# Existence and Uniqueness of Positive Solutions of Boundary-Value Problems for Fractional Differential Equations withp-Laplacian Operator and Identities on the Some Special Polynomials

Authors: Erdoğan Şen; Mehmet Acikgoz; Jong Jin Seo; Serkan Araci; Kamil Oruçoğlu
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We consider the following boundary-value problem of nonlinear fractional differential equation withp-Laplacian operator D0+β(ϕp(D0+αu(t)))+a(t)f(u)=0, 0<t<1, u(0)=γu(h)+λ, u′(0)=μ, ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(0)))′′=(ϕp(D0+αu(0)))′′′=0, where 1<α⩽2, 3<β⩽4 are real numbers, D0+α,D0+β are the standard Caputo fractional derivatives, ϕp(s)=|s|p-2s, p>1, ϕp-1=ϕq, 1/p+1/q=1,0⩽γ<1, 0⩽h⩽1, λ,μ>0 are parameters, a:(0,1)→[0,+∞), and f:[0,+∞)→[0,+∞) are continuous. By the properties of Green function and Schauder fixed point theorem, several existence and nonexistence results for positive solutions, in terms of the parameters λ and μ are obtained. The uniqueness of positive solution on the parameters λ and μ is also studied. In the final section of this paper, we derive not only new but also interesting identities related special polynomials by which Caputo fractional derivative.


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## Body

## 1. Introduction

In 1695, L’Hôpital asked Leibniz: what if the order of the derivative is1/2? To which Leibniz considered in a useful means, thus it follows that will be equal to xdx:x, an obvious paradox. In recent years, fractional calculus has been studied by many mathematicians from Leibniz’s time to the present.Also, fractional differential equations arise in many engineering and scientific disciplines as the mathematical modelling of systems and processes in the fields of physics, fluid flows, electrical networks, viscoelasticity, aerodynamics, and many other branches of science. For details, see [1–9].In the last few decades, fractional-order models are found to be more adequate than integer-order models for some real world problems. Recently, there have been some papers dealing with the existence and multiplicity of solutions (or positive solutions) of nonlinear initial fractional differential equations by the use of techniques of nonlinear analysis [10–21], upper and lower solutions method [22–24], fixed point index [25, 26], coincidence theory [27], Banach contraction mapping principle [28], and so forth.Chai [11] investigated the existence and multiplicity of positive solutions for a class of boundary-value problem of fractional differential equation with p-Laplacian operator
(1)D0+β(ϕp(D0+αu(t)))+f(t,u(t),D0+ρu(t))=0,<<<<<<<<<<<<<<<<<<<<<<<<<0<t<1,u(0)=0,u(1)+σD0+γu(1)=0,D0+αu(0)=0,
where 1<α≤2, 0<γ≤1, 0≤α-γ-1, σ is a positive constant number, and D0+α,D0+β,D0+γ are the standard Riemann-Liouville derivatives. By means of the fixed point theorem on cones, some existence and multiplicity results of positive solutions are obtained.Although the fractional differential equation boundary-value problems have been studied by several authors, very little is known in the literature on the existence and nonexistence of positive solutions of fractional differential equation boundary-value problems withp-Laplacian operator when a parameter λ is involved in the boundary conditions. We also mention that, there is very little known about the uniqueness of the solution of fractional differential equation boundary-value problems with p-Laplacian operator on the parameter λ. Han et al. [29] studied the existence and uniqueness of positive solutions for the fractional differential equation with p-Laplacian operator
(2)D0+β(ϕp(D0+αu(t)))+a(t)f(u)=0,0<t<1,u(0)=γu(ξ)+λ,ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(0)))′′=0,
where 0<α⩽1, 2<β⩽3 are real numbers; D0+α,D0+β are the standard Caputo fractional derivatives; ϕp(s)=|s|p-2s, p>1. Therefore, to enrich the theoretical knowledge of the above, in this paper, we investigate the following p-Laplacian fractional differential equation boundary-value problem:
(3)D0+β(ϕp(D0+αu(t)))+a(t)f(u)=0,0<t<1,u(0)=γu(h)+λ,u′(0)=μ,ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(0)))′′=(ϕp(D0+αu(0)))′′′=0,
where 1<α⩽2, 3<β⩽4 are real numbers, D0+α,D0+β are the standard Caputo fractional derivatives, ϕp(s)=|s|p-2s, p>1, ϕp-1=ϕq, 1/p+1/q=1, 0⩽γ<1, 0⩽h⩽1, λ,μ>0 are parameters, a:(0,1)→[0,+∞), and f:[0,+∞)→[0,+∞) are continuous. By the properties of Green function and Schauder fixed point theorem, several existence and nonexistence results for positive solutions, in terms of the parameters λ and μ are obtained. The uniqueness of positive solution on the parameters λ and μ, is also studied.


