On the Sixth-Order Beam Equation of Small Deflection with Variable Parameters

Abstract

This paper establishes an existence and uniqueness theorem for the nonlocal sixth-order nonlinear beam differential equations with four parameters of the form $u^{\left(6\right)}+A\left(x\right)u^{\left(4\right)}+B\left(x\right)u^{″}+C\left(x\right)u=\lambda f(x,u,u^{″},u^{\left(4\right)}),0<x<1,$ subject to the integral boundary conditions: $u\left(0\right)=u\left(1\right)={\int }_0^{1}p\left(x\right)u\left(x\right)dx,$ $u^{″}\left(0\right)=u^{″}\left(1\right)={\int }_0^{1}q\left(x\right)u^{″}\left(x\right)dx$ and $u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)={\int }_0^{1}s\left(x\right)u^{\left(4\right)}\left(x\right)dx$ such that $1-{\int }_0^1{p}^2\left(x\right)dx=\alpha >0,1-{\int }_0^1{q}^2\left(x\right)dx=\beta >0,1-{\int }_0^1{s}^2\left(x\right)dx=\gamma >0,$ under some growth condition on f, and provided that an upper bound exists for the flexural rigidity $\lambda$ to guarantee that no large deflections will occur.

1. Introduction

Fourth- and sixth-order differential equations of the form

$$u^{\left(6\right)}=g(x,u,u^{″},u^{\left(4\right)}),0<x<1$$

have wide applications in many physical and engineering problems. In elastic beam theory, the fourth-order equations model the deflection of sandwich beams, the sixth-order equations model circular ring beams [1,2,3,4], and they have connections with vibration problems in automotive industry [5]. The sixth-order BVPs also appear in astrophysics and hydrodynamics [6,7,8,9]. There have been several successful attempts to investigate the existence of positive solutions of the problem and finding them numerically [10,11,12,13,14,15,16,17,18,19]. In particular, the problem

$$u^{\left(6\right)}+A{u}^{\left(4\right)}+B{u}^{″}+Cu=f(x,u),0<x<1,$$

$$u\left(0\right)=u^{″}\left(0\right)=u^{\left(4\right)}\left(0\right)=u^{\left(6\right)}\left(0\right)=0,$$

$$u\left(1\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(1\right)=u^{\left(6\right)}\left(1\right)=0$$

has been investigated in [10,11,12,13,14] with $A=B=C=0$, and in [15] with $A=B=0,$ and C is a variable parameter. The authors in [20,21,22,23] investigated positive solutions of sixth-order boundary value problems with three variable coefficients. The authors in [20] investigated the solution using spectral theory of operators and fixed-point theorem in cones.

The present paper deals with the sixth-order spectral BVP

$$u^{\left(6\right)}+A\left(x\right)u^{\left(4\right)}+B\left(x\right)u^{″}+C\left(x\right)u=\lambda f(x,u,u^{″},u^{\left(4\right)}),0<x<1,$$

(1)

subject to the integral boundary conditions

$$u\left(0\right)=u\left(1\right)=\underset{0}{\overset{1}{\int }}p\left(x\right)u\left(x\right)dx,$$

(2)

$$u^{″}\left(0\right)=u^{″}\left(1\right)=\underset{0}{\overset{1}{\int }}q\left(x\right)u^{″}\left(x\right)dx,$$

(3)

$$u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=\underset{0}{\overset{1}{\int }}s\left(x\right)u^{\left(4\right)}\left(x\right)dx,$$

(4)

where $A,B,C\in C[0,1]$, $p,q,s\in {\mathbb{L}}^2[0,1],$

$$1-\underset{0}{\overset{1}{\int }}{p}^2\left(x\right)dx=\alpha >0,1-\underset{0}{\overset{1}{\int }}{q}^2\left(x\right)dx=\beta >0,1-\underset{0}{\overset{1}{\int }}{s}^2\left(x\right)dx=\gamma >0$$

(5)

and f is continuous on $[0,1]\times \mathbb{R}\times \mathbb{R}$ and satisfies a growth condition with variable parameters:

$$\left|f(x,u,v,w)\right|\le a\left(x\right)\left|u\right|+b\left(x\right)\left|v\right|+c\left(x\right)\left|w\right|+d\left(x\right),$$

(6)

where $a\left(x\right),b\left(x\right),c\left(x\right)$ and $d\left(x\right)$ are positive continuous functions on $[0,1].$ So, let $a\left(x\right)\le a,$ $b\left(x\right)\le b,$ and $c\left(x\right)\le c$ on $[0,1].$ In the beam equation, the external source f refers to the distributed load, and researchers used to assume it is bounded on $[0,1]\times \mathbb{R}\times \mathbb{R}.$ However, Y. Yang [24] assumed a growth condition on f, which is more general and provides flexibility and more efficiency in applications. Equation (1) is called a beam equation and is an Euler–Bernoulli Equation. Note here that Pr. (1)–(6) is a generalization of the problem investigated in [1] with $A=B=0,$ $s\left(x\right)=0$ and $C\left(x\right)=0.$ If $p=q=s=0$, then the end points are simply supported, so we can perturb the end points by making the functions $p,q,s$ relatively small.

