# Solutions to Dirichlet-Type Boundary Value Problems of Fractional Order in Banach Spaces

Authors: Jing-jing Tan; Cao-zong Cheng
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

We consider the boundary value problems with Dirichlet-type
boundary conditions of nonlinear fractional differential equation in Banach space. The
existence of the solution to the boundary value problems is established. Our analysis relies
on the Sadovskii fixed point theorem. As an application, we give an example to demonstrate
our results.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications in various sciences, such as physics, mechanics, chemistry, and engineering, (e.g., [1–9]).Consequently, the fractional calculus and its applications in various fields of science and engineering have received much attention and have developed very rapidly. Jiang and Yuan [10], by using fixed point theorem on the cone, discussed the existence and multiplicity of solutions of the nonlinear fractional differential equation boundary value problem as follows:
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,
where1<α≤2is a real number andD0+αis standard Riemann-Liouville fractional derivative. The authors in [11] consider the same boundary value problem. They derived the corresponding Green function and obtained the existence of solutions of this problem.As far as we know, the nonlinear integer order differential equation for the Dirichlet boundary value problem has been studied extensively (e.g., [12–17]). However, only a few papers have dealt with the boundary value problem for fractional differential equation, especially, in Banach spaces. The authors in [18] studied the existence of positive solutions of second-order two-point boundary value problem as follows:
(2)u′′(t)+f(t,u(t))=θ,0<t<1,u(0)=u(1)=θ,
in Banach spaces. The authors in [19], by using the Mönch fixed point theorems, obtained the same results.Motivated by the results mentioned above, we discuss the following boundary value problem (BVP for short):(3)D0+αu(t)+f(t,u(t))=θ,0<t<1,u(0)=u(1)=θ,
in Banach spaceE, whereθis the zero element ofE,1<α≤2is a real number,D0+αis standard Riemann-Liouville fractional derivative,I=[0,1], andf:I×E→Eis continuous. We establish an existence result of BVP in Banach spaces. The technique relies on the properties of the Kuratowski noncompactness measure and and Sadovskii fixed point theorem. To the best of our knowledge, this is the first paper considering the existence of solutions to Dirichlet-type value problems of fractional order in Banach spaces.