## 2. Preliminaries and Related Lemmas

Definition 1 (see [4]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:(0,+∞)→ℝ is given by
(4)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds
provided the right side is pointwise defined on (0,+∞).Definition 2 (see [4]).
The Caputo fractional derivative of orderα>0 of a continuous function y:(0,+∞)→ℝ is given by
(5)D0+αy(t)=1Γ(n-α)∫0t‍y(n)(s)(t-s)α-n+1ds,
where n is the smallest integer greater than or equal to α, provided that the right side is pointwise defined on (0,+∞).Remark 3 (see [8]).
By Definition2, under natural conditions on the function f(t), for α→n, the Caputo derivative becomes a conventional nth derivative of the function f(t).Remark 4 (see [4]).
As a basic example,(6)D0+αtμ=μ(μ-1)⋯(μ-n+1)×Γ(1+μ-n)Γ(1+μ-α)tμ-α,fort∈(0,∞).
In particular, D0+αtμ=0, μ=0,1,…,n-1, where D0+α is the Caputo fractional derivative, n is the smallest integer greater than or equal to α.From the definition of the Caputo derivative and Remark4, we can obtain the following statement.Lemma 5 (see [4]).
Letα>0. Then, the fractional differential equation
(7)D0+αu(t)=0
has a unique solution
(8)u(t)=c0+c1t+c2t2+⋯+cn-1tn-1,====ci∈ℝ,i=0,1,2,…,n-1,
where n is the smallest integer greater than or equal to α.Lemma 6 (see [4]).
Letα>0. Assume that u∈Cn[0,1]. Then
(9)I0+αD0+αu(t)=u(t)+c0+c1t+c2t2+⋯+cn-1tn-1,
for some ci∈ℝ, i=0,1,2,…,n-1, where n is the smallest integer greater than or equal to α.Lemma 7.
Lety∈C[0,1] and 1<α≤2. Then, fractional differential equation boundary-value problem
(10)D0+αu(t)=y(t),0<t<1,(11)u(0)=γu(h)+λ,u′(0)=μ
has a unique solution
(12)u(t)=∫0t‍(t-s)α-1Γ(α)y(s)ds+γ1-γ∫0h‍(h-s)α-1Γ(α)y(s)ds+λ+γμh1-γ.Proof.
We apply Lemma6 to reduce (10) to an equivalent integral equation,
(13)u(t)=I0+αy(t)+c0+c1t,c0,c1∈ℝ.
Consequently, the general solution of (10) is
(14)u(t)=∫0t‍(t-s)α-1Γ(α)y(s)ds+c0+c1t,c0,c1∈ℝ.
By (11), we have
(15)c0=γ1-γ∫0h‍(h-s)α-1Γ(α)y(s)ds+γc1h1-γ+λ1-γ,
and since u′(0)=c1, we have by (11)
(16)c1=μ.
Therefore, the unique solution of problem (10) and (11) is
(17)u(t)=∫0t‍(t-s)α-1Γ(α)y(s)ds+γ1-γ∫0h‍(h-s)α-1Γ(α)y(s)ds+λ1-γ+γμh1-γ.Lemma 8.
Lety∈C[0,1] and 1<α≤2, 3<β≤4. Then, fractional differential equation boundary-value problem
(18)D0+β(ϕp(D0+αu(t)))+y(t)=0,0<t<1,(19)u(0)=γu(h)+λ,u′(0)=μ,ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(0)))′′=(ϕp(D0+αu(0)))′′′=0,
has a unique solution
