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Properties of positive solutions for a fractional boundary value problem involving fractional derivative with respect to another function

  • In this article, we develop the existence and uniqueness of positive solutions to a class of fractional boundary value problems involving fractional order derivative with respect to another function for any given parameter. The analysis is based upon the fixed point theorems of concave operators in partial ordering Banach spaces. For the sake of discussing the existence of solutions for the problem, we first construct Green function and study its properties. Furthermore, some properties of positive solutions to the boundary value problem are proved under the different parameters. Examples illustrating the results are also presented.

    Citation: Yitao Yang, Dehong Ji. Properties of positive solutions for a fractional boundary value problem involving fractional derivative with respect to another function[J]. AIMS Mathematics, 2020, 5(6): 7359-7371. doi: 10.3934/math.2020471

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  • In this article, we develop the existence and uniqueness of positive solutions to a class of fractional boundary value problems involving fractional order derivative with respect to another function for any given parameter. The analysis is based upon the fixed point theorems of concave operators in partial ordering Banach spaces. For the sake of discussing the existence of solutions for the problem, we first construct Green function and study its properties. Furthermore, some properties of positive solutions to the boundary value problem are proved under the different parameters. Examples illustrating the results are also presented.


    In this work, we are concerned with the following fractional boundary value problem

    Dα, φ0+u(t)+λf(t,u(t))=0,   0<t<1, (1.1)
    u(0)=0,   u(1)=m2i=1βiu(ηi), (1.2)

    where 1<α2, Dα, φ0+ is the fractional derivative operator of order α with respect to a certain strictly increasing function φC2[0,1] with φ(x)>0 for all x[0,1] and f,φ,βi,ηi,λ satisfy:

    (H1) f(t,u):[0,1]×R+R+ is continuous and increasing in u for each t[0,1], the function φ:[0,1]R is a strictly increasing function such that φC2[0,1], with φ(x)>0 for all x[0,1];

    (H2) 0βi<1, 0<ηi<1 (i=1,2,,m2) satisfy 0m2i=1βi<1, 0<η1<η2<<ηm2<1,λ>0 is a parameter.

    The fractional calculus first began in 1965. After a long time, this subject was relevant only in pure mathematics, many mathematicians have studied these new fractional operators by presenting new definitions and studying their important properties, but in the last century, this subject showed its applications in the modeling of many phenomena in various fields of science and engineering, such as control theory, electric circuits, viscoelasticity, mechanics, physics, neural networks [1,2,3,4,5,6,7,8,9,10,11,12].

    In all those definitions there is a special kind of a kernel dependency. Therefore, in order to analyze fractional differential equations in a generic way, a fractional derivative with respect to another function called φ-Riemann-Liouville and φ-Caputo derivative was proposed [15] and for particular choices of φ, we obtain some well known fractional derivative. For example, if we choose φ(t)=t, the φ-Riemann-Liouville and φ-Caputo fractional derivative are reduced to the Riemann-Liouville and Caputo derivative respectively in traditional sense. Alternatively, if we choose φ(t)=logt, where log()=loge(), the φ-Riemann-Liouville and φ-Caputo fractional derivative are reduced to the Hadamard and Hadamard-Caputo derivative respectively [13,14]. Therefore, problem (1.1), (1.2) generates many types and also mixed types of fractional differential equations with boundary conditions.

    Let us mention some motivations for studying problem (1.1), (1.2). In the limit case λ=1 and βi=β,ηi=η,i=1,2,, (1.1), (1.2) reduces to

    Dα, φ0+u(t)+f(t,u(t))=0,   0<t<1, (1.3)
    u(0)=0,   u(1)=βu(η). (1.4)

    In [22], the authors studied the existence of positive solutions for the problem (1.3), (1.4). In [23], using the fixed point theorems of Banach and Schaefer, the authors studied the existence and uniqueness results of the boundary value problems for a fractional differential equation involving Caputo operator with respect to the new function ψ given by the form

    CDα, ψa+y(t)=f(t,y(t)),    t[a,b],
    y[k]ψ(a)=yka,   k=0,1,,n2;   y[n1]ψ(b)=yb,

    where CDα, ψa+ is the ψ-Caputo fractional derivative of order n1<αn (n=[α]+1).

