Problems with mixed boundary conditions in Banach spaces

Using Leray-Schauder degree or degree for $\alpha$-condensing maps we obtain the existence of at least one solution for the boundary value problem of the type \[ \left\{\begin{array}{lll} (\varphi(u' ))' = f(t,u,u')&&\\ u(T)=0=u'(0),&&\quad \quad \end{array}\right. \] where $\varphi: X\rightarrow X $ is a homeomorphism with reverse Lipschitz such that $\varphi(0)=0$, $f:\left[0, T\right]\times X \times X \rightarrow X $ is a continuous function, $T$ a positive real number and $X$ is a real Banach space.


Introduction
The purpose of this article is to obtain some existence results for the nonlinear boundary value problem of the form ( ( )) = ( , , ) , where : → is a homeomorphism such that (0) = 0 and −1 is Lipschitz constant, : [0, ] × × → is a continuous function, is a positive real number, and is a real Banach space. Of course, a solution of this problem is a function : [0, ] → of class 1 such that the function → ( ( )) is continuously differentiable, satisfying the boundary conditions, and ( ( ( ))) = ( , ( ), ( )) for all ∈ [0, ].
In particular, the authors in [3] have studied the following boundary value problem: = ( , , ) , where , , , , and + > 0. They obtained the existence of solutions of (2) using Darbo fixed point theorem and properties of the measure of noncompactness.
Recently, Zhou and Peng [5] have studied the following boundary value problem: − = ( , ) , where : [0, 1] × → is a continuous function and is a Banach space. They obtained the existence of solutions of (3), where the main tools used in the study are Sadovskii fixed point theorem and precise computation of measure of noncompactness.
Inspired by these results, the main aim of this paper is to study the existence of at least one solution for the boundary value problem (1) using Leray-Schauder degree or degree for -condensing maps. For this, we reduce the nonlinear boundary value problem to some fixed points problem. Next, we shall essentially consider two types of regularity assumptions for ( , , ). In Theorem 10 we suppose that is completely continuous, which allows us to prove that the associated fixed point operator is completely continuous required by a Leray-Schauder approach. In Theorem 11 we only assume some regularity conditions expressed in terms of the measure of noncompactness, which allows us to apply the methods of topological degree theory for -condensing maps.

Chinese Journal of Mathematics
The paper is organized as follows. In Section 2, we establish the notation, terminology, and various lemmas which will be used throughout this paper. In Section 3, we formulate the fixed point operator equivalent to problem (1). In Section 4, we give main results in this paper. In Section 5, we study the existence of at least one solution for (1) in Hilbert spaces. For these results, we adapt the ideas of [7][8][9] to the present situation.

Notations and Preliminary Results
We first introduce some notation. For fixed , we denote the usual norm in 1 We introduce the following applications: The Nemytskii operator the integration operator : → 1 is and the continuous linear application : → 1 is Throughout this paper, we denote by ( , ‖⋅‖) a real Banach space and = [0, ]. For ⊆ 1 , we use the notation   (g) (conv( )) = ( ).
The details of and its properties can be found in [10].
Definition 2 (see [11]). Assuming that ⊂ the mapping : → is said to be a condensing operator if is continuous and bounded (sends bounded sets into bounded sets), and, for any nonrelatively compact and bounded set ⊂ , The following lemmas are of great importance in the proof of our main results. The proofs can be found in [11].
In the following, we denote and 1 as the noncompactness measure in and 1 , respectively.

Lemma 3. Let be a bounded subset of real numbers and a bounded subset of . Then
Lemma 4. Let , be bounded subsets of Banach spaces and , respectively, with Then Lemma 5. If ⊂ is bounded and equicontinuous, then we have the following: (1) ( ) = ( ( )).

Lemma 7.
If is a bounded set in 1 and equicontinuous, then

Fixed Point Formulations
Let us consider the operator Chinese Journal of Mathematics 3 Here −1 is understood as the operator −1 : → defined for −1 (V)( ) = −1 (V( )). It is clear that −1 is continuous and sends bounded sets into bounded sets.

Lemma 8. ∈ 1 is a solution of (1) if and only if is a fixed point of the operator 1 .
Proof. Let be a solution of (1). This implies that ( ( )) = ( , , ) , Integrating of 0 to and using the fact that (0) = 0, we deduce that Applying −1 and to both of its members and using that ( ) = 0, we have that Conversely, since, by definition of the mapping 1 , it is a simple matter to see that if is such that = 1 ( ) then is a solution to (1).
Using the theorem of Arzelà-Ascoli we show that the operator 1 is completely continuous.

