Existence Results for a Coupled System of Nonlinear Fractional Hybrid Differential Equations with Homogeneous Boundary Conditions

Abstract

We study an existence result for the following coupled system of nonlinear fractional hybrid differential equations with homogeneous boundary conditions 𝐷^𝛼₀⁺[𝑥(𝑡)/𝑓(𝑡,𝑥(𝑡),𝑦(𝑡))]=𝑔(𝑡,𝑥(𝑡),𝑦(𝑡)), 𝐷^𝛼₀⁺[𝑦(𝑡)/𝑓(𝑡,𝑦(𝑡),𝑥(𝑡))]=𝑔(𝑡,𝑦(𝑡),𝑥(𝑡)),0<𝑡<1, and 𝑥(0)=𝑦(0)=0, where 𝛼∈(0,1) and 𝐷^𝛼₀⁺ denotes the Riemann-Liouville fractional derivative. The main tools in our study are the techniques associated to measures of noncompactness in the Banach algebras and a fixed point theorem of Darbo type.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modelling of a great number of processes which appear in physics, chemistry, aerodynamics, and so forth and involve also derivatives of fractional order. For details, see [1–5] and the references therein.

On the other hand, about the theory of hybrid differential equations, we refer to the paper [6] where the authors studied the hybrid differential equation of first order: 𝑑𝑥(𝑡)𝑑𝑡𝑓((𝑡((𝑡[),𝑥𝑡𝑡,𝑥))=𝑔𝑡,𝑥)),𝑡∈𝐽=0,𝑇₀=𝑥₀∈ℝ, (1) where 𝑓∈𝐶(𝐽×ℝ,ℝ⧵{0}) and 𝑔∈𝐶(𝐽×ℝ,ℝ).

In [7], the authors studied the fractional version of the abovementioned problem, that is, 𝐷^𝛼₀⁺𝑥(𝑡)𝑓((𝑡((𝑡𝑥(0)𝑡,𝑥))=𝑔𝑡,𝑥)),𝑡∈𝐽,0<𝛼<1,=0, (2) under the same assumptions on 𝑓 and 𝑔 in [6].

Recently, in [8], the authors studied the following fractional hybrid initial value problem with supremum: 𝐷^𝛼₀⁺𝑥(𝑡)𝑓𝑡,𝑥(𝑡),max_{0≤𝜏≤𝑡}|𝑥(𝜏((𝑡,𝑥(0))|=𝑔𝑡,𝑥))0<𝑡<1,=0, (3) where 0<𝛼<1, 𝑓∈𝐶([0,1]×ℝ×ℝ,ℝ⧵{0}), and 𝑔∈𝐶([0,1]×ℝ,ℝ).

The coupled systems involving fractional differential equations are very important because they occur in numerous problems of applied nature; for instance, see [9–13].

In this paper, we consider the following coupled system: 𝐷^𝛼₀⁺𝑥(𝑡)𝑓(𝑡)(𝑡)(𝑡)(𝑡),𝐷𝑡,𝑥,𝑦=𝑔𝑡,𝑥,𝑦_𝛼0+𝑦(𝑡)𝑓(𝑡)(𝑡)(𝑡)(𝑡),𝑥(0)(0)𝑡,𝑦,𝑥=𝑔𝑡,𝑦,𝑥0<𝑡<1,=𝑦=0, (4) where 𝛼∈(0,1) and 𝐷^𝛼₀⁺ is the standard Riemann-Liouville fractional derivative.

The main tool in our study is a fixed point theorem of Darbo type associated to measures of noncompactness.

2. Preliminaries

We begin this section with some definitions and results about fractional calculus.

Let 𝛼>0 and 𝑛=[𝛼]+1, where [𝛼] denotes the integer part of 𝛼. For a function 𝑓∶(0,∞)→ℝ, the Riemann-Liouville fractional integral of order 𝛼>0 of 𝑓 is defined as 𝐼^𝛼₀⁺𝑓(𝑡)=1Γ(𝛼)^𝑡₀𝑓(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠, (5) provided that the right side is pointwise defined on (0,∞).

The Riemann-Liouville fractional derivative of order 𝛼 of a continuous function 𝑓 is defined by 𝐷^𝛼₀⁺𝑓(𝑥)=1Γ()𝑑𝑛−𝛼𝑑𝑥^𝑛𝑛𝑥₀𝑓(𝑠)(𝑥−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (6) provided that the right side is pointwise defined on (0,∞).

The following lemma will be useful for our study, [14].

Lemma 1. Let ℎ∈𝐿¹(0,1) and 0<𝛼<1. Then,(a)𝐷^𝛼₀⁺𝐼^𝛼₀⁺ℎ(𝑥)=ℎ(𝑥); (7) (b)𝐼^𝛼₀⁺𝐷^𝛼₀⁺ℎ(𝑥)(𝑥)−𝐼=ℎ^01−𝛼+ℎ(𝑥)_𝑥=0Γ(𝛼)𝑥^𝛼−1𝑎.𝑒.𝑜𝑛(0,1). (8)

Lemma 2. Let 0<𝛼<1 and suppose that 𝑓∈𝐶([0,1],ℝ⧵{0}) and 𝑦∈𝐶[0,1]. Then, the unique solution of the fractional hybrid initial value problem 𝐷^𝛼₀⁺𝑥(𝑡)𝑓(𝑡)(𝑡)𝑥(0)=𝑦,0<𝑡<1=0, (9) is given by 𝑥(𝑡)=𝑓(𝑡)Γ(𝛼)^𝑡₀𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,𝑡∈[0,1]. (10)

