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Research article

Large-time behavior of cylindrically symmetric Navier-Stokes equations with temperature-dependent viscosity and heat conductivity

  • Received: 27 November 2023 Revised: 09 January 2024 Accepted: 17 July 2024 Published: 21 August 2024
  • 35Q35, 76N10

  • In this study, the initial-boundary value problem for cylindrically symmetric Navier-Stokes equations was considered with temperature-dependent viscosity and heat conductivity. Firstly, we established the existence and uniqueness of a strong solution when the viscosity and heat conductivity were both power functions of temperature. Moreover, the large-time behavior of the strong solution was obtained with large initial data, since all of the estimates in this paper were independent of time. It is worth noting that we identified the relationship between the initial data and the power of the temperature in the viscosity for the first time.

    Citation: Dandan Song, Xiaokui Zhao. Large-time behavior of cylindrically symmetric Navier-Stokes equations with temperature-dependent viscosity and heat conductivity[J]. Communications in Analysis and Mechanics, 2024, 16(3): 599-632. doi: 10.3934/cam.2024028

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  • In this study, the initial-boundary value problem for cylindrically symmetric Navier-Stokes equations was considered with temperature-dependent viscosity and heat conductivity. Firstly, we established the existence and uniqueness of a strong solution when the viscosity and heat conductivity were both power functions of temperature. Moreover, the large-time behavior of the strong solution was obtained with large initial data, since all of the estimates in this paper were independent of time. It is worth noting that we identified the relationship between the initial data and the power of the temperature in the viscosity for the first time.



    The models of the differential and integral equations have been appeared in different applications (see [1,2,3,4,6,7,9,12,13,14,15,16,17,18,19,20]).

    Boundary value problems involving fractional differential equations arise in physical sciences and applied mathematics. In some of these problems, subsidiary conditions are imposed locally. In some other cases, nonlocal conditions are imposed. It is sometimes better to impose nonlocal conditions since the measurements needed by a nonlocal condition may be more precise than the measurement given by a local condition. Consequently, a variety of excellent results on fractional boundary value problems (abbreviated BVPs) with resonant conditions have been achieved. For instance, Bai [4] studied a type of fractional differential equations with m-points boundary conditions. The existence of nontrivial solutions was established by using coincidence degree theory. Applying the same method, Kosmatov [17] investigated the fractional order three points BVP with resonant case.

    Although the study of fractional BVPs at resonance has acquired fruitful achievements, it should be noted that such problems with Riemann-Stieltjes integrals are very scarce, so it is worthy of further explorations. Riemann-Stieltjes integral has been considered as both multipoint and integral in a single framework, which is more common, see the relevant works due to Ahmad et al. [1].

    The boundary value problems with nonlocal, integral and infinite points boundary conditions have been studied by some authors (see, for example [8,10,11,12]).

    Here, we discuss the boundary value problem of the nonlinear differential inclusions of arbitrary (fractional) orders

    dxdtF1(t,x(t),Iγf2(t,Dαx(t))),α,γ(0,1]t(0,1) (1.1)

    with the nonlocal boundary condition

    mk=1akx(τk)=x0,ak>0τk[0,1], (1.2)

    the integral condition

    10x(s)dg(s)=x0 (1.3)

    and the infinite point boundary condition

    k=1akx(τk)=x0,ak>0andτk[0,1]. (1.4)

    We study the existence of solutions xC[0,1] of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). Then the existence of the maximal and minimal solutions will be proved. The sufficient condition for the uniqueness and continuous dependence of the solution will be studied.

    This paper is organised as: In Section 2, we prove the existence of continuous solutions of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). In Section 3, the existence of the maximal and minimal solutions is proved. In Section 4, the sufficient condition for the uniqueness and continuous dependence of the solution are studied. Next, in Section 5, we extend our results to the nonlocal problems (1.3) and (2.1). Finally, some existence results is proved for the nonlocal problems (1.4) and (2.1) in Section 6.

    Consider the following assumptions:

    (I) (i) The set F1(t,x,y) is nonempty, closed and convex for all (t,x,y)[0,1]×R×R.

    (ii) F1(t,x,y) is measurable in t[0,1] for every x,yR.

    (iii) F1(t,x,y) is upper semicontinuous in x and y for every t[0,1].

    (iv) There exist a bounded measurable function a1:[0,1]R and a positive constant K1, such that

    F1(t,x,y)=sup{|f1|:f1F1(t,x,y)}|a1(t)|+K1(|x|+|y|).

