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New sharp estimates of the interval length of the uniqueness results for several two-point fractional boundary value problems

  • This paper investigates the existence and uniqueness of solutions for several two-point fractional BVPs, including hybrid fractional BVP, sequential fractional BVP and so on. Using the Banach contraction mapping theorem, some sharp conditions that depend on the length of the given interval are presented, which ensure the uniqueness of solutions for the considered BVPs. Illustrative examples are also constructed. The results obtained in this study are noteworthy extensions of earlier works.

    Citation: Wei Zhang, Jinbo Ni. New sharp estimates of the interval length of the uniqueness results for several two-point fractional boundary value problems[J]. Electronic Research Archive, 2023, 31(3): 1253-1270. doi: 10.3934/era.2023064

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  • This paper investigates the existence and uniqueness of solutions for several two-point fractional BVPs, including hybrid fractional BVP, sequential fractional BVP and so on. Using the Banach contraction mapping theorem, some sharp conditions that depend on the length of the given interval are presented, which ensure the uniqueness of solutions for the considered BVPs. Illustrative examples are also constructed. The results obtained in this study are noteworthy extensions of earlier works.



    Bailey, Shampine and Waltman analyzed the following classical two-point boundary value problems (BVPs) [1]:

    {y(t)+f(t,y(t))=0,t(a,b),y(a)=A,y(b)=B,and (1.1)
    {y(t)+f(t,y(t),y(t))=0,t(a,b),y(a)=A,y(b)=B, (1.2)

    where A,BR. The authors presented the following results.

    Theorem 1.1. [1] Let f(t,y) be continuous on [a,b]×R and satisfy Lipschitz condition with Lipschitz constant K,

    |f(t,y)f(t,x)|K,

    for all (t,y),(t,x)[a,b]×R. Then BVP (1.1) has a unique solution whenever

    ba<22K. (1.3)

    Theorem 1.2. [1] Let f(t,y,y) be continuous on [a,b]×R2 and satisfy Lipschitz condition with Lipschitz constants K and L,

    |f(t,y,y)f(t,x,x)|K|yx|+L|yx|,

    for all (t,y,y),(t,x,x)[a,b]×R2. Then BVP (1.2) has a unique solution if

    K(ba)28+L(ba)2<1. (1.4)

    Fractional calculus has risen in many fields of science and engineering over the past few decades. Numerous problems in physics, chemistry, biology, economics, signal and image processing, fluid dynamics, economics and control theory can be modeled in the form of fractional models, especially to describe processes with memory effects [2,3,4,5,6]. Consequently, solving fractional BVPs has always received considerable attention, and various interesting results dealing with the existence and uniqueness results for fractional differential equations involving a variety of boundary conditions have been established [7,8,9,10,11].

    Recently, several scholars have proposed to extend the result of Theorem 1.1 by considering a fractional derivative. Additionally, based on different definitions of the fractional derivative, inequality (1.3) has been generalized to various forms. Examples include the Riemann-Liouville derivative [12], the Caputo fractional derivative [13], the Conformable fractional derivative [14], and the Hadamard fractional derivative [15]. For example, Ferreira [12] extended the result of Theorem 1.1 by using the Riemann-Liouville fractional derivative, that is, the following two-point fractional BVP was studied:

    {Dαa+y(t)+f(t,y(t))=0,a<t<b,y(a)=0,y(b)=B, (1.5)

    where 1<α2, Dαa+ is the Riemann-Liouville fractional derivative of order α. The following result was obtained.

    Theorem 1.3. [12] Let f:[a,b]×RR be continuous and satisfy the Lipschitz condition on [a,b]×R with Lipschitz constant K,

    |f(t,x)f(t,y)|K|xy|,

    for all (t,x),(t,y)[a,b]×R. Then BVP (1.5) has a unique solution if

    ba<(Γ(α))1/αα(α+1)/αK1/α(α1)(α1)/α.

    Laadjal, Abdeljawad and Jarad [14] extended the result of Theorem 1.1 involving a conformable fractional derivative. The author investigated the following two-point fractional BVP:

    {Taβu(t)+f(t,u(t))=0,t(a,b),u(a)=A,u(b)=B,A,BR, (1.6)

    where 1<β2 and Taβ is the conformable fractional derivative of order β. The following result was obtained.

    Theorem 1.4. [14] Let f:[a,b]×RR be continuous and satisfy the Lipschitz condition on [a,b]×R with Lipschitz constant K,

    |f(t,x)f(t,y)|K|xy|,

    for all (t,x),(t,y)[a,b]×R. Then BVP (1.6) has a unique solution if

    ba<β(2β1)/(2β1)β(β1)β(β1)K1/1ββ.

