# Existence and Uniqueness of Solutions for theCauchy-TypeProblems of Fractional Differential Equations

Authors: Chunhai Kou; Jian Liu; Yan Ye
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

By using the Banach fixed point theorem and step method, we study the existence and uniqueness of solutions for the Cauchy-type problems of fractional differential equations. Meanwhile, by citing some counterexamples, it is pointed out that there exist a few defects in the proofs of the known results.


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## Body

## 1. Introduction

Recently, fractional differential equations are applied widely in various fields of science and engineering. Regarding applications of fractional differential equations, we refer to [1–15] and references cited therein. However, the investigation of basic theory of fractional differential equations is still not complete, and there is a great deal of work which needs to be done. Most of the investigations in this field involve the existence and uniqueness of solutions to fractional differential equations on the finite interval [a,b]. In 1938, Pitcher and Sewell[16] first considered the nonlinear fractional differential equation(1.1)(Da+αy)(x)=f[x,y(x)],
with the following initial conditions:(1.2)(Da+α-ky)(a+)=bk,bk∈ℝ,(k=1,…,n,n=-[-α]),
where 0<α<1, and Da+α is Riemann-Liouville fractional derivative. Barrett [17], in 1954, first considered the Cauchy-type problem for the linear fractional differential equation(1.3)(Da+αy)(x)-λy(x)=f(x),(n-1≤R(α)<n,α≠n-1),
with the same initial conditions (1.2). Afterwards, there is a great deal of work about the basic theory [18–27]. In [28], Kilbas et al. summarized systematically the main results.In this paper we consider the cauchy problem (1.1)-(1.2); here Da+α can be Riemann-Liouville fractional derivative, and Hadamard-type fractional derivative. We establish some results about the existence and uniqueness of solution of (1.1)-(1.2). By the way, we will point out that there exist several defects in the proofs of the related theorems of [28].This paper is organized as follows: in Section2, we introduce some preliminaries and notations; main results are proved in Section 3; in Section 4, by citing several counterexamples, we will point out the defects in [28]; Section 5 is a brief summary of this paper.


