Citation: Bashir Ahmad, Madeaha Alghanmi, Sotiris K. Ntouyas, Ahmed Alsaedi. A study of fractional differential equations and inclusions involving generalized Caputo-type derivative equipped with generalized fractional integral boundary conditions[J]. AIMS Mathematics, 2019, 4(1): 26-42. doi: 10.3934/Math.2019.1.26
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We investigate a new class of boundary value problems for generalized Caputo-type fractional differential equations and inclusions supplemented with Katugampola type generalized fractional integral boundary conditions. Precisely, we study the following problems:
{ρcDα0+y(t)=f(t,y(t)), t∈J:=[0,T],y(T)=m∑i=1σiρIβ0+y(ηi)+κ, δy(0)=0, ηi∈(0,T), | (1.1) |
and
{ρcDα0+y(t)∈F(t,y(t)), t∈J:=[0,T],y(T)=m∑i=1σiρIβ0+y(ηi)+κ, δy(0)=0, ηi∈(0,T), | (1.2) |
where ρcDα0+ denotes the generalized Caputo-type fractional derivative of order 1<α≤2,ρ>0, ρIβ0+ is the Katugampola type fractional integral of order β>0,ρ>0, f:J×R→R is a continuous function, σi∈R,i=1,2,…,m,κ∈R, δ=t1−ρddt, and F:J×R→P(R) is a multivalued function (P(R) is the family of all nonempty subjects of R).
Here we emphasize that the problem considered in the present paper is motivated by Laskin's work [1] on the generalization of the Feynman and Wiener path integrals in the context of fractional quantum mechanics and fractional statistical mechanics. One can find more details in the articles [2, 3]. It is expected that the results obtained in this paper will provide more leverage in dealing with Feynman and Wiener path type integrals involving an index like ρ>0 in (1.1), instead of a fixed choice ρ=1 (Caputo fractional derivative case). Moreover, chaos for a fractional order differential equation involving two parameters (α and ρ>0) becomes more complicated than the one containing Caputo fractional derivative of order α (generalized Caputo-type fractional derivative with ρ=1); one can find more details in [4, 5]. It is worthwhile to notice that Katugampola fractional integral unifies the Riemann-Liouville and Hadamard integrals into a single integral [6]. Thus, our results are more general in the context of integral boundary conditions.
The topic of fractional-order differential equations and inclusions attracted significant attention in recent years and several results on fractional differential equations involving Riemann-Liouville, Caputo, Hadamard type derivatives, supplemented with a variety of boundary conditions, can be found in the related literature [7, 8, 9]. The interest in the subject owes to its extensive applications in various disciplines of science and engineering, for instance, see the papers [10, 11, 12, 13, 14, 15, 16, 17, 18], and the references cited therein. In a recent paper [19], the authors studied fractional differential equations involving Caputo-Katugampola derivative. In a more recent work [20], the authors studied a fractional order boundary value problem involving Katugampola-type generalized fractional derivative and generalized fractional integral.
We organize the rest of the paper as follows. Section 2 contains preliminary material related to our work. The existence and uniqueness results for the problem (1.1), obtained with the aid of the standard fixed point theorem, are presented in Section 3. The existence results for the inclusions problem (1.2) are derived in Section 4. Examples are provided to demonstrate the application of the main theorems.
For c∈R,1≤p≤∞, let Xpc(a,b) denote the space of all complex-valued Lebesgue measurable functions ϕ on (a,b) endowed with the norm:
‖ϕ‖Xpc=(∫ba|xcϕ(x)|pdxx)1/p<∞. |
We denote by L1(a,b) the space of all Lebesgue measurable functions φ on (a,b) equipped with the norm:
‖φ‖L1=∫ba|φ(x)|dx<∞. |
Let G=C(J,R) denote the Banach space of all continuous functions from [0,T] to R endowed with the norm defined by ‖y‖=supt∈[0,T]|y(t)|.