    However, no result exists in the literature that extends Theorem 1.2 to fractional differential equations. The main objective of this study is to bridge this gap. To this end, inspired by the above literature, the following fractional BVPs are considered.

    Motivated by the above-mentioned works, the sharp estimate for the unique solution of the following two-point hybrid fractional BVP is investigated:

    {Dαa+[x(t)f(t,x(t))]+g(t,x(t))=0,t(a,b),x(a)=0,x(b)=B, (1.7)

    where 1<α2, Dαa+ is the Riemann-Liouville fractional derivative of order α, BR is a constant, f:[a,b]×RR{0} and g:[a,b]×RR are two continuous functions.

    As a second problem, inspired by the above ideas and by [1], this paper aims to extend Theorem 1.2 by considering a fractional derivative. The sharp estimate for the unique solution of the following two-point fractional BVPs were studied with a non-linear term depending on the fractional derivative given by

    {Dαa+y(t)+f(t,(ta)2αy(t),Dα1a+y(t))=0,t(a,b),limta+(ta)2αy(t)=A,y(b)=B, (1.8)

    and the sequential fractional BVP

    {Dβa+CDγa+z(t)+g(t,z(t),(ta)1βCDγa+z(t))=0,t(a,b),z(a)=A,z(b)=B, (1.9)

    where 1<α2, 0<γ,β1, 1<γ+β2, Dκa+ is the Riemann-Liouville fractional derivative of order κ=α,β, CDγa+ is the Caputo fractional derivative of order γ; A,BR are two constants, and f,g:[a,b]×R2R are two continuous functions.

    The following assumptions were considered throughout the present analysis:

    (A1) There exist Lipschitz constants L1,L2, such that, for all (t,xi)[a,b]×R, (i=1,2),

    |f(t,x1)f(t,x2)|L1|x1x2|,|g(t,x1)g(t,x2)|L2|x1x2|.

    (A2) There exist Lipschitz constants K,L, such that, for any (t,ui,vi)[a,b]×R2, (i=1,2),

    |f(t,u1,v1)f(t,u2,v2)|K|u1u2|+L|v1v2|.

    (A3) There exist Lipschitz constants P,Q, such that, for any (t,wi,zi)[a,b]×R2, (i=1,2),

    |g(t,w1,z1)g(t,w2,z2)|P|w1w2|+Q|z1z2|.

    Based on the above interpretation, the contribution of this work is summarized as follows:

    A new condition in terms of the end points of the given interval is presented, which ensures the uniqueness of the solution for a two-point hybrid fractional BVP and generalizes the result of Theorem 1.3.

    The sharp estimate for the unique solution of the two-point fractional BVPs with a non-linear term depending on a lower fractional order derivative is established, which extends the classical integer order results of Theorem 1.2.

    The sequential two-point fractional BVP is considered, providing a means to solve the open problem (30) proposed in [14].

    The rest of the paper is organized as follows. In Section 2, some basic results related to the fractional calculus are given. In Section 3, by using the Banach contraction mapping theorem the estimate for the uniqueness results of the two-point fractional BVPs (1.7)–(1.9) are investigated. In Section 4, we present some examples which illustrate the efficiency of the main results. Finally, Section 5 addresses the conclusion of the work.

    In this section, we recall some basic definitions and lemmas on fractional calculus, we refer the reader to [2].

    Definition 2.1. [2] The Riemann-Liouville fractional integral of order α>0 for a function u:[a,b]R is defined as

    Iαa+u(t)=1Γ(α)ta(ts)α1u(s)ds,t[a,b].

    Definition 2.2. [2] The Riemann-Liouville fractional derivative of order α>0 for a function u:[a,b]R is defined as

    Dαa+u(t)=1Γ(nα)(ddt)nta(ts)nα1u(s)ds,t[a,b],n1<α<n,n=[α]+1,

    where [α] denotes the integer part of the real number α.

    Definition 2.3. [2] The Caputo fractional derivative of order α>0 for a (n1)-times absolutely continuous function u:[a,b]R is defined as

    CDαa+u(t)=1Γ(nα)ta(ts)nα1u(n)(s)ds,t[a,b],n1<α<n,n=[α]+1,

    where [α] denotes the integer part of the real number α.

    Lemma 2.1. [2,7] Let α>0. If uC(a,b)L1(a,b), then

    Iαa+Dαa+u(t)=u(t)c1(ta)α1c2(ta)α2cn(ta)αn,

    for some constants ciR,i=1,2,,n, and n=[α]+1.

    Lemma 2.2. [2] Let α>0. If x,CDαa+xL([a,b],R), then

    Iαa+CDαa+x(t)=x(t)c0c1(ta)cn1(ta)n1,

    for some constants ciR, i=0,1,,n1, and n=[α]+1.