Proof. Suppose that 𝑥(𝑡) is a solution of problem (9). Using the operator 𝐼^𝛼₀⁺ and taking into account Lemma 1, we get 𝐼^𝛼₀⁺𝐷^𝛼₀⁺𝑥(𝑡)𝑓(𝑡)=𝐼^𝛼₀⁺𝑦(𝑡), (11) or, equivalently, 𝑥(𝑡)𝑓(𝑡)−𝐼₀^1−𝛼+𝑥(𝑡)/𝑓(𝑡)_𝑡=0Γ(𝛼)𝑡^𝛼−1=𝐼^𝛼₀⁺𝑦(𝑡). (12) Since 𝑥(𝑡)/𝑓(𝑡)|_𝑡=0=𝑥(0)/𝑓(0)=0/𝑓(0)=0 (because 𝑓(0)≠0), we have 𝑥(𝑡)=𝑓(𝑡)𝐼^𝛼₀⁺𝑦(𝑡). (13) This means that 𝑥(𝑡)=𝑓(𝑡)Γ(𝛼)^𝑡₀𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠. (14) Conversely, suppose that 𝑥(𝑡) is given by 𝑥(𝑡)=𝑓(𝑡)Γ(𝛼)^𝑡₀𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,𝑡∈[0,1]. (15) This means that 𝑥(𝑡)=𝑓(𝑡)𝐼^𝛼₀⁺𝑦(𝑡),𝑡∈[0,1]. (16) Applying 𝐷^𝛼₀⁺ and taking into account Lemma 1 and that 𝑓(𝑡)≠0 for 𝑡∈[0,1], we obtain 𝐷^𝛼₀⁺𝑥(𝑡)𝑓(𝑡)=𝐷^𝛼₀⁺𝐼^𝛼₀⁺𝑦(𝑡)=𝑦(𝑡),0<𝑡<1. (17) Moreover, for 𝑡=0 in (16), we have 𝑥(0)=𝑓(0)⋅0=0. This completes the proof.

In the sequel, we recall some definitions and basic facts about measures of noncompactness.

Assume that 𝐸 is a real Banach space with norm ‖⋅‖ and zero element 𝜃. By 𝐵(𝑥,𝑟) we denote the closed ball in 𝐸 centered at 𝑥 with radius 𝑟. By 𝐵_𝑟 we denote the ball 𝐵(𝜃,𝑟). If 𝑋 is a nonempty subset of 𝐸, by the symbols 𝑋 and Conv𝑋 we denote the closure and the convex closure of 𝑋, respectively. By ‖𝑋‖ we denote the quantity ‖𝑋‖=sup{‖𝑥‖∶𝑥∈𝑋}. Finally, by 𝔐_𝐸 we will denote the family of all nonempty and bounded subsets of 𝐸 and by 𝔑_𝐸 we denote its subfamily consisting of all relatively compact subsets of 𝐸.

Definition 3. A mapping 𝜇∶𝔐_𝐸→ℝ₊=[0,∞) is said to be a measure of noncompactness in 𝐸 if it satisfies the following conditions.(a)The family ker𝜇={𝑋∈𝔐_𝐸∶𝜇(𝑋)=0} is nonempty and ker𝜇⊂𝔑_𝐸.(b)𝑋⊂𝑌⇒𝜇(𝑋)≤𝜇(𝑌).(c)𝜇(𝑋)=𝜇(Conv𝑋)=𝜇(𝑋).(d)𝜇(𝜆𝑋+(1−𝜆)𝑌)≤𝜆𝜇(𝑋)+(1−𝜆)𝜇(𝑌) for 𝜆∈[0,1].(e)If (𝑋_𝑛) is a sequence of closed subsets from 𝔐_𝐸 such that 𝑋_𝑛+1⊂𝑋_𝑛(𝑛≥1) and lim_𝑛→∞𝜇(𝑋_𝑛)=0, then 𝑋_∞=∩^∞_𝑛=1𝑋_𝑛≠𝜙.

The family ker𝜇 appearing in (a) is called the kernel of the measure of noncompactness 𝜇. Notice that the set 𝑋_∞ appearing in (e) is an element of ker𝜇. Indeed, since 𝜇(𝑋_∞)≤𝜇(𝑋_𝑛) for 𝑛=1,2,…, we infer that 𝜇(𝑋_∞)=0 and this says that 𝑋_∞∈ker𝜇.

An important theorem about fixed point theorem in the context of measures of noncompactness is the following Darbo’s fixed point theorem [15].

Theorem 4. Let 𝐶 be a nonempty, bounded, closed, and convex subset of a Banach space 𝐸 and let 𝑇∶𝐶→𝐶 be a continuous mapping. Suppose that there exists a constant 𝑘∈[0,1) such that 𝜇(𝑇(𝑋))≤𝑘𝜇(𝑋), (18) for any nonempty subset 𝑋 of 𝐶.
Then, 𝑇 has a fixed point.

A generalization of Theorem 4 which will be very useful in our study is the following theorem, due to Sadovskiǐ [16].

Theorem 5. Let 𝐶 be a nonempty, bounded, closed, and convex subset of a Banach space 𝐸 and let 𝑇∶𝐶→𝐶 be a continuous operator satisfying 𝜇(𝑇(𝑋))<𝜇(𝑋), (19) for any nonempty subset 𝑋 of 𝐶 with 𝜇(𝑋)>0.
Then, 𝑇 has a fixed point.

Next, we will assume that the space 𝐸 has structure of Banach algebra. By 𝑥𝑦 we will denote the product of two elements 𝑥,𝑦∈𝑋 and by 𝑋𝑌 we will denote the set defined by 𝑋𝑌={𝑥𝑦∶𝑥∈𝑋,𝑦∈𝑌}.

Definition 6. Let 𝐸 be a Banach algebra. We will say that a measure of noncompactness 𝜇 defined on 𝐸 satisfies condition (𝑚) if 𝜇(𝑋𝑌)≤‖𝑋‖𝜇(𝑌)+‖𝑌‖𝜇(𝑋), (20) for any 𝑋, 𝑌∈𝔐_𝐸.

This definition appears in [17].

In this paper, we will work in the space 𝐶[0,1] consisting of all real functions defined and continuous on [0,1] with the standard supremum norm ‖𝑥‖=sup{|𝑥(𝑡)|∶𝑡∈[0,1]}, (21) for 𝑥∈𝐶[0,1]. It is clear that (𝐶[0,1],‖⋅‖) is a Banach algebra, where the multiplication is defined as the usual product of real functions.

