A Second-Order Boundary Value Problem with Nonlinear and Mixed Boundary Conditions: Existence, Uniqueness, and Approximation

and Applied Analysis 3 In BVP 1.4 – 1.6 , it can be found that the boundary condition 1.5 is dependent on all the x a , x b , x′ a , and x′ b terms. First, existence and uniqueness of solutions of BVP 1.4 – 1.6 is established by combining the method of upper and lower solutions with LeraySchauder degree theory. Then, the general quasilinearization method is applied to construct the approximations of the unique solution. Twomonotone sequences of iterations converging to the unique solution quadratically are obtained. 2. Preliminaries In this section, several definitions and lemmas needed to the main results are given first. Definition 2.1. β t , α t ∈ C2 a, b are called the upper and lower solutions of BVP 1.4 – 1.6 , respectively, if Lβ ≥ ft, β, β′, t ∈ a, b , g ( β a , β b , β′ a , β′ b ) ≤ 0, β b β a , Lα ≤ ft, α, α′, t ∈ a, b , g ( α a , α b , α′ a , α′ b ) ≥ 0, α b α a . 2.1 Definition 2.2. Let E be a subset of a, b × R2; it is said that the right-hand side function of 1.4 satisfies Nagumo condition on E if ∣f ( t, x, x′ )∣ ≤ h∣x′∣ O ∣x′ ∣2 ) 2.2 holds for t, x, x′ ∈ E and |x′| → ∞. Lemma 2.3 see 28 . Let f : a, b × R2 → R be a continuous function satisfying Nagumo condition on E {( t, x, x′ ) ∈ a, b × R2 : α t ≤ x t ≤ β t } , 2.3 where α, β : a, b → R are continuous functions such that α t ≤ β t for all t ∈ a, b . Then there exists a constant N > 0 such that every solution x t of second-order equations x′′ f t, x, x′ with α t ≤ x t ≤ β t , t ∈ a, b 2.4 satisfies ‖x′‖∞ ≤ N, in which N is called the Nagumo constant. 4 Abstract and Applied Analysis Lemma 2.4. Boundary value problem as follows: Lx −x, t ∈ a, b , 2.5 x a 0, x b 0 2.6 has only the trivial solution. Proof. Assume that x0 t is an arbitrarily nontrivial solution of BVP 2.5 2.6 . From the boundary conditions 2.6 , it can be concluded that x0 t can achieve its positive maximum or negative minimum in the interior of a, b , suppose at t0, t0 ∈ a, b . If x0 t achieves its positive maximum, then x0 t0 > 0, x′ 0 t0 0, x ′′ 0 t0 ≤ 0 2.7 which means that Lx0 t0 −p t0 x′′ 0 t0 − p′ t0 x′ 0 t0 q t0 x t0 ≥ 0. 2.8 On the other hand, it can be derived from 2.5 that Lx0 t0 −x0 t0 < 0. 2.9 It is a contradiction. If x0 t achieves its negative minimum, similar arguments lead to a contradiction too. Hence, BVP 2.5 2.6 has only the trivial solution. Lemma 2.5 see 26 . Define a linear operator l : C1 a, b −→ C a, b × R × R 2.10