2S3(α)>S4(α)>αS4(α)>αS3(α),

    and

    1>10S2(ϖ)dϖ>10ϖS2(ϖ)dϖ>10ϖS1(ϖ)dϖ>0,1>10S2(ϖ)dϖ>10S1(ϖ)dϖ>10ϖS1(ϖ)dϖ>0,1>10S4(ϖ)dϖ>10ϖS4(ϖ)dϖ>10ϖS3(ϖ)dϖ>0,1>10S4(ϖ)dϖ>10S3(ϖ)dϖ>10ϖS3(ϖ)dϖ>0,102S1(ϖ)dϖ>10S2(ϖ)dϖ,102ϖS1(ϖ)dϖ>10ϖS2(ϖ)dϖ,10S3(ϖ)dϖ>10S4(ϖ)dϖ,102ϖS3(ϖ)dϖ>10ϖS4(ϖ)dϖ.

    Implies

    1>MS2>MS2>MS1>0,1>MS2>MS1>MS1>0,2MS1>MS2,2MS1>MS2, (2.7)
    1>MS4>MS4>MS3>0,1>MS4>MS3>MS3>0,2MS3>MS4,2MS3>MS4. (2.8)

    For our results we need an assumption and the lemma.

    (C2) δ1S1=12MS1+MS10, δ1S2=1+MS2MS20 and δS1δS2(2MS2MS2)(MS1MS1)1.

    Lemma 3.1. Let (C2) hold, n1<Z1n and h1:JR are is an appropriate function. The BVP:

    {cDZ1X(α)h1(α)=0,α(0,1),X(0)+X(0)=10S1(ϖ)X(ϖ)dϖ+10H1(ϖ)dϖ,X(1)+X(1)=10S2(ϖ)X(ϖ)dϖ+10H2(ϖ)dϖ,Xj(0)=0,j=¯2,n1, (3.1)

    has solution(unique) of the form

    X(α)=10GZ1(α,ϖ)h1(ϖ)dϖ+ψ1(α), (3.2)

    where

    GZ1(α,ϖ)=G1(α,ϖ)+G2(α,ϖ), (3.3)
    G1(α,ϖ)=1Γ(Z1){(Z1ϖ)(1α)(1ϖ)Z12+(αϖ)Z11,0ϖα1,(Z1ϖ)(1α)(1ϖ)Z12,0αϖ1, (3.4)

    and

    G2(α,ϖ)=η1(α)(10S1(θ)G1(ϖ,θ)dθ)+η2(α)(10S2(θ)G1(ϖ,θ)dθ). (3.5)

    Proof. Consider

    cDZ1X(α)=h1(α). (3.6)

    Using Lemma 2.2, we get

    X(α)=1Γ(Z1)α0(αϖ)Z11h1(ϖ)dϖ+e0+e1α++en1αn1. (3.7)

    Using boundary conditions of (3.1), we get

    e0=210H1(ϖ)dϖ10H2(ϖ)dϖ+210S1(ϖ)X(ϖ)dϖ10S2(ϖ)X(ϖ)dϖ+1Γ(Z1)10(1ϖ)Z11h1(ϖ)dϖ+1Γ(Z11)10(1ϖ)Z12h1(ϖ)dϖ,e1=10H1(ϖ)dϖ+10H2(ϖ)dϖ10S1(ϖ)X(ϖ)dϖ+10S2(ϖ)X(ϖ)dϖ1Γ(Z1)10(1ϖ)Z11h1(ϖ)dϖ1Γ(Z11)10(1ϖ)Z12h1(ϖ)dϖ,

    and e2=e3==en1=0. Putting e's in (3.7) and

    X(α)=(2α)10S1(ϖ)X(ϖ)dϖ+(α1)10S2(ϖ)X(ϖ)dϖ+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ+1αΓ(Z1)10(1ϖ)Z11h1(ϖ)dϖ+1αΓ(Z11)10(1ϖ)Z12h1(ϖ)dϖ+1Γ(Z1)t0(αϖ)Z11h1(ϖ)dϖ=10G1(α,ϖ)h1(ϖ)dϖ+(2α)10S1(ϖ)X(ϖ)dϖ+(α1)10S2(ϖ)X(ϖ)dϖ+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ=10G1(α,ϖ)h1(ϖ)dϖ+(2α)AS1+(α1)AS2+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ, (3.8)

    where G1(α,ϖ) is given in (3.4), AS1=10S1(ϖ)X(ϖ)dϖ and AS2=10S2(ϖ)X(ϖ)dϖ.

    In view of (3.8), we get

    S1(α)X(α)=S1(α)10G1(α,ϖ)h1(ϖ)dϖ+(2α)AS1S1(α)+(α1)AS2S1(α)+(2α)S1(α)10H1(ϖ)dϖ+(α1)S1(α)10H2(ϖ)dϖ. (3.9)

    Integrating (3.9) from 0 to 1, we obtain

    AS1=10S1(ϖ)X(ϖ)dϖ=10S1(ϖ)(10G1(ϖ,θ)h(θ)dθ)dϖ+(2AS1AS2)10S1(ϖ)dϖ+(AS2AS1)10ϖS1(ϖ)dϖ+210S1(ϖ)10H1(θ)dθdϖ10ϖS1(ϖ)10H1(θ)dθdϖ+10ϖS1(ϖ)10H2(θ)dθdϖ10S1(ϖ)10H2(θ)dθdϖ=10S1(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+(2AS1AS2)MS1+(AS2AS1)MS1+210S1(ϖ)10H1(θ)dθdϖ10ϖS1(ϖ)10H1(θ)dθdϖ+10ϖS1(ϖ)10H2(θ)dθdϖ10S1(ϖ)10H2(θ)dθdϖ.

    Implies that

    AS1=δS110S1(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS1AS2(MS1MS1)+2δS110S1(ϖ)10H1(θ)dθdϖδS110ϖS1(ϖ)10H1(θ)dθdϖ+δS110ϖS1(ϖ)10H2(θ)dθdϖδS110S1(ϖ)10H2(θ)dθdϖ. (3.10)

    Similarly, we obtain

    AS2=10S2(ϖ)X(ϖ)dϖ=10S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+(2AS1AS2)10S2(ϖ)dϖ+(AS2AS1)10ϖS2(ϖ)dϖ+210S2(ϖ)10H1(θ)dθdϖ10ϖS2(ϖ)10H1(θ)dθdϖ+10ϖS2(ϖ)10H2(θ)dθdϖ10S2(ϖ)10H2(θ)dθdϖ=10S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+(2AS1AS2)MS2+(AS2AS1)MS2+210S2(ϖ)10H1(θ)dθdϖ10ϖS2(ϖ)10H1(θ)dθdϖ+10ϖS2(ϖ)10H2(θ)dθdϖ10S2(ϖ)10H2(θ)dθdϖ.

