# A Study of Caputo-Hadamard-Type Fractional Differential Equations with Nonlocal Boundary Conditions

Authors: Wafa Shammakh
Journal: Journal of Function Spaces
 (2016)

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## Abstract

Existence and uniqueness results of positive solutions to nonlinear boundary value problems for Caputo-Hadamard fractional differential equations by using some fixed point theorems are presented. The related Green’s function for the boundary value problem is given, and some useful properties of Green’s function are obtained. Example is presented to illustrate the main results.


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## Body

## 1. Introduction

In this paper, we study the existence and uniqueness of positive solutions for them-point boundary value problems for Caputo-Hadamard fractional differential equations of the form(1)DαHCxt+ft,xt=0,t∈1,e,n-1<α≤n,n>2,DxHC1=∑i=1m-2βiDxHCηi,D2HCx1=D3HCx1=⋯=Dn-1HCx1=0,xe=∑i=1m-2γixηi,where DαHC is the Caputo-Hadamard fractional derivative of order n-1<α≤n, ∑i=1m-2γi<1, ∑i=1m-2βi<1, ηi∈1,e for all i=1,2,…,m-2, η1<η2<⋯<ηm-2 and f:1,e×0,∞→0,∞.The derivative is a kind of fractional derivatives attributed to Hadamard in 1892 [1]; this fractional derivative differs from the Riemann-Liouville and Caputo fractional derivatives in the sense that the kernel of the integral contains a logarithmic function of arbitrary exponent. The Riemann-Liouville and Hadamard derivative have their own disadvantages as well, one of which is the fact that the derivative of a constant is not equal to zero in general. The subject of Hadamard-type fractional differential equations has received much attention by many researchers. Some new results on the existence of solutions for a fractional boundary value problem involving Hadamard-type fractional differential inclusions and integral boundary conditions can be found in [2].In [3], Tariboon et al. studied the existence and uniqueness of solutions to the boundary value problems consisting of a fractional differential equation of Riemann-Liouville type, subject to the Hadamard fractional integral equations. Thiramanus et al. [4] investigated the existence and uniqueness of solutions for a Hadamard fractional differential equations with nonlocal fractional integral boundary conditions. Ahmad and Ntouyas [5, 6] studied the existence results for a boundary value problem of nonlinear fractional hybrid differential inclusions of Hadamard type with Dirichlet boundary condition and for a coupled system of Hadamard fractional differential equations and Hadamard-type integral boundary conditions, respectively.Jarad et al. [7] modified the Hadamard fractional derivative into a more suitable one having physically interpretable initial conditions similar to the ones in the Caputo setting.Basic definitions and properties of fractional calculus and Hadamard-type fractional calculus can be found in [8–10].The idea of this paper is to demonstrate sufficient conditions on existence and uniqueness of positive solutions to nonlinear boundary value problems (1) for modified Hadamard fractional (Caputo-Hadamard) differential equations, by using Banach’s fixed point theorem, Leray-Schauder nonlinear alternative theorem for single valued maps, Krasnoselskii’s fixed point theorem, and some properties of the Green function.