Next, we present the measure of noncompactness in 𝐶[0,1] which will be used later. Let us fix 𝑋∈𝔐_𝐶[0,1] and 𝜀>0. For 𝑥∈𝑋, we denote by 𝜔(𝑥,𝜀) the modulus of continuity of 𝑥; that is, 𝜔(𝑥,𝜀)=sup{|𝑥(𝑡)−𝑥(𝑠)|∶𝑡,𝑠∈[0,1],|𝑡−𝑠|≤𝜀}. (22) Put 𝜔(){𝜔()},𝜔𝑋,𝜀=sup𝑥,𝜀∶𝑥∈𝑋₀(𝑋)=lim_𝜀→0𝜔(𝑋,𝜀). (23) In [15], it is proved that 𝜔₀(𝑋) is a measure of noncompactness in 𝐶[0,1].

Proposition 7. The measure of noncompactness 𝜔₀ on 𝐶[0,1] satisfies condition (𝑚).

Proof. Fix 𝑋, 𝑌∈𝔐_𝐶[0,1], 𝜀>0, and 𝑡, 𝑠∈[0,1] with |𝑡−𝑠|≤𝜀. Then, we have 𝑥(𝑡)𝑦(𝑡)(𝑠)𝑦(𝑠≤𝑥(𝑡)𝑦(𝑡)(𝑡)𝑦(𝑠+𝑥(𝑡)𝑦(𝑠)(𝑠)𝑦(𝑠=|𝑥(𝑡𝑦(𝑡)(𝑠+𝑦(𝑠𝑥(𝑡)(𝑠≤‖𝑥‖𝜔+𝑦𝜔().−𝑥)−𝑥)−𝑥))|−𝑦))|−𝑥)|𝑦,𝜀𝑥,𝜀 (24) This means that 𝜔𝑥𝑦,𝜀≤‖𝑥‖𝜔𝑦,𝜀+𝑦𝜔(𝑥,𝜀), (25) and, therefore, 𝜔(𝑋𝑌,𝜀)≤‖𝑋‖𝜔(𝑌,𝜀)+‖𝑌‖𝜔(𝑋,𝜀). (26) Taking 𝜀→0, we get 𝜔₀(𝑋𝑌)≤‖𝑋‖𝜔₀(𝑌)+‖𝑌‖𝜔₀(𝑋). (27) This completes the proof.

Proposition 7 appears in [17] and we have given the proof for the paper is self-contained.

3. Main Results

We begin this section introducing the following class 𝒜 of functions: 𝒜=𝜑∶ℝ₊⟶ℝ₊∶𝜑isnondecreasingandlim_𝑛→∞𝜑^𝑛(𝑡)=0forany𝑡>0, (28) where 𝜑^𝑛 denotes the 𝑛-iteration of 𝜑.

Remark 8. Notice that if 𝜑∈𝒜, then 𝜑(𝑡)<𝑡, for any 𝑡>0. Indeed, in contrary case, we can find 𝑡₀>0 and 𝑡₀≤𝜑(𝑡₀). Since 𝜑 is nondecreasing, we have 0<𝑡₀≤𝜑𝑡₀≤𝜑²𝑡₀≤⋯≤𝜑^𝑛𝑡₀≤⋯, (29) and, therefore, 0<𝑡₀≤lim_𝑛→∞𝜑^𝑛(𝑡₀) and this contradicts the fact that 𝜑∈𝒜.
Moreover, the fact that 𝜑(𝑡)<𝑡 for any 𝑡>0 proves that if 𝜑∈𝒜, then 𝜑 is continuous at 𝑡₀=0.

Using Remark 8 and Theorem 5, we have the following fixed point theorem.

Theorem 9. Let 𝐶 be a nonempty, bounded, closed, and convex subset of a Banach space 𝐸 and let 𝑇∶𝐶→𝐶 be a continuous operator satisfying 𝜇(𝑇(𝑋))≤𝜑𝜇(𝑋), (30) for any nonempty subset 𝑋 of 𝐶, where 𝜑∈𝒜.
Then, 𝑇 has a fixed point.

Theorem 9 appears in [18], where the authors present a proof without using Theorem 5.

The following result which appears in [19] will be interesting in our study.

Theorem 10. Let 𝜇₁,𝜇₂,…,𝜇_𝑛 be measures of noncompactness in the Banach spaces 𝐸₁,𝐸₂,…,𝐸_𝑛, respectively. Suppose that 𝐹∶[0,∞)^𝑛→[0,∞) is a convex function such that 𝐹(𝑥₁,𝑥₂,…,𝑥_𝑛)=0 if and only if 𝑥_𝑖=0 for 𝑖=1,2,…,𝑛.
Then, 𝜇(𝑋)=𝐹𝜇₁𝑋₁,𝜇₂𝑋₂,…,𝜇_𝑛𝑋_𝑛 (31) defines a measure compactness in 𝐸₁×𝐸₂×⋯×𝐸_𝑛, where 𝑋_𝑖 denotes the natural projection of 𝑋 into 𝐸_𝑖, for 𝑖=1,2,…,𝑛.

Remark 11. As a consequence of Theorem 10, we have that if 𝜇 is a measure of noncompactness on a Banach space 𝐸 and we consider the function 𝐹∶[0,∞)×[0,∞)→[0,∞) defined by 𝐹(𝑥,𝑦)=max(𝑥,𝑦), then, since 𝐹 is convex and 𝐹(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0, 𝜇(𝑋)=max{𝜇(𝑋₁),𝜇(𝑋₂)} defines a measure of noncompactness in the space 𝐸×𝐸.

Next, we present the definition of a coupled fixed point.

Definition 12. An element (𝑥,𝑦)∈𝑋×𝑋 is said to be a coupled fixed point of a mapping 𝐺∶𝑋×𝑋→𝑋 if 𝐺(𝑥,𝑦)=𝑥 and 𝐺(𝑦,𝑥)=𝑦.

The following result is crucial for our study.

