Existence Results for a Coupled System of Nonlinear Fourth-Order Differential Equations

Abstract

Sufficient conditions are obtained for the existence of solutions to a coupled system of nonlinear fourth-order differential equations.

1. Introduction

The purpose of this paper is to study the existence of solutions to the following coupled system of nonlinear fourth-order differential equations: 𝑥^′′′′(𝑡)(𝑡)(𝑡(),𝑦=𝑓𝑡,𝑥,𝑦),𝑡∈0,1_′′′′(𝑡)(𝑡)(𝑡)(),𝑥(0)=𝑔𝑡,𝑥,𝑦,𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)𝑥=0,_′′′(1)(𝑥(1,𝑦(0)=𝜉))=𝑦^′′(0)=0,𝑦^′(1)𝑦=0,_′′′(1)=𝜁𝑦(1), (1) where 𝑓,𝑔∶[0,1]×ℝ×ℝ→ℝ and 𝜉,𝜁∶ℝ→ℝ are given continuous functions.

In recent years, boundary value problems for fourth-order ordinary differential equations have been extensively studied. Gupta [1] considered the boundary value problem 𝑥^′′′′(𝑡)(𝑡)𝑥(𝑡)(𝑡)(),𝑥+𝑓=𝑔,𝑡∈0,𝜋_′(0)=𝑥^′′′(0)=𝑥^′′′(𝜋)=𝑥^′(𝜋)=0, (2) which describes the unstable static equilibrium of an elastic beam which is supported by sliding clamps at both ends, 𝑡=0 and 𝑡=𝜋. Many authors (see [2–8] and references therein) studied the existence of solutions to the problem 𝑥^′′′′(𝑡)=𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)(),𝑥(0)(1),𝑡∈0,1=𝑥=𝑥^′′(0)=𝑥^′′(1)=0, (3) which describes the deformation of the equilibrium state of an elastic beam with its two ends being simply supported. For other results on the study of fourth-order ordinary differential equations, we refer to [9–13] and references therein. On the other hand, to our knowledge, there are only few papers concerning existence results for systems of fourth-order ordinary differential equations. For example, in [14], via the fixed point index theory, the authors studied the existence of solutions to the system of bending elastic beam equations 𝑥^′′′′(𝑡)=𝑓𝑡,𝑥(𝑡),𝑦(𝑡),𝑥^′′(𝑡),𝑦^′′(𝑡)(),𝑦,𝑡∈0,1_′′′′(𝑡)=𝑔𝑡,𝑥(𝑡),𝑦(𝑡),𝑥^′′(𝑡),𝑦^′′(𝑡)(),𝑥(0)(1),𝑡∈0,1=𝑥=𝑥^′′(0)=𝑥^′′(1)𝑦(0)(1)=0,=𝑦=𝑦^′′(0)=𝑦^′′(1)=0. (4)

In this work, we provide four sufficient conditions for the existence of solutions to the problem (1). Some illustrative examples are presented.

The paper is organized as follows. In Section 2, we present the main results. In Section 3, we give some illustrative examples. In Section 4, we present some basic lemmas that will be used for the proofs. Finally, in Section 5, we prove our results.

2. Main Results

We obtained the following existence results.

Theorem 1. Suppose the following conditions hold:(i)there exist 𝛼,𝛽,𝜆,𝜇>0 such that, for all 𝑡∈[0,1] and 𝑝,𝑞,𝑟,𝑒∈ℝ, one has 𝑓(,𝑔(,𝑡,𝑝,𝑞−𝑓𝑡,𝑟,𝑒)≤𝛼𝑝−𝑟+𝛽𝑞−𝑒𝑡,𝑝,𝑞−𝑔𝑡,𝑟,𝑒)≤𝜆𝑝−𝑟+𝜇𝑞−𝑒 (5)(ii)there exist 𝛾,𝜃>0 such that, for all 𝑝,𝑞∈ℝ, one has 𝜉𝑝𝑞,𝜁𝑝𝑞,−𝜉≤𝛾𝑝−𝑞−𝜁≤𝜃𝑝−𝑞 (6)(iii)𝜌(𝐴)<1, where 𝐴∈𝑀_2,2(ℝ₊) is the matrix given by 1𝐴=2𝛾+𝛼𝛽𝜃+𝜆𝜇 (7)and 𝜌(𝐴) denotes the spectral radius of 𝐴. Then(I)equation (1) has a unique solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ),(II)for any 𝑢₀,𝑣₀∈𝐶([0,1];ℝ), the sequences {𝑢_𝑛},{𝑣_𝑛}⊂𝐶([0,1];ℝ) defined by 𝑢_𝑛+1=𝐹₁𝑢_𝑛,𝑣_𝑛,𝑣_𝑛+1=𝐹₂𝑢_𝑛,𝑣_𝑛 (8)converge, respectively, to 𝑥^∗ and 𝑦^∗; that is, lim_𝑛→∞𝑢_𝑛−𝑥^∗_∞=lim_𝑛→∞𝑣_𝑛−𝑦^∗_∞=0. (9)

