A Study of Nonlinear Fractional Differential Equations of Arbitrary Order with Riemann-Liouville Type Multistrip Boundary Conditions

Abstract

We develop the existence theory for nonlinear fractional differential equations of arbitrary order with Riemann-Liouville type boundary conditions involving nonintersecting finite many strips of arbitrary length. Our results are based on some standard tools of fixed point theory. For the illustration of the results, some examples are also discussed.

1. Introduction

The subject of fractional calculus has recently developed into a hot topic for the researchers in view of its numerous applications in the field of physics, mechanics, chemistry, engineering, and so forth. One can find the systematic progress of the topic in the books ([1–6]). A significant characteristic of a fractional-order differential operator distinguishing it from the integer-order differential operator is that it is nonlocal in nature, that is, the future state of a dynamical system or process involving fractional derivative depends on its current state as well its past states. In fact, this feature of fractional-order operators has contributed towards the popularity of fractional-order models, which are recognized as more realistic and practical than the classical integer-order models. In other words, we can say that the memory and hereditary properties of various materials and processes can be described by differential equations of arbitrary order. There has been a rapid development in the theoretical aspects such as periodicity, asymptotic behavior, and numerical methods for fractional equations. For some recent work on the topic, see ([7–23]) and the references therein. In particular, Ahmad et al. [22] studied nonlinear fractional differential equations and inclusions of arbitrary order with multistrip boundary conditions.

In this paper, we continue the study initiated in [22] and consider a boundary value problem of fractional differential equations of arbitrary order 𝑞∈(𝑛−1,𝑛], 𝑛≥2 with finite many multistrip Riemann-Liouville type integral boundary conditions: ^𝑐𝐷^𝑞𝑥(𝑡)((𝑡[],𝑥(0)=𝑓𝑡,𝑥)),𝑡∈0,𝑇=0,𝑥^′(0)=0,…,𝑥^(𝑛−2)(0)𝑥(𝑇)==0,_𝑚𝑖=1𝛾_𝑖𝐼^𝛽𝑖𝑥𝜂_𝑖−𝐼^𝛽𝑖𝑥𝜁_𝑖, (1) where ^𝑐𝐷^𝑞 denotes the Caputo fractional derivative of order 𝑞, 𝑓 is a given continuous function, 𝐼^𝛽𝑖 is the Riemann-Liouville fractional integral of order 𝛽_𝑖>0, 𝑖=1,2,…,𝑚, 0<𝜁₁<𝜂₁<𝜁₂<𝜂₂<…<𝜁_𝑚<𝜂_𝑚<𝑇, and 𝛾_𝑖∈ℝ are suitable chosen constants.

Regarding the motivation of the problem, we know that the strip conditions appear in the mathematical modeling of certain real world problems, for instance, see [24, 25]. In [22], the authors considered the nonlocal strip conditions of the form: 𝑥(1)=^𝑛−2_𝑖=1𝛼_𝑖^𝜂𝑖_𝜁𝑖𝑥(𝑠)𝑑𝑠,0<𝜁_𝑖<𝜂_𝑖().,<1,𝑖=1,2,…,𝑛−2 (2) In the problem (1), we have introduced Riemann-Liouville type multistrip integral boundary conditions which can be interpreted as the controller at the right-end of the interval under consideration is influenced by a discrete distribution of finite many nonintersecting sensors (strips) of arbitrary length expressed in terms of Riemann-Liouville type integral boundary conditions. For some engineering applications of strip conditions, see ([26–32]).

The main objective of the present study is to develop some existence results for the problem (1) by using standard techniques of fixed point theory. The paper is organized as follows. In Section 2 we discuss a linear variant of the problem (1), which plays a key role in developing the main results presented in Section 3. For the illustration of the theory, we have also included some examples.

2. Preliminary Result

Let us begin this section with some basic definitions of fractional calculus [2–4].

