Lower and upper solutions method to the fully elastic cantilever beam equation with support

The aim of this paper is to consider a fully cantilever beam equation with one end fixed and the other connected to a resilient supporting device, that is, {u(4)(t)=f(t,u(t),u′(t),u″(t),u‴(t)),t∈[0,1],u(0)=u′(0)=0,u″(1)=0,u‴(1)=g(u(1)),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)), \quad t\in [0,1], \\ u(0)=u'(0)=0, \\ u''(1)=0,\qquad u'''(1)=g(u(1)), \end{cases} $$\end{document} where f:[0,1]×R4→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f:[0,1]\times \mathbb{R}^{4}\rightarrow \mathbb{R}$\end{document}, g:R→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$g: \mathbb{R}\rightarrow \mathbb{R}$\end{document} are continuous functions. Under the assumption of monotonicity, two existence results for solutions are acquired with the monotone iterative technique and the auxiliary truncated function method.


Introduction
In this paper, we investigate a fully fourth-order differential equation with nonlinear boundary condition (4) (t) = f (t, u(t), u (t), u (t), u (t)), t ∈ [0, 1], where f : [0, 1] × R 4 → R, g : R → R are continuous functions. The two-point boundary value problems (BVPs) of fourth-order differential equations are the mathematical models for describing the states of elastic beams. It is generally known that the elastic beams are one of the basic structures of modern architecture, aircraft, and ships. Due to their practical mathematical models and extensive application background, they have attracted the general attention of researchers, see  and the references therein.
It is common knowledge that in the force analysis of beams, the physical meaning of the derivatives u (t), u (t), u (t), and u (4) (t) of u(t) are slope, bending moment, shear force, and load density, respectively (see [1,6,7,[11][12][13][14]). Thus, the nonlinear boundary condition indicates that the shear force is equal to g(u(1)), which implies that there may be a nonlinear relationship between the displacement u(1) and the vertical force, as well as u (1) = 0 denotes that the beam has no bending moment at t = 1, so it is supported on the bearing g. Therefore, BVP (1.1) simulates the static deformation for a spring beam of length 1 in which the left end is fixed and the right end is linked together with an elastic bearing device described by g, which is the cantilever beam equation with support in engineering and mechanics. Owing to its specific boundary condition and realistic physical meaning, the solvability for cantilever equation (1.1) has been studied by some scholars, see [2-4, 7, 8, 13, 14, 17, 19]. In [4,13,14,17,19]. The existence theory of solutions for the cantilever beam equation with nonlinear boundary condition (1.3) was studied under the condition that the nonlinear term f does not involve the derivative terms of deformation function u. In [2,7,8], the solvability conclusions were obtained by some fixed point theorems and monotone iterative method under the conditions that the nonlinear function f only contains the firstorder derivative term. Recently, Azarnavid et al. [3] used the reproducing kernel space method to construct an analytical approximate solution for BVP (1.1). However, because of the influence of fully derivative terms in the nonlinear function f and the nonlinearity of the boundary condition, the solvability for BVP (1.1) has not been studied extensively. In particular, as far as we know, there are fewer results on the equations of fully cantilever beam with nonlinear boundary conditions by using the method involving lower and upper solutions.
Inspired by the above literature, in the present paper, we utilize the monotone iterative technique involving lower and upper solutions and the auxiliary truncation function method to discuss the existence of solutions for BVP (1.1) between lower solution and upper solution. In order to study BVP (1.1), we put forward some reasonable monotonicity assumptions for the nonlinear functions f and g. Indeed, in order to reasonably introduce the definition of lower or upper solutions, we also consider the function g. Therefore, our results generalize and improve many results in the existing literature, which are new and meaningful. Our main results and proof processes are presented in Sect. 3, and two examples are given to verify our results in Sect. 4. In the following section, we introduce the definitions of lower and upper solutions and provide some preliminary results, which are useful in the proof.

Preliminaries
Denote I = [0, 1]. Let C(I) be a continuous function space with the norm u C = max t∈I |u(t)|, and C n (I) (n = 1, 2, 3, 4) be an n-order continuous differentiable function space with the norm Firstly, with regard to the linear boundary value problem (LBVP) By [10], we know that, for given h ∈ C(I), LBVP (2.1) has a unique solution where the Green function is defined by Next, we consider the LBVP corresponding to BVP (1.1) With a simple calculation, for each h ∈ C(I), As a matter of fact, by (2.4) and (2.5), one can find that the solutions of BVP (1.1) can be expressed by which suggests that the solutions of BVP (1.1) are equivalent to the fixed points of the operator T. For u ∈ C 3 (I), define operators B : C 3 (I) → C 3 (I) and F : Thus, for every u ∈ C 3 (I), is a completely continuous operator and Bu C 3 = |g(u(1))|.
Proof By (2.6), for any t ∈ I, Hence, we can conclude that Then, from (2.8), it follows that, for any u ∈ C 3 (I), and B : C 3 (I) → C 3 (I) is completely continuous. Evidently, The proof is completed. Now, we establish the following comparison principle.

Lemma 2.2
Let γ ≤ 0 be a constant and u ∈ C (4) (I) satisfy Proof By the definition of integral, for any t ∈ I, The proof is completed.
So, for any t ∈ I, we can figure out that At this point, the proof is finished.
For convenience, we recommend a semi-ordering " " in C 3 (I): we note that β α is equal to α β. Let α, β ∈ C 3 (I) and α β, we also introduce the order-interval in C 3 (I): then D β α ⊂ C 3 (I) is a nonempty bounded convex closed set. At the end of this section, we present an important result, which will be used in Sect. 3. Proof Obviously, F : C 3 (I) → C(I) is bounded and continuous. Thus, S • F : C 3 (I) → C 3 (I) is completely continuous, which implies that T : C 3 (I) → C 3 (I) is a completely continuous operator. Here, we apply the Schauder fixed point theorem to testify that there is a fixed point for the operator T in C 3 (I).
By the boundedness of f and g, there exist constants M 1 , M 2 > 0 such that Choose R ≥ M 1 S + M 2 , and define a closed bounded convex set = {u ∈ C 3 (I) : u C 3 ≤ R}, where S is the norm of operator S in C 3 (I). For u ∈ , one can find it shows that T( ) ⊂ . Hence, the Schauder fixed point theorem guarantees that the operator T has a fixed point u ∈ , which is a solution of BVP (1.1).

Theorem 3.1 Let α(t) and β(t) be a pair of lower and upper solutions of BVP
for every t ∈ I. If the following conditions are established: (H1) f : I × R 4 → R is continuous and satisfies: Proof It is easy to know that α β by Lemma 2.4. Let F : C 3 (I) → C(I) be defined by (2.9), then F is continuous. Based on (H1), we can certify that Since S is completely continuous, thus S •F : D β α → C 3 (I) is completely continuous. Therefore, T : D β α → C 3 (I) is a completely continuous operator. Next, we will accomplish the proof in three steps: I. We prove that T : D β α → D β α is an order-increasing operator under the semi-ordering " ", namely To this end, let (1)), then x = Tu is the solution of LBVP (2.4) and satisfies x (1) = g(u(1)).
The proof is finished.