Theorem 2. Suppose the following conditions hold:(i)there exists a nonnegative function 𝑎∈𝐿¹(0,1) such that, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑓𝑡,𝑝,𝑞≤𝑎(𝑡)+𝜖₁𝑝^𝜌1+𝜖₂𝑞^𝜌2,𝜖₁,𝜖₂>0,0<𝜌₁,𝜌₂<1, (10)(ii)there exists a nonnegative function 𝑏∈𝐿¹(0,1) such that, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑔𝑡,𝑝,𝑞≤𝑏(𝑡)+𝛿₁p^𝜎1+𝛿₂𝑞^𝜎2,𝛿₁,𝛿₂>0,0<𝜎₁,𝜎₂<1, (11)(iii)for all 𝑝∈ℝ, one has max𝜉𝑝,𝜁𝑝≤𝑝. (12)Then (1) has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

Theorem 3. Suppose the following conditions hold:(i)for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑓𝑡,𝑝,𝑞≤𝑎+𝜖𝑝,𝑎>0,0≤𝜖<1, (13)(ii)for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑔𝑡,𝑝,𝑞≤𝑏+𝛿𝑞,𝑏>0,0≤𝛿<1, (14)(iii)for all 𝑝∈ℝ, one has max𝜉𝑝,𝜁𝑝≤𝑝. (15)Then (1) has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

Theorem 4. Suppose the following conditions hold:(i)for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑓𝑡,𝑝,𝑞≤𝑎+𝜖₁𝑝^𝜌1+𝜖₂𝑞^𝜌2,0≤2𝑎<1,𝜖₁,𝜖₂>0,2𝜖₁+𝜖₂≤1,𝜌₁,𝜌₂≥1, (16)(ii)for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, one has 𝑔𝑡,𝑝,𝑞≤𝑏+𝛿₁𝑝^𝜎1+𝛿₂𝑞^𝜎2,0≤2𝑏<1,𝛿₁,𝛿₂>0,2𝛿₁+𝛿₂≤1,𝜎₁,𝜎₂≥1, (17)(iii)for all 𝑝,𝑞∈ℝ, one has max𝜉𝑝−𝜉𝑞,𝜁𝑝−𝜁𝑞≤𝑘𝑝−𝑞,0<𝑘<2, (18)(iv)𝜉(0)=𝜁(0)=0.Then (1) has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

3. Illustrative Examples

Example 1. Consider the initial value problem 𝑥^′′′′(𝑡)=12(𝑥(𝑡+3√sin))3𝑦(𝑡)(𝑡),(),𝑦16sincos𝑡∈0,1_′′′′(𝑡)=43|𝑥(𝑡+√)|32𝑦(𝑡)𝑒^−𝑡(𝑡)(),𝑥(0)sin,𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)=0,4𝑥^′′′(1)((1,𝑦(0)=sin2𝑥))=𝑦^′′(0)=0,𝑦^′(1)𝑦=0,_′′′(1)=49sin3𝑦(1). (19) In this case, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, we have 𝑓=1𝑡,𝑝,𝑞2𝑝+3√sin3𝑞(𝑡),𝑔=416sincos𝑡,𝑝,𝑞3𝑝+√32𝑞𝑒^−𝑡(𝑡),𝜉𝑝=1sin4,𝜁𝑝=4sin2𝑝9sin3𝑝. (20) Conditions (i) and (ii) of Theorem 1 are satisfied with 1𝛼=23√,𝛽=3416,𝜆=3,√𝜇=321,𝛾=24,𝜃=3. (21) Moreover, the matrix 𝐴=𝛼+𝛾2𝛽2𝜆+𝜃2𝜇2=123√34323√34 (22) has the eigenvalues 𝜆₁=0 and 𝜆₂=0.9330. From Theorem 1, the considered problem has a unique solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

