# Existence and Stability of Implicit Fractional Differential Equations with Stieltjes Boundary Conditions Involving Hadamard Derivatives

Authors: Danfeng Luo; Mehboob Alam; Akbar Zada; Usman Riaz; Zhiguo Luo
Journal: Complexity
 (2021)

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## Abstract

In this article, we make analysis of the implicit fractional differential equations involving integral boundary conditions associated with Stieltjes integral and its corresponding coupled system. We use some sufficient conditions to achieve the existence and uniqueness results for the given problems by applying the Banach contraction principle, Schaefer’s fixed point theorem, and Leray–Schauder result of the cone type. Moreover, we present different kinds of stability such as Hyers–Ulam stability, generalized Hyers–Ulam stability, Hyers–Ulam–Rassias stability, and generalized Hyers–Ulam–Rassias stability by using the classical technique of functional analysis. At the end, the results are verified with the help of examples.


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## Body

## 1. Introduction

Fractional-order derivatives are the generalization of integer-order derivatives. The idea of fractional-order derivatives has been introduced at the end of the sixteen century when Leibniz used the notationdα/dσα for αth-order derivative. It is well known that if ϕσ=1/2σ2, then ϕ′σ=σ and ϕ″σ=1, respectively. But, if the order is in the fractional form, i.e., α=1/2,α=1/3, and so on, then what would be the result? This question was mentioned in a letter to L’Hospital by Leibniz in (1695) [1]. Since then, several mathematicians such as Fourier, Laplace, and Letnikov contributed to the development of the fractional calculus. Riemann and Liouville had worked on this problem and introduced the Riemann–Liouville fractional derivative, which was further generalized by Caputo and can be considered as the fundamental concept in fractional calculus. Fractional calculus has many applications in many scientific disciplines, e.g., in the fields of signal and image processing [2], mathematical biological systems [3], electronics, economics, control theory [4], chemistry, biophysics, and blood flow phenomena. For more applications of fractional differential equations, we refer the reader to [5, 6] and the references cited therein.In 1892 [7], Hadamard presented a concept of fractional derivative, which is different from Caputo- and Riemann–Liouville-type fractional derivatives. An important feature of the Hadamard fractional derivative is that it contains a logarithmic function of an arbitrary exponent in its definition. Here, we stress that the studies about Hadamard fractional differential equations are still at the early stage and need additional analysis. For more details and recent contributions to the topic, see [8–11].In the field of fractional differential equations, the area which received great attention from researchers is the existence of solutions of different boundary-value problems. Many researchers used different fixed-point theorems and developed different approaches for the existence of solutions of complicated boundary-value problems. For details, we refer the reader to [12–14]. The study of coupled systems of differential equations is also very significant because these types of systems appear naturally in many problems of applied nature. For details and examples, the reader may see [15–18] and the references cited therein.Another area of research which has received great attention from researchers in the field of fractional differential equations is the notion of stability in the sense of Ulam. There are different kinds of Ulam’s stabilities, i.e., Hyers–Ulam stabilityHUS, generalized Hyers–Ulam stability GHUS, Hyers–Ulam–Rassias stability HURS, and generalized Hyers–Ulam–Rassias stability GHURS. The Ulam-type stability was first introduced by Ulam [19] in 1940 and then was studied and generalized by many mathematicians with different approaches [10, 20–29].Ren and Zhai [30] discussed the existence of a unique solution and multiple positive solutions with nonlocal boundary conditions for the system involving standard Riemann–Liouville fractional q-derivatives:(1)Dqαuσ+ϕσ,uσ=0,σ∈J,uσ|σ=0=Dqα−2uσ|σ=0=0,Dqα−1uσ|σ=1=αu+∫0ζφσDqβuσdqσ,where J=0,1 and Dqα denote the standard Riemann–Liouville fractional q-derivatives of order α, with α∈2,3,φ∈L10,1 being nonnegative.Benchohra and Lazreg [8] studied the existence and Hyers–Ulam stability of the following implicit fractional differential equations FDℰ′s involving Hadamard derivatives:(2)DαHuσ=ϕ1σ,uσ,Dαuσ,α∈0,1,u1=u1,u1∈ℝ,where σ∈1,T,T>1 and DαH denotes the Hadamard fractional derivative of order α.Riaz et al. [10] studied the existence, uniqueness, and stability of the solution of a nonlinear coupled system of impulsive ℱDℰs using Hadamard derivatives of the following form:(3)DαHuσ−fσ,uσ,vσ=0,σ∈J,σ≠σi,i=1,2,…,m,DβHvσ−gσ,uσ,vσ=0,σ∈J,σ≠σj,j=1,2,…,m,Δuσi=ℐiuσi,Δu′σi=ℐ˜iuσi,i=1,2,…,m,Δvσj=ℐjvσj,Δv′σj=ℐ˜jvσj,j=1,2,…,m,μln2u2+νln2u′2=ϕu,μuT+νu′T=φu,μln2v2+νln2v′2=ϕv,μvT+νv′T=φv,where 1<α,β≤2,f,g:J×ℝ2⟶ℝ and ϕ,φ:CJ,ℝ⟶ℝ are continuous functions.Riaz et al. [31] studied the existence, uniqueness, and stability of solution of coupled implicit impulsive fractional differential equations using Hadamard derivatives:(4)DαHuσ−fσ,uσ,DβHvσ=0,σ∈J,σ≠σi,i=1,2,…,m,DβHvσ−gσ,vσ,DαHuσ=0,σ∈J,σ≠σj,j=1,2,…,m,Δuσi=ℐiuσi,Δu′σi=ℐ˜iuσi,i=1,2,…,m,Δvσj=ℐjvσj,Δv′σj=ℐ˜jvσj,j=1,2,…,m,uT=∫1TlnT/πα−1Γαϕπ,uπdππ,u′T=φu,vT=∫1TlnT/πβ−1Γβϕπ,vπdππ,v′T=φv,where 1<α,β≤2,f,g:J×ℝ2⟶ℝ and ϕ,φ:CJ,ℝ⟶ℝ are continuous functions.In this article, we extend the system of Ren and Zhai [30] to implicit Hadamard fractional derivatives having the Stieltjes integral condition instead of the Riemann–Liouville fractional q-derivative and present its existence, uniqueness, at least one solution, and different kinds of Ulam’s stabilities. We study the following system:(5)DαHuσ+ϕσ,uσ,DαHuσ=0,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφσDβHuσdσ,where α∈2,3,ϕ:J=1,2×ℝ×ℝ⟶ℝ is the continuous function and DαH,DβH are the Hadamard fractional derivatives of order α, with α−1−β>0,φ∈L11,2 being non-negative, and αu is a linear function given by(6)αu=∫12uσdμσ,involving the Stieltjes integral with respect to the function μ:1,2⟶ℝ,μσ is right continuous on [1, 2), left continuous at σ=2. Particularly, μ is a nondecreasing function with μ1=0; then, dμ is a positive Stieltjes measure.We also extend system (5) to the coupled system and discuss its existence, uniqueness, at least one solution, and different kinds of Ulam’s stabilities of the problem:(7)DαHuσ+ϕ1σ,uσ,DγHvσ=0,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ=0,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφασDβHuσdσ,vσ|σ=1=Dγ−2Hvσ|σ=1=0,Dγ−1Hvσ|σ=2=γv+∫1ξφγσDδHvσdσ,where α,γ∈2,3,ϕ1,ϕ2:J=1,2×ℝ×ℝ⟶ℝ are the continuous functions and DαH,DβH,DγH, and DδH are the Hadamard fractional derivatives of orders α,γ, respectively with α−1−β>0,γ−1−δ>0,φα,φγ∈L11,2 being nonnegative, and αu,γv are linear functions given by(8)αu=∫12uσdμσ,γv=∫12vσdμσ,involving the Stieltjes integral with respect to the function μ:1,2⟶ℝ,μσ is right continuous on [1, 2), left continuous at σ=2. Particularly, μ is a nondecreasing function with μ1=0; then, dμ is a positive Stieltjes measure.The rest of the article is arranged as follows: In Section2, we present some basic definitions, lemmas, and theorems that are used in our main results. In Section 3, we use different conditions and some standard fixed-point theorems for the existence and uniqueness of solutions to the given system (5) and its corresponding coupled system (7). In Section 4, we present Ulam’s stabilities for the given systems (5) and (7) under some specific conditions. At the end, examples are given to illustrate the main results.Throughout the paper, we assume thatM1: Δ=Γα−β−∫1ζφσlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ.M2: ϕ:J×ℝ×ℝ⟶ℝ is continuous.M3: for σ∈J and u,z∈X, there are π1,π2,π3∈CJ,ℝ+ such that(9)ϕσ,uσ,zσ≤π1σ+π2σuσ+π3σzσ,withπ1∗=supσ∈Jπ1σ,π2∗=supσ∈Jπ2σ, and π3∗=supσ∈Jπ3σ<1.M4: for all u,z,u^,z^∈ℝ and for each σ∈J, there exist constants Lϕ>0,0<Lϕ∗<1 such that(10)ϕσ,u,z−ϕσ,u^,z^≤Lϕu−u^+Lϕ∗z−z^.M5: for σ∈J, there exists a constant E such that(11)φσ≤E.M6: let Ψ∈CJ,ℝ be an increasing function; then, there is ΩΨ>0 such that, for each σ∈J, the inequality(12)ℑαHΨσ≤ΩΨΨσ,holds.M7: Δα=Γα−β−∫1ζφασlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ, and(13)Δγ=Γγ−ξ−∫1ξφγσlnσγ−1−ξdσ−Γγ−ξΓγ∫12lnσγ−1dμσ.M8: ϕ1,ϕ2:J×ℝ×ℝ⟶ℝ are continuous.M9: for σ∈J and u,z∈X, there are π4,π5,π6∈CJ,ℝ+ such that(14)ϕ1σ,uσ,zσ≤π4σ+π5σuσ+π6σzσ,withπ4∗=supσ∈Jπ4σ,π5∗=supσ∈Jπ5σ, and π6∗=supσ∈Jπ6σ<1.Similarly, forσ∈J and v,y∈X, there are π7,π8,π9∈CJ,ℝ+ such that(15)ϕ2σ,vσ,yσ≤π7σ+π8σvσ+π9σyσ,withπ7∗=supσ∈Jπ7σ,π8∗=supσ∈Jπ8σ, and π9∗=supσ∈Jπ9σ<1.M10: for all u,z,u^,z^,v,y,v^,y^∈ℝ and for each σ∈J, there exist constants Lϕ1,Lϕ2>0,0<Lϕ1∗,Lϕ2∗<1 such that(16)ϕ1σ,u,z−ϕ1σ,u^,z^≤Lϕ1u−u^+Lϕ1∗z−z^,ϕ2σ,v,y−ϕ2σ,v^,y^≤Lϕ2v−v^+Lϕ2∗y−y^.M11: for σ∈J, there exist constants Eα,Eγ such that(17)φασ≤Eα,φγσ≤Eγ.M12: let Ψα,Ψγ∈CJ,ℝ be increasing functions, and there exist ΩΨα,ΩΨγ>0 such that, for each σ∈J, the inequalities(18)ℑαHΨασ≤ΩΨαΨασ,ℑγHΨγσ≤ΩΨγΨγσ,hold.


## 2. Preliminaries

In this section, we present some useful definitions, lemmas, and theorems, which will be used throughout the manuscript\enleadertwodots.Definition 1 (see [31]).
The Hadamard fractional integral of order, of functionuσ, is defined by(19)ℑαHuσ=1Γα∫1σlnσηα−1uηdηη,1≤σ≤2,where Γ⋅ is the gamma function.Definition 2 (see [31]).
The Hadamard fractional derivative of orderα∈n−1,n, n∈ℤ+ of function uσ is defined by(20)DαHuσ=1Γn−ασddσn∫nσlnσηn−α+1uηdηη,1≤σ≤2,where Γ⋅ is the gamma function.Lemma 1 (see [31]).
Letα>0 and u be any function; then, the homogeneous differential equation along with Hadamard fractional order HDαuσ=0 has a solution(21)uσ=∑i=1ncilnσα−i,and the following formula holds:(22)ℑαHDαHuσ=uσ+∑i=1ncilnσα−i,where ci∈ℝ,i=1,2,…,n, and n−1<α<n.Theorem 1 (see [32]) (Arzela-Ascoli’s Theorem).
LetB⊂CJ,ℝ be relatively compact, and(A)
B is a uniformly bounded set such that there exists ϖ>0 with(23)ϕ=supy∈Jϕy<ϖ,for everyϕ∈B.(B)
B is an equicontinuous set, i.e., for every ϵ>0, there exists δ>0 such that, for any(24)z,z¯∈J,z−z¯≤δ⇒ϕz−ϕz¯≤ε,for everyϕ∈B.Theorem 2 (see [33]) (Banach Fixed-Point Theorem).
Letℬ be a nonempty closed subset of a Banach space X. Then, any contraction mapping T from X into itself has a unique fixed point.Theorem 3 (see [33]) (Schaefer’s Fixed-Point Theorem).
LetX be a Banach space. Suppose the operator T:X⟶X is a continuous compact mapping (or completely continuous). Moreover, suppose(25)ℬ=u∈X|u=δTu,0<δ<1,is a bounded set. Then, ℬ has at least one fixed point in X.Lemma 2.
LetX=uσ:u∈CJ,ℝ be a Banach space endowed with the norm uX=supuσ:σ∈J. Similarly, the norm defined on the product space X×X is u,vX×X=uX+vX. Obviously, X×X,u,vX×X is a Banach space. Also, the cone C´⊂X×X is defined as(26)C´=u,v∈X×X:uσ≥0,vσ≥0.Theorem 4 (see [34]).
LetX be a Banach space containing a cone C´. ℬ´⊂C´ with 0∈ℬ´ is a relatively open set, and the operator T:ℬ´⟶ℬ´ is completely continuous. Then, one of the following conditions holds:(1)
There existu,v∈∂ℬ´ and δ∈0,1 such that u,v=δTu,v(2)
T has a fixed point in ℬ´


## 3. Existence and Uniqueness

In this section, we give the existence and uniqueness of solutions of (5) and its coupled system (7).