Definition 1. If 𝑔(𝑡)∈𝐴𝐶^𝑛[𝑎,𝑏], then the Caputo derivative of fractional order 𝑞 is defined as ^𝑐𝐷^𝑞_𝑎+𝑔(𝑡)=1Γ𝑛−𝑞^𝑡_𝑎(𝑡−𝑠)^{𝑛−𝑞−1}𝑔^(𝑛)(𝑠)𝑑𝑠=𝐼_𝑎^𝑛−𝑞₊𝐷^𝑛𝑔(𝑥),𝑛−1<𝑞<𝑛,𝑛=𝑞+1, (3) where [𝑞] denotes the integer part of the real number 𝑞. For details, see Theorem 2.1 ([4, page 92]). Here 𝐴𝐶^𝑛[𝑎,𝑏] denote the space of real valued functions 𝑔(𝑡) which have continuous derivatives up to order 𝑛−1 on [𝑎,𝑏] such that 𝑔^𝑛−1(𝑡)∈𝐴𝐶[𝑎,𝑏].

Definition 2. The Riemann-Liouville fractional integral of order 𝑞 is defined as 𝐼^𝑞𝑔(𝑡)=1Γ𝑞^𝑡₀𝑔(𝑠)(𝑡−𝑠)^1−𝑞𝑑𝑠,𝑞>0, (4) provided the integral exists.

The following result associated with a linear variant of problem (1) plays a pivotal role in establishing the main results.

Lemma 3. For ℎ∈𝐶[0,𝑇], the fractional boundary value problem ^𝑐𝐷^𝑞𝑥(𝑡)(𝑡)[](]𝑥(0)=ℎ,𝑡∈0,𝑇,𝑞∈𝑛−1,𝑛=0,𝑥^′(0)=0,…,𝑥^(𝑛−2)(0)𝑥(𝑇)==0,_𝑚𝑖=1𝛾_𝑖𝐼^𝛽𝑖𝑥𝜂_𝑖−𝐼^𝛽𝑖𝑥𝜁_𝑖, (5) has a unique solution 𝑥(𝑡)∈𝐴𝐶^𝑛[0,𝑇] given by 𝑥(𝑡)=1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1ℎ(𝑠)−𝑡𝑑𝑠_𝑛−1𝜆Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1ℎ(𝑠)+𝑡𝑑𝑠_𝑛−1𝜆Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1ℎ(𝑢)−𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×ℎ(𝑢)𝑑𝑢𝑑𝑠, (6) where 𝜆=𝑇^𝑛−1−^𝑚_𝑖=1𝛾_𝑖𝜂^𝛽𝑖_𝑖^+𝑛−1−𝜁^𝛽𝑖_𝑖^+𝑛−1Γ(𝑛)Γ𝛽_𝑖+𝑛≠0. (7)

Proof. The general solution of fractional differential equations in (5) can be written as 𝑥(𝑡)=1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1ℎ(𝑠)𝑑𝑠−𝑐₀−𝑐₁𝑡−⋯−𝑐_𝑛−1𝑡^𝑛−1. (8) Using the given boundary conditions, it is found that 𝑐₀=0, 𝑐₁=0,…,𝑐_𝑛−2=0. Applying the Riemann-Liouville integral operator 𝐼^𝛽𝑖 on (8), we get 𝐼^𝛽𝑖𝑥(𝑡)=1Γ𝛽_𝑖^𝑡₀(𝑡−𝑠)^𝛽𝑖−1×1Γ𝑞^𝑠₀(𝑠−𝑢)^𝑞−1ℎ(𝑢)𝑑𝑢−𝑐_𝑛−1𝑠^𝑛−1=1𝑑𝑠Γ𝛽_𝑖Γ𝑞^𝑡₀^𝑠₀(𝑡−𝑠)^𝛽𝑖−1(𝑠−𝑢)^𝑞−1ℎ(𝑢)𝑑𝑢𝑑𝑠−𝑐_𝑛−11Γ𝛽_𝑖^𝑡₀(𝑡−𝑠)^𝛽𝑖−1𝑠^𝑛−1𝑑𝑠. (9) Using the condition 𝑥(𝑇)=∑^𝑚_𝑖=1𝛾_𝑖[𝐼^𝛽𝑖𝑥(𝜂_𝑖)−𝐼^𝛽𝑖𝑥(𝜁_𝑖)], together with the fact that 1Γ𝛽_𝑖^𝑡₀(𝑡−𝑠)^𝛽𝑖−1𝑠^𝑛−1𝑡𝑑𝑠=^{𝛽𝑖+𝑛−1}Γ(𝑛)Γ𝛽_𝑖,+𝑛 (10) we obtain 1Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1ℎ(𝑠)𝑑𝑠−𝑐_𝑛−1𝑇^𝑛−1=^𝑚_𝑖=1𝛾_𝑖Γ𝑞Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1ℎ(𝑢)𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1ℎ(𝑢)𝑑𝑢𝑑𝑠−𝑐^𝑚_{𝑛−1𝑖=1}𝛾_𝑖𝜂^𝛽𝑖_𝑖^+𝑛−1−𝜁^𝛽𝑖_𝑖^+𝑛−1Γ(𝑛)Γ𝛽_𝑖,+𝑛 (11) which yields 𝑐_𝑛−1=1𝜆Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1ℎ(𝑠)−1𝑑𝑠𝜆Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1ℎ(𝑢)−𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×ℎ(𝑢)𝑑𝑢𝑑𝑠, (12) where 𝜆 is given by (7). Substituting the values of 𝑐₀,𝑐₁,…,𝑐_𝑛−2,𝑐_𝑛−1 in (8), we obtain (6). This completes the proof.