Example 2. Consider the initial value problem 𝑥^′′′′(𝑡)=𝑒^𝑡+sin𝑥(𝑡)(𝑡(𝑡)(),𝑦+𝑦)cos,𝑡∈0,1_′′′′(𝑡)=3𝑡²−5+𝑒^−𝑡|𝑥(𝑡)|^2/5+𝑦(𝑡)21+𝑦(𝑡,(),𝑥(0))𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)𝑥=0,_′′′(1)(𝑥(1,𝑦(0)=sin))=𝑦^′′(0)=0,𝑦^′(1)𝑦=0,_′′′(1)=𝑦(1). (23) In this case, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, we have 𝑓𝑡,𝑝,𝑞=𝑒^𝑡+sin(𝑡),𝑔𝑝+𝑞cos𝑡,𝑝,𝑞=3𝑡²−5+𝑒^−𝑡𝑝^2/5+𝑞21+𝑞,𝜉𝑝𝑝,𝜁𝑝=𝑝.=sin (24) Conditions (i) and (ii) of Theorem 2 are satisfied with 𝑎(𝑡)=𝑒^𝑡,𝜖₁=𝜖₂=1,𝜌₁=𝜌₂=12,𝑏(𝑡)=3𝑡²−5,𝛿₁𝛿=1,₂=12,𝜎₁=25,𝜎₂=12. (25) From Theorem 2, the considered problem has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

Example 3. Consider the initial value problem 𝑥^′′′′(𝑡)=𝑒^−𝑡+12𝑥(𝑡)sin1+𝑦²(𝑡)(),𝑦,𝑡∈0,1_′′′′(𝑡)=11+𝑡²+𝑥²(𝑡)+𝑦²(𝑡)(),𝑥(0),𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)𝑥=0,_′′′(1)𝑥(1)=sin1+𝑥(1)²,𝑦(0)=𝑦^′′(0)=0,𝑦^′(1)𝑦=0,_′′′(1)=𝑦(1)1+𝑦(1)⁴. (26) In this case, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, we have 𝑓𝑡,𝑝,𝑞=𝑒^−𝑡+12𝑝sin1+𝑞²,𝑔=1𝑡,𝑝,𝑞1+𝑡²+𝑝²+𝑞²,𝜉𝑝𝑝=sin1+𝑝²,𝜁𝑝=𝑝1+𝑝⁴. (27) Conditions (i) and (ii) of Theorem 3 are satisfied with 1𝑎=1,𝜖=2,𝑏=1,𝛿=0. (28) From Theorem 3, the considered problem has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

Example 4. Consider the initial value problem 𝑥^′′′′(𝑡)=𝑡²5+𝑥(𝑡)²+𝑦(𝑡)²4(𝑡)(),𝑦cos,𝑡∈0,1_′′′′(𝑡)=𝑡𝑒^𝑡−13+sin𝑥(𝑡)+𝑦(𝑡)4(),𝑥(0),𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)𝑥=0,_′′′(1)(1)=sin3𝑥2,𝑦(0)=𝑦^′′(0)=0,𝑦^′(1)𝑦=0,_′′′(1)=32𝑦(1). (29) In this case, for all 𝑡∈[0,1] and 𝑝,𝑞∈ℝ, we have 𝑓=𝑡𝑡,𝑝,𝑞²5+𝑝²+𝑞²4(𝑡),𝑔=cos𝑡,𝑝,𝑞𝑡𝑒^𝑡−13+sin𝑝+𝑞4,𝜉𝑝=sin3𝑝2,𝜁𝑝=3𝑝2. (30) Conditions (i), (ii), and (iii) of Theorem 4 are satisfied with 1𝑎=5,𝜖₁=𝜖₂=14,𝜌₁=𝜌₂1=2,𝑏=3,𝛿₁=1,𝛿₂=14,𝜎₁=𝜎₂3=1,𝑘=2. (31) From Theorem 4, the considered problem has at least one solution (𝑥^∗,𝑦^∗)∈𝐶([0,1];ℝ)×𝐶([0,1];ℝ).

4. Preliminaries

In order to obtain an integral formulation of the problem (1), we need the following lemma. At first, we denote by 𝐶([0,1];ℝ) the set of continuous functions from [0,1] into ℝ.

Lemma 5. Let 𝜌∈𝐶([0,1];ℝ) and 𝜔∈ℝ. The problem 𝑥^′′′′(𝑡)(𝑡)(),𝑥(0)=𝜌,𝑡∈0,1=𝑥^′′(0)=0,𝑥^′(1)=0,𝑥^′′′(1)=𝜔 (32) has a unique solution 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝜌(𝑠)𝑑𝑠+𝜑_𝜔(𝑡), (33) where 𝜑_𝜔(𝑡)=𝜔2𝑡𝑡²3,𝐺()=1−1𝑡,𝑠6𝑡6𝑠−𝑡²−3𝑠²𝑠,0≤𝑡≤𝑠,6𝑡−𝑠²−3𝑡²,𝑠≤𝑡≤1. (34)

Proof. We omit the proof as it employs standard arguments.