### 3.1. Existence and Uniqueness Solution for System (5)

Our first result is stated as follows.Lemma 3.
Let us assume thatf∈CJ,ℝ and Δ>0; then, the fractional differential equation(27)DαHuσ+fσ=0,α∈2,3,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=∫12uσdμσ+∫1ζφσDβHuσdσ,has a solution(28)uσ=∫12Gσ,ηfηdηη,σ∈J,where(29)Gσ,η=G1σ,η+G2σ,η+G3σ,η,G1σ,η=lnσα−1−lnσ/ηα−1Γα,1≤η≤σ≤2,lnσα−1Γα,1≤σ≤η≤2,G2σ,η=lnσα−1Δ∫1ζφσℚ1σ,ηdσ,G3σ,η=Γα−βlnσα−1ΔΓα∫12G1σ,ηdμσ,ℚ1σ,η=lnσα−1−β−lnσ/ηα−1−βΓα,1≤η≤σ≤2,lnσα−1−βΓα,1≤σ≤η≤2.Proof.
Consider(30)DαHuσ+fσ=0,α∈2,3,σ∈J.
Forσ∈J, Lemma 1 gives(31)uσ=c1lnσα−1+c2lnσα−2+c3lnσα−3−1Γα∫1σlnσηα−1fηdηη,for some ci∈ℝ,i=1,2,3. By the condition uσ|σ=1=0, we have c3=0. Hence,(32)uσ=c1lnσα−1+c2lnσα−2−1Γα∫1σlnσηα−1fηdηη.
Now, considering the conditionDα−2Huσ|σ=1=0, we get(33)Dα−2Huσ=c1Dα−2Hlnσα−1+c2Dα−2Hlnσα−2−Dα−2HℑαHfσ,=c1ΓαΓ2lnσ+c2Γα−1Γ1−ℑ2Hfσ,in view of Hℑ2fσ⟶0 as σ⟶0, so we must set c2=0. Then, we have(34)uσ=c1lnσα−1−1Γα∫1σlnσηα−1fηdηη.
Furthermore, we obtain(35)Dα−1Huσ=c1Dα−1Hlnσα−1−Dα−1HℑαHfσ,=c1Γα−∫1σfηdηη,DβHuσ=c1DβHlnσα−1−DβHℑαHfσ,=c1ΓαΓα−βlnσα−1−β−1Γα−β∫1σlnσηα−1−βfηdηη.
Then, we obtain(36)∫12uσdμσ+∫1ζφσDβHuσdσ,=c1∫12lnσα−1dμσ+c1ΓαΓα−β∫1ζφσlnσα−1−βdσ−1Γα∫12∫1σlnσηα−1fηdηηdμσ−1Γα−β∫1ζφσ∫1σlnσηα−1−βfηdηηdσ.
Applying the 2nd condition, we get(37)c1=∫12fηdη/ηΓα−Γα/Γα−β∫1ζφσlnσα−1−βdσ−∫12lnσα−1dμσ−∫1ζφσ∫1σlnσ/ηα−1−βfηdη/ηdσΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ−∫12∫1σlnσ/ηα−1fηdη/ηdμσΓ2α−Γ2α/Γα−β∫1ζφσlnσα−1−βdσ−Γα∫12lnσα−1dμσ.
Substituting (37) into (34), we get(38)uσ=lnσα−1∫12fηdη/ηΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×Γα−β−∫1ζφσlnσα−1−βdσ−Γα−βΓα∫12lnσα−1dμσ+∫1ζφσlnσα−1−βdσ+Γα−βΓα∫12lnσα−1dμσ−1Γα∫1σlnσηα−1fηdηη−lnσα−1∫1ζφσ∫1σlnσ/ηα−1−βfηdη/ηdσΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ−lnσα−1∫12∫1σlnσ/ηα−1fηdη/ηdμσΓ2α−Γ2α/Γα−β∫1ζφσlnσα−1−βdσ−Γα∫12lnσα−1dμσ,=lnσα−1Γα−β−∫1ζφσlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ×1Γα∫1ζ∫12φσlnσα−1−βfηdηηdσ−∫1ζ∫1σφσlnσηα−1−βfηdηηdσ+Γα−βlnσα−1ΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×1Γα∫12lnσα−1∫12fηdηηdμσ−lnσα−1Γ2αΓα−β−Γ2α∫1ζφσlnσα−1−βdσ−ΓαΓα−β∫12lnσα−1dμσ×∫12∫1σlnσηα−1fηdηηdμσ+∫12G1σ,ηfηdηη,=lnσα−1Γα−β−∫1ζφσlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ×1Γα∫1ζ∫1σφσlnσα−1−β−lnσηα−1−βfηdηηdσ+∫1ζ∫σ2φσlnσα−1−βfηdηηdσ+Γα−βlnσα−1ΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×1Γα∫12∫1σlnσα−1−lnσηα−1fηdηηdμσ+∫12∫σ2lnσα−1fηdηηdμσ+∫12G1σ,ηfηdηη=lnσα−1Δ∫1ζφσ∫12ℚ1σ,ηfηdηηdσ+∫12G1σ,ηfηdηη+Γα−βlnσα−1ΔΓα∫12∫12G1σ,ηfηdηηdμσ,=∫12G1σ,η+G2σ,η+G3σ,ηfηdηη=∫12Gσ,ηfηdηη.
By replacingfσ=ϕσ,uσ,HDαuσ in Lemma 3, we get the integral equation of problem (5) as(39)uσ=∫12Gσ,ηϕη,uη,DαHuηdηη.Lemma 4.
Green’s functionGσ,η, which is obtained in (39), has the following properties:(1)
Gσ,η⩾0 for all σ,η∈J(2)
Gσ,η is continuous over J×J(3)
maxσ∈J∫12Gσ,ηdη/η⩽Y,
where(40)Y=ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2,∇=Γα−β+Eζ∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+lnρα−1PΓα−βΓαProof.
It is very easy to prove1 and 2, so we leave it.
3 Since we know that Green’s function of the considered problem is in the form(41)∫12Gσ,ηdηη=lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ−∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ−∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη−1Γα∫1σlnσηα−1dηη≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ+∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ+∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηη,using M5 and mean-value theorem [35] with ρ∈1,2andμ2=P>0, we get(42)maxσ∈J∫12Gσ,ηdηη≤ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2=Y.
Hence, the proof of3 is complete.
Ifu is the solution of the given system (5) and σ∈J, then(43)uσ=lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βϕη,uη,zηdηηdσ−lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βϕη,uη,zηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1ϕη,uη,zηdηηdμσ−lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1ϕη,uη,zηdηηdμσ+lnσα−1Γα∫12ϕη,uη,zηdηη−1Γα∫1σlnσηα−1ϕη,uη,zηdηη,where(44)zσ=DαHuσ.
We define an operatorT:X⟶X as(45)Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη,where z∈CJ,ℝ such that(46)zσ=−ϕσ,uσ,zσ.Theorem 5.
LetM2−M5 hold. Then, the operator T, defined in (45), is compact.Proof.
To show that the operatorT is compact, we follow several steps.Step 1: we consider a sequenceun such that un⟶u in X; then, for each σ∈J, we have(47)Tunσ−Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βznη−zηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βznη−zηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1znη−zηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1znη−zηdηηdμσ−lnσα−1Γα∫12znη−zηdηη+1Γα∫1σlnσηα−1znη−zηdηη≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βznη−zηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βznη−zηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1znη−zηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1znη−zηdηηdμσ+lnσα−1Γα∫12znη−zηdηη+1Γα∫1σlnσηα−1znη−zηdηη.
From (46), we can write(48)znσ=−ϕσ,unσ,znσ,zσ=−ϕσ,uσ,zσ.
Now, byM4, we have(49)znσ−zσ=−ϕσ,unσ,znσ−ϕσ,uσ,zσ≤Lϕunσ−uσ+Lϕ∗znσ−zσ,which implies(50)znσ−zσ≤Lϕ1−Lϕ∗unσ−uσ.
Since we supposed thatun⟶u, zn⟶z as n⟶∞ for each σ∈J. So, by Lebesgue dominated convergence theorem [36], (47) gives(51)Tunσ−Tuσ⟶0,asn⟶∞.
This implies(52)Tun−TuX⟶0,asn⟶∞.
Hence,T is continuous.Step 2: now, we are going to prove that the operatorT is bounded in set X. For this, we show that, for any p∘>0, there exists q∘>0 such that, for each(53)u∈Q=u∈X:uX≤p∘,we have(54)TuX≤q∘.
From (45), for each σ∈J, we have(55)Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη.
Now, byM3 and (46), we have(56)zσ=−ϕσ,uσ,zσ≤π1σ+π2σuσ+π3σzσ.
Takingsupσ>0, we get(57)zX≤π1∗+π2∗p∘+π3∗zX≤π1∗+π2∗p∘1−π3∗.
Thus, (55) becomes(58)TuX≤π1∗+π2∗p∘1−π3∗ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2≤π1∗+π2∗p∘1−π3∗Y=q∘.
Hence,TQ is uniformly bounded.Step 3: now, to show that the operatorT is equicontinuous in X. For this, let σ1,σ2∈J with σ1>σ2, since Q is a bounded set in X, and let u∈Q. Then,(59)Tuσ1−Tuσ2=−1ΔΓαlnσ1α−1∫1ζ∫12φσ1lnσ1α−1−βzηdηηdσ1−lnσ2α−1∫1ζ∫12φσ2lnσ2α−1−βzηdηηdσ2−lnσ1α−1∫1ζ∫1σ1φσ1lnσ1ηα−1−βzηdηηdσ1+lnσ2α−1∫1ζ∫1σ2φσ2lnσ2ηα−1−βzηdηηdσ2−Γα−βΔΓ2αlnσ1α−1∫12∫12lnσ1α−1zηdηηdμσ1−lnσ2α−1∫12∫12lnσ2α−1zηdηηdμσ2−lnσ1α−1∫12∫1σ1lnσ1ηα−1zηdηηdμσ1+lnσ2α−1∫12∫1σ2lnσ2ηα−1zηdηηdμσ2−lnσ1α−1−lnσ2α−1Γα∫12zηdηη+1Γα∫σ2σ1lnσ1ηα−1zηdηη−1Γα∫1σ2lnσ2ηα−1−lnσ1ηα−1zηdηη≤lnσ1α−1−lnσ2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+lnσ1α−lnσ2αΓα+1π1∗+π2∗p∘1−π3∗.The right-hand side of (59) approaches to zero as σ1⟶σ2. Hence, TQ is equicontinuous. As a consequence of Step 1 to 3, the operator T is completely continuous. Therefore, in view of the Arzelà-Ascoli theorem, the operator T is compact.Theorem 6.
Let the hypothesesM3 and M5 hold, and if N<1, the given problem (5) has at least one solution in X.Proof.
For the proof of this theorem, we are considering a setℬ⊂X which is defined in the following form:(60)ℬ=u∈X|u=δTu,0<δ<1.
We have to show that the setℬ is bounded. Let u∈X such that(61)uσ=δTuσ,where0<δ<1.
Then, for eachσ∈J, we have(62)uσ=−δlnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βzηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1zηdηηdμσ+lnσα−1Γα∫12zηdηη−1Γα∫1σlnσηα−1zηdηη≤ln2α−1∇Γα+1zηαEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
Now, byM3, we have(63)zσ=−ϕσ,uσ,zσ≤π1σ+π2σuσ+π3σzσ.
So, we get(64)zσ≤π1σ+π2σuσ1−π3σ.
Putting (64) in (62) and taking supσ>1, we get(65)uX≤ln2α−1∇Γα+1π1∗+π2∗uX1−π3∗αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
For simplicity, let(66)M=π1∗ln2α−1∇1−π3∗Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2,N=π2∗ln2α−1∇1−π3∗Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
So, (65) becomes(67)uX≤M+NuX,which implies(68)uX≤M1−N.
This shows that the setℬ is bounded. So, by Theorems 3 and 5, we get that the operator T has at least one fixed point. Therefore, the given problem (5) has at least one solution in X.Theorem 7.
Suppose that the hypothesisM2, M4, and M5 hold. Then, the given problem (5) has a unique solution in X if(69)Lϕ1−Lϕ∗Y<1.Proof.
We shall use the Banach contraction principle to prove that the operatorT has a unique fixed point, which will be the unique solution of the given system (5), by considering the operator T:X⟶X defined in (45).
Letu,u¯ be the solution of (5), and for σ∈J, we have(70)Tuσ−Tu¯σ≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzη−z¯ηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βzη−z¯ηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zη−z¯ηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1zη−z¯ηdηηdμσ+lnσα−1Γα∫12zη−z¯ηdηη+1Γα∫1σlnσηα−1zη−z¯ηdηη,where z,z¯∈CJ,ℝ such that(71)zσ=−ϕσ,uσ,zσ,z¯σ=−ϕσ,u¯σ,z¯σ.
Now, byM4, we have(72)zσ−z¯σ=−ϕσ,uσ,zσ−ϕσ,u¯σ,z¯σ≤Lϕu−u¯+Lϕ∗z−z¯,which implies(73)zσ−z¯σ≤Lϕ1−Lϕ∗u−u¯.
So, (70) becomes(74)Tuσ−Tu¯σ≤Lϕ1−Lϕ∗lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βuη−u¯ηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βuη−u¯ηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1uη−u¯ηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1uη−u¯ηdηηdμσ+lnσα−1Γα∫12uη−u¯ηdηη+1Γα∫1σlnσηα−1uη−u¯ηdηη.
UsingM5 and taking supσ>1 on both sides, we get(75)Tu−Tu¯X≤Lϕ1−Lϕ∗ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2u−u¯X.
This implies that(76)Tu−Tu¯X≤Lϕ1−Lϕ∗Yu−u¯X.
Hence, the operatorT is a contraction. Thus, by the Banach contraction principle, we get that T has a unique fixed point, which is a unique solution of the given problem (5).