    Almeida et al. [24] studied the initial value problem of a fractional differential equation including φ-Caputo fractional derivative

    CDα, φa+y(t)=f(t,y(t)),     t[a,b],
    y(a)=ya,   y[k]ψ(a)=yka,   k=1,2,,n1

    where CDα, φa+ is the φ-Caputo fractional derivative of order n1<α<n.

    On the other hand, most of the existing literature does not consider the properties of the solution. Well, here's the problem: How can we get the properties of the solutions when they are known to exist. So far, few papers can be found in the literature both on the existence and properties of solutions for fractional boundary value problem.

    Zhai et al. [16] studied the following nonlinear fractional four-point boundary value problem with a parameter:

    Dα0+u(t)+λf(t,u(t))=0,  0<t<1,
    u(0)μ1u(ξ)=0,  u(1)+μ2u(η)=0,

    where Dα0+ is the Riemann-Liouville fractional derivative. Unfortunately, the fractional derivative in this paper is just the Caputo derivative in the traditional sense.

    Inspired by the work in [16] and many known results, in this paper, we develop the existence and uniqueness of positive solutions to the fractional boundary value problems (1.1), (1.2). Using the fixed point theorems of concave operators in partial ordering Banach spaces, we not only show the existence of positive solution for the problem (1.1), (1.2), but also present some properties of positive solutions to the boundary value problem dependent on the parameter. Some definitions, lemmas and basic results about φ-Riemann-Liouville derivative are presented in Section 2. The main results are given in section 3. Some illustrative examples are constructed in Section 4.

    Let φ be a function of C2[a,b] with φ(t)>0 for all t[a,b].

    Definition 2.1 [21] The φ-Riemann-Liouville fractional integral of order α>0 of a function x:[a,b]R is given by

    Iα, φa+x(t):=1Γ(α)taφ(s)(φ(t)φ(s))α1x(s)ds.

    Definition 2.2 [21] The φ-Riemann-Liouville fractional derivative of order α of a function x is defined as

    Dα, φa+x(t):=(1φ(t)ddt)nInα, φa+x(t)=1Γ(nα)(1φ(t)ddt)ntaφ(s)(φ(t)φ(s))nα1x(s)ds,

    where n=[α]+1.

    The φ-Caputo fractional derivative of order α of a function x is defined as

    CDα, φa+x(t):=Dα, φa+[x(t)n1k=0x[k]φ(a)k!(φ(t)φ(a))k], n=[α]+1 for αN, n=α for αN,

    where x[k]φ(t):=(1φ(t)ddt\bigamma)kx(t).

    Theorem 2.1 [15] If xCn[a,b], then

    CDα, φa+x(t)=Inα, φa+(1φ(t)ddt)nx(t).

    Theorem 2.2 [15] Let x:[a,b]R. Then

    1. If xC[a,b], then CDα, φa+Iα, φa+x(t)=x(t);

    2. If xCn1[a,b], then

    Iα, φ Ca+Dα, φa+x(t)=x(t)n1k=0x[k]φ(a)k!(φ(t)φ(a))k.

    In the following, we give some basic definitions in ordered Banach spaces and two fixed point theorems of concave operators which will be used later, all the materials can be found in [17,18,19,20]. We denote θ the zero element of E.

    Definition 2.3 Let X be a semi-ordered real Banach space. The nonempty convex closed subset P of X is called a cone in X if axP for all xP and a0 and xX and xX imply x=θ.

    Let the real Banach space E be partially ordered by a cone P of E, that is xy if and only if yxP. If xy and xy, then we denote x<y or y>x.