From Lemma 5, problem (1) is equivalent to the following system of integral equations: 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜑_{𝜉(𝑥(1))}(𝑡),𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜑_{𝜁(𝑦(1))}(𝑡),𝑡∈[0,1]. (35) Let 𝑋=𝐶([0,1];ℝ). Define the mappings 𝐹₁,𝐹₂∶𝑋×𝑋→𝑋 by 𝐹₁(𝑢,𝑣)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠),𝑣(𝑠))𝑑𝑠+𝜑_{𝜉(𝑢(1))}(𝑡),𝐹₂(𝑢,𝑣)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠),𝑣(𝑠))𝑑𝑠+𝜑_{𝜁(𝑣(1))}(𝑡) (36) for every 𝑡∈[0,1]. Let 𝐹∶𝑋×𝑋→𝑋×𝑋 be the mapping defined by 𝐹(𝑢,𝑣)=𝐹₁(𝑢,𝑣),𝐹₂(𝑢,𝑣). (37) Observe that (𝑥,𝑦)∈𝑋×𝑋 is a solution to (35) if and only if (𝑥,𝑦) is a fixed point of 𝐹.

Lemma 6. One has 𝑡0≤²𝑠3()≤1≤𝐺𝑡,𝑠2𝑠2𝑡−𝑡²≤12,∀𝑡,𝑠∈[0,1]. (38)

Proof. For 0≤𝑡≤𝑠≤1, we have 𝜕𝐺(𝑡,𝑠)=()1𝜕𝑡1−𝑡𝑠−2(𝑠−𝑡)²≤(1−𝑡)𝑠. (39) Similarly, for 𝑠≤𝑡≤1, we have 𝜕𝐺(𝑡,𝑠)=()𝜕𝑡1−𝑡𝑠. (40) Then, for all 𝑡,𝑠∈[0,1], we have 𝐺(𝑡,𝑠)=^𝑡₀𝜕𝐺(𝑟,𝑠)≤𝜕𝑟𝑑𝑟_𝑡0()11−𝑟𝑠𝑑𝑟=2𝑠2𝑡−𝑡²≤12. (41) Now, for 0≤𝑡≤𝑠≤1, we have 𝐺()=𝑡𝑡,𝑠66𝑠−𝑡²−3𝑠²≥𝑡66𝑠−4𝑠²≥𝑡²𝑠3. (42) Similarly, for 𝑠≤𝑡≤1, we have 𝐺()=𝑠𝑡,𝑠66𝑡−𝑠²−3𝑡²≥𝑠66𝑡−4𝑡²≥𝑡²𝑠3. (43)

We endow the product set 𝑍=𝑋×𝑋 with the norm (𝑥,𝑦)_𝑍=max‖𝑥‖_∞,𝑦_∞, (44) where ‖𝑥‖_∞=max{|𝑥(𝑡)|∶𝑡∈[0,1]}. (45)

Lemma 7. The mapping 𝐹∶𝑍→𝑍 is continuous with respect to the norm ‖⋅‖_𝑍.

Proof. Let (𝑥,𝑦)∈𝑍 be an arbitrary point. Let 𝑅 be a nonnegative real number such that ‖(𝑥,𝑦)‖_Z≤𝑅. To show the continuity of 𝐹 at the point (𝑥,𝑦), let us consider a sequence {(𝑥_𝑛,𝑦_𝑛)}⊂𝑍 converging to (𝑥,𝑦); that is, lim_𝑛→∞𝑥_𝑛−𝑥_∞=lim_𝑛→∞𝑦_𝑛−𝑦_∞=0. (46) Without restriction to the generality, we can suppose that ‖(𝑥_𝑛,𝑦_𝑛)‖_𝑍≤𝑅, for every 𝑛∈ℕ (ℕ is the set of natural numbers). Since 𝑓 is continuous, we can define 𝑀=max𝑓𝑡,𝑝,𝑞∶0≤𝑡≤1,−𝑅≤𝑝,𝑞≤𝑅. (47) For all 𝑡∈[0,1], we have 𝐹₁𝑥_𝑛,𝑦_𝑛(𝑡)−𝐹₁(𝑡≤𝑥,𝑦)₁₀𝐺(𝑡,𝑠)𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠(𝑠)(𝑠+1)−𝑓𝑠,𝑥,𝑦)𝑑𝑠2𝜉𝑥_𝑛(1)−𝜉(𝑥(1)). (48) Using Lemma 6, for all 𝑡∈[0,1], we have 2𝐹₁𝑥_𝑛,𝑦_𝑛(𝑡)−𝐹₁(𝑡≤𝑥,𝑦)₁₀𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠(𝑠)(𝑠+𝜉𝑥)−𝑓𝑠,𝑥,𝑦)𝑑𝑠_𝑛(1)−𝜉(𝑥(1)), (49) which implies that 2𝐹₁𝑥_𝑛,𝑦_𝑛−𝐹₁𝑥,𝑦_∞≤¹₀𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠)(𝑠)(𝑠)+𝜉𝑥−𝑓𝑠,𝑥,𝑦𝑑𝑠_𝑛(1)−𝜉(𝑥(1)). (50) From the continuity of 𝜉, we have lim_𝑛→∞𝜉𝑥_𝑛(1)−𝜉(𝑥(1))=0. (51) The continuity of 𝑓 implies that, for all 𝑠∈[0,1], lim_𝑛→∞𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠)−𝑓𝑠,𝑥(𝑠),𝑦(𝑠)=0. (52) On the other hand, for all 𝑛∈ℕ, we have ¹₀𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠)−𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠≤2𝑀. (53) Then, by the dominated convergence theorem, we have lim_𝑛→∞¹₀𝑓𝑠,𝑥_𝑛(𝑠),𝑦_𝑛(𝑠)−𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠=0. (54) Consequently, we have lim_𝑛→∞𝐹₁𝑥_𝑛,𝑦_𝑛−𝐹₁𝑥,𝑦_∞=0. (55) Similarly, we can show that lim_𝑛→∞𝐹₂𝑥_𝑛,𝑦_𝑛−𝐹₂𝑥,𝑦_∞=0. (56) We deduce that lim_𝑛→∞𝐹𝑥_𝑛,𝑦_𝑛−𝐹𝑥,𝑦_𝑍=0, (57) which yields the desired result.