Recall that
ACn(J,R)={h:J→R:h,h′,…,h(n−1)∈C(J,R) and h(n−1)is absolutely continuous}. |
For 0≤ϵ<1, we define Cϵ,ρ(J,R)={f:J→R:(tρ−aρ)ϵf(t)∈C(J,R)} equipped with the norm ‖f‖Cϵ,ρ=‖(tρ−aρ)ϵf(t)‖C. Moreover, let us introduce ACnδ(J), which consists of the functions f that have absolutely continuous δn−1-derivative, where δ=t1−ρddt. Thus we define spaces ACnδ(J,R)={f:J→R:δn−1f∈AC(J,R), δ=t1−ρddt}, and Cnδ,ϵ(J,R)={f:J→R:δn−1f∈C(J,R),δnf∈Cϵ,ρ(J,R),δ=t1−ρddt} endowed with the norms ‖f‖Cnδ=∑n−1k=0‖δkf‖C and ‖f‖Cnδ,ϵ=∑n−1k=0‖δkf‖C+‖δnf‖Cϵ,ρ respectively. Here we use the convention Cnδ,0=Cnδ.
Definition 2.1. [6] The generalized fractional integral of order α>0 and ρ>0 of f∈Xpc(a,b) for −∞<a<t<b<∞, is defined by
(ρIαa+f)(t)=ρ1−αΓ(α)∫tasρ−1(tρ−sρ)1−αf(s)ds. | (2.1) |
Note that the integral in (2.1) is called the left-sided fractional integral. Similarly we can define right-sided fractional integral ρIαb−f as
(ρIαb−f)(t)=ρ1−αΓ(α)∫btsρ−1(sρ−tρ)1−αf(s)ds. | (2.2) |
Definition 2.2. [21] The generalized fractional derivative, associated with the generalized fractional integrals (2.1) and (2.2) for 0≤a<t<b<∞, are defined by
(ρDαa+f)(t)=(t1−ρddt)n(ρIn−αa+f)(t)=ρα−n+1Γ(n−α)(t1−ρddt)n∫tasρ−1(tρ−sρ)α−n+1f(s)ds | (2.3) |
and
(ρDαb−f)(t)=(−t1−ρddt)n(ρIn−αb−f)(t)=ρα−n+1Γ(n−α)(−t1−ρddt)n∫btsρ−1(sρ−tρ)α−n+1f(s)ds, | (2.4) |
if the integrals in (2.3) and (2.4) exist.
Definition 2.3. [22] For α≥0 and f∈ACnδ[a,b], the generalized Caputo-type fractional derivatives ρcDαa and ρcDαb are defined via the above generalized fractional derivatives as follows
ρcDαa+f(x)=ρDαa+[f(t)−n−1∑k=0δkf(a)k!(tρ−aρρ)k](x), δ=x1−ρddx, | (2.5) |
ρcDαb−f(x)=ρDαb−[f(t)−n−1∑k=0(−1)kδkf(b)k!(bρ−tρρ)k](x), δ=x1−ρddx, | (2.6) |
where n=[α]+1.
Lemma 2.1. [22] Let α≥0,n=[α]+1 and f∈ACnδ[a,b], where 0<a<b<∞. Then,
(1) for α∉N,
ρcDαa+f(t)=1Γ(n−α)∫ta(tρ−sρρ)n−α−1(δnf)(s)dss1−ρ=ρIn−αa+(δnf)(t), | (2.7) |
ρcDαb−f(t)=1Γ(n−α)∫bt(sρ−tρρ)n−α−1(−1)n(δnf)(s)dss1−ρ=ρIn−αb−(δnf)(t); | (2.8) |
(2) for α∈N,
ρcDαa+f=δnf, ρcDαb−f=(−1)nδnf. | (2.9) |
Lemma 2.2. [22] Let f∈ACnδ[a,b] or Cnδ[a,b] and α∈R. Then
ρIαa+ρcDαa+f(x)=f(x)−n−1∑k=0(δkf)(a)k!(xρ−aρρ)k, |
ρIαb−ρcDαb−f(x)=f(x)−n−1∑k=0(−1)k(δkf)(a)k!(bρ−xρρ)k. |
In particular, for 0<α≤1, we have
ρIαa+ρcDαa+f(x)=f(x)−f(a), |
ρIαb−ρcDαb−f(x)=f(x)−f(b). |
Next we define a solution for the problem (1.1).
Definition 2.4. A function y∈AC2δ([0,T],R) is said to be a solution of (1.1) if y satisfies the equation ρcDαy(t)=f(t,y(t)) on [0,T], and the conditions y(T)=∑mi=1σi ρIβy(ηi)+κ, δy(0)=0.
Relative to the problem (1.1), we consider the following lemma.