    Recall that cone P is said to be solid if the interior ˚P is nonempty and we denote x>>0 if x˚P. P is normal if there exists a positive constant N such that θxy implies xNy, N is called the normal constant of P.

    Definition 2.4 Let D be a convex subset in E. An operator T:DE is a concave operator if

    T(tx+(1t)y)tTx+(1t)Ty

    for all x,yD and t[0,1].

    Definition 2.5 Let P be a cone in real Banach space E and eP{θ}. Set

    Ee={xE:there exists λ>0 such thatλexλe}

    and define

    xe=inf{λ>0:λexλe},   xEe.

    It is easy to see that Ee becomes a normed linear space under the norm e. xe is called the e-norm of the element xEe.

    Lemma 2.3 [17] Suppose P is a normal cone. Then the following results hold:

    (i) Ee is a Banach space with e-norm, and there exists a constant m>0 such that xmxe,   xEe;

    (ii) Pe=EeP is a normal solid cone of Ee and ˚Pe={xEe:there exists δ=δ(x)>0 such that xδe}.

    Lemma 2.4 [19] Suppose P is a normal solid cone and operator T:PP is a concave operator. Assume that Tθ>>θ. Then the following results hold:

    (ⅰ) there exists 0<λ such that the equation

    u=λTu (2.1)

    has a unique solution u(λ) in P for 0λ<λ, when λλ, the operator equation (2.1) has no solution in P;

    (ⅱ) for any u0P, let u0(λ)=u0,  un(λ)=λTun1(λ), n=1,2,3,. Then for 0<λ<λ, we have un(λ)u(λ) as n;

    (ⅲ) u():[0,λ)P is continuous and strongly increasing (0λ1<λ2<λu(λ1)<<u(λ2)). Furthermore, u(tλ)tu(λ) for 0λ<λ, 0t1.

    (ⅳ) λλ0 implies u(λ);

    (ⅴ) if there exist v0P and λ0>0 such that λ0Tv0v0, then λ>λ0.

    Lemma 2.5 [19] If T:PP is concave, then T is increasing.

    Lemma 2.6 [20] Suppose P is a normal solid cone and operator T:˚P˚P is an increasing operator. We presume there exists a constant 0<r<1 such that

    T(tu)trTu,   u˚P, 0<t<1.

    Let uλ be the unique solution in ˚P of the equation Tu=λu(λ>0). Then

    (ⅰ) uλ is strongly decreasing (0<λ1<λ2uλ1>>uλ2);

    (ⅱ) uλ is continuous (λλ0(λ0>0)uλuλ00);

    (ⅲ) limλ+uλ=0,   limλ0+uλ=+.

    Denote

    μ=(φ(1)φ(0))α1m2i=1βi(φ(ηi)φ(0))α1.

    Lemma 3.1 Assume that (H1),(H2) hold. Then the fractional boundary value problem (1.1), (1.2) is equivalent to the following integral equation

    u(t)=10G(t,s)φ(s)λf(s,u(s))ds, (3.1)

    where

    G(t,s)={m2j=1βj(φ(ηj)φ(0))α1(φ(t)φ(s))α1(φ(t)φ(s))α1(φ(1)φ(0))α1μΓ(α)+(φ(t)φ(0))α1(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1(φ(t)φ(0))α1μΓ(α)0t1, ηi1smin{ηi,t},i=1,2,,m1;(φ(t)φ(0))α1(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1(φ(t)φ(0))α1μΓ(α)0t1, max{ηi1,t}sηi, i=1,2,,m1. (3.2)

    Here for the sake of convenience, we write η0=0,ηm1=1 and m2i=m1fi=0 for m2<m1.

    Proof. Use y(t) to replace f(t,u) in (1.1). Let

    Dα, φ0+u(t)+λy(t)=0.

