Monotonically Iterative Method for the Cantilever Beam Equations

Abstract

In this paper, we consider the existence of extremal solutions for the nonlinear fourth-order differential equation. By use of a new comparison result, some sufficient conditions for the existence of extremal solutions are established by combining the monotone iterative technique and the methods of lower and upper solutions. Finally, an example is given to illustrate the validity of our main results.

1. Introduction

In this paper, we shall establish the existence of extremal solutions for the nonlinear fourth-order differential equation

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right),u^{\prime }\left(t\right)),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0,}\hfill \end{array}\right.$$

(1)

where $f\in C\left(\right[0,1]\times \mathbb{R}\times \mathbb{R},\mathbb{R})$.

Recently, differential equations of fourth-order have received more and more attention due to their various applications in science and engineering such as physics, control of dynamical systems etc. For example, The cantilever beam equation of problem (1) is a simplified mechanical model. This cantilever beam equation models the deformations of an elastic beam in equilibrium state, whose one end-point is fixed and the other is free [1,2]. Owing to its significance in physics, a number of works are devoted to the existence of solutions of fourth-order differential equations with different boundary conditions [3,4,5,6,7,8,9,10,11,12,13,14]. The methods used in these works are the Krasnosel’skii’s fixed point theorem [3,4], critical point theorem [5], the contraction mapping principle [6,7,8], the topological degree theory [9,10] the fixed point index [10,11,12], the Ekeland variational principle [13], and bifurcation theory [14].

The existence of positive solutions for the simply fourth-order boundary value problem

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0,}\hfill \end{array}\right.$$

(2)

which f does not contain any derivative terms has been discussed by several authors, see [2,15,16,17]. In References [15,16,17], (2) appears as a special case of the $(p,n-p)$ focal boundary value problems for $p=2$ and $n=4$. In all these works the Krasnoselskii’s fixed point theorem are applied.

For the cantilever beam equation with a nonlinear boundary condition of third-order derivative

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right),u^{‴}\left(1\right)=g\left(u\left(1\right)\right),}\hfill \end{array}\right.$$

(3)

the existence of solutions was considered by Ma [18] and Ma et al. [19] respectively based on variational methods and the contraction principle. The boundary condition in (3) may be interpreted in a material sense as the beam having a clamped end at $x=0$ and a shear force resting on the bearing g at $x=1$.

For the nonlinear fourth-order boundary value problem

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right),u^{\prime }\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0,}\hfill \end{array}\right.$$

the existence of positive solutions has also been discussed by making use of the monotonically iterative technique and applying the successively approximate method, see [20].

Alves et al. [21] considered the cantilever beam equation

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right),u^{\prime }\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right),u^{‴}\left(1\right)=g\left(u\left(1\right)\right),}\hfill \end{array}\right.$$

where $f:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ is continuous. The existence of monotone positive solutions is obtained by using the monotone iteration method.

Many scholars have considered the case of the fourth-order boundary value problem that f contains the fully derivative terms

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=f(t,u,u^{\prime },u^{″},u^{‴}),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0.}\hfill \end{array}\right.$$

(4)

In [22], Li used the fixed point index theory in cones to obtain the existence results of problem (16) when $f(t,u_0,u_1,u_2,u_3)$ is superlinear or sublinear growth on $u_0,u_1,u_2,u_3$. In [23], Li and Chen extended the existence result by letting f may be superlinear growth and have negative value. Using the method of lower and upper solutions and the monotone iterative technique, some existence results are obtained in [24]. For fully fourth-order nonlinear BVPs with other boundary conditions, the existence of solutions has been discussed by the use of nonlinear analysis method such as the lower and upper solution method [25], Rus’s contraction mapping [26], the monotone iterative technique [27], the Fourier analysis method and Leray-Schauder fixed point theorem [28]. However, the key to the application of the monotone iterative technique use in [21,24,27] is the monotonicity assumptions on nonlinearity f.

Inspired by the work mentioned above, the aim of this paper is to discuss the existence of extremal solutions to the boundary value problem of the nonlinear differential Equation (1) by the monotone iterative technique and the upper and lower solution method. According to the author’s knowledge, it is the first application of this method to such problems under nomonotonicity assumptions on unknown function and monotonicity assumptions on the first order derivative of unknown function in nonlinearity. The paper is organized as follows. In Section 2, we present here the necessary lemmas and establish two new comparison results. In Section 3, we give the definitions of the upper and lower solutions and obtain the existence results of extremal solutions of the problems (1) and (2).

2. Preliminaries

In this sections, we present Green’s function, some lemmas and comparison results that will be used to prove our main results.

Let $E=C[0,1]$ be a Banach space endowed with the maximum norm $\parallel u\parallel ={max}_{0\le t\le 1}\left|u\left(t\right)\right|$.

Lemma 1

([21]). For $\sigma \in C[0,1]$, the linear boundary value problem

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=\sigma \left(t\right),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=a,u^{\prime }\left(0\right)=b,u^{″}\left(1\right)=c,u^{‴}\left(1\right)=-d,}\hfill \end{array}\right.$$

(5)

has the unique solution

$$u\left(t\right)={\int }_0^{1}G(t,s)\sigma \left(s\right)ds+d(\frac{t^2}{2}-\frac{t^3}{6})+\frac{c{t}^2}{2}+bt+a,$$

where $G(t,s)$ is the Green’s function defined by

$$G(t,s)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{6}{s}^2(3t-s),}\hfill & {\displaystyle 0\le s\le t\le 1,}\hfill \\ {\displaystyle \frac{1}{6}{t}^2(3s-t),}\hfill & {\displaystyle 0\le t<s\le 1.}\hfill \end{array}\right.$$

From the expression of G, we easily verify that $G_t(t,s)$, the partial derivative of $G(t,s)$ to t, is given by

$$G_t(t,s)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2}{s}^2,}\hfill & {\displaystyle 0\le s\le t\le 1,}\hfill \\ {\displaystyle ts-\frac{1}{2}{t}^2,}\hfill & {\displaystyle 0\le t<s\le 1.}\hfill \end{array}\right.$$

Lemma 2

([19,22]). The following inequalities hold true.

(1)

${\displaystyle G(t,s)\le \frac{1}{2}{s}^{2}tandG(t,s)\le \frac{1}{2}{t}^{2}s,\forall t,s\in [0,1],}$

(2)

${\displaystyle G(t,s)\ge \frac{1}{3}{t}^2{s}^2,\forall t,s\in [0,1],}$

(3)

${\displaystyle G_t(t,s)\le ts,\forall t,s\in [0,1],}$

(4)

${\displaystyle G_t(t,s)\ge \frac{1}{2}t{s}^2,\forall t,s\in [0,1].}$

Lemma 3.