(20)u(t)=∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+γ1-γ∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ,
where
(21)H(t,s)={t(β-1)(1-s)β-2-(t-s)β-1Γ(β),0≤s≤t≤1,t(β-1)(1-s)β-2Γ(β),0≤t≤s≤1.Proof.
From Lemma6, the boundary-value problems (18) and (19) are equivalent to the integral equation
(22)ϕp(D0+αu(t))=-I0+βy(t)+c0+c1t+c2t2+c3t3,
for some c0,c1,c2,c3∈ℝ; that is,
(23)ϕp(D0+αu(t))=-∫0t‍(t-τ)β-1Γ(β)y(τ)dτ+c0+c1t+c2t2+c3t3.
By the boundary conditions ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(0)))′′=(ϕp(D0+αu(0)))′′′=0, we have
(24)c0=c2=c3=0,c1=∫01‍(β-1)(1-τ)β-2Γ(β)y(τ)dτ.
Therefore, the solution u(t) of fractional differential equation boundary-value problems (18) and (19) satisfies
(25)ϕp(D0+αu(t))=-∫0t‍(t-τ)β-1Γ(β)y(τ)dτ+t∫01‍(β-1)(1-τ)β-2Γ(β)y(τ)dτ=∫01‍H(t,τ)y(τ)dτ.
Consequently, D0+αu(t)=ϕq(∫01‍H(t,τ)y(τ)dτ). Thus, fractional differential equation boundary-value problem (18) and (19) is equivalent to the problem
(26)D0+αu(t)=ϕq(∫01‍H(t,τ)y(τ)dτ),0<t<1,u(0)=γu(h)+λ,u′(0)=μ.
Lemma 7 implies that the fractional differential equation boundary-value problems (18) and (19) have a unique solution,
(27)u(t)=∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+γ1-γ∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ.
The proof is complete.Lemma 9 (see [15]).
Let1<α⩽2,3<β⩽4. The function H(t,s) is continuous on [0,1]×[0,1] and satisfies  (1)
H(t,s)⩾0,H(t,s)⩽H(1,s), for t,s∈[0,1];(2)
H(t,s)⩾tβ-1H(1,s), for t,s∈(0,1).Lemma 10 (Schauder fixed point theorem [30]).
Let(E,d) be a complete metric space, U be a closed convex subset of E, and A:U→U be a mapping such that the set {Au:u∈U} is relatively compact in E. Then, A has at least one fixed point.To prove our main results, we use the following assumptions.(H1)
0<∫01‍H(1,τ)a(τ)dτ<+∞;(H2)
there exist0<σ<1 and c>0 such that
(28)f(x)⩽σLϕp(x),for0⩽x⩽c,whereL satisfies
(29)0<L⩽[ϕp(1+γ(hα-1)Γ(α+1)(1-γ))∫01‍H(1,τ)a(τ)dτ]-1;(H3)
there existd>0 such that
(30)f(x)⩽Mϕp(x),ford<x<+∞,whereM satisfies
(31)0<M<[ϕp(1+γ(hα-1)Γ(α+1)(1-γ)2q-1)∫01‍H(1,τ)a(τ)dτ]-1;(H4)
there exist0<δ<1 and e>0 such that
(32)f(x)⩾Nϕp(x),fore<x<+∞,whereN satisfies
(33)N>[ϕp(cδ∫01‍(1-s)α-2Γ(α)ϕq(sβ-1)ds)f×∫δ1‍H(1,τ)a(τ)dτ(cδ∫01‍(1-s)α-2Γ(α)ϕq(sβ-1)ds)]-1;with(34)cδ=∫0δ‍α(1-s)α-2ϕq(sβ-1)ds∈(0,1);(H5)
f(x) is nondecreasing in x;(H6)
there exist0⩽θ<1 such that
(35)f(kx)⩾(ϕp(k))θf(x),forany0<k<1,0<x<+∞.Remark 11.
Let(36)f0=limx→0+f(x)ϕp(x),f∞=limx→+∞f(x)ϕp(x).
Then, (H2) holds if f0=0, (H3) holds if f∞=0, and (H4) holds if f∞=+∞.


