Existence of Solution to a Second-Order Boundary Value Problem via Noncompactness Measures

The existence and uniqueness of the solutions to the Dirichlet boundary value problem in the Banach spaces is discussed by using the fixed point theory of condensing mapping, doing precise computation of measure of noncompactness, and calculating the spectral radius of linear operator.


Introduction
This paper is mainly concerned with the following second-order Dirichlet boundary value problem: in a Banach space E, where f t, x ∈ C I × E, E , θ is the zero element of E.
In the last several decades, there has been much attention focused on the boundary value problems for various nonlinear ordinary differential equations, difference equations, and functional differential equations, see 1-20 and the references therein.The existence of solutions for Neumann boundary value problems has been considerably investigated in many publications such as 2-5, 8-10 .Dirichlet boundary value problems have deserved the attention of many researchers, see 11-20 and the references therein.
In particular, the authors in 11 have studied the following two-point boundary value problem: where a, b, c, d ≥ 0 and ad bc > 0. They obtained the existence of solutions by means of the Darbo fixed point theorem and properties of the measure of noncompactness.
We would like to mention the results due to 11 .First, we point out that many authors applied the famous Sadovskii's fixed point theorem to investigate similar problems and used the following hypothesis with respect to the Kuratowski measure of noncompactness α • : there exists a constant k > 0 such that for any bounded and equicontinuous set A, B ⊂ C I, E and t ∈ I, α H I × A × B ≤ k max{α A , α B }. What is more, they required a stronger condition, that is, H t, x, y ≤ L for t, x, y ∈ I × E × E and the constant k satisfies 0 < k < 1/2 see Remarks 3.2-3.6 .
The authors in 15, 18 have studied the following boundary value problem: where They obtained the existence of solutions by means of Sadovskii's fixed point theorem and properties of the measure of noncompactness.
Motivated by the above-mentioned work 11, 15, 18 , the main aim of this paper is to study the existence and uniqueness of solutions for the problem 1.1 under the new conditions.The main new features presented in this paper are as follows First, the existence and uniqueness of solutions to Banach space's Dirichlet boundary value problem is proved precisely calculating the spectral radius of linear operation.Second, the conditions imposed on the BVP 1.1 are weak.Third, the main tools used in the analysis are Sadovskii's fixed point theorem and precise computation of measure of noncompactess.Our results can be seen as a supplement of the results in 11 see Remarks 3.2-3.6 .
This paper is organized as follows.In Section 2, we provide some basic definitions, preliminaries facts, and various lemmas which will be used throughout this paper.In Section 3, we give main results in this paper.

Preliminaries and Lemmas
Let E be a real Banach space and P be a cone in E which defines a partial ordering in E by x ≤ y if and only if y − x ∈ P • P is said to be normal if there exists a positive constant N such that θ ≤ x ≤ y implies x ≤ N y , where θ denotes the zero element of E, and the smallest N is called the normal constant of P it is clear, N ≥ 1 .If x ≤ y and x / y, we write x < y.For details on cone theory, see the monograph 7 .

2.5
Case 2. if M 0, we have

2.7
After direct computations, it is easy to see that It is easy to see that T M is bounded linear operator.This completes the proof.
Lemma 2.5.Assume that M > −π 2 , and T M : C I, E → C I, E is given by 2.8 .Then where By the spectral mapping theorem 21 , we get then we get G M t, s ≥ 0, so by 2.8 , we have T M h ≥ 0, that is, T M is a positive operator.This achieves the proof.
Remark 2.6.In particular.If E R 1 , M 0, then by 1 of Lemma 2.5, we get r T 0 1/π 2 , where T 0 is an operator in C I : 12 and T 0 c 1/8.In fact:
Proof.If b a ϕ s ds 0, then 2.14 is true.We suppose that b a ϕ s ds > 0, and take a partition of a, b : 2.17 This finishes the proof.
Lemma 2.8.Suppose that D is a bounded set in E, then there exists a countable subset D 1 of D, such that

2.22
For any u, v ∈ B i and t, s ∈ I j , it follows from 2.20 and 2.21 that Proof.For any ε > 0, it follows from the equicontinuity of Ω 1 that there exists δ > 0 such that For any h ∈ co Ω 1 , by virtue of definition of co Ω 1 , we have

2.27
Thus, we get

Main Results
In this section, we present and prove our main results.

3.4
Hence, Using the properties of the noncompactness measure together with H 2 , we obtain Lα B I .
Let Ω : B c θ, R {u ∈ C I, E : u < R}, we will prove that u − λAu / θ, 0 < λ ≤ 1, u ∈ ∂Ω for R sufficiently large.By means of the homotopy invariance theorem, we have deg I − A, Ω, θ 1.By virtue of the solvability of Kronecker 6 , we know that A has a fixed point in Ω, and the fixed point of A is a solution of the problem 1.1 .

3.11
Thus, 12 continuing this process, by induction, we obtain which implies that By the Gelfand theorem 22 , we have

3.15
Set c 2 c 1 π 2 /2 ∈ c 1 , π 2 , then we have 1/c 2 > 1/π 2 .By 3.15 , there exists an integer N 0 , such that n T n 0 < 1/c 2 as n ≥ N 0 , that is, Take R > R 0 , then we have u−λAu / θ, for all λ ∈ 0, 1 , u ∈ ∂Ω.Thus problem 1.1 has at least one solution.This proves the theorem.Remark 3.2.In 11 , the nonlinear term f t, u, u is bounded, if f t, u, u f t, u , in our result, the nonlinear term f t, u may no more than a linear growth.