Theorem 13. Let Ω be a nonempty, bounded, closed, and convex subset of a Banach space 𝐸, and let 𝜇 be a measure of noncompactness in 𝐸. Suppose that 𝐺∶Ω×Ω→Ω is a continuous operator satisfying 𝜇𝐺𝑋₁×𝑋₂≤𝜑max𝜇𝑋₁,𝜇𝑋₂ (32) for all nonempty subsets 𝑋₁ and 𝑋₂ of Ω, where 𝜑∈𝒜.
Then, 𝐺 has at least a coupled fixed point.

Proof. Notice that, by Remark 11, 𝜇(𝑋)=max{𝜇(𝑋₁),𝜇(𝑋₂)} is a measure of noncompactness in the space 𝐸×𝐸, where 𝑋₁ and 𝑋₂ are the projections of 𝑋 into 𝐸.
Now, we consider the mapping 𝐺∶Ω×Ω→Ω×Ω defined by 𝐺(𝑥,𝑦)=(𝐺(𝑥,𝑦),𝐺(𝑦,𝑥)). It is easily seen that Ω×Ω is a nonempty, bounded, closed, and convex subset of 𝐸×𝐸. Since 𝐺 is continuous, it is clear that 𝐺 is also continuous.
Next, we take a nonempty 𝑋 of Ω×Ω. Then, 𝜇𝐺(𝑋)=𝜇𝐺𝑋₁×𝑋₂=𝜇𝐺𝑋₁×𝑋₂×𝐺𝑋₂×𝑋₁=max𝜇𝐺𝑋₁×𝑋₂,𝜇𝐺𝑋₂×𝑋₁≤max𝜑max𝜇𝑋₁,𝜇𝑋₂,𝜑max𝜇𝑋₂,𝜇𝑋₁=𝜑max𝜇𝑋₁,𝜇𝑋₂=𝜑𝜇𝑋₁×𝑋₂. (33) Since 𝜑∈𝒜, by Theorem 9, the mapping 𝐺 has at least one fixed point. This means that there exists (𝑥₀,𝑦₀)∈Ω×Ω such that 𝐺(𝑥₀,𝑦₀)=(𝑥₀,𝑦₀) or, equivalently, 𝐺(𝑥₀,𝑦₀)=𝑥₀ and 𝐺(𝑦₀,𝑥₀)=𝑦₀. This proves that 𝐺 has at least a coupled fixed point.

Now, we consider the coupled system of integral equations: 𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝑦(𝑡)Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑦(𝑡)=𝑓(𝑡)(𝑡)𝑑𝑠,𝑡,𝑦,𝑥Γ(𝛼)^𝑡₀𝑔𝑠,𝑦(𝑠),𝑥(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,𝑡∈[0,1]. (34)

Lemma 14. Assume that 𝑓∈𝐶([0,1]×ℝ×ℝ,ℝ⧵{0}) and 𝑔∈𝐶([0,1]×ℝ×ℝ,ℝ). Then, (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] is a solution of (34) if and only if (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] is a solution of (4).

Proof. The proof is an immediate consequence of Lemma 2, so we omit it.

Next, we will study problem (4) under the following assumptions.(H₁)𝑓∈𝐶([0,1]×ℝ×ℝ,ℝ⧵{0}) and 𝑔∈𝐶([0,1]×ℝ×ℝ,ℝ).(H₂)The functions 𝑓 and 𝑔 satisfy 𝑓𝑡,𝑥₁,𝑦₁−𝑓𝑡,𝑥₂,𝑦₂≤𝜑₁max𝑥₁−𝑥₂,𝑦₁−𝑦₂,𝑔𝑡,𝑥₁,𝑦₂−𝑔𝑡,𝑥₂,𝑦₂≤𝜑₂max𝑥₁−𝑦₁,𝑥₂−𝑦₂, (35) respectively, for any 𝑡∈[0,1] and 𝑥₁, 𝑥₂, 𝑦₁, 𝑦₂∈ℝ, where 𝜑₁, 𝜑₂∈𝒜 and 𝜑₁ is continuous. Notice that assumption (H₁) gives us the existence of two nonnegative constants 𝑘₁ and 𝑘₂ such that |𝑓(𝑡,0,0)|≤𝑘₁ and |𝑔(𝑡,0,0)|≤𝑘₂, for any 𝑡∈[0,1].(H₃)There exists 𝑟₀>0 satisfying the inequalities 𝜑₁(𝑟)+𝑘₁⋅𝜑₂(𝑟)+𝑘₂(),𝜑≤𝑟Γ𝛼+1₂(𝑟)+𝑘₂≤Γ(𝛼+1). (36)

Theorem 15. Under assumptions (𝐻₁)–(𝐻₃), problem (4) has at least one solution in 𝐶[0,1]×𝐶[0,1].