    Implies that

    AS2=δS210S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS2AS1(2MS2MS2)+2δS210S2(ϖ)10H1(θ)dθdϖδS210ϖS2(ϖ)10H1(θ)dθdϖ+δS210ϖS2(ϖ)10H2(θ)dθdϖδS210S2(ϖ)10H2(θ)dθdϖ. (3.11)

    From (3.10) and (3.12), we get

    AS1=δS11δS1δS2(2MS2MS2)(MS1MS1)10S1(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS1δS2(MS1MS1)1δS1δS2(2MS2MS2)(MS1MS1)10S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS11δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S1(ϖ)dϖ+δS1δS2(MS1MS1)1δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S2(ϖ)dϖ (3.12)

    and

    AS2=δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1)10S1(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS21δS1δS2(2MS2MS2)(MS1MS1)10S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S1(ϖ)dϖ+δS21δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S2(ϖ)dϖ. (3.13)

    Putting (3.12) and (3.13) in (3.8), we get

    X(α)=10G1(α,ϖ)h1(ϖ)dϖ+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ+(2α)δS1+(α1)δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1)10S1(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+(2α)δS1δS2(MS1MS1)+(α1)δS21δS1δS2(2MS2MS2)(MS1MS1)10S2(ϖ)(10G1(ϖ,θ)h1(θ)dθ)dϖ+(2α)δS1+(α1)δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S1(ϖ)dϖ+(2α)δS1δS2(MS1MS1)+(α1)δS21δS1δS2(2MS2MS2)(MS1MS1)10[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S2(ϖ)dϖ. (3.14)

    From (2.3) and (2.4), we can write (3.14) as

    X(α)=10G1(α,ϖ)h1(ϖ)dϖ+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ+10η1(α)(10S1(θ)G1(θ,ϖ)dθ)h1(ϖ)dϖ+10η2(α)(10S2(θ)G1(θ,ϖ)dθ)h1(ϖ)dϖ+10η1(α)[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S1(ϖ)dϖ+10η2(α)[(2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ]S2(ϖ)dϖ=10G1(α,ϖ)h(ϖ)dϖ+(2α)10H1(ϖ)dϖ+(α1)10H2(ϖ)dϖ+10[η1(α)(10S1(θ)G1(θ,ϖ)dθ)h1(ϖ)+η2(α)(10S2(θ)G1(θ,ϖ)dθ)h1(ϖ)+(η1(α)S1(ϖ)+η2(α)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)]dϖ=10G1(α,ϖ)h1(ϖ)dϖ+10G2(α,ϖ)h1(ϖ)dϖ+ψ1(α)=10GZ1(α,ϖ)h1(ϖ)dϖ+ψ1(α).

    Lemma 3.2. The coupled BVP (1.2) is equivalent to the following system of integral equations

    {X(α)=10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α),Y(α)=10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ2(α), (3.15)

    where G(α,ϖ) and K(α,ϖ) are given by (3.3) and (2.2).

    Proof. Using Lemmas 2.3 and 3.1, set ω(α)=ϑp(DZ1X(α)) and h1(α)=ϝ1(α,Dγ1X(α),Dγ2Y(α)), and we have

    X(α)=10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α).

    Similarly, we can set ω(α)=ϑp(DZ2Y(α)) and h2(α)=ϝ2(α,Dγ1X(α),Dγ2Y(α)), we obtain

    Y(α)=10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ2(α).

    On the other hand, if (X,Y) satisfy (3.15), we can easily prove that (X,Y) satisfy the pair of boundary value problem (1.2). Hence, the proof is completed.

    Lemma 3.3. We use GZ(α,ϖ)=(GZ1(α,ϖ),GZ2(α,ϖ)) and KB(α,ϖ)=(KB1(α,ϖ),KB2(α,ϖ)) as the Green's functions of the proposed system (1.2) having the following properties:

    (I) KB(α,ϖ)0 is continuous for all α,ϖ[0,1];

    (II) KB(α,ϖ)KB(s,ϖ) for all α,ϖ[0,1];

    (III) 10KB1(α,ϖ)dϖ=1αB1Γ(B1+1)1Γ(B1+1) for all α,ϖ[0,1];

    10KB2(α,ϖ)dϖ=1αB2Γ(B2+1)1Γ(B2+1) for all α,ϖ[0,1];

    (IV) GZ(α,ϖ)0 is continuous for all α,ϖ[0,1];

    (V) ψ(α)=(ψ1(α),ψ2(α))0 for all α[0,1];

    Proof. The proofs of (I) and (II) can be seen in [33].

    (III) For α[0,1], we have

    10KB1(α,ϖ)dϖ=y0((1ϖ)B11+(αϖ)B11)dϖ+1y(1ϖ)B11dϖ=1αB1Γ(B1+1)1Γ(B1+1).

    Also,

    10KB2(α,ϖ)dϖ=y0((1ϖ)B21+(αϖ)B21)dϖ+1y(1ϖ)B21dϖ=1αB2Γ(B2+1)1Γ(B2+1).

    (IV) Firstly, from (3.4), one get G1(α,ϖ)0, α,ϖ[0,1], and from (C1), δ>0 and G1(α,ϖ)0, we have

    G2(α,ϖ)t=δS1+δS1δS2(2MS2MS2)1δS1δS2(2MS2MS2)(MS1MS1)10S1(θ)G1(θ,ϖ)dθ+δS1δS2(MS1MS1)+δS21δS1δS2(2MS2MS2)(MS1MS1)10S2(θ)G1(θ,ϖ)dθ=δ10δS1δS2[(2MS2MS2δ1S2)S1(θ)(MS1MS1+δ1S1)S2(θ)]G1(θ,ϖ)dθ>0.

    Thus G2(α,ϖ) is a increasing on α[0,1].

