# Existence of Solutions for Fractional Boundary Value Problem with Nonlinear Derivative Dependence

Authors: Wenzhe Xie; Jing Xiao; Zhiguo Luo
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We investigate the existence of solutions for fractional boundary value problem including both left and right fractional derivatives by using variational methods and iterative technique.


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## Body

## 1. Introduction

Fractional differential equations appear naturally in a number of fields such as physics, chemistry, biology, economics, control theory, signal and image processing, and blood flow phenomena. During last decades, the theory of fractional differential equations is an area intensively developed, due mainly to the fact that fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes (see [1–4] and the references therein). Therein, the composition of fractional differential operators has got much attention from many scientists, mainly due to its wide applications in modeling physical phenomena exhibiting anomalous diffusion. Specifically, the models involving a fractional differential oscillator equation, which contains a composition of left and right fractional derivatives, are proposed for the description of the processes of emptying the silo [5] and the heat flow through a bulkhead filled with granular material [6], respectively. Their studies show that the proposed models based on fractional calculus are efficient and describe well the processes.In the aspect of theory, the study of fractional boundary value problem including both left and right fractional derivatives has attracted much attention by using variational methods [7–12]. It is not easy to use the critical point theory to study the fractional differential equations including both left and right fractional derivatives, since it is often very difficult to establish a suitable space and a variational functional for the fractional boundary value problem.For the first time, Jiao and Zhou in [7] showed that the critical point theory is an effective approach to tackle the existence of solutions for the following fractional boundary value problem:
(1)DTαt(Dtα0u(t))=∇F(t,u(t)),a.e.t∈[0,T],u(0)=u(T)=0,
where Dtα0 and DTαt are the left and right Riemann-Liouville fractional derivatives of order 0<α≤1, respectively, F:[0,T]×RN→R is a given function satisfying some assumptions, and ∇F(t,x) is the gradient of F at x.In [8], by performing variational methods combined with iterative technique, Sun and Zhang investigated the solvability of the following fractional boundary value problem:
(2)ddx(p0Dx-β(u′(x))+qxD1-β(u′(x)))+f(x,u(x))=0,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhix∈(0,1),u(0)=u(1)=0,
where β∈(0,1), 0<p=1-q<1, Dx-β0, and D1-βx denote left and right Riemann-Liouville fractional integrals of order β, respectively, and f:[0,1]×R→R is continuous.Motivated by the above works and [13, 14], in this paper, we attempt to use Mountain Pass theorem and iterative technique to study the existence of solutions for the following nonlinear fractional boundary value problem with dependence on fractional derivative:
(3)DTαt(p(t)Dtα0u(t))=f(t,u(t),Dtα0u(t)),t∈[0,T],u(0)=u(T)=0,
where Dtα0 and DTαt are the left and right Riemann-Liouville fractional derivatives of order 1/2<α≤1, respectively, and f∈C([0,T]×R×R,R) and p∈C1([0,T],R) with p(t)>0, for t∈[0,T].In particular, ifα=1, problem (3) reduces to the standard second order boundary value problem of the following form:
(4)(p(t)u′(t))′+f(t,u(t),u′(t))=0,t∈[0,T],u(0)=u(T)=0,
where f:[0,T]×R×R→R is a given function.In fact, for problem (3), due to the appearance of left and right Riemann-Liouville fractional integral, we cannot deal with problem (3) by using fixed point theorems, because it is difficult to find the equivalent integral equation corresponding to problem (3). And since problem (3) is not variational, we cannot find some functional such that its critical point is the solution corresponding to problem (3). However, when there is no presence of the fractional derivative of the solution in the nonlinearity term and p(t) is a constant function, then problem (3) is studied by establishing corresponding variational structure in some suitable fractional space and applying the critical point theorems [7, 10, 12].In order to use variational methods, we consider a family of the following fractional boundary value problem with no dependence on the fractional derivative of the solution; that is, for eachw∈E0α (which is defined in Section 2), we consider the following problem:
(5)DTαt(p(t)Dtα0u(t))=f(t,u(t),Dtα0w(t)),t∈[0,T],u(0)=u(T)=0.
