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Positive solutions for fractional iterative functional differential equation with a convection term

  • In this paper, we deal with the fractional iterative functional differential equation nonlocal boundary value problem with a convection term. By using the fixed point theorems, some results about existence, uniqueness, continuous dependence and multiplicity of positive solutions are derived.

    Citation: Qingcong Song, Xinan Hao. Positive solutions for fractional iterative functional differential equation with a convection term[J]. Electronic Research Archive, 2023, 31(4): 1863-1875. doi: 10.3934/era.2023096

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  • In this paper, we deal with the fractional iterative functional differential equation nonlocal boundary value problem with a convection term. By using the fixed point theorems, some results about existence, uniqueness, continuous dependence and multiplicity of positive solutions are derived.



    In this paper, we study the fractional iterative functional differential equation with a convection term and nonlocal boundary condition

    {CDα0+u(t)+λu(t)=f(u[0](t),u[1](t),,u[N](t)),0<t<1,u(0)=0,u(1)=φ(u), (1.1)

    where CDα0+ denotes the Caputo derivative of order α, 1<α<2, λR, u[0](t)=t, u[1](t)=u(t),,u[N](t)=u[N1](u(t)). φ(u)=10u(s)dA(s) is a Stieltjes integral with a signed measure, that is, A is a function of bounded variation.

    During the recent few decades, a vast literature on fractional differential equations has emerged, see [1,2,3,4,5,6] and the references therein. On the excellent survey of these related documents it is pointed out that the applicability of the theoretical results to fractional differential equations arising in various fields, for instance, chaotic synchronization [3], signal propagation [4], viscoelasticity [5], dynamical networks with multiple weights [6], and so on. Recently, we notice that the study of Caputo fractional differential equations with a convection term has become a heat topic (see [7,8,9,10]).

    In [7], Meng and Stynes considered the Green function and maximum principle for the following Caputo fractional boundary value problem (BVP)

    {CDα0+u(t)+bu(t)=f(t),0<t<1,u(0)β0u(0)=γ0,u(1)+β1u(1)=γ1,

    where 1<α<2, b,β0,β1,γ0,γ1R and fC[0,1] are given. Bai et al. [8] studied the Green function of the above problem, and the results obtained improve some conclusions of [7] to some degree.

    Wang et al. [9] used operator theory to establish the Green function for the following problem

    {CDαa+u(t)+λu(t)=h(t),a<t<b,u(a)β0u(a)=γ0,u(b)+β1u(b)=γ1,

    where 1<α<2, the constants λ,β0,β1,γ0,γ1 and the function hC[a,b] are given. The methods are entirely different from those used in [7,8], and the results generalize corresponding ones in [7,8].

    In [10], Wei and Bai investigated the following fractional order BVP

    {CDα0+u(t)+bu(t)=f(t,u(t)),x(0,1),u(0)β0u(0)=0,u(1)+β1u(1)=0,

    where 1<α2 and b,β0,β1R are constants. By employing the Guo-Krasnoselskii fixed point theorem and Leggett-Williams fixed point theorem, the existence and multiplicity results of positive solutions are presented.

    Now, not only fractional differential equations have been studied constantly (see [11,12,13,14,15,16,17]), but also iterative functional differential equations have been discussed extensively as valuable tools in the modeling of many phenomena in various fields of scientific and engineering disciplines, for example, see [18,19,20,21,22,23,24,25,26] and the references cited therein. In [22], Zhao and Liu used the Krasnoselskii fixed point theorem to discuss the existence of periodic solutions of an iterative functional differential equation

    u(t)=c1(t)u(t)+c1(t)u[2](t)++cn(t)u[N](t)+F(t).

    For the general iterative functional differential equation

    u(t)=f(u[0](t),u[1](t),,u[N](t)),

    the existence, uniqueness, boundedness and continuous dependence on initial data of positive solutions was considered in [23].

    In [24], the authors studied the following BVP

    {u(t)+h(u[0](t),u[1](t),,u[N](t))=0,btb,u(b)=η1,u(b)=η2,η1,η2[b,b],

    where h:[b,b]×RNR is continuous. By using the fixed point theorems, the authors established the existence, uniqueness and continuous dependence of a bounded solution.

    To the best of our knowledge, there are few researches on fractional iterative functional differential equations integral boundary value problem with a convection term. Motivated by the above works and for the purpose to contribute to filling these gaps in the literature, this paper mainly focuses on handing with the existence, uniqueness, continuous dependence and multiplicity of positive solutions for the fractional iterative functional differential equation nonlocal BVP (1.1).