Lemma 8. Let 𝐷 be the subset of 𝑍 defined by 𝐷=𝑥,𝑦∈𝑍∶𝑥,𝑦_𝑍≤𝑅, (58) for some 𝑅>0. Under condition (iii) of Theorem 2 (or conditions (iii)-(iv) of Theorem 4), 𝐹(𝐷) is an equicontinuous set.

Proof. Suppose that (iii) of Theorem 2 holds. Since 𝑓 and 𝑔 are continuous, we can define 𝑓,𝑔.𝑀=max𝑡,𝑝,𝑞∶0≤𝑡≤1,−𝑅≤𝑝,𝑞≤𝑅𝑁=max𝑡,𝑝,𝑞∶0≤𝑡≤1,−𝑅≤𝑝,𝑞≤𝑅 (59) Let 𝜀>0. Since 𝐺 is continuous in [0,1]×[0,1], there exists 𝛼>0 such that 𝐺(𝑡,𝑠)−𝐺𝑡^′𝜀,𝑠<min,𝜀2𝑀2𝑁 (60) whenever |𝑡−𝑡^′|<𝛼. Let 𝜅=min{𝛼,𝜀/2𝑅}. For (𝑥,𝑦)∈𝐷, for |𝑡−𝑡^′|<𝜅, we have 𝐹₁𝑥,𝑦(𝑡)−𝐹₁𝑥,𝑦𝑡^′≤¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′𝑓(𝑠)(𝑠)+𝜉(𝑥(1,𝑠𝑠,𝑥,𝑦𝑑𝑠))𝑡−𝑡^′𝜀≤𝑀𝜀2𝑀+𝑅(())𝐹2𝑅fromiiiofTheorem2=𝜀,₂𝑥,𝑦(𝑡)−𝐹₂𝑥,𝑦𝑡^′≤¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′𝑔(𝑠)(𝑠)+𝜁𝑦(1),𝑠𝑠,𝑥,𝑦𝑑𝑠𝑡−𝑡^′𝜀≤𝑁𝜀2𝑁+𝑅(())2𝑅fromiiiofTheorem2=𝜀. (61) Then, the equicontinuity of 𝐹(𝐷) is proved. If conditions (iii)-(iv) of Theorem 4 hold, using the same technique with 𝜅=min{𝛼,𝜀/2𝑘𝑅}, we get the desired result.

Now, we are ready to prove our main results.

5. The Proofs

5.1. Proof of Theorem 1

The proof is based on Perov fixed point theorem [15]. At first, let us recall some concepts.

Let 𝑚≥1 be a positive integer. We endow the set ℝ^𝑚 with the partial order ⪯_𝑚 defined by 𝛼=𝛼₁,…,𝛼_𝑚,𝛽=𝛽₁,…,𝛽_𝑚∈ℝ^𝑚,𝛼⪯_𝑚𝛽⟺𝛼_𝑖≤𝛽_𝑖,∀𝑖=1,…,𝑚. (62) We denote by 0_ℝ^𝑚 the zero vector of ℝ^m.