From [7, 10, 12], we know that problem (5) is variational and we can treat it by variational methods. Thus, for each w∈E0α, we can find a solution uw∈E0α with some bounds. Next, by iterative methods we can show that there exists a solution for problem (3).


## 2. Preliminaries and Several Lemmas

In this section, we introduce some basic definitions of fractional calculus and several lemmas which are used further in this paper.Definition 1 (see [3]).
Letf be a function defined on [a,b]. The left and right Riemann-Liouville fractional integrals of order α for function f denoted by Dt-αaf(t) and Db-αtf(t), respectively, are defined by
(6)aDt-αf(t)=1Γ(α)∫at(t-s)α-1f(s)ds,t∈[a,b],α>0,tDb-αf(t)=1Γ(α)∫tb‍(s-t)α-1f(s)ds,t∈[a,b],α>0,
provided in both cases that the right-hand side is pointwise defined on [a,b].Definition 2 (see [3]).
Letf be a function defined on [a,b]. The left and right Riemann-Liouville fractional derivatives of order α for function f denoted by Dtαaf(t) and Dbαtf(t), respectively, exist almost everywhere on [a,b], Dtαaf(t), and Dbαtf(t) and are represented by
(7)Dtαaf(t)=1Γ(n-α)dndtn∫at‍(t-s)n-α-1f(s)ds,hhhhhhhhhhhhhhhhhhhhhhhht∈[a,b],α>0,Dbαtf(t)=(-1)nΓ(n-α)dndtn∫tb‍(s-t)n-α-1f(s)ds,hhhhhhhhhhhhhhhhhhhhhhhht∈[a,b],α>0.Proposition 3 (see [7, 10]).
Iff∈Lp([a,b],RN), g∈Lq([a,b],RN) and p≥1, q≥1, 1/p+1/q≤1+α or p≠1, q≠1, 1/p+1/q=1+α, then
(8)∫ab‍[Dt-αaf(t)]g(t)dt=∫ab‍[Db-αtg(t)]f(t)dt,α>0.Proposition 4 (see [7, 10]).
Iff(a)=f(b)=0, f′∈L∞([a,b],RN), and g∈L1([a,b],RN), or g(a)=g(b)=0, g′∈L∞([a,b],RN), and f∈L1([a,b],RN), then
(9)∫ab‍[Dtαaf(t)]g(t)dt=∫ab‍[Dbαtg(t)]f(t)dt,0<α≤1.In order to establish a variational structure which enables us to reduce the existence of solutions of problem (5) to one of finding critical points of corresponding functional, it is necessary to construct appropriate function spaces.Let us recall that for any fixedt∈[0,T] and 1≤p<∞,
(10)∥u∥∞=max⁡t∈[0,T]|u(t)|,∥u∥Lp=(∫0T‍|u(s)|pds)1/p.Definition 5.
Let0<α≤1. The fractional derivative space E0α is defined by the closure of C0∞([0,T],R) with respect to the weighted norm
(11)∥u∥α=(∫0T‍p(t)|Dtα0u(t)|2dt+∫0T‍|u(t)|2dt)1/2,hhhhhhhhhhhhhhhhhhhhhhhhhhhihhhh∀u∈E0α,
where denote by C0∞([0,T],R) the set of all functions u∈C∞([0,T],R) with u(0)=u(T)=0.Lemma 6 (see [7]).
Let1/2<α≤1; for all u∈E0α, one has(i)
(12)∥u∥L2≤TαΓ(α+1)∥Dtα0u(t)∥L2,(ii)
(13)∥u∥∞≤Tα-1/2Γ(α)2α-1∥Dtα0u(t)∥L2.Remark 7.