## 2. Preliminaries

For the convenience of the reader, we present here the necessary definitions and preliminary facts which are used throughout this paper.Definition 1 (see [1]).
The Riemann-Liouville fractional integral of orderα>0of a functionf:(0,∞)→Ris given by
(4)I0+αf(x)=1Γ(α)∫0x‍f(t)(x-t)1-αdt,
provided that the right side is pointwise defined on(0,∞), whereΓ(α)=∫0∞‍tα-1e-tdt.Definition 2 (see [1]).
The Riemann-liouville fractional derivative of orderα>0of a continuous functionf:(0,∞)→Ris given by
(5)D0+αf(x)=1Γ(n-α)(ddx)n∫0x‍f(t)(x-t)α-n-1dt,
wheren=[α]+1,[α]denotes the integer part of the numberα, provided that the right side is pointwise defined on(0,∞).Lemma 3 (see [2]).
Letα>0. If we assume thatu∈C(0,1)∩L(0,1), then the fractional differential equation
(6)D0+αu(t)=0
hasu(t)=c1tα-1+c2tα-2+⋯+cntα-N(ci∈R,i=1,2,3,…,N)as unique solutions, whereNis the smallest integer greater than or equal toα.Lemma 4 (see [2]).
Assume thatu∈C(0,1)∩L(0,1)with a fractional derivative of orderα>0that belongs toC(0,1)∩L(0,1). Then
(7)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cNtα-N,
for someci∈R,i=1,2,3,…,N, whereNis the smallest integer greater than or equal toα.Definition 5 (Kuratowski noncompactness measure).
LetE be a real Banach space, let Sbe a bounded set inE. We denote
(8)α(S)=inf{δ>0:S=⋃i=1m‍Si,2222diam(Si)<δ,i=1,2,…,m⋃i=1m‍}.α(S)is called Kuratowski noncompactness measure ofS, wherediam(Si)denote the diameters ofSi. Obviously0≤α(S)<∞.Definition 6.
LetE1andE2be real Banach spaces,D⊂E1, andA:D→E2be a continuous and bounded operator.
(
1
)
Ais called ak-set contraction operator if there exists a constantk, such thatα(A(S))≤kα(S)for any bounded setSinE. Whenk<1,Ais called a strict set contraction operator.
(
2
)LetSbe a nonrelative compact, bounded subset inE.Ais called a condensation ifα(A(S))<α(S).Remark 7.
A strict set contraction operator is condensation.
Now, we denote the Banach space of continuous functionsu:I→EbyC[I,E]with the maximal norm∥u∥=maxt∈I∥u(t)∥.The basic space used in this paper is:
(9)X[I,E]={u∈C[I,E]:supt∈I∥u(t)∥1+tα-1<+∞}
equipped with norm∥u∥X=supt∈I(∥u(t)∥/(1+tα-1)). It is easy to see thatX[I,E]is a Banach space.A mapx∈X[I,E]∩C1[I,E]is called a solution of BVP if it satisfies (3). For a bounded subsetDof Banach spaceE, letα(D)be the Kuratowski noncompactness measure ofD. In this paper, the Kuratowski noncompactness measure inE, C[I,E], and X[I,E] is denoted byαE(·),αC(·),  and  αX(·), respectively. The following properties of the Kuratowski noncompactness measure and Sadovskii fixed point theorem are needed for our discussion.Lemma 8 (see [20]).
IfH⊂C[I,E]is bounded and equicontinuous. ThenαE(H(t))is continuous onIandαC(H)=maxt∈IαE(H(t)),αE(∫I‍x(t)dt:x∈H)≤∫I‍αE(H(t))dt, whereH(t)={x(t):x∈H}for each  t∈I.Lemma 9 (see [21] (Sadovskii)).
LetDbe a bounded, closed, and convex subset of the Banach spaceE. If the operatorA:D→Dis condensing, thenAhas a fixed point inD.