## 2. Preliminaries

In this section, we introduce some notations and definitions of Hadamard-type fractional calculus.Definition 1 (see [11, 12]).
The Hadamard derivative of fractional orderα for a function g:[1,∞)→R is defined as(2)Dαgt=1Γn-αtddtn∫1tlog⁡tsn-α-1gssds,n-1<α<n,n=α+1,where α denotes the integer part of the real number α and log⁡·=loge·.Definition 2 (see [11, 12]).
The Hadamard fractional integral of orderα∈R+ of a function gx,∀x>0, is defined as(3)Ia+αgx=1Γα∫axlog⁡xtα-1gttdt,Ib-αgx=1Γα∫xblog⁡txα-1gttdt.Definition 3 (see [1, 7]).
LetRα≥0,n=Rα+1, and g∈ACδna,b,0<a<b<∞. Then(4)Da+αCgx=Da+αgx-∑k=0n-1δkgak!log⁡takx,Db-αCgx=Db-αgx-∑k=0n-1-1kδkgbk!log⁡btkx.Here Rα≥0,n=Rα+1,0<a<b<∞ and (5)g∈ACδna,b=g:a,b⟶C:δn-1gx∈ACa,b,δ=xddx.In particular, if 0<Rα<1, then(6)Da+αCgx=Da+αgx-gax,Db-αCgx=Db-αgx-gbx.Theorem 4 (see [1, 7]).
LetRα≥0,n=Rα+1 and g∈ACδna,b,0<a<b<∞. Then Da+αCgx and Db-αCgx exist everywhere on [a,b] and
(a) ifα∉N0,(7)Da+αCgx=1Γn-α∫axlog⁡xtn-α-1δngtdtt=Ia+n-αδngx,Db-αCgx=-1nΓn-α∫xblog⁡txn-α-1δngtdtt=-1nIa+n-αδngx,
(b) ifα=n∈N0,(8)Da+αCgx=δngx,Db-αCgx=-1nδngx.In particular, (9)Da+0Cgx=Db-0Cgx=gx.Lemma 5 (see [1, 7]).
LetRα≥0,n=Rα+1, and g∈Ca,b.
IfRα≠0 or α∈N, then (10)Da+αCIa+αgx=gx,Db-αCIb-αgx=gx.Lemma 6 (see [1, 7]).
Letg∈ACδna,b or Cδna,b and α∈C, and then (11)Ia+αDa+αCgx=gx-∑k=0n-1δkgak!log⁡tak,Ib-αDb-αCgx=gx-∑k=0n-1-1kδkgak!log⁡btk.Lemma 7 (see [1, 7]).
Forn-1<α≤n,n>2, and h∈C1,e,R, the unique solution of the problem(12)DαHCxt+ht=0,t∈1,e,DxHC1=∑i=1m-2βiDxHCηi,D2HCx1=D3HCx1=⋯=Dn-1HCx1=0,xe=∑i=1m-2γixηiis given by (13)xt=∫1eGt,s+Ht,s;η1,η2,…,ηm-2hsdss,where(14)Gt,s=1Γαlog⁡esα-1-log⁡tsα-1,1≤s≤t≤e,log⁡esα-1,1≤t≤s≤e,Ht,s;η1,η2,…,ηm-1=∑i=1m-2γilog⁡e/sα-1-log⁡ηi/sα-1δ2Γα+μ-δ2log⁡t∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1,1≤s≤ηi,i=1,2,…,m-1,∑i=1m-2γilog⁡e/sα-1δ2Γα,ηi≤s≤e,i=1,2,…,m-1,δ1=1-∑i=1m-2βi,δ2=1-∑i=1m-2γi,μ=1-∑i=1m-2γilog⁡ηi.Proof.
In view of Lemma6, the solution of the Hadamard differential equation (12) can be written as (15)xt=-IαHht+c0+c1log⁡t+c2log⁡t2+c3log⁡t3+⋯+cn-1log⁡tn-1,DxHCt=-Iα-1Hht+c1+2c2log⁡t+3c3log⁡t2+⋯+n-1cn-1log⁡tn-2,D2HCxt=-Iα-2Hht+2c2+6c3log⁡t+⋯+n-1n-2cn-1log⁡tn-3,⋮Dn-1HCxt=-Iα-n+1Hht+n-1!cn-1.In view of the boundary conditions, we conclude that (16)c2=c3=⋯=cn-1=0,c1=-1δ1∑i=1m-2βiIα-1hηi,c0=1δ2Iαhe-∑i=1m-2γiIαhηi+μδ1δ2∑i=1m-2βiIα-1hηi.Substituting the values of ci,i=0,1,…,n-1, we obtain (17)xt=-Iαht+1δ2Iαhe-∑i=1m-2γiIαhηi+μ-δ2log⁡tδ1δ2∑i=1m-2βiIα-1hηi=-Iαht+Iαhe+1δ21-δ2Iαhe-∑i=1m-2γiIαhηi+μ-δ2log⁡tδ1δ2∑i=1m-2βiIα-1hηi=-Iαht+Iαhe+1δ2∑i=1m-2γiIαhe-∑i=1m-2γiIαhηi+μ-δ2log⁡tδ1δ2∑i=1m-2βiIα-1hηi=-1Γα∫1tlog⁡tsα-1hsdss+1Γα∫1elog⁡esα-1hsdss+1δ2Γα∑i=1m-2γi∫1elog⁡esα-1hsdss-∫1ηilog⁡ηisα-1hsdss+μ-δ2log⁡tδ1δ2Γα-1∑i=1m-2βi∫1ηilog⁡ηisα-2hsdss.Lemma 8.