    Utilizing (3.5), we get

    G2(α,ϖ)G2(0,ϖ)=η1(0)(10S1(θ)G1(θ,ϖ)dθ)+η2(0)(10S2(θ)G1(θ,ϖ)dθ)10(η1(0)S1(θ)+η2(0)12S1(θ))G1(θ,ϖ)dθ=10(η1(0)+η2(0)2)S1(θ)G1(θ,ϖ)dθ0.

    So that GZ1(α,ϖ)0. Similarly GZ2(α,ϖ)0. Hence, GZ(α,ϖ)0.

    (V) From (C1) and (2.5), we know

    ψ1(α)=10H1(ϖ)dϖ+10H2(ϖ)dϖ+η1(α)10S1(ϖ)((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ+η2(α)10S2(ϖ)((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ>0,α[0,1].

    Thus ψ1(α), is increasing on α[0,1].

    From (2.5), we have

    ψ1(α)ψ1(0)=210H1(ϖ)dϖ10H2(ϖ)dϖ+10(η1(0)S1(ϖ)+η2(0)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ>0,α[0,1].

    Similarly ψ1(α)0, and hence ψ(α)0.

    Lemma 3.4. For κ(0,12), let

    maxα[0,1]GZ1(α,ϖ)G1(0,ϖ)+G2(1,ϖ),maxα[0,1]GZ2(α,ϖ)G3(0,ϖ)+G4(1,ϖ),minα[0,κ]GZ1(α,ϖ)ρZ1(G1(0,ϖ)+G2(1,ϖ)),minα[0,κ]GZ2(α,ϖ)ρZ2(G3(0,ϖ)+G4(1,ϖ)),

    where ρZ1=2MS1MS21+MS2+MS1MS1 and ρZ2=2MS3MS41+MS4+MS3MS3.

    Proof. First Step. We need to get

    minα[0,κ]G1(α,ϖ)(1κ)G1(0,ϖ)>12maxα[0,1]G1(α,ϖ). (3.16)

    For ϖ<α, where ϖ[0,1) and α[0,κ], we have

    Γ(Z1)G1(α,ϖ)α=(Z1ϖ)(1ϖ)Z12+(Z11)(αϖ)Z12<0,

    which implies that G1(α,ϖ) is decreasing function monotonically with respect to α[ϖ,κ], so that

    G1(α,ϖ)G1(s,ϖ)=(Z1ϖ)(1ϖ)Z11<(Z1ϖ)(1ϖ)Z12=G1(0,ϖ)

    and

    G1(α,ϖ)G1(0,ϖ)=(Z1ϖ)(1α)(1ϖ)Z12+(αϖ)Z11(Z1ϖ)(1ϖ)Z12=1α+(αϖ)Z11(Z1ϖ)(1ϖ)Z121α1κ>12.

    Now if ϖα and α[0,κ],

    Γ(Z1)G1(α,ϖ)α=(Z1ϖ)(1ϖ)Z12<0,

    which implies that G1(α,ϖ) is a monotone decreasing function with respect to α[0,ϖ] and α[0,κ], so that

    G1(α,ϖ)G1(0,ϖ)=(Z1ϖ)(1ϖ)Z12

    and

    G1(α,ϖ)G1(0,ϖ)=(Z1ϖ)(1α)(1ϖ)Z12(Z1ϖ)(1ϖ)Z12=1α1κ>12.

    Thus, (3.16) is satisfied.

    Similarly, we get

    minα[0,κ]G3(α,ϖ)(1κ)G3(0,ϖ)>12maxα[0,1]G3(α,ϖ). (3.17)

    Step 2. We prove

    minα[0,κ]G2(α,ϖ)ρZ1G2(1,ϖ)=ρZ1maxα[0,1]G2(α,ϖ). (3.18)

    From Lemma 3.3, we know that G2(α,ϖ) is increasing for α[0,1], in such a way that

    minα[0,κ]G2(α,ϖ)=G2(0,ϖ),maxα[0,1]G2(α,ϖ)=G2(1,ϖ).

    By (C1) and (2.7), we have

    G2(0,ϖ)G2(0,ϖ)=10(η1(0)S1(θ)+η2(0)S2(θ))G1(θ,ϖ)dθ10(η1(1)S1(θ)+η2(1)S2(θ))G1(θ,ϖ)dθ10(12η1(0)S2(θ)+η2(0)S2(θ))G1(θ,ϖ)dθ10(12η1(1)S2(θ)+η2(1)S2(θ))G1(θ,ϖ)dθ=(12η1(0)+η2(0))10S2(θ)G1(θ,ϖ)dθ(12η1(1)+η2(1))10S2(θ)G1(θ,ϖ)dθ=2MS1MS21+MS2+MS1MS1=ρZ1.

    Obviously, ρZ1>0, and

    ρZ112=2MS1MS21+MS2+MS1MS112=MS1+3MS1MS22MS212(1+MS2+MS1MS1)<0.

    Thus 0<ρZ1<12 and (3.18) is satisfied.

    By the same procedure,

    minα[0,κ]G4(α,ϖ)ρZ2G4(1,ϖ)=ρZ2maxα[0,1]G4(α,ϖ). (3.19)

    Finally, from (3.16) and (3.18), we can easily show that the following results hold:

    maxα[0,1]GZ1(α,ϖ)maxα[0,1]G1(α,ϖ)+maxα[0,1]G2(α,ϖ)=G1(0,ϖ)+G2(1,ϖ),

    and

    minα[0,κ]GZ1(α,ϖ)minα[0,κ]G1(α,ϖ)+minα[0,κ]G2(α,ϖ)12G1(0,ϖ)+ρZ1G2(1,ϖ)>ρZ1(G1(0,ϖ)+G2(1,ϖ))ρZ1maxα[0,1]GZ1(α,ϖ).

    Similarly, from (3.17) and (3.19), we can also easily show that

    maxα[0,1]GZ2(α,ϖ)G3(0,ϖ)+G4(1,ϖ),

    and

    minα[0,κ]GZ2(α,ϖ)>ρZ2maxα[0,1]GZ2(α,ϖ).

    Lemma 3.5. If (C1) is satisfied, then ψ(α) holds the following properties:

    (I) ψ1(α)ψ1(1)=maxα[0,1]ψ1(α) and ψ2(α)ψ2(1)=maxα[0,1]ψ2(α);

    (II) minα[0,κ]ψ1(α)ρZ1maxα[0,1]ψ1(α) and minα[0,κ]ψ2(α)ρZ2maxα[0,1]ψ2(α).

