Unique Solution of a Coupled Fractional Differential System Involving Integral Boundary Conditions from Economic Model

Abstract

We study the existence and uniqueness of the positive solution for the fractional differential system involving the Riemann-Stieltjes integral boundary conditions −𝓓^𝛼_𝑡𝑥(𝑡)=𝑓(𝑡,𝑦(𝑡)), −𝓓^𝛽_𝑡𝑦(𝑡)=𝑔(𝑡,𝑥(𝑡)), 𝑡∈(0,1), 𝑥(0)=𝑦(0)=0, 𝑥(1)=∫¹₀𝑥(𝑠)𝑑𝐴(𝑠), and 𝑦(1)=∫¹₀𝑦(𝑠)𝑑𝐵(𝑠), where 1<𝛼, 𝛽≤2, and 𝓓^𝛼_𝑡 and 𝓓^𝛽_𝑡 are the standard Riemann-Liouville derivatives, 𝐴 and 𝐵 are functions of bounded variation, and ∫¹₀𝓓^𝛽_𝑡𝑥(𝑠)𝑑𝐴(𝑠) and ∫¹₀𝓓^𝛽_𝑡𝑦(𝑠)𝑑𝐵(𝑠) denote the Riemann-Stieltjes integral. Our results are based on a generalized fixed point theorem for weakly contractive mappings in partially ordered sets.

1. Introduction

Mathematical model is an important tool designed to describe the operation of the economy of a country or a region. Because fractional operators are non-local, they are more suitable for constructing models possessing memory effect with the long time periods, and then fractional differential equations possess large advantage in describing economic phenomena over the time periods. In this paper, we focus on a fractional model arising from economy. Assuming that 𝑓,𝑔∶[0,1]×[0,+∞)→[0,+∞) are continuous and nondecreasing with respect to the second variable on [0,+∞), we discuss the existence and uniqueness of positive solutions for the following system of fractional differential equation with nonlocal Riemann-Stieltjes integral boundary conditions: −𝓓^𝛼_𝑡𝑥(𝑡)=𝑓𝑡,𝑦(𝑡),−𝓓^𝛽_𝑡𝑦(𝑡)((𝑡(),𝑥(0)(0)(1)==𝑔𝑡,𝑥)),𝑡∈0,1=𝑦=0,𝑥¹₀𝑥(𝑠)(𝑠),𝑦(1)=𝑑𝐴₁₀𝑦(𝑠)𝑑𝐵(𝑠), (1) where 1<𝛼, 𝛽≤2, and  𝓓^𝛼_𝑡 and 𝓓^𝛽_𝑡 are the standard Riemann-Liouville derivatives, 𝐴 and 𝐵 are functions of bounded variation, and ∫¹₀𝓓^𝛽_𝑡𝑥(𝑠)𝑑𝐴(𝑠) and ∫¹₀𝓓^𝛽_𝑡𝑦(𝑠)𝑑𝐵(𝑠) denote the Riemann-Stieltjes integral.

Boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. The idea of using a Riemann-Stieltjes integral with a signed measure is due to Webb and Infante in [1, 2]. The papers [1–3] contain several new ideas and give a unified approach to many BVPs. This implies that the Riemann-Stieltjes integral boundary value problem is a more generalized case which includes multipoints, integral boundary conditions and many nonlocal boundary conditions, as special cases. For some recent work on boundary value problems of fractional differential equation, we refer the reader to some recent papers (see [1–14]).

