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Dynamic fracture with continuum-kinematics-based peridynamics

  • This contribution presents a concept to dynamic fracture with continuum-kinematics-based peridynamics. Continuum-kinematics-based peridynamics is a geometrically exact formulation of peridynamics, which adds surface- or volume-based interactions to the classical peridynamic bonds, thus capturing the finite deformation kinematics correctly. The surfaces and volumes considered for these non-local interactions are constructed using the point families derived from the material points' horizon. For fracture, the classical bond-stretch damage approach is not sufficient in continuum-kinematics-based peridynamics. Therefore it is here extended to the surface- and volume-based interactions by additional failure variables considering the loss of strength in the material points' internal force densities. By numerical examples, it is shown that the presented approach can correctly handle crack growth, impact damage, and spontaneous crack initiation under dynamic loading conditions with large deformations.

    Citation: Kai Friebertshäuser, Christian Wieners, Kerstin Weinberg. Dynamic fracture with continuum-kinematics-based peridynamics[J]. AIMS Materials Science, 2022, 9(6): 791-807. doi: 10.3934/matersci.2022049

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  • This contribution presents a concept to dynamic fracture with continuum-kinematics-based peridynamics. Continuum-kinematics-based peridynamics is a geometrically exact formulation of peridynamics, which adds surface- or volume-based interactions to the classical peridynamic bonds, thus capturing the finite deformation kinematics correctly. The surfaces and volumes considered for these non-local interactions are constructed using the point families derived from the material points' horizon. For fracture, the classical bond-stretch damage approach is not sufficient in continuum-kinematics-based peridynamics. Therefore it is here extended to the surface- and volume-based interactions by additional failure variables considering the loss of strength in the material points' internal force densities. By numerical examples, it is shown that the presented approach can correctly handle crack growth, impact damage, and spontaneous crack initiation under dynamic loading conditions with large deformations.



    In this paper, we consider the following p-Laplacian fractional differential equation involving Riemann-Stieltjes integral boundary condition

    {Dβt(φp(Dαtz(t)g(t,z(t),Dγtz(t))))=f(t,z(t),Dγtz(t)), 0<t<1,Dαtz(0)=Dα+1tz(0)=Dγtz(0)=0,Dαtz(1)=0,Dγtz(1)=10Dγtz(s)dA(s),  (1.1)

    where Dαt,Dβt,Dγt are the Riemann-Liouville fractional derivatives of orders α,β,γ with 0<γ1<α2<β<3,αγ>1,10Dγtz(t)dA(s) denotes a Riemann-Stieltjes integral, and A is a function of bounded variation. The p-Laplacian operator is defined as φp(s)=|s|p2s, p>2, φp(s) is invertible and its inverse operator is φq(s), where q=pp1 is the conjugate index of p.

    Fractional calculus and fractional differential equations arise in many fields, such as, mathematics, physics, economics, engineering, biology, electroanalytical chemistry, capacitor theory, electrical circuits, control theory, and fluid dynamics, see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49]. The problem (1.1) can be regarded as a fractional order model for the turbulent flow in a porous medium, see [8,50]. As we know, integral boundary value problems have different applications in applied fields such as blood flow problems, chemical engineering, underground water flow and population dynamics. For example, in [31], Meng and Cui considered the following fractional differential equation involving integral boundary condition

    {Dαx(t)=f(t,x(t)), 0<t<1,x(0)=10x(t)dA(t),  (1.2)

    where fC([0,1]×R,R), 10x(t)dA(t) denotes the Riemann-Stieltjes integral with positive Stieltjes measure. Dα is the conformable fractional derivative of order 0<α1 at t>0. By topological degree theory, the method of lower and upper solutions and a fixed point theorem, they discussed the existence of at least three solutions to the problem (1.2).