    Lemma 2.3. [2,7] Let α,β>0, n=[α]+1. Then,

    Iαa+(ta)β1=Γ(β)Γ(β+α)(ta)α+β1,t>a; (2.1)
    Dαa+(ta)β1=Γ(β)Γ(βα)(ta)βα1,t>a; (2.2)
    Dαa+(ta)αj=0,t>a,j=1,2,,n; (2.3)
    CDαa+(ta)β1=Γ(β)Γ(βα)(ta)βα1,t>a; (2.4)
    CDαa+(ta)k=0,t>a,k=0,1,2,,n1,andCDαa+1=0. (2.5)

    Lemma 2.4. [2] Let α>β>0, u(t)C(a,b). Then,

    Iαa+Iβa+u(t)=Iα+βa+u(t),Dαa+Iαa+x(t)=u(t)=CDαa+Iαa+u(t), (2.6)
    Dβa+Iαa+u(t)=Iαβa+u(t). (2.7)

    Define Banach space X of continuous functions on [a,b] with the norm ||x||=maxt[a,b]|x(t)|. Let α,β,γR, 1<α2, 0<γ,β1, 1<γ+β2 be fixed and I=[a,b]. For any y:(a,b]R and z:[a,b]R, we define functions yα:IR and ˙zβ:IR by

    yα(t)={(ta)2αy(t),ift(a,b],limta+(ta)2αy(t),ift=a,˙zβ(t)={(ta)1βCDγa+z(t),ift(a,b],limta+(ta)1βCDγa+z(t),ift=a,

    given that the right-hand limits are exist. Define spaces

    Y:={y:(a,b]R|yα,Dα1a+yC[a,b]},Z:={z:[a,b]R|z,˙zβ(t)C[a,b]}.

    It is not difficult to show that Y and Z are two Banach spaces equipped with the norms

    ||y||α=maxt[a,b](K|yα|+L|Dα1a+y|),||z||β=maxt[a,b](P|z|+Q|˙zβ|),

    respectively.

    Lemma 3.1. Assume that f and g are continuous functions. Then, a function xC[a,b] is a solution of Eq (1.7) if and only if x(t) satisfies

    x(t)=f(t,x(t))baG(t,s)g(s,x(s))ds+B(ta)α1f(t,x(t))(ba)α1f(b,x(b)), (3.1)

    where

    G(t,s)=1Γ(α){(ta)α1(ba)α1(bs)α1(ts)α1,astb,(ta)α1(ba)α1(bs)α1,atsb. (3.2)

    Proof. According to Lemma 2.1, applying operator Iαa+ on both sides of Eq (1.7) yields

    x(t)f(t,x(t))=Iαa+g(t,x(t))+c1(ta)α1+c2(ta)α2,

    where c1,c2R are arbitrary constants. Therefore,

    x(t)=f(t,x(t))[Iαa+g(t,x(t))+c1(ta)α1+c2(ta)α2].

    Using boundary condition x(a)=0, we get c2=0. Then,

    x(t)=f(t,x(t))[Iαa+g(t,x(t))+c1(ta)α1]. (3.3)

    From boundary condition x(b)=B, it follows that

    Bf(b,x(b))=Iαa+g(t,x(t))|t=b+c1(ba)α1,

    that is,

    c1=B(ba)α1f(b,x(b))+Iαa+g(t,x(t))|t=b(ba)α1.

    Substituting the value of c1 in Eq (3.3) yields

    x(t)=f(t,x(t))(1Γ(α)ta(ts)α1g(s,x(s))ds+(ta)α1Γ(α)(ba)α1ba(bs)α1g(s,x(s))ds)+B(ta)α1f(t,x(t))(ba)α1f(b,x(b))=f(t,x(t))baG(t,s)g(s,x(s))ds+B(ta)α1f(t,x(t))(ba)α1f(b,x(b)).

    Conversely, by direct computation, it can be established that (3.1) satisfies problem (1.7). This completes the proof.