    Then from Theorem 2.2, we have

    u(t)=c1(φ(t)φ(0))α1+c2(φ(t)φ(0))α21Γ(α)t0φ(s)(φ(t)φ(s))α1λy(s)ds. (3.3)

    Note that u(0)=0 implies c2=0. Therefore, we obtain

    u(t)=c1(φ(t)φ(0))α11Γ(α)t0φ(s)(φ(t)φ(s))α1λy(s)ds. (3.4)

    In particular,

    u(1)=c1(φ(1)φ(0))α11Γ(α)10φ(s)(φ(1)φ(s))α1λy(s)ds

    and

    u(ηi)=c1(φ(ηi)φ(0))α11Γ(α)ηi0φ(s)(φ(ηi)φ(s))α1λy(s)ds.

    Note that u(1)=m2i=1βiu(ηi), and hence we obtain

    c1=1μΓ(α)10φ(s)(φ(1)φ(s))α1λy(s)dsm2i=1βiμΓ(α)ηi0φ(s)(φ(ηi)φ(s))α1λy(s)ds.

    As a result, from (3.4) we have

    u(t)=1Γ(α)t0φ(s)(φ(t)φ(s))α1λy(s)ds+(φ(t)φ(0))α1μΓ(α)10φ(s)(φ(1)φ(s))α1λy(s)dsm2i=1βi(φ(t)φ(0))α1μΓ(α)ηi0φ(s)(φ(ηi)φ(s))α1λy(s)ds=10G(t,s)φ(s)λy(s)ds. (3.5)

    This completes the proof. Lemma 3.2 Assume that (H1),(H2) hold. Then the Green's function G(t,s) defined by (3.2) has the following properties:

    (ⅰ) G(t,s) is a continuous function on [0,1]×[0,1];

    (ⅱ) G(t,s)>0 for all (t,s)(0,1);

    (ⅲ)

    G(t,s)(φ(1)φ(0))α1(φ(1)φ(s))α1μΓ(α),  s(0,1). (3.6)

    Proof. (ⅰ) From the expression of G(t,s), it is easy to get that G(t,s) is continuous.

    (ⅱ) From (H1),(H2), we can get μ>0.

    For 0t1, ηi1smin{ηi,t},i=1,2,,m1,

    G(t,s)=1μΓ(α)[(φ(t)φ(0))α1(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1(φ(t)φ(0))α1μ(φ(t)φ(s))α1]=(φ(t)φ(0))α1μΓ(α)[(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1μ(φ(t)φ(s)φ(t)φ(0))α1].

    Consider

    e(t)=(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1μ(φ(t)φ(s)φ(t)φ(0))α1.

    Then we have

    e(t)=μ(α1)(φ(t)φ(s)φ(t)φ(0))α2φ(t)(φ(s)φ(0))(φ(t)φ(0))20,

    which implies that e(t) is decreasing for 0t1, ηi1smin{ηi,t},i=1,2,,m1.

    Furthermore, we have

    e(1)=(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1μ(φ(1)φ(s)φ(1)φ(0))α1=m2j=1βj(φ(ηj)φ(0))α1(φ(1)φ(s))α1(φ(1)φ(0))α1m2j=iβj(φ(ηj)φ(s))α1=m2j=1βj(φ(ηj)φ(0))α1(φ(1)φ(s))α1(φ(1)φ(0))α1m2j=iβj(φ(ηj)φ(0))α1(φ(ηj)φ(s))α1(φ(ηj)φ(0))α1.

    As we know,

    (φ(1)φ(s))α1(φ(1)φ(0))α1>(φ(ηj)φ(s))α1(φ(ηj)φ(0))α1,

    consequently, we have e(1)>0.

    As a result, G(t,s)>0 for 0t1, ηi1smin{ηi,t},i=1,2,,m1.

    For 0t1, max{ηi1,t}sηi,i=1,2,,m1,

    G(t,s)=1μΓ(α)[(φ(t)φ(0))α1(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1(φ(t)φ(0))α1]=1μΓ(α)(φ(t)φ(0))α1[(φ(1)φ(s))α1m2j=iβj(φ(ηj)φ(s))α1]>0.