    In [8], the authors studied the following fractional differential boundary value problem

    {Dβt(φp(Dαtx))(t)=f(x(t),Dγtx(t)), t(0,1),Dαtx(0)=Dα+1tx(0)=Dαtx(1)=0,Dγtx(0)=0, Dγtx(1)=10Dγtx(s)dA(s),  (1.3)

    where Dαt, Dβt, Dγt are the Riemann-Liouville derivatives, 10x(s)dA(s) is the Riemann-Stieltjes integral and 0<γ1<α2<β<3, αγ>1, A is a function bounded variation, φp is the p-Laplacian operator. By employing a fixed point theorem for mixed monotone operator, they obtained the existence and uniqueness of positive solutions for the problem (1.3).

    When g0 in (1.1), the author in [21] gave the existence and uniqueness of positive solutions by using monotone iterative technique. Motivated by the results mentioned above and wide applications of different boundary value conditions, we consider the existence and uniqueness of positive solutions for p-Laplacian fractional order differential equation involving Riemann-Stieltjes integral boundary condition (1.1). In Section 2, we present some preliminaries that can be used to prove our main results. The main theorems are formulated and proved in Section 3. Two simple examples are given to illustrate the main results in Section 4.

    In the following, we start with some basic concepts and lemmas.

    Definition 2.1. [1] For a function x:(0,+)R, the Riemann-Liouville fractional integral of order α>0 is

    Iαx(t)=1Γ(α)t0(ts)α1x(s)ds,

    provide that the right-hand side is pointwise defined on (0,+).

    Definition 2.2. [1] For a function x:(0,+)R, the Riemann-Liouville fractional derivative of order α>0 is

    Dαtx(t)=1Γ(nα)(ddt)nt0(ts)nα1x(s)ds,

    where n=[α]+1, [α] denotes the integer part of the number α, provided that the right-hand side is pointwise defined on (0,+).

    To reduce the p-Laplacian fractional order differential equation (1.1) to a convenient form, for xC[0,1], making a change of variable z(t)=Iγx(t). By the definitions of the Riemann-Liouville fractional integral and derivative, we can see that Iγx(t)0,Dαtx(t)0 as t0+. So we first get x(0)=0. From [8,21], the problem (1.1) reduces to an equivalent boundary value problem as follows:

    {Dβtφp(Dαγtx(t)g(t,Iγx(t),x(t)))=f(t,Iγx(t),x(t))Dαγtx(0)=Dαγ+1tx(0)=Dαγtx(0)=0,x(0)=0,x(1)=10x(s)dA(s).  (2.1)

    So, to get the existence and uniqueness of positive solutions for the problem (1.1), we only need to condiser the equivalent problem (2.1). To do this, we fist give an important function

    Gβ(t,s)=1Γ(β){[t(1s)]β1, 0ts1,1[t(1s)]β1(ts)β1, 0st1. (2.2)

    From Lemma 2.2 in [8], we have the following conclusion:

    Lemma 2.1. Given f, gL1[0,1], 0<γ1<α2<β<3 and αγ>1, the fractional order p-Laplacian differential equation

    {Dβt(φp(Dαγtx(t)g(t))=f(t),Dαγtx(0)=Dαγ+1tx(0)=Dαγtx(1)=0,x(0)=0,x(1)=10x(s)dA(s)  (2.3)

    has a unique solution

    x(t)=10H(t,s)φq(10Gβ(s,τ)f(τ)dτ)ds+10H(t,s)g(s)ds,

    where

    H(t,s)=GA(s)1Atαγ1+Gαγ(t,s) (2.4)

    with

    A=10tαγ1dA(t), GA(s)=10Gαβ(t,s)dA(t).

    Lemma 2.2. [21] Let 0A<1 and GA(s)0 for s[0,1], then the functions Gβ(t,s) and H(t,s) satisfy:

    (1) Gβ(t,s)>0,H(t,s)>0, for t,s(0,1);

    (2) tβ1(1t)s(1s)β1Γ(β)Gβ(t,s)β1Γ(β)tβ1(1t) for t,s[0,1];

    (3) There exist two positive constants d,e such that

    dtαγ1GA(s)H(t,s)etαγ1, t,s[0,1].