    Lemma 3.2. Assume that f is a continuous function. Then, a function yY is a solution of Eq (1.8) if and only if y(t) satisfies the integral equations

    y(t)=baG(t,s)f(s,(sa)2αy(s),Dα1a+y(s))ds+(ta)α1(ba)α1[BA(ba)α2]+A(ta)α2, (3.4)

    and

    Dα1a+y(t)=Γ(α)[BA(ba)α2](ba)α1+baH(t,s)f(s,(sa)2αy(s),Dα1a+y(s))ds, (3.5)

    where G(t,s) defined the same as in (3.2),

    G(t,s)=1Γ(α){(ta)α1(ba)α1(bs)α1(ts)α1,astb,(ta)α1(ba)α1(bs)α1,atsb,

    and

    H(t,s)=1(ba)α1{(bs)α1(ba)α1,astb,(bs)α1,atsb. (3.6)

    Proof. In view of Lemma 2.1, a general solution of the fractional equation in (1.8) is given by

    y(t)=Iαa+f(t,(ta)2αy(t),Dα1a+y(t))+c0(ta)α1+c1(ta)α2, (3.7)

    where c0,c1R are arbitrary constants. From the first boundary condition limta+(ta)2αy(t)=A, we obtain c1=A, which yields

    y(t)=Iαa+f(t,(ta)2αy(t),Dα1a+y(t))+c0(ta)α1+A(ta)α2. (3.8)

    From y(b)=B and by using (3.8), we derive

    c0=1(ba)α1[Iαa+f(t,(ta)2αy(t),Dα1a+y(t))|t=b+BA(ba)α2].

    Substituting the value of c0 in (3.8) yields solution (3.4),

    y(t)=Iαa+f(t,(ta)2αy(t),Dα1a+y(t))+A(ta)α2+(ta)α1(ba)α1[Iαa+f(t,(ta)2αy(t),Dα1a+y(t))|t=b+BA(ba)α2]=baG(t,s)f(s,(sa)2αy(s),Dα1a+y(s))ds+(ta)α1(ba)α1[BA(ba)α2]+A(ta)α2.

    The converse follows by direct computation. On the other hand, according to (2.2), (2.3) and (2.7), taking the (α1)-th Riemann-Liouville fractional derivative on the both sides of (3.4) yields (3.5),

    Dα1a+y(t)=Dα1a+Iαa+f(t,(ta)2αy(t),Dα1a+y(t))+A[Dα1a+(ta)α2]+Dα1a+(ta)α1(ba)α1[Iαa+f(t,(ta)2αy(t),Dα1a+y(t))|t=b+BA(ba)α2]=taf(s,(sa)2αy(s),Dα1a+y(s))ds+Γ(α)(ba)α1[BA(ba)α2]+1(ba)α1[ba(bs)α1f(s,(sa)2αy(s),Dα1a+y(s))ds]=Γ(α)[BA(ba)α2](ba)α1+baH(t,s)f(s,(sa)2αy(s),Dα1a+y(s))ds.

    This proves the lemma.

    Lemma 3.3. Assume that g is a continuous function. Then, a function zY is a solution of equation (1.9) if and only if z(t) satisfies the integral equations

    z(t)=baG(t,s)g(s,z(s),(sa)1βCDγa+z(s))ds+(ta)γ+β1(ba)γ+β1(BA)+A, (3.9)

    and

    CDγa+z(t)=(BA)Γ(γ+β)(ba)γ+β1Γ(β)(ta)β1+baH(t,s)g(s,z(s),(sa)1βCDγa+z(s))ds, (3.10)

    where kernel functions G(t,s) and H(t,s) are defined as

    G(t,s)=1Γ(γ+β){(ta)γ+β1(ba)γ+β1(bs)γ+β1(ts)γ+β1,astb,(ta)γ+β1(ba)γ+β1(bs)γ+β1,atsb, (3.11)

    and

    H(t,s)=1Γ(β){(ta)β1(ba)γ+β1(bs)γ+β1(ts)β1,astb,(ta)β1(ba)γ+β1(bs)γ+β1,atsb. (3.12)

    Proof. Applying operators Iβa+ and Iγa+ on the fractional equation in (1.9) and then using Lemmas 2.1–2.4 yields

    z(t)=Iγ+βa+g(t,z(t),(ta)1βCDγa+z(t))+Γ(β)Γ(γ+β)c0(ta)γ+β1+c1, (3.13)

    where c0,c1R are arbitrary constants. Applying the boundary conditions z(a)=A,z(b)=B in (3.13) yields

    c1=A,c0=Γ(γ+β)(ba)γ+β1Γ(β)[BA+Iγ+βa+g(t,z(t),(ta)1βCDγa+z(t))|t=b].

    Substituting the above values into (3.13), the solution given by (3.9) is obtained. The converse of the lemma can be obtained by direct computation. On the other hand, by Lemma 2.3 and Lemma 2.4, taking the γ-th Caputo fractional derivative on both sides of (3.9) yields (3.10). The proof is completed.