3. Main Results

Let 𝒞∶=𝐶([0,𝑇],ℝ) denotes the Banach space of all continuous functions defined on [0,𝑇]×ℝ endowed with a topology of uniform convergence with the norm ‖𝑥‖=sup_{𝑡∈[0,𝑇]}|𝑥(𝑡)|.

By Lemma 3, we define an operator 𝒫∶𝒞→𝒞 as (𝑡)=1𝒫𝑥)(Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠−𝑡𝑠,𝑥))𝑑𝑠_𝑛−1𝜆Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1𝜆Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1((𝑢,[].×𝑓𝑢,𝑥))𝑑𝑢𝑑𝑠𝑡∈0,𝑇 (13)

Observe that the problem (1) has a solution if and only if the associated fixed point problem 𝒫𝑥=𝑥 has a fixed point.

In the first result we prove an existence and uniqueness result by means of Banach’s contraction mapping principle. For the sake of convenience, we set 𝑇Λ=^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1_𝑛−1|𝜆|^𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖.+1 (14)

Theorem 4. Suppose that 𝑓∶[0,𝑇]×ℝ→ℝ is a continuous function and satisfies the following assumption:(A₃)𝑓(),[]𝑡,𝑥−𝑓𝑡,𝑦≤𝐿𝑥−𝑦∀𝑡∈0,1,𝐿>0,𝑥,𝑦∈ℝ. (15)Then the boundary value problem (1) has a unique solution provided 1𝐿<Λ, (16) where Λ is given by (14).

Proof. With 𝑟≥𝑀Λ/(1−𝐿Λ), we define 𝐵_𝑟={𝑥∈𝒞∶‖𝑥‖≤𝑟}, where 𝑀=sup_{𝑡∈[0,𝑇]}|𝑓(𝑡,0)|<∞ and Λ is given by (14). Then we show that 𝒫𝐵_𝑟⊂𝐵_𝑟. For 𝑥∈𝐵_𝑟, by means of the inequality |𝑓(𝑠,𝑥(𝑠))|≤|𝑓(𝑠,𝑥(𝑠))−𝑓(𝑠,0)|+|𝑓(𝑠,0)|≤𝐿‖𝑥‖+𝑀≤𝐿𝑟+𝑀, it can easily be shown that ‖𝒫𝑥‖=(𝐿𝑟+𝑀)Λ≤𝑟. (17) Now, for 𝑥,𝑦∈𝒞 and for each 𝑡∈[0,𝑇], we obtain (𝒫𝑥)−𝒫𝑦≤sup_{𝑡∈[0,𝑇]}1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠(𝑠)+𝑡𝑠,𝑥))−𝑓𝑠,𝑦𝑑𝑠_𝑛−1𝜆Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1×𝑓((𝑠(𝑠)+𝑡𝑠,𝑥))−𝑓𝑠,𝑦𝑑𝑠_𝑛−1𝑞×𝜆Γ_𝑚𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1×𝑓((𝑢(𝑢)𝑑𝑢𝑢,𝑥))−𝑓𝑢,𝑦𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓((𝑢(𝑢).𝑢,𝑥))−𝑓𝑢,𝑦𝑑𝑢𝑑𝑠≤𝐿Λ𝑥−𝑦 (18) Note that Λ depends only on the parameters involved in the problem. As 𝐿Λ<1, therefore 𝒫 is a contraction. Hence, by Banach’s contraction mapping principle, the problem (1) has a unique solution on [0,𝑇].