    Let (E,) be a real Banach space and θ be the zero element of E. E is partially ordered by a cone PE, i.e., xy if and only if yxP. A cone P is called normal if there exists a constant N>0 such that, for all x,yE, θxyxNy; in this case, N is called the normality constant of P. We say that an operator A:EE is increasing (decreasing) if xy implies AxAy(AxAy).

    For x,yE, the notation xy denotes that there exist λ>0 and μ>0 such that λxyμx. Clearly, is an equivalence relation. Given h>θ (i.e., hθ and hθ), define Ph={xE:xh}. It is clear to see that PhP.

    Definition 2.3. [51] Let 0<δ<1. An operator A:PP is said to be δconcave if A(tx)tδAx for t(0,1), xP. An operator A:PP is called to be sub-homogeneous if A(tx)tAx for t>0, xP.

    In papers [52,53], the authors investigated a sum operator equation

    Ax+Bx=x, (2.5)

    where A,B are monotone operators. They gave the existence and uniqueness of positive solutions for (2.5) and obtained some interesting theorems.

    Lemma 2.3. [52] Let E be a real Banach space. P is a normal cone in E, A:PP is an increasing δ-concave operator and B : PP is an increasing sub-homogeneous operator. Suppose that

    (ⅰ) there is h>θ such that AhPh and BhPh;

    (ⅱ) there exists a constant δ0>0 such that Axδ0Bx for all xP.

    Then the operator equation (2.5) has a unique solution x in Ph. Further, making the sequence yn=Ayn1+Byn1, n=1,2 for any initial value y0Ph, one has ynx as n.

    Lemma 2.4. [53] Let E be a real Banach space. P is a normal cone in E, A:PP is an increasing operator, and B:PP is a decreasing operator. In addition,

    (ⅰ) for xP and t(0,1), there exist ϕi(t)(t,1),i=1,2 such that

    A(tx)ϕ1(t)Ax,  B(tx)1ϕ2(t)Bx; (2.6)

    (ⅱ) there is h0Ph such that Ah0+Bh0Ph.

    Then the operator equation (2.5) has a unique solution x in Ph. Further, for any initial values x0,y0Ph, making the sequences

    xn=Axn1+Byn1,yn=Ayn1+Bxn1,n=1,2,

    one has xnx,ynx as n.

    Remark 2.1. If B is a null operator, the conclusions in Lemmas 2.1 and 2.2 are still right.

    In this section, we intend to obtain some results on the existence and uniqueness of positive solutions for the problem (1.1) by using Lemmas 2.3 and 2.4.

    We work in a Banach space E=C[0,1] with the usual norm x=sup{|x(t)|:t[0,1]}. Let P={xC[0,1]:x(t)0,t[0,1]}, then it is a normal cone in C[0,1]. Hence this space is equipped with a partial order

    xy, x,yC[0,1]x(t)y(t), t[0,1].

    Theorem 3.1. Let 0A<1 and GA(s)0 for s[0,1]. Assume

    (H1) f,g:[0,1)×[0,+)×[0,+)[0,+) are continuous and increasing with respect to the second and third arguments, g(t,0,0)0, t[0,1];

    (H2) for λ(0,1), f(t,λx,λy)λ1q1f(t,x,y) for x,y[0,+) and there exists a constant δ(0,1) such that g(t,λx,λy)λδg(t,x,y) for all t[0,1], x,y[0,+);

    (H3) There exists a constant δ0>0 such that f(t,x,y)δ0g(t,0,0), t[0,1], x0, y0.

    Then there is a unique yPh, where h(t)=tαγ1, t[0,1], such that the problem (1.1) has a unique positive solution z(t)=Iγy(t) in set Ω:={Iγy(t)|yPh}. And for any initial value y0Ph, making sequences

    yn(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγyn1(τ),yn1(τ))dτ)ds+10H(t,s)g(s,Iγyn1(s),yn1(s))ds

    and zn(t)=Iγyn(t), n=1,2, we have yn(t)y(t) and zn(t)z(t) as n, where Gβ(s,τ),H(t,s) are given as in (2.2), (2.4) respectively.