    Lemma 3.4. The Green's functions G(t,s), H(t,s), G(t,s) and H(t,s) given by Lemmas 3.1–3.3, respectively, satisfy the following properties:

    (i) G(t,s), H(t,s), G(t,s) and H(t,s) are continuous functions in [a,b]×[a,b];

    (ii) G(t,s) and G(t,s) are two nonnegative functions in [a,b]×[a,b];

    (iii) baG(t,s)ds(α1)α1Γ(α)αα+1(ba)α, for any t[a,b];

    (iv) ba(ta)2αG(t,s)ds(ba)24Γ(α+1), for any t[a,b];

    (v) ba|H(t,s)|ds1α(ba), for any t[a,b];

    (vi) baG(t,s)ds(γ+β1)γ+β1Γ(γ+β)(γ+β)γ+β+1(ba)γ+β, for any t[a,b];

    (vii) ba(ta)1β|H(t,s)|dsmax{β,γ}(ba)(γ+β)Γ(β+1), for any t[a,b].

    Proof. It is obvious that (i) is true. For (ii), in view of the definition of G(t,s), let

    G1(t,s)=1Γ(α)(ta)α1(ba)α1(bs)α1(ts)α1,astb,G2(t,s)=1Γ(α)(ta)α1(ba)α1(bs)α1,atsb.

    Then, we can easily obtain that

    G2(t,s)0,(t,s)[a,b]×[a,b].

    Differentiating G1(t,s) with respect to s for every fixed t[a,b],

    G1(t,s)s=α1Γ(α)[(ta)α1(ba)α1(bs)α2+(ts)α2]=α1Γ(α)(ts)α2[1(taba)α1(tsbs)2α]0,

    that is, G1(t,s) is increasing with respect to s[a,t] for any fixed t[a,b]. Therefore,

    G1(t,s)G1(t,a)=0,t[a,b].

    Thus, we have derived that G(t,s) ia nonnegative in [a,b]×[a,b]. Let α=β+γ, then we can also get G(t,s) is nonnegative in [a,b]×[a,b]. For (iii) and (iv), by the expression for the function G(t,s), we obtain

    baG(t,s)ds=1Γ(α){ta[(ta)α1(ba)α1(bs)α1(ts)α1]ds+bt(ta)α1(ba)α1(bs)α1ds}=1Γ(α){1α[(ta)α1(ba)α1(bs)α+(ts)α]|s=ts=a1α(ta)α1(ba)α1(bs)α|s=bs=t}=1αΓ(α)(ta)α1(bt).

    It follows that

    ba(ta)2αG(t,s)ds=(ta)2αbaG(t,s)ds=1αΓ(α)(ta)(bt).

    Define

    g(t)=(ta)(bt),t[a,b],

    and

    ˜g(t)=(ta)α1(bt),t[a,b].

    Differentiating the functions g(t) and ˜g(t) on (a,b), we immediately find that g(t) and ˜g(t) are achieved their maximum at the following points, respectively,

    t=a+b2,˜t=1α[a+(α1)b].

    This yields,

    maxt[a,b]g(t)=14(ba)2,maxt[a,b]˜g(t)=(α1)α1(ba)ααα,

    which completes the proof of (iii) and (iv). Let α=β+γ, then property (vi) can be obtained directly from (iii). We will now show that properties (v) and (vii) are true. First, for (v), in view of the definition of H(t,s), we have

    ba|H(t,s)|ds=1(ba)α1[ta((ba)α1(bs)α1)ds+bt(bs)α1ds]=1(ba)α1[(ba)α1(ta)+2α(bt)α1α(ba)α].

    Define

    h(t)=(ba)α1(ta)+2α(bt)α1α(ba)α,t[a,b].

    Then,

    h(a)=1α(ba)αh(b)=(ba)α1α(ba)α>0.

    Taking the second-order derivative of function h(t) on (a,b), we obtain

    h(t)=2(α1)(bt)α20,t[a,b].

    Therefore, h(t) is convex on (a,b). Hence,

    maxt[a,b]h(t)=maxt[a,b]{h(a),h(b)}=h(a)=1α(ba)α.

    This completes the proof of (v). Finally, for (vii), for astb, we can derive that

    (bs)γ+β1(ta)β1(ba)γ+β1(ts)β1=(ba)γ+β1(ts)β1[(bsba)γ+β1(tsta)1β1]0.

    Hence, it follows from the definition of H(t,s) that

    ba|H(t,s)|ds=1Γ(β)ta[(ts)β1(ta)β1(ba)γ+β1(bs)γ+β1]ds+1Γ(β)bt(ta)β1(ba)γ+β1(bs)γ+β1ds=1(γ+β)Γ(β)(ta)β1(ba)γ+β1[2(bt)γ+β(ba)γ+β]+1Γ(β+1)(ta)β.

    Consequently,

    ba(ta)1β|H(t,s)|ds=2(bt)γ+β(ba)γ+β(ba)γ+β1(γ+β)Γ(β)+(ta)Γ(β+1).