    Proof. Using Lemma 3.3 and (2.5), implies ψ1(α) is increasing on α[0,1], and thus

    minα[0,κ]ψ1(α)=ψ1(0)=210H1(ϖ)dϖ10H2(ϖ)dϖ+10(η1(0)S1(ϖ)+η2(0)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ,
    maxα[0,1]ψ1(α)=ψ1(1)=10H1(ϖ)dϖ+10(η1(1)S1(ϖ)+η2(1)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ

    and

    ψ1(0)ψ1(1)=210H1(ϖ)dϖ10H2(ϖ)dϖ+10(η1(0)S1(ϖ)+η2(0)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ10H1(ϖ)dϖ+10(η1(1)S1(ϖ)+η2(1)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ (3.20)
    210H1(ϖ)dϖ10H2(ϖ)dϖ+10(12η1(0)S2(ϖ)+η2(0)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ10H1(ϖ)dϖ+10(12η1(1)S2(ϖ)+η2(1)S2(ϖ))((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ=210H1(ϖ)dϖ10H2(ϖ)dϖ+(12η1(0)+η2(0))10S2(ϖ)((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ10H1(ϖ)dϖ+(12η1(1)+η2(1))10S2(ϖ)((2ϖ)10H1(θ)dθ+(ϖ1)10H2(θ)dθ)dϖ. (3.21)

    Setting H2(0)=2H1(0) and H1(1)=0, then (3.20) implies

    ψ1(0)ψ1(1)2MS1MS21+MS2+MS1MS1=ρZ1.

    Also,

    ψ2(0)ψ2(1)ρZ2.

    Thus,

    ψ1(α)ψ1(1)=maxα[0,1]ψ1(α),minα[0,κ]ψ1(α)ρZ1maxα[0,1]ψ1(α),ψ2(α)ψ2(1)=maxα[0,1]ψ2(α),minα[0,κ]ψ2(α)ρZ2maxα[0,1]ψ2(α).

    Suppose that μ1 and μ2 be nonnegative convex and continuous functionals on Θ, μ3 be nonnegative concave and continuous functional on Θ and μ4 be a nonnegative continuous functional on Θ. For M1,M2,M3,M4>0, let us introduce the convex sets:

    Θ(μ1,M1)={(X,Y)Θ:μ1(X,Y)<M1},Θ(μ1,μ3;M3,M1)={(X,Y)Θ:M3μ3(X,Y),μ1(X,Y)M1},Θ(μ1,μ2,μ3;M3,M2,M1)={(X,Y)Θ:M3μ3(X,Y),μ2(X,Y)M2,μ1(X,Y)M1},

    and a closed set

    Θ(μ1,μ4;M4,M1)={(X,Y)Θ:M4μ4(X,Y),μ1(X,Y)M1}.

    Lemma 4.1. Let Θ be a cone in a real Banach space χ. Let μ1 and μ2 be nonnegative continuous convex functionals on Θ, μ3 be nonnegative continuous concave functional on Θ and μ4 be a nonnegative continuous functionals on Θ satisfying μ4(λ(X,Y))λμ4(X,Y) for 0λ1, such that for some positive numbers N and M1,

    μ3(X,Y)μ4(X,Y),(X,Y)Nμ1(X,Y), (4.1)

    for all (X,Y)¯Θ(μ1,M1). Suppose

    T:¯Θ(μ1,M1)¯Θ(μ1,M1)

    is completely continuous and there exist positive numbers M2, M3 and M4 with M4<M3 such that

    (C3): {(X,Y)Θ(μ1,μ2,μ3; M3,M2,M1):μ3(X,Y)>M3} and μ3(X,Y)>M3 for (X,Y)Θ(μ1,μ2,μ3; M3,M2,M1);

    (C4): μ3(T(X,Y))>M3 for (X,Y)Θ(μ1,μ3;M3,M1) with μ2(T(X,Y))>M4;

    (C5): (0,0)Θ(μ1,μ4;M4,M1) and μ4(T(X,Y))<M4 for (X,Y)Θ(μ1,μ4;M4,M1) with μ4(X,Y)=M4.

    Then, T has at least three fixed points (X1,Y1),(X2,Y2),(X3,Y3)¯Θ(μ1,M1) such that

    μ1(Xj,Yj)M1,j=1,2,3;μ3(X1,Y1)>M3,M4<μ3(X2,Y2),μ4(X2,Y2)<M3,μ3(X2,Y2)<M4.

    Let χ={(X,Y)C([0,1],R)×C([0,1],R)} be endowed with the norm (X,Y)=maxα[0,1]|(X,Y)|, then χ is a Banach space.

    We define a set Θχ by

    Θ={(X,Y)χ:(X(α),Y(α))0,minα[0,1](X,Y)(α)ρmaxα[0,1](X,Y)(α)}.

    For (X,Y),(˜X,˜Y)Θ and ν1,ν20, it is easy to obtain that ν1(X,Y)(α)+ν2(˜X,˜Y)(α)0, and

    minα[0,κ]{ν1(X,Y)(α)+ν2(˜X,˜Y)(α)}minα[0,κ]{ν1(X,Y)(α)}+minα[0,κ]{ν2(˜X,˜Y)(α)}ρmaxα[0,1]{ν1(X,Y)(α)}+ρmaxα[0,1]{ν2(˜X,˜Y)(α)}=ρ(maxα[0,1]{ν1(X,Y)(α)}+maxα[0,1]{ν2(˜X,˜Y)(α)})ρmaxα[0,1]{ν1(X,Y)(α)+ν2(˜X,˜Y)(α)}.

    Thus, for (X,Y),(˜X,˜Y)Θ and ν1,ν20, ν1(X,Y)(α)+ν2(˜X,˜Y)(α)Θ. And if (X,Y)Θ, (X,Y)0, it can be seen that (X,Y)Θ. Thus, Θ is a cone in χ.

    Let T:Θχ be an operator, i.e., T=(T1,T2), defined by

    T1(X,Y)(α)=10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α),T2(X,Y)(α)=10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ2(α).

    Lemma 4.2. If (C1) is true, then T:ΘΘ is a completely continuous operator.

    Proof. For (X,Y)Θ, it is easy to know that T is a continuous operator, and T(X,Y)(α)0. By (3.15), we have

    minα[0,κ]T1(X,Y)(α)=minα[0,κ]10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+minα[0,κ]ψ1(α)=10minα[0,κ]GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+minα[0,κ]ψ1(α)10ρZ1minα[0,κ]GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ρZ1minα[0,κ]ψ1(α)ρZ1maxα[0,1]T1(X,Y)(α).