Lemma 2.3. Let h∈C(0,T)∩L1(0,T), y∈AC2δ(J) and
Ω=1−m∑i=1σiηρβiρβΓ(β+1)≠0. | (2.10) |
Then the integral solution of the linear boundary value problem (BVP):
{ρcDα0+y(t)=h(t), t∈J:=[0,T],y(T)=m∑i=1σiρIβ0+y(ηi)+κ, δy(0)=0, ηi∈(0,T), | (2.11) |
is given by
y(t)=ρIα0+h(t)+1Ω{−ρIα0+h(T)+m∑i=1σiρIα+β0+h(ηi)+κ}. | (2.12) |
Proof. Applying ρIα0+ on both sides of the fractional differential equation in (2.11) and using Lemma 2.2, we get
y(t)=ρIα0+h(t)+c1+c2tρρ=ρ1−αΓ(α)∫t0sρ−1(tρ−sρ)α−1h(s)ds+c1+c2tρρ, | (2.13) |
for some c1,c2∈R. Taking δ-derivative of (2.13), we get
δy(t)=ρIα−10+h(t)+c2=ρ2−αΓ(α−1)∫t0sρ−1(tρ−sρ)α−2h(s)ds+c2. | (2.14) |
Using the boundary condition δy(0)=0 in (2.14), we get c2=0. Applying the generalized integral ρIβ0+ to (2.13) after inserting the value of c2 in it, we get
ρIβ0+y(t)=ρIα+β0+h(t)+c1tρβρβΓ(β+1). | (2.15) |
Making use of the first boundary condition y(T)=∑mi=1σi ρIβy(ηi)+κ in (2.15), we get
ρIα0+h(T)+c1=m∑i=1σi ρIα+β0+h(ηi)+m∑i=1σic1ηρβiρβΓ(β+1)+κ, |
which, on solving for c1 together with (2.10), yields
c1=1Ω{−ρIα0+h(T)+m∑i=1σi ρIα+β0+h(ηi)+κ}. |
Substituting the values of c1 and c2 in (2.13), we obtain the solution (2.12). The converse follows by direct computation. The proof is completed.
Using Lemma 2.3, we define an operator N:G→G by
Ny(t)=ρIα0+f(t,y(t))+1Ω{−ρIα0+f(T,y(T))+m∑i=1σiρIα+β0+f(ηi,y(ηi))+κ}. | (3.1) |
In the following, for brevity, we set the notation:
Λ=TραραΓ(α+1)+1|Ω|{TραραΓ(α+1)+m∑i=1|σi|ηρ(α+β)iρα+βΓ(α+β+1)}. | (3.2) |
Our first existence result for the problem (1.1) relies on Leray-Schauder nonlinear alternative [23].
Theorem 3.1. Assume that
(A1) |f(t,y)|≤p(t)ψ(‖y‖),∀(t,y)∈[0,T]×R, where p∈L1([0,T],R+) and ψ:R+→ R+ is a nondecreasing function;
(A2) we can find a positive constant W satisfying the inequality:
Wψ(W)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|})>1. |
Then there exists at least one solution for the problem (1.1) on [0,T].
Proof. Consider the operator N:G→G defined by (3.1) and show that it is continuous and completely continuous. We establish in four steps.
(ⅰ) N is continuous. Let {yn} be a sequence such that yn→y in G. Then
|N(yn)(t)−N(y)(t)|≤ρIα0+|f(t,yn(t))−f(t,y(t))|+1Ω{ρIα0+|f(T,yn(T))−f(T,y(T))|+m∑i=1|σi| ρIα+β0+|f(ηi,yn(ηi))−f(ηi,y(ηi))|}≤Λ‖f(⋅,yn)−f(⋅,y)‖. |
Since f is a continuous function, therefore, we have
‖N(yn)−N(y)‖≤Λ‖f(⋅,yn)−f(⋅,y)‖→0,asn→∞. |
(ⅱ) The operator N maps bounded sets into bounded sets in G.
For any ˉr>0, it is indeed enough to show that there exists a positive constant ℓ such that ‖N(y)‖≤ℓ for y∈Bˉr={y∈G:‖y‖≤ˉr}. By the assumption (A1), for each t∈J, we have
|N(y)(t)|≤ρIα0+|f(t,y(t))|+1|Ω|{ρIα0+|f(T,y(T))|+m∑i=1|σi| ρIα+β0+|f(ηi,y(ηi))|+|κ|}≤ρIα0+p(T)Ω(‖y‖)+1|Ω|{ρIα0+p(T)ψ(‖y‖)+m∑i=1|σi| ρIα+β0+p(ηi)ψ(‖y‖)+|κ|}≤ψ(‖y‖)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|}). |
Thus
‖N(y)‖≤ψ(ˉr)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|}):=ℓ. |
(ⅲ) N maps bounded sets into equicontinuous sets of G.
