The continuity of solution set of a multivalued equation and applications in control problem

In this paper, we prove the existence, unbounded continuity of positive set for a multivalued equation containing a parameter of the form x ∈ A ◦ F (λ, x) and give applications in the control problem with multi-point boundary conditions and second order derivative operator  u′′(t) + g(λ, t)f(u(t)) = 0, t ∈ (0, 1), g(λ, t) ∈ F (λ, u(t)) a.e. on J u(0) = 0, u(1) = ∑m i=1 αiu(ηi) (1)


Introduction
The single-valued equation of the form x = F (λ, x) in ordered spaces has been studied for a long time by many mathematician researchers and has found many successful results (see [2,3,4,11,13,17]). It was naturally generalized to multivalued form x ∈ F (λ, x).
The solution set of (2) is well-known in two following forms In this paper, we prove the existence, unbounded continuity of the solution set S of the form (4), with the equation x ∈ A • F (λ, x), where F is a multivalued function containing a parameter λ and A is a linear mapping. We establish sucient conditions for the set S to be unbounded continuous branch and emanating from zero, i.e., it contains the elements of S on the boundary of any set Ω, which is open, bounded, and contains zero.
Our method is to combine the method of using the denition of topological degree for multivalued mapping and the method of evaluating solutions. The others authors using only one. This is the fundamental dierence between our work and the authors mentioned above.
By the abstract result obtained, we apply for the control problem with second order derivative and multipoint boundary conditions. The such problem has attracted the increasing attention of many researchers. We will solve this problem to illustrate the method.
The paper is organized as follows. In Section 2, we recall some notations and useful lemma. In Section 3, the main results are stated. In Section 4, we present the existence of solutions to the control problem.

Preliminaries
Let (E, K, . ) be a real Banach space ordering by the cone K, i.e., K is a closed convex subset of E such that λK ⊂ K for λ ≥ 0, K ∩ (−K) = {0}, and x ≤ y i y − x ∈ K for x, y ∈ X. For nonempty subsets A, B of E we write A 2 B (or, B 2 A) i for every x ∈ A, there exists y ∈ B satisfying x ≥ y (or, y ≤ x) and we also write A 1 B i for every x ∈ A, there exists y ∈ B such that x ≤ y. The cone K is said to be normal if there exists a constant N > 0 such that 0 ≤ x ≤ y implies x ≤ N y . Throughout this article we always assume that K is normal cone with N = 1. For A ⊂ E, the all nonempty closed convex (resp., closed) subsets of A is denoted by cc(A) (resp., c(A)). Let Ω be an open subset of E, denote Ω K = K ∩ Ω, for all x ∈ ∂ K Ω, the xed point index of T in Ω with respect to K is dened which is an integer denoted by i K (T, Ω) (see e.g. [5]). The following lemma on the computation of the index were taken in [5, proof of Theorem 3.2]. Lemma 2.1. [8, 5, proof of Theorem 3.2] Let T : K ∩ Ω → cc(K) be an u.s.c. compact multivalued operator. Then We review the results using to prove our abstract results.

Abstract results
Lemma 3.1. Let T : [0, ∞) × K → cc(K) be an u.s.c compact operator and Ω 0 be an open bounded subset of E. Assume that the following conditions are satised It is clear that α > 0. Indeed, assume that the following assertion holds Since T is compact, without loss of generality we may assume that This contradicts the rst condition. Thus, there exists > 0 such that By Lemma 2.1 from the rst condition it follows i K (T (0, .), Ω) = 1. Thus i K (T ( , .), Ω) = 1, we deduce λ 0 > α ≥ > 0.
This is a contradiction. Therefore (7) is impossible, i.e., there is By an argument analogous to the previous one we can nd x ∈ ∂ K Ω with x ∈ T (α, x). This contradicts (6). The proof is complete.
is continuous, and F : [0, ∞) × K → cc(K Y ) is u.s.c. compact multivalued operator. Let A : Y → X be a compact linear operator satisfying A(K Y ) ⊂ K. Theorem 3.1. Assume that the following conditions are satised 1. kx ∈ A • F (0, x) for some x ∈ K implies k < 1; 2. there are positive numbers a, b, c and a linear function L : Y → R + with L(y) = 0 for some y ∈ K such that (a) LAx 2 {aLx} and LAx 2 {a. Ax Y } for all x ∈ K Y , (b) L(F (λ, x)) 2 {bλLx − c} for all x ∈ K, and (c) there exists a function h : R + × R + → R increasing on the second variable with Then, S = {x ∈ K : ∃λ > 0, x ∈ A • F (λ, x)} is unbounded continuous branch emanating from 0.

Applications
Let F : [0, ∞) × R + → cc(R + ) be an u.s.c. compact multivalued operator and f : R + → R + be a continuous function. Denote J = [0, 1]. We consider control problem which contains a parameter of the form Let C(J) and C 1 (J), resp., be the Banach spaces of all continuous and continuous dierentiable function Instead of solving problem (20) we shall consider its equivalent form where the multivalued operator T is dened by . Assume that there exist numbers α > 0, β > 0, γ ∈ (0, ρ) and r ∈ (0, 2) such that Then, the set S of positive solutions for (21) is unbounded continuous in C 1 (J), emanating from 0.
Proof. We shall apply Theorem 3.1 with the cone Then, Y and E, resp., are Banach spaces with the norms Suppose x ∈ K and k satises kx ∈ A • T (0, x), we can nd u(s) ∈ F (0, x(s)) such that This implies k < 1. From the well-known results in [17], the compact linear operator A has an eigenvalue µ 0 > 0 with respect to a positive eigen-function u 0 . We dene the linear operator L on Y , by Lx = 1 0 x(s)u 0 (s)ds. From the condition 2, we have where c = β 1 0 u 0 (s)ds. If y is non-negative continuous concave function on J satisfying y(0) = 0 and y(1) ≥ 0, there exists number ξ > 0 such that y(t) ≥ ξ y Y u 0 (t) on J. For x ∈ K Y , Ax is concave function with Ax(0) = 0 and Ax(1) ≥ 0, we have Ax(t) ≥ ξ Ax Y u 0 (t). From Fubini's Theorem it follows that L(Ax) = This implies − x ∈ kT (λ, x) + (1 − k)αλx.
In the following the numbers m j , j = 0, 1, 2, .., 6 and m are constant numbers, not depending on λ, x and t ∈ J. By a similar argument as the proof of Theorem 3.1 we obtain Therefore we can choose m 1 such that λ x Y ≤ m 1 From (26), the well-known inequality and (24) we obtain Further, for x ∈ K, we have x Y ≤ m 0 x Y . Combining this inequality, (26), (27), (29) and (29) we have From (27) we can choose m such that x Y ≤ m x 1 2 Y . Since x = x Y , the condition (2c) of Theorem 3.1 are satised with function h(λ, t) = mt 1 2 .

Conclusion
In this paper, the unbounded continuity of positive solution set for a multivalued equation containing a parameter has established and given the application in the control problem with multi-point boundary conditions.