    Similarly, for minα[0,κ]T2(X,Y)(α)ρZ2maxα[0,1]T2(X,Y)(α). So, T(Θ)Θ.

    Set ΘτΘ be bounded, i.e., a constant τ>0 in such a way that (X,Y)τ, for every (X,Y)Θτ. Let M0=maxα[0,1],(X,Y)|ϝ1(α,Dγ1X(α),Dγ2Y(α))|>0 and N0=maxα[0,1],(X,Y)|ϝ2(α,Dγ1X(α),Dγ2Y(α))|>0. For (X,Y)Θτ, from Lemmas 3.3 and 3.4, we have

    ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)ϑq(M010KB1(ϖ,θ)dθ)ϑq(M0Γ(B1+1))

    and

    |T1(X,Y)(α)|=|10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α)|10|GZ1(α,ϖ)|ϑq(M0Γ(B1+1))dϖ+|ψ1(α)|ϑq(M0Γ(B1+1))10GZ1(α,ϖ)dϖ+ψ1(1)ϑq(M0Γ(B1+1))10(G1(α,ϖ)+G2(α,ϖ))dϖ+ψ1(1):=M01. (4.2)

    Similarly,

    ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)ϑq(N010KB2(ϖ,θ)dθ)ϑq(N0Γ(B2+1))

    and

    |T2(X,Y)(α)|=|10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ2(α)|10|GZ2(α,ϖ)|ϑq(N0Γ(B2+1))dϖ+|ψ2(α)|ϑq(N0Γ(B2+1))10GZ2(α,ϖ)dϖ+ψ2(1)ϑq(N0Γ(B2+1))10(G3(α,ϖ)+G4(α,ϖ))dϖ+ψ2(1):=N01. (4.3)

    So that T(X,Y)max{M01,N01}, which implies T(Θτ) is uniformly bounded. Next, we will show that T(Θτ) is equicontinuous. Since GZ(α,ϖ), ψ(α) are continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1]. Thus, for any ϵ>0, there exists a constant ς1>0, such that

    |GZ1(α1,ϖ)GZ1(α2,ϖ)|<ϵ4ϑq(M0Γ(B1+1)),|ψ1(α1)ψ1(α2)|<ϵ4,

    for α1,α2[0,1] with |α1α2|<ς1. Therefore,

    |T1(X,Y)(α1)T1(X,Y)(α2)|=|10GZ1(α1,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α1) (4.4)
    10GZ1(α2,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(α2)|10|GZ1(α1,ϖ)GZ1(α2,ϖ)|ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+|ψ1(α1)ψ1(α2)|ϵ4ϑq(M0Γ(B1+1))ϑq(M0Γ(B1+1))+ϵ4=ϵ2. (4.5)

    Similarly

    |T2(X,Y)(α1)T2(X,Y)(α2)|ϵ2. (4.6)

    Combining (4.4) and (4.6), we obtain

    |T(X,Y)(α1)T(X,Y)(α2)|ϵ.

    Implies, T(Θτ) is equicontinuous.

    Thus Arzelˊa-Ascoli Theorem implies that T is a completely continuous operator.

    Define the functionals

    μ1(X,Y)=(X,Y),μ2(X,Y)=μ4(X,Y)=maxα[0,1]|(X,Y)(α)|,μ3(X,Y)=minα[0,κ]|(X,Y)(α)|,

    then μ1 and μ2 are continuous nonnegative convex functionals, μ3 is a continuous nonnegative concave functional, μ4 is a continuous nonnegative functional, and

    ρμ2(X,Y)μ3(X,Y)μ2(X,Y)=μ4(X,Y),(X,Y)Nμ1(X,Y).

    Thus, (4.1) in Lemma 4.1 is fulfilled.

    Let

    l1=10GZ1(0,ϖ)GZ1(1,ϖ)ϑq(10KB1(ϖ,θ)dθ)dϖ,l2=10GZ2(0,ϖ)GZ2(1,ϖ)ϑq(κ0KB2(ϖ,θ)dθ)dϖ,l3=10GZ1(0,ϖ)GZ1(1,ϖ)ϑq(10KB1(ϖ,θ)dθ)dϖ,l4=10GZ2(0,ϖ)GZ2(1,ϖ)ϑq(κ0KB2(ϖ,θ)dθ)dϖ.

    Theorem 4.1. Let (C1) is true, and constants M1,M3,M4ψ(1) with M4<M3<ρM1minl2l4l1l3 and M2=M3ρ, in such a way that

    (C6) ϝ1(α,X,Y)ϑp(M1ψ1(1)l1) and ϝ2(α,X,Y)ϑp(M1ψ2(1)l2), (α,X,Y)[0,1]×[0,M1]×[0,M1];

    (C7) ϝ1(α,X,Y)>ϑp(M3ρZ1ψ1(1)ρZ1l3) and ϝ2(α,X,Y)>ϑp(M3ρZ2ψ2(1)ρZ2l4), (α,X,Y)[0,κ]×[M3,M3ρZ1]×[0,M1];

    (C8) ϝ1(α,X,Y)<ϑp(M4ψ1(1)l1) and ϝ2(α,X,Y)<ϑp(M4ψ2(1)l2), (α,X,Y)[0,1]×[0,M4]×[0,M1].

    Then, (1.2) has at least three positive solutions (X1,Y1), (X2,Y2), (X3,Y3) satisfying

    (Xi,Yi)M1(i=1,2,3), (4.7)
    minα[0,κ](X1,Y1)(α)M3,M4<minα[0,κ](X2,Y2)(α),maxα[0,1](X2,Y2)(α)<M3,maxα[0,1](X3,Y3)(α)<M4. (4.8)

    Proof. Obviously, the fixed points of T are equivalent to the solutions of (1.2). For (X,Y)¯Θ(μ1,M1), we get

    μ1(X,Y)=(X,Y)M1,

    thus

    maxα[0,1]|(X,Y)|M1,

    and then

    0(X,Y)M1.