Let t1,t2∈(0,T],t1<t2, Bˉr be a bounded set of G as in (ii) and let y∈Bˉr. Then
|N(y)(t2)−N(y)(t1)|≤|ρIα0+f(t2,y(t2))−ρIα0+f(t1,y(t1))|≤ρ1−αψ(ˉr)Γ(α)|∫t10[sρ−1(tρ2−sρ)1−α−sρ−1(tρ1−sρ)1−α]p(s)ds+∫t2t1sρ−1(tρ2−sρ)1−αp(s)ds|→0 as t1⟶t2, independent ~of y. |
From the steps (i)−(iii), we deduce by the Arzelá-Ascoli theorem that N:G⟶G is completely continuous.
(ⅳ) There exists an open set V⊆G with y≠νN(y) for ν∈(0,1) and y∈∂V.
Let y∈G be a solution of y−νNy=0 for ν∈[0,1]. Then, for t∈[0,T], we obtain
|y(t)|=|ν(Ny)(t)|≤ρIα0+|f(t,y(t))|+1|Ω|{ρIα0+|f(T,y(T))|+m∑i=1|σi| ρIα+β0+|f(ηi,y(ηi))|+|κ|}≤ρIα0+p(T)ψ(‖y‖)+1|Ω|{ρIα0+p(T)ψ(‖y‖)+m∑i=1|σi| ρIα+β0+p(ηi)ψ(‖y‖)+|κ|}≤ψ(‖y‖)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|}), |
which, on taking the norm for t∈[0,T], implies that
‖y‖ψ(‖y‖)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|})≤1. |
By the assumption (A2), there exists a positive constant W such that ‖y‖≠W. Next we define V={y∈G:‖y‖<W} and note that the operator N:¯V→G is continuous and completely continuous. By the choice of V, there does not exist any y∈∂V satisfying y=νN(y) for some ν∈(0,1). In consequence, by the nonlinear alternative of Leray-Schauder type [23], we deduce that there exists a fixed point y∈¯V for the operator N, which is a solution of the problem (1.1).
In our next result, we make use of Banach contraction mapping principle to establish the uniqueness of solutions for the problem (1.1).
Theorem 3.2. Suppose that
(A3) there exists a nonnegative constant L such that
|f(t,u)−f(t,v)|≤L‖u−v‖,fort∈[0,T]and everyu,v∈R. |
Then the problem (1.1) has a unique solution on [0,T] if
LΛ<1, | (3.3) |
where Λ is defined by (3.2).
Proof. Consider the operator N:G→G associated with the problem (1.1) defined by (3.1). With Λ given by (3.2), we fix
r≥Λf0+|κ|/|Ω|1−LΛ,f0=supt∈[0,T]|f(t,0)|, |
and show that FBr⊂Br, where Br={y∈G:‖y‖≤r}. For y∈Br, using (A3), we get
|N(y)(t)|≤ρIα0+[|f(t,y(t))−f(t,0)|+|f(t,0)|]+1|Ω|{ρIα0+[|f(T,y(T))−f(T,0)|+|f(T,0)|]+m∑i=1|σi| ρIα+β0+[|f(ηi,y(ηi))−f(ηi,0)|+|f(ηi,0)|]+|κ|}≤(L‖y‖+f0)[TραραΓ(α+1)+1|Ω|{TραραΓ(α+1)+m∑i=1|σi|ηρ(α+β)iρα+βΓ(α+β+1)}]+|κ||Ω|≤(Lr+f0)Λ+|κ||Ω|≤r, |
which, on taking the norm for t∈[0,T], yields ‖N(y)‖≤r. This shows that N maps Br into itself. Now we show that the operator N is a contraction. Let y,u∈G. Then we get
|N(y)(t)−N(u)(t)|≤ρIα0+|f(t,y(t))−f(t,u(t))|+1|Ω|{ρIα0+|f(T,y(T))−f(T,u(T))|+m∑i=1|σi| ρIα+β0+|f(ηi,y(ηi))−f(ηi,u(ηi))|}≤LΛ‖y−u‖. |
Consequently we obtain ‖N(y)−N(u)‖≤LΛ‖y−u‖, which shows that N is a contraction by means of (3.3). Thus the contraction mapping principle applies and the operator N has a unique fixed point. This shows that there exists a unique solution for the problem (1.1) on [0,T].