Example 5. Let us consider the following 4-strip nonlocal boundary value problem: ^𝑐𝐷^9/2𝑥(𝑡)((𝑡[],𝑥(0)=𝑓𝑡,𝑥)),𝑡∈0,2=0,𝑥^′(0)=0,𝑥^′′(0)=0,𝑥^′′′(0)𝑥(𝑇)==0,_4𝑖=1𝛾_𝑖𝐼^𝛽𝑖𝑥𝜂_𝑖−𝐼^𝛽𝑖𝑥𝜁_𝑖, (19) where 𝑞=9/2, 𝑛=5, 𝜁₁=1/4, 𝜂₁=1/2, 𝜁₂=2/3, 𝜂₂=1, 𝜁₃=5/4, 𝜂₃=4/3, 𝜁₄=3/2, 𝜂₄=7/4, 𝛾₁=5, 𝛾₂=10, 𝛾₃=15, 𝛾₄=25, 𝛽₁=5/4, 𝛽₂=7/4, 𝛽₃=9/4, 𝛽₄=11/4.

With the given values of the parameters involved, we find that 𝜆=𝑇^𝑛−1−^𝑚_𝑖=1𝛾_𝑖𝜂^𝛽𝑖_𝑖^+𝑛−1−𝜁^𝛽𝑖_𝑖^+𝑛−1Γ(𝑛)Γ𝛽_𝑖𝑇+𝑛≃9.334784,Λ=^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1|𝜆|^𝑛−1𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖+1≃1.406972. (20) Let us choose 𝑓((𝑡=1𝑡,𝑥))₃(𝑡+8)tan⁻¹𝑥+√4+3sin2𝑡. (21) Clearly 𝐿=1/2 as |𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)|≤(1/2)|𝑥−𝑦| and 𝐿<1/Λ, where Λ≃1.406972. Therefore all the conditions of Theorem 4 hold and consequently there exists a unique solution for the problem (19) with 𝑓(𝑡,𝑥(𝑡)) given by (21).

In case of the following unbounded nonlinear function: 𝑓((𝑡=𝑥𝑡,𝑥))7+₃1(𝑡+8)tan⁻¹𝑥+√4+3sin2𝑡, (22) we have 𝐿=9/14 and 𝐿<1/Λ (Λ≃1.406972). As before, the problem (19) with 𝑓(𝑡,𝑥(𝑡)) given by (22) has a unique solution.

In the second result we use the Leray-Schauder alternative.

Theorem 6 ((Leray-Schauder alternative) [33, page 4]). Let 𝑋 be a Banach space. Assume that 𝑇∶𝑋→𝑋 is completely continuous operator and the set 𝑉=𝑢∈𝑋∣𝑢=𝜇𝑇𝑢,0<𝜇<1 (23) is bounded. Then 𝑇 has a fixed point in 𝑋.

Theorem 7. Assume that there exists a positive constant 𝐿₁ such that |𝑓(𝑡,𝑥)|≤𝐿₁ for 𝑡∈[0,𝑇], 𝑥∈ℝ. Then the problem (1) has at least one solution.