## 3. Existence

Theorem 12.
Assume that (H1), (H2) hold. Then, the fractional differential equation boundary-value problem (3) has at least one positive solution for 0<λ+γμ≤(1-γ)(1-ϕq(σ))c.Proof.
Letc>0 be given in (H2). Define
(37)K1={u∈C[0,1]:0⩽u(t)⩽con[0,1]}
and an operator Tλ:K1→C[0,1] by
(38)Tλu(t)=∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(s,τ)a(τ)f(u(τ))dτ)+γ1-γ∫0h‍(h-s)α-1Γ(α)ϕqffffff×(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+λ+γμh1-γ.
Then, K1 is a closed convex set. From Lemma 8, u is a solution of fractional differential equation boundary-value problem (3) if and only if u is a fixed point of Tλ. Moreover, a standard argument can be used to show that Tλ is compact.
For anyu∈K1, from (28) and (29), we obtain
(39)f(u(t))⩽σLϕp(u(t))⩽σLϕp(c),on[0,1],1+γ(hα-1)Γ(α+1)(1-γ)ϕq(L)ϕq(∫01‍H(1,τ)a(τ)dτ)⩽1.
Let 0<λ+γμ⩽(1-γ)(1-ϕq(σ))c. Then, from Lemma 6 and (38), it follows that
(40)0⩽Tλu(t)⩽∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)ds+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)ds+λ+γμh1-γ⩽1Γ(α+1)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)+γhαΓ(α+1)(1-γ)ϕq×(∫01‍H(1,τ)a(τ)f(u(τ))dτ)+(1-ϕq(σ))c=1+γhα-γΓ(α+1)(1-γ)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)+(1-ϕq(σ))c⩽1+γ(hα-1)Γ(α+1)(1-γ)ϕq(L)ϕq(∫01‍H(1,τ)a(τ)dτ)ϕq(σ)c+(1-ϕq(σ))c⩽ϕq(σ)c+(1-ϕq(σ))c=c,t∈[0,1].
Thus, Tλ(K1)⊆K1, by Schauder fixed point theorem, Tλ has a fixed point u∈K1; that is, the fractional differential equation boundary-value problem (3) has at least one positive solution. The proof is complete.Corollary 13.
Assume that (H1) holds andf0=0. Then, the fractional differential equation boundary-value problem (3) has at least one positive solution for sufficiently small λ>0.Theorem 14.
Assume that (H1), (H3) hold. Then, the fractional differential equation boundary-value problem (3) has at least one positive solution for all λ>0.Proof.
Letλ>0 be fixed and d>0 be given in (H3). Define D=max0⩽x⩽df(x). Then
(41)f(x)⩽D,for0⩽x⩽d.
From (31), we have
(42)1+γ(hα-1)Γ(α+1)(1-γ)2q-1ϕq(M)ϕq(∫01‍H(1,τ)a(τ)dτ)<1.
Thus, there exists d*>d large enough so that
(43)1+γ(hα-1)Γ(α+1)(1-γ)2q-1(ϕq(D)+ϕq(M)d*)ϕq×(∫01‍H(1,τ)a(τ)dτ)+λ+γμh1-γ⩽d*.
Let
(44)K2={u∈C[0,1]:0⩽u(t)⩽d*on[0,1]}.
For u∈K2, define
(45)I1u={t∈[0,1]:0⩽u(t)⩽d},I2u={t∈[0,1]:d<u(t)⩽d*}.
Then, I1u∪I2u=[0,1], I1u∩I2u=∅, and in view of (30), we have
(46)f(u(t))⩽Mϕp(u(t))⩽Mϕp(d*),fort∈I2u.
Let the compact operator Tλ be defined by (38). Then, from Lemma 6, (30), and (41), we have
(47)0⩽Tλu(t)⩽∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)ds+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)ds+λ+γμh1-γ⩽1Γ(α+1)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)+γhαΓ(α+1)(1-γ)ϕq(∫01‍H(1,τ)a(τ)f(u(τ))dτ)+λ+γμh1-γ=1+γ(hα-1)Γ(α+1)(1-γ)ϕq×(∫I1u‍H(1,τ)a(τ)f(u(τ))dτffff+∫I2u‍H(1,τ)a(τ)f(u(τ))dτ)+λ+γμh1-γ⩽1+γ(hα-1)Γ(α+1)(1-γ)ϕq×(D∫I1u‍H(1,τ)a(τ)dτffff+Mϕp(d*)∫I2u‍H(1,τ)a(τ)dτ)+λ+γμh1-γ⩽1+γ(hα-1)Γ(α+1)(1-γ)ϕq(D+Mϕp(d*))ϕq×(∫01‍H(1,τ)a(τ)dτ)+λ+γμh1-γ.
From (43) and the inequality (a+b)r⩽2r(ar+br) for any a,b,r>0 (see, e.g., [31]), we obtain
(48)0⩽Tλu(t)⩽1+γ(hα-1)Γ(α+1)(1-γ)2q-1(ϕq(D)+ϕq(M)d*)ϕq×(∫01‍H(1,τ)a(τ)dτ)+λ+γμh1-γ⩽d*.
Thus, Tλ:K2→K2. Consequently, by Schauder fixed point theorem, Tλ has a fixed point u∈K2, that is, the fractional differential equation boundary-value problem (3) has at least one positive solution. The proof is complete.Corollary 15.
Assume that (H1) holds andf∞=0. Then, the fractional differential equation boundary-value problem (3) has at least one positive solution for all λ>0.