### 3.2. Existence and Uniqueness Solution for System (7)

In this section, we show the existence and uniqueness of the solution of the system (7). First, we have the following:Lemma 5.
The system(77)DαHuσ+ϕ1σ,uσ,DγHvσ=0,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ=0,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφασDβHuσdσ,vσ|σ=1=Dγ−2Hvσ|σ=1=0,Dγ−1Hvσ|σ=2=γv+∫1ξφγσDδHvσdσ,has a solution (u, v) if and only if(78)uσ=∫12Gασ,ηϕ1η,uη,DγHvηdηη,vσ=∫12Gγσ,ηϕ2η,vη,DαHuηdηη,where(79)Gασ,η=G1σ,η+G2σ,η+G3σ,η,Gγσ,η=G4σ,η+G5σ,η+G6σ,η,G1σ,η=lnσα−1−lnσ/ηα−1Γα,1≤η≤σ≤2,lnσα−1Γα,1≤σ≤η≤2,G2σ,η=lnσα−1Δα∫1ζφασℚ1σ,ηdσ,G3σ,η=Γα−βlnσα−1ΔαΓα∫12G1σ,ηdμσ,G4σ,η=lnσγ−1−lnσ/ηγ−1Γγ,1≤η≤σ≤2,lnσγ−1Γγ,1≤σ≤η≤2,G5σ,η=lnσγ−1Δγ∫1ξφγσℚ1σ,ηdσ,G6σ,η=Γγ−δlnσγ−1ΔγΓγ∫12G4σ,ηdμσ,ℚ1σ,η=lnσα−1−β−lnσ/ηα−1−βΓα,1≤η≤σ≤2,lnσα−1−βΓα,1≤σ≤η≤2,ℚ2σ,η=lnσγ−1−δ−lnσ/ηγ−1−δΓγ,1≤η≤σ≤2,lnσγ−1−δΓγ,1≤σ≤η≤2.Proof.
The proof is similar to that given in Lemma3 and, hence, is not included here.We use the following notations for convenience:(80)yσ≔−ϕ1σ,uσ,HDγvσ=−ϕ1σ,uσ,zσ,zσ≔−ϕ2σ,vσ,HDαuσ=−ϕ2σ,vσ,yσ.
Hence, forσ∈J, (78) becomes(81)uσ=∫12Gασ,ηyηdηη,vσ=∫12Gγσ,ηzηdηη,where y,z∈X satisfying the functional equation.Lemma 6.
Green’s functionGα,γσ,η=Gασ,η,Gγσ,η of the system (7) have the following properties:(1
)Gα,γσ,η is continuous over J×J(2)
maxσ∈J∫12Gασ,ηdη/η⩽Yαandmaxσ∈J∫12Gγσ,ηdη/η⩽Yγ,
where(82)Yα=ln2α−1∇αΓα+1αEαζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEαζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2,Yγ=ln2γ−1∇γΓγ+1γEγξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEγξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γγln2+ln2,∇α=Γα−β+Eαζ∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+lnρα−1PΓα−βΓα,∇γ=Γγ−δ+Eγξ∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+lnργ−1PΓγ−δΓγ.Proof.
(1)
It is easy to prove thatGα,γσ,η is continuous, so we leave it.(2)
Since we know that Green’s functions of the considered problem (7) is in the form(83)∫12Gασ,ηdηη=lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ−∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ−∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη−1Γα∫1σlnσηα−1dηη≤lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ+∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ+∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηη,using M5 and the mean value theorem [35] with ρ∈1,2andμ2=P>0, we get(84)maxσ∈J∫12Gασ,ηdηη≤ln2α−1∇αΓα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2=Yα.
Similarly, we can obtain(85)maxσ∈J∫12Gγσ,ηdηη≤ln2γ−1∇γΓγ+1γEξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γγln2+ln2=Yγ.
Hence, the proof of 2 is complete.
Ifu,v are the solutions of the given system (7) and σ∈J; then,(86)uσ=lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βϕ1η,uη,zηdηηdσ−lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βϕ1η,uη,zηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1ϕ1η,uη,zηdηηdμσ−lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1ϕ1η,uη,zηdηηdμσ+lnσα−1Γα∫12ϕ1η,uη,zηdηη−1Γα∫1σlnσηα−1ϕ1η,uη,zηdηη,vσ=lnσγ−1ΔγΓγ∫1ξ∫12φγσlnσγ−1−δϕ2η,vη,yηdηηdσ−lnσγ−1ΔγΓγ∫1ξ∫1σφγσlnσηγ−1−δϕ2η,vη,yηdηηdσ+lnσγ−1Γγ−δΔγΓ2γ∫12∫12lnσγ−1ϕ2η,vη,yηdηηdμσ−lnσγ−1Γγ−δΔγΓ2γ∫12∫1σlnσηγ−1ϕ2η,vη,yηdηηdμσ+lnσγ−1Γγ∫12ϕ2η,vη,yηdηη−1Γγ∫1σlnσηγ−1ϕη,vη,yηdηη.
Now, we transform the given system (7) into a fixed-point problem. Let an operator T:X×X⟶X×X be defined as(87)Tu,vσ=∫12Gασ,ηϕ1η,uη,zηdηη∫12Gγσ,ηϕ2η,vη,yηdηη=Tαu,zσTγv,yσ.
Then, the solution of (7) coincides with the fixed point of T, where(88)Tαu,zσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη,Tγv,yσ=−lnσγ−1ΔγΓγ∫1ξ∫12φγσlnσγ−1−δzηdηηdσ−∫1ξ∫1σφγσlnσηγ−1−δzηdηηdσ−lnσγ−1Γγ−δΔγΓ2γ∫12∫12lnσγ−1zηdηηdμσ−∫12∫1σlnσηγ−1zηdηηdμσ−lnσγ−1Γγ∫12zηdηη+1Γγ∫1σlnσηγ−1zηdηη.Theorem 8.
Letϕ1,ϕ2:J×ℝ×ℝ⟶ℝ and M9, M11 hold. Then, the operator T:C´⟶C´ defined in (87) is completely continuous.Proof.
In view of continuity ofϕ1,ϕ2 and Gα,γσ,η, T is also continuous for all u,z∈C´. Suppose ℬ´⊆C´ is a bounded set. So, for every u∈ℬ´, we have(89)Tαu,vσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ+∫1ζ∫1σφασlnσηα−1−βyηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ+∫12∫1σlnσηα−1yηdηηdμσ+lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη.
Now, byM9, we have(90)yσ=−ϕ1σ,uσ,zσ⩽π4σ+π5σuσ+π6σzσ,=π4σ+π5σuσ+π6σϕ2σ,vσ,yσ⩽π4σ+π5σuσ+π6σπ7σ+π6σπ8σvσ+π6σπ9σyσ.
So, we obtain(91)yσ⩽π4σ+π6σπ7σ1−π6σπ9σ+π5σuσ+π6σπ8σvσ1−π6σπ9σ.
Now, using 2 of Lemma6, M11, and (91) in (89), we get(92)Tαu,vX×X⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗×ln2α−1∇αΓα+1αEαζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEαζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗Yα.
In the same way, we obtain(93)Tγu,vX×X⩽π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗Yγ.
Thus, from (92) and (93), we get(94)Tu,vX×X⩽Tαu,vX×X+Tγu,vX×X⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗Yα+π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗Yγ=ℳ.
Thus,T is uniformly bounded. Now, we prove the operator T is equicontinuous. For this, suppose σ1>σ2∈J and u,v∈ℬ´; then,(95)Tαu,vσ1−Tαu,vσ2,=−lnσ1α−1ΔαΓα∫1ζ∫12φασ1lnσ1α−1−βyηdηηdσ1+lnσ2α−1ΔαΓα∫1ζ∫12φασ2lnσ2α−1−βyηdηηdσ2−lnσ2α−1ΔαΓα∫1ζ∫1σ2φασ2lnσ2ηα−1−βyηdηηdσ2+lnσ1α−1ΔαΓα∫1ζ∫1σ1φασ1lnσ1ηα−1−βyηdηηdσ1−lnσ1α−1Γα−βΔαΓ2α∫12∫12lnσ1α−1yηdηηdμσ1+lnσ2α−1Γα−βΔαΓ2α∫12∫12lnσ2α−1yηdηηdμσ2−lnσ2α−1Γα−βΔαΓ2α∫12∫1σ2lnσ2ηα−1yηdηηdμσ2+lnσ1α−1Γα−βΔαΓ2α∫12∫1σ1lnσ1ηα−1yηdηηdμσ1−lnσ1α−1−lnσ2α−1Γα∫12yηdηη+1Γα∫σ2σ1lnσ1ηα−1yηdηη−1Γα∫1σ2lnσ2ηα−1−lnσ1ηα−1zηdηη≤π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗lnσ1α−1−lnσ2α−1∇Γα+1×αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+lnσ1α−lnσ2αΓα+1.
In the same way, we can show that(96)Tγu,vσ1−Tγu,vσ2⩽π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗lnσ1γ−1−lnσ2γ−1∇Γγ+1×γEξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γln2+lnσ1γ−lnσ2γΓγ+1.
The right-hand sides of (94) and (96) approache to zero as σ1⟶σ2. Hence, by the Arzelà-Ascoli theorem, T is equicontinuous and uniformly equicontinuous. Also, it is very easy to prove that Tℬ´⊂ℬ´. Therefore, T is completely continuous.Theorem 9.
Under the hypothesisM10, M11 and(97)YαLϕ1+Lϕ1∗Lϕ2+YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗<1.
The coupled system (7) has a unique solution.Proof.
Letu,v,u^,v^∈C´, and we consider(98)Tαu,vσ−Tαu^,v^σ⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηηyη−y^η,where(99)yσ=−ϕ1σ,uσ,zσ,y^σ=−ϕ1σ,u^σ,z^σ.
Now, usingM10,(100)yσ−y^σ=−ϕ1σ,uσ,zσ−ϕ1σ,u^σ,z^σ⩽Lϕ1uσ−u^σ+Lϕ1∗zσ−z^σ,=Lϕ1uσ−u^σ+Lϕ1∗ϕ2σ,vσ,yσ−ϕ2σ,u^σ,y^σ⩽Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−u^σ+Lϕ2∗yσ−y^σ,=Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−u^σ1−Lϕ1∗Lϕ2∗.
Substituting (100) in (98) and taking supσ∈J, we get(101)Tαu,v−Tαu^,v^X×X⩽YαLϕ1+Lϕ1∗Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
In the same way, we can obtain(102)Tγu,v−Tγu^,v^X×X⩽YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
So, from (101) and (102), we get(103)Tu,v−Tu^,v^X×X⩽YαLϕ1+Lϕ1∗Lϕ2+YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
Thus,T is a contraction. Therefore, by the Banach contraction principle, T has a fixed point. So, we infer that the given coupled system (7) has a unique solution.Theorem 10.
In view of the continuity of the functionsϕ1,ϕ2 and supposing M9 and M13 with(104)N11∗=∫12|Gα2,η|π4∗+π6∗π7∗1−π6∗π9∗dηη<1,N12∗=∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη<1,N21∗=∫12Gγ2,ηπ7∗+π4∗π9∗1−π6∗π9∗dηη<1,N22∗=∫12Gγ2,ηπ5∗π9∗+π8∗1−π6∗π9∗dηη<1hold, the coupled system (7) has at least one solution.Proof.
Let a setℬ´ be defined as(105)ℬ´=u,v∈X×X:u,vX×X<ℛℬ´,where max2N11∗/1−2N12∗,2N21∗/1−2N22∗<ℛℬ´. Furthermore, the operator defined by T:ℬ´¯⟶C´ in (87) is completely continuous. Suppose u,v∈ℬ´; then, by definition of ℬ´, we have u,vX×X<ℛℬ´;(106)Tαu,zX×X⩽maxσ∈J∫12Gασ,ηϕ1η,uη,zηdηη⩽maxσ∈J∫12Gασ,ηπ4η+π6ηπ7η1−π6ηπ9ηdηη+maxσ∈J∫12Gασ,ηπ5ηuη+π6ηπ8ηvη1−π6ηπ9ηdηη⩽∫12Gα2,ηπ4∗+π6∗π7∗1−π6∗π9∗dηη+ℛℬ´∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη,=N11∗+ℛℬ´N12∗⩽ℛℬ´2.
Similarly,(107)Tγv,yX×X⩽ℛℬ´2.
Therefore,(108)Tu,vX×X⩽ℛℬ´,so Tu,z∈ℬ´¯. Thus, in view of Theorem 8, T:ℬ´¯⟶ℬ´¯ is completely continuous.
Now, we consider an eigenvalue problem defined as(109)u,v=δTu,v,δ∈0,1.
So, in view of the solution (u, v) of (109), we obtain(110)uX=δTαu,zX⩽maxσ∈J∫12Gασ,ηϕ1η,uη,zηdηη⩽maxσ∈J∫12Gασ,ηπ4η+π6ηπ7η1−π6ηπ9ηdηη+maxσ∈J∫12Gασ,ηπ5ηuη+π6ηπ8ηvη1−π6ηπ9ηdηη⩽∫12Gα2,ηπ4∗+π6∗π7∗1−π6∗π9∗dηη+ℛℬ´∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη,=N11∗+ℛℬ´N12∗⩽ℛℬ´2.
Similarly,(111)vX⩽ℛℬ´2.
Thus,(112)u,vX×X⩽ℛℬ´.
From equation (112), we get u,v∉∂ℬ´. So, in view of Theorem 4, T has at least one fixed point which lies in ℬ´¯. This shows there is at least one solution of the coupled system (7).