    Proof. From Lemma 2.1, we know that the problem (2.1) has an integral formulation give by

    y(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds+10H(t,s)g(s,Iγy(s),y(s))ds.

    Define two operators A:PE and B:PE by

    Ay(t)=10H(t,s)g(s,Iγy(s),y(s))ds,By(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds.

    Then we see that y is the solution of the problem (2.1) if and only if y=Ay+By. From (H1), (2.4) and Lemma 2.2, we can easily get A:PP and B:PP. In the following, we show that A,B satisfy all assumptions of Lemma 2.3.

    Firstly, we prove that A,B are two increasing operators. For y1,y2P with y1y2, we have y1(t)y2(t), t[0,1] and thus Iγy1(t)Iγy2(t). By (H1), Lemma 2.2,

    Ay1(t)=10H(t,s)g(s,Iγy1(s),y1(s))ds10H(t,s)g(s,Iγy2(s),y2(s))ds=Ay2(t).

    Further, noting that φp(t) is increasing in t, we obtain

    By1(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy1(τ),y1(τ))dτ)ds10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy2(τ),y2(τ))dτ)ds=By2(t).

    That is, Ay1Ay2 and By1By2.

    Secondly, we claim that operator A is δconcave and operator B is sub-homogeneous. For any λ(0,1) and yP, from (H2),

    A(λy)(t)=10H(t,s)g(s,Iγ(λy)(s),λy(s))ds=10H(t,s)g(s,λIγy(s),λy(s))dsλδ10H(t,s)g(s,Iγy(s),y(s))ds=λδAy(t),

    that is, A(λy)λδAy for λ(0,1), yP. So operator A is δconcave. Also, for any λ(0,1) and yP, by (H2),

    B(λy)(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγ(λy)(τ),λy(τ))dτ)ds=10H(t,s)φq(10Gβ(s,τ)f(τ,λIγy(τ),λy(τ))dτ)ds10H(t,s)φq(λ1q110Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds=λ10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds=λBy(t),

    that is, B(λy)λBy for λ(0,1), yP. So operator B is sub-homogeneous.

    Thirdly, we show AhPh and BhPh. Let

    m1=d10GA(s)g(s,0,0)ds, m2=e10g(s,1Γ(γ+1),1)ds,
    l1=d(Γ(β))q110GA(s)[sβ1(1s)]q1ds[10τβ1(1τ)f(τ,0,0)dτ]q1,
    l2=e(β1Γ(β))q1[10f(τ,1Γ(γ+1),1)dτ]q1.

    From (H1) and Lemma 2.2,

    Ah(t)=10H(t,s)g(s,Iγh(s),h(s))dse10g(s,Iγ1,1)dstαγ1=e10g(s,tγΓ(γ+1),1)dsh(t)e10g(s,1Γ(γ+1),1)dsh(t)=m2h(t).

    Also,

    Ah(t)=10H(t,s)g(s,Iγh(s),h(s))dsd10GA(s)g(s,Iγ0,0)dstαγ1=d10GA(s)g(s,0,0)dsh(t)=m1h(t).

    By similar discussion, it follows from (H1) and Lemma 2.2 that

    Bh(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγh(τ),h(τ))dτ)ds10etαγ1(10β1Γ(β)sβ1(1s)f(τ,Iγh(τ),h(τ))dτ)q1dse(β1Γ(β))q1(10f(τ,Iγh(τ),h(τ))dτ)q1tαγ1e(β1Γ(β))q1(10f(τ,Iγ1,1)dτ)q1tαγ1=e(β1Γ(β))q1(10f(τ,τγΓ(γ+1),1)dτ)q1h(t)e(β1Γ(β))q1(10f(τ,1Γ(γ+1),1)dτ)q1h(t)=l2h(t)

    and

    Bh(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγh(τ),h(τ))dτ)dsdtαγ110GA(s)(10τβ1(1τ)s(1s)β1Γ(β)f(τ,Iγh(τ),h(τ))dτ)q1ds=d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,Iγh(τ),h(τ))dτ)q1tαγ1d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,Iγ0,0)dτ)q1tαγ1=d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,0,0)dτ)q1h(t)=l1h(t).