## 3. Main Result

In order to discuss the BVP, the preliminary lemmas are given in this section.For convenience, let us list some conditions.(H1)
There exist nonnegative functionsa,b∈C[0,1]such that
(10)∥f(t,x)∥≤a(t)∥x∥+b(t),∀t∈I,x∈E,∫01‍(1+tα-1)a(t)dt<Γ(α)2,∫01‍b(t)dt<+∞.(H2)
For anyr>0,[α,β]⊂I,f(t,x)is uniformly continuous on[α,β]×BE[θ,r], whereθis the zero element ofEandBE[θ,r]={x∈E:∥x∥≤r}.(H3)
There exists a positive constantkwith(8/Γ(α)(1+tα-1))k<1,  such thatαE(f(t,W))≤kαE(W)for allt∈[0,1]and all bounded subsetWinE.Lemma 10 (see [2]).
Giveny∈C[0,1]and1<α≤2, then the problem
(11)D0+αu(t)+y(t)=0,0<t<1,u(0)=u(1)=0,
has a unique solution satisfying
(12)u(t)=-∫0t(t-s)α-1Γ(α)y(s)ds+∫01‍(1-s)α-1tα-1Γ(α)y(s)ds.Lemma 11 (see [2]).
Suppose that condition(H1)is satisfied, then BVP is equivalent to integral equation:
(13)u(t)=-∫0t‍(t-s)α-1Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1tα-1Γ(α)f(s,u(s))ds.
For anyu∈X[I,E], we define the operatorTby
(14)(Tu)(t)=-∫0t‍(t-s)α-1Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1tα-1Γ(α)f(s,u(s))ds.Remark 12.
Lemma11 indicates that the solution of BVP coincides with the fixed point of the operatorT.Tuis well defined andTu∈X[I,E].Indeed, from condition(H1), we have
(15)∫01‍∥f(t,u(t))∥dt≤∫01‍[a(t)∥u∥+b(t)]dt=∫01‍[a(t)(1+tα-1)∥u∥1+tα-1+b(t)]dt≤∫01‍[(1+tα-1)a(t)∥u∥X+b(t)]dt=∥u∥X∫01‍(1+tα-1)a(t)dt+∫01‍b(t)dt<∞.
By (14) and (15), we get that
(16)∥(Tu)(t)1+tα-1∥=11+tα-1×∥-∫0t‍(t-s)α-1Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1tα-1Γ(α)f(s,u(s))ds∥≤2Γ(α)∫01‍∥f(s,u(s))∥ds<∞.Lemma 13.
Suppose conditions(H1)and(H2)are satisfied, thenT:X[I,E]→X[I,E]is continuous and bounded.Proof.
First, by (14) and (16), we get(Tu)(t)∈X[I,E]for anyu(t)∈X[I,E]. Thus,T:X[I,E]→X[I,E]is bounded. Next, we will prove thatTis continuous onX[I,E]. Take{un},{u}∈X[I,E]and∥un-u∥X→0(n→+∞).Hence,{un}is a bounded subset ofX[I,E], that is, there existsM>0such that∥un∥X≤Mfor alln≥1. Taking the limit, we have∥u∥X≤M.
In addition, by (14) we have
(17)∥(Tun)(t)1+tα-1-(Tu)(t)1+tα-1∥=∥11+tα-1[-∫0t‍(t-s)α-1Γ(α)f(s,un(s))ds+∫01‍(1-s)α-1tα-1Γ(α)f(s,un(s))ds]-11+tα-1[-∫0t‍(t-s)α-1Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1tα-1Γ(α)f(s,u(s))ds]∥≤∥11+tα-1[-∫0t‍(t-s)α-1Γ(α)f(s,un(s))ds+∫0t‍(t-s)α-1Γ(α)f(s,u(s))ds]∥+∥11+tα-1[∫01‍(1-s)α-1tα-1Γ(α)f(s,un(s))ds-∫01‍(1-s)α-1tα-1Γ(α)f(s,u(s))ds]∥≤2Γ(α)(1+tα-1)∫01∥f(s,un(s))-f(s,u(s))∥ds.‍
It follows from condition(H2)that for anyε>0, there existsN>0such that
(18)∥f(s,un(s))-f(s,u(s))∥≤ε3Γ(α),∀n≥N,s∈I.
Therefore, for anyε>0,t∈[0,1],  andn>N, by (17) and (18), we have