## 3.1. Existence and Uniqueness Solution for System (5)

Our first result is stated as follows.Lemma 3.
Let us assume thatf∈CJ,ℝ and Δ>0; then, the fractional differential equation(27)DαHuσ+fσ=0,α∈2,3,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=∫12uσdμσ+∫1ζφσDβHuσdσ,has a solution(28)uσ=∫12Gσ,ηfηdηη,σ∈J,where(29)Gσ,η=G1σ,η+G2σ,η+G3σ,η,G1σ,η=lnσα−1−lnσ/ηα−1Γα,1≤η≤σ≤2,lnσα−1Γα,1≤σ≤η≤2,G2σ,η=lnσα−1Δ∫1ζφσℚ1σ,ηdσ,G3σ,η=Γα−βlnσα−1ΔΓα∫12G1σ,ηdμσ,ℚ1σ,η=lnσα−1−β−lnσ/ηα−1−βΓα,1≤η≤σ≤2,lnσα−1−βΓα,1≤σ≤η≤2.Proof.
Consider(30)DαHuσ+fσ=0,α∈2,3,σ∈J.
Forσ∈J, Lemma 1 gives(31)uσ=c1lnσα−1+c2lnσα−2+c3lnσα−3−1Γα∫1σlnσηα−1fηdηη,for some ci∈ℝ,i=1,2,3. By the condition uσ|σ=1=0, we have c3=0. Hence,(32)uσ=c1lnσα−1+c2lnσα−2−1Γα∫1σlnσηα−1fηdηη.
Now, considering the conditionDα−2Huσ|σ=1=0, we get(33)Dα−2Huσ=c1Dα−2Hlnσα−1+c2Dα−2Hlnσα−2−Dα−2HℑαHfσ,=c1ΓαΓ2lnσ+c2Γα−1Γ1−ℑ2Hfσ,in view of Hℑ2fσ⟶0 as σ⟶0, so we must set c2=0. Then, we have(34)uσ=c1lnσα−1−1Γα∫1σlnσηα−1fηdηη.
Furthermore, we obtain(35)Dα−1Huσ=c1Dα−1Hlnσα−1−Dα−1HℑαHfσ,=c1Γα−∫1σfηdηη,DβHuσ=c1DβHlnσα−1−DβHℑαHfσ,=c1ΓαΓα−βlnσα−1−β−1Γα−β∫1σlnσηα−1−βfηdηη.
Then, we obtain(36)∫12uσdμσ+∫1ζφσDβHuσdσ,=c1∫12lnσα−1dμσ+c1ΓαΓα−β∫1ζφσlnσα−1−βdσ−1Γα∫12∫1σlnσηα−1fηdηηdμσ−1Γα−β∫1ζφσ∫1σlnσηα−1−βfηdηηdσ.
Applying the 2nd condition, we get(37)c1=∫12fηdη/ηΓα−Γα/Γα−β∫1ζφσlnσα−1−βdσ−∫12lnσα−1dμσ−∫1ζφσ∫1σlnσ/ηα−1−βfηdη/ηdσΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ−∫12∫1σlnσ/ηα−1fηdη/ηdμσΓ2α−Γ2α/Γα−β∫1ζφσlnσα−1−βdσ−Γα∫12lnσα−1dμσ.
Substituting (37) into (34), we get(38)uσ=lnσα−1∫12fηdη/ηΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×Γα−β−∫1ζφσlnσα−1−βdσ−Γα−βΓα∫12lnσα−1dμσ+∫1ζφσlnσα−1−βdσ+Γα−βΓα∫12lnσα−1dμσ−1Γα∫1σlnσηα−1fηdηη−lnσα−1∫1ζφσ∫1σlnσ/ηα−1−βfηdη/ηdσΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ−lnσα−1∫12∫1σlnσ/ηα−1fηdη/ηdμσΓ2α−Γ2α/Γα−β∫1ζφσlnσα−1−βdσ−Γα∫12lnσα−1dμσ,=lnσα−1Γα−β−∫1ζφσlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ×1Γα∫1ζ∫12φσlnσα−1−βfηdηηdσ−∫1ζ∫1σφσlnσηα−1−βfηdηηdσ+Γα−βlnσα−1ΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×1Γα∫12lnσα−1∫12fηdηηdμσ−lnσα−1Γ2αΓα−β−Γ2α∫1ζφσlnσα−1−βdσ−ΓαΓα−β∫12lnσα−1dμσ×∫12∫1σlnσηα−1fηdηηdμσ+∫12G1σ,ηfηdηη,=lnσα−1Γα−β−∫1ζφσlnσα−1−βdσ−Γα−β/Γα∫12lnσα−1dμσ×1Γα∫1ζ∫1σφσlnσα−1−β−lnσηα−1−βfηdηηdσ+∫1ζ∫σ2φσlnσα−1−βfηdηηdσ+Γα−βlnσα−1ΓαΓα−β−Γα∫1ζφσlnσα−1−βdσ−Γα−β∫12lnσα−1dμσ×1Γα∫12∫1σlnσα−1−lnσηα−1fηdηηdμσ+∫12∫σ2lnσα−1fηdηηdμσ+∫12G1σ,ηfηdηη=lnσα−1Δ∫1ζφσ∫12ℚ1σ,ηfηdηηdσ+∫12G1σ,ηfηdηη+Γα−βlnσα−1ΔΓα∫12∫12G1σ,ηfηdηηdμσ,=∫12G1σ,η+G2σ,η+G3σ,ηfηdηη=∫12Gσ,ηfηdηη.
By replacingfσ=ϕσ,uσ,HDαuσ in Lemma 3, we get the integral equation of problem (5) as(39)uσ=∫12Gσ,ηϕη,uη,DαHuηdηη.Lemma 4.
Green’s functionGσ,η, which is obtained in (39), has the following properties:(1)
Gσ,η⩾0 for all σ,η∈J(2)
Gσ,η is continuous over J×J(3)
maxσ∈J∫12Gσ,ηdη/η⩽Y,
where(40)Y=ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2,∇=Γα−β+Eζ∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+lnρα−1PΓα−βΓαProof.
It is very easy to prove1 and 2, so we leave it.
3 Since we know that Green’s function of the considered problem is in the form(41)∫12Gσ,ηdηη=lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ−∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ−∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη−1Γα∫1σlnσηα−1dηη≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ+∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ+∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηη,using M5 and mean-value theorem [35] with ρ∈1,2andμ2=P>0, we get(42)maxσ∈J∫12Gσ,ηdηη≤ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2=Y.
Hence, the proof of3 is complete.
Ifu is the solution of the given system (5) and σ∈J, then(43)uσ=lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βϕη,uη,zηdηηdσ−lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βϕη,uη,zηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1ϕη,uη,zηdηηdμσ−lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1ϕη,uη,zηdηηdμσ+lnσα−1Γα∫12ϕη,uη,zηdηη−1Γα∫1σlnσηα−1ϕη,uη,zηdηη,where(44)zσ=DαHuσ.
We define an operatorT:X⟶X as(45)Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη,where z∈CJ,ℝ such that(46)zσ=−ϕσ,uσ,zσ.Theorem 5.
LetM2−M5 hold. Then, the operator T, defined in (45), is compact.Proof.
To show that the operatorT is compact, we follow several steps.Step 1: we consider a sequenceun such that un⟶u in X; then, for each σ∈J, we have(47)Tunσ−Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βznη−zηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βznη−zηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1znη−zηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1znη−zηdηηdμσ−lnσα−1Γα∫12znη−zηdηη+1Γα∫1σlnσηα−1znη−zηdηη≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βznη−zηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βznη−zηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1znη−zηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1znη−zηdηηdμσ+lnσα−1Γα∫12znη−zηdηη+1Γα∫1σlnσηα−1znη−zηdηη.
From (46), we can write(48)znσ=−ϕσ,unσ,znσ,zσ=−ϕσ,uσ,zσ.
Now, byM4, we have(49)znσ−zσ=−ϕσ,unσ,znσ−ϕσ,uσ,zσ≤Lϕunσ−uσ+Lϕ∗znσ−zσ,which implies(50)znσ−zσ≤Lϕ1−Lϕ∗unσ−uσ.
Since we supposed thatun⟶u, zn⟶z as n⟶∞ for each σ∈J. So, by Lebesgue dominated convergence theorem [36], (47) gives(51)Tunσ−Tuσ⟶0,asn⟶∞.
This implies(52)Tun−TuX⟶0,asn⟶∞.
Hence,T is continuous.Step 2: now, we are going to prove that the operatorT is bounded in set X. For this, we show that, for any p∘>0, there exists q∘>0 such that, for each(53)u∈Q=u∈X:uX≤p∘,we have(54)TuX≤q∘.
From (45), for each σ∈J, we have(55)Tuσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη.
Now, byM3 and (46), we have(56)zσ=−ϕσ,uσ,zσ≤π1σ+π2σuσ+π3σzσ.
Takingsupσ>0, we get(57)zX≤π1∗+π2∗p∘+π3∗zX≤π1∗+π2∗p∘1−π3∗.
Thus, (55) becomes(58)TuX≤π1∗+π2∗p∘1−π3∗ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2≤π1∗+π2∗p∘1−π3∗Y=q∘.
Hence,TQ is uniformly bounded.Step 3: now, to show that the operatorT is equicontinuous in X. For this, let σ1,σ2∈J with σ1>σ2, since Q is a bounded set in X, and let u∈Q. Then,(59)Tuσ1−Tuσ2=−1ΔΓαlnσ1α−1∫1ζ∫12φσ1lnσ1α−1−βzηdηηdσ1−lnσ2α−1∫1ζ∫12φσ2lnσ2α−1−βzηdηηdσ2−lnσ1α−1∫1ζ∫1σ1φσ1lnσ1ηα−1−βzηdηηdσ1+lnσ2α−1∫1ζ∫1σ2φσ2lnσ2ηα−1−βzηdηηdσ2−Γα−βΔΓ2αlnσ1α−1∫12∫12lnσ1α−1zηdηηdμσ1−lnσ2α−1∫12∫12lnσ2α−1zηdηηdμσ2−lnσ1α−1∫12∫1σ1lnσ1ηα−1zηdηηdμσ1+lnσ2α−1∫12∫1σ2lnσ2ηα−1zηdηηdμσ2−lnσ1α−1−lnσ2α−1Γα∫12zηdηη+1Γα∫σ2σ1lnσ1ηα−1zηdηη−1Γα∫1σ2lnσ2ηα−1−lnσ1ηα−1zηdηη≤lnσ1α−1−lnσ2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+lnσ1α−lnσ2αΓα+1π1∗+π2∗p∘1−π3∗.The right-hand side of (59) approaches to zero as σ1⟶σ2. Hence, TQ is equicontinuous. As a consequence of Step 1 to 3, the operator T is completely continuous. Therefore, in view of the Arzelà-Ascoli theorem, the operator T is compact.Theorem 6.
Let the hypothesesM3 and M5 hold, and if N<1, the given problem (5) has at least one solution in X.Proof.
For the proof of this theorem, we are considering a setℬ⊂X which is defined in the following form:(60)ℬ=u∈X|u=δTu,0<δ<1.
We have to show that the setℬ is bounded. Let u∈X such that(61)uσ=δTuσ,where0<δ<1.
Then, for eachσ∈J, we have(62)uσ=−δlnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βzηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1zηdηηdμσ+lnσα−1Γα∫12zηdηη−1Γα∫1σlnσηα−1zηdηη≤ln2α−1∇Γα+1zηαEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
Now, byM3, we have(63)zσ=−ϕσ,uσ,zσ≤π1σ+π2σuσ+π3σzσ.
So, we get(64)zσ≤π1σ+π2σuσ1−π3σ.
Putting (64) in (62) and taking supσ>1, we get(65)uX≤ln2α−1∇Γα+1π1∗+π2∗uX1−π3∗αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
For simplicity, let(66)M=π1∗ln2α−1∇1−π3∗Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2,N=π2∗ln2α−1∇1−π3∗Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2.
So, (65) becomes(67)uX≤M+NuX,which implies(68)uX≤M1−N.
This shows that the setℬ is bounded. So, by Theorems 3 and 5, we get that the operator T has at least one fixed point. Therefore, the given problem (5) has at least one solution in X.Theorem 7.
Suppose that the hypothesisM2, M4, and M5 hold. Then, the given problem (5) has a unique solution in X if(69)Lϕ1−Lϕ∗Y<1.Proof.
We shall use the Banach contraction principle to prove that the operatorT has a unique fixed point, which will be the unique solution of the given system (5), by considering the operator T:X⟶X defined in (45).
Letu,u¯ be the solution of (5), and for σ∈J, we have(70)Tuσ−Tu¯σ≤lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzη−z¯ηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βzη−z¯ηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zη−z¯ηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1zη−z¯ηdηηdμσ+lnσα−1Γα∫12zη−z¯ηdηη+1Γα∫1σlnσηα−1zη−z¯ηdηη,where z,z¯∈CJ,ℝ such that(71)zσ=−ϕσ,uσ,zσ,z¯σ=−ϕσ,u¯σ,z¯σ.
Now, byM4, we have(72)zσ−z¯σ=−ϕσ,uσ,zσ−ϕσ,u¯σ,z¯σ≤Lϕu−u¯+Lϕ∗z−z¯,which implies(73)zσ−z¯σ≤Lϕ1−Lϕ∗u−u¯.
So, (70) becomes(74)Tuσ−Tu¯σ≤Lϕ1−Lϕ∗lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βuη−u¯ηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βuη−u¯ηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1uη−u¯ηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1uη−u¯ηdηηdμσ+lnσα−1Γα∫12uη−u¯ηdηη+1Γα∫1σlnσηα−1uη−u¯ηdηη.
UsingM5 and taking supσ>1 on both sides, we get(75)Tu−Tu¯X≤Lϕ1−Lϕ∗ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2u−u¯X.
This implies that(76)Tu−Tu¯X≤Lϕ1−Lϕ∗Yu−u¯X.
Hence, the operatorT is a contraction. Thus, by the Banach contraction principle, we get that T has a unique fixed point, which is a unique solution of the given problem (5).