    Note that g(t,0,0)0, GA(s)0 and f(τ,1Γ(γ+1))f(τ,0,0), we can easily prove 0<m1m2 and 0<l1l2, and thus m1hAhm2h, l1hBhl2h. So we have Ah,BhPh. It means that the first condition of Lemma 2.3 holds.

    Next we prove that the second condition of Lemma 2.3 is also satisfied. For yP, by (H3),

    By(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds10H(t,s)φq(10β1Γ(β)sq1(1s)f(τ,Iγy(τ),y(τ))dτ)ds(β1Γ(β))q110H(t,s)dsφq(10f(τ,Iγy(τ),y(τ))dτ)(β1Γ(β))q110δ0q1H(t,s)ds=(β1Γ(β))q1δq2010δ0H(t,s)ds(β1Γ(β))q1δq2010H(t,s)g(s,0,0)ds(1Γ(β1))q1δq2010H(t,s)g(s,Iγy(s),y(s))ds=(1Γ(β1))q1δq20Ay(t).

    Let δ0=[Γ(β1)]q1δ2q0, so we obtain Ay(t)δ0By(t), t[0,1]. Therefore, Ayδ0By for yP.

    By the above discussion and Lemma 2.3, we know that operator equation Ay+By=y has a unique solution y in Ph; for any initial value y0Ph, making a sequence yn=Ayn1+Byn1, n=1,2,, we have yny as n. Evidently, z(t):=Iγy(t) is the unique solution of the problem (1.1) in Ω={Iγy(t)|yPh}. And for any initial value y0Ph, the sequences

    yn+1(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγyn(τ),yn(τ))dτ)ds+10H(t,s)g(s,Iγyn(s),yn(s))ds

    and zn(t)=Iγyn(t), n=1,2 satisfy yn(t)y(t) and zn(t)z(t) as n.

    Corollary 3.1. Let 0A<1 and GA(s)0 for s[0,1]. Assume that f satisfies (H1) and for λ(0,1), there exists a constant δ(0,1) such that f(t,λx,λy)λδf(t,x,y) for all t[0,1], x,y[0,+).

    Then there is a unique yPh, where h(t)=tαγ1, t[0,1], such that the following problem

    {Dβt(φp(Dαtz(t))=f(t,z(t),Dγtz(t)), 0<t<1,Dαtz(0)=Dα+1tz(0)=Dγtz(0)=0,Dαtz(1)=0,Dγtz(1)=10Dγtz(s)dA(s), 

    has a unique positive solution z=Iγy in Ω={Iγy(t)|yPh}. And for any initial value y0Ph, making the sequences

    yn+1(t)=10H(t,s)φq(10Gβ(s,τ)g(τ,Iγyn(τ),yn(τ))dτ)ds, n=0,1,2

    and zn(t)=Iγyn(t), n=1,2, we have yn(t)y(t) and zn(t)z(t) as n, where Gβ(s,τ),H(t,s) are given as in (2.2), (2.4) respectively.

    Proof. From Remark 2.1 and Theorem 3.1, the conclusion holds.

    Theorem 3.2. Let 0A<1 and GA(s)0 for s[0,1]. Assume f satisfies (H1) and

    (H4) g:[0,1]×[0,+)×[0,+) is continuous and decreasing with respect to second and third arguments, g(t,1Γ(γ+1),1)0,t[0,1];

    (H5) for λ(0,1), there exist ϕi(λ)(λ,1)(i=1,2) such that

    f(t,λx,λy)ϕ11q1(λ)f(t,x,y), g(t,λx,λy)1ϕ2(λ)g(t,x,y)

    for t[0,1],x,y[0,+).