    Remark 2.1. From the assumptions (i)–(iv) we can deduce that (see [3,6,7,13]) there exists f1F1(t,x,y), such that

    (v) f1:[0,1]×R×RR is measurable in t for every x,yR and continuous in x,y for t[0,1], and there exist a bounded measurable function a1:[0,1]R and a positive constant K1>0 such that

    |f1(t,x,y)||a1(t)|+K1(|x|+|y|),

    and the functional f1 satisfies the differential equation

    dxdt=f1(t,x(t),Iγf2(t,Dαx(t))),α,γ(0,1]andt(0,1]. (2.1)

    (II) f2:[0,1]×RR is measurable in t for any xR and continuous in x for t[0,1], and there exist a bounded measurable function a2:[0,1]R and a positive constant K2>0 such that

    |f2(t,x)||a2(t)|+K2|x|,t[0,1]andxR

    and

    supt[0,1]|ai(t)|ai,i=1,2.

    (III) 2K1γ+K1K2α<αγΓ(2α),α,γ(0,1].

    Remark 2.2. From (I) and (v) we can deduce that every solution of (1.1) is a solution of (2.1). Now, we shall prove the following lemma.

    Lemma 2.1. If the solution of the problems (1.2)–(2.1) exists then it can be expressed by the integral equation

    y(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds. (2.2)

    Proof. Consider the boundary value problems (1.2)–(2.1) be satisfied. Operating by I1α on both sides of (2.1) we can obtain

    Dαx(t)=I1αdxdt=I1αf1(t,x(t),Iγf2(t,Dαx(t))). (2.3)

    Taking

    Dαx(t)=y(t), (2.4)

    then we obtain

    x(t)=x(0)+Iαy(t). (2.5)

    Putting t=τ and multiplying (2.5) by Σmk=1ak, then we get

    Σmk=1akx(τk)=Σmk=1akx(0)+Σmk=1akIαy(τk), (2.6)
    x0=Σmk=1akx(0)+Σmk=1akIαy(τk) (2.7)

    and

    x(0)=1Σmk=1ak[x0Σmk=1akIαy(τk)]. (2.8)

    Then

    x(t)=1Σmk=1ak[x0Σmk=1akIαy(τk)]+Iαy(t). (2.9)

    Substituting (2.8) and (2.9) in (2.5), which completes the proof.

    Theorem 2.1. Let assumptions (I)–(III) be satisfied. Then the integral equation (2.2) has at least one continuous solution.

    Proof. Define a set Qr as

    Qr={yC[0,1]:yr},
    r=Γ(γ+1)[a1+K1A|x0|]+k1a2αγΓ(2α)[2K1α+K1K2γ],(mk=1ak)1=A,

    and the operator F by

    Fy(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds.

    For yQr, then

    |Fy(t)|=|t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds|K1t0(ts)αΓ(1α)[|x0|mk=1ak+mk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθmk=1ak+s0(sθ)α1Γ(α)|y(θ)|dθ+s0(sϕ)γ1Γ(γ)(K2|y(θ)|+|a2(t)|)dθ]+|a1(t)|)ds1Γ(2α)[K1A|x0|+K1yΓ(α+1)+K1yΓ(α+1)+K1K2yΓ(γ+1)+K1a2Γ(γ+1)+a1]1Γ(2α)[K1A|x0|+K1rΓ(α+1)+K1rΓ(α+1)+K1K2rΓ(γ+1)+K1a2Γ(γ+1)+a1]1Γ(2α)[K1A|x0|+2K1rΓ(α+1)+K1K2rΓ(γ+1)+K1a2Γ(γ+1)+a1]r.

    Thus, the class of functions {Fy} is uniformly bounded on Qr and F:QrQr. Let yQr and t1,t2[0,1] such that |t2t1|<δ, then

    |Fy(t2)Fy(t1)|=|t20(t2s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))dst10(t1s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))ds||t20(t2s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))dst10(t2s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))ds+t10(t2s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))dst10(t1s)αΓ(1α)(f1(s,x(s),Iγf2(s,y(s)))ds|t2t1(t2s)αΓ(1α)|(f1(s,x(s),Iγf2(s,y(s)))|ds+t10(t2s)α(t1s)αΓ(1α)|(f1(s,x(s),Iγf2(s,y(s)))ds|t2t1(t2s)αΓ(1α)|(f1(s,x(s),Iγf2(s,y(s)))|ds+t10(t2s)α(t1s)αΓ(1α)(t2s)α(t1s)α.|(f1(s,x(s),Iγf2(s,y(s)))|ds.