Proof. In virtue of Lemma 14, a solution (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] of problem (4) satisfies 𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝑦(𝑡)Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑦(𝑡)=𝑓(𝑡)(𝑡)𝑑𝑠,𝑡,𝑦,𝑥Γ(𝛼)^𝑡₀𝑔𝑠,𝑦(𝑠),𝑥(𝑠)(𝑡−𝑠)^1−𝛼[].𝑑𝑠,𝑡∈0,1 (37) We consider the space 𝐶[0,1]×𝐶[0,1] equipped with the norm ‖(𝑥,𝑦)‖_{𝐶[0,1]×𝐶[0,1]}=max{‖𝑥‖,‖𝑦‖}, for any (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1].
In 𝐶[0,1]×𝐶[0,1], we define the operator 𝐺(𝑡)=𝑓(𝑡)(𝑡)𝑥,𝑦𝑡,𝑥,𝑦Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼[].𝑑𝑠,𝑡∈0,1 (38) Let ℱ and 𝒢 be the operators given by ℱ(𝑡)(𝑡)(𝑡),𝒢(𝑡)=1𝑥,𝑦=𝑓𝑡,𝑥,𝑦𝑥,𝑦Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠 (39) for any (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] and any 𝑡∈[0,1]. Then, 𝐺𝑥,𝑦=ℱ𝑥,𝑦⋅𝒢𝑥,𝑦. (40) Firstly, we will prove that 𝐺 applies 𝐶[0,1]×𝐶[0,1] into 𝐶[0,1]. To do this, it is sufficient to prove that ℱ(𝑥,𝑦), 𝒢(𝑥,𝑦)∈𝐶[0,1] for any (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] since the product of continuous functions is continuous.
In virtue of assumption (H₁), it is clear that ℱ(𝑥,𝑦)∈𝐶[0,1] for (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1]. In order to prove that 𝒢(𝑥,𝑦)∈𝐶[0,1] for (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1], we fix 𝑡₀∈[0,1] and we consider a sequence (𝑡_𝑛)∈[0,1] such that 𝑡_𝑛→𝑡₀, and we have to prove that 𝒢(𝑥,𝑦)(𝑡_𝑛)→𝒢(𝑥,𝑦)(𝑡₀). Without loss of generality, we can suppose that 𝑡_𝑛>𝑡₀. Then, we have 𝒢𝑥,𝑦𝑡_𝑛−𝒢𝑥,𝑦𝑡₀=1Γ(𝛼)^𝑡𝑛₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡_𝑛−𝑠^1−𝛼−𝑑𝑠_𝑡00𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₀−𝑠^1−𝛼≤1𝑑𝑠Γ(𝛼)^𝑡𝑛₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡_𝑛−𝑠^1−𝛼−𝑑𝑠_𝑡𝑛0𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₀−𝑠^1−𝛼+1𝑑𝑠Γ(𝛼)^𝑡𝑛₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₀−𝑠^1−𝛼−𝑑𝑠_𝑡00𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₀−𝑠^1−𝛼≤1𝑑𝑠Γ(𝛼)^𝑡𝑛₀𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1×𝑔(𝑠)(𝑠)+1𝑠,𝑥,𝑦𝑑𝑠Γ(𝛼)^𝑡𝑛_𝑡0𝑡₀−𝑠^𝛼−1𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠. (41) By assumption (H₁), since 𝑔∈𝐶([0,1]×ℝ×ℝ,ℝ), 𝑔 is bounded on the compact set [0,1]×[−‖𝑥‖,‖𝑥‖]×[−‖𝑦‖,‖𝑦‖]. Denote by 𝑀=sup𝑔𝑠,𝑥₁,𝑦₁∶𝑠∈[0,1],𝑥₁∈[−‖𝑥‖,‖𝑥,𝑦‖]₁∈−𝑦,𝑦. (42) From the last estimate, we obtain 𝒢𝑥,𝑦𝑡_𝑛−𝒢𝑥,𝑦𝑡₀≤𝑀Γ(𝛼)^𝑡𝑛₀𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1+𝑀𝑑𝑠Γ(𝛼)^𝑡𝑛_𝑡0𝑡₀−𝑠^𝛼−1𝑑𝑠. (43) As 0<𝛼<1 and 𝑡_𝑛>𝑡₀, we infer that 𝒢𝑥,𝑦𝑡_𝑛−𝒢𝑥,𝑦𝑡₀≤𝑀Γ(𝛼)^𝑡0₀𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1+𝑑𝑠_{𝑡𝑛𝑡0}𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1+𝑀𝑑𝑠Γ(𝛼)^𝑡𝑛_𝑡01𝑠−𝑡₀^1−𝛼=𝑀𝑑𝑠Γ(𝛼)^𝑡0₀𝑡₀−𝑠^𝛼−1−𝑡_𝑛−𝑠^𝛼−1+𝑑𝑠_{𝑡𝑛𝑡0}𝑑𝑠𝑡_𝑛−𝑠^1−𝛼+^𝑡𝑛_𝑡0𝑑𝑠𝑠−𝑡₀^1−𝛼+𝑀Γ(𝛼)^𝑡𝑛_𝑡01𝑠−𝑡₀^1−𝛼≤𝑀𝑑𝑠Γ()𝑡𝛼+1^𝑛−𝑡₀^𝛼+𝑡^𝛼₀−𝑡^𝛼_𝑛+𝑡_𝑛−𝑡₀^𝛼+𝑡_𝑛−𝑡₀^𝛼+𝑀Γ()𝑡𝛼+1^𝑛−𝑡₀^𝛼=4𝑀Γ()𝑡𝛼+1^𝑛−𝑡₀^𝛼+𝑀Γ()𝑡𝛼+1^𝛼0−𝑡^𝛼_𝑛<4𝑀Γ()𝑡𝛼+1^𝑛−𝑡₀^𝛼, (44) where the last inequality has been obtained by using the fact that 𝑡^𝛼₀−𝑡^𝛼_𝑛<0.
Therefore, since 𝑡_𝑛→𝑡₀, from the last estimate, we deduce that 𝒢(𝑥,𝑦)(𝑡_𝑛)→𝒢(𝑥,𝑦)(𝑡₀). This proves that 𝒢(𝑥,𝑦)∈𝐶[0,1]. Consequently, 𝒢∶𝐶[0,1]×𝐶[0,1]→𝐶[0,1]. On the other hand, for (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] and 𝑡∈𝐶[0,1], we have 𝐺(𝑡=ℱ𝑡)𝑡=𝑓(𝑡)(𝑡)1𝑥,𝑦)𝑥,𝑦(⋅𝒢𝑥,𝑦()𝑡,𝑥,𝑦Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼≤𝑓(𝑡)(𝑡)(+𝑓(×1𝑑𝑠𝑡,𝑥,𝑦−𝑓𝑡,0,0)𝑡,0,0)Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)−𝑔(𝑠,0,0)(𝑡−𝑠)^1−𝛼𝑑𝑠+^𝑡₀𝑔(𝑠,0,0)(𝑡−𝑠)^1−𝛼≤1𝑑𝑠Γ(𝛼)𝜑₁max|𝑥(𝑡)|,𝑦(𝑡)+𝑘₁×^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)−𝑔(𝑠,0,0)(𝑡−𝑠)^1−𝛼+𝑑𝑠_𝑡0𝑔(𝑠,0,0)(𝑡−𝑠)^1−𝛼≤1𝑑𝑠Γ(𝛼)𝜑₁max‖𝑥‖,𝑦+𝑘₁×^𝑡₀𝜑₂max|𝑥(𝑠)|,𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠+𝑘₂^𝑡₀𝑑𝑠(𝑡−𝑠)^1−𝛼≤1Γ(𝛼)𝜑₁max‖𝑥‖,𝑦+𝑘₁⋅𝜑₂max‖𝑥‖,𝑦+𝑘₂^𝑡₀𝑑𝑠(𝑡−𝑠)^1−𝛼≤1Γ()𝜑𝛼+1¹𝑥,𝑦+𝑘₁⋅𝜑₂𝑥,𝑦+𝑘₂. (45) Now, taking into account assumption (H₃), we infer that the operator 𝐺 applies 𝐵_𝑟0×𝐵_𝑟0 into 𝐵_𝑟0. Moreover, from the last estimates, it follows that ℱ𝐵_𝑟0×𝐵_𝑟0≤𝜑₁𝑟₀+𝑘₁,𝒢𝐵_𝑟0×𝐵_𝑟0≤𝜑₂𝑟₀+𝑘₂Γ().𝛼+1 (46)