## 2. Preliminaries and Notations

In this section, we introduce some basic definitions and notations about fractional calculus. Meanwhile, several known theorems are given, which are useful in this paper.Definition 2.1 (see [28]).
LetΩ=(a,b](-∞<a<b<∞) be a finite interval on the real axis ℝ. The Riemann-Liouville left-sided fractional integral Ia+αg of the function g with order α∈ℝ(α>0) is defined by
(2.1)(Ia+αg)(x)=1Γ(α)∫axg(t)dt(x-t)1-α,(x>a),
where the real function g is defined on the interval Ω and the right-side integral of the above equality is assumed to make sense.Definition 2.2 (see [28]).
LetΩ=(a,b](-∞<a<b<∞) be a finite interval on the real axis ℝ. The Riemann-Liouville left-sided fractional derivative Da+αg of the function g with order α∈ℝ(α≥0) is defined by
(2.2)(Da+αg)(x)=(ddx)n(Ia+n-αg)(x)=1Γ(n-α)(ddx)n∫axg(t)dt(x-t)α-n+1,(x>a;n=-[-α]),
where the real function g is defined on the interval Ω and the right side of the above equality is assumed to make sense.Definition 2.3.
Assume thatf[x,y] is defined on the set (a,b]×G(G⊂ℝ). f[x,y] is said to satisfy Lipschitzian condition with respect to the second variable, if for all x∈(a,b] and for any y1,y2∈G one has
(2.3)|f[x,y1]-f[x,y2]|≤A|y1-y2|,
where A>0 does not depend on x∈(a,b].Definition 2.4 (see [28]).
Letn-1<α≤n(n∈N), then the space Cn-αα[a,b] is defined by
(2.4)Cn-αα[a,b]={y(x)∈Cn-α[a,b]:(Da+αy)(x)∈Cn-α[a,b]}.
Here Cn-α[a,b] is a weighted space of continuous functions
(2.5)Cn-α[a,b]={g:(a,b]→R:(x-a)n-αg(x)∈C[a,b]},
and Da+α is the Riemann-Liouville fractional derivative.
In the spaceCn-α[a,b], we define the norm ∥g∥Cn-α=∥(x-a)n-αg(x)∥C.Definition 2.5 (see [28]).
Let(a,b)(0<a<b≤∞) be a finite or infinite interval of the half-axis ℝ+. The Hadamard type left-sided fractional integral 𝒥a+αh of the function h with order α∈ℝ(α>0) is defined by
(2.6)(𝒥a+αh)(x)=1Γ(α)∫ax(lnxt)α-1h(t)dtt,(a<x<b),
where h:(a,b)→ℝ and the right-side integral of the above equality is assumed to make sense.Definition 2.6 (see [28]).
Letδ=xD(D=d/dx) be the δ-derivative. The Hadamard left-sided fractional derivative 𝒟a+αy of the function y on (a,b) with order α∈ℝ(α≥0) is defined by
(2.7)(𝒟a+αy)(x)=δn(𝒥a+n-αy)(x)=1Γ(n-α)(xddx)n∫ax(lnxt)n-α-1y(t)dtt,(a<x<b;n=-[-α]),
where y:(a,b)→ℝ, δn=δ⋯δ︸n, and the right side of the above equality is assumed to make sense.Definition 2.7 (see [28]).
Letn-1<α≤n(n∈N),0<a<b<+∞, and 0≤γ<1. The space Cδ;n-α,γα[a,b] is defined by
(2.8)Cδ;n-α,γα[a,b]={y(x)∈Cn-α,ln[a,b]:(𝒟a+αy)∈Cγ,ln[a,b]},
where 𝒟a+α is a Hadamard left-sided fractional derivative, and Cγ,ln[a,b] is a weighted space of continuous functions
(2.9)Cγ,ln[a,b]={g:(a,b]→ℝ:(lnxa)γg(x)∈C[a,b]}.
In the spaceCγ,ln[a,b], we define the norm ∥g∥Cγ,ln=∥(ln(x/a))γg(x)∥C.Theorem 2.8 (see [28]).
Letα>0, n=-[-α]. Let f:(a,b]×ℝ→ℝ be a function such that f[x,y(x)]∈Cn-α[a,b] for any y(x)∈Cn-α[a,b]. If y(x)∈Cn-α[a,b], then y(x) satisfies the relations:
(2.10)(Da+αy)(x)=f[x,y(x)],(α>0),(2.11)(Da+α-ky)(a+)=bk,bk∈ℝ,(k=1,…,n=-[-α]),
if and only if y(x) satisfies the Volterra integral equation
(2.12)y(x)=y0(x)+1Γ(α)∫axf[t,y(t)]dt(x-t)1-α,(x>a),
where
(2.13)y0(x)=∑j=1nbjΓ(α-j+1)(x-a)α-j,
where Da+α is a Riemann-Liouville left-sided fractional derivative.Theorem 2.9 (see [28]).
(Banach Fixed Point Theorem) Let (U,d) be a nonempty complete metric space, let0≤ω<1, and let T:U→U be a map such that, for every u,v∈U, the relation
(2.14)d(Tu,Tv)≤ωd(u,v),(0≤ω<1)
holds. Then the operator T has a unique fixed point u*∈U.Theorem 2.10 (see [28]).
Let0<a<b<∞,α>0, n=-[-α], and 0≤γ<1. Let f:(a,b]×ℝ→ℝ be a function such that f[x,y(x)]∈Cγ,ln[a,b] for any y(x)∈Cγ,ln[a,b]. If y(x)∈Cn-α,ln[a,b], then y(x) satisfies
(2.15)(𝒟a+αy)(x)=f[x,y(x)],(x>a),(𝒟a+α-ky)(a+)=bk,bk∈ℝ,(k=1,…,n;n=-[-α]),
if and only if y(x) satisfies the Volterra integral equation
(2.16)y(x)=∑j=1nbjΓ(n-j+1)(lnxa)α-j+1Γ(α)∫ax(lnxt)α-1f[t,y(t)]dtt,(x>a),
where 𝒟a+α is a Hadamard-type left-sided fractional derivative.Remark 2.11.
It should be worthy noting that the conditions in Theorems2.8 and 2.10 are a little different from the ones in [28, pages 163, 213]. In [28], G is an open set in R and f is assumed to be a function such that f[x,y]∈Cn-α[a,b](Cγ,ln[a,b]) for any y∈G. In fact, we think that such assumption is not complete for the proof of the related conclusion.