Assume that the nonnegative constant M satisfies

$$\frac{1}{8}M<1,$$

(6)

the boundary value problem

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=-Mu\left(t\right)+\sigma \left(t\right),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=a,u^{\prime }\left(0\right)=b,u^{″}\left(1\right)=c,u^{‴}\left(1\right)=-d}\hfill \end{array}\right.$$

(7)

has the unique solution

$$u\left(t\right)=\psi \left(t\right)+{\int }_0^{1}H(t,s)\sigma \left(s\right)ds+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds,$$

where

$$\begin{array}{c}{\displaystyle \psi \left(t\right)=d(\frac{t^2}{2}-\frac{t^3}{6})+\frac{c{t}^2}{2}+bt+a,}\hfill \\ {\displaystyle K_1(t,s)=-MG(t,s),}\hfill \\ {\displaystyle K_m(t,s)=-M{\int }_0^{1}G(t,r_{m-1})K_{m-1}(r_{m-1},s)ds}\hfill \\ {\displaystyle ={(-M)}^m{\int }_0^1\cdots {\int }_0^{1}G(t,r_{m-1})\cdots G(r_1,s)d{r}_1\cdots d{r}_{m-1},m\ge 2,}\hfill \\ {\displaystyle Q(t,s)=\sum _{m=1}^{+\infty }{K}_m(t,s),}\hfill \\ {\displaystyle H(t,s)={\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau +G(t,s).}\hfill \end{array}$$

Proof. 

By Lemma 1, we know the solution of problem (7) as follows

$$u\left(t\right)={\int }_0^{1}G(t,s)(-Mu\left(s\right)+\sigma \left(s\right))ds+d(\frac{t^2}{2}-\frac{t^3}{6})+\frac{c{t}^2}{2}+bt+a.$$

(8)

Define the operator $T:E\to E$ given by

$$\left(Tu\right)\left(t\right)=M{\int }_0^{1}G(t,s)u\left(s\right)ds,$$

(9)

and let

$$\phi \left(t\right)={\int }_0^{1}G(t,s)\sigma \left(s\right)ds.$$

It is clear that the operator T is a positive linear continuous operator, and we can rewrite (8) as

$$(I+T)u=\phi +\psi ,$$

(10)

where I stands for the identity operator. For any $u\in E$, by the definition of operator norm, it follows that

$$\begin{array}{cc}\parallel Tu\parallel \hfill & {\displaystyle =\underset{t\in [0,1]}{max}\left|Tu\left(t\right)\right|\le \underset{t\in [0,1]}{max}M{\int }_0^{1}G(t,s)\left|u\left(s\right)\right|ds}\hfill \\ & {\displaystyle \le \underset{t\in [0,1]}{max}M{\int }_0^{1}G(t,s)ds\parallel u\parallel =\underset{t\in [0,1]}{max}M\left(\frac{1}{24}{t}^4-\frac{1}{6}{t}^3+\frac{1}{4}{t}^2\right)\parallel u\parallel }\hfill \\ & {\displaystyle =\frac{1}{8}M\parallel u\parallel .}\hfill \end{array}$$

Note that $0<\frac{1}{8}M<1$, then we get

$$\parallel T\parallel \le \frac{1}{8}M<1.$$

Thus, the operator T is a contraction mapping. By Banach fixed-point theorem, T has a unique fixed point in E, or equivalently, the problem (7) has a unique solution $u\in E$.

It follows from the perturbation theorem of identity operator that $I+T$ has a bounded inverse operator

$$\begin{array}{cc}{(I+T)}^{-1}\hfill & {\displaystyle =\sum _{i=0}^{+\infty }{(-1)}^i{\left(T\right)}^i}\hfill \\ & {\displaystyle =I-T+T^2-\cdots +{(-1)}^i{\left(T\right)}^i+\cdots .}\hfill \end{array}$$

Though direct calculation, we have

$$\begin{array}{cc}{\displaystyle \left(I\phi \right)\left(t\right)}\hfill & {\displaystyle ={\int }_0^{1}G(t,s)\sigma \left(s\right)ds,}\hfill \\ \left(T\phi \right)\left(t\right)\hfill & {\displaystyle =M{\int }_0^{1}G(t,\tau )d\tau {\int }_0^{1}G(\tau ,s)\sigma \left(s\right)ds=(-1){\int }_0^1\left[{\int }_0^1{K}_1(t,\tau )G(\tau ,s)d\tau \right]\sigma \left(s\right)ds}\hfill \\ \left(T^2\phi \right)\left(t\right)\hfill & {\displaystyle =M{\int }_0^{1}G(t,r_1)\left(T\phi \right)\left(r_1\right)d{r}_1}\hfill \\ & {\displaystyle =M^2{\int }_0^{1}G(t,r_1)d{r}_1{\int }_0^{1}G(r_1,\tau )d\tau {\int }_0^{1}G(\tau ,s)\sigma \left(s\right)ds}\hfill \\ & {\displaystyle =M^2{\int }_0^1\left[{\int }_0^1\left[{\int }_0^{1}G(t,r_1)G(r_1,\tau )d{r}_1\right]G(\tau ,s)d\tau \right]\sigma \left(s\right)ds}\hfill \\ & {\displaystyle ={(-1)}^2{\int }_0^1\left[{\int }_0^1{K}_2(t,\tau )G(\tau ,s)d\tau \right]\sigma \left(s\right)ds,}\hfill \end{array}$$

then, we can obtain

$$\begin{array}{cc}\left(T^m\phi \right)\left(t\right)\hfill & {\displaystyle =M{\int }_0^{1}G(t,r_{m-1})\left(T^{m-1}\phi \right)\left(r_{m-1}\right)d{r}_{m-1}}\hfill \\ & {\displaystyle =M^m{\int }_0^{1}G(t,r_{m-1})d{r}_{m-1}\cdots {\int }_0^{1}G(r_2,r_1)d{r}_1{\int }_0^{1}G(r_1,\tau )d\tau {\int }_0^{1}G(\tau ,s)\sigma \left(s\right)ds}\hfill \\ & {\displaystyle ={(-1)}^m{\int }_0^1\left[{\int }_0^1{K}_m(t,\tau )G(\tau ,s)d\tau \right]\sigma \left(s\right)ds.}\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}& {\displaystyle \left[(I-T+T^2+\cdots +{(-1)}^m{T}^m+\cdots )\phi \right]\left(t\right)}\hfill \\ {\displaystyle =}\hfill & {\displaystyle {\int }_0^1\left[G(t,s)+{\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,\tau )G(\tau ,s)d\tau \right]\sigma \left(s\right)ds}\hfill \\ {\displaystyle =}\hfill & {\displaystyle {\int }_0^{1}H(t,s)\sigma \left(s\right)ds.}\hfill \end{array}$$