    Define

    h(t)=2(bt)γ+β(ba)γ+β(ba)γ+β1(γ+β)Γ(β)+(ta)Γ(β+1),t[a,b].

    Then,

    h(a)=β(ba)(γ+β)Γ(β+1)>0,h(b)=γ(ba)(γ+β)Γ(β+1)>0.

    Taking the second derivative of function h(t) on (a,b) leads to

    h(t)=(γ+β1)2(bt)γ+β2(ba)γ+β1Γ(β)0,t(a,b).

    This implies that h(t) is convex on (a,b). Therefore,

    maxt[a,b]h(t)=max{h(a),h(b)}=max{β,γ}(ba)(γ+β)Γ(β+1).

    The proof is completed.

    Based on Lemmas 3.1 to 3.3, operators T1:XX, T2:YY and T3:ZZ are defined as

    T1x(t)=f(t,x(t))baG(t,s)g(s,x(s))ds+B(ta)α1f(t,x(t))(ba)α1f(b,x(b)),x(t)X,T2y(t)=baG(t,s)f(s,(sa)2αy(s),Dα1a+y(s))ds+(ta)α1(ba)α1[BA(ba)α2]+A(ta)α2,y(t)Y,T3z(t)=baG(t,s)g(s,z(s),(sa)1βCDγa+z(s))ds+(ta)γ+β1(ba)γ+β1(BA)+A,z(t)Z.

    Theorem 3.1. Assume that f,g:[a,b]×RR are continuous functions and satisfy condition (A1). Let

    M1=supt[a,b]|f(t,x(t))|,m1=inft[a,b]|f(t,x(t))|,M2=supt[a,b]|g(t,x(t))|.

    If

    ba<{αα+1[m21BL1(m1+M1)]Γ(α)m21(M2L1+M1L2)(α1)α1}1/1αα, (3.14)

    then problem (1.7) has a unique solution.

    Proof. For x1(t),x2(t)X, using condition (A1), Lemma 3.4 (iii) and together with (3.1), yields

    |Tx1(t)Tx2(t)||B(ta)α1f(t,x1(t))(ba)α1f(b,x1(b))B(ta)α1f(t,x2(t))(ba)α1f(b,x2(b))|+|f(t,x1(t))baG(t,s)g(s,x1(s))dsf(t,x2(t))baG(t,s)g(s,x2(s))ds|B(ta)α1(ba)α1(|f(t,x1(t))f(b,x1(b))f(t,x2(t))f(b,x1(b))|+|f(t,x2(t))f(b,x1(b))f(t,x2(t))f(b,x2(b))|)+|f(t,x1(t))baG(t,s)g(s,x1(s))dsf(t,x2(t))baG(t,s)g(s,x1(s))ds|+|f(t,x2(t))baG(t,s)g(s,x1(s))dsf(t,x2(t))baG(t,s)g(s,x2(s))ds|B(ta)α1(ba)α1|f(b,x1(b))||f(t,x1(t))f(t,x2(t))|+B(ta)α1|f(t,x2(t))|(ba)α1|f(b,x1(b))||f(b,x2(b))||f(b,x2(b))f(b,x1(b))|+|f(t,x1(t))f(t,x2(t))|baG(t,s)|g(s,x1(s))|ds+baG(t,s)|g(s,x1(s))g(s,x2(s))|ds|f(t,x2(t))|(Bm1+BM1m21)L1|x1x2|+(M2L1+M1L2)(α1)α1(ba)αΓ(α)αα+1|x1x2|
    [BL1(m1+M1)m21+(M2L1+M1L2)(α1)α1(ba)αΓ(α)αα+1]||x1x2||.

    Therefore, we conclude from (3.14) that operator T1 is a contraction mapping. Hence problem (1.7) has a unique solution.

    As special cases of Theorem 3.1, we have the following corollary:

    Corollary 3.1. Let g(t,x) be continuous on [a,b]×R and satisfy Lipschitz condition

    |g(t,x1)g(t,x2)|K|x1x2|,foranyx1,x2R,K>0.

    Then the BVP

    {Dαa+x(t)+g(t,x(t))=0,t(a,b),1<α2,x(a)=0,x(b)=B,BR,

    has a unique solution whenever

    ba<[Γ(α)αα+1K(α1)α1]1/1αα. (3.15)

    Proof. By Theorem 3.1, let f(t,x)1, (t,x)[a,b]×R, and L2=K. Then,

    M1=supt[a,b]|f(t,x(t))|=inft[a,b]|f(t,x(t))|=m1=1,L1=0.

    Substituting the above values into (3.14), the desired result (3.15) is obtained. As such, our results match the results of Theorem 2.3 in [12].