    Therefore, G(t,s)>0 for all t,s(0,1).

    (ⅲ) From (H1),(H2) and the expression of G(t,s), we can easily obtain this property. Let E=C[0,1] be a Banach space equipped with the norm

    uE=maxt[0,1]|u(t)|.

    Define the cone KE by

    K={uE:u(t)0, 0t1},

    then K is a normal solid cone of E. Let

    g(t)=10G(t,s)φ(s)ds,   0t1. (3.7)

    From (H1) and Lemma 3.2, we have g(t)E and g(t)K{θ}.

    Let X=Ee={uE:τ>0, such that τg(t)u(t)τg(t), 0t1} be a Banach space equipped with the norm uX=inf{τ>0:τg(t)u(t)τg(t),  0t1}.

    Define ˜K=XK, then ˜K is a normal solid cone of X and ˚˜K={uX:there exists δ>0 such that u(t)δg(t),  0t1,  uX}. Moreover, there exists a constant m>0 such that uEmuX.

    Theorem 3.3 Assume that (H1),(H2) hold. Furthermore, f(t,) is concave and there exist constants χ>0,β>0 such that

    f(t,0)χ,   f(t,1)β,   0t1. (3.8)

    Then the following results hold:

    (ⅰ) there exists 0<λ such that the fractional boundary value problem (1.1), (1.2) has a unique positive solution uλ in ˜K for 0<λ<λ, the fractional boundary value problem(1.1), (1.2) has no positive solution in ˜K for λλ;

    (ⅱ) for any u0˜K, let

    un(t)=λ10G(t,s)φ(s)f(s,un1(s))ds, n=1,2,3,.

    Then for 0<λ<λ, we have

    max0t1|un(t)uλ(t)|0 as  n;

    (ⅲ) if 0<λ<λ, max0t1|uλ(t)uλ0(t)|0 as λλ0 and if 0<λ1<λ2<λ, then uλ1(t)uλ2(t),  0t1 and uλ1(t)uλ2(t);

    (ⅳ) uvλ(t)vuλ(t) for 0<λ<λ,  0v1,  0t1.

    (ⅴ) if limuf(t,u)u=0 uniformly on [0,1], then λ=.

    Proof. Define an operator T by

    (Tu)(t)=10G(t,s)φ(s)f(s,u(s))ds. (3.9)

    Thus we know that u is a solution of (1.1), (1.2) if and only if u=λTu. For uK, in view of (H1),(H2) and Lemma 3.2, we can get TuK. Owing to the concavity of f(t,), for u>1, we have

    f(t,1)=f(t,1uu+(11u)0)1uf(t,u)+(11u)f(t,0),

    from (3.8), we obtain

    f(t,u)uf(t,1)(u1)f(t,0)uf(t,1)uβ.

    So, for uK, we have

    010G(t,s)φ(s)f(s,u(s))ds10G(t,s)φ(s)f(s,uE)dsN1g(t), 0t1,

    where N1=max0t1f(t,uE)max0t1f(t,uE+1)β(1+uE), so, we have 0(Tu)(t)N1g(t), which implies TuX. Thus, Tu˜K, this means that T:˜K˜K. In view of the concavity of f(t,), we have the operator T is concave.