## 3.2. Existence and Uniqueness Solution for System (7)

In this section, we show the existence and uniqueness of the solution of the system (7). First, we have the following:Lemma 5.
The system(77)DαHuσ+ϕ1σ,uσ,DγHvσ=0,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ=0,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφασDβHuσdσ,vσ|σ=1=Dγ−2Hvσ|σ=1=0,Dγ−1Hvσ|σ=2=γv+∫1ξφγσDδHvσdσ,has a solution (u, v) if and only if(78)uσ=∫12Gασ,ηϕ1η,uη,DγHvηdηη,vσ=∫12Gγσ,ηϕ2η,vη,DαHuηdηη,where(79)Gασ,η=G1σ,η+G2σ,η+G3σ,η,Gγσ,η=G4σ,η+G5σ,η+G6σ,η,G1σ,η=lnσα−1−lnσ/ηα−1Γα,1≤η≤σ≤2,lnσα−1Γα,1≤σ≤η≤2,G2σ,η=lnσα−1Δα∫1ζφασℚ1σ,ηdσ,G3σ,η=Γα−βlnσα−1ΔαΓα∫12G1σ,ηdμσ,G4σ,η=lnσγ−1−lnσ/ηγ−1Γγ,1≤η≤σ≤2,lnσγ−1Γγ,1≤σ≤η≤2,G5σ,η=lnσγ−1Δγ∫1ξφγσℚ1σ,ηdσ,G6σ,η=Γγ−δlnσγ−1ΔγΓγ∫12G4σ,ηdμσ,ℚ1σ,η=lnσα−1−β−lnσ/ηα−1−βΓα,1≤η≤σ≤2,lnσα−1−βΓα,1≤σ≤η≤2,ℚ2σ,η=lnσγ−1−δ−lnσ/ηγ−1−δΓγ,1≤η≤σ≤2,lnσγ−1−δΓγ,1≤σ≤η≤2.Proof.
The proof is similar to that given in Lemma3 and, hence, is not included here.We use the following notations for convenience:(80)yσ≔−ϕ1σ,uσ,HDγvσ=−ϕ1σ,uσ,zσ,zσ≔−ϕ2σ,vσ,HDαuσ=−ϕ2σ,vσ,yσ.
Hence, forσ∈J, (78) becomes(81)uσ=∫12Gασ,ηyηdηη,vσ=∫12Gγσ,ηzηdηη,where y,z∈X satisfying the functional equation.Lemma 6.
Green’s functionGα,γσ,η=Gασ,η,Gγσ,η of the system (7) have the following properties:(1
)Gα,γσ,η is continuous over J×J(2)
maxσ∈J∫12Gασ,ηdη/η⩽Yαandmaxσ∈J∫12Gγσ,ηdη/η⩽Yγ,
where(82)Yα=ln2α−1∇αΓα+1αEαζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEαζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2,Yγ=ln2γ−1∇γΓγ+1γEγξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEγξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γγln2+ln2,∇α=Γα−β+Eαζ∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+lnρα−1PΓα−βΓα,∇γ=Γγ−δ+Eγξ∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+lnργ−1PΓγ−δΓγ.Proof.
(1)
It is easy to prove thatGα,γσ,η is continuous, so we leave it.(2)
Since we know that Green’s functions of the considered problem (7) is in the form(83)∫12Gασ,ηdηη=lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ−∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ−∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη−1Γα∫1σlnσηα−1dηη≤lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ+∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ+∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηη,using M5 and the mean value theorem [35] with ρ∈1,2andμ2=P>0, we get(84)maxσ∈J∫12Gασ,ηdηη≤ln2α−1∇αΓα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2=Yα.
Similarly, we can obtain(85)maxσ∈J∫12Gγσ,ηdηη≤ln2γ−1∇γΓγ+1γEξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γγln2+ln2=Yγ.
Hence, the proof of 2 is complete.
Ifu,v are the solutions of the given system (7) and σ∈J; then,(86)uσ=lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βϕ1η,uη,zηdηηdσ−lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βϕ1η,uη,zηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1ϕ1η,uη,zηdηηdμσ−lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1ϕ1η,uη,zηdηηdμσ+lnσα−1Γα∫12ϕ1η,uη,zηdηη−1Γα∫1σlnσηα−1ϕ1η,uη,zηdηη,vσ=lnσγ−1ΔγΓγ∫1ξ∫12φγσlnσγ−1−δϕ2η,vη,yηdηηdσ−lnσγ−1ΔγΓγ∫1ξ∫1σφγσlnσηγ−1−δϕ2η,vη,yηdηηdσ+lnσγ−1Γγ−δΔγΓ2γ∫12∫12lnσγ−1ϕ2η,vη,yηdηηdμσ−lnσγ−1Γγ−δΔγΓ2γ∫12∫1σlnσηγ−1ϕ2η,vη,yηdηηdμσ+lnσγ−1Γγ∫12ϕ2η,vη,yηdηη−1Γγ∫1σlnσηγ−1ϕη,vη,yηdηη.
Now, we transform the given system (7) into a fixed-point problem. Let an operator T:X×X⟶X×X be defined as(87)Tu,vσ=∫12Gασ,ηϕ1η,uη,zηdηη∫12Gγσ,ηϕ2η,vη,yηdηη=Tαu,zσTγv,yσ.
Then, the solution of (7) coincides with the fixed point of T, where(88)Tαu,zσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη,Tγv,yσ=−lnσγ−1ΔγΓγ∫1ξ∫12φγσlnσγ−1−δzηdηηdσ−∫1ξ∫1σφγσlnσηγ−1−δzηdηηdσ−lnσγ−1Γγ−δΔγΓ2γ∫12∫12lnσγ−1zηdηηdμσ−∫12∫1σlnσηγ−1zηdηηdμσ−lnσγ−1Γγ∫12zηdηη+1Γγ∫1σlnσηγ−1zηdηη.Theorem 8.
Letϕ1,ϕ2:J×ℝ×ℝ⟶ℝ and M9, M11 hold. Then, the operator T:C´⟶C´ defined in (87) is completely continuous.Proof.
In view of continuity ofϕ1,ϕ2 and Gα,γσ,η, T is also continuous for all u,z∈C´. Suppose ℬ´⊆C´ is a bounded set. So, for every u∈ℬ´, we have(89)Tαu,vσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ+∫1ζ∫1σφασlnσηα−1−βyηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ+∫12∫1σlnσηα−1yηdηηdμσ+lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη.
Now, byM9, we have(90)yσ=−ϕ1σ,uσ,zσ⩽π4σ+π5σuσ+π6σzσ,=π4σ+π5σuσ+π6σϕ2σ,vσ,yσ⩽π4σ+π5σuσ+π6σπ7σ+π6σπ8σvσ+π6σπ9σyσ.
So, we obtain(91)yσ⩽π4σ+π6σπ7σ1−π6σπ9σ+π5σuσ+π6σπ8σvσ1−π6σπ9σ.
Now, using 2 of Lemma6, M11, and (91) in (89), we get(92)Tαu,vX×X⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗×ln2α−1∇αΓα+1αEαζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEαζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇ααln2+ln2⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗Yα.
In the same way, we obtain(93)Tγu,vX×X⩽π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗Yγ.
Thus, from (92) and (93), we get(94)Tu,vX×X⩽Tαu,vX×X+Tγu,vX×X⩽π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗Yα+π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗Yγ=ℳ.
Thus,T is uniformly bounded. Now, we prove the operator T is equicontinuous. For this, suppose σ1>σ2∈J and u,v∈ℬ´; then,(95)Tαu,vσ1−Tαu,vσ2,=−lnσ1α−1ΔαΓα∫1ζ∫12φασ1lnσ1α−1−βyηdηηdσ1+lnσ2α−1ΔαΓα∫1ζ∫12φασ2lnσ2α−1−βyηdηηdσ2−lnσ2α−1ΔαΓα∫1ζ∫1σ2φασ2lnσ2ηα−1−βyηdηηdσ2+lnσ1α−1ΔαΓα∫1ζ∫1σ1φασ1lnσ1ηα−1−βyηdηηdσ1−lnσ1α−1Γα−βΔαΓ2α∫12∫12lnσ1α−1yηdηηdμσ1+lnσ2α−1Γα−βΔαΓ2α∫12∫12lnσ2α−1yηdηηdμσ2−lnσ2α−1Γα−βΔαΓ2α∫12∫1σ2lnσ2ηα−1yηdηηdμσ2+lnσ1α−1Γα−βΔαΓ2α∫12∫1σ1lnσ1ηα−1yηdηηdμσ1−lnσ1α−1−lnσ2α−1Γα∫12yηdηη+1Γα∫σ2σ1lnσ1ηα−1yηdηη−1Γα∫1σ2lnσ2ηα−1−lnσ1ηα−1zηdηη≤π4∗+π6∗π7∗1−π6∗π9∗+π5∗uX+π6∗π8∗vX1−π6∗π9∗lnσ1α−1−lnσ2α−1∇Γα+1×αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+lnσ1α−lnσ2αΓα+1.
In the same way, we can show that(96)Tγu,vσ1−Tγu,vσ2⩽π7∗+π4∗π9∗1−π6∗π9∗+π5∗π9∗uX+π8∗vX1−π6∗π9∗lnσ1γ−1−lnσ2γ−1∇Γγ+1×γEξln2∑k=0γ−1−δ−1γ−1−δ−kγ−1−δ!k!lnξk+γEξγ−δ∑k=0γ−δ−1γ−δ−kγ−δ!k!lnξk+Plnργ−1Γγ−δγln2+lnρ+∇γln2+lnσ1γ−lnσ2γΓγ+1.
The right-hand sides of (94) and (96) approache to zero as σ1⟶σ2. Hence, by the Arzelà-Ascoli theorem, T is equicontinuous and uniformly equicontinuous. Also, it is very easy to prove that Tℬ´⊂ℬ´. Therefore, T is completely continuous.Theorem 9.
Under the hypothesisM10, M11 and(97)YαLϕ1+Lϕ1∗Lϕ2+YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗<1.
The coupled system (7) has a unique solution.Proof.
Letu,v,u^,v^∈C´, and we consider(98)Tαu,vσ−Tαu^,v^σ⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1dηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηηyη−y^η,where(99)yσ=−ϕ1σ,uσ,zσ,y^σ=−ϕ1σ,u^σ,z^σ.
Now, usingM10,(100)yσ−y^σ=−ϕ1σ,uσ,zσ−ϕ1σ,u^σ,z^σ⩽Lϕ1uσ−u^σ+Lϕ1∗zσ−z^σ,=Lϕ1uσ−u^σ+Lϕ1∗ϕ2σ,vσ,yσ−ϕ2σ,u^σ,y^σ⩽Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−u^σ+Lϕ2∗yσ−y^σ,=Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−u^σ1−Lϕ1∗Lϕ2∗.
Substituting (100) in (98) and taking supσ∈J, we get(101)Tαu,v−Tαu^,v^X×X⩽YαLϕ1+Lϕ1∗Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
In the same way, we can obtain(102)Tγu,v−Tγu^,v^X×X⩽YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
So, from (101) and (102), we get(103)Tu,v−Tu^,v^X×X⩽YαLϕ1+Lϕ1∗Lϕ2+YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗u,v−u^,v^X×X.
Thus,T is a contraction. Therefore, by the Banach contraction principle, T has a fixed point. So, we infer that the given coupled system (7) has a unique solution.Theorem 10.
In view of the continuity of the functionsϕ1,ϕ2 and supposing M9 and M13 with(104)N11∗=∫12|Gα2,η|π4∗+π6∗π7∗1−π6∗π9∗dηη<1,N12∗=∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη<1,N21∗=∫12Gγ2,ηπ7∗+π4∗π9∗1−π6∗π9∗dηη<1,N22∗=∫12Gγ2,ηπ5∗π9∗+π8∗1−π6∗π9∗dηη<1hold, the coupled system (7) has at least one solution.Proof.
Let a setℬ´ be defined as(105)ℬ´=u,v∈X×X:u,vX×X<ℛℬ´,where max2N11∗/1−2N12∗,2N21∗/1−2N22∗<ℛℬ´. Furthermore, the operator defined by T:ℬ´¯⟶C´ in (87) is completely continuous. Suppose u,v∈ℬ´; then, by definition of ℬ´, we have u,vX×X<ℛℬ´;(106)Tαu,zX×X⩽maxσ∈J∫12Gασ,ηϕ1η,uη,zηdηη⩽maxσ∈J∫12Gασ,ηπ4η+π6ηπ7η1−π6ηπ9ηdηη+maxσ∈J∫12Gασ,ηπ5ηuη+π6ηπ8ηvη1−π6ηπ9ηdηη⩽∫12Gα2,ηπ4∗+π6∗π7∗1−π6∗π9∗dηη+ℛℬ´∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη,=N11∗+ℛℬ´N12∗⩽ℛℬ´2.
Similarly,(107)Tγv,yX×X⩽ℛℬ´2.
Therefore,(108)Tu,vX×X⩽ℛℬ´,so Tu,z∈ℬ´¯. Thus, in view of Theorem 8, T:ℬ´¯⟶ℬ´¯ is completely continuous.
Now, we consider an eigenvalue problem defined as(109)u,v=δTu,v,δ∈0,1.
So, in view of the solution (u, v) of (109), we obtain(110)uX=δTαu,zX⩽maxσ∈J∫12Gασ,ηϕ1η,uη,zηdηη⩽maxσ∈J∫12Gασ,ηπ4η+π6ηπ7η1−π6ηπ9ηdηη+maxσ∈J∫12Gασ,ηπ5ηuη+π6ηπ8ηvη1−π6ηπ9ηdηη⩽∫12Gα2,ηπ4∗+π6∗π7∗1−π6∗π9∗dηη+ℛℬ´∫12Gα2,ηπ5∗+π6∗π8∗1−π6∗π9∗dηη,=N11∗+ℛℬ´N12∗⩽ℛℬ´2.
Similarly,(111)vX⩽ℛℬ´2.
Thus,(112)u,vX×X⩽ℛℬ´.
From equation (112), we get u,v∉∂ℬ´. So, in view of Theorem 4, T has at least one fixed point which lies in ℬ´¯. This shows there is at least one solution of the coupled system (7).