    By a positive solution u of (1.1), we mean u(t)>0 for t[0,1] and satisfies (1.1).

    Definition 2.1.([1], [2]) The Riemann-Liouville fractional integral of order α>0 of a function f:(0,+)R is given by

    Iα0+f(t)=1Γ(α)t0(ts)α1f(s)ds,

    provided the right-hand side is pointwise defined on (0,+), where Γ(x)=+0tx1etdt (x>0) is the gamma function.

    Definition 2.2.([1], [2]) The Caputo fractional derivative of order α>0 of a function f:(0,)R is given by

    CDα0+f(t)=1Γ(mα)t0(ts)mα1f(m)(s)ds,

    provided the right-hand side is pointwise defined on (0,+), where m=[α]+1.

    Definition 2.3.[1] The two-parameter Mittag-Leffler function is defined by

    Eα,γ(x):=k=0xkΓ(αk+γ), for  α>0, γ>0  and  xR.

    Lemma 2.1. [7] Let Fβ(x)=xβ1Eα1,β(λxα1). Then Fβ has the following properties:

    (P1):[Fβ+1(x)]=Fβ(x) for β0 and x0;

    (P2):F1(0)=1, Fβ(0)=0 for β>1;

    (P3):F1(x)>0 for x>0, F2(x) is increasing for x0;

    (P4):Fα1(x)>0 for x>0, Fα(x) is increasing for x>0;

    (P5):F1(x)=λFα(x)+1 for 0x1.

    For β>0 and ν>0 one has by [1]

    (Iβ0+Fν)(t)=1Γ(β)t0(ts)β1sν1Eα1,ν(λsα1)ds=tβ+ν1Eα1,β+ν(λtα1).

    That is to say,

    (Iβ0+Fν)(t)=Fβ+ν(t),0<t1. (2.1)

    Lemma 2.2. Suppose that hAC[0,1] and φ(1)1. Then uAC2[0,1] is the solution of

    {CDα0+u(t)+λu(t)=h(t),0<t<1, 1<α<2,u(0)=0, u(1)=φ(u), (2.2)

    if and only if the function u satisfies u(t)=10H(t,s)h(s)ds, where

    H(t,s)=11φ(1)GA(s)+G(t,s),
    GA(s)=10G(t,s)dA(t),G(t,s)={Fα(1s)Fα(ts),0st1,Fα(1s),0ts1.

    Proof. Applying Iα0+ to the both sides of the Eq (2.2), we know by simple calculation that the general solution of (2.2) is given by

    u(t)=C0+C1F2(t)t0Fα(ts)h(s)ds,t[0,1]. (2.3)

    Then

    u(t)=C1F1(t)t0Fα1(ts)h(s)ds.

    In view of u(0)=0 and u(1)=φ(u), we deduce by (2.3) that

    C1=0,C0=φ(u)+10Fα(1s)h(s)ds.

    Therefore,

    u(t)=φ(u)+10Fα(1s)h(s)dst0Fα(ts)h(s)ds=φ(u)+10G(t,s)h(s)ds,t[0,1]. (2.4)

    Direct computations yield

    φ(u)=10φ(u)dA(t)+1010G(t,s)h(s)dsdA(t)=φ(u)φ(1)+1010G(t,s)dA(t)h(s)ds=φ(u)φ(1)+10GA(s)h(s)ds.

    It follows that

    φ(u)=11φ(1)10GA(s)h(s)ds.

    Substituting it to (2.4), we have u(t)=10H(t,s)h(s)ds, t[0,1].

    Conversely, due to

    u(t)=10H(t,s)h(s)ds=10(11φ(1)GA(s)+G(t,s))h(s)ds,

    we obtain

    φ(u)=φ(1)1φ(1)10GA(s)h(s)ds+10GA(s)h(s)ds=11φ(1)10GA(s)h(s)ds,

    and

    u(t)=φ(u)+10G(t,s)h(s)ds=φ(u)+10Fα(1s)h(s)dst0Fα(ts)h(s)ds.

    Then, u(1)=φ(u) and u(0)=0.

    Let

    H(t)=t0Fα(ts)h(s)ds=t0Fα(s)h(ts)ds,0t1.