    From (3.9), we have (Tθ)(t)=10G(t,s)φ(s)f(s,0)dsχg(t), 0t1. So, Tθ˚˜K, this means that Tθ>>θ. Then all the conditions of Lemma 2.4 are satisfied. Thus from Lemma 2.4, there exists 0<λ such that the equation u=λTu has a unique positive solution uλ in ˜K for 0<λ<λ, the equation u=λTu has no positive solution in ˜K for λλ.;

    For any u0˜K, let un=λTun1,n=1,2,3,, then we have unuλ as  n for 0<λ<λ, uλ:(0,λ)˜K is continuous and strongly increasing. Furthermore, utλtuλ for 0λ<λ, 0t1; λλ0 implies uλX; if there exist v0˜K and λ0>0 such that λ0Tv0v0, then λ>λ0. From this results, we can get

    (ⅰ) the fractional boundary value problem (1.1), (1.2) has a unique positive solution uλ in ˜K for 0<λ<λ, the fractional boundary value problem (1.1), (1.2) has no positive solution in ˜K for λλ;

    (ⅱ) for any u0˜K, let

    un(t)=λ10G(t,s)φ(s)f(s,un1(s))ds, n=1,2,3,.

    Then for 0<λ<λ, we have

    max0t1|un(t)uλ(t)|0 as  n;

    (ⅲ)max0t1|uλ(t)uλ0(t)|0 as λλ0, where 0<λ0<λ and if 0<λ1<λ2<λ, then there exists δ>0 such that uλ2(t)uλ1(t)δg(t),  0t1, this means that uλ1(t)uλ2(t), 0t1 and uλ1(t)uλ2(t);

    (ⅳ) for 0<λ<λ, uδλ(t)δuλ(t),  0δ1,  0t1.

    (ⅴ) Denote η=gE, for any given λ>0, by limuf(t,u)u=0, there exists a large T>0, such that f(t,T)(ηλ)1T,  0t1. Let v0(t)=η1Tg(t), from Lemma 2.5, we have

    λ(Tv0)(t)=λ10G(t,s)φ(s)f(s,η1Tg(s))dsλ10G(t,s)φ(s)f(s,T)ds(ηλ)1Tλg(t)=η1Tg(t)=v0(t),
    v0(t)λ(Tv0)(t)=10G(t,s)φ(s)(η1Tλf(s,η1Tg(s)))dsM2g(t),  0t1,

    where M2=sup0t1[η1T+λf(s,T)]η1T+λβ(T+1), this means that v0(t)λTv0˜K. That is λTv0v0. By Lemma 2.4 (v), we have λ>λ, because of the arbitrariness of λ, we can get λ=. Theorem 3.4 Assume that (H1),(H2) hold. Furthermore, the following conditions hold:

    (ⅰ) there exists a number 0<r<1 such that

    f(t,ωu)ωrf(t,u),  0t1, u0, 0<ω<1;

    (ⅱ) there exists a number ξ>0 such that f(t,1)ξ;

    (ⅲ) min0t1f(t,g(t))>0, where g(t) is given by (3.7).

    Then the following results hold:

    (ⅰ) for any given λ>0, the fractional boundary value problem (1.1), (1.2) has a unique positive solution uλ in ˚˜K;

    (ⅱ) if 0<λ1<λ2, then uλ1(t)uλ2(t), 0t1 and uλ1(t)uλ2(t);

    (ⅲ) max0t1|uλ(t)uλ0(t)|0 as λλ0;

    (ⅳ) max0t1|uλ(t)|+ as λ+, max0t1|uλ(t)|0 as λ0+.

    Proof. For u>1, from the condition (ⅰ), we have

    f(t,1)=f(t,1uu)(1u)rf(t,u),

    thus, we get

    f(t,u)urf(t,1)ξur.

    So, for uK, we have

    010G(t,s)φ(s)f(s,u(s))ds10G(t,s)φ(s)f(s,uE)dsM3g(t), 0t1,

    where M3=max0t1f(t,uE)max0t1f(t,uE+1)ξ(1+uE)r,

    Thus, 0Tu(t)M3g(t),  0t1, this means that T:˜K˜K.

    For u˚˜K, there exists δ>0 such that u(t)δg(t)0,  0t1. Let ω(0,1) such that ω<δ, we have

    (Tu)(t)=10G(t,s)φ(s)f(s,u(s))ds10G(t,s)φ(s)f(s,δg(s))ds10G(t,s)φ(s)f(s,ωg(s))dsωr10G(t,s)φ(s)f(s,g(s))ds.