The functionsGt,s, Ht,s;η1,η2,…,ηm-2 defined by (15) satisfyi
Gt,s≥0,Ht,s;η1,η2,…,ηm-2≥0,∀t,s∈1,e,ii
minτ1≤t≤τ2⁡Gt,s≥τ0max1≤t≤e⁡Gt,s=τ0Gs,s,∀t,s∈1,e,1<τ1<τ2<e,(18)Gs,s=1Γαlog⁡esα-1,τ0=minτ1≤t≤τ2⁡ψt=log⁡eτ2,iii
N2qs≤Ht,s;η1,η2,…,ηm-2≤N1qs, where(19)qs=log⁡e/sα-2δ1δ2Γα,N1=δ1∑i=1m-2γi+α-1μ∑i=1m-2βi,N2=min∑i=1m-2γilog⁡eηi,δ1∑i=1m-2γiα-1,iv
minτ1≤t≤τ2⁡Ht,s;η1,η2,…,ηm-2≥τ∗max1≤t≤e⁡Ht,s;η1,η2,…,ηm-2,s∈1,e, where(20)τ∗=1μμ-δ2log⁡τ2<1,1<τ1<τ2<e.Proof.
It is clear that (i) holds. So we prove that (ii) is true.
(ii) In view of the expression forGt,s, it follows that Gt,s≤Gs,s for all s,t∈1,e.
If1≤s≤t≤e, we have(21)Gt,sGs,s=log⁡e/sα-1-log⁡t/sα-1log⁡e/sα-1=log⁡e/slog⁡e/sα-2-log⁡t/slog⁡t/sα-2log⁡e/sα-1≥log⁡e/sα-2log⁡e/s-log⁡t/slog⁡e/sα-1=log⁡e/tlog⁡e/s≥log⁡et≔ψt.If 1≤t≤s≤e, we have(22)Gt,sGs,s=1≥ψt.Thus(23)max1≤t≤e⁡Gt,s=Gs,s,ψtGs,s≤Gt,s≤Gs,s,∀t,s∈1,e.Therefore,(24)minτ1≤t≤τ2⁡Gt,s≥τ0max1≤t≤e⁡Gt,s=τ0Gs,s,∀t,s∈1,e,1<τ1<τ2<e. (iii) If 1≤s≤ηi,i=1,2,…,m-2, (25)Ht,s;η1,η2,…,ηm-2=∑i=1m-2γilog⁡e/sα-1-logηi/sα-1δ2Γα+μ-δ2log⁡t∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1≤∑i=1m-2γilog⁡e/sα-1δ2Γα+μ-δ2log⁡t∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1=1δ1δ2Γαδ1∑i=1m-2γilog⁡esα-1+α-1μ∑i=1m-2βilog⁡ηisα-2≤log⁡e/sα-2δ1δ2Γαδ1log⁡es∑i=1m-2γi+α-1μ∑i=1m-2βi≤log⁡e/sα-2δ1δ2Γαδ1∑i=1m-2γi+α-1μ∑i=1m-2βi≔N1qs,Ht,s;η1,η2,…,ηm-2=∑i=1m-2γilog⁡e/sα-1-log⁡ηi/sα-1δ2Γα+μ-δ2log⁡t∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1≥∑i=1m-2γilog⁡e/slog⁡e/sα-2-log⁡ηi/slog⁡ηi/sα-2δ2Γα≥log⁡e/sα-2δ2Γα∑i=1m-2γilog⁡es-log⁡ηis=log⁡e/sα-2δ2Γα∑i=1m-2γilogeηi≔N2qs.
Ifηi≤s≤e,i=1,2,…,m-2, we have (26)Ht,s;η1,η2,…,ηm-2=∑i=1m-2γilog⁡e/sα-1δ2Γα≤log⁡e/sα-2δ1δ2Γαδ1∑i=1m-2γi≤δ1∑i=1m-2γiqs<N1qs,Ht,s;η1,η2,…,ηm-2=∑i=1m-2γilog⁡e/sα-1δ2Γα=∑i=1m-2γiα-1log⁡e/sα-1δ2α-1Γα≥log⁡e/sα-2δ1δ2Γαδ1∑i=1m-2γiα-1≥N2qs.(iv) Since ∂Ht,s;η1,η2,…,ηm-2/∂t=-∑i=1m-2βilogηi/sα-2/δ1Γα-1t≤0, then Ht,s;η1,η2,…,ηm-2 is nonincreasing in t, so (27)minτ1≤t≤τ2⁡Ht,s;η1,η2,…,ηm-2=∑i=1m-2γilog⁡e/sα-1-log⁡ηi/sα-1δ2Γα+μ-δ2log⁡τ2∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1=∑i=1m-2γilog⁡e/sα-1-log⁡ηi/sα-1δ2Γα+τ∗μ∑i=1m-2βilog⁡ηi/sα-2δ1δ2Γα-1≥τ∗max1≤t≤e⁡Ht;η1,η2,…,ηm-2,s.