(19)∥(Tun)(t)1+tα-1-(Tu)(t)1+tα-1∥≤2Γ(α)(1+tα-1)∫01‍ε3Γ(α)ds<ε.
Thus, we conclude that∥Tun-Tu∥X<ε; namely,T:X[I,E]→X[I,E]is continuous, and the conclusion of lemma follows.Lemma 14.
Let condition(H1)be satisfied and letVbe a bounded subset ofX[I,E]. Then(TV)(t)/(1+tα-1)is equicontinuous on[0,1].Proof.
In order to show that(TV)(t)/(1+tα-1)is equicontinuous on[0,1], we only need to testify the following conclusion.
For anyε>0,u∈V, there exists aδ>0such that
(20)∥(Tu)(t2)1+t2α-1-(Tu)(t1)1+t1α-1∥<ε,∀t1,t2∈[0,1],2222222222222222222222222222|t1-t2|<δ.
In fact, from condition(H1), it follows that(a1)
nonnegative functionsa(t),b(t)∈C[0,1]are bounded in[0,1], that is, there exist positive constantsAandB, such thatmaxt∈Ia(t)≤Aandmaxt∈Ib(t)≤B,(a2)
∥f(t,u(t))∥≤(1+tα-1)a(t)∥u∥X+b(t).
In view of the boundedness ofV, there existsR>0such that∥u∥X≤Rfor anyu∈V. Without loss of generality, for anyu∈V,t1,t2∈Iwitht1<t2, by (14), we can find that
(21)∥(Tu)(t2)1+t2α-1-(Tu)(t1)1+t1α-1∥=∥+∫01‍(1-s)α-1t1α-1(1+t1α-1)Γ(α)f(s,u(s))ds∫0t1‍(t1-s)α-1(1+t1α-1)Γ(α)]-∫0t2‍(t2-s)α-1(1+t2α-1)Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1t2α-1(1+t2α-1)Γ(α)f(s,u(s))ds-[-∫0t1‍(t1-s)α-1(1+t1α-1)Γ(α)f(s,u(s))ds+∫01‍(1-s)α-1t1α-1(1+t1α-1)Γ(α)f(s,u(s))ds∫0t1‍(t1-s)α-1(1+t1α-1)Γ(α)]∥≤∥∫0t2‍(t2-s)α-1(1+t2α-1)Γ(α)f(s,u(s))ds-∫0t1‍(t1-s)α-1(1+t1α-1)Γ(α)f(s,u(s))ds∥+∥∫01‍(1-s)α-1t2α-1(1+t2α-1)Γ(α)f(s,u(s))ds-∫01‍(1-s)α-1t1α-1(1+t1α-1)Γ(α)f(s,u(s))ds∥≤1Γ(α)∥∫0t2‍(t2-s)α-11+t2α-1f(s,u(s))ds-∫0t1‍(t2-s)α-11+t2α-1f(s,u(s))ds∥+1Γ(α)∥∫0t1‍(t2-s)α-11+t2α-1f(s,u(s))ds-∫0t1‍(t1-s)α-11+t1α-1f(s,u(s))ds∥+1Γ(α)∥∫01‍[t2α-11+t2α-1-t1α-11+t1α-1]22222222222×(1-s)α-1f(s,u(s))ds∫01‍[t2α-11+t2α-1-t1α-11+t1α-1]∥≤1Γ(α)∫t1t2‍(t2-s)α-11+t2α-1∥f(s,u(s))∥ds+1Γ(α)∫0t1‍[(t2-s)α-11+t2α-1-(t1-s)α-11+t1α-1]22222222222×∥f(s,u(s))∥ds+1Γ(α)∫01‍[t2α-11+t2α-1-t1α-11+t1α-1](1-s)α-1×∥f(s,u(s))∥ds≤2AR+BΓ(α)(t2-t1)+2AR+BΓ(α+1)[t2α1+t2α-1-t1α1+t1α-1-(t2-t1)α2]+2AR+BΓ(α)(t2α-11+t2α-1-t1α-11+t1α-1).
This ensures that(TV)(t)/(1+tα-1)is equicontinuous on[0,1]. Ift1≥t2, we can also get the same result. Thus,(TV)(t)/(1+tα-1)is equicontinuous on[0,1].The main result of this paper is as follows.Theorem 15.
Let conditions(H1)–(H3)be satisfied. Then the BVP has at least one solution belonging toX[I,E].Proof.
We only need to prove that the the operatorThas a fixed point inX[I,E]. By condition(H1), we can choose a real numberRsuch that
(22)R>∫01‍b(s)ds[Γ(α)2-∫01‍(1+sα-1)a(s)ds]-1,
and let
(23)B=:BX(θ,R)={u∈X[I,E]:∥u∥X≤R}.
Frist we prove thatTB⊂B. In fact, for anyu∈B, by (16), we have