    Thus, the class of functions {Fy} is equicontinuous on Qr and {Fy} is compact operator by the Arzela-Ascoli Theorem [5].

    Now we prove that F is continuous operator. Let ynQr be convergent sequence such that yny, then

    Fyn(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yn(θ)dθ]+s0(sθ)α1Γ(α)yn(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,yn(θ)))ds.

    Using Lebesgue dominated convergence Theorem [5] and assumptions (iv)–(II) we have

    limnFyn(t)=limnt0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yn(θ)dθ]+s0(sθ)α1Γ(α)yn(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,yn(θ)))ds=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)limnyn(θ)dθ]+s0(sθ)α1Γ(α)limnyn(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,limnyn(θ)))ds=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ)))ds=Fy(t).

    Then F:QrQr is continuous, and by Schauder Fixed Point Theorem[5] there exists at least one solution yC[0,1] of (2.2). Now

    dxdt=ddt[x(0)+Iαy(t)]=ddtIαI1αf1(t,x(t),Iγf2(t,y(t)))=ddtIf1(t,x(t),Iγf2(t,y(t)))=f1(t,x(t),Iγf2(t,y(t))).

    Putting t=τ and using (2.9), we obtain

    x(t)=1Σmk=1ak[x0Σmk=1akIαy(τk)]+Iαy(t),Σmk=1akx(τk)=Σmk=1ak1Σmk=1ak[x0Σmk=1akIαy(τk)]+Σmk=1akIαy(τk),

    then

    Σmk=1akx(τk)=x0,ak>0andτk[0,1].

    This proves the equivalence between the problems (1.2)–(2.1) and the integral equation (2.2). Then there exists at least one solution yC[0,1] of the problems (1.2)–(2.1).

    Here, we shall study the maximal and minimal solutions for the problems (1.2) and (2.1). Let y(t) be any solution of (2.2), let u(t) be a solution of (2.2), then u(t) is said to be a maximal solution of (2.2) if it satisfies the inequality

    y(t)u(t),t[0,1].

    A minimal solution can be defined by similar way by reversing the above inequality.

    Lemma 3.1. Let the assumptions of Theorem 2.1 be satisfied. Assume that x(t) and y(t) are two continuous functions on [0,1] satisfying

    y(t)t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ)))dst[0,1],x(t)t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)x(θ)dθ]+s0(sθ)α1Γ(α)x(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,x(θ)))dst[0,1],

    where one of them is strict.

    Let functions f1 and f2 be monotonic nondecreasing in y, then

    y(t)<x(t),t>0. (3.1)

    Proof. Let the conclusion (3.1) be not true, then there exists t1 with

    y(t1)<x(t1),t1>0andy(t)<x(t),0<t<t1.

    Since f1 and f2 are monotonic functions in y, then we have

    y(t1)t10(t1s)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ)))ds<t10(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)x(θ)dθ]+s0(sθ)α1Γ(α)x(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,x(θ)))ds<x(t1),t1[0,1].

    This contradicts the fact that y(t1)=x(t1), then y(t)<x(t). This completes the proof.

    For the existence of the continuous maximal and minimal solutions for (2.1), we have the following theorem.

    Theorem 3.1. Let the assumptions of Theorem 2.1 be hold. Moreover, if f1 and f2 are monotonic nondecreasing functions in y for each t[0,1], then Eq (2.1) has maximal and minimal solutions.

    Proof. First, we should demonstrate the existence of the maximal solution of (2.1). Let ϵ>0 be given. Now consider the integral equation

    yϵ(t)=t0(ts)αΓ(1α)f1,ϵ(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ(θ)dθ]+s0(sθ)α1Γ(α)yϵ(θ)dθ,s0(sθ)γ1Γ(γ)f2,ϵ(θ,yϵ(θ))dθ)ds,t[0,1],

    where

    f1,ϵ(s,xϵ(s),yϵ(s))=f1(s,xϵ(s),yϵ(s))+ϵ,
    f2,ϵ(s,xϵ(s))=f2(s,xϵ(s))+ϵ.