Similarly, we can obtain

$$\begin{array}{cc}{\displaystyle \left(I\psi \right)\left(t\right)}\hfill & {\displaystyle =d\left(\frac{t^2}{2}-\frac{t^3}{6}\right)+\frac{c{t}^2}{2}+bt+a,}\hfill \\ {\displaystyle \left(T\psi \right)\left(t\right)}\hfill & {\displaystyle =M{\int }_0^{1}G(t,s)\psi \left(s\right)ds=(-1){\int }_0^1{K}_1(t,s)\psi \left(s\right)ds,}\hfill \\ {\displaystyle \left(T^2\psi \right)\left(t\right)}\hfill & {\displaystyle =M^2{\int }_0^1{\int }_0^{1}G(t,t_1)G(t_1,s)\psi \left(s\right)d{t}_{1}ds={(-1)}^2{\int }_0^1{K}_2(t,s)\psi \left(s\right)ds}\hfill \end{array}$$

and

$$\begin{array}{cc}{\displaystyle \left(T^m\psi \right)\left(t\right)}\hfill & {\displaystyle =M^m{\int }_0^1\cdots {\int }_0^{1}G(t,t_{m-1})\cdots G(t_1,s)\psi \left(s\right)d{t}_{m-1}\cdots d{t}_{1}ds}\hfill \\ & {\displaystyle ={(-1)}^m{\int }_0^1{K}_m(t,s)\psi \left(s\right)ds,m\ge 2.}\hfill \end{array}$$

This is,

$$\begin{array}{cc}& {\displaystyle \left[(I-T+T^2+\cdots +{(-1)}^m{T}^m+\cdots )\psi \right]\left(t\right)}\hfill \\ =\hfill & {\displaystyle \psi \left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,s)\psi \left(s\right)ds,}\hfill \\ =\hfill & {\displaystyle \psi \left(t\right)+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds.}\hfill \end{array}$$

Thus, we get the solution of problem (4)

$$u\left(t\right)=\psi \left(t\right)+{\int }_0^{1}H(t,s)\sigma \left(s\right)ds+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds.$$

□

Remark 1.

It follows from the proof of Lemma 3 that the series ${\displaystyle \sum _{m=1}^{+\infty }}{K}_m(t,s)$ converges uniformly on $[0,1]\times [0,1]$ and all functions $K_n(t,s)$, $Q(t,s)$, $H(t,s)$ are continuous on $[0,1]\times [0,1]$. Furthermore, by the differentiability of parametrized integrals, we obtain

$$\begin{array}{cc}{\displaystyle Q_t(t,s)}\hfill & {\displaystyle =\sum _{m=1}^{+\infty }\frac{\partial K_m}{\partial t}(t,s)}\hfill \\ & {\displaystyle =-M{G}_t(t,s)-M\sum _{m=2}^{+\infty }{\int }_0^1{G}_t(t,r_{m-1})K_{m-1}(r_{m-1},s)ds}\hfill \end{array}$$

and

$$H_t(t,s)={\int }_0^1{Q}_t(t,\tau )G(\tau ,s)d\tau +G_t(t,s).$$

This together with the expression of $G_t$ implies that $Q_t(t,s)$, $H_t(t,s)$ are continuous on $[0,1]\times [0,1]$.

Define $F:C[0,1]\to C^1[0,1]$ by

$$\left(Fu\right)\left(t\right)={\int }_0^{1}H(t,s)u\left(s\right)ds.$$

(11)

Based on the continuity of functions H and $H_t$, standard arguments show that the following lemma hold.

Lemma 4.

F is complete continuous.

Lemma 5.

(Comparison result) Assume $u\in C^4[0,1]$ satisfies

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)\ge -Mu\left(t\right),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=0,u^{\prime }\left(0\right)\ge 0,u^{″}\left(1\right)\ge 0,u^{‴}\left(1\right)\le 0,}\hfill \end{array}\right.$$

where the nonnegative constant M satisfying (6) and

$$\frac{1}{3}-\frac{M}{12}-\frac{2{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{{\left(\frac{M}{15}\right)}^2}{3[1-{\left(\frac{M}{15}\right)}^2]}\ge 0,$$

(12)

$$\frac{1}{3}-\frac{M}{6}-\frac{4{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{13{\left(\frac{M}{15}\right)}^2}{36[1-{\left(\frac{M}{15}\right)}^2]}\ge 0,$$

(13)

$$\frac{1}{2}-\frac{M}{6}-\frac{4{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{{\left(\frac{M}{15}\right)}^2}{2[1-{\left(\frac{M}{15}\right)}^2]}\ge 0,$$

(14)

$$\frac{1}{2}-\frac{M}{3}-\frac{8{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{13{\left(\frac{M}{15}\right)}^2}{24[1-{\left(\frac{M}{15}\right)}^2]}\ge 0,$$

(15)

then $u\left(t\right)\ge 0$ and $u^{\prime }\left(t\right)\ge 0$ for $t\in [0,1]$.

Proof. 

Let $\sigma \left(t\right)=u^{\left(4\right)}\left(t\right)+Mu\left(t\right)$ and $a=u\left(0\right)$, $b=u^{\prime }\left(0\right)$, $c=u^{″}\left(1\right)$, $d=-u^{‴}\left(1\right)$, then $\sigma \left(t\right)\ge 0$ for $t\in [0,1]$ and $a=0,b\ge 0,c\ge 0,d\ge 0$. By Lemma 3, the linear problem (7) has a unique solution

$$u\left(t\right)=\psi \left(t\right)+{\int }_0^{1}H(t,s)\sigma \left(s\right)ds+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds.$$

Moreover,

$$u^{\prime }\left(t\right)={\psi }^{\prime }\left(t\right)+{\int }_0^1{H}_t(t,s)\sigma \left(s\right)ds+{\int }_0^1{Q}_t(t,s)\psi \left(s\right)ds.$$

Now, we consider ${\int }_0^{1}H(t,s)\sigma \left(s\right)ds$ for $t\in [0,1]$. Let $m=2k+1,k=1,2,\cdots$, by Lemma 2, we have

$$\begin{array}{cc}{K}_{2k+1}(t,s)\hfill & {\displaystyle =-M^{2k+1}{\int }_0^1\cdots {\int }_0^{1}G(t,r_{2k})\cdots G(r_1,s)d{r}_1\cdots d{r}_{2k}}\hfill \\ & {\displaystyle \ge -M^{2k+1}{\int }_0^1\cdots {\int }_0^1(\frac{1}{2}{t}^2·r_{2k})(\frac{1}{2}{r}_{2k}^2·r_{2k-1})\cdots (\frac{1}{2}{r}_2^2·r_1)·(\frac{1}{2}{r}_1·s^2)d{r}_1\cdots d{r}_{2k}}\hfill \\ & {\displaystyle =-\frac{16M^{2k+1}}{3·8^{2k+1}}{t}^2{s}^2.}\hfill \end{array}$$