Recently, motivated by [1–3], Hao et al. [4] studied the existence of positive solutions for the𝑛th-order singular nonlocal boundary value problem 𝑥^(𝑛)(𝑡)(𝑡)𝑓((𝑡𝑥+𝑎𝑡,𝑥))=0,0<𝑡<1,_(𝑘)(0)=0,0≤𝑘≤𝑛−2,𝑥(1)=¹₀𝑥(𝑠)𝑑𝐴(𝑠), (2) where 𝑎 may be singular at 𝑡=0,1 and 𝑓 also may be singular at 𝑥=0, but there is no singularity at 𝑡=0,1. The existence of positive solutions of the BVP (2) is obtained by means of the fixed point index theory in cones. In [5], Zhang and Han considered the existence of positive solutions of the following singular fractional differential equation: 𝓓^𝛼_𝑡𝑥(𝑡)((𝑡𝑥+𝑓𝑡,𝑥))=0,0<𝑡<1,𝑛−1<𝛼≤𝑛,_(𝑘)(0)=0,0≤𝑘≤𝑛−2,𝑥(1)=¹₀𝑥(𝑠)𝑑𝐴(𝑠), (3) where 𝛼≥2, and 𝑑𝐴(𝑠) can be a signed measure. Some growth conditions were adopted to guarantee that (3) has a unique positive solution. Recently, by using a new fixed point theorem for weakly contractive mappings in partially ordered sets, Tao et al. [6] considered the existence and uniqueness of positive solution for the following problem: −𝓓^𝛼_𝑡𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),−𝓓^𝛽_𝑡𝑥(𝑡)(),𝓓,𝑡∈0,1_𝛽𝑡𝑥(0)=𝓓_𝑡^𝛽+1𝑥(0)𝓓=0,_𝛽𝑡𝑥(1)=¹₀𝓓^𝛽_𝑡𝑥(𝑠)𝑑𝐴(𝑠), (4) where 2<𝛼≤3, 0<𝛽<1, and 𝛼−𝛽>2, and 𝓓_𝑡 is the standard Riemann-Liouville derivative.

Motivated by the above work, in this paper, we study the existence and uniqueness of positive solution for the system of fractional differential equations with nonlocal Riemann-Stieltjes boundary integral conditions. Our main tool is the fixed point theorem for weakly contractive mappings in partially ordered sets, which is obtained by papers [6, 15].

2. Preliminaries and Lemmas

Definition 1 (see [16–18]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝐼^𝛼𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠 (5) provided that the right-hand side is pointwise defined on (0,+∞).

Definition 2 (see [16–18]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝓓^𝛼_𝑡𝑥(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑥(𝑠)𝑑𝑠, (6) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number 𝛼, provided that the right-hand side is pointwise defined on (0,+∞).

Lemma 3 (see [16–18]). (1) If 𝑥∈𝐿¹(0,1), 𝜈>𝜎>0, then 𝐼^𝜈𝐼^𝜎𝑥(𝑡)=𝐼^𝜈+𝜎𝑥(𝑡),𝓓^𝜎_𝑡𝐼^𝜈𝑥(𝑡)=𝐼^𝜈−𝜎𝑥(𝑡),𝓓^𝜎_𝑡𝐼^𝜎𝑥(𝑡)=𝑥(𝑡). (7)
(2) If 𝜈>0, 𝜎>0, then 𝓓^𝑣_𝑡𝑡^𝜎−1=Γ(𝜎)Γ()𝑡𝜎−𝜈^{𝜎−𝜈−1}. (8)
(3) If 𝛼>0 and 𝑓(𝑥) is integrable, then 𝐼^𝛼𝓓^𝛼_𝑡𝑥(𝑡)=𝑓(𝑥)+𝑐₁𝑥^𝛼−1+𝑐₂𝑥^𝛼−2+⋯+𝑐_𝑛𝑥^𝛼−𝑛, (9) where 𝑐_𝑖∈ℝ(𝑖=1,2,…,𝑛), 𝑛 is the smallest integer greater than or equal to 𝛼.

Lemma 4 (see [19]). Given ℎ∈𝐿¹(0,1). Then, the problems 𝓓^𝛼_𝑡𝑥(𝑡)(𝑡)()(0)(1)𝓓+ℎ=0,𝑡∈0,1,𝑥=0,𝑥=0,_𝛽𝑡𝑦(𝑡)+ℎ(𝑡)=0,𝑡∈(0,1),𝑦(0)=0,𝑦(1)=0 (10) have the unique solution 𝑥(𝑡)=¹₀𝐺_𝐴(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,𝑦(𝑡)=¹₀𝐺_𝐵(𝑡,𝑠)ℎ(𝑠)𝑑𝑠, (11) where 𝐺_𝐴(𝑡,𝑠) and 𝐺_𝐵(𝑡,𝑠) are given by 𝐺_𝐴()=1𝑡,𝑠Γ(𝛼)[𝑡(1−𝑠)]^𝛼−1[𝑡(,𝑡≤𝑠,1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1𝐺,𝑠≤𝑡,_𝐵()=1𝑡,𝑠Γ𝛽[𝑡(1−𝑠)]^𝛽−1[],𝑡≤𝑠,𝑡(1−𝑠)^𝛽−1−(𝑡−𝑠)^𝛽−1,𝑠≤𝑡, (12) which are the Green function of the BVP (10), respectively.