    For ϵ2>ϵ1, we have

    yϵ2(t)=t0(ts)αΓ(1α)f1,ϵ2(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ2(θ)dθ]+s0(sθ)α1Γ(α)yϵ2(θ)dθ,s0(sθ)γ1Γ(γ)f2,ϵ2(θ,yϵ2(θ))dθ)ds,t[0,1],
    yϵ2(t)=t0(ts)αΓ(1α)[f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ2(θ)dθ]+s0(sθ)α1Γ(α)yϵ2(θ)dθ,s0(sθ)γ1Γ(γ)(f2(θ,yϵ2(θ))+ϵ2)dθ)+ϵ2]ds,t[0,1].

    Also

    yϵ1(t)=t0(ts)αΓ(1α)f1,ϵ1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ1(θ)dθ]+s0(sθ)α1Γ(α)yϵ1(θ)dθ,s0(sθ)γ1Γ(γ)f2,ϵ1(θ,yϵ1(θ))dθ)ds,yϵ1(t)=t0(ts)αΓ(1α)[f1(s,1mk=1ak[x0mk=1akτk0(τkϕ)α1Γ(α)yϵ1(θ)dθ]+s0(sθ)α1Γ(α)yϵ1(θ)dθ,s0(sθ)γ1Γ(γ)(f2(θ,yϵ1(θ))+ϵ1)dθ)+ϵ1]ds>t0(ts)αΓ(1α)[f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ2(θ)dθ]+s0(sθ)α1Γ(α)yϵ2(θ)dθ,s0(sθ)γ1Γ(γ)(f2(θ,yϵ2(τ))+ϵ2)dθ)+ϵ2]ds.

    Applying Lemma 3.1, we obtain

    yϵ2<yϵ1,t[0,1].

    As shown before, the family of function yϵ(t) is equi-continuous and uniformly bounded, then by Arzela Theorem, there exist decreasing sequence ϵn, such that ϵn0 as n, and u(t)=limnyϵn(t) exists uniformly in [0,1] and denote this limit by u(t). From the continuity of the functions, f2,ϵn(t,yϵn(t)), we get f2,ϵn(t,yϵn(t))f2(t,y(t)) as n and

    u(t)=limnyϵn(t)=t0(ts)αΓ(1α)[f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵn(θ)dθ]+s0(sθ)α1Γ(α)yϵn(θ)dθ,s0(sθ)γ1Γ(γ)(f2(θ,yϵn(θ))+ϵn))+ϵn]ds,t[0,1].

    Now we prove that u(t) is the maximal solution of (2.1). To do this, let y(t) be any solution of (2.1), then

    y(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds,yϵ(t)=t0(ts)αΓ(1α)[f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ(θ)dθ]+s0(sθ)α1Γ(α)yϵ(θ)dθ,s0(sθ)γ1Γ(γ)(f2(θ,yϵ(θ))+ϵ)dθ)+ϵ]ds

    and

    yϵ(t)>t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)yϵ(θ)dθ]+s0(sθ)α1Γ(α)yϵ(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,yϵ(θ)))ds.

    Applying Lemma 3.1, we obtain

    y(t)<yϵ(t),t[0,1].

    From the uniqueness of the maximal solution it clear that yϵ(t) tends to u(t) uniformly in [0,1] as ϵ0.

    By a similar way as done above, we can prove the existence of the minimal solution.

    Here, we study the sufficient condition for the uniqueness of the solution yC[0,1] of problems (1.2) and (2.1). Consider the following assumptions:

    (I) (i) The set F1(t,x,y) is nonempty, closed and convex for all (t,x,y)[0,1]×R×R.

    (ii) F1(t,x,y) is measurable in t[0,1] for every x,yR.

    (iii) F1 satisfies the Lipschitz condition with a positive constant K1 such that

    H(F1(t,x1,y1),F1(t,x2,y2)|K1(|x1x2|+|y1y2|),

    where H(A,B) is the Hausdorff metric between the two subsets A,B[0,1]×E.

    Remark 4.1. From this assumptions we can deduce that there exists a function f1F1(t,x,y), such that

    (iv) f1:[0,1]×R×RR is measurable in t[0,1] for every x,yR and satisfies Lipschitz condition with a positive constant K1 such that (see [3,7])

    |f1(t,x1,y1)f1(t,x2,y2)|K1(|x1x2|+|y1y2|).

    (II)f2:[0,1]×RR is measurable in t[0,T] and satisfies Lipschitz condition with positive constant K2, such that

    |f2(t,x)f2(t,y)|K2|xy|.