    Then there is a unique yPh, where h(t)=tαγ1, t[0,1], such that the problem (1.1) has a unique positive solution z(t)=Iγy(t) in set Ω={Iγy(t)|yPh}. And for any initial values x0,y0Ph, putting the sequences

    xn+1(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγxn(τ),xn(τ))dτ)ds+10H(t,s)g(s,Iγyn(s),yn(s))dsyn+1(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγyn(τ),yn(τ))dτ)ds+10H(t,s)g(s,Iγxn(s),xn(s))ds

    and ¯zn(t)=Iγxn(t), zn(t)=Iγyn(t), n=0,1,2, we have xn(t)y(t),yn(t)y(t),¯zn(t)z(t), zn(t)z(t) as n, where Gβ(s,τ),H(t,s) are given as in (2.2), (2.4) respectively.

    Proof. Similar to the proof of Theorem 3.1, we still consider two operators A:PE and B:PE given by

    Ay(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds,
    By(t)=10H(t,s)g(s,Iγy(s),y(s))ds.

    It follows from Lemma 2.2, (H1) and (H4) that A:PP is increasing and B:PP is decreasing.

    Further, from (H5), for λ(0,1),

    A(λy)(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγ(λy)(τ),λy(τ))dτ)ds=10H(t,s)φq(10Gβ(s,τ)f(τ,λIγ(y)(τ),λy(τ))dτ)ds10H(t,s)(10Gβ(s,τ)ϕ11q1(λ)f(τ,Iγy(τ),y(τ))dτ)q1ds=ϕ1(λ)10H(t,s)φq(10Gβ(s,τ)f(τ,Iγy(τ),y(τ))dτ)ds=ϕ1(λ)Ay(t)

    and

    B(λy)(t)=10H(t,s)g(s,Iγ(λy(s)),λy(s))ds=10H(t,s)g(s,λIγ(y(s)),λy(s))ds10H(t,s)1ϕ2(λ)g(s,Iγy(s),y(s))ds=1ϕ2(λ)10H(t,s)g(s,Iγy(s),y(s))ds=1ϕ2(λ)By(t),

    that is, A,B satisfy (2.6). Next, we prove Ah+BhPh. Let

    n1=d10GA(s)g(s,1Γ(γ+1),1)ds, n2=e10g(s,0,0)ds.

    By Lemma 2.2,

    Ah(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγh(τ),h(τ))dτ)ds10etαγ1(10β1Γ(β)sβ1(1s)f(τ,Iγh(τ),h(τ))dτ)q1dse(β1Γ(β))q1(10f(τ,Iγh(τ),h(τ))dτ)q1tαγ1e(β1Γ(β))q1(10f(τ,Iγ1,1)dτ)q1h(t)e(β1Γ(β))q1(10f(τ,1γ+1,1)dτ)q1h(t)=l2h(t),Ah(t)=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγh(τ),h(τ))dτ)dsdtαγ110GA(s)(10τβ1(1τ)s(1s)β1Γ(β)f(τ,Iγh(τ),h(τ))dτ)q1ds=d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,Iγh(τ),h(τ))dτ)q1tαγ1d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,Iγ0,0)dτ)q1h(t)=d(Γ(β))q110GA(s)[sβ1(1s)]q1ds(10τβ1(1τ)f(τ,0,0)dτ)q1h(t)=l1h(t)

    and

    Bh(t)=10H(t,s)g(s,Iγh(s),h(s))dsdtαγ110GA(s)g(s,Iγh(s),h(s))dsdtαγ110GA(s)g(s,Iγ1,1)ds=dtαγ110GA(s)g(s,1Γ(γ+1),1)ds=n1h(t),
    Bh(t)=10H(t,s)g(s,Iγh(s),h(s))dsetαγ110g(s,Iγ0,0)ds=etαγ110g(s,0,0)ds=n2h(t).

    Hence, Ah(t)+Bh(t)l2h(t)+n2h(t)=(l2+n2)h(t) and

    Ah(t)+Bh(t)l1h(t)+n1h(t)=(l1+n1)h(t).