    Then, for almost all t[0,1],

    H(t)=h(0)Fα(t)+t0Fα(s)h(ts)ds=h(0)Fα(t)+t0Fα(ts)h(s)ds,

    and

    H(t)=h(0)Fα1(t)+t0Fα1(ts)h(s)ds.

    Then using (2.1), we calculate

    CDα0+H(t)=(I2α0+H)(t)=h(0)F1(t)+1Γ(2α)t0(tr)1αr0Fα1(rs)h(s)dsdr=h(0)F1(t)+t0h(s)[1Γ(2α)ts0(tsτ)1αFα1(τ)dτ]ds=h(0)F1(t)+t0h(s)(I2αFα1)(ts)ds=h(0)F1(t)+t0h(s)F1(ts)ds, (2.5)

    and

    λH(t)=λh(0)Fα(t)+λt0Fα(ts)h(s)ds=λh(0)Fα(t)+t0[F1(ts)1]h(s)ds. (2.6)

    Combining (2.5) with (2.6), we obtain

    CDα0+H(t)λH(t)=h(0)F1(t)λh(0)Fα(t)+t0h(s)ds=h(t)+h(0)[F1(t)λFα(t)1]=h(t),0<t<1,

    where we utilize hAC[0,1] and (P5). Consequently, we obtain CDα0+u(t)+λu(t)=h(t). The proof is finished.

    As a direct consequence of the previous results, we deduce the following properties that, as we will see, will be fundamental for subsequent studies.

    Lemma 2.3. Assume that 0φ(1)<1 and GA(s)0 for s[0,1]. Then for t,s[0,1],

    1). G(t,s) and H(t,s) are continuous;

    2). 0G(t,s)Fα(1s), and G(t,s) is decreasing with respect to t;

    3). 0H(t,s)ω(s), where ω(s)=11φ(1)GA(s)+Fα(1s), and H(t,s) is decreasing with respect to t.

    Let E=C[0,1]. Then E is a Banach space with the usual maximum norm ||u||=maxt[0,1]|u(t)|. For 0P1 and L>0, define

    Ω(P,L)={uE:0u(t)P, |u(t2)u(t1)|L|t2t1|, t,t1,t2[0,1]}.

    It is easy to show that Ω(P,L) is a convex and compact set.

    Define an operator Tλ:EE as follows:

    (Tu)(t)=10H(t,s)f(u[0](s),u[1](s),,u[N](s))ds,t[0,1].

    By Lemma 2.2, we can easily know that u is a solution of BVP (1.1) iff u is the fixed point of the operator T.

    Suppose that

    (H1) for the function f:[0,1]N+1[0,+), there exist constants 0<k0,k1,,kN<+ such that

    |f(t,u1,u2,,uN)f(s,v1,,vN)|k0|ts|+Nj=1kj|ujvj|.

    (H2) 0φ(1)<1, and A is a function of bounded variation such that GA(s)0 for s[0,1].

    Clearly, using (H1), we obtain

    |f(t,u1,u2,,uN)|k0|t|+β+Nj=1kj|uj|, (2.7)

    where β=|f(0,0,,0)|.

    Lemma 2.4.[22] For any u,vΩ(P,L),

    u[n]v[n]n1i=0Liuv,n=1,2,,N.

    Lemma 2.5.[27] Let P be a cone in a real Banach space E, Pc={u P:uc}, P(θ,a,b)={uP:aθ(u), ub}. Suppose A:PcPc is completely continuous, and suppose there exists a concave positive functional θ with θ(u)u (u P) and numbers a,b and d with 0<d<a<bc, satisfying the following conditions:

    (C1) {uP(θ,a,b):θ(u)>a} and θ(Au)>a if uP(θ,a,b);

    (C2) Au<d if uPd;

    (C3) θ(Au)>a for all u  P(θ,a,c) with Au > b.

    Then A has at least three fixed points u1,u2,u3Pc with

    u1<d, θ(u2)>a, u3>d, θ(u3)<a.

    Remark. If b=c, then (C1) implies (C3).

    Theorem 3.1. Suppose that (H1) and (H2) hold. If

    (k0+β+Nj=1j1i=0kjLiP)10ω(s)dsP, (3.1)

    and

    (k0+β+Nj=1j1i=0kjLiP)Fα(1)L, (3.2)

    then problem (1.1) has a unique nonnegative solution. If in addition A is an increasing function, and there exists t0[0,1] such that f(t0,0,,0)>0, then problem (1.1) has a unique positive solution.