    Theorem 3.2. Let f(t,(ta)2αy(t),Dα1a+y(t)) be continuous on [a,b]×R2 and satisfy condition (A2). If

    K(ba)24Γ(α+1)+Lα(ba)<1, (3.16)

    then problem (1.8) has a unique solution.

    Proof. To see when T2 is contracting, we again from

    (ta)2α|T2y1(t)T2y2(t)|ba(ta)2αG(t,s)|f(s,(sa)2αy1(s),Dα1a+y1(s))f(s,(sa)2αy2(s),Dα1a+y2(s))|ds,

    and from (3.5)

    |Dα1a+T2y1(t)Dα1a+T2y2(t)|ba|H(t,s)||f(s,(sa)2αy1(s),Dα1a+y1(s))f(s,(sa)2αy2(s),Dα1a+y2(s))|ds.

    Using Lipschitz condition (A2) and Lemma 3.4 (iv), (v),

    (ta)2α|T2y1(t)T2y2(t)|ba(ta)2αG(t,s)(K(sa)2α|y1(s)y2(s)|+L|Dα1a+y1(s)Dα1a+y2(s)|)ds||y1y2||αba(ta)2αG(t,s)ds(ba)24Γ(α+1)||y1y2||α,

    and

    |Dα1a+T2y1(t)Dα1a+T2y2(t)|ba|H(t,s)|(K(sa)2α|y1(s)y2(s)|+L|Dα1a+y1(s)Dα1a+y2(s)|)ds||y1y2||αba|H(t,s)|ds1α(ba)||y1y2||α.

    Together these imply

    ||T2y1T2y2||α[K(ba)24Γ(α+1)+Lα(ba)]||y1y2||α.

    It follows from (3.16) that operator T2 is a contraction mapping. Hence, problem (1.8) has a unique solution.

    Remark 3.1. Let α2. Then, Theorem 3.2 can be reduced to Theorem 1.2.

    Theorem 3.3. Let g(t,z(t),(ta)1βCDγa+z(t)) be continuous on [a,b]×R2 and satisfy condition (A3). If

    P(γ+β1)γ+β1(ba)γ+βΓ(γ+β)(γ+β)γ+β+1+Qmax{β,γ}(ba)(γ+β)Γ(β+1)<1, (3.17)

    then problem (1.9) has a unique solution.

    Proof. This result will follow from the Banach contraction mapping theorem if we can show that operator T3 is a contraction mapping. In fact, for any z1(t),z2(t)Z, we have

    |T3z1(t)T3z2(t)|baG(t,s)|g(s,z1(s),(sa)1βCDγa+z1(s))g(s,z2(s),(sa)1βCDγa+z2(s))|ds,

    and in view of (3.10),

    |(ta)1βCDγa+T3z1(t)(ta)1βCDγa+T3z2(t)|ba(ta)1β|H(t,s)||g(s,z1(s),(sa)1βCDγa+z1(s))g(s,z2(s),(sa)1βCDγa+z2(s))|ds.

    Using Lipschitz condition (A3) and Lemma 3.4 (vi), (vii),

    |T3z1(t)T3z2(t)|baG(t,s)P|z1(s)z2(s)|+Q(sa)1β|CDγa+z1(s)CDγa+z2(s)|ds||z1z2||βbaG(t,s)ds(γ+β1)γ+β1(ba)γ+βΓ(γ+β)(γ+β)γ+β+1||z1z2||β,

    and

    (ta)1β|CDγa+T3z1(t)CDγa+T3z2(t)|ba(ta)1β|H(t,s)|P|z1(s)z2(s)|+Q(sa)1β|CDγa+z1(s)CDγa+z2(s)|ds||z1z2||βba(ta)1β|H(t,s)|dsmax{β,γ}(ba)(γ+β)Γ(β+1)||z1z2||β,

    which, by taking the norm for t[a,b], implies that

    ||T3z1T3z2||β[P(γ+β1)γ+β1(ba)γ+βΓ(γ+β)(γ+β)γ+β+1+Qmax{β,γ}(ba)(γ+β)Γ(β+1)]||z1z2||β.

    From (3.17) we conclude that operator T3 is a contraction mapping. Thus, problem (1.9) has a unique solution.

    Remark 3.2. Let γ,β1. Then Theorem 3.3 can be reduced to Theorem 1.2.

    Example 4.1. Consider the following two-point fractional BVP

    {D4/31+[x(t)(6t/5)+(cosx(t))/10]+lnt5sinx(t)=0,t(1,2),x(1)=0,x(2)=1. (4.1)

    Corresponding to BVP (1.7), here

    α=4/3,a=B=1,b=2,
    f(t,x(t))=6t/5+cosx(t)/10,g(t,x(t))=((lnt)/5)sinx(t).