## 3. Main Results

In this section, we will establish several useful lemmas. It should be pointed out that, in [28], some analogous lemmas play important roles in the proofs of the related results. However, we have found out that there exist a few defects in these lemmas of [28], which means that the proofs of the related results in [28] are not complete. Several counterexamples will be given in Section 4. In a sense, our lemmas are to mend these cracks. Furthermore, several theorems about the existence and uniqueness of solution for the cauchy-type problem (2.10)-(2.11) will be given; then, in the sense of Hadamard fractional derivative, we have the similar result.Lemma 3.1.
Letγ∈[0,∞),a<c<b,g∈Cγ[a,c], and g∈C[c,b]. Then g∈Cγ[a,b] and
(3.1)∥g∥Cγ[a,b]≤max{∥g∥Cγ[a,c],(b-a)γ∥g∥C[c,b]}.Proof.
Sinceg∈Cγ[a,c] and g∈C[c,b], then g∈C(a,b] and g∈Cγ[a,b]. Now we prove the estimate. Because g∈Cγ[a,b], there exists x0∈[a,b] such that
(3.2)∥g∥Cγ[a,b]=|(x0-a)γg(x0)|.
Ifx0∈[a,c], then
(3.3)∥g∥Cγ[a,b]≤∥g∥Cγ[a,c].
Ifx0∈[c,b], then
(3.4)∥g∥Cγ[a,b]≤(b-a)γ∥g∥C[c,b].
Hence we have
(3.5)∥g∥Cγ[a,b]≤max{∥g∥Cγ[a,c],(b-a)γ∥g∥C[c,b]}.
This completes the proof of Lemma3.1.Lemma 3.2 (see [28]).
Ifγ∈ℝ(0≤γ<1), then the fractional integration operator Ia+α with order α∈ℝ(α>0) is a mapping from Cγ[a,b] to Cγ[a,b], and
(3.6)∥Ia+αg∥Cγ≤(b-a)αΓ(1-γ)Γ(1+α-γ)∥g∥Cγ,
here Ia+α is a Riemann-Liouville fractional integral operator and g∈Cγ[a,b].Furthermore, we have the following conclusion.Lemma 3.3.
The fractional integration operatorIa+α with order α∈ℝ(α>0) is a mapping from C[a,b] to C[a,b], and
(3.7)∥Ia+αg∥C≤(b-a)ααΓ(α)∥g∥C,
where Ia+α is a Riemann-Liouville fractional integral operator and g∈C[a,b].Proof.
Firstly we prove that ifg∈C[a,b], then (Ia+αg)(x)∈C[a,b]. For any x∈[a,b] and Δx>0,x+Δx≤b, we have
(3.8)|(Ia+αg)(x+Δx)-(Ia+αg)(x)|=|1Γ(α)∫ax+Δxg(t)dt(x+Δx-t)1-α-1Γ(α)∫axg(t)dt(x-t)1-α|≤1Γ(α){|∫axg(t)[1(x+Δx-t)1-α-1(x-t)1-α]dt|+|∫xx+Δxg(t)(x+Δx-t)1-αdt|}≤∥g∥C[a,b]αΓ(α){[(x+Δx-a)α-(x-a)α]+(Δx)α+(Δx)α}.
It is easy to see that asΔx→0+, we have
(3.9)|(Ia+αg)(x+Δx)-(Ia+αg)(x)|→0.
Similarly, we can prove that asΔx→0-, we have
(3.10)|(Ia+αg)(x+Δx)-(Ia+αg)(x)|→0.
Thus Ia+αg∈C[a,b].
Now we prove the estimate. In fact
(3.11)∥Ia+αg∥C[a,b]=maxx∈[a,b]|1Γ(α)∫axg(t)dt(x-t)1-α|≤∥g∥C[a,b]Γ(α)∫ax(x-t)α-1dt≤(b-a)ααΓ(α)∥g∥C.
This completes the proof of Lemma3.3.Lemma 3.4.
Letγ∈[0,∞), 0<a<c<b<∞, g∈Cγ,ln[a,c] and g∈C[c,b]. Then g∈Cγ,ln[a,b] and
(3.12)∥g∥Cγ,ln[a,b]≤max{∥g∥Cγ,ln[a,c],(lnba)γ∥g∥C[c,b]}.Proof.
The proof is similar to the proof of Lemma3.1. Since g∈Cγ,ln[a,c] and g∈C[c,b], we have g∈C(a,b], that is, g∈Cγ,ln[a,b].
Next we give the estimate. Becauseg∈Cγ,ln[a,b], there exists at least x*∈[a,b] such that