(24)∥(Tu)(t)1+tα-1∥≤2Γ(α)∫01‍∥f(s,u(s))∥ds≤2Γ(α){∫01‍[(1+sα-1)a(s)]ds∥u∥X+∫01‍b(s)ds}≤2Γ(α){[Γ(α)2-∫01‍(1+sα-1)a(s)ds]∫01‍[(1+sα-1)a(s)]dsR+R[Γ(α)2-∫01‍(1+sα-1)a(s)ds]}<R,(∀t∈I).
From Lemma 13, it follows thatTB⊂B.
ChooseD=c-oX(TB), that is,Dis the convex closure ofTBinX[I,E]. Clearly,Dis nonempty, bounded, convex, and closed subset ofB. By Lemma 14, it follows that(TB)(t)/(1+tα-1)is equicontinuous on[0,1], together with the definition of D, it follows thatD(t)/(1+tα-1)are equicontinuous on[0,1].
Now we show thatTis a strict set contraction operator fromDtoD.
Observing thatD⊂BandTB⊂D, together with Lemma 13 we know thatT:D→Dis bounded and continuous. Finally, we prove that there exists a constant0≤a<1such thatαX(TV)≤aαX(V), forV⊂D.
In fact, by (14), condition(H2)and Lemma 14, applying [22, Lemma 2.6], we have
(25)αX(TV)=supt∈IαE((TV)(t)1+tα-1),∀V⊂D,
where(TV)(t)/(1+tα-1)={(Tu)(t)/(1+tα-1):u∈V,tisfixed}⊂Dfor eacht∈I. Thus, by (25), we need only to prove thatsupt∈IαE((Tu)(t)/(1+tα-1))≤aαX(V), forV⊂D.By condition(H2)and the definition ofD, we know{f(s,u(s)):u∈D}are equicontinuous on[0,1]. Thus, by virtue of Lemma 8 and condition(H3), we get
(26)αE((TV)(t)1+tα-1)≤-11+tα-1∫0t‍(t-s)α-1Γ(α)αE({f(s,u(s)):u∈V})ds+11+tα-1∫01‍(1-s)α-1tα-1Γ(α)222222222222222×αE({f(s,u(s)):u∈V})ds≤2Γ(α)(1+tα-1)∫01‍αE({f(s,u(s)):u∈V})ds=2Γ(α)(1+tα-1)∫01‍αE(f(s,W))ds≤2Γ(α)(1+tα-1)kαE(W),
whereW={u(s):s∈I,u∈V}. For any givenε>0, there exists a partitionV=⋃j=1nVj with
(27)diam(Vj)<αX(V)+ε5,j=1,2,…,n.
Now forj=1,2,…,n, chooseuj∈Vjand a partition0=t0<t1<t2<⋯<tm=1such that
(28)∥uj(t)-uj(t-)∥<ε5,∀j=1,2,…,n,222222t,t-∈[ti-1,ti],∀i=1,2,…,m.
Clearly,W=⋃i=1m⋃j=1nWij, whereWij={u(t):t∈[ti-1,ti],u∈Vj}. For anyu(t),u-(t-)∈Wij(t,t-∈[ti-1,ti],u,u-∈Vj), by (27) and (28), we have
(29)∥u(t)-u-(t-)∥≤∥u(t)-uj(t)∥+∥uj(t)-uj(t-)∥+∥uj(t-)-u-(t-)∥=(1+tα-1)∥u(t)-uj(t)∥1+tα-1+∥uj(t)-uj(t-)∥+(1+t-α-1)∥uj(t-)-u-(t-)∥1+t-α-1≤(1+tα-1)∥u-uj∥X+ε5+(1+t-α-1)∥uj-u-∥X≤4diam(Vj)+ε5<4αX(V)+ε,
which impliesdiamWij≤4αX(V)+εand, thus,αE(W)≤4αX(V)+ε.Sinceεis arbitrary, we get
(30)αE(W)≤4αX(V).
It follows from (25), (26), and (30) that
(31)αE((TV)(t)1+tα-1)≤8Γ(α)(1+tα-1)kαX(V),∀V⊂D.
Thus
(32)supt∈IαE((TV)(t)1+tα-1)≤8Γ(α)(1+tα-1)kαX(V),∀V⊂D.
Consequently
(33)αX(TV)≤aαX(V),
wherea=(8/Γ(α)(1+tα-1))k. From condition(H3), it follows that0≤a<1. Therefore,Tis a strict set contraction operator fromDtoD, obviouslyTis condensing too. It follows from Lemma 9 thatThas at least one fixed point inD, that is, the BVP has at least one solution inX[I,E].