## 3. Existence Results

Let us denote byE=C(1,e,R) the Banach space of all continuous functions from 1,e to R endowed with the norm x=Supt∈[1,e]x(t) and let P be the cone P=x∈E:xt≥0,t∈1,e.Through this paper, we assume that the functionf:1,e×0,∞→0,∞ satisfies the following conditions of Carathéodory type:H
1: (i) f(t,x) is Lebesgue measurable with respect to t on 1,e,(ii)f(t,x) is continuous with respect to x on 0,∞.By Lemma6, we obtain an operator F:E→E as (28)Fxt=xt=∫1eGt,s+Ht,s;η1,η2,…,ηm-2hsdss,t∈1,e.It should be noticed that problem (1) has solutions if and only if the operator F has fixed points.The first existence and uniqueness result is based on the Banach contraction principle.Theorem 9.
Assume that the conditionH1 holds and there exists a real-valued function g(t)∈L1,e such that(29)ft,xt-ft,yt≤gtx-y,∀t∈1,e,x,y∈R.Then problem (1) has a unique solution provided gΨ<1, where(30)Ψ=1Γα+1+N1δ1δ2α-1Γα.Proof.
We setSupt∈1,efs,0=p<∞ and choose ρ≥Ψp/(1-Ψg).
Now, we show thatFBρ⊂Bρ, where Bρ=x∈E:x≤ρ. For any x∈Bρ, we have (31)Fxt=Supt∈1,e⁡∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xsdss≤∫1eGs,s+N1qsfs,xs-fs,0+fs,0dss≤∫1eGs,s+N1qsgsρ+pdss≤gρ+p1Γα∫1elog⁡esα-1dss+N1δ1δ2Γα∫1elog⁡esα-2dss≤gρ+p1Γα+1+N1δ1δ2α-1Γα≤Ψgρ+p≤ρ.It follows that FBρ⊂Bρ. For x,y∈E and for each t∈1,e, we have (32)Fxt-Fyt=∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xs-fs,ysdss≤∫1eGs,s+N1qsgsxs-ysdss≤gx-y∫1eGs,s+N1qsdss≤Ψgx-y.Hence it follows that Fxt-Fyt≤Ψgx-y, where Ψg<1. Therefore F is a contraction. Hence by the contraction mapping principle, problem (1) has a uniqueness solution.Theorem 10 (nonlinear alternative for single valued maps [13]).
LetE be a Banach space, C a closed, convex subset of E, U an open subset of C, and 0∈U. Suppose that F:U-→C is a continuous, compact (i.e., F(U-) is a relatively compact subset of C) map. Then either(i)
F has a fixed point in U- or(ii)
There isx∈∂U (the boundary of U in C) and λ∈1,e with x=λFx.Theorem 11.
Assume thatH1 and the following conditions hold:
H
2 There exist two nonnegative real-valued functions h1,h2∈1,e such that (33)ft,x≤h1t+h2tx,foreveryt∈1,e,allx∈0,∞.H3 There exists a constant M>0, such that(34)Mh1+h2MΨ>1.Then the boundary value problem (1) has at least one solution on 1,e.Proof.
First, we show that the operatorF:P→P is continuous.
For anyxn,x∈P,n=1,2,… with limn→∞⁡xnt=xt,t∈1,e. Thus, by condition (ii) of H1, we have limn→∞⁡f(t,xnt)=f(t,xt),t∈1,e. So, we can conclude that (35)Supt∈1,e⁡ft,xnt-ft,xt⟶0,asn⟶∞.On the other hand,(36)Fxnt-Fxt=∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xs-fs,ysdss≤Supt∈1,e⁡ft,xnt-ft,xt∫1eGs,s+N1qsdss≤1Γα+1+N1δ1δ2α-1ΓαSupt∈1,e⁡ft,xnt-ft,xt.Hence, (37)Fxnt-Fxt⟶0asn⟶∞.This means that F is continuous.