(3.13)∥g∥Cγ,ln[a,b]=|(lnx*a)γg(x*)|.
Ifx*∈[a,c], then
(3.14)∥g∥Cγ,ln[a,b]≤∥g∥Cγ,ln[a,c].
Ifx*∈[c,b], then
(3.15)∥g∥Cγ,ln[a,b]≤(lnba)γ∥g∥C[c,b].
Hence we have
(3.16)∥g∥Cγ,ln[a,b]≤max{∥g∥Cγ,ln[a,c],(lnba)γ∥g∥C[c,b]}.
This completes the proof of Lemma3.4.Next, on the basis of above lemmas, we establish the results about the existence and uniqueness of solution for the cauchy-type problem (2.10)-(2.11) in the sense of Riemann-Liouville fractional derivative and Hadamard fractional derivative.Theorem 3.5.
Letα>0 and n=-[-α]. Let f:(a,b]×ℝ→ℝ be a function such that f[x,y(x)]∈Cn-α[a,b] for any y(x)∈Cn-α[a,b] and the Lipschitzian condition holds with respect to the second variable y. Then there exists a unique solution y(x)∈Cn-αα[a,b] for the cauchy-type problem (2.10)-(2.11).Proof.
First we prove the existence of a unique solutiony(x)∈Cn-α[a,b]. According to Theorem 2.8, it is sufficient to prove the existence of a unique solution y(x)∈Cn-α[a,b] to the nonlinear Volterra integral equation (2.12). Equation (2.12) makes sense in any interval (a,x1]⊂(a,b](a<x1<b). Choose x1 such that
(3.17)A(x1-a)αΓ(α-n+1)Γ(2α-n+1)<1,
where A>0 is the Lipschitzian coefficient. Next we prove the existence of a unique solution y(x)∈Cn-α[a,x1] to (2.12) on the interval (a,x1]. For this, we use the Banach fixed point theorem for the space Cn-α[a,x1], which is a complete metric space with the distance given by
(3.18)d(y1,y2)=∥y1-y2∥Cn-α[a,x1]=maxx∈[a,x1]|(x-a)n-α[y1(x)-y2(x)]|.
We rewrite the integral (2.12) in the form
(3.19)y(x)=(Ty)(x),
where
(3.20)(Ty)(x)=y0(x)+1Γ(α)∫axf[t,y(t)]dt(x-t)1-α.
To apply Theorem2.9, we have to prove the following: (1) if y(x)∈Cn-α[a,x1], then (Ty)(x)∈Cn-α[a,x1]; (2) for any y1,y2∈Cn-α[a,x1] the following estimate holds:
(3.21)∥Ty1-Ty2∥Cn-α[a,x1]≤ω∥y1-y2∥Cn-α[a,x1],ω=A(x1-a)αΓ(α-n+1)Γ(2α-n+1).
It follows from (2.13) that y0(x)∈Cn-α[a,x1]. Since f[x,y(x)]∈Cn-α[a,x1] for any y(x)∈Cn-α[a,x1], then, by Lemma 3.2 [28] (with γ=n-α,b=x1, and g(x)=f[x,y(x)]), the integral in the right-hand side of (3.19) also belongs to Cn-α[a,x1], and hence (Ty)(x)∈Cn-α[a,x1]. Now we prove the estimate in (3.21). By (3.20), using the Lipschitzian condition and applying the relation (3.6) (with γ=n-α, b=x1, and g(x)=f[x,y1(x)]-f[x,y2(x)]), we have
(3.22)∥Ty1-Ty2∥Cn-α[a,x1]≤∥Ia+α[|f[t,y1(t)]-f[t,y2(t)]|]∥Cn-α[a,x1]≤A∥Ia+α[|y1(t)-y2(t)|]∥Cn-α[a,x1]≤A(x1-a)αΓ(α-n+1)Γ(2α-n+1)∥y1-y2∥Cn-α[a,x1],
which yields the estimate (3.21). In accordance with (3.17), 0<ω<1, and hence, by Theorem 2.9, there exists a unique solution y*(x)∈Cn-α[a,x1] to (2.12) on the interval [a,x1].
By Theorem2.9, this solution y*(x) is a limit of a convergent sequence (Tmy0*)(x):