3.19
By H 3 , we get 0 < L/2 < 1, therefore A is condensing.By using the same arguments of Theorem 3.1, we can obtain the conclusion of Theorem 3.4.The detailed proof is omitted here.The proof is achieved.
Next, we establish a uniqueness of solution for the problem 1.1 .
Theorem 3.5.Let E be a Banach space.Suppose that f t, x ∈ C I × E, E and that there exists a constant L with 0 < L < π 2 such that

3.20
Then problem 1.1 has a unique solution.
Proof.Assume that operator A is defined the same as in Theorem 3.1, and the fixed point of A is a solution of the problem 1.1 .We will prove that for sufficiently large n the operator A n is a contraction operator.Indeed, by the definition of A and 3.20 , we have the estimate

3.21
By induction, we have

3.23
Moreover, we can choose n to be sufficiently large such that n T n 0 tends to r T 0 1/π 2 .
Further, take L 1 ∈ L, π 2 , there exists an integer N 0 , such that T 0 ≤ 1/L n 1 as n ≥ N 0 .By 3.23 , we obtain which implies that A n is a contraction mapping by L/L 1 < 1.By the contraction mapping principle, we conclude that there exists a unique fixed point for A, this proves that problem 1.1 has a unique solution.This completes the proof.
Remark 3.6.By the direct application of the Banach contraction mapping principle, the conclusion of Theorem 3.5 holds true under the condition 0 < L < 8.However, we require the condition 0 < L < π 2 , here π 2 is optimum.
The following theorem is concerned with the existence of positive solutions for problem 1.1 .Theorem 3.7.Let E be an ordered Banach space, K be a normal cone with positive elements.Suppose that f t, x ∈ C I × E, E satisfy the following conditions: P 1 there exists a constant c with 0 < c < π 2 , and h 0 ∈ C I, K , such that θ ≤ f t, x ≤ cx h 0 t , ∀x ≥ θ, 3.25 P 2 for any a bounded set D in E, there exists a constant L with 0 < L < 8 such that α f I × D ≤ Lα D .

3.34
This means that α A B ≤ L 8 • α B .

3.35
By P 2 , we have 0 < L/8 < 1, so A is condensing.Applying Lemma 2.14, we conclude that A has a fixed point which is a solution of problem 1.1 .The proof of the theorem is completed.

2 . 16 6
Discrete Dynamics in Nature and SocietyWe take n sufficiently large, such that Δ n → 0, then we get b a ϕ s u s ds b a ϕ s ds

G 33 Further, we obtain α A B t ≤ 1 0G 2 t 1 −
0 t, s ds • co f I × B I .3.0 t, s ds • α f I × B I ≤ L t • α B I ≤ L 8 • α B .
C I, E is continuous, and it is clear that u is a solution of the problem 1.1 if and only if u is a fixed point of A. We now show that A is a condensing operator.Let B be bounded in C I, E , by H 1 , we claim that {− Au | u ∈ B} is bounded.Since Au 0 0, we know that { Au | u ∈ B} is bounded, this means that A B is equicontinuous.Therefore, it follows from Lemma 2.13 that α A B max x∈I α A B t .For any u ∈ B, t ∈ I, from Lemma 2.7, we have Theorem 3.1.LetE be a Banach space.Suppose that f t, x ∈ C I × E, E and the following conditions hold:H 1 there exist two positive numbers c 0 and c 1 , such that f t, x ≤ c 0 c 1 x , ∀t ∈ I, x ∈ E, 3.1H 2 for any a bounded set D in E, there exists a constant L > 0 such thatα f I × D ≤ Lα D , 3.2H 3 there exist two positive numbers c 1 and L with c 1 < π 2 , L < 4. 0 t, s f s, u s ds T 0 f •, u .3.3ThenA : C I, E → Remark 3.3.In 11 , if f t, u, u f t, u , the growth restriction of L for H 2 is 0 < L < 1/2.However, in our result, L satisfies 0 < L < 4. Let E be a Banach space, and f t, x ∈ C I × E, E .Assume that condition H 1 and the following conditions hold:H 2 for all t ∈ I, for any a bounded set D in E, there exists a constant L > 0 such thatα f t, D ≤ Lα D , 3.16H 3 there exist two positive numbers c 1 and L with 0 < c 1 < π 2 , L < 2. Assume that the operator A is defined the same as in Theorem 3.1.We show that the operator A is condensing.In fact, for a bounded set B ∈ C I, E , there exists a countable subset B 1 {u n }, such that α A B ≤ 2α A B 1 .However, on the other hand, we have 1 has at least one positive solution.⊂CI,E, then D is bounded and convex closed set in C I, E .For any u ∈ D, by P 1 , we haveθ ≤ f t, u t ≤ cu t h 0 t ≤ cu 0 t h 0 t .3.28Multiply by G 0 t, s and integrate from 0 to 1, we obtain Au ∈ D, so A D ⊂ D. By the proof of Theorem 3.1, it follows that A is equicontinuous.Thus, by Lemma 2.12, we know that Ω 0 co A D is equicontinuous.Next we show that A : Ω 0 → Ω 0 is condensing.For any B ⊂ Ω 0 , then B is bounded and equicontinuous, therefore A B ⊂ Ω 0 is bounded and equicontinuous.By Lemma 2.13, we have Thus, For any u ∈ B, t ∈ I, we acquire