Now, we show thatF maps bounded sets into bounded sets in P. It suffices to show that, for any σ>0, there exists a positive constant ρ1>0 such that, for each x∈Bρ∗=x∈P:x≤ρ∗, we have Fx≤ρ1. By (33), for each t∈1,e, we have (38)Fxt=∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xsdss≤∫1eGs,s+N1qsh1s+h2sxsdss≤h1+h2ρ∗∫1eGs,s+N1qsdss≤h1+h2ρ∗1Γα+1+N1δ1δ2α-1Γα≤h1+h2ρ∗Ψ≔ρ1,which implies that Fx≤ρ1. Further, we let t1,t2∈1,e with t1<t2 and x∈Bρ∗, where Bρ∗ is a bounded set of P, and then we find that(39)DHCFxt=-1Γα-1∫1elog⁡tsα-2+1δ1∑i=1m-2βilog⁡ηisα-2fs,xsdss≤h1+h2ρ∗δ1+∑i=1m-2βiδ1Γα-1≔ρ2.Hence, for t1,t2∈1,e, we have (40)Fxt2-Fxt1≤∫t1t2DHCFxsds≤ρ2t2-t1.This implies that F maps bounded sets into equicontinuous sets of P.
Thus, by the Arzelà-Ascoli theorem, the operatorF:P→P is completely continuous.
Next, we consider the setV=u∈E∣x=μFx,0<μ<1 and show that the set V is bounded, let x∈V, and then x=μFx,0<μ<1. For any t∈1,e, we have(41)xt=μFxt≤∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xsdss≤ρ1.Thus, xt≤ρ1 for any t∈1,e, so that set V is bounded. Thus, by the conclusion of Theorem 10, the operator F has at least one fixed point, which implies that problem (1) has at least one solution.Theorem 12 (Krasnoselskii fixed point theorem [14]).
LetE be a Banach space and K⊂E is a cone in E. Assume that Ω1 and Ω2 are open subsets of E with 0∈Ω1 and Ω-1⊂Ω2. Let T:K∩Ω-2∖Ω1→K be completely continuous operator. In addition suppose that either(i)
Tu≤u,∀u∈K∩∂Ω1 and Tu≥u,∀u∈K∩∂Ω2 or(ii)
Tu≤u,∀u∈K∩∂Ω2 and Tu≥u,∀u∈K∩∂Ω1
holds. ThenT has a fixed point in K∩Ω-2∖Ω1.
To state the last result of this section, we set(42)M1=Ψ-1,M2=τ0Γα+1+α-1τ∗μ∑i=1m-2βiδ1δ2Γα-∑i=1m-2γiδ2Γα+1-1.Theorem 13.
Suppose that there exist two positive constantsr2>r1>0 and M1∈0,M2:(i)
ft,x≤M1r2, for t,x∈1,e×0,r2.(ii)
ft,x≥M2r1, for t,x∈1,e×0,r1.
Then (1) has at least a positive solution.Proof.
LetΩi=x∈E∣x<ri,i=1,2. From the proof of Theorem 11, we know that the operator F defined by (28) is completely continuous on P.
For anyx∈P∩∂Ω1, it follows that (43)Fxt=Supt∈1,e⁡∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xsdss≥∫1eτ0Gs,s+τ∗H1,s;η1,η2,…,ηm-2M2r1dss≥M2r1τ0Γα+1+α-1τ∗μ∑i=1m-2βiδ1δ2Γα-∑i=1m-2γiδ2Γα+1≥r1=x;that is, Fxt≥x,x∈P∩∂Ω1.
On the other hand, for anyx∈P∩∂Ω2, it follows that(44)Fxt=Supt∈1,e⁡∫1eGt,s+Ht,s;η1,η2,…,ηm-2fs,xsdss≤∫1eGs,s+N1qsM1r2dss≤M1r2Ψ=r2=x;that is, Fxt≤x,x∈P∩∂Ω2.
In view of Theorem12, F has a fixed point in P∩Ω-2/Ω1, which is a positive solution to (1).