## 4. Example

Now we consider the system of scalar nonlinear fractional differential equations to illustrate our results.Let(34)E=l∞={[0,1]supn|un|<+∞,t∈[0,1]u=(u1,u2,…,un,…),22supn|un|<+∞,t∈[0,1]},
with the norm∥u∥=supn|un|.EvidentlyEis a Banach space. Consider the boundary value problem:
(35)D3/2un=un16(1+t)(1+t2)+sint+|sinun+12|42n2et,2222222222222222222222222222222222t∈(0,1),un(0)=un(1)=0.
System (35) can be regard as a boundary value problem of the form (3), where
(36)α=32,f(t,u)=(f1(t,u),f2(t,u),…,fn(t,u),…)withfn(t,u)=un16(1+t)(1+t2)+sint+|sinun+12|42n2et.Next, we show that conditions(H1)–(H3)are satisfied. Clearly,f(t,u)∈C(I×E,E)and
(37)∥f(t,u)∥≤[116(1+t)(1+t2)+142et]∥u∥+142et.
With the aid of simple computation, we have
(38)∫01‍(1+t)[116(1+t)(1+t2)+142et]dt=π64+17-13e≈0.0693,∫01‍142etdt≈0.0126.
Note thatΓ(3/2)≈0.8862.Takea(t)=(1/16(1+t)(1+t2))+(1/42et)andb(t)=1/42et. Clearly, condition(H1)is satisfied. It is easy to see that condition(H2)is also satisfied.Finally, we verify condition(H3). Denotef=f(1)+f(2), where
(39)fn(1)=un(t)16(1+t)(1+t2),fn(2)=sint+|sinun+12(t)|42n2et.
Then we can obtain thatα(f(2)(t,D))=0for any bounded setD⊂E. Indeed, let{u(m)}⊂Ebe bounded, that is, there existsM>0, such that∥u(m)∥≤M,m=1,2,3,…, whereu(m)=(u1(m),…,un(m),…). Then we have, for eacht∈I,
(40)∥fn(2)(t,u(m))∥≤1+M42n2,n,m=1,2,3,…
which implied that{fn(2)(t,u(m))}is bounded. And by the diagonal method, we can choose a subsequence{u(mk)}⊂{u(m)}such that
(41)fn(2)(t,u(mk))⟶vn,k⟶∞,n=1,2,3,….
From (40), it follows that
(42)∥vn∥≤1+M42n2,n=1,2,3,…
that is,v={v1,…,vn,…}∈E.For anyε>0, (40) and (42) imply that there existsN>0such that
(43)∥fn(2)(t,u(mk))∥<ε2,∥vn∥<ε2,n>N,k=1,2,3,….
On the other hand, from (41) we obtain that there existsK>0such that
(44)∥fn(2)(t,u(mk))-vn∥<ε,k>K,n=1,2,…N.
It follows from (43), (44), and the definition of the norm inEthat
(45)∥f(2)(t,u(mk))-v∥<ε,k>K.
This means that∥f(2)(t,u(mk))-v∥→0ask→∞ and sof(2)(t,D)is relatively compact for any boundedD⊂E. Hence
(46)α(f(2)(t,D))=0,∀t∈I,D⊂E.
Consequently, we arrive at
(47)α(f(t,D))≤α(f(1)(t,D))=α(D)16(1+t)(1+t2)≤116α(D),∀t∈I,D⊂E,
since
(48)8(1+t)Γ(2/3)116<1,
we conclude that condition(H3)is satisfied fork=1/16.Therefore, all the conditions of Theorem15 are satisfied. An application of Theorem 15 implies that problem (35) has a solution.


## 5. Conclusion

In this paper, we present some sufficient conditions which ensure the existence of solutions to fractional differential equation for Dirichlet-type boundary value problems. Applying the Sadovskii fixed point theorem, we establish some new existence criteria for boundary value problems (3) in Banach space. Although, for the fractional differential equation for the Dirichlet boundary value problem (3), only a few papers have dealt with the boundary value problem for fractional differential equations, especially in Banach space. In this aspect, our work fills up the deficiency. As applications, examples are presented to illustrate the main results.



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