It follows from Lemma 3 that the linear problems 𝓓^𝛼_𝑡𝑥(𝑡)(0)(1)𝓓=0,0<𝑡<1,𝑥=0,𝑥=1,_𝛽𝑡𝑦(𝑡)=0,0<𝑡<1,𝑦(0)=0,𝑦(1)=1 (13) have the unique solutions 𝑡^𝛼−1 and 𝑡^𝛽−1, respectively. Letting 𝒜=¹₀𝑡^𝛼−1𝑑𝐴(𝑡),ℬ=¹₀𝑡^𝛽−1𝑑𝐵(𝑡) (14) and defining 𝒢_𝐴(𝑠)=¹₀𝐺(𝑡,𝑠)𝑑𝐴(𝑡),𝒢_𝐵(𝑠)=¹₀𝐺(𝑡,𝑠)𝑑𝐵(𝑡) (15) as in [3–5], we can get that the Green functions for nonlocal BVP (1) are given by 𝐻_𝐴()=𝑡𝑡,𝑠^𝛼−1𝒢1−𝒜^𝐴(𝑠)+𝐺_𝐴(),𝐻𝑡,𝑠_𝐵()=𝑡𝑡,𝑠^𝛽−1𝒢1−ℬ^𝐵(𝑠)+𝐺_𝐵(𝑡,𝑠). (16)

To ensure the nonnegativity of Green functions, we use the following elementary assumption in this paper.(H0)𝐴,𝐵 are increasing functions of bounded variation such that 𝒢_𝐴(𝑠)≥0, 𝒢_𝐵(𝑠)≥0 for 𝑠∈[0,1] and 0≤𝒜, ℬ<1, where 𝒜,ℬare defined by (14).

Lemma 5. Let 1<𝛼, 𝛽≤2, and let (𝐻0) hold; then 𝐻_𝐴(𝑡,𝑠) and 𝐻_𝐵(𝑡,𝑠) satisfy 0≤𝐻_𝐴()≤1𝑡,𝑠(1−𝒜)Γ(𝛼−1)=𝒜^∗,0≤𝐻_𝐵()≤1𝑡,𝑠(1−ℬ)Γ𝛽−1=ℬ^∗. (17) 

The main tool of this paper is the following fixed point theorem, which was firstly obtained by Harjani and Sadarangani [15] and was improved by Tao et al. [6].

Lemma 6 (see [6]). Let (𝑋,≥) be a partially ordered set, and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space. Assume that 𝑋 satisfies the following condition: if 𝑥_𝑛 is a nondecreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥, then 𝑥_𝑛≤𝑥 for all 𝑛∈ℕ. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping, and there exists a constant 𝜆∈(0,1) such that 𝑑𝑇𝑥,𝑇𝑦≤𝜆𝑑𝑥,𝑦−𝜓𝜆𝑑𝑥,𝑦,𝑓𝑜𝑟𝑥≥𝑦, (18) where 𝜓∶[0,+∞)→[0,+∞) is a nondecreasing function. If there exists 𝑥∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a fixed point.

If the space (𝑋,≤) satisfies the following condition: for𝑥,𝑦∈𝑋,thereexists𝑧∈𝑋whichiscomparableto𝑥and𝑦, (19) then we have the following theorem; see [6, 15].

Lemma 7. Adding condition (19) to the hypotheses of Lemma 6, one obtains uniqueness of the fixed point of 𝑇.