## 4. Hyers–Ulam Stability

In this section, we provide novel characterizations of the Hyers–Ulam stability for systems (5) and (7). For the various concepts of Hyers–Ulam stability, see, for example, [37].


### 4.1. Hyers–Ulam Stability Concepts for System (5)

Definition 3.
The problem (5) is said to be Hyers–Ulam stable if there exists some constant K>0 such that, for any ϵ>0 and for any solution u∈X of the inequality(113)DαHuσ+ϕσ,uσ,DαHuσ⩽ε,σ∈J,there exists a solution u^∈X of (5) with(114)uσ−u^σ⩽Kε,σ∈J.Definition 4.
The problem (5) is said to be generalized Hyers–Ulam stable if there exists Ψ∈Cℝ+,ℝ+ with Ψ0=0 such that, for any solution u∈X of the inequality (113), there exists a solution u^∈X of (5) satisfying(115)uσ−u^σ⩽Ψε,σ∈J.Definition 5.
The problem (5) is said to be Hyers–Ulam–Rassias stable with respect to Ψ∈CJ,ℝ if there exists some constant KΨ>0 such that, for any ϵ>0 and for any solution u∈X of the inequality(116)DαHuσ+ϕσ,uσ,DαHuσ⩽Ψσε,σ∈J,there exists a solution u^∈X of (5) with(117)uσ−u^σ⩽KΨΨσε,σ∈J.Definition 6.
The problem (5) is said to be generalized Hyers–Ulam–Rassias stable with respect to Ψ∈CJ,ℝ if there exists some constant KΨ>0 such that, for any solution u∈X of the inequality (116), there exists a solution u^∈X of (5) satisfying(118)uσ−u^σ⩽KΨΨσ,σ∈J.Remark 1.
Clearly,(1)
Definition3⇒ Definition 4(2)
Definition5⇒ Definition 6Remark 2.
Letu∈X be a solution of the inequality (113); then, there exists a function Uϕ∈CJ,ℝ depending on u such that(1
)Uϕσ⩽ϵ,for allσ∈J(2)
DαHuσ+ϕσ,uσ,DαHuσ=Uϕσ,σ∈JLemma 7.
Let2<α⩽3; if u∈C1J,ℝ is the solution of the inequality (113), then u will be the solution of the following integral inequality:(119)uσ−ω1σ⩽Yϵ.Proof.
Letu be the solution of the inequality (113). So, in view 2 of Remark 2, we have(120)DαHuσ+ϕσ,uσ,DαHuσ=Uϕσ,σ∈J,uσ|σ=1=HDα−2uσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφσDβHuσdσ.
So, forσ∈J, the solution of (120) will be in the form(121)uσ=∫12Gσ,ηϕη,uη,HDαuηdηη+∫12Gσ,ηUϕηdηη.
From equation (121), we have(122)uσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη+lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ−∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ−∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη−1Γα∫1σlnσηα−1Uϕηdηη.
For computational convenience, we useω1σ for the sum of terms which are free of Uϕ, so we have(123)ω1σ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη.
From above, we have(124)uσ−ω1σ⩽lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη+1Γα∫1σlnσηα−1Uϕηdηη⩽ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2ϵ.
Using (3) of Lemma4 and 1 of Remark 2, we get(125)uσ−ω1σ⩽Yε.Theorem 11.
Under the hypothesisM2, M4, and M5 and if(126)YLϕ1−Lϕ∗<1holds, then the given system (5) is HUS and, consequently, GHUS.Proof.
Letu∈CJ,ℝ be the solution of (113) and u^ be the unique solution of the system given by(127)DαHu^σ+ϕσ,u^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=HDα−2u^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφσDβHu^σdσ.
Then, forσ∈J, the solution of (127) is(128)u^σ=∫12Gσ,ηϕη,u^η,HDαu^ηdηη.
Consider(129)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ.
Using Lemma7 in (129), we have(130)uσ−u^σ⩽Yϵ+lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηηzuη−zu^η,where zu,zu^∈CJ,ℝ are of the form(131)zuσ=−ϕσ,uσ,zuσ,zu^σ=−ϕσ,u^σ,zu^σ.
ByM4, we get(132)zuσ−zu^σ=−ϕσ,uσ,zuσ−ϕσ,u^σ,zu^σ⩽Lϕuσ−u^σ+Lϕ∗zuσ−zu^σ,which implies(133)zuσ−zu^σ⩽Lϕ1−Lϕ∗uσ−u^σ.
Using (3) of Lemma4 and (133) in (130), we get(134)u−u^X⩽Yϵ+YLϕ1−Lϕ∗u−u^X,u−u^X⩽Y1−Lϕ∗1−Lϕ∗−YLϕϵ=Kϵ,where(135)K=Y1−Lϕ∗1−Lϕ∗−YLϕ,such that(136)YLϕ1−Lϕ∗<1.
Thus, problem (5) is HUS.Remark 3.
By settingΨϵ=Kϵ, Ψ0=0 in (134), by Definition 4, the given system (5) is GHUS.Lemma 8.
Let the hypothesisM6 hold, and suppose u∈CJ,ℝ is the solution of the inequality (116); then, u is a solution of the following inequality:(137)uσ−ω1σ⩽YΩΨΨσϵ.Proof.
From Lemma7, we have(138)uσ−ω1σ⩽lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη+1Γα∫1σlnσηα−1Uϕηdηη.
By using (3) of Lemma4, 1 of Remark 2, and M6, we get(139)uσ−ω1σ⩽ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2ΩΨΨσϵ⩽YΩΨΨσϵ.Theorem 12.
Under the hypothesisM2 and M4–M6 and if the inequality(140)1−Lϕ∗−YLϕ1−Lϕ∗>0holds, then the given system (5) is stable in the sense of ℋUℛ.Proof.
Letu∈CJ,ℝ be the solution of (113) and u^ be the unique solution of the system given by(141)DαHu^σ+ϕσ,u^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=HDα−2u^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφσDβHu^σdσ.
Then, forσ∈J, the solution of (141) is(142)u^σ=∫12Gσ,ηϕη,u^η,HDαu^ηdηη.
Consider(143)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ.
UsingM4 in a similar way as used in Theorem 11, we get(144)ω1σ−u^σ⩽YLϕ1−Lϕ∗uη−u^η.
Now, by Lemma8 and by (144), (143) becomes(145)u−u^X⩽YΩΨΨσϵ+YLϕ1−Lϕ∗u−u^X,which implies(146)u−u^X⩽1−Lϕ∗1−Lϕ∗−YLϕYΩΨΨσϵ.
Thus, we have(147)u−u^X⩽KΨΨσϵ,where(148)KΨ=1−Lϕ∗1−Lϕ∗−YLϕYΩΨ,1−Lϕ∗−YLϕ1−Lϕ∗>0.
Hence, the given system (5) is ℋUℛ stable.Remark 4.
Ifϵ=1 in (147), then by Definition 6, the given system (5) is GHURS.


### 4.2. Hyers–Ulam Stability Concepts for System (7)

Definition 7 (see [38]).
The given system (7) has HUS if there exists ℭα,γ=maxℭα,ℭγ>0 such that there exist some εα,γ=maxεα,εγ>0, and for every solution u,v∈X×X of the inequality(149)DαHuσ+ϕ1σ,uσ,DγHvσ⩽ϵα,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ⩽ϵγ,σ∈J,there exists a solution u^,v^∈X×X with(150)u,vσ−u^,v^σ⩽ℭα,γϵα,γ,σ∈J.Definition 8 (see [38]).
The given system (7) has GHUS if there exists ℭ′∈Cℝ+,ℝ+ with ℭ′0=0 such that, for any solution u,v∈X×X of the inequality (149), there exists a solution u^,v^∈X×X of (7) satisfying(151)u,vσ−u^,v^σ⩽ℭ′ϵα,γ,σ∈J.Definition 9 (see [38]).
The given system (7) has HURS with respect to Ψα,γ=maxΨα,Ψγ with Ψα,γ∈CJ,ℝ if there exists some constant ℭΨα,γ=maxℭΨα,ℭΨγ>0 such that, for any εα,γ=maxεα,εγ>0 and for any solution u,v∈X×X of the inequalities(152)DαHuσ+ϕ1σ,uσ,DγHvσ⩽Ψασϵα,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ⩽Ψγσϵγ,σ∈J,there exists a solution u^,v^∈X×X with(153)u,vσ−u^,v^σ⩽ℭΨα,γΨα,γσϵα,γ,σ∈J.Definition 10 (see [38]).
The given system (7) has GHURS with respect to Ψα,γ=maxΨα,Ψγ with Ψα,γ∈CJ,ℝ if there exists some constant ℭΨα,γ=maxℭΨα,ℭΨγ>0 such that, for any solution u^,v^∈X×X of the inequality (152), there exists a solution u,v∈X×X of (7) satisfying(154)u,vσ−u^,v^σ⩽ℭΨα,γΨα,γσ,σ∈J.Remark 5.
Letu,v∈X×X be a solution of the inequality (149), if there exist functions Uϕ1,Uϕ2∈CJ,ℝ depending on u,v, respectively, such that(1
)Uϕ1σ⩽ϵα,Uϕ2σ⩽ϵγ,σ∈J(2)
DαHuσ+ϕ1σ,uσ,DγHvσ=Uϕ1σ,σ∈JDγHvσ+ϕ2σ,vσ,DαHuσ=Uϕ2σ,σ∈JLemma 9.
Letu,v∈X×X be the solution of (149); then, for σ∈J, we have(155)uσ−ω1σ⩽Yαϵα,vσ−ω2σ⩽Yγϵγ.Proof.
By 2 of Remark5 and for σ∈J, we have(156)DαHuσ+ϕ1σ,uσ,DγHvσ=Uϕ1σ,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ=Uϕ2σ,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφασDβHuσdσ,vσ|σ=1=Dγ−2Hvσ|σ=1=0,Dγ−1Hvσ|σ=2=γv+∫1ξφγσDβHvσdσ.
So, forσ∈J, the solution of (156) will be in the form(157)uσ=∫12Gασ,ηϕ1η,uη,HDγvηdηη+∫12Gασ,ηUϕ1ηdηη,vσ=∫12Gγσ,ηϕ2η,vη,HDαuηdηη+∫12Gγσ,ηUϕ2ηdηη.
From the first equation of system (157), we have(158)uσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη+lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βUϕ1ηdηηdσ−∫1ζ∫1σφασlnσηα−1−βUϕ1ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1Uϕ1ηdηηdμσ−∫12∫1σlnσηα−1Uϕ1ηdηηdμσ+lnσα−1Γα∫12Uϕ1ηdηη−1Γα∫1σlnσηα−1Uϕ1ηdηη.
For computational convenience, we useω1σ for the sum of terms which are free of Uϕ1, so we have(159)ω1σ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη.
So, from the above and taking the absolute value, (158) becomes(160)uσ−ω1σ⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βUϕ1ηdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βUϕ1ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1Uϕ1ηdηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1Uϕ1ηdηηdμσ+lnσα−1Γα∫12Uϕ1ηdηη+1Γα∫1σlnσηα−1Uϕ1ηdηη.
Using2 of Lemma 6 and 1 of Remark 5, we get(161)uσ−ω1σ⩽Yαϵα.
In the same way, we have(162)vσ−ω2σ⩽Yγϵγ.Theorem 13.
Under the hypothesisM10 and if(163)1−K1K3>0holds, then the given system (7) is stable in the sense of ℋU.Proof.
Letu,v∈X×X be the solution of (149) and u^,v^∈X×X be the solution to the system(164)DαHu^σ+ϕ1σ,u^σ,DγHv^σ=0,σ∈J,DγHv^σ+ϕ2σ,v^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=Dα−2Hu^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφασDβHu^σdσ,v^σ|σ=1=Dγ−2Hv^σ|σ=1=0,Dγ−1Hv^σ|σ=2=γv^+∫1ξφγσDβHv^σdσ.
Then, forσ∈J, u^,v^ is the solution of (164), i.e.,(165)u^σ=∫12Gασ,ηϕ1η,u^η,DγHv^ηdηη,v^σ=∫12Gγσ,ηϕ2η,v^η,DαHu^ηdηη.
Consider(166)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ⩽Yαϵα+lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyη−yu^ηdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βyη−yu^ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yη−yu^ηdηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1yη−yu^ηdηηdμσ+lnσα−1Γα∫12yη−yu^ηdηη+1Γα∫1σlnσηα−1yη−yu^ηdηη,where y,yu^∈X are of the form(167)yσ=−ϕ1σ,uσzσ,yu^σ=−ϕ1σ,u^σ,zv^σ.
ByM10, we get(168)yσ−yu^σ=−ϕ1σ,uσ,zσ−ϕ1σ,u^σ,zv^σ⩽Lϕ1uσ−u^σ+Lϕ1∗zσ−zv^σ,and we obtain(169)yσ−yu^σ⩽Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−v^σ1−Lϕ1∗Lϕ2∗.
Using2 of Lemma 6 and (168) in (166), we get(170)u−u^X⩽Yαϵα+YαLϕ1u−u^X+Lϕ1∗Lϕ2v−v^X1−Lϕ1∗Lϕ2∗.
Similarly, we have(171)v−v^X⩽Yγϵγ+YγLϕ1Lϕ2∗u−u^X+Lϕ2v−v^X1−Lϕ1∗Lϕ2∗,where z,zv^∈X in the form(172)zσ=−ϕ2σ,vσ,yσ,zv^σ=−ϕ2σ,v^σ,yu^σ.
We write (170) and (171) as(173)u−u^X−YαLϕ1∗Lϕ21−Lϕ1∗Lϕ2∗−YαLϕ1v−v^X⩽Yαϵα1−YαLϕ1/1−Lϕ1∗Lϕ2∗,v−v^X−YγLϕ1Lϕ2∗1−Lϕ1∗Lϕ2∗−YγLϕ2u−u^X⩽Yγϵγ1−YγLϕ2/1−Lϕ1∗Lϕ2∗,respectively. Let(174)K1=YαLϕ1∗Lϕ21−Lϕ1∗Lϕ2∗−YαLϕ1,K2=Yα1−YαLϕ1/1−Lϕ1∗Lϕ2∗,K3=YγLϕ1Lϕ2∗1−Lϕ1∗Lϕ2∗−YγLϕ2andK4=Yγ1−YγLϕ2/1−Lϕ1∗Lϕ2∗.
In the matrix form, the abovementioned inequalities can be written as(175)1−K1−K31u−u^Xv−v^X⩽K2ϵαK4ϵγ.
Solving the abovementioned inequality, we have(176)u−u^Xv−v^X⩽11−K1K3K11−K1K3K31−K1K311−K1K3K2ϵαK4ϵγ.
Further simplification gives(177)u−u^X=K2ϵα1−K1K3+K1K4ϵγ1−K1K3,v−v^X=K4ϵγ1−K1K3+K2K3ϵα1−K1K3,from which we have(178)u−u^X+v−v^X=K2ϵα1−K1K3+K4ϵγ1−K1K3+K1K4ϵγ1−K1K3+K2K3ϵα1−K1K3.
Letεα,γ=maxεα,εγ; then, from (178), we have(179)u,v−u^,v^X×X⩽ℭα,γϵα,γ,where(180)ℭα,γ=K21−K1K3+K41−K1K3+K1K41−K1K3+K2K31−K1K3.Remark 6.
By settingℭ′ϵα,γ=ℭα,γϵα,γ, ℭ′0=0 in (179), by Definition 8, the given system (7) is GHUS.Remark 7.
Under the hypothesisM12 and (163) and by using Definitions 9 and 10, one can repeat the process of Lemma 9 and the Theorem 13, and the system (7) will be ℋUℛ and GHURS.