Let $m=2k,k=1,2,\cdots$, we have

$$\begin{array}{cc}{K}_{2k}(t,s)\hfill & {\displaystyle =M^{2k}{\int }_0^1\cdots {\int }_0^{1}G(t,r_{2k-1})\cdots G(r_1,s)d{r}_1\cdots d{r}_{2k-1}}\hfill \\ & {\displaystyle \ge M^{2k}{\int }_0^1\cdots {\int }_0^1(\frac{1}{3}{t}^2·r_{2k-1}^2)\cdots (\frac{1}{3}{r}_2^2·r_1^2)(\frac{1}{3}{r}_1^2·s^2)d{r}_1\cdots d{r}_{2k-1}}\hfill \\ & {\displaystyle =\frac{5M^{2k}}{{15}^{2k}}{t}^2{s}^2.}\hfill \end{array}$$

Thus, we can gain

$$\begin{array}{cc}H(t,s)\hfill & {\displaystyle ={\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau +G(t,s)}\hfill \\ & {\displaystyle ={\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,\tau )G(\tau ,s)d\tau +G(t,s)}\hfill \\ & {\displaystyle \ge G(t,s)-M{\int }_0^{1}G(t,\tau )G(\tau ,s)d\tau +\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m+1}(t,\tau )G(\tau ,s)d\tau }\hfill \\ & {\displaystyle +\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m}(t,\tau )G(\tau ,s)d\tau }\hfill \\ & {\displaystyle \ge \frac{1}{3}{t}^2{s}^2-M{\int }_0^1\left(\frac{1}{2}{t}^2\tau \right)\left(\frac{1}{2}\tau s^2\right)d\tau -\sum _{m=1}^{+\infty }\frac{16M^{2m+1}}{3·8^{2m+1}}{\int }_0^1{t}^2{\tau }^2\left(\frac{1}{2}\tau s^2\right)d\tau }\hfill \\ & {\displaystyle +\sum _{m=1}^{+\infty }\frac{5M^{2m}}{{15}^{2m}}{\int }_0^1{t}^2{\tau }^2\left(\frac{1}{3}{\tau }^2{s}^2\right)d\tau }\hfill \\ & {\displaystyle =\frac{1}{3}{t}^2{s}^2-\frac{M}{12}{t}^2{s}^2-\frac{2}{3}\sum _{m=1}^{+\infty }\frac{M^{2m+1}}{8^{2m+1}}{t}^2{s}^2+\frac{1}{3}\sum _{m=1}^{+\infty }\frac{M^{2m}}{{15}^{2m}}{t}^2{s}^2.}\hfill \end{array}$$

Since $0<\frac{1}{8}M<1$ and ${\displaystyle \sum _{m=1}^{+\infty }}\frac{M^{2m+1}}{8^{2m+1}},{\displaystyle \sum _{m=1}^{+\infty }}\frac{M^{2m}}{{15}^{2m}}$ is convergence, we get

$$H(t,s)\ge \left\{\frac{1}{3}-\frac{M}{12}-\frac{2{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{{\left(\frac{M}{15}\right)}^2}{3[1-{\left(\frac{M}{15}\right)}^2]}\right\}{t}^2{s}^2.$$

By (12), we know ${\int }_0^{1}H(t,s)\sigma \left(s\right)ds\ge 0$ for $t\in [0,1]$.

Next, we claim that $\psi \left(t\right)+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds\ge 0$ for $t\in [0,1]$. Let $n\left(t\right)=t,p\left(t\right)=\frac{t^2}{2},q\left(t\right)=\frac{t^2}{2}-\frac{t^3}{6}$, then we have $\psi \left(t\right)=dq\left(t\right)+cp\left(t\right)+bn\left(t\right).$ By simple calculation and deduction, we can get

$$\begin{array}{cc}& {\displaystyle n\left(t\right)+{\int }_0^{1}Q(t,s)n\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle n\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,s)n\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle t-M{\int }_0^{1}G(t,s)n\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m+1}(t,s)n\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m}(t,s)n\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle t^2-M(\frac{t^3}{120}-\frac{t}{12}+\frac{1}{6})t^2-\sum _{m=1}^{+\infty }\frac{16M^{2m+1}}{3·8^{2m+1}}\frac{t^2}{4}+\sum _{m=1}^{+\infty }\frac{5M^{2m}}{{15}^{2m}}\frac{t^2}{4}}\hfill \\ \ge \hfill & {\displaystyle \left\{1-\frac{M}{6}-\frac{4{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{5{\left(\frac{M}{15}\right)}^2}{4[1-{\left(\frac{M}{15}\right)}^2]}\right\}{t}^2}\hfill \\ =\hfill & {\displaystyle n_1\left(t\right)\ge 0,}\hfill \end{array}$$

$$\begin{array}{cc}& {\displaystyle p\left(t\right)+{\int }_0^{1}Q(t,s)p\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle p\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,s)p\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle \frac{1}{2}{t}^2-M{\int }_0^{1}G(t,s)p\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m+1}(t,s)p\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m}(t,s)p\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle \frac{1}{2}{t}^2-M(\frac{t^4}{720}-\frac{t}{36}+\frac{1}{16})t^2-\sum _{m=1}^{+\infty }\frac{16M^{2m+1}}{3·8^{2m+1}}\frac{t^2}{10}+\sum _{m=1}^{+\infty }\frac{5M^{2m}}{{15}^{2m}}\frac{t^2}{10}}\hfill \\ \ge \hfill & {\displaystyle \left\{\frac{1}{2}-\frac{M}{16}-\frac{8{\left(\frac{M}{8}\right)}^3}{15[1-{\left(\frac{M}{8}\right)}^2]}+\frac{{\left(\frac{M}{15}\right)}^2}{2[1-{\left(\frac{M}{15}\right)}^2]}\right\}{t}^2}\hfill \\ =\hfill & {\displaystyle p_1\left(t\right)\ge 0,}\hfill \end{array}$$

and

$$\begin{array}{cc}& {\displaystyle q\left(t\right)+{\int }_0^{1}Q(t,s)q\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle q\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }{K}_m(t,s)q\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle (\frac{t^2}{2}-\frac{t^3}{6})-M{\int }_0^{1}G(t,s)q\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m+1}(t,s)q\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1{K}_{2m}(t,s)q\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle \frac{3-t}{6}{t}^2-\frac{M}{120}(-\frac{t^5}{42}+\frac{t^4}{6}-\frac{5t}{2}+\frac{11}{2})t^2-\sum _{m=1}^{+\infty }\frac{16M^{2m+1}}{3·8^{2m+1}}\frac{13t^2}{180}+\sum _{m=1}^{+\infty }\frac{5M^{2m}}{{15}^{2m}}\frac{13t^2}{180}}\hfill \\ \ge \hfill & {\displaystyle \left\{\frac{1}{3}-\frac{11M}{240}-\frac{52{\left(\frac{M}{8}\right)}^3}{135[1-{\left(\frac{M}{8}\right)}^2]}+\frac{13{\left(\frac{M}{15}\right)}^2}{36[1-{\left(\frac{M}{15}\right)}^2]}\right\}{t}^2}\hfill \\ =\hfill & {\displaystyle q_1\left(t\right)\ge 0.}\hfill \end{array}$$