Lemma 9. If is completely continuous, then the operator
Next, we show that 1 (Λ) ⊂ 1 is a compact set. Let (V ) be a sequence in 1 (Λ), and let ( ) be a sequence in Λ such that V = 1 ( ). Using (19), we have that there exists a constant > 0 such that, for all ∈ N, which implies that Hence the sequence ( ( ( ))) is bounded in . Moreover, for , 1 ∈ [0, ] and for all ∈ N, we have that which implies that ( ( ( ))) is equicontinuous.
On the other hand, for ∈ [0, ], where Recalling that the convex hull of a set ⊆ is given by it follows that which implies that Using the fact that : [0, ] × × → is completely continuous, we deduce that ( ( )) = 0. Hence, ( ) is a relatively compact set in . Thus, by the Arzelà-Ascoli theorem there is a subsequence of ( ( ( ))) , which we call ( ( ( ))) , which is convergent in . Using the fact that −1 : → is continuous it follows from that the sequence ( 1 ( ) ) is convergent in and hence (V ) = ( 1 ( )) is convergent in 1 . Finally, let (V ) be a sequence in 1 (Λ). Let ( ) ⊆ 1 (Λ) be such that Let ( ) be a subsequence of ( ) such that it converges to . It follows that ∈ 1 (Λ) and (V ) converge to . This concludes the proof.
In order to apply Leray-Schauder degree to the operator Notice that (30) coincide, for = 1, with (1). So, for each ∈ [0, 1], the operator associated with (30) for Lemma 8 is the operator ( , ⋅), where is defined on [0, 1] × 1 by Using the same arguments as in the proof of Lemma 9 we show that the operator is completely continuous. Moreover, using the same reasoning as above, system (30) (see Lemma 8) is equivalent to the problem = ( , ) . (32)

Main Results
In this section, we present and prove our main results.
Then problem (1) has at least one solution.

Theorem 11. Let be a Banach space and −1 a homeomorphism with Lipschitz constant . Assume that is continuous and satisfies the following conditions:
(1) There exist two numbers 0 , 1 ≥ 0 such that (2) For all bounded subsets , in , where 0 < 1 < 1 2 .
Proof. Observe that 1 maps bounded sets into bounded sets. Furthermore, its continuity follows by the continuity of the operators which compose 1 . We show that the operator 1 is condensing ( -condensing). In fact, for a bounded set Λ in 1 , there exists a constant 1 > 0 such that For , 1 ∈ [0, 1] we have that Let us consider the first case.
Using the properties of , we see that    This implies, by Lemma 6, that Consider the alternative case. Proceeding as before, we obtain Therefore, in either case, we obtain By (41), we get 0 < 2 1 < 1, and therefore 1 iscondensing. Let us consider the functioñ Let ( , ) ∈ [0, 1]× 1 be such that =̃( , ). Using the fact that −1 is a homeomorphism with Lipschitz constant and Gronwall's Inequality, we deduce that there exists a constant > 0 such that ‖ ‖ 1 < . Finally, we show the existence of at least one solution of (1) using the homotopy invariance of the degree forcondensing maps. Let be bounded in 1 Then, from the existence property of degree, there exists ∈ (0) such that =̃(1, ) = 1 ( ) = , which is a solution for (1).
Remark 12. In [3], the nonlinear term ( , , ) is bounded; in our result, the nonlinear term ( , , ) may possess no more than linear growth. Now we consider an example to illustrate our results.
and hence So, by Theorem 11 we get one solution.

Boundary Value Problems in Hilbert Spaces
Throughout this section, let ( , ⟨⋅, ⋅⟩) denote a real Hilbert space. Assume that : → satisfies the following conditions: (1) −1 is a homeomorphism with Lipschitz constant .
Using a proof similar to that of Theorem 11, we obtain the following existence result.
Then, from the existence property of degree, there exists ∈ (0) such that = (1, ), which is a solution for (1).
The following corollary is concerned with the existence of one solution for (1).
By using the arguments of Theorem 16, we can obtain the conclusion of Corollary 17.

Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of this article.