(3.23)limm→∞∥Tmy0*-y*∥Cn-α[a,x1]=0,
where y0*(x) is any function in Cn-α[a,x1]. If there is at least one bk≠0 in the initial condition (2.11), then we can take y0*(x)=y0(x) with y0(x) defined by (2.13). The last relation can be rewritten into the form
(3.24)limm→∞∥ym-y*∥Cn-α[a,x1]=0,
where
(3.25)ym(x)=(Tmy0*)(x)=y0(x)+1Γ(α)∫axf[t,(Tm-1y0*)(t)]dt(x-t)1-α,(m∈N).
Next we consider the interval[x1,b]. Rewrite (2.12) in the form
(3.26)y(x)=y01(x)+1Γ(α)∫x1xf[t,y(t)]dt(x-t)1-α,
where y01(x) is defined by
(3.27)y01(x)=∑j=1nbjΓ(α-j+1)(x-a)α-j+1Γ(α)∫ax1f[t,y(t)]dt(x-t)1-α.
We obtain y01(x)∈C[x1,b]. Next we prove the existence of a unique solution y(x)∈C[x1,b] to (2.12) on the interval [x1,b]. For this, we also use Banach fixed point theorem for the space C[x1,x2], where x2 satisfies
(3.28)A(x2-x1)ααΓ(α)<1.C[x1,x2] is a complete metric space with the distance given by
(3.29)d(y1-y2)=∥y1-y2∥C[x1,x2]=maxx∈[x1,x2]|y1(x)-y2(x)|.
We rewrite the integral equation (3.26) into the form
(3.30)y(x)=(Ty)(x),
where
(3.31)(Ty)(x)=y01(x)+1Γ(α)∫x1xf[t,y(t)]dt(x-t)1-α.
To apply Theorem2.9, we have to prove the following: (1) if y(x)∈C[x1,x2], then (Ty)(x)∈C[x1,x2]; (2) for any y1,y2∈C[x1,x2], the following estimate holds:
(3.32)∥Ty1-Ty2∥C[x1,x2]≤ω∥y1-y2∥C[x1,x2],ω=A(x2-x1)ααΓ(α).
Sincef[x,y(x)]∈Cn-α[a,b] for any y(x)∈Cn-α[a,b], then, by Lemma 3.3, the integral in the right-hand side of (3.31) also belongs to C[x1,x2], and hence (Ty)(x)∈C[x1,x2]. Now we prove the estimate in (3.32) as follows:
(3.33)∥Ty1-Ty2∥C[x1,x2]≤∥Ia+α[|f[t,y1(t)]-f[t,y2(t)]|]∥C[x1,x2]≤A(x2-x1)ααΓ(α)∥y1-y2∥C[x1,x2],
which yields the estimate (3.32). In accordance with (3.28), then 0<ω<1, and hence by Theorem 2.9, there exists a unique solution y1*(x)∈C[x1,x2] to (2.12) on the interval [x1,x2]. By Theorem 2.9, this solution in C[x1,x2] is a limit of a convergent sequence (Tmy01*)(x):
(3.34)limm→∞∥Tmy01*-y1*∥C[x1,x2]=0,
where y01*(x) is any function in C[x1,x2]. If y0(x)≠0 on [x1,x2], then we can take y01*(x)=y0(x) with y0(x) defined by (2.13). The last relation can be rewritten in the form
(3.35)limm→∞∥ym-y1*∥C[x1,x2]=0,
where
(3.36)ym(x)=(Tmy01*)(x)=y01(x)+1Γ(α)∫x1xf[t,(Tm-1y01*)]dt(x-t)1-α,(m∈N).
Next we consider the interval[x2,x3], where x3=x2+h2 such that x3≤b and (A(x3-x2)α/αΓ(α))<1. Using the same arguments as the above, we derive that there exists a unique solution y2*(x)∈C[x2,x3] to (2.12) on the interval [x2,x3]. If x3≠b, then take the next interval [x3,x4], where x4=x3+h3 and h3>0 such that x4≤b and (A(x4-x3)α/αΓ(α))<1. If x4<b, repeating the above process, then we find that there exists a unique solution y(x) to (2.12), y(x)=yk*(x), and yk*(x)∈C[xk,xk+1](k=1,…,L), where x0<x1<⋯<xL+1=b and (A(xk+1-xk)α/αΓ(α))<1, and we take y0(x)=y0k(x), and y0*(x)=y0k*(x)(k=1,…,L) on each interval [xk,xk+1]. By (A(xk+1-xk)α/αΓ(α))<1, we know that by finite steps we can arrive at xL+1=b.