Next, we will prove that the operators ℱ and 𝒢 are continuous on the ball 𝐵_𝑟0×𝐵_𝑟0 and, consequently, 𝐺 will be also continuous.
In fact, we fix 𝜀>0 and we take (𝑥₀,𝑦₀), (𝑥,𝑦)∈𝐵_𝑟0×𝐵_𝑟0 with ‖(𝑥,𝑦)−(𝑥₀,𝑦₀)‖ = ‖(𝑥−𝑥₀,𝑦−𝑦₀)‖ = max{‖𝑥−𝑥₀‖,‖𝑦−𝑦₀‖}≤𝜀. Then, for 𝑡∈[0,1], we have ℱ𝑥,𝑦(𝑡)−ℱ𝑥₀,𝑦₀(𝑡=𝑓(𝑡)(𝑡))𝑡,𝑥,𝑦−𝑓𝑡,𝑥₀(𝑡),𝑦₀(𝑡)≤𝜑₁max𝑥(𝑡)−𝑥₀(𝑡),𝑦(𝑡),𝑦₀(𝑡)≤𝜑₁max𝑥−𝑥₀,𝑦−𝑦₀≤𝜑₁(𝜀)<𝜀, (47) where we have used Remark 8. This proves the continuity of ℱ on 𝐵_𝑟0×𝐵_𝑟0.
In order to prove the continuity of 𝒢 on 𝐵_𝑟0×𝐵_𝑟0, we have 𝒢𝑥,𝑦(𝑡)−𝒢𝑥₀,𝑦₀(𝑡=1)Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠−^𝑡₀𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)(𝑡−𝑠)^1−𝛼≤1𝑑𝑠Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)(𝑡−𝑠)^1−𝛼≤1𝑑𝑠Γ(𝛼)^𝑡₀𝜑₂max𝑥(𝑠)−𝑥₀(𝑠),𝑦(𝑠)−𝑦₀(𝑠)(𝑡−𝑠)^1−𝛼≤1𝑑𝑠Γ(𝛼)𝜑₂max𝑥−𝑥₀,𝑦−𝑦₀^𝑡₀𝑑𝑠(𝑡−𝑠)^1−𝛼≤1Γ()𝜑𝛼+1²(𝜀)<𝜀Γ().𝛼+1 (48) Therefore, 𝒢𝑥,𝑦−𝒢𝑥₀,𝑦₀<𝜀Γ(𝛼+1) (49) and, consequently, 𝒢 is a continuous operator on 𝐵_𝑟0×𝐵_𝑟0.
In order to prove that 𝒢 satisfies assumptions of Theorem 13, only we have to check the condition 𝜇𝐺𝑋₁×𝑋₂≤𝜑max𝜇𝑋₁,𝜇𝑋₂ (50) for any subsets 𝑋₁ and 𝑋₂ of 𝐵_𝑟0.
To do this, we fix 𝜀>0, 𝑡₁, 𝑡₂∈[0,1] with |𝑡₁−𝑡₂|≤𝜀 and (𝑥,𝑦)∈𝑋₁×𝑋₂; then, we have ℱ𝑥,𝑦𝑡₁−ℱ𝑥,𝑦𝑡₂=𝑓𝑡₁,𝑥𝑡₁,𝑦𝑡₁−𝑓𝑡₂,𝑥𝑡₂,𝑦𝑡₂≤𝑓𝑡₁,𝑥𝑡₁,𝑦𝑡₁−𝑓𝑡₁,𝑥𝑡₂,𝑦𝑡₂+𝑓𝑡₁,𝑥𝑡₂,𝑦𝑡₂−𝑓𝑡₂,𝑥𝑡₂,𝑦𝑡₂≤𝜑₁max𝑥𝑡₁−𝑥𝑡₂,𝑦𝑡₁−𝑦𝑡₂+𝜔𝑓,𝜀≤𝜑₁max𝜔(𝑥,𝜀),𝜔𝑦,𝜀+𝜔𝑓,𝜀, (51) where 𝜔(𝑓,𝜀) denotes the quantity 𝜔𝑓[],||𝑓,𝜀=sup𝑡,𝑥,𝑦−𝑓𝑠,𝑥,𝑦∶𝑡,𝑠∈0,1𝑡−𝑠≤𝜀,𝑥,𝑦∈−𝑟₀,𝑟₀. (52) From the last estimate, we infer that 𝜔ℱ𝑋₁×𝑋₂,𝜀≤𝜑₁max𝜔𝑋₁,𝜀,𝜔𝑋₂,𝜀+𝜔𝑓,𝜀. (53) Since 𝑓(𝑡,𝑥,𝑦) is uniformly continuous on bounded subsets of [0,1]×ℝ×ℝ, we deduce that 𝜔(𝑓,𝜀)→0 as 𝜀→0 and, therefore, 𝜔₀ℱ𝑋₁×𝑋₂≤lim_𝜀→0𝜑₁max𝜔𝑋₁,𝜀,𝜔𝑋₂,𝜀. (54) By assumption (H₂), since 𝜑₁ is continuous, we infer 𝜔₀ℱ𝑋₁×𝑋₂≤𝜑₁max𝜔₀𝑋₁,𝜔₀𝑋₂. (55) Now, we estimate the quantity 𝜔₀(𝒢(𝑋₁×𝑋₂)).
Fix 𝜀>0, 𝑡₁, 𝑡₂∈[0,1] with |𝑡₁−𝑡₂|≤𝜀 and (𝑥,𝑦)∈𝑋₁×𝑋₂. Without loss of generality, we can suppose that 𝑡₁<𝑡₂; then, we have 𝒢𝑥,𝑦𝑡₂−𝒢𝑥,𝑦𝑡₁=1Γ(𝛼)^𝑡2₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₂−𝑠^1−𝛼−𝑑𝑠_𝑡10𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑡₁−𝑠^1−𝛼≤1𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₂−𝑠^𝛼−1−𝑡₁−𝑠^𝛼−1×𝑔((𝑠)(+𝑠,𝑥,𝑥𝜎𝑠))𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝛼−1𝑔(𝑠)(𝑠)≤1𝑠,𝑥,𝑦𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₁−𝑠^𝛼−1−𝑡₂−𝑠^𝛼−1×𝑔((𝑠)(+𝑠,𝑥,𝑥𝜎𝑠))𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝛼−1𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠. (56) Since 𝑔∈𝐶([0,1]×ℝ×ℝ,×ℝ) is bounded on the compact subsets of [0,1]×ℝ×ℝ, particularly, on [0,1]×[−𝑟₀,𝑟₀]×[−𝑟₀,𝑟₀]. Put 𝐿=sup{|𝑔(𝑡,𝑥,𝑦)|∶𝑡∈[0,1], 𝑥, 𝑦∈[−𝑟₀,𝑟₀]}. Then, from the last inequality, we infer that 𝒢𝑥,𝑦𝑡₂−𝒢𝑥,𝑦𝑡₁≤𝐿Γ(𝛼)^𝑡1₀𝑡₁−𝑠^𝛼−1−𝑡₂−𝑠^𝛼−1+𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝛼−1≤𝐿𝑑𝑠Γ()𝑡𝛼+1²−𝑡₁^𝛼+𝑡^𝛼₁−𝑡^𝛼₂+𝑡₂−𝑡₁^𝛼≤2𝐿Γ()𝑡𝛼+1²−𝑡₁^𝛼≤2𝐿Γ()𝜀𝛼+1^𝛼, (57) where we have used the fact that 𝑡^𝛼₁−𝑡^𝛼₂≤0. Therefore, 𝜔𝒢𝑋₁×𝑋₂,𝜀≤2𝐿Γ()𝜀𝛼+1^𝛼. (58) From this, it follows that 𝜔₀𝒢𝑋₁×𝑋₂=0. (59)
Next, by Proposition 7, (46), (55), and (59), we have 𝜔₀𝐺𝑋₁×𝑋₂=𝜔₀ℱ𝑋₁×𝑋₂⋅𝒢𝑋₁×𝑋₂≤ℱ𝑋₁×𝑋₂𝜔₀𝒢𝑋₁×𝑋₂+𝒢𝑋₁×𝑋₂𝜔₀ℱ𝑋₁×𝑋₂≤ℱ𝐵_𝑟0×𝐵_𝑟0𝜔₀𝒢𝑋₁×𝑋₂+𝒢𝐵_𝑟0×𝐵_𝑟0𝜔₀ℱ𝑋₁×𝑋₂≤𝜑₂𝑟₀+𝑘₂Γ()𝜑𝛼+1¹max𝜔₀𝑋₁,𝜔₀𝑋₂. (60) By assumption (H₃), since 𝜑₂(𝑟₀)+𝑘₂≤Γ(𝛼+1) and since it is easily proved that if 𝛼∈[0,1] and 𝜑∈𝒜, then 𝛼𝜑∈𝒜, we deduce that 𝜔₀𝐺𝑋₁×𝑋₂≤𝜑max𝜔₀𝑋₁,𝜔₀𝑋₂, (61) where 𝜑∈𝒜.
Finally, by Theorem 13, the operator 𝒢 has at least a coupled fixed point and this is the desired result. This completes the proof.