## 4. Uniqueness

Definition 16 (see [32]).
A coneP in a real Banach space X is called solid if its interior Po is not empty.Definition 17 (see [32]).
LetP be a solid cone in a real Banach space X,T:Po→Po be an operator, and 0⩽θ<1. Then, T is called a θ-concave operator if
(49)T(ku)⩾kθTuforany0<k<1,u∈Po.Lemma 18 (see [32, Theorem 2.2.6]).
Assume thatP is a normal solid cone in a real Banach space X, 0⩽θ<1, and T:Po→Po is a θ-concave increasing operator. Then, T has only one fixed point in Po.Theorem 19.
Assume that (H1), (H5), (H6) hold. Then, the fractional differential equation boundary-value problem (3) has a unique positive solution for any λ>0.Proof.
DefineP={u∈C[0,1]:u(t)⩾0on[0,1]}. Then, P is a normal solid cone in C[0,1] with
(50)Po={u∈C[0,1]:u(t)>0on[0,1]}.
For any fixed λ>0, let Tλ:P→C[0,1] be defined by (38). Define T:P→C[0,1] by
(51)Tu(t)=∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+γ1-γ∫0h‍(h-s)α-1Γ(α)ϕqfffffff×(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds.
Then, from (H5), we have T increasing in u∈Po and
(52)Tλu(t)=Tu(t)+λ+γμh1-γ.
Clearly, Tλ:Po→Po. Next, we prove that Tλ is a θ-concave increasing operator. In fact, for u1,u2∈P with u1(t)⩾u2(t) on [0,1], we obtain
(53)Tλu1(t)⩾Tu2(t)+λ+γμh1-γ=Tλu2(t);
that is, Tλ is increasing. Moreover, (H6) implies
(54)Tλ(ku)(t)⩾kθ∫0t‍(t-s)α-1Γ(α)fffff×ϕq(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+kθγ1-γ×∫0h‍(h-s)α-1Γ(α)fffff×ϕq(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+λ+γμh1-γ=kθTu(t)+λ+γμh1-γ⩾kθ(Tu(t)+λ+γμh1-γ)=kθTλu(t),
that is, Tλ is θ-concave. By Lemma 8, Tλ has a unique fixed point uλ in Po, that is, the fractional differential equation boundary-value problem (3) has a unique positive solution. The proof is complete.


