On a nonlinear differential equation with two-point nonlocal condition with parameters

On a nonlinear differential equation with two-point nonlocal condition with parameters A. M. A. El-Sayed1,∗, M. SH. Mohamed1 and E. M. Al-Barg2 1 Faculty of Science, Alexandria University, Alexandria, Egypt. 2 Faculty of Science, Sirt University, Libya. * Correspondence: eman.albarg@gmail.com Received: 2 March 2020; Accepted: 7 July 2020; Published: 31 August 2020. Abstract: Here we study the existence of solutions of a nonlocal two-point, with parameters, boundary value problem of a first order nonlinear differential equation. The maximal and minimal solutions will be proved. The continuous dependence of the unique solution on the parameters of the nonlocal condition will be proved. The anti-periodic boundary value problem will be considered as an application.


Existence of solutions
Consider the Problem (1)-(2) under the following assumptions; (i) f : [0.T] × R → R is measurable in t ∈ [0, T] for every x ∈ R and continuous in x ∈ R for every t ∈ [0, T].

Integral equation representation
Here we give the integral representation of the solution of the Problem (1)-(2) if it exists. We have the following lemma.

Proof. Define the operator F by
Define the set Then the class of functions {Fx} is uniformly bounded on Q r , and F : Q r → Q r . Let x ∈ Q r and t 1 , So, the class of functions {Fx} is equi-continuous on Q r . From Arzela Theorem [18] we deduce that the class of functions {Fx} is compact, and F : Q r → Q r is compact. Now we prove that F is continuous operator.
Applying Lebesgue dominated convergence Theorem [18], we have Hence, F : Q r → Q r is continuous. Now by Schauder fixed point Theorem [19] there exists at least one solution x ∈ C[0, T] of the Problem (1)- (2). Let x ∈ C[0, T] be a solution of the Problem (1)- (2). Differentiating the integral Equation (7), we obtain Since f is measurable in t ∈ [0, T] and bounded by integrable function, then f ∈ L 1 [0, T], and Putting t = τ in the integral Equation (7), we get

Maximal and minimal solution
Let u(t) be a solution of the integral Equation (7), then u(t) is said to be a maximal solution of (7), if for every solution of (7) satisfies the inequality: A minimal solution v(t) can be defined by similar way by reversing the above inequality i.e., We will use the following lemma to prove the existence of the maximal and minimal solutions.

Lemma 2.
Let the assumption of Theorem 1 are satisfied and x(t) and y(t) are two continuous functions on [0, T] satisfying and one of them is strict. If f is monotonic nondecreasing in x, then Proof. Let the conclusion (8) is false, then there exist t 1 such that From the monotonicity of f (t, x(t)) in x, we have This contradicts the fact that x(t 1 ) = y(t 1 ), then For the existence of the maximal and minimal solutions we have the following theorem.

Theorem 2.
Let the assumptions of Theorem 1 are satisfied. If f (t, x(t)) is monotonic nondecreasing in x for each t ∈ [0, T], then the Equation (7) (consequently the Problem (1)- (2)) has maximal and minimal solutions.
Proof. Firstly we shall prove the existence of the maximal solution of (7). Let > 0 be given then consider the integral equation where It is clear that the Equation (9) has at least one solution x (t) ∈ C[0, T]. Now, let 1 and 2 be such that 0 < 2 < 1 < , then Applying Lemma 2, we obtain As shown before the family of function x (t) is equi-continuous and uniformly bounded. Then, by Arzela Theorem [18], there exist a decreasing sequence n such that 0 → 0 as n → ∞, and u(t) = lim n→∞ x n (t) exists uniformly in [0, T] and denote his limit by u(t). From the continuity of the functions f (t, x (t)), we get f (t, x (t)) → f (t, x(t)) as n → ∞ and Now we prove that u(t) is the maximal solution of (7). To do this, let x(t) be any solution of (7), then and Applying Lemma 2, we obtain From the uniqueness of the maximal solution, it is clear that x (t) tends to u(t) uniformly in [0, T] as → 0. By similar way as done above we can prove the existence of the minimal solution.

Uniqueness of the solution
Consider the problem (1)-(2) under the following assumptions Proof. From assumption (i * ) we get Then the assumptions (ii) is satisfied, so there exists at least one solution x ∈ AC[0, T] of the Problem (1)- (2). Let x and y be two solutions of the Problem (1)-(2), then we have Hence, the solution of the integral Equation (7) This prove the continuous dependence of solution of the nonlocal two-point boundary value Problem (1)-(2) on x 0 . Definition 2. The solution of the nonlocal two-point boundary value Problem (1)-(2) depends continuously on α and β, if ∀ > 0, ∃ δ > 0, we have where x * is the unique solution of the nonlocal two-points boundary value Problem (1)-(2).

Conclusions
We proved here, under certain conditions, the existence of at least one absolutely continuous solution x ∈ AC[0, T] of the nonlocal two-point, with parameters ( α, β and x o ) boundary value Problem (1)- (2). The maximal and minimal solutions of the Problem (1)- (2) have been proved. The continuous dependence of the unique solution on the parameters α, β and x o ) have been also proved. The anti-periodic boundary value problem have been considered as an application.