On a Boundary Value Problem of Hybrid Functional Differential Inclusion with Nonlocal Integral Condition

In this work, we present a boundary value problem of hybrid functional differential inclusion with nonlocal condition. The boundary conditions of integral and infinite points will be deduced. The existence of solutions and its maximal and minimal will be proved. A sufficient condition for uniqueness of the solution is given. The continuous dependence of the unique solution will be studied.

Here, we assess the boundary value problem of hybrid nonlinear functional differential inclusion with nonlocal condition.
d dt x(t) − x(0) g(t, x(φ 1 (t))) ∈ F(t, x(φ 2 (t))), t ∈ (0, 1) with the nonlocal boundary condition The existence of solutions x ∈ C[0, 1] will be proved. The maximal and minimal solutions will be studied. A sufficient condition for uniqueness of the solution will be given. The continuous dependence of the unique solution on x o and on ∑ m k=1 a k will be proved. Additionally, we deduce the same results for the boundary value problem of hybrid nonlinear functional differential inclusion (1) with a nonlocal integral condition 1 0 x(s)dh(s) = x 0 (3) and infinite point boundary conditions ∞ ∑ k=1 a k x(τ k ) = x 0 , a k > 0 τ k ∈ [0, 1]. (4) and that the function f satisfies the differential equation d dt x(t) − x(0) g(t, x(φ 1 (t))) = f (t, x(φ 2 (t))), t ∈ (0, 1).
Therefore, any solution of the nonlocal problem of the hybrid functional differential Equation (5) with any of the nonlocal boundary conditions (2)-(4) is a solution of the nonlocal problem of the hybrid nonlinear functional differential inclusion with any one of the nonlocal conditions (1)-(4).
(II) g : I × R −→ R is measurable in t for any x ∈ R and Lipschitz in x for t ∈ [0, 1], and there exists a positive constant K 1 > 0 such that |g(t, x) − g(t, y)| ≤ K 1 |x − y|, ∀t ∈ I, and x, y ∈ R.

Definition 1. x of the problem in Equations
satisfies (5). Now, we have the following lemma.

Lemma 1.
If the solution of the problems in Equations (2) and (5) exists, then it can be expressed by the integral equation Proof. Let the boundary value problem in Equations (2) and (5) be satisfied; then, we can obtain Putting t = τ and multiplying both sides of (7) by a k , we obtain Substituting (8) in (7), we obtain (6).

Existence of Solutions
Theorem 1. Assume that assumptions (I)-(III) are valid. Then, the integral Equation (6) has at least one solution x ∈ C[0, 1].
Proof. Define the set Q r by Define the operator F by Thus, the class of functions {F x} is uniformly bounded on Q r and F : Q r → Q r . Let x ∈ Q r and t 1 , Thus, the class of functions {F x} is equicontinuous on Q r and {F y} is a compact operator by the Arzela-Ascoli Theorem [19]. Now, we prove that F is a continuous operator. Let x n ⊂ Q r be a convergent sequence such that x n → x; then, Using Lebesgue-dominated convergence Theorem [19] and assumptions (iv)-(III), we have Then, F : Q r → Q r is continuous, and by Schauder fixed point Theorem [19], there exists at least one solution x ∈ C[0, 1] of (6). Now, putting t = τ k and multiplying by Σ m k=1 a k in (6), we obtain This proves the equivalence between the problem in Equations (2) and (5) and the integral Equation (6). Then, there exists at least one solution x ∈ C[0, 1] of the hybrid nonlinear functional differential Equation (5) with the nonlocal condition (2). Consequently, there exists at least one solution x ∈ C[0, 1] of the nonlocal problem of the hybrid nonlinear functional differential inclusion (1) with the nonlocal condition (2).

Maximal and Minimal Solutions
Here, we study the maximal and minimal solutions for the problem in Equations (2) and (5). Let u(t) be a solution of (6); then, u(t) is said to be a maximal solution of (6) if it satisfies the inequality A minimal solution v(t) can be defined in a similar way by reversing the above inequality i.e.,

Lemma 2.
Let the assumptions of Theorem 1 be satisfied. Assume that x and y are two continuous functions on [0, 1] satisfying.
where one of them is strict.
Let the functions f and g be monotonically nondecreasing in x; then, Proof. Let the conclusion (9) be untrue; then, there exists t 1 with If f and g are monotonic functions in x, we have This contradicts the fact that x(t 1 ) = y(t 1 ). This completes the proof.
For the continuous maximal and minimal solutions for (6), we have the following theorem. Proof. First, we must demonstrate the existence of the maximal solution of (6). Let > 0 be given. Now, consider the integral equation Let 1 , 2 be such that 0 < 2 < 1 ; then, Additionally, Then, Applying Lemma 2, we obtain As shown before, the family of function x (t) is equi-continuous and uniformly bounded; then, by the Arzela Theorem, there exists a decreasing sequence n such that 0 → 0 as n → ∞, and u(t) = lim n→∞ x n (t) exists uniformly in [0, 1], and denote this limit by u(t). From the continuity of the functions, f (t, x (φ 2 (t))), we get f (t, x (φ 2 (t))) −→ f (t, x(φ 2 (t))) as n → ∞, g (t, x (φ 2 (t))) −→ g(t, x(φ 2 (t))) as n → ∞ and Now, we prove that u(t) is the maximal solution of (6). To do this, let x(t) be any solution of (6); then, and Then, Applying Lemma 2, we obtain From the uniqueness of the maximal solution, it clear that x (t) tends to u(t) uniformly in [0, 1] as → 0 in a similar way as above and we can prove the existence of the minimal solution.

Uniqueness of the Solution
Here, we study a sufficient condition for the uniqueness of the solution x ∈ C[0, 1] of the problem in Equations (2) and (5).
(ii) The set F(t, x) is nonempty, compact, and convex for all (t, 1] for every x ∈ R and satisfies the Lipschitz condition with a positive constant K 2 such that where H (A, B) is the Hausdorff metric between the two subsets A, B ∈ I × E (see [16]).

Remark 2.
From this assumptions, we can deduce that there exists a function f ∈ F(t, x) such that (iv) f : I × R → R is measurable in t ∈ [0, 1] for every x ∈ R and satisfies the Lipschitz condition with a positive constant K 1 such that (see [19][20][21]) (I I * )g : I × R −→ R is continuous and satisfies the Lipschitz condition with positive constant K 1 such that |g(t, x) − g(t, y)| ≤ K 1 |x − y|.

Definition 2. The unique solution of the problem in Equations
Then,

Riemann-Stieltjes Integral Condition
Let x ∈ C[0, 1] be the solution of the nonlocal boundary value problem in Equations (2) and (5).