## 5. Nonexistence

In this section, we let the Banach spaceC[0,1] be endowed with the norm ∥u∥=max0⩽t⩽1|u(t)|.Lemma 20.
Assume (H1) holds and let0<δ<1 be given in (H4). Then, the unique solution u(t) of fractional differential equation boundary-value problem (18) and (19) satisfies
(55)u(t)⩾cδ∥u∥forδ⩽t⩽1,
where cδ is defined by (34).Proof.
In view of Lemma9 and (19), we have
(56)u(t)⩽∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍H(1,τ)y(τ)dτ)ds+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ⩽1Γ(α+1)ϕq(∫01‍H(1,τ)y(τ)dτ)+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ
for t∈[0,1], and
(57)u(t)⩾∫0t‍(t-s)α-1Γ(α)ϕq(∫01‍sβ-1H(1,τ)y(τ)dτ)ds+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ=∫0t‍α(t-s)α-2ϕq(sβ-1)ds1Γ(α+1)ϕq×(∫01‍H(1,τ)y(τ)dτ)+γ1-γ∫0h‍(h-s)α-1Γ(α)ϕqfffffff×(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ⩾cδ1Γ(α+1)ϕq(∫01‍H(1,τ)y(τ)dτ)+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)y(τ)dτ)ds+λ+γμh1-γ⩾cδ[1Γ(α+1)ϕq(∫01‍H(1,τ)y(τ)dτ)ffff+γ1-γ∫0h‍(h-s)α-1Γ(α)fffffffffff×ϕq(∫01‍H(s,τ)y(τ)dτ)dsffff+λ+γμh1-γ]
for t∈[δ,1]. Therefore, u(t)⩾cδ∥u∥ for δ⩽t⩽1. The proof is complete.Theorem 21.
Assume that (H1), (H4) hold. Then, the fractional differential equation boundary-value problem (3) has no positive solution for λ+γμh>(1-γ)e.Proof.
Assume, to the contrary, the fractional differential equation boundary-value problem (3) has a positive solution u(t) for λ+γμh>(1-γ)e. Then, by Lemma 8, we have
(58)u(t)=∫0t‍(t-s)α-1Γ(α)ϕq×(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+γ1-γ×∫0h‍(h-s)α-1Γ(α)ϕq(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds+λ+γμh1-γ.
Therefore, u(t)>e on [0,1]. In view of (32) and (33), we obtain
(59)f(u(t))⩾Nϕp(u(t))on[0,1],cδϕq(N)ϕq(∫δ1‍H(1,τ)a(τ)dτ)×∫01‍(1-s)α-1Γ(α)ϕq(sβ-1)ds>1.
Then, by Lemmas 6 and 9, we obtain
(60)∥u∥≥u(1)>∫01‍(1-s)α-1Γ(α)ϕq×(∫01‍H(s,τ)a(τ)f(u(τ))dτ)ds⩾∫01‍(1-s)α-1Γ(α)ϕq(sβ-1)dsϕq×(∫01‍H(1,τ)a(τ)f(u(τ))dτ)⩾∫01‍(1-s)α-1Γ(α)ϕq(sβ-1)dsϕq(N)ϕq×(∫δ1‍H(1,τ)a(τ)ϕp(u(τ))dτ)⩾∥u∥cδ∫01‍(1-s)α-1Γ(α)ϕq(sβ-1)dsϕq(N)ϕq×(∫δ1‍H(1,τ)a(τ)dτ)>∥u∥.
This contradiction completes the proof.Corollary 22.
Assume that (H1) holds andf∞=+∞. Then, the fractional differential equation boundary-value problem (3) has no positive solution for sufficiently large λ>0.