Letm:=min⁡t∈[0,T]p(t), by (12) and (13); then
(14)∥u∥L2≤TαΓ(α+1)m(∫0T‍p(t)|Dtα0u(t)|2dt)1/2,(15)∥u∥∞≤Tα-1/2Γ(α)m(2α-1)(∫0T‍p(t)|Dtα0u(t)|2dt)1/2.At this point, by (14), we can consider E0α with respect to the norm
(16)∥u∥α=(∫0T‍p(t)|Dtα0u(t)|2dt)1/2,∀u∈E0α,
which is equivalent to (11).Similarly to some results in [7], the properties of the fractional derivative space E0α are given, as follows.Lemma 8.
Let0<α≤1. The fractional derivative space E0α is a reflexive and a separable Banach space.Lemma 9.
Assume that1/2<α≤1 and the sequence {uk} converges weakly to u in E0α; that is, uk⇀u. Then uk→u in C([0,T],RN); that is, ∥u-uk∥∞→0, as k→∞.One is now in a position to give the definition for the solution of problem (3).Definition 10.
A functionu∈E0α is called a solution of problem (3), if(i)
DTα-1t(p(t)Dtα0u(t)) and Dtα-10u(t) are derivatives for all t∈[0,T],(ii)
u satisfies (3).Definition 11.
A functionu∈E0α is called a weak solution of problem (3), if
(17)∫0T‍p(t)Dtα0u(t)Dtα0v(t)dt-∫0T‍f(t,u(t),Dtα0u(t))v(t)dt=0,
for all v(t)∈E0α.Associated to the boundary value problem (5), for given w(t)∈E0α, we have the functional Iw(u):E0α→R defined by
(18)Iw(u)=12∫0T‍p(t)|Dtα0u(t)|2dt-∫0T‍F(t,u(t),Dtα0w(t))dt,
where F(t,ξ,ζ)=∫0ξ‍f(t,s,ζ)ds and f∈C([0,T]×R×R,R). Clearly, by the continuity hypothesis on f and p, we have Iw∈C1(E0α,R) and for all v(t)∈E0α,
(19)Iw′(u)v=∫0T‍p(t)Dtα0u(t)Dtα0v(t)dt-∫0T‍f(t,u(t),Dtα0w(t))v(t)dt.
Moreover, similarly to the proof of [7, Theorem 5.1], we know that the critical point of Iw is a solution of problem (5).


## 3. Main Results

First, we make the following assumptions:(H1)
f(t,ξ,ζ)=g(t,ξ,ζ)+h(t)ζ, where g(t,ξ,ζ)=o(|ξ|) as ξ→0 uniformly for t∈[0,T],ζ∈R, and h(t)∈C[0,T]; denote Ch=max⁡t∈[0,T]|h(t)|.(H2)
There exist constantsC0>0 and r>1 such that
(20)|g(t,ξ,ζ)|≤C0(1+|ξ|r),∀t∈[0,T],ξ∈R,ζ∈R.(H3)
There exist constantsμ>2 and ξ0>0 such that
(21)μG(t,ξ,ζ)≤ξg(t,ξ,ζ),∀t∈[0,T],|ξ|≥ξ0,ζ∈R.(H4)
There exist positive constantsC1,C2>0 such that
(22)G(t,ξ,ζ)≥C1|ξ|μ-C2,∀t∈[0,T],ξ∈R,ζ∈R,whereG(t,ξ,ζ)=∫0ξ‍g(t,s,ζ)ds and g∈C([0,T]×R×R,R).(H5)
Fort∈[0,T], ξ1,ξ2∈[-ρ1,ρ1], ζ1,ζ2∈R, the function g satisfies the following Lipschitz conditions:
(23)|g(t,ξ1,ζ1)-g(t,ξ2,ζ2)|≤L1|ξ1-ξ2|+L2|ζ1-ζ2|,whereL1,L2>0 and ρ1 is a positive constant, which is given in (49).We note that it obviously follows from(H1) and (H2) that for all ε>0, there exists a positive constant Cε, independent of w, such that
(24)|g(t,ξ,ζ)|≤ε|ξ|+Cε|ξ|r.The main result of this paper is the following.Theorem 12.