Now we prove the uniqueness of solutions for the problem (1.1) by applying a fixed point theorem for nonlinear contractions due to Boyd and Wong [24].
Definition 3.1. A mapping H:E→E is called a nonlinear contraction if we can find a continuous nondecreasing function ϕ:R+→R+ such that ϕ(0)=0, ϕ(ξ)<ξ for all ξ>0 and ‖Hy−Hu‖≤ϕ(‖y−u‖),∀y,u∈E (E is a Banach space).
Lemma 3.1. (Boyd and Wong) [24] Let E be a Banach space and let N:E→E be a nonlinear contraction. Then N has a unique fixed point in E.
Theorem 3.3. Assume that
(A4) |f(t,y)−f(t,u)|≤g(t)|y−u|G∗+|y−u|,t∈[0,T],y,u≥0, where g:[0,T]→R+ is continuous and
G∗=ρIα0+g(T)+1|Ω|{ρIα0+g(T)+m∑i=1|σi|ρIα+β0+g(ηi)}. | (3.4) |
Then the problem (1.1) has a unique solution on [0,T].
Proof. Let Ψ:R+→R+ be a continuous nondecreasing function such that Ψ(0)=0 and Ψ(ξ)<ξ for all ξ>0, defined by
Ψ(ξ)=G∗ξG∗+ξ,∀ξ≥0. |
Let y,u∈G. Then
|f(s,y(s))−f(s,u(s))|≤g(s)G∗Ψ(‖y−u‖), |
so that
|N(y)(t)−N(u)(t)|≤ρIα0+(g(t)|y(t)−u(t)|G∗+|y(t)−u(t)|)+1|Ω|{ρIα0+(g(T)|y(T)−u(T)|G∗+|y(T)−u(T)|)+m∑i=1|σi|ρIα+β0+(g(ηi)|y(ηi)−u(ηi)|G∗+|y(ηi)−u(ηi)|)}≤|y(t)−u(t)|G∗+|y(t)−u(t)|{ρIα0+g(T)+1|Ω|{ρIα0+g(T)+m∑i=1|σi|ρIα+β0+g(ηi)}}, |
for t∈[0,T]. By the condition (3.4), we deduce that ‖N(y)−N(u)‖≤Ψ(‖y−u‖) and hence N is a nonlinear contraction. Thus it follows from the fixed point theorem due to Boyd and Wong [24] that the operator N has a unique fixed point in G, which is indeed a unique solution of problem (1.1).
Example 3.1. Let us consider the following boundary value problem
{1/3 cD5/40+y=f(t,y), t∈[0,2],y(2)=21/3I3/4y(1/2)+1/2 1/3I3/4y(3/2)+1/4, δy(0)=0, | (3.5) |
where ρ=1/3,α=5/4,σ1=2,σ2=1/2,β=3/4,η1=1/2,η2=3/2,κ=1/4, T=2 and f(t,y(t)) will be fixed later.
Using the given data, we find that |Ω|=4.543695998 and Λ=7.572001575, where Ω and Λ are given by (2.10) and (3.2) respectively.
For illustrating Theorem 3.1, we take
f(t,y)=(1+t)30(|y||y|+1+y+18), | (3.6) |
and find that p(t)=(1+t)30 and ψ(‖y‖)=||y||+98. By condition (A2), we have W>0.7066246467. Thus, the hypothesis of Theorem 3.1 holds true, which implies that the problem (3.5) has at least one solution.
Furthermore, for the uniqueness results, Theorem 3.2 can be illustrated by choosing
f(t,y)=tan−1y+e−t2√81+sint. | (3.7) |
Clearly the condition (A3) is satisfied with L=1/18. Also
LΛ≈0.4206667542<1. |
Obviously all the conditions of Theorem 3.2 hold and consequently the problem (3.5) with f(t,y) given by (3.7) has a unique solution on [0,2] by the conclusion of Theorem 3.2.
Finally, for illustrating Theorem 3.3, we take
f(t,y)=t(|y||y|+11+18). | (3.8) |
Here we choose g(t)=(1+t) and find that
G∗=ρIα0+g(T)+1|Ω|{ρIα0+g(T)+m∑i=1|σi|ρIα+β0+g(ηi)}≈9.923097014, |
and
|f(t,y)−f(t,u)|=t(|y|−|u|11+|y|+|u|+|y||u|11)≤(1+t)|y−u|9.923097014+|y−u|. |
So, the conclusion of Theorem 3.3 applies to the problem (3.5) with f(t,y) given by (3.8).