Then there exists a unique solutiony(x)∈C[x1,b] to (2.12) on the interval [x1,b]. By Lemma 3.1, we obtain that there exists a unique solution y(x)∈Cn-α[a,b] to the Volterra integral equation (2.12) on the whole interval [a,b], and hence y(x)∈Cn-α[a,b] is the unique solution to the cauchy-type problem (2.10)-(2.11).
To complete the proof of Theorem3.5, we must show that such a unique solution y(x)∈Cn-α[a,b] belongs to the the space Cn-αα[a,b]; it is sufficient to prove that (Da+αy)(x)∈Cn-α[a,b]. By the above proof, the solution y(x)∈Cn-α[a,b] is a limit of the sequence ym(x), where ym(x)=(Tmy0*)∈Cn-α[a,b]:
(3.37)limm→∞∥ym-y∥Cn-α[a,b]=0,
with the choice of certain y0*(x) on each [a,x1],…,[xL,b].
Ify0(x)≠0, then we can take y0*(x)=y0(x).
By (2.10) and the Lipschitzian-condition, we have
(3.38)∥Da+αym-Da+αy∥Cn-α=∥f[x,ym]-f[x,y]∥Cn-α≤A∥ym-y∥Cn-α.
Thus
(3.39)limm→∞∥Da+αym-Da+αy∥Cn-α=0.
By(Da+αym)(x)=f[x,ym-1(x)] and f[x,y(x)]∈Cn-α[a,b] for any y(x)∈Cn-α[a,b], we have f[x,ym-1(x)]∈Cn-α[a,b], that is, (Da+αym)(x)∈Cn-α[a,b]. Hence (Da+αy)(x)∈Cn-α[a,b].
This completes the proof of Theorem3.5.Corollary 3.6.
Letα>0 and n=-[-α]. Let f:(a,b]×ℝ→ℝ be a function such that f(x,y)∈Cn-α[a,b] for any y∈ℝ; the Lipschitzian condition holds with respect to y and limx→a(x-a)n-αf(x,y(x)), and limx→a(x-a)n-αf(x,y(x)) exist for any y(x)∈Cn-α[a,b]. Then there exists a unique solution y(x)∈Cn-αα[a,b] for the cauchy-type problem (2.10)-(2.11).Remark 3.7.
It should be pointed out that the conditions in Theorem3.5 are different from the ones in [28, Theorem 3.11, page 165]. In [28], G is an open set in R and f is assumed to be a function such that f[x,y]∈Cn-α[a,b] for any y∈G. In fact, such assumptions are not complete for the proof of the related conclusion. A counterexample will be given in Section 4. There exists the similar problem in [28, Theorem 3.29, page 213]. By applying Lemma 3.4, modifying the conditions in [28, Theorem 3.29, page 213] and using the similar arguments to the proof of Theorem 3.5, we arrive at the following result.Theorem 3.8.
Letα>0,n=-[-α], and 0≤γ<1 such that γ≥n-α. Let f:(a,b]×ℝ→ℝ(a>0) be a function such that f[x,y(x)]∈Cγ,ln[a,b] for any y(x)∈Cγ,ln[a,b] and the Lipschitzian condition holds with respect to y. Then there exists a unique solution y(x) for the Cauchy-type problem
(3.40)(𝒟a+αy)(x)=f[x,y(x)],α>0;x>a,(𝒟a+α-ky)(a+)=bk,bk∈ℝ(k=1,…,n,n=-[-α]).
in the space Cδ;n-α,γα[a,b], where 𝒟a+α is a Hadamard fractional derivative.