    By (C6), we have

    maxα[0,1]|T1(X,Y)(α)|maxα[0,1]10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+maxα[0,1]ψ1(α)10(G1(0,ϖ)G2(1,ϖ))ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+ψ1(1)10(G1(0,ϖ)G2(1,ϖ))ϑq(ϑp(M1l1)10KB1(ϖ,θ)dθ)dϖ+ψ1(1)=M1ψ1(1)l110(G1(0,ϖ)G2(1,ϖ))ϑq(10KB1(ϖ,θ)dθ)dϖ+ψ1(1)=M1.

    Similarly,

    maxα[0,1]|T2(X,Y)(α)|M1.

    So,

    μ1(T(X,Y))=T(X,Y)=maxα[0,1]|T(X,Y)(α)|M1.

    Thus, T:¯Θ(μ1,M1)¯Θ(μ1,M1).

    For (X,Y)=M3ρ, (X,Y)(α)Θ(μ1,μ2,μ3;M3,M2,M1), μ3(M3ρ)>M3, we get

    {(X,Y)Θ(μ1,μ2,μ3;M3,M2,M1):μ3(X,Y)>M3}.

    For (X,Y)(α)Θ(μ1,μ2,μ3;M3,M2,M1), we know that M3(X,Y)(α)M2=M3ρ for α[0,κ].

    In view of (C7),

    μ3(T1(X,Y))=minα[0,κ]|T1(X,Y)(α)|minα[0,κ]10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+minα[0,κ]ψ1(α)10ρZ1(G1(0,ϖ)G2(1,ϖ))ϑq(κ0KB1(ϖ,θ)ϑp(M3ρZ1ψ1(1)ρZ1l3)dθ)dϖ+ρZ1ψ1(1)=ρZ1ϑp(M3ρZ1ψ1(1)ρZ1l3)κ0(G1(0,ϖ)G2(1,ϖ))ϑq(10KB1(ϖ,θ)dθ)dϖ+ρZ1ψ1(1)=M3.

    Similarly,

    μ3(T2(X,Y))M3.

    So, μ3(T(X,Y))>M3 for all (X,Y)(α)Θ(μ1,μ2,μ3;M3,M2,M1). Hence, (C3) of Lemma 4.1 is fulfilled.

    By (4.7), for all (X,Y)Θ(μ1,μ3;M3,M1) with μ2(T(X,Y))>M2=M3ρ, one get

    μ2(T(X,Y))>ρM2=ρM3ρ=M3.

    Implies (C4) of Lemma 4.1 is true.

    As μ4(0,0)=0<M4, thus 0Θ(μ1,μ4;M4,M1). If (X,Y)Θ(μ1,μ4;M4,M1) with μ4(X,Y)=M4, we deduced μ1(X,Y)M1. Thus, maxα[0,1](X,Y)(α)=M4 and 0(X,Y)(α)M1.

    Using (C8), we get

    μ4(T1(X,Y))=maxα[0,1]|T1(X,Y)(α)|minα[0,1]10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ+maxα[0,1]ψ1(α)<10ρZ1(G1(0,ϖ)G2(1,ϖ))ϑq(κ0KB1(ϖ,θ)ϑp(M4ψ1(1)l1)dθ)dϖ+ψ1(1)=M4ψ1(1)l1κ0(G1(0,ϖ)G2(1,ϖ))ϑq(10KB1(ϖ,θ)dθ)dϖ+ψ1(1)=M4.

    Similarly,

    μ4(T2(X,Y))M4.

    So, μ3(T(X,Y))<M3 for all (X,Y)(α)Θ(μ1,μ4;M4,M1). Therefore, the condition (C4) of Lemma 4.1 hold.

    To sum up, all the conditions of Lemma 4.1 are verified, and it was noticed that (Xi,Yi)(α)ψ(0)>0. Hence, the coupled system (1.2) has at least three positive solutions (X1,Y1), (X2,Y2), (X3,Y3) satisfying (4.7) and (4.8).

    Definition 5.1. The system (1.2) is HU stable if there exist πZ1,Z2=max{πZ1,πZ2}>0 in such a way that, for ε=max{εZ1,εZ2}>0 and for any (X,Y)χ satisfying

    {|DB1ϑp(DZ1X(α))+ϝ1(α,Dγ1X(α),Dγ2Y(α))|εZ1,   α[0,1],|DB2ϑp(DZ2)Y(α)+ϝ2(α,Dγ1X(α),Dγ2Y(α))|εZ2,   α[0,1], (5.1)

    (ˆX,ˆY)χ satisfying (1.2) such that

    (X,Y)(ˆX,ˆY)πZ1,Z2ε,  α[0,1].

    Definition 5.2. The coupled implicit FDEs (1.2) is GHU stable if there exist ΦC(R+,R+) with Φ(0)=0 such that, for any solution (X,Y)χ of inequality (5.1), there exists a solution (ˆX,ˆY)χ of (1.2) fulfilling

    (X,Y)(ˆX,ˆY)Φ(ε),  α[0,1].

    Let ΨZ1,Z2=max{ΨZ1,ΨZ2}C([0,1],R), and πΨZ1,ΨZ2=max{πΨZ1,πΨZ2}>0.

    Definition 5.3. The coupled system of implicit FDEs (1.2) is said to be HUR stable with respect to ΨZ1,Z2 if there exists constants πΨZ1,ΨZ2 such that, for some ε>0 and for any solution (X,Y)χ of the inequality

    {|DB1ϑp(DZ1X(α))+ϝ1(α,Dγ1X(α),Dγ2Y(α))|ΨZ1(α)εZ1,   α[0,1],|DB2ϑp(DZ2)Y(α)+ϝ2(α,Dγ1X(α),Dγ2Y(α))|ΨZ2(α)εZ2,   α[0,1], (5.2)

    there exists (ˆX,ˆY)χ satisfying (1.2) in such a way that

    (X,Y)(ˆX,ˆY)πΨZ1,ΨZ2ΨZ1,Z2ε,   α[0,1]. (5.3)

    Definition 5.4. The system (1.2) is said to be HUR stable with respect to ΨZ1,Z2 if there exists constants πΨZ1,ΨZ2 such that, for any approximate solution (X,Y)χ of inequality (5.2), there exists a solution (ˆX,ˆY)χ of (1.2) fulfilling

    (X,Y)(ˆX,ˆY)πΨZ1,ΨZ2ΨZ1,Z2(α),   α[0,1]. (5.4)

    Remark 5.1. We say that (X,Y)χ is a solution of the system of inequalities (5.1) if there exist functions Λf, ΛgC([0,1],R) depending upon X,Y, respectively, such that

    (I)

    |Λf(α)|εZ1,|Λg(α)|εZ2,α[0,1];

    (II)

    {DB1ϑp(DZ1X(α))+ϝ1(α,Dγ1X(α),Dγ2Y(α))=Λf(α),DB2ϑp(DZ2)Y(α)+ϝ2(α,Dγ1X(α),Dγ2Y(α))=Λg(α).