The nonoscillary character of the solutions of problem (4) seems to be an interesting question from the practical point of view. This means that the solutions of problem (4) have a constant sign on the interval (0,1). In connection with this question, we notice that if 𝑓(𝑡,𝑥,𝑦) and 𝑔(𝑡,𝑥,𝑦) have constant sign and are equal (this means that 𝑓(𝑡,𝑥,𝑦)>0 and 𝑔(𝑡,𝑥,𝑦)≥0 or 𝑓(𝑡,𝑥,𝑦)<0 and 𝑔(𝑡,𝑥,𝑦)≤0 for any 𝑡∈[0,1] and 𝑥, 𝑦∈ℝ) and under assumptions of Theorem 15, then the solution (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] of problem (4) given by Theorem 15 satisfies 𝑥(𝑡)≥0 and 𝑦(𝑡)≥0 for 𝑡∈[0,1], since the solution (𝑥,𝑦) satisfies the system of integral equations 𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝑦(𝑡)Γ(𝛼)^𝑡₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)(𝑡−𝑠)^1−𝛼𝑦(𝑡)=𝑓(𝑡)(𝑡)𝑑𝑠,𝑡,𝑦,𝑥Γ(𝛼)^𝑡₀𝑔𝑠,𝑦(𝑠),𝑥(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,0≤𝑡≤1. (62) On the other hand, if we perturb the data function in problem (4) of the following manner: 𝐷^𝛼₀⁺𝑥(𝑡)𝑓(𝑡)(𝑡)(𝑡)(𝑡)(𝑡),𝐷𝑡,𝑥,𝑦=𝑔𝑡,𝑥,𝑦+𝜂_𝛼0+𝑦(𝑡)𝑓(𝑡)(𝑡)(𝑡)(𝑡)(𝑡),𝑡,𝑦,𝑥=𝑔𝑡,𝑦,𝑥+𝜂hh0<𝑡<1, (63) where 0<𝛼<1, 𝑓∈𝐶([0,1]×ℝ×ℝ,ℝ⧵{0}), 𝑔∈𝐶([0,1]×ℝ×ℝ,ℝ), and 𝜂∈𝐶[0,1], then, under assumptions of Theorem 15, problem (63) can be studied by using Theorem 15, where assumptions (H₁) and (H₂) are automatically satisfied and we only have to check assumption (H₃). This fact gives a great applicability to Theorem 15.