    From the assumption (I), we have

    |f1(t,x,y)||f1(t,0,0)||f1(t,x,y)f1(t,0,0)|K1(|x|+|y|).

    Then

    |f1(t,x,y)|K1(|x|+|y|)+|f1(t,0,0)|K1(|x|+|y|)+a1(t),

    where |a1(t)|=suptI|f1(t,0,0)|.

    From the assumption (II), we have

    |f2(t,y)||f2(t,0)||f2(t,y)f2(t,0)|K2|y|.

    Then

    |f2(t,x)|K2|x|+|f2(t,0)|K2|x|+a2(t),

    where |a2(t)|=suptI|f1(t,0)|.

    Theorem 4.1. Let the assumptions (I) and (II) be satisfied. Then the solution of the problems (1.2) and (2.1) is unique.

    Proof. Let y1(t) and y2(t) be solutions of the problems (1.2) and (2.1), then

    Then

    y1y2[αγΓ(2α)(2K1α+K1K2γ)]<0.

    Since (αγΓ(2α)(2K1α+K1K2γ))<1, then y1(t)=y2(t) and the solution of (1.2) and (2.1) is unique.

    Definition 4.1. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data x0, if ϵ>0, δ>0, such that

    |x0x0|δyyϵ,

    where y is the unique solution of the integral equation

    y(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds|.

    Theorem 4.2. Let the assumptions (I) and (II) be satisfied, then the unique solution of (1.2) and (2.1) depends continuously on x0.

    Proof. Let y(t) and y(t) be the solutions of problems (1.2) and (2.1), then

    then we obtain

    |y(t)y(t)|(AK1δ)(αγΓ(2α)(2K1α+K1K2γ))1ϵ

    and

    y(t)y(t)ϵ.

    Definition 4.2. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data ak, if ϵ>0, δ>0, such that

    mk=1|akak|δyyϵ,

    where y is the unique solution of the integral equation

    y(t)=t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds|.

    Theorem 4.3. Let the assumptions (I) and (II) be satisfied, then the unique solution of problems (1.2) and (2.1) depends continuously on ak.

    Proof. Let y(t) and y(t) be the solutions of problems (1.2) and (2.1) and (mk=1ak)1=A,

    |y(t)y(t)|=|t0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)dst0(ts)αΓ(1α)f1(s,1mk=1ak[x0mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(θ,y(θ))dθ)ds|t0(ts)αΓ(1α)[|x0|(mk=1akmk=1ak)mk=1akmk=1ak+mk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθmk=1akmk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθmk=1ak+K1s0(sθ)α1Γ(α)|y(θ)y(θ)|dθ+K1K2s0(sθ)γ1Γ(γ)|y(ϕ)y(θ)|dθ]ds1Γ(2α)[K1|x0|mk=1|akak|mk=1akmk=1ak+K1mk=1ak(mk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθmk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθ)mk=1akmk=1ak+K1mk=1ak(mk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθmk=1akτk0(τkθ)α1Γ(α)|y(θ)|dθ)mk=1akmk=1ak+K1s0(sθ)α1Γ(α)|y(θ)y(θ)|dθ+K1K2s0(sθ)γ1Γ(γ)|y(ϕ)y(θ)|dθ]ds1Γ(2α)[K1|x0|δmk=1akmk=1ak+K1(mk=1akδrΓ(α+1)+mk=1akδrΓ(α+1))mk=1akmk=1ak+K1yyΓ(α+1)+K1K2yyΓ(γ+1)K1δ|x0|Γ(2α)mk=1akmk=1ak+K1(mk=1akδrΓ(α+1)+mk=1akδrΓ(α+1))Γ(2α)Γ(α+1)mk=1akmk=1ak+K1yyΓ(2α)Γ(α+1)+K1K2yyΓ(γ+1)Γ(2α),

    then we obtain

    |y(t)y(t)|(AAK1δ[|x0|+r(mk=1ak+mk=1ak)]).(αγΓ(2α)(2K1α+K1K2γ))1ϵ

    and

    y(t)y(t)ϵ.

    Let yC[0,1] be the solution of the nonlocal boundary value problems (1.2) and (2.1). Let ak=(g(tk)g(tk1), g is increasing function, τk(tk1tk), 0=t0<t1<t2,...<tm=1, then, as m the nonlocal condition (1.2) will be

    mk=1(g(tk)g(tk1)y(τk)=x0.