    Lemma 5.1. Under the assumptions given in Remark 5.1, the solution (X,Y)χ of coupled system

    {DB1ϑp(DZ1X(α))+ϝ1(α,Dγ1X(α),Dγ2Y(α))=Λf(α),α[0,1],DB2ϑp(DZ2)Y(α)+ϝ2(α,Dγ1X(α),Dγ2Y(α))=Λg(α),α(0,1),(ϑp(DZ1X(0)))i=ϑp(DZ1X(1))=0,i=¯1,m1,(ϑp(DZ2Y(0)))i=ϑp(DZ2Y(1))=0,i=¯1,m1,X(0)+X(0)=10S1(ϖ)X(ϖ)dϖ+10H1(ϖ)dϖ,X(1)+X(1)=10S2(ϖ)X(ϖ)dϖ+10H2(ϖ)dϖ,Y(0)+Y(0)=10S3(ϖ)Y(ϖ)dϖ+10H3(ϖ)dϖ,Y(1)+Y(1)=10S4(ϖ)Y(ϖ)dϖ+10H4(ϖ)dϖ,Xj(0)=0,Yj(0)=0,j=¯2,n1, (5.5)

    satisfies the system of inequalities

    |X(α)10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖψ1(α)|ϑq(εZ1)M01ψ1(1)ϑq(M0),|Y(α)10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖψ2(α)|ϑq(εZ2)N01ψ2(1)ϑq(N0).

    Proof. In light of Lemma 3.2, a solution of the coupled system (5.1) is

    {X(α)=10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)Λf(θ)dθ)dϖ+ψ1(α),Y(α)=10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)Λg(θ)dθ)dϖ+ψ2(α), (5.6)

    Using (4.2) in (5.6), we have

    |X(α)10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖψ1(α)|10|GZ1(α,ϖ)|ϑq(εZ110|KB1(ϖ,θ)|dθ)dϖϑq(εZ1)M01ψ1(1)ϑq(M0).

    Similarly, using (4.3) in (5.6), we get

    |Y(α)10GZ2(α,ϖ)ϑq(10KB2(ϖ,θ)ϝ2(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖψ2(α)|ϑq(εZ2)N01ψ2(1)ϑq(N0).

    For the next result, we suppose the following condition.

    (C9) Let N1,N2,S1,S2C([0,1],R), and there exists L1,L2,L3,L4>0 such that

    |ϝ1(α,N1,S1)ϝ1(α,N2,S2)|L1|N1N2|+L2|S1S2|,|ϝ2(α,N1,S1)ϝ2(α,N2,S2)|L3|N1N2|+L4|S1S2|.

    Theorem 5.1. Suppose the condition (C9) and Lemma 5.1 satisfies, then (1.2) is HU stable, if υ1υ4υ2υ3=υ4υ2>0, where

    υ1=1(M01ψ1(1))ϑq(L1Γ(B1γ1)Γ(1γ1))ϑq(M0Γ(B1+1)),υ2=(M01ψ1(1))ϑq(L2Γ(B1γ2)Γ(1γ2))ϑq(M0Γ(B1+1)),υ3=1(N01ψ2(1))ϑq(L3Γ(B2γ1)Γ(1γ1))ϑq(N0Γ(B2+1)),υ4=(N01ψ2(1))ϑq(L4Γ(B2γ2)Γ(1γ2))ϑq(N0Γ(B2+1)).

    Proof. Consider (X,Y)χ to be any solution of (5.5), and (ˆX,ˆY)χ is a solution of the coupled system (1.2), then we take

    |(X(α)ˆX(α))||X(α)10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖψ1(α)+10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))dθ)dϖ10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)ϝ1(θ,Dγ1ˆX(θ),Dγ2ˆY(θ))dθ)dϖ|ϑq(εZ1)M01ψ1(1)ϑq(M0)+10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)|ϝ1(θ,Dγ1X(θ),Dγ2Y(θ))ϝ1(θ,Dγ1ˆX(θ),Dγ2ˆY(θ))|dθ)dϖϑq(εZ1)M01ψ1(1)ϑq(M0)+10GZ1(α,ϖ)ϑq(10KB1(ϖ,θ)(L1Dγ1|X(θ)ˆX(θ)|+L2Dγ2|Y(θ)ˆY(θ)|)dθ)dϖ(M01ψ1(1))(ϑq(εZ1)ϑq(M0)+ϑq(L1Γ(B1γ1)Γ(1γ1))ϑq(M0Γ(B1+1))|X(α)ˆX(α)|+ϑq(L2Γ(B1γ2)Γ(1γ2))ϑq(M0Γ(B1+1))|Y(α)ˆY(α)|)

    implies that

    υ1uˆXυ2vˆY(M01ψ1(1))ϑq(εZ1)ϑq(M0). (5.7)

    Similarly, we have

    υ4vˆYυ3uˆX(N01ψ2(1))ϑq(εZ2)ϑq(N0). (5.8)

    From (5.7) and (5.8), we get

    [υ1υ2υ3υ4][XˆXYˆY][(M01ψ1(1))ϑq(εZ1)ϑq(M0)(N01ψ2(1))ϑq(εZ2)ϑq(N0)]

    implies

    [XˆXYˆY][υ4υ4υ2υ2υ4υ2υ3υ4υ2υ1υ4υ2][(M01ψ1(1))ϑq(εZ1)ϑq(M0)(N01ψ2(1))ϑq(εZ2)ϑq(N0)]. (5.9)

    From system (5.9), we have

    XˆXυ4υ4υ2(M01ψ1(1))ϑq(εZ1)ϑq(M0)+υ2υ4υ2(N01ψ2(1))ϑq(εZ2)ϑq(N0),YˆYυ3υ4υ2(M01ψ1(1))ϑq(εZ1)ϑq(M0)+υ1υ4υ2(N01ψ2(1))ϑq(εZ2)ϑq(N0),

    which implies that

    XˆX+YˆYυ1υ4υ2(N01ψ2(1))ϑq(εZ2)ϑq(N0)+υ2υ4υ2(N01ψ2(1))ϑq(εZ2)ϑq(N0)+υ3υ4υ2(M01ψ1(1))ϑq(εZ1)ϑq(M0)+υ4υ4υ2(M01ψ1(1))ϑq(εZ1)ϑq(M0).