## 4. Counterexamples

In this section, by citing some counterexamples we would like to point out that, in [28, Lemmas 3.4, 3.9, and 3.10, pages 165, 202, and 213] are not complete.Example 4.1.
Let one consider the function(4.1)g(x)={μ(x-a)γx∈(a,c],μ(x-c)γx∈(c,b],
where μ≠0 is a constant number and γ∈(0,∞).From the above definition ofg(x), we know that g(x)∈Cγ[a,c] and g(x)∈Cγ[c,b], but we cannot get the conclusion g(x)∈Cγ[a,b] and ∥g∥Cγ[a,b]≤max{∥g∥Cγ[a,c],∥g∥Cγ[c,b]}. Hence the conclusion of [28, Lemma 3.4, page 165] does not hold. We cannot apply it to prove [28, Theorem 3.11, page 165]. Furthermore, there also exists a problem about f in [28, Theorem 3.11, page 165]. For example, choosing f(x,y)=ysin(1/(x-a)), we know that f(x,y)∈Cγ[a,b] for any y, f(x,y) satisfies Lipschitz condition with respect to the second variable y. However, choosing y(x)=1/(x-a)γ, we can not arrive at (x-a)γf(x,y(x))=sin(1/(x-a))∈C[a,b]. Hence the condition of f in [28, Theorem 3.11, page 165] is not proper.The next example illustrates that there also exists a problem in [28, Lemma 3.9, page 202].Example 4.2.
Consider the function(4.2)ϕ(x)=lnx,x∈[a,b],1<a<b.
It is evident thatϕ(x) belongs to the space C1[a,b] = {ψ(x):ψ(1)(x)∈C[a,b], ∥ψ∥C1[a,b] = ∥ψ∥C[a,b]+∥ψ(1)∥C[a,b]}.
Settingc∈(a,b), that is, a<c<b, then
(4.3)ϕ(x)∈C1[a,b],ϕ(x)∈C1[a,c],ϕ(x)∈C1[c,b].
We could not conclude that
(4.4)∥ϕ∥C1[a,b]≤max{∥ϕ∥C1[a,c],∥ϕ∥C1[c,b]},
because
(4.5)∥ϕ∥C1[a,b]=lnb+1a,∥ϕ∥C1[a,c]=lnc+1a,∥ϕ∥C1[c,b]=lnb+1c.
However, we have
(4.6)∥ϕ∥C1[a,b]≤∥ϕ∥C1[a,c]+∥ϕ∥C1[c,b].The following example is for [28, Lemma 3.10, page 213].Example 4.3.
Let one consider the function(4.7)h(x)={μ(lnx/a)γx∈(a,c],μ(lnx/c)γx∈(c,b],
where μ≠0 is a constant number and γ∈(0,∞).The same problem exists in [28, Lemma 3.10, page 213]. From the definition of h(x), we have h(x)∈Cγ,ln[a,c] and h(x)∈Cγ,ln[c,b]. However, the conclusion that  h(x)∈Cγ,ln[a,b] and ∥h∥Cγ[a,b]≤max{∥h∥Cγ[a,c],∥h∥Cγ[c,b]} is still not correct. This defect means that [28, Lemma 3.10] could not be applied to prove [28, Theorem 3.29].In a sense, our lemmas and main results have remedied these defects.


## 5. Conclusion

In this paper, we first get several useful lemmas, especially Lemmas3.1 and 3.4, which have improved the corresponding lemmas in [28]. By modifying the conditions on f and improving the method used in [28], we have established the results of existence and uniqueness of solution for the cauchy-type problems involving the Riemann-Liouville fractional derivative and the Hadamard fractional derivative in the weight space of continuous functions. Meanwhile, we have given some counterexamples to prove that [28, Lemmas 3.4, 3.9, and 3.10, pages 165, 203, and 213] are not complete, which means that there exist some defects in the proofs of the related results in [28].



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