By (13), we have

$$\begin{array}{cc}& {\displaystyle \psi \left(t\right)+{\int }_0^{1}Q(t,s)\psi \left(s\right)ds}\hfill \\ =\hfill & {\displaystyle [dq\left(t\right)+cp\left(t\right)+bn\left(t\right)]+{\int }_0^{1}Q(t,s)[dq\left(s\right)+cp\left(s\right)+bn\left(s\right)]ds}\hfill \\ =\hfill & {\displaystyle d[q\left(t\right)+{\int }_0^{1}Q(t,s)q\left(s\right)ds]+c[p\left(t\right)+{\int }_0^{1}Q(t,s)p\left(s\right)ds]+b[n\left(t\right)+{\int }_0^{1}Q(t,s)n\left(s\right)ds]}\hfill \\ \ge \hfill & {\displaystyle b{n}_1\left(t\right)+c{p}_1\left(t\right)+d{q}_1\left(t\right)\ge 0.}\hfill \end{array}$$

Thus, we can obtain that $u\left(t\right)\ge 0$ for $t\in [0,1]$.

Next, we consider ${\int }_0^1{H}_t(t,s)\sigma \left(s\right)ds$ for $t\in [0,1]$. Let $m=2k+1,k=1,2,\cdots$, we gain

$$\begin{array}{cc}{\displaystyle \frac{\partial K_{2k+1}(t,s)}{\partial t}}\hfill & {\displaystyle =-M^{2k+1}{\int }_0^1\cdots {\int }_0^1{G}_t(t,r_{2k})\cdots G(r_1,s)d{r}_1\cdots d{r}_{2k}}\hfill \\ & {\displaystyle \ge -M^{2k+1}{\int }_0^1\cdots {\int }_0^1(t·r_{2k})(\frac{1}{2}{r}_{2k}·r_{2k-1}^2)\cdots (\frac{1}{2}{r}_2·r_1^2)(\frac{1}{2}{r}_1·s^2)d{r}_1\cdots d{r}_{2k}}\hfill \\ & {\displaystyle =-\frac{32}{3}\frac{M^{2k+1}}{8^{2k+1}}t{s}^2.}\hfill \end{array}$$

Let $m=2k,k=1,2,\cdots$, we obtain

$$\begin{array}{cc}\frac{\partial K_{2k}(t,s)}{\partial t}\hfill & {\displaystyle =M^{2k}{\int }_0^1\cdots {\int }_0^1{G}_t(t,r_{2k-1})\cdots G(r_1,s)d{r}_1\cdots d{r}_{2k-1}}\hfill \\ & {\displaystyle \ge M^{2k}{\int }_0^1\cdots {\int }_0^1(\frac{1}{2}t·r_{2k-1}^2)(\frac{1}{3}{r}_{2k-1}^2·r_{2k-2}^2)\cdots (\frac{1}{3}{r}_1^2·s^2)d{r}_1\cdots d{r}_{2k-1}}\hfill \\ & {\displaystyle =\frac{15M^{2k}}{2·{15}^{2k}}t{s}^2.}\hfill \end{array}$$

Therefore, we know

$$\begin{array}{cc}{H}_t(t,s)\hfill & {\displaystyle ={\int }_0^1{Q}_t(t,\tau )G(\tau ,s)d\tau +G_t(t,s)}\hfill \\ & {\displaystyle ={\int }_0^1\sum _{m=1}^{+\infty }\frac{\partial K_m(t,\tau )}{\partial t}G(\tau ,s)d\tau +G_t(t,s)}\hfill \\ & {\displaystyle \ge G_t(t,s)-M{\int }_0^1{G}_t(t,\tau )G(\tau ,s)d\tau +\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m+1}(t,\tau )}{\partial t}G(\tau ,s)d\tau }\hfill \\ & {\displaystyle +\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m}(t,\tau )}{\partial t}G(\tau ,s)d\tau }\hfill \\ & {\displaystyle \ge \frac{1}{2}t{s}^2-\frac{M}{6}t{s}^2-\frac{4}{3}\sum _{m=1}^{+\infty }\frac{M^{2m+1}}{8^{2m+1}}t{s}^2+\frac{1}{2}\sum _{m=1}^{+\infty }\frac{M^{2m}}{{15}^{2m}}t{s}^2.}\hfill \end{array}$$

Since $0<\frac{1}{8}M<1$ and ${\displaystyle \sum _{m=1}^{+\infty }}\frac{M^{2m+1}}{8^{2m+1}},{\displaystyle \sum _{m=1}^{+\infty }}\frac{M^{2m}}{{15}^{2m}}$ is convergence, we get

$$H_t(t,s)\ge \left\{\frac{1}{2}-\frac{M}{6}-\frac{4{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{{\left(\frac{M}{15}\right)}^2}{2[1-{\left(\frac{M}{15}\right)}^2]}\right\}t{s}^2.$$

By (14), we know ${\int }_0^1{H}_t(t,s)\sigma \left(s\right)ds\ge 0$ for $t\in [0,1]$.

Lastly, we consider ${\psi }^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)\psi \left(s\right)ds$ for $t\in [0,1]$, where ${\psi }_t\left(t\right)=d{q}^{\prime }\left(t\right)+c{p}^{\prime }\left(t\right)+b{n}^{\prime }\left(t\right).$ Though calculation and deduction, we have

$$\begin{array}{cc}& {\displaystyle n^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)n\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle n^{\prime }\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }\frac{\partial K_m(t,s)}{\partial t}n\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle 1-M{\int }_0^1{G}_t(t,s)sds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m+1}(t,s)}{\partial t}n\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m}(t,s)}{\partial t}n\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle 1-M(\frac{t^3}{24}-\frac{t}{4}+\frac{1}{3})t-\sum _{m=1}^{+\infty }\frac{32M^{2m+1}}{3·8^{2m+1}}\frac{t}{4}+\sum _{m=1}^{+\infty }\frac{15M^{2m}}{2·{15}^{2m}}\frac{t}{4}}\hfill \\ \ge \hfill & {\displaystyle \left\{1-\frac{M}{3}-\frac{8{\left(\frac{M}{8}\right)}^3}{3[1-{\left(\frac{M}{8}\right)}^2]}+\frac{15{\left(\frac{M}{15}\right)}^2}{8[1-{\left(\frac{M}{15}\right)}^2]}\right\}t}\hfill \\ =\hfill & {\displaystyle n_2\left(t\right)\ge 0,}\hfill \end{array}$$