    In addition, it is easy to show l2+n2l1+n1>0. Therefore, Ah+BhPh.

    Consequently, by using Lemma 2.4, operator equation Ay+By=y has a unique solution y in Ph; for given initial values x0, y0Ph, putting the sequences

    xn=Axn1+Byn1,  yn=Ayn1+Bxn1,  n=1,2,

    we have xny, yny as n. Evidently, z(t)=Iγy(t) is the unique solution of the problem (1.1) in Ω={Iγy(t)|yPh}. And for given initial values x0,y0Ph, the following sequences

    xn+1=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγxn(τ),xn(τ))dτ)ds+10H(t,s)g(s,Iγyn(s),yn(s))ds,yn+1=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγyn(τ),yn(τ))dτ)ds+10H(t,s)g(s,Iγyn(s),yn(s))ds

    and ¯zn(t)=Iγxn(t), zn(t)=Iγyn(t), n=0,1,2, satisfy xn(t)y(t),yn(t)y(t),¯zn(t)z(t), zn(t)z(t) as n.

    Corollary 3.2. Let 0A<1 and GA(s)0 for s[0,1]. Assume f satisfies (H1),(H5). Then there is a unique yPh, where h(t)=tαγ1, t[0,1], such that the following problem

    {Dβt(φp(Dαtz))(t)=f(t,z(t),Dγtz(t)), 0<t<1,Dαtz(0)=Dα+1tz(0)=Dγtz(0)=0,Dαtz(1)=0,Dγtz(1)=10Dγtz(s)dA(s), 

    has a unique positive solution z=Iγy in Ω={Iγy(t)|yPh}, where h(t)=tαγ1,t[0,1]. And for any initial value y0Ph, putting the sequences

    yn+1=10H(t,s)φq(10Gβ(s,τ)f(τ,Iγyn(τ),yn(τ))dτ)ds,n=0,1,2,

    and zn(t)=Iγyn(t), n=1,2, we have yn(t)y(t) and zn(t)z(t) as n, where Gβ(s,τ),H(t,s) are given as in (2.2), (2.4) respectively.

    Proof. From Remark 2.1 and Theorem 3.2, the conclusions hold.

    In this section, two examples are given to illustrate our main results.

    Example 4.1. Consider the following 3-Laplacian fractional differential equation with Riemann-Stieltjes integral boundary conditions

    {D94t(φ3(D74tz(t)t14(z14(t)+(D12tz(t))13)3))=cos2t+z13(t)1+z13(t)+(D12tz(t))141+(D12tz(t))14, 0<t<1,D74tz(0)=D114tz(0)=D12tz(0)=0,D74tz(1)=0,D12tz(1)=10D12tz(s)dA(s),  (4.1)

    where α=74, β=94, γ=12, p=3, q=32, A is a function of bounded variation by

    A(t)={0, 0t<14,13, 14t<12,2, 12t1.

    Further, f(t,x,y)=cos2t+x131+x13+y141+y14, g(t,x,y)=t14(x14+y13)+3, clearly, f, gC([0,1]×[0,+)×[0,+), [0,+)),g(t,0,0)0. For fixed t(0,1), f(t,x,y), g(t,x,y) are increasing in x and y. So, the condition (H1) is satisfied.

    In addition, take δ=12, for t[0,1], λ(0,1), x, y[0,+), we have

    g(t,λx,λy)=t14(λ14x14+λ13y13)+3λ12[t14(x14+y13)+3]=λδg(t,x,y).

    On the other hand, for t[0,1], λ(0,1), x,y[0,+),

    f(t,λx,λy)=cos2t+(λx)131+(λx)13+(λy)141+(λy)14λ2cos2t+λ2x131+x13+λ2y141+y14=λ1q1f(t,x,y).

    Hence, the condition (H2) is satisfied.

    Take δ0=3, then

    f(t,x,y)=cos2t+x131+x13+y141+y14δ0=3g(t,0,0).