## 4.1. Hyers–Ulam Stability Concepts for System (5)

Definition 3.
The problem (5) is said to be Hyers–Ulam stable if there exists some constant K>0 such that, for any ϵ>0 and for any solution u∈X of the inequality(113)DαHuσ+ϕσ,uσ,DαHuσ⩽ε,σ∈J,there exists a solution u^∈X of (5) with(114)uσ−u^σ⩽Kε,σ∈J.Definition 4.
The problem (5) is said to be generalized Hyers–Ulam stable if there exists Ψ∈Cℝ+,ℝ+ with Ψ0=0 such that, for any solution u∈X of the inequality (113), there exists a solution u^∈X of (5) satisfying(115)uσ−u^σ⩽Ψε,σ∈J.Definition 5.
The problem (5) is said to be Hyers–Ulam–Rassias stable with respect to Ψ∈CJ,ℝ if there exists some constant KΨ>0 such that, for any ϵ>0 and for any solution u∈X of the inequality(116)DαHuσ+ϕσ,uσ,DαHuσ⩽Ψσε,σ∈J,there exists a solution u^∈X of (5) with(117)uσ−u^σ⩽KΨΨσε,σ∈J.Definition 6.
The problem (5) is said to be generalized Hyers–Ulam–Rassias stable with respect to Ψ∈CJ,ℝ if there exists some constant KΨ>0 such that, for any solution u∈X of the inequality (116), there exists a solution u^∈X of (5) satisfying(118)uσ−u^σ⩽KΨΨσ,σ∈J.Remark 1.
Clearly,(1)
Definition3⇒ Definition 4(2)
Definition5⇒ Definition 6Remark 2.
Letu∈X be a solution of the inequality (113); then, there exists a function Uϕ∈CJ,ℝ depending on u such that(1
)Uϕσ⩽ϵ,for allσ∈J(2)
DαHuσ+ϕσ,uσ,DαHuσ=Uϕσ,σ∈JLemma 7.
Let2<α⩽3; if u∈C1J,ℝ is the solution of the inequality (113), then u will be the solution of the following integral inequality:(119)uσ−ω1σ⩽Yϵ.Proof.
Letu be the solution of the inequality (113). So, in view 2 of Remark 2, we have(120)DαHuσ+ϕσ,uσ,DαHuσ=Uϕσ,σ∈J,uσ|σ=1=HDα−2uσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφσDβHuσdσ.
So, forσ∈J, the solution of (120) will be in the form(121)uσ=∫12Gσ,ηϕη,uη,HDαuηdηη+∫12Gσ,ηUϕηdηη.
From equation (121), we have(122)uσ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη+lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ−∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ−∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη−1Γα∫1σlnσηα−1Uϕηdηη.
For computational convenience, we useω1σ for the sum of terms which are free of Uϕ, so we have(123)ω1σ=−lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βzηdηηdσ−∫1ζ∫1σφσlnσηα−1−βzηdηηdσ−lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1zηdηηdμσ−∫12∫1σlnσηα−1zηdηηdμσ−lnσα−1Γα∫12zηdηη+1Γα∫1σlnσηα−1zηdηη.
From above, we have(124)uσ−ω1σ⩽lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη+1Γα∫1σlnσηα−1Uϕηdηη⩽ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2ϵ.
Using (3) of Lemma4 and 1 of Remark 2, we get(125)uσ−ω1σ⩽Yε.Theorem 11.
Under the hypothesisM2, M4, and M5 and if(126)YLϕ1−Lϕ∗<1holds, then the given system (5) is HUS and, consequently, GHUS.Proof.
Letu∈CJ,ℝ be the solution of (113) and u^ be the unique solution of the system given by(127)DαHu^σ+ϕσ,u^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=HDα−2u^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφσDβHu^σdσ.
Then, forσ∈J, the solution of (127) is(128)u^σ=∫12Gσ,ηϕη,u^η,HDαu^ηdηη.
Consider(129)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ.
Using Lemma7 in (129), we have(130)uσ−u^σ⩽Yϵ+lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1dηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1dηηdμσ+lnσα−1Γα∫12dηη+1Γα∫1σlnσηα−1dηηzuη−zu^η,where zu,zu^∈CJ,ℝ are of the form(131)zuσ=−ϕσ,uσ,zuσ,zu^σ=−ϕσ,u^σ,zu^σ.
ByM4, we get(132)zuσ−zu^σ=−ϕσ,uσ,zuσ−ϕσ,u^σ,zu^σ⩽Lϕuσ−u^σ+Lϕ∗zuσ−zu^σ,which implies(133)zuσ−zu^σ⩽Lϕ1−Lϕ∗uσ−u^σ.
Using (3) of Lemma4 and (133) in (130), we get(134)u−u^X⩽Yϵ+YLϕ1−Lϕ∗u−u^X,u−u^X⩽Y1−Lϕ∗1−Lϕ∗−YLϕϵ=Kϵ,where(135)K=Y1−Lϕ∗1−Lϕ∗−YLϕ,such that(136)YLϕ1−Lϕ∗<1.
Thus, problem (5) is HUS.Remark 3.
By settingΨϵ=Kϵ, Ψ0=0 in (134), by Definition 4, the given system (5) is GHUS.Lemma 8.
Let the hypothesisM6 hold, and suppose u∈CJ,ℝ is the solution of the inequality (116); then, u is a solution of the following inequality:(137)uσ−ω1σ⩽YΩΨΨσϵ.Proof.
From Lemma7, we have(138)uσ−ω1σ⩽lnσα−1ΔΓα∫1ζ∫12φσlnσα−1−βUϕηdηηdσ+lnσα−1ΔΓα∫1ζ∫1σφσlnσηα−1−βUϕηdηηdσ+lnσα−1Γα−βΔΓ2α∫12∫12lnσα−1Uϕηdηηdμσ+lnσα−1Γα−βΔΓ2α∫12∫1σlnσηα−1Uϕηdηηdμσ+lnσα−1Γα∫12Uϕηdηη+1Γα∫1σlnσηα−1Uϕηdηη.
By using (3) of Lemma4, 1 of Remark 2, and M6, we get(139)uσ−ω1σ⩽ln2α−1∇Γα+1αEζln2∑k=0α−1−β−1α−1−β−kα−1−β!k!lnζk+αEζα−β∑k=0α−β−1α−β−kα−β!k!lnζk+Plnρα−1Γα−βαln2+lnρ+∇αln2+ln2ΩΨΨσϵ⩽YΩΨΨσϵ.Theorem 12.
Under the hypothesisM2 and M4–M6 and if the inequality(140)1−Lϕ∗−YLϕ1−Lϕ∗>0holds, then the given system (5) is stable in the sense of ℋUℛ.Proof.
Letu∈CJ,ℝ be the solution of (113) and u^ be the unique solution of the system given by(141)DαHu^σ+ϕσ,u^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=HDα−2u^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφσDβHu^σdσ.
Then, forσ∈J, the solution of (141) is(142)u^σ=∫12Gσ,ηϕη,u^η,HDαu^ηdηη.
Consider(143)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ.
UsingM4 in a similar way as used in Theorem 11, we get(144)ω1σ−u^σ⩽YLϕ1−Lϕ∗uη−u^η.
Now, by Lemma8 and by (144), (143) becomes(145)u−u^X⩽YΩΨΨσϵ+YLϕ1−Lϕ∗u−u^X,which implies(146)u−u^X⩽1−Lϕ∗1−Lϕ∗−YLϕYΩΨΨσϵ.
Thus, we have(147)u−u^X⩽KΨΨσϵ,where(148)KΨ=1−Lϕ∗1−Lϕ∗−YLϕYΩΨ,1−Lϕ∗−YLϕ1−Lϕ∗>0.
Hence, the given system (5) is ℋUℛ stable.Remark 4.
Ifϵ=1 in (147), then by Definition 6, the given system (5) is GHURS.