$$\begin{array}{cc}& {\displaystyle p^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)p\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle p^{\prime }\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }\frac{\partial K_m(t,s)}{\partial t}p\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle t-M{\int }_0^1{G}_t(t,s)\frac{s^2}{2}ds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m+1}(t,s)}{\partial t}p\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m}(t,s)}{\partial t}p\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle t-M(\frac{t^4}{120}-\frac{t}{12}+\frac{1}{8})t-\sum _{m=1}^{+\infty }\frac{32M^{2m+1}}{3·8^{2m+1}}\frac{t}{10}+\sum _{m=1}^{+\infty }\frac{15M^{2m}}{2·{15}^{2m}}\frac{t}{10}}\hfill \\ \ge \hfill & {\displaystyle \{1-\frac{M}{8}-\frac{16{\left(\frac{M}{8}\right)}^3}{15[1-{\left(\frac{M}{8}\right)}^2]}+\frac{3{\left(\frac{M}{15}\right)}^2}{4[1-{\left(\frac{M}{15}\right)}^2]}\}t}\hfill \\ =\hfill & {\displaystyle p_2\left(t\right)\ge 0}\hfill \end{array}$$

and

$$\begin{array}{cc}& {\displaystyle q^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)q\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle q^{\prime }\left(t\right)+{\int }_0^1\sum _{m=1}^{+\infty }\frac{\partial K_m(t,s)}{\partial t}q\left(s\right)ds}\hfill \\ =\hfill & {\displaystyle (t-\frac{t^2}{2})-M{\int }_0^1{G}_t(t,s)ds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m+1}(t,s)}{\partial t}q\left(s\right)ds+\sum _{m=1}^{+\infty }{\int }_0^1\frac{\partial K_{2m}(t,s)}{\partial t}q\left(s\right)ds}\hfill \\ \ge \hfill & {\displaystyle (1-\frac{t}{2})t-\frac{M}{6}(-\frac{t^5}{120}+\frac{t^4}{20}-\frac{3t}{8}+\frac{11}{20})t-\sum _{m=1}^{+\infty }\frac{32M^{2m+1}}{3·8^{2m+1}}\frac{13t}{180}+\sum _{m=1}^{+\infty }\frac{15M^{2m}}{2·{15}^{2m}}\frac{13t}{180}}\hfill \\ \ge \hfill & {\displaystyle \{\frac{1}{2}-\frac{11M}{120}-\frac{104{\left(\frac{M}{8}\right)}^3}{135[1-{\left(\frac{M}{8}\right)}^2]}+\frac{13{\left(\frac{M}{15}\right)}^2}{24[1-{\left(\frac{M}{15}\right)}^2]}\}t}\hfill \\ =\hfill & {\displaystyle q_2\left(t\right)\ge 0.}\hfill \end{array}$$

By (15), we get

$$\begin{array}{cc}& {\displaystyle {\psi }^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)\psi \left(s\right)ds}\hfill \\ & {\displaystyle =[d{q}^{\prime }\left(t\right)+c{p}^{\prime }\left(t\right)+b{n}^{\prime }\left(t\right)]+{\int }_0^1{Q}_t(t,s)[d{q}^{\prime }\left(t\right)+c{p}^{\prime }\left(t\right)+b{n}^{\prime }\left(t\right)]ds}\hfill \\ & {\displaystyle =d[q^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)q^{\prime }\left(t\right)ds]+c[p^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)p^{\prime }\left(t\right)ds]+b[n^{\prime }\left(t\right)+{\int }_0^1{Q}_t(t,s)n^{\prime }\left(t\right)ds]}\hfill \\ & {\displaystyle \ge d{q}_2\left(t\right)+c{p}_2\left(t\right)+b{n}_2\left(t\right)\ge 0.}\hfill \end{array}$$

Thus, we can obtain that $u^{\prime }\left(t\right)\ge 0$ for $t\in [0,1]$. □

If the condition $u\left(0\right)=0$ replaced by $u\left(0\right)\ge 0$, the result in Lemma 5 may be invalid. However, similar to the proof of Lemma 5, we have the following comparison result.

Lemma 6.

Assume $u\in C^4[0,1]$ satisfies

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)\ge -Mu\left(t\right),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)\ge 0,u^{\prime }\left(0\right)\ge 0,u^{″}\left(1\right)\ge 0,u^{‴}\left(1\right)\le 0,}\hfill \end{array}\right.$$

where the nonnegative constant M satisfying (6), (12) and

$$\frac{1}{3}-\frac{M}{4}-\frac{16{\left(\frac{M}{8}\right)}^3}{9[1-{\left(\frac{M}{8}\right)}^2]}+\frac{13{\left(\frac{M}{15}\right)}^2}{36[1-{\left(\frac{M}{15}\right)}^2]}\ge 0,$$

(16)

then $u\left(t\right)\ge 0$ for $t\in [0,1]$.

3. Main Results

Definition 1.

A function $v\in C^4[0,1]$ is called a lower solution of problem (1) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle v^{\left(4\right)}\left(t\right)\le f(t,v\left(t\right),v^{\prime }\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle v\left(0\right)=0,v^{\prime }\left(0\right)\le 0,v^{″}\left(1\right)\le 0,v^{‴}\left(1\right)\ge 0.}\hfill \end{array}\right.$$

Definition 2.

A function $w\in C^4[0,1]$ is called a upper solution of problem (1) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle w^{\left(4\right)}\left(t\right)\ge f(t,w\left(t\right),w^{\prime }\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle w\left(0\right)=0,w^{\prime }\left(0\right)\ge 0,w^{″}\left(1\right)\ge 0,w^{‴}\left(1\right)\le 0.}\hfill \end{array}\right.$$

For $v_0,w_0\in C^1[0,1]$, we write $v_0\le w_0$ if and only if $v_0\left(t\right)\le w_0\left(t\right)$ and $v_0^{\prime }\left(t\right)\le w_0^{\prime }\left(t\right)$ for all $t\in [0,1]$. In such a case, we denote

$$[v_0,w_0]=\{u\in C^1[0,1]:v_0\left(t\right)\le u\left(t\right)\le w_0\left(t\right),v_0^{\prime }\left(t\right)\le u^{\prime }\left(t\right)\le w_0^{\prime }\left(t\right)t\in [0,1]\}.$$

In the following, we list the assumptions to be used throughout our main results.

$\left(H_1\right)$ Assume that the functions $v_0,w_0$ are lower and upper solutions of the problem (1) respectively, and $v_0\le w_0$.

$\left(H_2\right)$ For fixed $(t,x)\in [0,1]\times [{min}_{t\in [0,1]}{v}_0\left(t\right),{max}_{t\in [0,1]}{w}_0\left(t\right)]$, $f(t,x,y)$ is monotone nondecreasing to y.