In this section, we present existence results for the problem (1.2).
Definition 4.1. A function y∈AC2δ([0,T],R) is called a solution of the problem (1.2) if y(T)=∑mi=1σi ρIβ0+y(ηi)+κ,δy(0)=0 and there exists function v∈L1([0,T],R) such that v(t)∈F(t,y(t)) a.e. on [0,T] and
y(t)=ρIα0+v(t)+1Ω{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ}. |
Here we prove an existence result for the problem (1.2) by applying nonlinear alternative for Kakutani maps [23] when F has convex values and is of Carathéodory type.
Theorem 4.1. Assume that
(B1) F:[0,T]×R→Pcp,c(R) is L1-Carathéodory, where Pcp,c(R) ={Y∈P(R): Y is compact and convex};
(B2) there exist a continuous nondecreasing function φ:[0,∞)→(0,∞) and a function p∈L1([0,T],R+) such that
‖F(t,y)‖P:=sup{|x|:x∈F(t,y)}≤p(t)φ(‖y‖) for each (t,y)∈[0,T]×R; |
(B3) there exists a constant ˆW>0 satisfying
ˆWφ(ˆW)(ρIα0+p(T)+1Ω( ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+κ))>1. |
Then there exists at least one solution for the problem (1.2) on [0,T].
Proof. Define an operator M:C([0,T],R)⟶P(C([0,T],R)) by
M(y)={h∈C([0,T],R):h(t)=F(y)(t)}, | (4.1) |
where
F(y)(t)=ρIα0+v(t)+1Ω{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ}, |
for v∈SF,y. Here SF,y denotes the set of selections of F and is defined by
SF,y:={v∈L1([0,T],R):v(t)∈F(t,y(t)) a.e. on [0,T]}, |
for each y∈C([0,T],R). Notice that the fixed points of the operator M are solutions of the problem (1.2).
To show that M satisfies the assumptions of Leray-Schauder nonlinear alternative [23], we split the proof in several steps.
Step 1. M(y) is convex for each y∈C([0,T],R) as SF,y is convex (F has convex values).
Step 2. Let Br={y∈C([0,T],R):‖y‖≤r} be a bounded ball in C([0,T],R), where r is a positive number. Then, for each h∈M(y),y∈Br, there exists v∈SF,y such that
h(t)=ρIα0+v(t)+1Ω{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ} |
with
‖h‖≤φ(r)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|}):=ℓ1. |
This shows that M maps bounded sets (balls) into bounded sets in C([0,T],R).
Step 3. In order to show that M maps bounded sets into equicontinuous sets of C([0,T],R), we take t1,t2∈(0,T],t1<t2, and y∈Br. Then we find that
|h(t2)−h(t1)|≤ρ1−αφ(r)Γ(α)|∫t10[sρ−1(tρ2−sρ)1−α−sρ−1(tρ1−sρ)1−α]p(s)ds+∫t2t1sρ−1(tρ−sρ)1−αp(s)ds|, |
which tends to zero independently of y∈Br as t2−t1→0. In view of the foregoing steps, it follows by the Arzelá-Ascoli theorem that M:C([0,T],R)→P(C([0,T],R)) is completely continuous.