Let Σ be a nonempty set. A mapping 𝑑∶Σ×Σ→𝑅^𝑚 is called a vector valued metric on Σ if the following properties are satisfied:(d1)𝑑(𝑥,𝑦)=0_ℝ^𝑚⇔𝑥=𝑦, for all 𝑥,𝑦∈Σ;(d2)𝑑(𝑥,𝑦)⪰_𝑚0_ℝ^𝑚, for all 𝑥,𝑦∈Σ;(d3)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), for all 𝑥,𝑦∈Σ;(d4)𝑑(𝑥,𝑦)⪯_𝑚𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦), for all 𝑥,𝑦,𝑧∈Σ.A set Σ equipped with a vector valued metric 𝑑 is called a generalized metric space. We will denote such a space by (Σ,𝑑). For the generalized metric spaces, the notions of convergent sequence, Cauchy sequence, completeness, open subset, and closed subset are similar to those for usual metric spaces.

We denote by 𝑀_𝑚,𝑚(ℝ₊) the set of all 𝑚×𝑚 matrices with positive elements. For 𝐴∈𝑀_𝑚,𝑚(ℝ₊), we denote by 𝜌(𝐴) the spectral radius of 𝐴, that is, the supremum among the absolute values of the elements in its spectrum.

Lemma 9. Let (𝑀,𝑑) be a complete generalized metric space and the mapping 𝑇∶𝑀→𝑀 with the property that there exists a matrix 𝐴∈𝑀_𝑚,𝑚(ℝ₊) such that 𝑑𝑇𝑥,𝑇𝑦⪯_𝑚𝐴𝑑𝑥,𝑦,∀𝑥,𝑦∈𝑀. (63) If 𝜌(𝐴)<1, then 𝑇 has a unique fixed point 𝑥^∗∈𝑀. Moreover, for any 𝑥₀∈𝑀, the iterative sequence {𝑥_𝑛}⊂𝑀 defined by 𝑥_𝑛+1=𝑇𝑥_𝑛 converges to 𝑥^∗.

Proof. We endow 𝑍=𝑋×𝑋 with the vector valued metric 𝑑∶𝑍×𝑍→ℝ² defined by 𝑑,()=‖‖𝑥,𝑦𝑢,𝑣𝑥−𝑢_∞𝑦−𝑣_∞,∀𝑥,𝑦,(𝑢,𝑣)∈𝑍. (64) It is easy to show that (𝑍,𝑑) is a complete generalized metric space. Let (𝑥,𝑦),(𝑢,𝑣)∈Z. For any 𝑡∈[0,1], we have 𝐹₁𝑥,𝑦(𝑡)−𝐹₁(𝑡)=𝑢,𝑣)(₁₀𝐺(𝑡,𝑠)𝑓𝑠,𝑥(𝑠),𝑦(𝑠)−𝑓(𝑠,𝑢(𝑠),𝑣(𝑠))𝑑𝑠+𝜑_{𝜉(𝑥(1))}(𝑡)−𝜑_{𝜉(𝑢(1))}(𝑡)=¹₀𝐺()𝑓(𝑠)(𝑠)((𝑠)(𝑠𝑡𝑡,𝑠𝑠,𝑥,𝑦−𝑓𝑠,𝑢,𝑣))𝑑𝑠+𝑡²3𝜉(𝑥(1(𝑢(1−1))−𝜉))2. (65) Using Lemma 6, for all 𝑡∈[0,1], we have 2𝐹₁𝑥,𝑦(𝑡)−𝐹₁(𝑡≤𝑢,𝑣)()₁₀𝑓(𝑠)(𝑠)(𝑠)(𝑠+𝜉(𝑥(1(𝑢(1.𝑠,𝑥,𝑦−𝑓𝑠,𝑢,𝑣)𝑑𝑠))−𝜉)) (66) Using conditions (i) and (ii), we get that 𝐹₁𝑥,𝑦(𝑡)−𝐹₁(𝑢,𝑣)(𝑡)≤𝛾+𝛼2‖𝑥−𝑢‖_∞+𝛽2𝑦−𝑣_∞, (67) for all 𝑡∈[0,1]. This implies that 𝐹₁𝑥,𝑦−𝐹₁(𝑢,𝑣)_∞≤𝛾+𝛼2‖𝑥−𝑢‖_∞+𝛽2𝑦−𝑣_∞. (68) Using similar calculations, we obtain that 𝐹₂𝑥,𝑦−𝐹₂(𝑢,𝑣)_∞≤𝜃+𝜆2‖𝑥−𝑢‖_∞+𝜇2𝑦−𝑣_∞. (69) It follows from (68) and (69) that 𝑑𝐹𝑥,𝑦,𝐹(𝑢,𝑣)⪯₂,(),∀,()𝐴𝑑𝑥,𝑦𝑢,𝑣𝑥,𝑦𝑢,𝑣∈𝑍, (70) where 1𝐴=2𝛾+𝛼𝛽𝜃+𝜆𝜇. (71) Since 𝜌(𝐴)<1, the desired result follows from Lemma 9.

5.2. Proof of Theorem 2

The proof is based on Schauder fixed point theorem [16].