Now, we mean by 𝐶[0,1] the Banach space of all continuous functions on [0,1] with the usual ‖𝑥‖=max_0≤𝑡≤1|𝑥(𝑡)|. Note that this space can be equipped with a partial order given by [](𝑡)(𝑡),[].𝑥,𝑦∈𝐶0,1,𝑥≤𝑦⟺𝑥≤𝑦for𝑡∈0,1 (20) It has been proved in [20, 21] that (𝐶[0,1],≤) with the classic metric given by 𝑑𝑥,𝑦=max_0≤𝑡≤1𝑥(𝑡)−𝑦(𝑡) (21) satisfies the following condition.

If 𝑥_𝑛 is a nondecreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥, then 𝑥_𝑛≤𝑥 for all 𝑛∈ℕ. Moreover, for 𝑥,𝑦∈𝐶[0,1], the function max{𝑥,𝑦} is continuous in [0,1], and (𝐶[0,1],≤) satisfies condition (19).

Next, define a subcone of 𝐶[0,1] as follows: 𝑃={𝑥∈𝐶[0,1]∶𝑥(𝑡)≥0}. (22) Note that 𝑃 is a closed set of 𝐶[0,1]; 𝑃 is a complete metric space.

Clearly, (𝑥,𝑦) is a solution of system (1) if and only if (𝑥,𝑦)∈𝐶[0,1]×𝐶[0,1] is a solution of the following nonlinear integral system of equations: 𝑥(𝑡)=¹₀𝐻_𝐴()𝑓(𝑠)𝑦(𝑡)=𝑡,𝑠𝑠,𝑦𝑑𝑠,₁₀𝐻_𝐵(𝑡,𝑠)𝑔(𝑠,𝑥(𝑠))𝑑𝑠. (23) Consequently, system (23) is equivalent to the following integral equation: 𝑥(𝑡)=¹₀𝐻_𝐴(𝑡,𝑠)𝑓𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠. (24) Thus, for 𝑥∈𝑃, define the operator 𝑇 by (𝑇𝑥)(𝑡)=¹₀𝐻_𝐴(𝑡,𝑠)𝑓𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠. (25) Then from the assumption on 𝑓,𝑔, and Lemma 5, we have 𝑇(𝑃)⊂𝑃.

3. Main Results

Now, define the class of functions 𝒳, [)⟶[)(𝑥)(𝑥)[)⟶[).𝒳=𝜙∣𝜙∶0,+∞0,+∞isdifferentialandnondecreasing,and𝜓=𝑥−𝜙satisfies𝜓∶0,+∞0,+∞isnondecreasing (26)

Remark 8. The standard function 𝜙∈𝒳; for example, 𝜙(𝑥)=arctan𝑥, 𝜙(𝑥)=(1/2)𝑥, 𝜙(𝑥)=ln(1+𝑥), 𝜙(𝑥)=𝑥/(1+𝑥), and so forth.

Remark 9. Clearly, given that𝜙₁,𝜙₂∈𝒳, then𝜙₁(𝜙₂(𝑥))∈𝒳.

Theorem 10. Suppose that (𝐻0) holds, and there exist two functions𝜙₁,𝜙₂∈𝒳and constants 𝜌₁, 𝜌₂, 𝜃₁, and 𝜃₂ which satisfy 0<𝜌₁≤𝒜^∗−1,0<𝜌₂≤ℬ^∗−1,0<𝜃₁<𝜌₂⁻¹ℬ^∗−1,0<𝜃₂<1 (27) such that 𝑓𝑡,𝑥₁−𝑓𝑡,𝑥₂≤𝜌₁𝜙₁𝜃₁𝑥₁−𝑥₂,𝑔𝑡,𝑦₁−𝑔𝑡,𝑦₂≤𝜌₂𝜙₂𝜃₂𝑦₁−𝑦₂, (28) for𝑥₁,𝑥₂,𝑦₁,𝑦₂∈[0,+∞)with𝑥₁≥𝑥₂, 𝑦₁≥𝑦₂. Then, problem (1) has a unique nonnegative solution.