## 4.2. Hyers–Ulam Stability Concepts for System (7)

Definition 7 (see [38]).
The given system (7) has HUS if there exists ℭα,γ=maxℭα,ℭγ>0 such that there exist some εα,γ=maxεα,εγ>0, and for every solution u,v∈X×X of the inequality(149)DαHuσ+ϕ1σ,uσ,DγHvσ⩽ϵα,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ⩽ϵγ,σ∈J,there exists a solution u^,v^∈X×X with(150)u,vσ−u^,v^σ⩽ℭα,γϵα,γ,σ∈J.Definition 8 (see [38]).
The given system (7) has GHUS if there exists ℭ′∈Cℝ+,ℝ+ with ℭ′0=0 such that, for any solution u,v∈X×X of the inequality (149), there exists a solution u^,v^∈X×X of (7) satisfying(151)u,vσ−u^,v^σ⩽ℭ′ϵα,γ,σ∈J.Definition 9 (see [38]).
The given system (7) has HURS with respect to Ψα,γ=maxΨα,Ψγ with Ψα,γ∈CJ,ℝ if there exists some constant ℭΨα,γ=maxℭΨα,ℭΨγ>0 such that, for any εα,γ=maxεα,εγ>0 and for any solution u,v∈X×X of the inequalities(152)DαHuσ+ϕ1σ,uσ,DγHvσ⩽Ψασϵα,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ⩽Ψγσϵγ,σ∈J,there exists a solution u^,v^∈X×X with(153)u,vσ−u^,v^σ⩽ℭΨα,γΨα,γσϵα,γ,σ∈J.Definition 10 (see [38]).
The given system (7) has GHURS with respect to Ψα,γ=maxΨα,Ψγ with Ψα,γ∈CJ,ℝ if there exists some constant ℭΨα,γ=maxℭΨα,ℭΨγ>0 such that, for any solution u^,v^∈X×X of the inequality (152), there exists a solution u,v∈X×X of (7) satisfying(154)u,vσ−u^,v^σ⩽ℭΨα,γΨα,γσ,σ∈J.Remark 5.
Letu,v∈X×X be a solution of the inequality (149), if there exist functions Uϕ1,Uϕ2∈CJ,ℝ depending on u,v, respectively, such that(1
)Uϕ1σ⩽ϵα,Uϕ2σ⩽ϵγ,σ∈J(2)
DαHuσ+ϕ1σ,uσ,DγHvσ=Uϕ1σ,σ∈JDγHvσ+ϕ2σ,vσ,DαHuσ=Uϕ2σ,σ∈JLemma 9.
Letu,v∈X×X be the solution of (149); then, for σ∈J, we have(155)uσ−ω1σ⩽Yαϵα,vσ−ω2σ⩽Yγϵγ.Proof.
By 2 of Remark5 and for σ∈J, we have(156)DαHuσ+ϕ1σ,uσ,DγHvσ=Uϕ1σ,σ∈J,DγHvσ+ϕ2σ,vσ,DαHuσ=Uϕ2σ,σ∈J,uσ|σ=1=Dα−2Huσ|σ=1=0,Dα−1Huσ|σ=2=αu+∫1ζφασDβHuσdσ,vσ|σ=1=Dγ−2Hvσ|σ=1=0,Dγ−1Hvσ|σ=2=γv+∫1ξφγσDβHvσdσ.
So, forσ∈J, the solution of (156) will be in the form(157)uσ=∫12Gασ,ηϕ1η,uη,HDγvηdηη+∫12Gασ,ηUϕ1ηdηη,vσ=∫12Gγσ,ηϕ2η,vη,HDαuηdηη+∫12Gγσ,ηUϕ2ηdηη.
From the first equation of system (157), we have(158)uσ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη+lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βUϕ1ηdηηdσ−∫1ζ∫1σφασlnσηα−1−βUϕ1ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1Uϕ1ηdηηdμσ−∫12∫1σlnσηα−1Uϕ1ηdηηdμσ+lnσα−1Γα∫12Uϕ1ηdηη−1Γα∫1σlnσηα−1Uϕ1ηdηη.
For computational convenience, we useω1σ for the sum of terms which are free of Uϕ1, so we have(159)ω1σ=−lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyηdηηdσ−∫1ζ∫1σφασlnσηα−1−βyηdηηdσ−lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yηdηηdμσ−∫12∫1σlnσηα−1yηdηηdμσ−lnσα−1Γα∫12yηdηη+1Γα∫1σlnσηα−1yηdηη.
So, from the above and taking the absolute value, (158) becomes(160)uσ−ω1σ⩽lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βUϕ1ηdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βUϕ1ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1Uϕ1ηdηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1Uϕ1ηdηηdμσ+lnσα−1Γα∫12Uϕ1ηdηη+1Γα∫1σlnσηα−1Uϕ1ηdηη.
Using2 of Lemma 6 and 1 of Remark 5, we get(161)uσ−ω1σ⩽Yαϵα.
In the same way, we have(162)vσ−ω2σ⩽Yγϵγ.Theorem 13.
Under the hypothesisM10 and if(163)1−K1K3>0holds, then the given system (7) is stable in the sense of ℋU.Proof.
Letu,v∈X×X be the solution of (149) and u^,v^∈X×X be the solution to the system(164)DαHu^σ+ϕ1σ,u^σ,DγHv^σ=0,σ∈J,DγHv^σ+ϕ2σ,v^σ,DαHu^σ=0,σ∈J,u^σ|σ=1=Dα−2Hu^σ|σ=1=0,Dα−1Hu^σ|σ=2=αu^+∫1ζφασDβHu^σdσ,v^σ|σ=1=Dγ−2Hv^σ|σ=1=0,Dγ−1Hv^σ|σ=2=γv^+∫1ξφγσDβHv^σdσ.
Then, forσ∈J, u^,v^ is the solution of (164), i.e.,(165)u^σ=∫12Gασ,ηϕ1η,u^η,DγHv^ηdηη,v^σ=∫12Gγσ,ηϕ2η,v^η,DαHu^ηdηη.
Consider(166)uσ−u^σ⩽uσ−ω1σ+ω1σ−u^σ⩽Yαϵα+lnσα−1ΔαΓα∫1ζ∫12φασlnσα−1−βyη−yu^ηdηηdσ+lnσα−1ΔαΓα∫1ζ∫1σφασlnσηα−1−βyη−yu^ηdηηdσ+lnσα−1Γα−βΔαΓ2α∫12∫12lnσα−1yη−yu^ηdηηdμσ+lnσα−1Γα−βΔαΓ2α∫12∫1σlnσηα−1yη−yu^ηdηηdμσ+lnσα−1Γα∫12yη−yu^ηdηη+1Γα∫1σlnσηα−1yη−yu^ηdηη,where y,yu^∈X are of the form(167)yσ=−ϕ1σ,uσzσ,yu^σ=−ϕ1σ,u^σ,zv^σ.
ByM10, we get(168)yσ−yu^σ=−ϕ1σ,uσ,zσ−ϕ1σ,u^σ,zv^σ⩽Lϕ1uσ−u^σ+Lϕ1∗zσ−zv^σ,and we obtain(169)yσ−yu^σ⩽Lϕ1uσ−u^σ+Lϕ1∗Lϕ2vσ−v^σ1−Lϕ1∗Lϕ2∗.
Using2 of Lemma 6 and (168) in (166), we get(170)u−u^X⩽Yαϵα+YαLϕ1u−u^X+Lϕ1∗Lϕ2v−v^X1−Lϕ1∗Lϕ2∗.
Similarly, we have(171)v−v^X⩽Yγϵγ+YγLϕ1Lϕ2∗u−u^X+Lϕ2v−v^X1−Lϕ1∗Lϕ2∗,where z,zv^∈X in the form(172)zσ=−ϕ2σ,vσ,yσ,zv^σ=−ϕ2σ,v^σ,yu^σ.
We write (170) and (171) as(173)u−u^X−YαLϕ1∗Lϕ21−Lϕ1∗Lϕ2∗−YαLϕ1v−v^X⩽Yαϵα1−YαLϕ1/1−Lϕ1∗Lϕ2∗,v−v^X−YγLϕ1Lϕ2∗1−Lϕ1∗Lϕ2∗−YγLϕ2u−u^X⩽Yγϵγ1−YγLϕ2/1−Lϕ1∗Lϕ2∗,respectively. Let(174)K1=YαLϕ1∗Lϕ21−Lϕ1∗Lϕ2∗−YαLϕ1,K2=Yα1−YαLϕ1/1−Lϕ1∗Lϕ2∗,K3=YγLϕ1Lϕ2∗1−Lϕ1∗Lϕ2∗−YγLϕ2andK4=Yγ1−YγLϕ2/1−Lϕ1∗Lϕ2∗.
In the matrix form, the abovementioned inequalities can be written as(175)1−K1−K31u−u^Xv−v^X⩽K2ϵαK4ϵγ.
Solving the abovementioned inequality, we have(176)u−u^Xv−v^X⩽11−K1K3K11−K1K3K31−K1K311−K1K3K2ϵαK4ϵγ.
Further simplification gives(177)u−u^X=K2ϵα1−K1K3+K1K4ϵγ1−K1K3,v−v^X=K4ϵγ1−K1K3+K2K3ϵα1−K1K3,from which we have(178)u−u^X+v−v^X=K2ϵα1−K1K3+K4ϵγ1−K1K3+K1K4ϵγ1−K1K3+K2K3ϵα1−K1K3.
Letεα,γ=maxεα,εγ; then, from (178), we have(179)u,v−u^,v^X×X⩽ℭα,γϵα,γ,where(180)ℭα,γ=K21−K1K3+K41−K1K3+K1K41−K1K3+K2K31−K1K3.Remark 6.
By settingℭ′ϵα,γ=ℭα,γϵα,γ, ℭ′0=0 in (179), by Definition 8, the given system (7) is GHUS.Remark 7.
Under the hypothesisM12 and (163) and by using Definitions 9 and 10, one can repeat the process of Lemma 9 and the Theorem 13, and the system (7) will be ℋUℛ and GHURS.


## 5. Examples

In this section, we present three examples to demonstrate the existence and stability of the obtained results.Example 1.
(181)D8/3Huσ+2+uσ+D8/3Huσ10eσ+25+uσ+D8/3Huσ=0,uσ|σ=1=D2/3Huσ|σ=1=0,D5/3Huσ|σ=2=αu+∫15σ8/3D2/3Huσdσ,where α=8/3,β=2/3,ζ=5,P=20,E=20,ρ=5/3,φσ=σ8/3⩾1, for σ∈1,2. Moreover, αu=u1/5/25.
Set(182)ϕσ,y,z=2+y+z10eσ+25+y+z,σ∈1,2,y,z∈ℝ.
Now, for anyy,z,y^,z^∈ℝ and σ∈1,2, we have(183)ϕσ,y,z−ϕσ,y^,z^⩽110e2y−y^+110e2z−z^.
Hence,M4 is satisfied with Lϕ=Lϕ∗=1/10e2.
Also, for anyy,z∈ℝ, we have(184)ϕσ,y,z⩽110eσ+22+y+z.
Hence,M3 is satisfied with(185)π1ϱ=15eσ+2,π2ϱ=π3ϱ=110eσ+2,where π1∗=1/5e2 and π2∗=π3∗=1/10e2.
From Theorem7, we use the inequality which are found as(186)Lϕ1−Lϕ∗Y≈0.0134252<1.
Hence, (181) has a unique solution.
Furthermore,K>0 with condition (126) holds, and(187)Lϕ1−Lϕ∗Y≈0.0134252<1.
Hence, with the help of Theorem11, the given system (181) is HUS and, hence, GHUS. Also, by checking the conditions of Theorem 12, it can be easily verified that the considered problem (181) is HURS and GHURS.Example 2.
(188)D5/2Huσ+1+uσ90σ+41+uσ+sinD5/2Huσ90+σ2=0,uσ|σ=1=D1/2Huσ|σ=1=0,D3/2Huσ|σ=2=αu+∫150σ5/2D1/2Huσdσ,where α=5/2,β=1/2,ζ=50,P=50,E=50,ρ=3/2,φσ=σ5/2⩾0, for σ∈1,2. Moreover, αu=u1/5/15.
Set(189)ϕσ,uσ,zσ=1+uσ90σ+41+uσ+sinD5/2Huσ90+σ2.
Now, for anyy,z,y^,z^∈ℝ and σ∈1,2, we have(190)ϕσ,y,z−ϕσ,y,z^≤190y−y^+190z−z^.
Hence, M4is satisfied withLϕ=Lϕ∗=1/90.
Also, for anyy,z∈ℝ, we have(191)ϕσ,y,z≤190σ+41+y+190+σ2z.
Hence,M3 is satisfied with(192)π1σ=π2σ=190σ+4,π3σ=190+σ2,where π1∗=π2∗=π3∗=1/90.
From Theorem7, we use the inequality which are found as(193)Lϕ1−Lϕ∗Y≈0.0160445<1.
Hence, (188) has a unique solution.
Furthermore,K>0 with condition (126) holds, and(194)Lϕ1−Lϕ∗Y≈0.0160445<1.
Hence, with the help of Theorem11, the given system (188) is HUS and, hence, GHUS. Also, by checking the conditions of Theorem 11, we can find that the considered problem (188) is HURS and GHURS.Example 3.
(195)DH8/3uσ+5+uσ+DH7/3vσ50eσ+452+uσ+DH7/3vσ=0,DH7/3vσ+σcosuσ−uσsinσ80+DH8/3vσ45+DH8/3vσ=0,uσ|σ=1=DH2/3uσ|σ=1=0,DH5/3uσ|σ=2=αu+∫12σ8/3DH2/3uσdσ,vσ|σ=1=DH1/3vσ|σ=1=0,DH4/3vσ|σ=2=αv+∫12σ7/3DH2/3vσdσ,where α=8/3,β=2/3,γ=7/3,δ=1/3,ζ=ξ=2,P=5,Eα=Eγ=1,ρ=3/2,φασ=σ8/3⩾0,φγσ=σ7/3⩾0, for σ∈1,2. Moreover, αu=αv=u1/5/25.
Set(196)ϕ1σ,uσ,DHγvσ=5+uσ+DH7/3vσ50eσ+452+uσ+DH7/3vσ,ϕ2σ,vσ,DHαuσ=σcosuσ−uσsinσ80+DH8/3vσ45+DH8/3vσ.
By simple computations, we found that(197)Lϕ1=Lϕ1∗=150e45,Lϕ2=Lϕ2∗=145.
From Theorem9, we use the inequality which are found as(198)YαLϕ1+Lϕ1∗Lϕ2+YγLϕ1Lϕ2∗+Lϕ21−Lϕ1∗Lϕ2∗≈0.0254987<1.
Hence, (195) has a unique solution, and(199)1−K1K3≈1>0.
Hence. with the help of Theorem13, the given system (195) is HUS. Also, by using M12 and Remark 6 and 7, the given system is GHUS, HURS, and GHURS.


## 6. Conclusions

We have obtained some appropriate conditions for the existence, uniqueness, and Ulam’s stabilities of the system (5) and its corresponding coupled system (7). The required results are obtained using the Banach contraction principle, Schaefer’s fixed point theorem, Arzela-Ascoli theorem, and Leray–Schauder of the cone type. Also, with the help of some sufficient conditions, we have derived different kinds of Ulam’s stabilities for the solution of the problem (5) and the coupled problem (7). Additionally, examples are given to support the main results.



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