Lemma 10. Let 𝐷 be a closed bounded convex subset of a normed space 𝐸. If 𝑇∶𝐷→𝐸 is a continuous compact map such that 𝑇(𝐷)⊆𝐷, then 𝑇 has a fixed point.

Proof. Let 𝐷⊂𝑍 be the set defined by 𝐷=𝑥,𝑦∈𝑍∶𝑥,𝑦_𝑍≤𝑅, (72) where 𝑅≥max3¹₀𝑎(𝑠)𝑑𝑠,3¹₀𝑏(𝑠)𝑑𝑠,3𝜖₁^{1/(1−𝜌1)},3𝜖₂^{1/(1−𝜌2)},3𝛿₁^{1/(1−𝜎1)},3𝛿₂^{1/(1−𝜎2)}. (73)
We will prove that 𝐹(𝐷)⊆𝐷. (74) Let (𝑥,𝑦)∈𝐷; that is, ‖𝑥‖_∞≤𝑅 and ‖𝑦‖_∞≤𝑅. For all 𝑡∈[0,1], we have 𝐹₁𝑥,𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜑_{𝜉(𝑥(1))}(𝑡). (75) Using Lemma 6 and conditions (i) and (iii), we obtain that 𝐹₁(𝑡≤1𝑥,𝑦)2¹₀𝑎(𝑠)+𝜖𝑑𝑠₁2𝑅^𝜌1+𝜖₂2𝑅^𝜌2+𝑅2≤𝑅6+𝑅6+𝑅6+𝑅2=𝑅. (76) Thus, we have 𝐹₁(𝑥,𝑦)_∞≤𝑅. (77) Similarly, for all 𝑡∈[0,1], we have 𝐹₂𝑥,𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜑_{𝜁(𝑦(1))}(𝑡). (78) Using Lemma 6 and conditions (ii) and (iii), we obtain that 𝐹₂(𝑡≤1𝑥,𝑦)2¹₀𝑏(𝑠)+𝛿𝑑𝑠₁2𝑅^𝜎1+𝛿₂2𝑅^𝜎2+𝑅2≤𝑅6+𝑅6+𝑅6+𝑅2=𝑅. (79) Thus, we have 𝐹₂(𝑥,𝑦)_∞≤𝑅. (80) Then, (74) follows from (77) and (80).
Now, using Lemmas 7 and 8, Arzelá-Ascoli theorem, and Lemma 10, we obtain the desired result.

5.3. Proof of Theorem 3

The proof is based on Schaefer fixed point theorem [17].

Lemma 11. Let 𝐸 be a Banach space. Assume that 𝑇∶𝐸→𝐸 is a continuous compact operator and the set 𝑉={𝑢∈𝐸∶𝑢=𝜆𝑇𝑢,0≤𝜆≤1} (81) is bounded. Then 𝑇 has a fixed point in 𝐸.

Proof. The continuity and the compactness of the mapping 𝐹∶𝑍→𝑍 follow from Lemmas 7 and 8. We have just to show that the set 𝑉=𝑥,𝑦∈𝑍∶𝑥,𝑦=𝜆𝐹₁𝑥,𝑦,𝜆𝐹₂𝑥,𝑦,0≤𝜆≤1 (82) is bounded with respect to the norm ‖⋅‖_𝑍. Then, the result will be obtained by applying Lemma 11.
Let (𝑥,𝑦)∈𝑍 such that 𝑥=𝜆𝐹₁(𝑥,𝑦) and 𝑦=𝜆𝐹₂(𝑥,𝑦) for some 𝜆∈[0,1]. For all 𝑡∈[0,1], using (i), (iii), and Lemma 6, we obtain that |𝑥(𝑡)|=𝜆¹₀𝐺()𝑓(𝑠)(𝑠)𝜉(𝑥(1𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠+))2𝑡𝑡²3≤𝑎−12+𝜖2‖𝑥‖_∞+‖𝑥‖_∞2, (83) which implies that ‖𝑥‖_∞≤𝑎.1−𝜖 (84) Similarly, using (ii), (iii), and Lemma 6, for all 𝑡∈[0,1], we have 𝑦(𝑡)=𝜆¹₀𝐺()𝑔(𝑠)(𝑠)𝜁𝑦(1)𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠+2𝑡𝑡²3≤𝑏−12+𝛿2𝑦_∞+𝑦_∞2, (85) which implies that 𝑦_∞≤𝑏.1−𝛿 (86) Consequently, 𝑉 is a bounded set.

5.4. Proof of Theorem 4

The following fixed point theorem due to Agarwal et al. [18] will be used in the proof.