Assume that(H1)–(H5) hold and that 1/2-1/μ>ChTα/mΓ(α+1). If there exist b¯∈(0,1/2) and ε∈(0,(1-2b¯)m[Γ(α+1)]2/T2α) such that
(25)[12-b¯-εT2α2m[Γ(α+1)]2](r+1)[Γ(α)m(2α-1)]r+1CεT(α-1/2)(r+1)+1>[ChTαRb¯mΓ(α+1)]r-1,(26)L∶=Tα(L2+Ch)Γ(α+1)m[Γ(α+1)]2-L1T2α∈(0,1),
then problem (3) has one nontrivial solution.Proof.
In order to prove Theorem12, we proceed by three steps.
Step
1. Let w∈E0α; with ∥w∥α≤K, which is given in (48), we show that Iw has a nontrivial critical point in E0α by the Mountain Pass theorem.
Firstly, by(H1) and (24) we have
(27)|f(t,ξ,ζ)|≤ε|ξ|+Cε|ξ|r+Ch|ζ|,(28)|F(t,ξ,ζ)|≤ε2|ξ|2+Cεr+1|ξ|r+1+Ch|ζ||ξ|.
By Remarks (15), (18), (28), and Hölder inequality
(29)Iw(u)≥12∫0T‍p(t)|Dtα0u(t)|2dt-ε2∫0T‍|u(t)|2dt-Cεr+1∫0T‍|u(t)|r+1dt-Ch∫0T‍|Dtα0w(t)|·|u(t)|dt≥12∫0T‍p(t)|Dtα0u(t)|2dt-εT2α2m[Γ(α+1)]2∫0T‍p(t)|Dtα0u(t)|2dt-CεT(α-1/2)(r+1)+1(r+1)[Γ(α)m(2α-1)]r+1×(∫0T‍p(t)|Dtα0u|2dt)(r+1)/2-ChTαmΓ(α+1)∥w∥α·∥u∥α≥[CεT(α-(1/2))(r+1)+1(r+1)[Γ(α)m(2α-1)]r+112-b¯-εT2α2m[Γ(α+1)]2-CεT(α-1/2)(r+1)+1(r+1)[Γ(α)m(2α-1)]r+1∥u∥αr-1]∥u∥α2+[b¯∥u∥α-ChTαKmΓ(α+1)]∥u∥α.
By (25) we can choose ρ>0 such that
(30)12-b¯-εT2α2m[Γ(α+1)]2>CεT(α-1/2)(r+1)+1(r+1)[Γ(α)m(2α-1)]r+1ρr-1,ρ>ChTαKb¯mΓ(α+1).
Hence, let u∈E0α; with ∥u∥α=ρ, we know that there exist α1>0, such that for ∥u∥α=ρ, Iw(u)≥α1 uniformly for w∈E0α.
Secondly, for givenu*∈E0α, with ∥u*∥α=1, it follows from (H1) and (H4) that
(31)F(t,ξ,ζ)≥C1|ξ|μ+h(t)ζξ-C2.
Then we have that for τ>0(32)Iw(τu*)=τ22∥u*∥α2-∫0T‍F(t,τu*,Dtα0w(t))dt≤τ22-C1τμ∫0T‍|u*|μdt+τCh∫0T‍|Dtα0w|·|u*|dt+C2T≤τ22-C1τμ∫0T‍|u*|μdt+τChTαKmΓ(α+1)+C2T.
Since μ>2, taking τ large enough and letting e=τu*, then Iw(e)<0 with ∥e∥α>ρ.
Thirdly, we show thatIw satisfies the Palais-Smale condition.
Let{uk}⊂E0α, such that
(33)|Iw(uk)|≤K¯,lim⁡k→∞Iw′(uk)=0,forsomeK¯>0.