Proof. First of all, we show that the operator 𝒫 is completely continuous. Note that the operator 𝒫 is continuous in view of the continuity of 𝑓. Let ℬ⊂𝒞 be a bounded set. By the assumption that |𝑓(𝑡,𝑥)|≤𝐿₁, for 𝑥∈ℬ, we have 𝑡≤1|(𝒫𝑥)()|Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠≤𝐿₁1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑡𝑑𝑠+^𝑛−1|𝜆|Γ𝑞^𝑡₀(𝑇−𝑠)^𝑞−1+𝑡𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1−𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1𝑑𝑢𝑑𝑠≤𝐿₁𝑇^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1_𝑛−1|𝜆|^𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖+1=𝐿₂, (24) which implies that ‖(𝒫𝑥)‖≤𝐿₂. Further, we find that (𝒫𝑥)^′(𝑡=1)Γ𝑞−1^𝑡₀(𝑡−𝑠)^𝑞−2𝑓((𝑠+()𝑡𝑠,𝑥))𝑑𝑠𝑛−1^𝑛−2|𝜆|Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+()𝑡𝑠,𝑥))𝑑𝑠𝑛−1^𝑛−2|𝜆|Γ𝑞×^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠≤𝐿₁1Γ𝑞−1^𝑡₀(𝑡−𝑠)^𝑞−2+()𝑡𝑑𝑠𝑛−1^𝑛−2|𝜆|Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1+()𝑡𝑑𝑠𝑛−1^𝑛−2|𝜆|Γ𝑞×^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1×𝑓((𝑢−𝑢,𝑥))𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))×𝑑𝑢𝑑𝑠≤𝐿₁𝑇^𝑞−1Γ𝑞+(𝑛−1)𝑇^{𝑞+𝑛−2}|𝜆|Γ+()𝑇𝑞+1𝑛−1^𝑛−2|𝜆|^𝑚_𝑖=1𝛾_𝑗𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖+1=𝐿₃. (25) Hence, for 𝑡₁,𝑡₂∈[0,𝑇], we have (𝒫𝑥)𝑡₂−(𝒫𝑥)𝑡₁≤^𝑡2_𝑡1(𝒫𝑥)^′(𝑠)𝑑𝑠≤𝐿₃𝑡₂−𝑡₁. (26) This implies that 𝒫 is equicontinuous on [0,𝑇]. Thus, by the Arzelá-Ascoli theorem, the operator 𝒫∶𝒞→𝒞 is completely continuous.
Next, we consider the set 𝑉=𝑥∈𝒞∣𝑥=𝜇𝒫𝑥,0<𝜇<1, (27) and show that the set 𝑉 is bounded. Let 𝑥∈𝑉, then 𝑥=𝜇𝒫𝑥,0<𝜇<1. For any 𝑡∈[0,𝑇], we have |𝑥(𝑡𝑡≤1)|=𝜇|(𝒫𝑥)()|Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠≤𝐿₁𝑇^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1_𝑛−1|𝜆|^𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖+1=𝑀₁. (28)
Thus, ‖𝑥‖≤𝑀₁ for any 𝑡∈[0,𝑇]. So, the set 𝑉 is bounded. Thus, by the conclusion of Theorem 6, the operator 𝒫 has at least one fixed point, which implies that (1) has at least one solution.

Example 8. Consider the boundary value problem of Example 5 with 𝑓(𝑡,𝑥(𝑡))=3𝑒^{(2−|𝑥(𝑡)|)3}cos4𝑡+2ln1+4sin²𝑥(𝑡)((𝑡.10+cos𝑥)) (29) Observe that |𝑓(𝑡,𝑥)|≤𝐿₁ with 𝐿₁=𝑒^2√2(1+ln25). Thus the hypothesis of Theorem 7 is satisfied. Hence by the conclusion of Theorem 7, the problem (19) with 𝑓(𝑡,𝑥(𝑡)) given by (29) has at least one solution.

In the next we prove one more existence result for problem (1), based on the following known result.

Theorem 9 (see [34]). Let 𝑋 be a Banach space. Assume that Ω is an open bounded subset of 𝑋 with 𝜃∈Ω and let 𝑇∶Ω→𝑋 be a completely continuous operator such that ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω. (30) Then 𝑇 has a fixed point in Ω.

Theorem 10. Let there exist a small positive number 𝜏 such that |𝑓(𝑡,𝑥)|≤𝜈|𝑥| for 0<|𝑥|<𝜏, with 0<𝜈≤1/Λ, where Λ is given by (14). Then the problem (1) has at least one solution.

Proof. Let us define ℬ_𝜏={𝑥∈𝒞∣‖𝑥‖<𝜏} and take 𝑥∈𝒞 such that ‖𝑥‖=𝜏, that is, 𝑥∈𝜕ℬ_𝜏. As before, it can be shown that 𝒫 is completely continuous and ‖𝒫𝑥‖≤sup_{𝑡∈[0,𝑡]}1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞×^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞×^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓((𝑢−𝑢,𝑥))𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓((𝑢‖𝑥‖,𝑢,𝑥))𝑑𝑢𝑑𝑠≤Λ𝜈 (31) which in view of the given condition (𝜈Λ≤1), gives ‖𝒫𝑥‖≤‖𝑥‖, 𝑥∈𝜕ℬ_𝜏. Therefore, by Theorem 9, the operator 𝒫 has at least one fixed point, which in turn implies that the problem (1) has at least one solution.