Lemma 12. Let 𝐸 be a Banach space, let 𝐷 be a bounded closed convex subset of 𝐸, and let 𝐴,𝐵∶𝐷→𝐸 such that 𝐴𝑢+𝐵𝑣∈𝐷 for every (𝑢,𝑣)∈𝐷. If 𝐴 is a contraction and 𝐵 is continuous and compact, then the equation 𝐴𝑤+𝐵𝑤=𝑤 has a solution 𝑤∈𝐷.

Proof. We define the operators 𝐴,𝐵∶𝑍→𝑍 as follows: for all (𝑥,𝑦)∈𝑍, 𝐴𝑥,𝑦=𝐴₁𝑥,𝑦,𝐴₂,𝐵=𝐵𝑥,𝑦𝑥,𝑦₁𝑥,𝑦,𝐵₂𝑥,𝑦, (87) where, for every 𝑡∈[0,1], 𝐴₁(𝑡)=𝜉(𝑥(1𝑥,𝑦))2𝑡𝑡³3,𝐴−1₂(𝑡)=𝜁𝑦(1)𝑥,𝑦2𝑡𝑡³3,𝐵−1₁𝑥,𝑦(𝑡)=¹₀𝐺()𝑓(𝑠)(𝑠)𝐵𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠,₂𝑥,𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠. (88) Clearly, for every (𝑥,𝑦)∈𝑍, we have 𝐹𝑥,𝑦=𝐴𝑥,𝑦+𝐵𝑥,𝑦. (89) Let 2max{𝑎,𝑏}<𝑅<1. We define the set 𝐷⊂𝑍 as follows: 𝐷=𝑥,𝑦∈𝑍∶𝑥,𝑦_𝑍≤𝑅. (90) Let (𝑥,𝑦),(𝑢,𝑣)∈𝐷. For every 𝑡∈[0,1], we have 𝐴₁𝑥,𝑦(𝑡)+𝐵₁(𝑡≤𝑢,𝑣)()₁₀𝐺(𝑓((𝑠)(𝑠𝜉(𝑥(1𝑡,𝑠)𝑠,𝑢,𝑣))𝑑𝑠+))2((i),()(≤1fromLemma6andconditionsiii,andiv))2𝑎+𝜖₁𝑅^𝜌1+𝜖₂𝑅^𝜌2+𝑅2≤𝑎+𝜖₁+𝜖₂+12𝑅≤1/2+𝜖₁+𝜖₂+12𝑅≤𝑅. (91) Similarly, for every 𝑡∈[0,1], we have 𝐴₂𝑥,𝑦(𝑡)+𝐵₂(𝑡≤𝑢,𝑣)()₁₀𝐺(𝑔((𝑠)(𝑠𝜁(𝑢(1𝑡,𝑠)𝑠,𝑢,𝑣))𝑑𝑠+))2((),()(≤1fromLemma6andconditionsiiiii,andiv))2𝑏+𝛿₁𝑅^𝜎1+𝛿₂𝑅^𝜎2+𝑅2≤𝑏+𝛿₁+𝛿₂+12𝑅≤1/2+𝛿₁+𝛿₂+12𝑅≤𝑅. (92) Thus, we proved that, for every (𝑥,𝑦),(𝑢,𝑣)∈𝐷, we have 𝐴𝑥,𝑦+𝐵(𝑢,𝑣)∈𝐷. (93) Now, we will prove that 𝐴∶𝐷→𝑍 is a contraction. Let (𝑥,𝑦),(𝑢,𝑣)∈𝐷. For every 𝑡∈[0,1], we have 𝐴₁𝑥,𝑦(𝑡)−𝐴₁(𝑡≤𝜉(𝑥(1(𝑢(1𝑢,𝑣)()))−𝜉))2((≤𝑘fromconditioniii))2|𝑥(1)(1≤𝑘−𝑢)|2‖𝑥−𝑢‖_∞. (94) Similarly, for every 𝑡∈[0,1], we have 𝐴₂𝑥,𝑦(𝑡)−𝐴₂(𝑡≤𝜁𝑦(1)(𝑣(1𝑢,𝑣)()−𝜁))2((≤𝑘fromconditioniii))2𝑦(1)(1≤𝑘−𝑣)2𝑦−𝑣_∞. (95) Since 0<𝑘<2, then 𝐴 is a contraction (with respect to the norm ‖⋅‖_𝑍). Finally, the continuity and the compactness of the operator 𝐵 follow by using the same arguments of the proof of Theorem 2. Applying Lemma 12, we obtain the existence of (𝑥,𝑦)∈𝐷 such that (𝑥,𝑦) is a fixed point of 𝐴+𝐵=𝐹.

Acknowledgment

The author would like to extend his sincere appreciation to the Deanship of Scientific Research at King Saud University for the funding of this research through the Research Group Project no. RGP-VPP-237.