    Proof. For any uΩ(P,L), in view of (2.7) and Lemma 2.4, we deduce that

    |f(u[0](s),u[1](s),u[N](s))|k0|s|+β+Nj=1j1i=0kjLiuk0+β+Nj=1j1i=0kjLiP,0s1,

    and hence

    |(Tu)(t)|10ω(s)|f(u[0](s),u[1](s),,u[N](s))|ds(k0+β+Nj=1j1i=0kjLiP)10ω(s)dsP,t[0,1].

    Therefore, 0(Tu)(t)P for t[0,1].

    On the other hand, for any t1,t2[0,1] and t1<t2, by means of Lagrange mean value theorem, there exists ξ(t1,t2)(0,1) such that

    10|H(t2,s)H(t1,s)|ds=10(G(t1,s)G(t2,s))ds=t20Fα(t2s)dst10Fα(t1s)ds=Fα+1(t2)Fα+1(t1)=Fα(ξ)(t2t1)Fα(1)(t2t1).

    It follows from (3.2) that

    |(Tu)(t2)(Tu)(t1)|(k0+β+Nj=1j1i=0kjLiP)10|H(t2,s)H(t1,s)|ds(k0+β+Nj=1j1i=0kjLiP)Fα(1)(t2t1)L(t2t1).

    Therefore, T(Ω(P,L))Ω(P,L).

    Next, we show that T is a contraction mapping on Ω(P,L). Indeed, let u,vΩ(P,L). Then

    |f(u[0](s),u[1](s),,u[N](s))f(v[0](s),v[1](s),,v[N](s))|Nj=1kju[j]v[j]Nj=1j1i=0kjLiuv,

    and

    TuTv10ω(s)|f(u[0](s),u[1](s),,u[N](s))f(v[0](s),v[1](s),,v[N](s))|dsNj=1j1i=0kjLiuv10ω(s)ds.

    It follows from (3.1) that

    Nj=1j1i=0kjLi10ω(s)ds<1.

    This shows that T is a contraction mapping on Ω(P,L). It follows from the contraction mapping theorem that T has a unique fixed point u in Ω(P,L). In other words, problem (1.1) has a unique nonnegative solution.

    Suppose u is the nonnegative solution to problem (1.1). Then

    u(t)=10H(t,s)f(u[0](s),,u[N](s))ds,t[0,1].

    By the monotonicity of H(t,s), we have u(t)u(1)0 for t[0,1]. If A is an increasing function, and there exists t0[0,1] such that f(t0,0,,0)>0, we must have u(1)>0. Otherwise, u(1)=0 and we have 10u(s)dA(s)=φ(u)=u(1)=0. Then u(t)0 for t[0,1]. By the equation of (1.1), we conclude f(t,0,,0)0 for t[0,1], which is a contradiction. We have thus proved u(1)>0 and u(t)u(1)>0 for t[0,1].

    Theorem 4.1. Assume that the conditions of Theorem 3.1 are satisfied. Then the unique positive solution of problem (1.1) continuously depends on function f.

    Proof. For two continuous functions f1,f2:[0,1]N+1[0,+), they correspond respectively to unique solutions u1 and u2 in Ω(P,L) such that

    ui(t)=10H(t,s)fi(u[0]i(s),u[1]i(s),,u[N]i(s))ds,t[0,1], i=1,2.

    By (H1), we find that

    |f2(u[0]2(s),u[1]2(s),,u[N]2(s))f1(u[0]1(s),u[1]1(s),,u[N]1(s))||f2(u[0]2(s),u[1]2(s),,u[N]2(s))f2(u[0]1(s),u[1]1(s),,u[N]1(s))|+|f2(u[0]1(s),u[1]1(s),,u[N]1(s))f1(u[0]1(s),u[1]1(s),,u[N]1(s))|f2f1+Nj=1j1i=0kjLiu2u1.

    It follows from Lemma 2.3 that

    |u2(t)u1(t)|(f2f1+Nj=1j1i=0kjLiu2u1)10ω(s)ds,t[0,1].

    Then

    u2u110ω(s)ds1Nj=1j1i=0kjLi10ω(s)dsf2f1.

    The proof is complete.

    Define

    Ω={uE:u(t)0, t[0,1]},Ωc={uΩ:u<c},
    M=(10ω(s)ds)1, m=(5616mint[16,56]H(t,s)ds)1,
    θ(u)=min1/6t5/6|u(t)|,Ω(θ,b,d)={uΩ:bθ(u), ud}.