    Set a=min0t1f(t,g(t)), then a>0 and (Tu)(t)aωrg(t), 0t1, so T:˚˜K˚˜K. From f(t,) is increasing, we can get that T is increasing. Moreover, for u˚˜K and ω(0,1), we have

    T(ωu)(t)=10G(t,s)φ(s)f(s,ωu(s))ds10G(t,s)φ(s)ωrf(s,u(s))ds=ωr(Tu)(t).

    In the following, we consider the operator equation

    (Tu)(t)=ρu(t). (3.10)

    Then all the conditions of Lemma 2.6 are satisfied. Thus, from Lemma 2.6, for any given ρ>0, (3.10) has a unique solution uρ in ˚˜K, uρ is strongly decreasing, uρ is continuous, this means that uρuρ00 as ρρ0 and furthermore, limρ+uρ=0,   limρ0+uρ=+. Set λ=1ρ, λ0=1ρ0, λ1=1ρ1, λ2=1ρ2. Then (3.10) is equivalent to u(t)=λ(Tu)(t). As a result, we can get

    (ⅰ) for any given λ>0, the fractional boundary value problem (1.1), (1.2) has a unique positive solution uλ in ˚˜K;

    (ⅱ) if 0<λ1<λ2, there exists δ>0 such that uλ2(t)uλ1(t)δg(t) which implies uλ1(t)uλ2(t) and uλ1(t)uλ2(t), 0t1;

    (ⅲ) max0t1|uλ(t)uλ0(t)|0 as λλ0;

    (ⅳ) max0t1|uλ(t)|+ as λ+,  max0t1|uλ(t)|0 as λ0+.

    Example 4.1 Consider the fractional boundary value problem

    D32, t20+u(t)+λ(t2+1+u13)=0,   0<t<1, (4.1)
    u(0)=0,   u(1)=12u(12). (4.2)

    We note that α=32,β=12,η=12,φ(t)=t2,f(t,u)=t2+1+u13. Thus, we have f(t,u):[0,1]×R+R+ is continuous and increasing in u,f(t,u) is concave. We can check that conditions (H1),(H2) are satisfied. Choose α=1,β=3 and f(t,u) satisfy

    f(t,0)=t2+11,  f(t,1)=t2+23,  0t1,
    limuf(t,u)u=limut2+1+u13u=0.

    Then all the assumptions of Theorem 3.3 are satisfied. Thus from Theorem 3.3, we have λ=, and the following conclusions hold:

    (ⅰ) for λ>0, the fractional boundary value problem (4.1), (4.2) has a unique positive solution uλ in ˜K;

    (ⅱ) for any u0˜K, let

    un(t)=λ10G(t,s)φ(s)f(s,un1(s))ds, n=1,2,3,.

    Then for λ>0, we have

    max0t1|un(t)uλ(t)|0 as  n;

    (ⅲ) max0t1|uλ(t)uλ0(t)|0 as λλ0 and if 0<λ1<λ2<+, then uλ1(t)uλ2(t),  0t1 and uλ1(t)uλ2(t);

    (ⅳ) uvλ(t)vuλ(t) for 0<λ<+,  0v1,  0t1.

    In this paper, we discussed a class of fractional derivative with respect to another function called φ-Riemann-Liouville fractional derivative. The new derivative herein generalizes the classical definitions of derivatives by choosing particular forms of φ(t). The main advantage of our paper is we not only develop the existence and uniqueness of positive solutions to the fractional boundary value problem, but also give some properties of positive solutions to the problem.

    We express our sincere thanks to the anonymous reviewers for their valuable comments and suggestions. This work is supported by the Natural Science Foundation of Tianjin (No.(19JCYBJC30700)).

    The authors declare no conflict of interest in this paper.



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