We have
(34)Iw(uk)=12∥uk∥α2-∫0T‍F(t,uk(t),Dtα0w(t))dt,Iw′(uk)uk=∥uk∥α2-∫0T‍f(t,uk(t),Dtα0w(t))ukdt.
Then by (33) and (H3),
(35)(12-1μ)∥uk∥α2=Iw(uk)-1μIw′(uk)uk+μ-1μ∫0T‍h(t)uk(t)·Dtα0w(t)dt+∫{|uk|≥ξ0}‍G(t,uk(t),Dtα0w(t))+∫{|uk|≥ξ0}‍Gv-ukμg(t,uk(t),Dtα0w(t))dt+∫{|uk|≤ξ0}‍G(t,uk(t),Dtα0w(t))+∫{|uk|≥ξ0}‍Gv-ukμg(t,uk(t),Dtα0w(t))dt≤Iw(uk)-1μIw′(uk)uk+ChTαmΓ(α+1)∥w∥α·∥uk∥α+C3≤K¯+[1μ∥Iw′(uk)∥+ChTαKmΓ(α+1)]∥uk∥α+C3,
where C3>0. Combining with Iw′(uk)→0, as k→∞, we know that {uk} is bounded in E0α.
SinceE0α is a reflexive space, we can assume that uk⇀u. In E0α, according to Remark 7 and Lemma 9 we have that {uk} is bounded in C[0,T] and lim⁡k→∞∥uk-u∥∞=0. By the assumption (H2), one gets
(36)∫0T‍[f(t,uk(t),Dtα0w(t))-f(t,u(t),Dtα0w(t))]×(uk-u)dt=∫0T‍[g(t,uk(t),Dtα0w(t))-g(t,u(t),Dtα0w(t))]×(uk-u)dt⟶0,k⟶∞.
Notice that
(37)[Iw′(uk)-Iw′(u)](uk-u)=Iw′(uk)(uk-u)-Iw′(u)(uk-u)≤∥Iw′(uk)∥·∥uk-u∥α-Iw′(u)(uk-u)⟶0,k⟶∞.
Moreover,
(38)∥uk-u∥α2=∫0T[f(t,uk(t),Dtα0w(t))‍-f(t,u(t),Dtα0w(t))](uk-u)dt+[Iw′(uk)-Iw′(u)](uk-u),
so ∥uk-u∥α→0, as k→∞. That is, {uk} converges strongly to u in E0α.
Obviously,Iw(0)=0; therefore, by Mountain Pass theorem, Iw has a nontrivial critical point uw in E0α, with
(39)Iw(uw)=inf⁡u∈Γmax⁡t∈[0,1]Iw(u(t))≥α1>0,
where Γ={u∈C([0,1],E0α)∣u(0)=0,u(1)=e}.
Step
2. We construct iterative sequence {un} and estimate its norm in E0α.
We consider the solutions{un} of the following problem:
(40)DTαt(p(t)Dtα0un(t))=f(t,un(t),Dtα0un-1(t)),hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhit∈[0,T],un(0)=un(T)=0,
starting with u1≡0. By iterative technique, we can get a sequence of {un}, the nontrivial critical points obtained by Step 1, provided that there exists some constant K>0, such that ∥un∥α≤K,n≥1.
In the following, we estimate the norm of{un} by the induction method. In fact, we need to prove that if we assume that for some K>0, ∥un-1∥α≤K, then un, the nontrivial critical point of Iun-1, also satisfies ∥un∥α≤K.
By the Mountain Pass characterization of the critical level, and(H4) and Cauchy’s inequality with positive constant ε¯, which is given in the sequel, it follows that
(41)|Iun-1(un)|≤max⁡τ∈[0,∞)Iun-1(τu*)≤max⁡τ∈[0,∞){τ22-C1τμ∫0T‍|u*|μdtvvvvvvvvvv+τCh∫0T‍|Dtα0un-1|·|u*|dt+C2T}≤max⁡τ∈[0,∞){τ22-C1τμ∥u*∥Lμμvvvvvvvvvv+τChTαmΓ(α+1)∥un-1∥α+C2T}≤max⁡τ∈[0,∞){[12+[ChTα]24ε¯[mΓ(α+1)]2]τ2vvvvvvvvv-C1τμ∥u*∥Lμμ+C2T[12+[ChTα]24ε¯[mΓ(α+1)]2]}+ε¯∥un-1∥α2.