    The condition (H3) is also satisfied. So Theorem 3.1 shows that the problem (4.1) has a unique positive solution in Ω={Iγy(t)|yPh}, where h(t)=t14, t[0,1].

    Example 4.2. Consider the following 3-Laplacian fractional differential equation with Riemann-Stieltjes integral boundary conditions:

    {D94t(φ3(D74tz(t)[t14(z14(t)+(D12tz(t))13)+1]1=t13[z13(t)+(D12tz(t))14]+2,t(0,1),D74tz(0)=D114tz(0)=D12tz(0)=0,D74tz(1)=0,D12tz(1)=10D12tz(s)dA(s),  (4.2)

    where α=74, β=94, γ=12, p=3, q=32, A is a function of bounded variation by

    A(t) = \left\{ \begin{array}{l} 0, \ 0\leq t \lt \frac{1}{4}, \\ \frac{1}{3}, \ \frac{1}{4}\leq t \lt \frac{1}{2}, \\2, \ \frac{1}{2}\leq t\leq1. \end{array}\right.

    Let f(t, x, y) = t^{\frac13}(x^{\frac13}+y^{\frac14})+2, \ g(t, x, y) = [t^{\frac14}(x^{\frac14}+y^{\frac13})+1]^{-1} , clearly, f, \ g \in C([0, 1)\times [0, +\infty) \times [0, +\infty), \ [0, +\infty)) , f(t, 0, 0)\not\equiv 0 , g(t, \frac{1}{\Gamma(\gamma+1)}, 1)\not\equiv 0 . For fixed t\in [0, 1), \ f(t, x, y) is increasing in x and y , g(t, x, y) is decreasing in x and y . So, the conditions (H_4) and (H_5) are satisfied. Take \phi_1(\lambda) = \lambda^{\frac16} , \phi_2(\lambda) = \lambda^{\frac13} , then \phi_1(\lambda), \ \phi_2(\lambda) \in(\lambda, 1) for \lambda\in (0, 1). Thus,

    f(t, \lambda x, \lambda y) = t^{\frac13}(\lambda^{\frac13}x^{\frac13}+\lambda^{\frac14}y^{\frac14})+2\geq {\lambda^\frac13} [t^{\frac13}(x^{\frac13}+y^{\frac14})+2] = {(\phi_1(\lambda))}^{\frac{1}{q-1}} f(t, x, y),
    g(t, \lambda x, \lambda y) = {[t^{\frac{1}{4}}(\lambda^{\frac{1}{4}}x^{\frac{1}{4}}+\lambda^{\frac{1}{3}}y^{\frac{1}{3}})+1]}^{-1}\geq {\lambda^{-\frac13 } [t^{\frac{1}{4}}(x^{\frac{1}{4}}+y^{\frac{1}{3}})+1]}^{-1} = \frac{1}{\phi_2(\lambda)} g(t, x, y).

    So Theorem 3.2 implies that the problem (4.2) has a unique positive solution in \Omega = \{I^\gamma y(t)|y\in P_h\} , where h(t) = t^{\frac{1}{4}}, \ t\in [0, 1] .

    Integral boundary value problems have many applications in applied fields such as blood flow problems, chemical engineering, underground water flow and population dynamics. For nonlinear fractional differential equations with p -Laplacian operator subject to different boundary conditions, there are many works reported on the existence or multiplicity of positive solutions. But the unique results are very rare. In this paper, we study a p -Laplacian fractional order differential equation involving Riemann-Stieltjes integral boundary condition (1.1). By means of the properties of Green's function and two fixed point theorems of a sum operator in partial ordering Banach spaces, we establish some new existence and uniqueness criteria for (1.1). Our result shows that the unique positive solution exists in a special set P_h and can be approximated by constructing an iterative sequence for any initial point in P_h . Finally, two interesting examples are given to illustrate the application of our main results.

    This paper was supported by the Youth Science Foundation of China (11801322), Shanxi Province Science Foundation (201901D111020) and Graduate Science and Technology Innovation Project of Shanxi (2019BY014).

    All authors declare no conflicts of interest in this paper.



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