Step 4. In our next step, we show that M is upper semi-continuous (u.s.c.). Since M is completely continuous, it is enough to establish that it has a closed graph (see [25, Proposition 1.2]). For that, let yn→y∗,hn∈M(yn) and hn→h∗. Then we have to show that h∗∈M(y∗). Associated with hn∈M(yn), we can find vn∈SF,yn such that for each t∈[0,T],
hn(t)=ρIα0+vn(t)+1Ω{−ρIα0+vn(s)vn(T)+m∑i=1σi ρIα+β0+vn(ηi)+κ}. |
Next, for each t∈[0,T], we establish that there exists v∗∈SF,y∗ satisfying
h∗(t)=ρIα0+v∗(t)+1Ω{−ρIα0+v∗(T)+m∑i=1σi ρIα+β0+v∗(ηi)+κ}. |
Consider the linear operator Θ:L1([0,T],R)→C([0,T],R) given by
v↦Θv(t)=ρIα0+v(t)+1Ω{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ}. |
Notice that ‖hn(t)−h∗(t)‖→0 as n→∞. Thus we deduce by the closed graph theorem [26] that Θ∘SF is a closed graph operator. Furthermore, we have hn(t)∈Θ(SF,yn). As yn→y∗, we have
h∗(t)=ρIα0+v∗(t)+1Ω{−ρIα0+v∗(T)+m∑i=1σi ρIα+β0+v∗(ξ)+κ},for ~some v∗∈SF,y∗. |
Step 5. Finally, we show the existence of an open set U⊆C([0,T],R) such that y∉λM(y) for any λ∈(0,1) and all y∈∂U. For that we take λ∈(0,1) and y∈λM(y). Then there exists v∈L1([0,T],R) with v∈SF,y such that, for t∈[0,T], we have
y(t)=λρIα0+v(t)+λΩ{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ}. |
As in the second step, one can obtain
|y(t)|≤ρIα0+|v(T)|+1|Ω|{ρIα0+|v(T)|+m∑i=1|σi| ρIα+β0+|v(ηi)|+|κ|}≤φ(‖y‖)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|}), |
which implies that
‖y‖φ(‖y‖)(ρIα0+p(T)+1|Ω|{ρIα0+p(T)+m∑i=1|σi| ρIα+β0+p(ηi)+|κ|})≤1. |
By the hypothesis (B3), we can find ˆW such that ‖y‖≠ˆW. Setting Y={y∈C(J,R):‖y‖<ˆW}, we notice that the operator M:¯Y→P(C(J,R)) is compact multi-valued, u.s.c. with convex closed values. From the choice of Y, there does not exist any y∈∂Y satisfying y∈λM(y) for some λ∈(0,1). In consequence, we deduce by the nonlinear alternative of Leray-Schauder type [23] that M has a fixed point y∈¯Y which is a solution of the problem (1.2). This completes the proof.
Consider a mapping Hd:P(X)×P(X)→R∪{∞} defined by
Hd(S,V)=max{sups∈Sd(a,V),supv∈Vd(S,v)}, |
where d(S,v)=infs∈Sd(s;v), d(s,V)=infv∈Vd(s;v) and (X,d) is a metric space induced from the normed space (X;‖⋅‖). Note that (Pcl,b(X),Hd) is a metric space (see [27]), where Pcl,b(X)={Y∈P(X):Y is closed and bounded}.
The following result, dealing with the existence of solutions for the problem (1.2) with nonconvex valued right hand side of the inclusion, relies on Covitz and Nadler's fixed point theorem for multivalued maps [28].
Theorem 4.2. Assume that
(C1) F:[0,T]×R→Pcp(R) is such that F(⋅,y):[0,T]→Pcp(R) is measurable for each y∈R, where Pcp(R)={Y∈P(R):Y is compact};
(C2) Hd(F(t,y),F(t,ˉy))≤μ(t)|y−ˉy| for almost all t∈[0,T] and y,ˉy∈R with μ∈C([0,T],R+) and d(0,F(t,0))≤μ(t) for almost all t∈[0,T].
Then the problem (1.2) has at least one solution on [0,T] provided that
ϑ=‖μ‖Λ<1, | (4.2) |
where Λ is given by (3.2).
Proof. By the assumption (C1), the set SF,y is nonempty for each y∈C([0,T],R) and F has a measurable selection by Theorem Ⅲ.6 in [29]. Now we proceed to show that the operator M:(C[0,T],R)→Pcl(C([0,T],R)) (Pcl(C([0,T],R))={Y∈P(C([0,T],R)):Y is closed}) is a contraction so that Covitz and Nadler's Theorem [28] is applicable.
In the first step, we show that M(y)∈Pcl((C[0,T],R)) for each y∈C([0,T],R). Let {un}n≥0∈M(y) with un→u (n→∞) in C([0,T],R). Then u∈C([0,T],R) and there exists vn∈SF,yn satisfying
un(t)=ρIα0+vn(t)+1Ω{−ρIα0+vn(T)+m∑i=1σi ρIα+β0+vn(ηi)+κ}for each t∈[0,T]. |
In view of the compact values of F, we pass onto a subsequence (if necessary) to find that vn converges to v in L1([0,T],R). For v∈SF,y and for each t∈[0,T], we have
un(t)→u(t)=ρIα0+v(t)+1Ω{−ρIα0+v(T)+m∑i=1σi ρIα+β0+v(ηi)+κ}. |
Thus u∈M(y).