## 4. Examples

In this section, we exemplify our theoretical results obtained in Section3.Example 1.
Consider the following BVP for Hadamard fractional differential equation:(45)D7/3xt+HCft,xt=0,1≤t≤e,DHCx1=12DHCx43,D2HCx1=0,xe=14x43.Here α=7/3, β=0.5, γ=0.25, η=4/3, and(46)ft,x=et2et+1x2x+1+x4x+1+34,t,x∈1,e×0,∞.Using the given data, we find that Ψ=1.609, δ1=0.5, δ2=0.75, μ=0.928, and (47)ft,x-ft,y≤et2et+1x-y≤12x-y,withgt=et2et+1.Hence, we obtain gΨ=0.805<1. Therefore, by Theorem 9, problem (45) has a unique solution on 1,e.Example 2.
Consider the following BVP for Hadamard fractional differential equation:(48)D5/2xt+HCft,xt=0,1≤t≤e,DHCx1=14DHCx32,D2HCx1=0,xe=13x32.Here α=5/2, β=0.25, γ=1/3, η=3/2. Let(49)ft,x=9sin2⁡t+etx12et+1x2+1,t,x∈1,e×0,∞, with h1t=9sin2t, h2t=et/12et+1. Here, ft,x≤h1t+h2tx.
It is easy to verify thatM/h1+h2MΨ=1.147>1.
Then, by Theorem11, problem (48) has at least one solution on 1,e.Example 3.
Consider the following BVP for Hadamard fractional differential equation:(50)D7/2xt+HCft,xt=0,1≤t≤e,DHCx1=12DHCx43+14DHCx52,D2HCx1=0,xe=13x43+15x52,where α=7/2, β1=0.5, β2=0.25, γ1=1/3, γ2=1/5, η1=4/3, η2=5/2. Let(51)ft,x=14x+et5et+1,t,x∈1,e×0,∞. It is easy to verify that (52)M1=Ψ-1=0.6196,M2=τ0Γα+1+α-1τ∗μ∑i=1m-2βiδ1δ2Γα-∑i=1m-2γiδ2Γα+1-1=0.521.Choosing r1=1/5, r2=e, we have (53)ft·x≤1.2≤M1r2,t,x∈1,e×0,r2,ft·x>0.5≥M2r1,t,x∈1,e×0,r1.Hence all conditions of Theorem 13 are satisfied; then problem (50) has at least one positive solution x such that 1/5<x<e.



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