Proof. To prove that the problem (1) has a unique nonnegative solution, it is sufficient to check that the hypotheses of Lemma 6 are satisfied.
By the monotonicity of𝑓and𝑔, we know that the operator𝑇is nondecreasing. Now, denote 𝜙(𝑥)=𝜙₁𝜙₂(𝑥). (29) Then,𝜙∈𝒳. For any𝑢≥𝑣∈𝑃, from (27) and (28), we have 𝑑(𝑇𝑢,𝑇𝑣)=max_[]𝑡∈0,1|𝑇𝑢(𝑡)−𝑇𝑣(𝑡)|=max_[]𝑡∈0,1¹₀𝐻_𝐴()×𝑓𝑡,𝑠𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑢(𝜏))𝑑𝜏−𝑓𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑣(𝜏))𝑑𝜏𝑑𝑠≤max_[]𝑡∈0,1¹₀𝐻_𝐴()×𝜌𝑡,𝑠₁𝜙₁𝜃₁¹₀𝐻_𝐵()×𝑔((𝜏((𝜏𝑠,𝜏𝜏,𝑢))−𝑔𝜏,𝑣))𝑑𝑠≤𝒜^∗𝜌₁𝜙₁ℬ^∗𝜃₁𝜌₂𝜙₂𝜃₂(𝑢(𝜏))−𝑣(𝜏)≤𝒜^∗𝜌₁𝜙₁𝜙₂𝜃₂𝑑(𝑢,𝑣)≤𝜙𝜃₂𝑑(𝑢,𝑣). (30)
Since𝜙∈𝒳, which implies that 𝜓(𝑥)=𝑥−𝜙(𝑥), and𝜓∶[0,+∞)→[0,+∞)is nondecreasing. Thus, for𝑢≥𝑣, we find𝜓and𝜃₂such that 𝑑(𝑇𝑢,𝑇𝑣)≤𝜙𝜃₂𝑑(𝑢,𝑣)=𝜃₂𝑑(𝑢,𝑣)−𝜃₂𝑑(𝑢,𝑣)−𝜙𝜃₂𝑑(𝑢,𝑣)=𝜃₂𝑑(𝑢,𝑣)−𝜓𝜃₂𝑑(𝑢,𝑣). (31)
Finally, taking into account the zero function,0≤(𝑇0)(𝑡), by Lemma 6, problem (1) has a unique nonnegative solution.

In the following, we consider the positive solution of the problem (1). A positive solution(𝑥,𝑦)of the problem (1) means a solution of the problem (1) satisfying𝑥(𝑡)>0, 𝑦(𝑡)>0for𝑡∈(0,1).

Theorem 11. Suppose that there exists 𝑡̂∈[0,1] such that 𝑓(𝑡,̂0)≠0and𝑔(𝑡,̂0)≠0, and the assumptions of Theorem 10 also hold; then, the unique solution of (1) is positive.

Proof. It follows from Theorem 10 that the problem (1) has a unique nonnegative solution. We prove that it is also a positive solution of the problem (1).
Otherwise, there exists 0<𝑡^∗<1such that 𝑥(𝑡^∗)=0; that is; 𝑥𝑡^∗=¹₀𝐻_𝐴𝑡^∗𝑓×,𝑠𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠=0. (32) Then, 0=𝑥𝑡^∗=¹₀𝐻_𝐴𝑡^∗𝑓×,𝑠𝑠,¹₀𝐻_𝐵(𝑠,𝜏)𝑔(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠≥¹₀𝐻_𝐴𝑡^∗,𝑠𝑓(𝑠,0)𝑑𝑠≥0. (33) Consequently, ¹₀𝐻_𝐴𝑡^∗,𝑠𝑓(𝑠,0)𝑑s=0. (34) Note that 𝐻(𝑡^∗,𝑠)>0, 𝑠∈(0,1); then, we have 𝑓(𝑠,0)=0,a.e.(0,1). (35)
In addition, it follows from 𝑓(𝑡,̂0)≠0, 𝑡̂∈[0,1] that 𝑓(𝑡,̂0)>0, and by the continuity of𝑓, we can find a set Ω⊂[0,1]such that 𝑡̂∈Ωand𝑓(𝑡,0)>0for any𝑡∈Ω, where𝜇(Ω)>0and𝜇(⋅)stands for the Lebesgue measure, which contradicts with (35). Consequently, we have𝑥(𝑡)>0, 𝑡∈(0,1). In the same way, we also have 𝑦(𝑡)=¹₀𝐻_𝐵(𝑡,𝑠)𝑔(𝑠,𝑥(𝑠))𝑑𝑠 (36) and𝑦(𝑡)>0, 𝑡∈(0,1).Hence, the problem (1) has a unique positive solution.