## 6. Conclusion: Identities on the Special Polynomials whereby Caputo Fractional Derivative

In this final part, we will focus on the new interesting identities related to special polynomials by means of Caputo fractional derivative.As well known, the Bernoulli polynomials may be defined to be(61)F(t,z)=zez-1etz=eBz=∑n=0∞‍Bn(t)znn!,
where usual convention about replacing Bn by Bn in is used. Also, we note that the Bernoulli polynomials is analytic on the region D={z∈ℂ∣|z|<2π} (see [33]).Letd/dt be familiar normal derivative, then we can obtain the following identity
(62)ddttn=ntn-1.Differentiating in both sides of (61), we have
(63)ddtBn(t)=nBn-1(t)
(see [33]).Whent=0 in (61), we have Bn(0):=Bn are called Bernoulli numbers, which can be generated by
(64)F(z)=zez-1=∑n=0∞‍Bnznn!.By (61) and (64), we have the following functional equation:
(65)F(t,z)=etzF(z)
and this equation yields to
(66)Bm(t)=∑k=0m‍(mk)tm-kBk=∑k=0m‍(mk)tkBm-k
(see [33]).Let us now takey(t)=Bm(t) in Definition 2 leads to
(67)D0+αBm(t)=1Γ(n-α)∫0t‍(dn/dtn)Bm(t)|t=s(t-s)α-n+1ds=m(m-1)⋯(m-n+1)×∑k=0m-n‍(m-nk)Bm-n-k×[1Γ(n-α)∫0t‍sk(t-s)α-n+1ds]=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Bm-n-kΓ(n+k-α+1)tk-α+n.Therefore, we procure the following theorem.Theorem 23.
The following identity holds true:(68)D0+αBm(t)=Γ(m+1)Γ(m-n+1)×∑k=0m-n‍k!(m-nk)Bm-n-kΓ(n+k-α+1)tk-α+n.In [34], the Bernoulli polynomials of higher order are defined by
(69)zez-1zez-1⋯zez-1︸l-timesetz=∑m=0∞‍Bm(l)(t)znn!,
we note that Bm(l)(t) is analytic on D. It follows from (69), we have
(70)ddtBm(l)(t)=mBm-1(l)(t),dndtnBm(l)(t)=Γ(m+1)Γ(m-n+1)Bm-n(l)(t)
(see [34]).Substitutingx=0 into (69), Bm(l)(0):=Bm(l) are called Bernoulli polynomials of higher order.Owing to (69) and (70), we readily see that
(71)D0+αBm(l)(t)=1Γ(n-α)∫0t‍(dn/dtn)Bm(l)(t)|t=s(t-s)α-n+1ds=m(m-1)⋯(m-n+1)×∑k=0m-n‍(m-nk)Bm-n-k(l)×[1Γ(n-α)∫0t‍sk(t-s)α-n+1ds]=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Bm-n-k(l)Γ(n+k-α+1)tk-α+n=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Γ(n+k-α+1)tk-α+n×(∑s1+s2+⋯+sl=m-n-ksl≥0‍(m-n-ks1,s2,…,sl)(∏j=1lBsj)).Therefore, we can state the following theorem.Theorem 24.
The following identity holds true:(72)D0+αBm(l)(t)=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Γ(n+k-α+1)tk-α+n×(∑s1+s2+⋯+sl=m-n-ksl≥0‍(m-n-ks1,s2,…,sl)(∏j=1lBsj)),
in which Bsj and (m-n-ks1,s2,…,sl) are Bernoulli numbers and multibinomial coefficients.In the regionT={z∈ℂ∣|z|<π}, the Euler polynomials and the Euler polynomials of higher order are given, respectively, with the help of the following generating functions:
(73)2ez+1etz=∑m=0∞‍Em(t)zmm!,(74)2ez+12ez+1⋯2ez+1︸l-timesetz=∑m=0∞‍(∑s1+s2+⋯+sl=m‍(ms1,s2,…,sl)(∏j=1l-1Esj)tsl)zmm!=∑m=0∞‍Em(l)(t)znn!,
where Esj are Euler numbers in (see [33–36]). From the last equation, we discover the following:
(75)ddtEm(t)=mEm-1(t),ddtEm(l)(t)=mEm-1(l)(t)
(see [35]).Obviously, we have that(76)Em(1)(t):=Em(t).Takingy(t)=Em(l)(t) in Definition 2, by (73) and (75), we compute
(77)D0+αEm(l)(t)=1Γ(n-α)∫0t‍(dn/dtn)Em(l)(t)|t=s(t-s)α-n+1ds=m(m-1)⋯(m-n+1)×∑k=0m-n‍(m-nk)Em-n-k(l)×[1Γ(n-α)∫0t‍sk(t-s)α-n+1ds]=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Em-n-k(l)Γ(n+k-α+1)tk-α+n=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Γ(n+k-α+1)tk-α+n×(∑s1+s2+⋯+sl=m-n-ksl≥0‍(m-n-ks1,s2,…,sl)(∏j=1lEsj)).Therefore, we obtain the following theorem.Theorem 25.
The following identity(78)D0+αEm(l)(t)=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Γ(n+k-α+1)×(∑s1+s2+⋯+sl=m-n-ksl≥0‍(m-n-ks1,s2,…,sl)(∏j=1lEsj))tk-α+n.
is true. Obviously, we have that
(79)D0+αEm(t)=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Em-n-kΓ(n+k-α+1)tk-α+n.In the regionT={z∈ℂ∣|z|<π}, Genocchi polynomials, Gm(x), and Genocchi polynomials of higher order, Gm(l)(x), are defined as an extension of Genocchi numbers Gm defined in [33, 37, 38], respectively,
(80)2zez+1etz=∑m=0∞‍Gm(t)zmm!,2zez+12zez+1⋯2zez+1︸l-timesetz=∑m=0∞‍Gm(l)(t)znn!.In this final section, by the similar method, we arrive at the following theorem.Theorem 26.
The following identity(81)D0+αGm(l)(t)=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Γ(n+k-α+1)×(∑s1+s2+⋯+sl=m-n-ksl≥0‍(m-n-ks1,s2,…,sl)(∏j=1lGsj))tk-α+n
is true. Obviously, we have that,
(82)D0+αGm(t)=Γ(m+1)Γ(m-n+1)∑k=0m-n‍k!(m-nk)Gm-n-kΓ(n+k-α+1)tk-α+n.



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