Before presenting an example illustrating our results, we need some facts about the functions involving this example. The following lemma appears in [18].

Lemma 16. Let 𝜑∶ℝ₊→ℝ₊ be a nondecreasing and upper semicontinuous function. Then, the following conditions are equivalent:(i)lim_𝑛→∞𝜑^𝑛(𝑡)=0 for any 𝑡≥0;(ii)𝜑(𝑡)<𝑡 for any 𝑡>0.

Particularly, the functions 𝛼₁, 𝛼₂∶ℝ₊→ℝ₊ given by 𝛼₁(𝑡)=arctan𝑡 and 𝛼₂(𝑡)=𝑡/(1+𝑡) belong to the class 𝒜, since they are nondecreasing and continuous, and, as it is easily seen, they satisfy (ii) of Lemma 16.

On the other hand, since the function 𝛼₁(𝑡)=arctan𝑡 is concave (because 𝛼^′′(𝑡)≤0) and 𝛼₁(0)=0, we infer that 𝛼₁ is subadditive and, therefore, for any 𝑡,𝑡^′∈ℝ₊, we have 𝛼₁(𝑡)−𝛼₁𝑡^′=arctan𝑡−arctan𝑡^′≤arctan𝑡−𝑡^′. (64) Moreover, it is easily seen that max(𝛼₁,𝛼₂) is a nondecreasing and continuous function because 𝛼₁ and 𝛼₂ are nondecreasing and continuous and max(𝛼₁,𝛼₂) satisfies (ii) of Lemma 16. Therefore, max(𝛼₁,𝛼₂)∈𝒜.

Now, we are ready to present an example where our results can be applied.

Example 17. Consider the following coupled system of fractional hybrid differential equations: 𝐷₀^1/2₊𝑥(𝑡)×14+1|𝑥(𝑡+110arctan)|𝑦(𝑡20)1+𝑦(𝑡)⁻¹=17+19𝑥(𝑡)+1𝑦(𝑡),𝐷10_01/2+𝑦(𝑡)×14+1𝑦(𝑡+110arctan)|𝑥(𝑡20)|(1+|𝑥(𝑡)|)⁻¹=17+19𝑦(𝑡)+1𝑥(𝑡),𝑥(0)(0)100<𝑡<1,=𝑦=0. (65)

Notice that problem (65) is a particular case of problem (4), where 𝛼=1/2, 𝑓(𝑡,𝑥,𝑦)=1/4+(1/10)arctan|𝑥|+(1/20)(|𝑦|/(1+|𝑦|)), and 𝑔(𝑡,𝑥,𝑦)=1/7+(1/9)𝑥+(1/10)𝑦.

It is clear that 𝑓∈𝐶([0,1]×ℝ×ℝ,ℝ⧵{0}) and 𝑔∈𝐶([0,1]×ℝ×ℝ,ℝ) and, moreover, 𝑘₁=sup{|𝑓(𝑡,0,0)|∶𝑡∈[0,1]}=1/4 and 𝑘₂=sup{|𝑔(𝑡,0,0)|∶𝑡∈[0,1]}=1/7. Therefore, assumption (H₁) of Theorem 15 is satisfied.

Moreover, for 𝑡∈[0,1] and 𝑥₁, 𝑥₂, 𝑦₁, 𝑦₂∈ℝ, we have 𝑓𝑡,𝑥₁,𝑦₁−𝑓𝑡,𝑥₂,𝑦₂≤1𝑥10arctan₁−arctan𝑥₂+1𝑦201+¹𝑦₁−𝑦₂1+𝑦₂≤1𝑥10arctan₁−𝑥₂+1𝑦20¹−𝑦₂1+𝑦₁1+𝑦₂≤1𝑥10arctan₁−𝑥₂+1𝑦20¹−𝑦₂1+𝑦₁−𝑦₂=1𝛼10¹𝑥₁−𝑥₂+1𝛼20²𝑦₁−𝑦₂≤1𝛼10max₁,𝛼₂𝑥₁−𝑥₂+1𝛼10max₁,𝛼₂𝑦₁−𝑦₂≤1𝛼102max₁,𝛼₂×max𝑥₁−𝑥₂,𝑦₁−𝑦₂=15max𝛼₁,𝛼₂max𝑥₁−𝑥₂,𝑦₁−𝑦₂. (66) Therefore, 𝜑₁(𝑡)=(1/5)max(𝛼₁(𝑡),𝛼₂(𝑡)) and 𝜑₁∈𝒜.

On the other hand, 𝑔𝑡,𝑥₁,𝑦₁−𝑔𝑡,𝑥₂,𝑦₂≤19𝑥₁−𝑥₂+1𝑥10²−𝑦₂≤19𝑥₁−𝑦₁+𝑦₂−𝑦₂≤192max𝑥₁−𝑦₁,𝑦₂−𝑦₂=29max𝑥₁−𝑦₁,𝑦₂−𝑦₂, (67) and 𝜑₂(𝑡)=(2/9)𝑡. It is clear that 𝜑₂∈𝒜. Therefore, assumption (H₂) of Theorem 15 is satisfied.

In our case, the inequality appearing in assumption (H₃) of Theorem 15 has the expression 15𝑟maxarctan𝑟,+11+𝑟4291𝑟+73≤𝑟Γ2. (68) It is easily seen that 𝑟₀=1 satisfies the last inequality. Moreover, 29𝑟₀+17=29+173≤Γ2≅0.88623. (69) Finally, Theorem 15 says that problem (65) has at least one solution (𝑥,𝑦)∈𝐶[0,1] such that max(‖𝑥‖,‖𝑦‖)≤1.

Conflict of Interests

The authors declare that there is no conflict of interests in the submitted paper.