Example 11. Consider the boundary value problem of Example 5 and let us consider 𝑓(𝑡,𝑥(𝑡))=𝑥𝑏⁵+𝑥⁴(𝑡)^1/5+21+cos𝑡⁴+3⁵×(1−cos𝑥(𝑡)),𝑥≠0,𝑏>0. (32) For sufficiently small 𝑥 (ignoring 𝑥² and higher powers of 𝑥), we have 𝑥𝑏⁵+𝑥⁴(𝑡)^1/5+21+cos𝑡⁴+3⁵((𝑡|𝑥|.1−cos𝑥))≤𝑏 (33) Choosing 𝑏≤1/Λ, all the assumptions of Theorem 10 hold. Therefore, the conclusion of Theorem 10 implies that the problem (19) with 𝑓(𝑡,𝑥(𝑡)) given by (32) has at least one solution.

Our final existence result is based on Leray-Schauder nonlinear alternative.

Lemma 12 ((Nonlinear alternative for single valued maps) [33, page 135]). Let 𝐸 be a Banach space, 𝐶 a closed, convex subset of 𝐸, 𝑈 an open subset of 𝐶 and 0∈𝑈. Suppose that 𝐹∶𝑈→𝐶 is a continuous, compact (i.e., 𝐹(𝑈) is a relatively compact subset of 𝐶) map. Then either(i)𝐹has a fixed point in 𝑈, or(ii)there is a 𝑢∈𝜕𝑈 (the boundary of 𝑈 in 𝐶) and 𝜆∈(0,1) with 𝑢=𝜆𝐹(𝑢).

Theorem 13. Assume that(A₁) there exist a function 𝜎∈𝐶([0,1],ℝ⁺), and a nondecreasing function 𝜓∶ℝ⁺→ℝ⁺ such that |𝑓(𝑡,𝑥)|≤𝜎(𝑡)𝜓(‖𝑥‖), for all (𝑡,𝑥)∈[0,𝑇]×ℝ;(A₂) there exists a constant 𝑀>0 such that 𝑀𝜓(𝑀)Λ‖𝜎‖>1. (34)

Then the boundary value problem (1) has at least one solution on [0,𝑇].