    If max{ϑq(εZ1),ϑq(εZ2)}=ϑq(εZ) and υ1υ4υ2N01ψ2(1)ϑq(N0)+υ2υ4υ2N01ψ2(1)ϑq(N0)+υ3υ4υ2M01ψ1(1)ϑq(M0)+υ4υ4υ2M01ψ1(1)ϑq(M0)=υZ1,Z2, then

    (X,Y)(ˆX,ˆY)υZ1,Z2ϑq(εZ).

    Hence, system (1.2) is HU stable. Also, if

    (X,Y)(ˆX,ˆY)υZ1,Z2Φ(ϑq(εZ)),

    with Φ(0)=0, then the solution of system (1.2) is GHU stable.

    Theorem 5.2. Suppose the conditions (C9) and Lemma 5.1 holds, then (1.2) is GHU stable, if υ1υ4υ2υ3=υ4υ2>0, where υ1, υ2, υ3 and υ4 is defined in Theorem 5.1.

    Proof. By applying steps of Theorem 5.1, we can easily show that system (1.2) is GHU stable, by using Definition 5.4.

    For the next theorem, we assume that

    (C10) non-decreasing functions ˉwZ,ˉwBC([0,1],R+) in such a way that

    IZˉwZ(α)LZˉwZ(α)

    and

    IBˉwB(α)LBˉwB(α),

    where LZ,LB>0.

    Theorem 5.3. Suppose the conditions (C9), (C10) and Lemma 5.1 satisfies, then system (1.2) is HUR stable if υ1υ4υ2υ3=υ4υ2>0.

    Proof. By applying steps of Theorem 5.1, we can easily show that system (1.2) is HUR stable, by using Definition 5.3.

    Theorem 5.4. Suppose the conditions (C9), (C10) and Lemma 5.1 holds, then system (1.2) is GHUR stable if υ1υ4υ2υ3=υ4υ2>0.

    Proof. By applying steps of Theorem 5.1, we can easily show that system (1.2) is GHUR stable, by using Definition 5.4.

    In this section, we demonstrate an example to illustrate the main results.

    Example 6.1. Consider the following system of implicit FDEs:

    {D94ϑ32(D72X(α))+ϝ1(α,D12X(α),D13Y(α))=0,α(0,1),D115ϑ32(D103)Y(α)+ϝ2(α,D12X(α),D13Y(α))=0,α(0,1),(ϑ32(D72X(0)))=ϑ32(D72X(1))=0,(ϑ32(D103Y(0)))=ϑ32(D103Y(1))=0,X(0)+X(0)=10(ϖ+1)X(ϖ)dϖ+10ϖdϖ4,X(1)+X(1)=10(3ϖ2+ϖ)X(ϖ)dϖ+10ϖdϖ3,Y(0)+Y(0)=10(2ϖ+1)Y(ϖ)dϖ+10ϖdϖ2,Y(1)+Y(1)=10(5ϖ2+2ϖ)Y(ϖ)dϖ+10ϖdϖ,Xj=Yj,wherej=2,3. (6.1)

    Choosing M1=25000, M3=65, M4=3 and κ=13. Reminding ψ1, ψ2 from (2.5), (2.6) and ρZ1 and ρZ2 from Lemma 3.4, we get

    ψ1(1)=1.87144,ψ2(1)=2.32681,ρZ1=0.046572<italic>and</italic>ρZ2=0.0097325.

    In (6.1), we see that

    ϝ1(α,D12X(α),D13Y(α))=tan(α150)+D12X(α)+cos(D13Y(α)),ϝ2(α,D12X(α),D13Y(α))=sin(D12X(α))D13Y(α)160,

    which satisfies the following conditions:

    (C6) supϝ1(α,X,Y)246.534minϑp(M1ψ1(1)l1)291.732,

    supϝ2(α,X,Y)241.492minϑp(M1ψ2(1)l2)284.618, (α,X,Y)[0,1]×[0,25000]×[0,25000];

    (C7) infϝ1(α,X,Y)240.631>ϑp(M3ρZ1ψ1(1)ρZ1l3)201.981 (α,X,Y)[0,13]×[65,M3ρZ1]×[0,25000];

    infϝ2(α,X,Y)271>ϑp(M3ρZ2ψ2(1)ρZ2l4)201.843, (α,X,Y)[0,13]×[65,M3ρZ2]×[0,25000];

    (C8) supϝ2(α,X,Y)198.938<ϑp(M4ψ1(1)l1)243.861,

    supϝ2(α,X,Y)183.861<ϑp(M4ψ2(1)l2)202.991, (α,X,Y)[0,1]×[0,3]×[0,25000].

    Then, all assumptions of Theorem 4.1 are satisfied. Thus, BVP (6.1) has at least three positive solutions (X1,Y1), (X2,Y2), (X3,Y3) satisfying

    (Xi,Yi)25000(i=1,2,3),minα[0,κ]|(X1,Y1)|>65,3<minα[0,κ]|(X2,Y2)|,maxα[0,1]|(X2,Y2)|<65,maxα[0,1]|(X3,Y3)|<3.

    For HUS, we found L1=L2=1150, L3=L4=1160, υ10.02961, υ20.9721, υ30.03291 and υ40.9948. Hence, by Theorem 5.1, we have υ4υ20.99480.0227>0.

    In this paper, we considered an implicit coupled BVP of p-Laplacian FDEs (1.2), which involved disturbing functions. The relating fractional order derivative is taken in Caputo sense. By using the Avery-Peterson fixed point theorem for the proposed problem, we found at least three solutions, under sufficient conditions. In addition, we presented four types of Ulam's stability, i.e., Hyers-Ulam stability, generalized Hyers-Ulam stability, Hyers-Ulam-Rassias stability and generalized Hyers-Ulam-Rassias stability, for the coupled implicit fractional p-Laplacian system, and an example is provided to authenticate the theoretical results.

    This research is supported by the scientific research project of Anhui Provincial Department of Education (KJ2021A1155; KJ2020A0780).

    The authors declare no conflict of interest.



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