$\left(H_3\right)$ The function $f\in C\left(\right[0,1]\times \mathbb{R}\times \mathbb{R},\mathbb{R})$ satisfies

$$f(t,x,z)-f(t,y,z)\ge -M(x-y)$$

where $M>0$ satisfying Lemma 5 and $v_0\left(t\right)\le y\le x\le w_0\left(t\right),v_0^{\prime }\left(t\right)\le z\le w_0^{\prime }\left(t\right),t\in [0,1]$.

Theorem 1.

Assume that M satisfies $\left(H_1\right)$, $\left(H_2\right)$ and $\left(H_3\right)$. Then there exist monotone sequences $\left\{v_n\left(t\right)\right\},\left\{w_n\left(t\right)\right\}$ which converge in $C^1[0,1]$ to the extremal solutions of the problem (1) in $[v_0,w_0]$, respectively.

Proof. 

For any $\alpha \in [v_0,w_0]$, we consider the following problem:

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}\left(t\right)=M(\alpha \left(t\right)-u\left(t\right))+f(t,\alpha \left(t\right),{\alpha }^{\prime }\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0.}\hfill \end{array}\right.$$

(17)

From the proof of Lemma 1, the problem (17) has a unique solution $u\in E$, which can be expressed as

$$u\left(t\right)={\int }_0^{1}H(t,s)[f(s,\alpha \left(s\right),{\alpha }^{\prime }\left(s\right))+M\alpha \left(s\right)]ds.$$

Define an operator $A:[v_0,w_0]\to C^1[0,1]$ written as

$$A\alpha \left(t\right)={\int }_0^{1}H(t,s)[f(s,\alpha \left(s\right),{\alpha }^{\prime }\left(s\right))+M\alpha \left(s\right)]ds.$$

So, $u\in [v_0,w_0]$ is a solution of the problem (17) if and only if $u\in [v_0,w_0]$ is the fixed point of A.

Define a Nemytsky operator $Q:[v_0,w_0]\to C[0,1]$ written as

$$Q\alpha \left(t\right)=f(t,\alpha \left(t\right),{\alpha }^{\prime }\left(t\right))+M\alpha \left(t\right),\alpha \in [v_0,w_0].$$

Obviously, $A=F\circ Q$ and A is compact. Moreover, the operator A has the following properties:

(i)

$v_0\le A{v}_0$, $A{w}_0\le w_0$;

(ii)

$A{y}_1\le A{y}_2$, if $v_0\le y_1\le y_2\le w_0$.

To prove (i), let $\alpha =v_0,v_1=A{v}_0$, and $p=v_1-v_0$. Then from condition $\left(H_1\right)$ and the definition of the lower solution, we obtain

$$\left\{\begin{array}{cc}\hfill {\displaystyle p^{\left(4\right)}\left(t\right)}& \ge -M{v}_1\left(t\right)+f(t,v_0\left(t\right),v_0^{\prime }\left(t\right))+M{v}_0-f(t,v_0\left(t\right),v_0^{\prime }\left(t\right))\hfill \\ \hfill {\displaystyle }& {\displaystyle =-Mp\left(t\right),}\hfill \\ \hfill {\displaystyle p\left(0\right)}& {\displaystyle =0,p^{\prime }\left(0\right)\ge 0,p^{″}\left(1\right)\ge 0,p^{‴}\left(1\right)\le 0.}\hfill \end{array}\right.$$

Then, from Lemma 5, we get $p\left(t\right)\ge 0,p^{\prime }\left(t\right)\ge 0$ for $t\in [0,1]$, that is, $v_0\le A{v}_0$. Similarly, we can prove that $A{w}_0\le w_0$.

To prove (ii), let $y_1,y_2\in [v_0,w_0]$ with $y_1\le y_2$. Suppose that $u_1=A{y}_1,u_2=A{y}_2$. Let $p=u_2-u_1$. By condition $\left(H_3\right)$, we get

$$\left\{\begin{array}{ccc}\hfill {\displaystyle p^4\left(t\right)}& {\displaystyle =}\hfill & {\displaystyle f(t,y_2\left(t\right),y_2^{\prime }\left(t\right))-f(t,y_1\left(t\right),y_1^{\prime }\left(t\right))+M(y_2\left(t\right)-y_1\left(t\right))}\hfill \\ \hfill {\displaystyle }& & {\displaystyle -M{u}_2\left(t\right)+M{u}_1\left(t\right)}\hfill \\ & {\displaystyle \ge }\hfill & {\displaystyle f(t,y_2\left(t\right),y_1^{\prime }\left(t\right))-f(t,y_1\left(t\right),y_1^{\prime }\left(t\right))+M(y_2\left(t\right)-y_1\left(t\right))}\hfill \\ \hfill {\displaystyle }& & {\displaystyle -M{u}_2\left(t\right)+M{u}_1\left(t\right)\ge -Mp\left(t\right),}\hfill \\ \hfill {\displaystyle p\left(0\right)}& {\displaystyle =}\hfill & {\displaystyle p^{\prime }\left(0\right)=p^{″}\left(1\right)=p^{‴}\left(1\right)=0.}\hfill \end{array}\right.$$

By Lemma 5, we deduce $p\left(t\right)\ge 0,p^{\prime }\left(t\right)\ge 0$ which implies $A{y}_1\le A{y}_2$. Therefore, A is a monotone operator on $[v_0,w_0]$.

Let $v_m=A{v}_{m-1},w_m=A{w}_{m-1}$, by (i) and (ii), we have

$$v_0\left(t\right)\le v_1\left(t\right)\le v_2\left(t\right)\le \cdots \le v_m\left(t\right)\le \cdots \le w_m\left(t\right)\le \cdots \le w_2\left(t\right)\le w_1\left(t\right)\le w_0\left(t\right),$$

and

$$v_0^{\prime }\left(t\right)\le v_1^{\prime }\left(t\right)\le v_2^{\prime }\left(t\right)\le \cdots \le v_m^{\prime }\left(t\right)\le \cdots \le w_m^{\prime }\left(t\right)\le \cdots \le w_2^{\prime }\left(t\right)\le w_1^{\prime }\left(t\right)\le w_0^{\prime }\left(t\right).$$

Note that $\left\{v_m\left(t\right)\right\}$ and $\left\{v_m^{\prime }\left(t\right)\right\}$ are monotone nondecreasing and are bounded from above, and that $\left\{w_m\left(t\right)\right\}$ and $\left\{w_m^{\prime }\left(t\right)\right\}$ are monotone nonincreasing and are bounded from below. Then, by the completely continuity of operator A and $v_m\left(0\right)=w_m\left(0\right)=0$ for all $m\in \mathbb{N}$, we obtain

$$\underset{n\to \infty }{lim}{v}_n\left(t\right)=v_{*}\left(t\right),\underset{n\to \infty }{lim}{w}_n\left(t\right)=w^{*}\left(t\right)$$

$$\underset{n\to \infty }{lim}{v}_n^{\prime }\left(t\right)=v_{*}^{\prime }\left(t\right),\underset{n\to \infty }{lim}{w}_n^{\prime }\left(t\right)=w^{*}{}^{\prime }\left(t\right)$$

uniformly on $[0,1]$, respectively. And the limit functions $v_{*},w^{*}\in [v_0,w_0]$ are solutions of the problem (1).