Let
(42)H(τ)∶=[12+[ChTα]24ε¯[mΓ(α+1)]2]τ2-C1τμ∥u*∥Lμμ+C2T,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhiτ≥0,
and since μ>2, H(τ) can achieve its maximum at some τ0. Hence
(43)|Iun-1(un)|≤H(τ0)+ε¯K2.
By (35), ∥un-1∥α≤K, and Iun-1′(un)un=0, we have
(44)(12-1μ)∥un∥α2≤Iun-1(un)-1μIun-1′(un)un+ChTαmΓ(α+1)∥un-1∥α·∥un∥α+C4≤H(τ0)+ε¯K2+ChTαKmΓ(α+1)∥un∥α+C4,
where C4>0 is independent of K. For convenience, we denote a=1/2-1/μ, b=ChTα/mΓ(α+1), and c=H(τ0)+C4; then (44) can be written as
(45)a∥un∥α2-bK∥un∥α-(c+ε¯K2)≤0,
which implies
(46)∥un∥α≤bK+b2K2+4a(c+ε¯K2)2a.
Let
(47)bK+b2K2+4a(c+ε-K2)2a≤K,
together with the assumption a>b; by simple calculation, for enough small ε¯∈(0,a-b) we obtain
(48)K≥ca-b-ε¯.
Hence we can choose K∈[c/(a-b-ε¯),∞), such that ∥un∥α≤K,n≥1.
Step
3. We show that the iterative sequence {un} constructed in Step 2 is convergent to a nontrivial solution of problem (3).
By Step 2, we know∥un∥α≤K, and by Remark 7, there exists positive constant ρ1, such that
(49)∥un∥∞≤Tα-1/2KΓ(α)m(2α-1)=∶ρ1.
By (19), Iun′(un+1)(un+1-un)=0, and Iun-1′(un)(un+1-un)=0, we obtain
(50)∫0T‍p(t)Dtα0un+1·Dtα0(un+1-un)dt-∫0T‍f(t,un+1,Dtα0un)(un+1-un)dt=0,∫0T‍p(t)Dtα0un·Dtα0(un+1-un)dt-∫0T‍f(t,un,Dtα0un-1)(un+1-un)dt=0;
hence
(51)∫0T‍p(t)[Dtα0(un+1-un)]2dt=∫0T‍[f(t,un+1,Dtα0un)-f(t,un,Dtα0un-1)](un+1-un)dt.
By (H5) it follows that
(52)∥un+1-un∥α2=∫0T‍[g(t,un+1,Dtα0un)-g(t,un,Dtα0un-1)](un+1-un)dt+∫0T‍h(t)Dtα0(un-un-1)·(un+1-un)dt≤L1∫0T‍|un+1-un|2dt+(L2+Ch)∫0T‍|Dtα0(un-un-1)|·|un+1-un|dt≤L1T2αm[Γ(α+1)]2∥un+1-un∥α2+(L2+Ch)∥Dtα0(un-un-1)∥L2·∥un+1-un∥L2≤L1T2αm[Γ(α+1)]2∥un+1-un∥α2+(L2+Ch)TαmΓ(α+1)∥un-un-1∥α·∥un+1-un∥α;
hence
(53)(1-L1T2αm[Γ(α+1)]2)∥un+1-un∥α≤(L2+Ch)TαmΓ(α+1)∥un-un-1∥α.
Since 0<L<1, we know that {un} is a Cauchy sequence in E0α, so there exists a u∈E0α such that {un} converges strongly to u in E0α.