Example 12. Consider the following boundary value problem with fractional order𝛼=1.5, 𝛽=1.2: −𝓓_𝑡^1.5𝑥(𝑡)=𝑡²𝑦(𝑡)sin𝑡+(𝑡),1001+𝑦−𝓓_𝑡^1.2𝑦(𝑡)=𝑡𝑒^𝑡+1𝑥(𝑡)𝑥(0)(0)(1)=300,0<𝑡<1,=𝑦=0,𝑥¹₀𝑥(𝑠)(𝑠),𝑦(1)=𝑑𝐴₁₀𝑦(𝑠)𝑑𝐵(𝑠), (37) where 𝐴(𝑡)=10,𝑡∈0,3,161,𝑡∈3,23,122,𝑡∈3,𝐵(𝑡)=1,10,𝑡∈0,4,121,𝑡∈4,34,133,𝑡∈4,1. (38) The BVP (37) becomes the 4-point BVP with coefficients: −𝓓_𝑡^1.5𝑥(𝑡)=𝑡²𝑦(𝑡)sin𝑡+(𝑡),1001+𝑦−𝓓_𝑡^1.2𝑦(𝑡)=𝑡𝑒^𝑡+1𝑥(𝑡)𝑥(0)(0)(1)=1300,0<𝑡<1,=𝑦=0,𝑥6𝑥23+13𝑥13,𝑦(1)=12𝑦14+16𝑦34. (39) By Theorem 11, the BVP (37) has a unique positive solution.

Proof. Obviously,𝛼=1.5, 𝛽=1.2, and 𝒜=¹₀𝑡^0.5(𝑡)=1𝑑𝐴613^0.5+1323^0.5≈0.3684<1,ℬ=¹₀𝑡^0.2(𝑡)=1𝑑𝐵214^0.2+1634^0.2𝒜≈0.8171<1,_∗−1=()Γ()ℬ1−𝒜0.5=1.1195,_∗−1=(1−ℬ)Γ(0.2)=0.8397. (40) By simple calculation, we know that 𝒢_𝐴(𝑠)≥0, 𝒢_𝐵(𝑠)≥0, and 𝐴(𝑠), 𝐵(𝑠)are all increasing. So, (H0) holds.
Take 𝑓(𝑡,𝑢)=𝑡²𝑢sin𝑡+(),𝑔()1001+𝑢𝑡,𝑢=𝑡𝑒^𝑡+1()∈[]×[),𝜙300𝑢,𝑡,𝑢0,10,+∞₁(𝑥)=𝑥1+𝑥,𝜙₂(𝑥)=12𝑥. (41) Then, for any𝑥₁≥𝑥₂, 𝑦₁≥𝑦₂, 𝑓𝑡,𝑥₁−𝑓𝑡,𝑥₂=𝑥₁1001+𝑥₁−𝑥₂1001−𝑥₂≤𝑥₁−𝑥₂1001+𝑥₁−𝑥₂≤1×𝑥100¹−𝑥₂1+(1/2)𝑥₁−𝑥₂+1×𝑦300₂−𝑦₁1+(1/3)𝑦₂−𝑦₁=1×()𝑥501/2₁−𝑥₂1+(1/2)𝑥₁−𝑥₂=1𝜙50¹12𝑥₁−𝑥₂,𝑔𝑡,𝑦₁−𝑔𝑡,𝑦₂=1𝑦300¹−𝑦₂≤1211502𝑦₁−𝑦₁, (42) where 𝜌₁=150<1.1195=𝒜^∗−1,𝜌₂=12<0.8397=ℬ^∗−1,𝜃₁=12<1.6794=𝜌₂⁻¹ℬ^∗−1,𝜃₂=150<1. (43) Thus,𝜙₁,𝜙₂∈𝒳, and all of the conditions of Theorem 10 are satisfied.
On the other hand,𝑓(𝜋/6,0)=𝜋²/36≠0, 𝑔(1/2,0)=(1/2)𝑒^1/2≠0, and by Theorem 11, the BVP (37) has a unique positive solution.