In the following, we prove $v_{*}$, $w^{*}$ are extremal solutions of the problem (1) in $[v_0,w_0]$. Let $u\in [v_0,w_0]$ be a solution of the problem (1). In view of the monotonicity of A and $u=Au$, we conclude

$$v_0\le v_1=A{v}_0\le Au=u\le w_1=A{w}_0\le w_0,$$

which yields

$$v_0\le v_m\le u\le w_m\le w_0,m=1,2,\cdots .$$

Therefore, we have $v_0\le v_{*}\le u\le w^{*}\le w_0$. This shows $v_{*}$ and $w^{*}$ are minimal solution and maximal solution of the problem (1) in $[v_0,w_0]$, respectively. This ends the proof. □

For the boundary value problem (2), appears as the special case of problem (1) that f does not contain first-order derivative term, the definition of the upper and lower solutions can be weakened.

Definition 3

([24]). A function $v\in C^4[0,1]$ is called a lower solution of problem (2) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle v^{\left(4\right)}\left(t\right)\le f(t,v\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle v\left(0\right)\le 0,v^{\prime }\left(0\right)\le 0,v^{″}\left(1\right)\le 0,v^{‴}\left(1\right)\ge 0.}\hfill \end{array}\right.$$

Definition 4

([24]). A function $w\in C^4[0,1]$ is called a upper solution of problem (2) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle w^{\left(4\right)}\left(t\right)\ge f(t,w\left(t\right)),t\in (0,1),}\hfill \\ {\displaystyle w\left(0\right)\ge 0,w^{\prime }\left(0\right)\ge 0,w^{″}\left(1\right)\ge 0,w^{‴}\left(1\right)\le 0.}\hfill \end{array}\right.$$

Based on Lemma 6, we present the existence of extremal solutions for problem (2).

Theorem 2.

Assume that $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ is continuous, problem (2) has a lower solution $v_0$ and an upper solution $w_0$ with $v_0\left(t\right)\le w_0\left(t\right)$ for $t\in [0,1]$, and f satisfies the following condition:

$$f(t,x)-f(t,y)\ge -M(x-y)$$

where $M>0$ satisfying Lemma 6 and $v_0\left(t\right)\le y\le x\le w_0\left(t\right),t\in [0,1]$. Then there exist monotone sequences $\left\{v_n\left(t\right)\right\},\left\{w_n\left(t\right)\right\}$ which converge in $C[0,1]$ to the extremal solutions of the problem (2) in $[v_0,w_0]$, respectively.

4. Example

Consider the problem

$$\left\{\begin{array}{c}{\displaystyle u^{\left(4\right)}=-\frac{7}{15}{(\frac{t^3}{6}+u)}^3+\frac{7}{30}{t}^{4}sinu+\frac{7}{45}{\left(\frac{u^{\prime }}{2}\right)}^6+\frac{7}{90}{t}^6,}\hfill \\ {\displaystyle u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{‴}\left(1\right)=0,}\hfill \end{array}\right.$$

(18)

Take $f(t,u,v)=-\frac{7}{15}{(\frac{t^3}{6}+u)}^3+\frac{7}{30}{t}^{4}sinu+\frac{7}{45}{\left(\frac{v}{2}\right)}^6+\frac{7}{90}{t}^6$, $v_0\left(t\right)=0$ and $w_0\left(t\right)=t^2-\frac{t^3}{6}$, then we have

$$\left\{\begin{array}{c}{\displaystyle v_0^{\left(4\right)}\left(t\right)=0\le \frac{7}{90}{t}^6(1-t^3)=f(t,v_0\left(t\right),v_0^{\prime }\left(t\right)),}\hfill \\ {\displaystyle v_0\left(0\right)=0,v_0^{\prime }\left(0\right)\le 0,v_0^{″}\left(1\right)\le 0,v_0^{‴}\left(1\right)\ge 0}\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{\displaystyle w_0^{\left(4\right)}\left(t\right)=0\ge -\frac{7}{15}{t}^6+\frac{7}{30}{t}^{4}sin(t^2-\frac{t^3}{6})+\frac{7}{45}{(t-\frac{t^2}{4})}^6+\frac{7}{90}{t}^6=f(t,w_0\left(t\right),w_0^{\prime }\left(t\right)),}\hfill \\ {\displaystyle w_0\left(0\right)=0,w_0^{\prime }\left(0\right)\ge 0,w_0^{″}\left(1\right)\ge 0,w_0^{‴}\left(1\right)\le 0.}\hfill \end{array}\right.$$

It shows that condition $\left(H_1\right)$ of Theorem 1 holds.

Note that the definition of f, f is monotone nondecreasing to $y\in [v_0^{\prime },w_0^{\prime }]$ for fixed $(t,x)\in [0,1]\times [v_0,w_0]$. Therefore, the condition $\left(H_2\right)$ of Theorem 1 holds.

Let $M=\frac{7}{5}$. Then, for $v_0\left(t\right)\le y\le x\le w_0\left(t\right)$, $v_0^{\prime }\left(t\right)\le z\le w_0^{\prime }\left(t\right)$, $t\in [0,1]$,

$$\begin{array}{c}{\displaystyle f(t,x,z)-f(t,y,z)}\hfill \\ {\displaystyle =-\frac{7}{15}[{(\frac{t^3}{6}+x)}^3-{(\frac{t^3}{6}+y)}^3]+\frac{7}{30}{t}^4(sinx-siny)}\hfill \\ {\displaystyle \ge -\frac{7}{5}(x-y),}\hfill \end{array}$$

where $M>0$ satisfying Lemma 5. Thus, the condition $\left(H_3\right)$ of Theorem 1 holds.

In consequence, the problem (18) has the extremal solutions in $[v_0,w_0]$.

5. Conclusions

In this article, on a cantilever beam equation models the deformations of an elastic beam, we use the monotone iterative technique and the methods of lower and upper solutions to investigate the existence results for extremal solutions for problems (1) and (2). At the same time, two sequences are obtained for approximating the extremal solutions of the nonlinear fourth-order differential equation. It should be noted that the proof of comparison result does depend on the the perturbation theorem of identity operator. In the future, we will continue to use the monotone iterative technique to investigate problems (1) under nomonotonicity assumptions on u and v in nonlinearity $f(t,u,v)$.