In order to show thatu is a solution of problem (3), we need to prove that
(54)∫0T‍p(t)Dtα0u(t)Dtα0v(t)dt=∫0T‍f(t,u(t),Dtα0u(t))v(t)dt,∀v(t)∈E0α.
It suffices to show that
(55)∫0T‍f(t,un,Dtα0un-1)v(t)dt⟶∫0T‍f(t,u,Dtα0u)v(t)dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhiasn⟶∞.
Indeed, it follows from the assumption (H5) that
(56)∫0T‍[f(t,un,Dtα0un-1)-f(t,u,Dtα0u)]v(t)dt=∫0T‍[g(t,un,Dtα0un-1)-g(t,u,Dtα0u)]v(t)dt+∫0T‍h(t)Dtα0(un-1-u)v(t)dt≤L1∫0T‍|un-u|·|v(t)|dt+(L2+Ch)∫0T‍|0Dtα(un-1-u)|·|v(t)|dt≤[L1T2αm[Γ(α+1)]2∥un-u∥α+(L2+Ch)TαmΓ(α+1)∥un-1-u∥α]·∥v∥α⟶0,n⟶∞.
Then, u(t) is a solution of (3), and since Iun-1(un)≥α1>0 for n=2,3,…, we obtain that u(t) is a nontrivial solution of problem (3).


## 4. Example

In this section, we provide an explicit example to illustrate our main results.Example 1.
Consider the following fractional boundary value problem:(57)D10.75t(Dt0.750u(t))=u3(t)(2+sinDt0.750u(t)),hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht∈[0,1],u(0)=u(1)=0.
Obviously, α=0.75, f(t,ξ,ζ)=ξ3(2+sinζ), p(t)=1, and h(t)=0. Let C0=r=3, μ=4, and C1=1/4, for all ξ0, C2>0; it is not difficult to verify that function f satisfies the assumptions (H1)–(H4).
For(H5), for all t∈[0,T], ξ1,ξ2∈[-ρ1,ρ1], ζ1,ζ2∈R, it follows that
(58)|f(t,ξ1,ζ1)-f(t,ξ2,ζ2)|≤|ξ13(2+sinζ1)-ξ13(2+sinζ2)|+|ξ13(2+sinζ2)-ξ23(2+sinζ2)|≤9ρ12|ξ1-ξ2|+ρ13|ζ1-ζ2|.
Then we choose L1=9ρ12, L2=ρ13, where ρ1=2K/Γ(0.75). Since h(t)=0, it obviously follows that 1/2-1/μ>0 and that (25) is valid. It suffices to verify that
(59)L=Γ(1.75)L2[Γ(1.75)]2-L1=1.52K3(0.75)2[Γ(0.75)]4-18K2∈(0,1).
From (48), now we estimate the value of c=H(τ0)+C4, where C4 is dependent on ξ0 and
(60)H(τ)=12τ2-14τ4∫01‍|u*(t)|4dt+C2,τ≥0.
Since ∫01‍|Dt0.750(t2-t)|2dt=(1/[Γ(1.25)]2)(2/3-96/175), we can choose u*(t)=(Γ(1.25)/2/3-96/175)(t2-t)≈2.6375745(t2-t), such that u*(t)∈E00.75 with ∥u*(t)∥0.75=1.
By direct calculation, we have thatτ0=1/∫01‍|u*(t)|4dt≈0.1853281; moreover,
(61)H(τ0)=14τ02+C2≈0.008586625+C2.
Then c≈0.008586625+C2+C4; by the arbitrariness of ξ0 and C2, we can choose enough small ξ0, C2>0, such that c=0.01, a=1/4, b=0, and ε¯=0.1.
LetK=c/(a-b-ε¯)=1/15≈0.2581989; we get
(62)L=0.036514841.2684044-1.2000001≈0.5338091∈(0,1).
Then all conditions of Theorem 12 are satisfied. Consequently we obtain that problem (57) has a nontrivial solution.



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*Source: 812910-2014-04-27.xml*