    As the limit m, we obtain

    limmmk=1(g(tk)g(tk1)y(τk)=10y(s)dg(s)=x0.

    Theorem 5.1. Let the assumptions (I)–(III) be satisfied. If mk=1ak be convergent, then the nonlocal boundary value problems of (1.3) and (2.1) have at least one solution given by

    y(t)=t0(ts)αΓ(1α)f1(s,1g(1)g(0)[x010s0(sθ)α1Γ(α)y(θ)dθdg(s)]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds.

    Proof. As m, the solution of the nonlocal boundary value problem (2.1) will be

    limmy(t)=t0(ts)αΓ(1α)f1(s,(g(1)g(0))1x0(g(1)g(0))1.limmmk=1[g(tk)g(tk1]τk0(τkθ)α1Γ(α)y(θ)dθ+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds=t0(ts)αΓ(1α)f1(s,1g(1)g(0)[x010s0(sθ)α1Γ(α)y(θ)dθdg(s)]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds.

    Theorem 6.1. Let the assumptions (I)–(III) be satisfied, then the nonlocal boundary value problems of (1.4) and (2.1) have at least one solution given by

    y(t)=t0(ts)αΓ(1α)f1(s,1k=1ak[x0k=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds.

    Proof. Let the assumptions of Theorem 2.1 be satisfied. Let mk=1ak be convergent, then take the limit to (1.4), we have

    limmy(t)=t0(ts)αΓ(1α)f1(s,limmm1mk=1ak[x0limmmk=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds.

    Now

    |akτk0(τkθ)α1Γ(α)y(θ)dθ||ak|τk0(τkθ)α1Γ(α)|y(θ)|dθ|ak|yΓ(α+1)|ak|rΓ(α+1)

    and by the comparison test (mk=1akτk0(τkθ)α1Γ(α)y(θ)dθ) is convergent,

    y(t)=t0(ts)αΓ(1α)f1(s,1k=1ak[x0k=1akτk0(τkθ)α1Γ(α)y(θ)dθ]+s0(sθ)α1Γ(α)y(θ)dθ,s0(sθ)γ1Γ(γ)f2(ϕ,y(θ))dθ)ds.

    Furthermore, from (2.9) we have

    k=1akx(τk)=k=1ak[1k=1ak(x0k=1akτk0(τks)α1Γ(α)y(s)ds)]+τk0(τks)α1Γ(α)y(s)ds)]=x0.

    Example 6.1. Consider the following nonlinear integro-differential equation

    dxdt=t4et+x(t)t+2+13Iγ(cos(5t+1)+19[t5sinD13x(t)+e3tx(t)]) (6.1)

    with boundary condition

    mk=1[1k1k+1]x(τk)=x0,ak>0τk[0,1]. (6.2)

    Let

    f1(t,x(t),Iγf2(t,Dαx(t)))=t4et+x(t)t+2+13Iγ(cos(5t+1)+19[t5sinD13x(t)+e3tx(t)]),

    then

    |f1(t,x(t),Iγf2(t,Dαx(t)))=|t4et+x(t)2t+4+14(43Iγ(cos(5t+1)+19[t5sinD13x(t)+e3tx(t)]))|

    and also

    |Iγf2(t,Dαx(t))|14(43Iγ|cos(5t+1)+13[t5sinD13x(t)+e3tx(t)]|.

    It is clear that the assumptions (I) and (II) of Theorem 2.1 are satisfied with a1(t)=t4etL1[0,1], a2(t)=43Iγ|(cos(5t+1)|L1[0,1], and let α=13, γ=23, then 2K1γ+K1K2α=12<αγΓ(2α)=12Γ(2α). Therefore, by applying Theorem 2.1, the nonlocal problems (6.1) and (6.2) has a continuous solution.

    In this paper, we have studied a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. We have prove some existence results for that a single nonlocal boundary value problem, in of proving some existence results for a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. Next, we have proved the existence of maximal and minimal solutions. Then we have established the sufficient conditions for the uniqueness of solutions and continuous dependence of solution on some initial data and on the coefficients ak are studied. Finally, we have proved the existence of a nonlocal boundary value problem with Riemann-Stieltjes integral condition and with infinite-point boundary condition. An example is given to illustrate our results.

    The authors declare no conflict of interest.



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