    Obviously, θ is a continuous concave functional and θ(u)u for uΩ.

    Theorem 5.1.Assume that fAC([0,1]N+1,[0,+)) and (H2) hold. If there exist constants 0<a<16b<c56 such that

    (D1) f(t,u1,u2,,uN)<Ma,(t,u1,u2,,uN)[0,1]×[0,a]N;

    (D2) f(t,u1,u2,,uN)>mb,(t,u1,u2,,uN)[16,56]×[b,c]N;

    (D3) f(t,u1,u2,,uN)Mc,(t,u1,u2,,uN)[0,1]×[0,c]N,

    then problem (1.1) has three non-negative solutions u1,u2,u3¯Ωc with

    u1<a, θ(u2)>b, u3>a, θ(u3)<b.

    Proof. We first prove T:¯Ωc¯Ωc is completely continuous. For u¯Ωc, we have uc<1. Then 0u[j](s)c for j=1,2,,N and 0s1. It follows from (D3) that

    Tu10ω(s)f(u[0](s),u[1](s),,u[N](s))dsMc10ω(s)ds=c.

    Therefore, T(¯Ωc)¯Ωc and TD is uniformly bounded for any bounded set D¯Ωc. We denote ¯M as the maximum of f on [0,1]N+1. Since H(t,s) is uniformly continuous on [0,1]×[0,1], for any ε>0, there exists δ>0 such that for any uD,t1,t2[0,1] and |t2t1|<δ, we have |H(t2,s)H(t1,s)|<ε¯M. Then,

    |Tu(t2)Tu(t1)|¯M10|H(t2,s)H(t1,s)|ds<ε,

    which implies that TD is equicontinuous. Clearly, the fact that f is continuous implies that T is continuous. Hence, T:¯Ωc¯Ωc is completely continuous.

    For any u¯Ωa, due to (D1), we conclude

    Tu10ω(s)f(u[0](s),u[1](s),,u[N](s))ds10ω(s)Mads=a.

    Then, T(¯Ωa)¯Ωa, which implies that (C2) in Lemma 2.5 holds.

    Choose v(t)=c+b2, 0t1. Obviously, v{uΩ(θ,b,c):θ(u)>b}. Then {uΩ(θ,b,c):θ(u)>b}. For uΩ(θ,b,c), we have 16bu(t)c56 for 16t56. Then bu[j](s)c for j=1,2,,N and 16s56. It follows from (D2) that

    θ(Tu)=min1/6t5/6|(Tu)(t)|>5616min1/6t5/6H(t,s)mbds=b,

    which implies that (C1) in Lemma 2.5 holds. By remark under Lemma 2.5, we know that (C3) in Lemma 2.5 holds. Then according to Lemma 2.5, problem (1.1) has three nonnegative solutions u1,u2,u3¯Ωc with

    u1<a, θ(u2)>b, u3>a, θ(u3)<b.

    We consider the following BVP

    {CD320+u(t)+u(t)=f(t,u(t),u[2](t),u[3](t)),0<t<1,u(0)=0, u(1)=10u(s)dA(s), (6.1)

    where f(t,u(t),u[2](t),u[3](t))=1200t+1100sin(u(t))+1100sin(u[2](t))+1100sin(u[3](t)), A(s)=12s. It follows that φ(1)=10d(12s)=12 and GA(s)=10G(t,s)d(12t)0, which implies (H2) holds. Since

    |f(t,u(t),u[2](t),u[3](t))f(s,u(s),u[2](s),u[3](s))|1200|ts|+1100|u(t)u(s)|+1100|u[2](t)u[2](s)|+1100|u[3](t)u[3](s)|,

    we obtain k0=1200, k1=1100, k2=1100, k3=1100 and β=0, which implies (H1) holds.

    Direct computation shows that 10ω(s)ds=10[11φ(1)GA(s)+F32(1s)]ds<15950. Choose P=34 and L=1, we have

    (k0+β+3j=1j1i=0kjLiP)10ω(s)ds<1591000,

    and

    (k0+β+3j=1j1i=0kjLiP)F32(1)<310.

    Taking t0=12[0,1], we find that f(t0,0,0,0)=1400>0. By Theorem 3.1 we conclude that BVP (6.1) has a unique positive solution.

    This work is supported by the Natural Science Foundation of Shandong Province of China (ZR2022MA049) and the National Natural Science Foundation of China (11871302, 11501318).

    The authors declare there is no conflicts of interest.



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