Now, for each y,ˉy∈C([0,T],R), we establish that there exists ϑ<1 (defined by (4.2)) satisfying
Hd(M(y),M(ˉy))≤ϑ‖y−ˉy‖. |
Let y,ˉy∈C([0,T],R) and h1∈M(y). Then there exists v1(t)∈F(t,y(t)) satisfying
h1(t)=ρIα0+v1(t)+1Ω{−ρIα0+v1(T)+m∑i=1σi ρIα+β0+v1(ηi)+κ}, |
for each t∈[0,T]. By (C2), we have
Hd(F(t,y),F(t,ˉy))≤μ(t)|y(t)−ˉy(t)|. |
Therefore, we can find w∈F(t,ˉy(t)) satisfying
|v1(t)−w|≤μ(t)|y(t)−ˉy(t)|, t∈[0,T]. |
Introduce U:[0,T]→P(R) by
U(t)={w∈R:|v1(t)−w|≤μ(t)|y(t)−ˉy(t)|}. |
As U(t)∩F(t,ˉy(t)) is measurable (Proposition Ⅲ.4 [29]), we can find a measurable selection v2(t) for U such that v2(t)∈F(t,ˉy(t)) satisfying |v1(t)−v2(t)|≤μ(t)|y(t)−ˉy(t)| for each t∈[0,T].
Define
h2(t)=ρIα0+v2(t)+1Ω{−ρIα0+v2(T)+m∑i=1σi ρIα+β0+v2(ηi)+κ}, |
for each t∈[0,T]. Then
|h1(t)−h2(t)|≤ρIα0+|v1(t)−v2(t)|+1|Ω|{ρIα0+|v1(T)−v2(T)|+m∑i=1|σi| ρIα+β0+|v1(ηi)−v2(ηi)|}≤‖μ‖[TραραΓ(α+1)+1|Ω|{TραραΓ(α+1)+m∑i=1|σi|ηρ(α+β)iρα+βΓ(α+β+1)}]‖y−ˉy‖. |
Hence
‖h1−h2‖≤‖μ‖[TραραΓ(α+1)+1|Ω|{TραραΓ(α+1)+m∑i=1|σi|ηρ(α+β)iρα+βΓ(α+β+1)}]‖y−ˉy‖. |
Analogously, switching the roles of y and ¯y, we can obtain
Hd(M(y),M(ˉy))≤‖μ‖[TραραΓ(α+1)+1|Ω|{TραραΓ(α+1)+m∑i=1|σi|ηρ(α+β)iρα+βΓ(α+β+1)}]‖y−ˉy‖. |
So M is a contraction. Thus, by Covitz and Nadler's fixed point theorem [28], the operator M has a fixed point y, which corresponds to a solution of (1.2).
Example 4.1. Consider the following boundary value problem
{1/3 cD5/40+y∈F(t,y), t∈[0,2],y(2)=21/3I3/4y(1/2)+1/2 1/3I3/4y(3/2)+1/4, δy(0)=0, | (4.3) |
where F(t,y) will be fixed later.
For illustrating Theorem 4.1, we take
F(t,y)=[e−t√900+t(siny+12) ,(1+t)30(|y||y|+1+y+18)]. | (4.4) |
Using the given data, we find p(t)=(1+t)30,φ(‖y‖)=||y||+98, and by condition (B3), we have ˆW>0.7066246467. Thus all conditions of Theorem 4.1 are satisfied and consequently, there exists at least one solution for the problem (4.3) with F(t,y) given by (4.4) on [0,2].
In order to demonstrate the application of Theorem 4.2, let us choose
F(t,y)=[e−t√900+t(tan−1y+12) ,(1+t)30(|y||y|+1+18)]. | (4.5) |
Clearly
Hd(F(t,y),F(t,ˉy))≤(t+1)30‖y−ˉy‖. |
Letting μ(t)=(t+1)30, it is easy to check that d(0,F(t,0))≤μ(t) holds for almost all t∈[0,2] and ϑ≈0.7572001575<1 (ϑ is given by 4.2). As the hypotheses of Theorem 4.2 are satisfied, we conclude that the problem (4.3) with F(t,y) given by (4.5) has at least one solution on [0,2].
We have developed the existence theory for fractional differential equations and inclusions involving Caputo-type generalized fractional derivative equipped with generalized fractional integral boundary conditions (in the sense of Katugampola). Standard fixed point theorems for single-valued and multi-valued maps are employed to obtain the desired results, which are well illustrated with the aid of examples. Our results are new in the given configuration and contribute significantly to the existing literature on the topic.
The authors gratefully acknowledge the referees for their useful comments on their paper.
The authors declare that they have no conflict of interests.
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