Proof. Consider the operator 𝒫∶𝒞→𝒞 defined by (13). We show that 𝒫  maps bounded sets into bounded sets in 𝐶([0,𝑇],ℝ). For a positive number 𝑟, let 𝐵_𝑟={𝑥∈𝐶([0,𝑇],ℝ)∶‖𝑥‖≤𝑟} be a bounded set in 𝐶([0,𝑇],ℝ). Then ‖𝒫𝑥‖≤sup_{[]𝑡∈0,𝑇}1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_𝑛−1|𝜆|Γ𝑞×^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1×𝑓((𝑢−𝑢,𝑥))𝑑𝑢𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1×(𝑠−𝑢)^𝑞−1×𝑓((𝑢(𝑟𝑇𝑢,𝑥))×𝑑𝑢𝑑𝑠≤𝜓)^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1_𝑛−1|𝜆|^𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖𝜎‖.+1‖ (35) Next we show that 𝐹 maps bounded sets into equicontinuous sets of 𝐶([0,1],ℝ). Let 𝑡^′,𝑡^′′∈[0,1] with 𝑡^′<𝑡^′′ and 𝑥∈𝐵_𝑟, where 𝐵_𝑟 is a bounded set of 𝐶([0,1],ℝ). Then we obtain (𝒫𝑥)𝑡^′′−(𝒫𝑥)𝑡^′=1Γ𝑞^𝑡′′₀𝑡^′′−𝑠^𝑞−1𝑓((𝑠−1𝑠,𝑥))𝑑𝑠Γ𝑞^𝑡′₀𝑡^′−𝑠^𝑞−1𝑓((𝑠−𝑡𝑠,𝑥))𝑑𝑠_{′′𝑛−1}−𝑡^′𝑛−1𝜆Γ𝑞^𝑇₀(𝑇−𝑠)^𝑞−1𝑓((𝑠+𝑡𝑠,𝑥))𝑑𝑠_{′′𝑛−1}−𝑡^{′′𝑛−1}𝑞×𝜆Γ_𝑚𝑖=1𝛾_𝑖Γ𝛽_𝑖^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓(𝑢,𝑥(𝑢))𝑑𝑢𝑑𝑠−^𝜁𝑖₀^𝑠₀𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1×𝑓((𝑢≤1𝑢,𝑥))𝑑𝑢𝑑𝑠Γ𝑞^𝑡′₀𝑡^′′−𝑠^𝑞−1−𝑡^′−𝑠^𝑞−1𝜓(𝑟)𝜎(𝑠)+1𝑑𝑠Γ𝑞^𝑡′′_𝑡^′𝑡^′′−𝑠^𝑞−1𝜓(𝑟)𝜎(𝑠)+𝑡𝑑𝑠_{′′𝑛−1}−𝑡^′𝑛−1|𝜆|Γ𝑞^𝑇₀|𝑇−𝑠|^𝑞−1𝜓(𝑟)𝜎(𝑠)+𝑡𝑑𝑠_{′′𝑛−1}−𝑡^′𝑛−1|𝜆|Γ𝑞×^𝑚_𝑖=1𝛾_𝑖Γ𝛽_𝑖×^𝜂𝑖₀^𝑠₀𝜂_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1(𝑟)𝜎(𝑠)−𝑑𝑢𝜓𝑑𝑠_{𝜁𝑖0𝑠0}𝜁_𝑖−𝑠^𝛽𝑖−1(𝑠−𝑢)^𝑞−1𝜓(𝑟)𝜎(𝑠)𝑑𝑢𝑑𝑠. (36) Obviously the right hand side of the above inequality tends to zero independently of 𝑥∈𝐵_𝑟 as 𝑡^′′−𝑡^′→0. As 𝒫∶𝐶([0,𝑇],ℝ)→𝐶([0,𝑇],ℝ) satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that 𝒫 is completely continuous.
Let 𝑥 be a solution. Then, for 𝑡∈[0,𝑇], and following the similar computations as before, we find that |𝑥(𝑡=𝜇(𝑡(𝑟𝑇)|𝒫𝑥)()≤𝜓)^𝑞Γ+𝑇𝑞+1|𝜆|Γ^{𝑞+𝑛−1}+𝑇𝑞+1_𝑛−1|𝜆|^𝑚_𝑖=1𝛾_𝑖𝜂^{𝑞+𝛽𝑖}_𝑖−𝜁^{𝑞+𝛽𝑖}_𝑖Γ𝑞+𝛽_𝑖𝜎‖.+1‖ (37) In consequence, we have ‖𝑥‖𝜓(‖𝑥‖)Λ‖𝜎‖≤1. (38) Thus, by (A₂), there exists 𝑀 such that ‖𝑥‖≠𝑀. Let us set 𝑉={𝑥∈𝐶([0,𝑇],ℝ)∶‖𝑥‖<𝑀+1}. (39) Note that the operator 𝒫∶𝑉→𝐶([0,𝑇],ℝ) is continuous and completely continuous. From the choice of 𝑉, there is no 𝑥∈𝜕𝑉 such that 𝑥=𝜇𝒫(𝑥) for some 𝜇∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 12), we deduce that 𝒫 has a fixed point 𝑥∈𝑉 which is a solution of the problem (1). This completes the proof.

Example 14. Consider the boundary value problem of Example 5 with 𝑓((𝑡=1𝑡,𝑥))√|𝑥|𝑡+41+|𝑥|(𝑡)𝜓𝑥.1+≤𝜎(‖‖) (40) Then 𝜎(𝑡)=1/√𝑡+4 and 𝜓(‖𝑥‖)=2. Using ‖𝜎‖=1/2,Λ≃1.406972, we find by the condition (A₂) that 𝑀>Λ. Thus all the assumptions of Theorem 13 are satisfied. Hence, it follows by Theorem 13 that the problem (19) with 𝑓(𝑡,𝑥(𝑡)) defined by (40) has at least one solution.
If we choose an unbounded nonlinearity as follows: 𝑓((𝑡=1𝑡,𝑥))√|𝑥|𝑡+41+|𝑥|+|𝑥|1+2. (41) Then 𝑓(𝑡,𝑥(𝑡))≤𝜎(𝑡)𝜓(‖𝑥‖) with 𝜎(𝑡)=1/√𝑡+4 and 𝜓(‖𝑥‖)=2+‖𝑥‖/2. Using the earlier arguments, with ‖𝜎‖=1/2, Λ≃1.406972, we find that 𝑀>𝑀₁, 𝑀₁≈2.170392. Hence the problem (19) with 𝑓(𝑡,𝑥(𝑡)) given by (41) has at least one solution.

Acknowledgment

The authors are grateful to the anonymous referees for their valuable comments.