    Obviously, we have

    |f(t,x)f(t,y)|(1/10)|cosxcosy|(1/10)|xy|,forx,yR,|g(t,x)g(t,y)|((lnt)/5)|sinxsiny|((ln2)/5)|xy|,forx,yR.

    It is easy to find that L1=1/10,L2=M2=(ln2)/5,M1=2.5,m1=1.1. Thus,

    BL1(m1+M1)m21+(M2L1+M1L2)(α1)α1(ba)αΓ(α)αα+10.4405<1.

    Clearly, all assumptions of Theorem 3.1 are satisfied. Therefore, BVP (4.1) has a unique solution on [1,2].

    Example 4.2. Consider the following two-point fractional BVP

    {D3/3220+y(t)+34πsint1/122y(t)+34sinD1/1220+y(t)+52=0,t(0,1),limt0+t1/122y(t)=y(1)=0, (4.2)

    Corresponding to BVP (1.8), here

    α=32,a=A=B=0,b=1,
    f(t,(ta)2αy(t),Dα1a+y(t))=34πsint1/122y(t)+34sinD1/1220+y(t)+52,

    Clearly,

    |f(t,(ta)2αy1(t),Dα1a+y1(t))f(t,(ta)2αy2(t),Dα1a+y2(t))|34πt1/122|y1(t)y2(t)|+34|Dα1a+y1(t)Dα1a+y2(t)|.

    Choosing K=34π,L=34 and consequently, we obtain

    K(ba)24Γ(α+1)+Lα(ba)=3π16Γ(5/522)+12=34<1.

    Thus, all the conditions of Theorem 3.2 are satisfied. Hence, BVP (4.2) has a unique solution on [0,1].

    Example 4.3. Consider the following two-point fractional BVP

    {D1/1220+CD3/3440+z(t)+et1+t+310sinz(t)+25|t1/122CD3/3440+z(t)|1+|t1/122CD3/3440+z(t)|=0,t(0,2),z(0)=z(2)=0, (4.3)

    Corresponding to BVP (1.9), here

    β=12,γ=34,a=A=B=0,b=2,
    g(t,z(t),(ta)1βCDγa+z(t))=et1+t+310sinz(t)+25|t1/122CD3/3440+z(t)|1+|t1/122CD3/3440+z(t)|.

    Clearly,

    |g(t,z1(t),(ta)1βCDγa+z1(t))g(t,z2(t),(ta)1βCDγa+z2(t))|310|z1(t)z2(t)|+25t1/122|CD3/3440+z1(t)CD3/3440+z2(t)|.

    Taking P=310,Q=25, we can find that

    P(γ+β1)γ+β1(ba)γ+βΓ(γ+β)(γ+β)γ+β+1+Qmax{β,γ}(ba)(γ+β)Γ(β+1)=310(1/144)1/14425/544(5/544)9/944Γ(5/544)+253/322(5/544)Γ(3/322)0.9628<1.

    Thus, the hypotheses of Theorem 3.3 are satisfied. Therefore, the BVP (4.3) has a unique solution on [0,2].

    In this article, we discussed the uniqueness results for several two-point fractional BVPs. By using the Banach contraction mapping theorem, we obtained the sharp conditions in terms of the end-points of the given interval which ensures the uniqueness of solutions for these fractional BVPs. This seems to have something in common with studying of the Lyapunov inequality for BVPs. In terms of methods, both of them are converted the BVPs into the equivalent integral equations with corresponding Green's functions. By estimating the upper bound of Green's function, the existence of solutions to BVPs is finally characterized. The difference is that Lyapunov inequality directly estimates the upper bound of the Green's function G(t,s) on the interval [a,b]×[a,b], while this paper estimates the upper bound of baG(t,s)ds for any t[a,b]. Our work is an extension of the classical results of Theorem 1.1 and Theorem 1.2. It is also an extension and supplement to some recent work [10,11,12,13,14]. Compared with the paper [10,11,12,13,14], we discuss the BVP where the nonlinear term of the differential equation has the fractional derivative of unknown function, and obtain new interesting results. Based on the this study, in the forthcoming paper, we will investigate the sharp estimate for the unique solution of the two-point Ψ-Hilfer fractional hybird-Sturm-Liouville equations.

    The authors wish to express their sincere appreciation to the editor and the anonymous referees for their valuable comments and suggestions. This research is supported by Anhui Provincial Natural Science Foundation (2208085QA05), National Natural Science Foundation of China (11601007) and the Key Program of the University Natural Science Research Fund of Anhui Province (KJ2020A0291).

    The authors declare that there is no conflict of interest.



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