Pr. (1)–(6) models the deformation of an elastic beam, where u represents the deflection of the beam, and $\lambda$ the reciprocal of the flexural rigidity which measures the resistance to bend.

In much the same way as in the fourth-degree case, we are interested in the case of small deflections, which means that Hook’s law is applicable, and this will give rise to the Euler–Bernoulli Equation (1). In construction engineering, it becomes increasingly important to consider beams with small deflections since large deflections can cause cracks in the beams. In automotive engineering, large deflections lead to vibrations that can cause noise and discomfort, and shorten the lifespan of the car, so reducing damping automotive vibrations is of utmost importance for car engineers. One of the technical solutions they offer is by enforcing an upper bound for the flexural rigidity $\lambda$ to guarantee that only small deflections will occur. Thus, we propose the following natural assumption:

$$\lambda <{\displaystyle \frac{64\alpha \beta \gamma -16A_1-4B_1-C_1}{21k}},$$

(7)

where $k=max\{a,b,c\},A_1={\displaystyle \underset{[0,1]}{sup}}A\left(x\right),B_1={\displaystyle \underset{[0,1]}{sup}}B\left(x\right),C_1={\displaystyle \underset{[0,1]}{sup}}C\left(x\right).$

It is important to note that if (7) holds and assuming that $A_1=B_1=C_1=0,$ then

$$\lambda <{\displaystyle \frac{64\alpha \beta \gamma }{21k}}<{\displaystyle \frac{16\alpha \beta \gamma }{5k}},$$

which is the upper estimate of $\lambda$ for the fourth-order beam equation (see [1]). It turns out the above estimate works well for both types of beams modeled by the fourth-order and sixth-order equations.

2. Existence Theorem

Pr. (1)–(4) can be converted into the following system:

$$\left\{\begin{array}{ccc}\hfill u^{″}& =& v,u\left(0\right)=u\left(1\right)={\int }_0^{1}p\left(x\right)u\left(x\right)dx,\hfill \\ \hfill v^{″}& =& w,v\left(0\right)=v\left(1\right)={\int }_0^{1}q\left(x\right)v\left(x\right)dx,\hfill \\ \hfill w^{″}& =& -A\left(x\right)w-B\left(x\right)v-C\left(x\right)u+\lambda f(x,u,v,w),\hfill \\ \hfill w\left(0\right)& =& w\left(1\right)={\int }_0^{1}s\left(x\right)w\left(x\right)dx.\hfill \end{array}\right.$$

(8)

We prove the following statement

Proposition 1. 

If the conditions (6) and (7) hold with (5), then there exists a constant $M>0,$ which is independent of the solution u such that for any $x\in [0,1]$ and any solution u to Pr. (1)–(4), we have

$$\begin{array}{c}\hfill {\Vert u\Vert }_{\rho ,0}+{\Vert u\Vert }_{\rho ,1}+{\Vert u\Vert }_{\rho ,2}\le M,\end{array}$$

(9)

where ${\Vert u\Vert }_{\rho ,0}={\displaystyle \underset{[0,1]}{max}}\mid \rho \left(x\right)u\left(x\right)\mid ,$ ${\Vert u\Vert }_{\rho ,1}={\Vert u^{″}\Vert }_{\rho ,0},$ ${\Vert u\Vert }_{\rho ,2}={\Vert u^{\left(4\right)}\Vert }_{\rho ,0}$ and $\rho \left(x\right)=x(1-x),x\in [0,1].$

Proof. 

Multiplying both sides of the second equation of (8) by $\rho \left(x\right)v$ and integrating the resulting equation from 0 to $1,$ then using integration by parts gives

$$2{\int }_0^1{v}^2\left(x\right)dx+2{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^2\left(x\right)dx=\left[v^2\left(1\right)+v^2\left(0\right)\right]-2{\int }_0^1\rho \left(x\right)v\left(x\right)w\left(x\right)dx.$$

(10)

Given that $v\left(0\right)=v\left(1\right)={\int }_0^{1}q\left(x\right)u\left(x\right)dx,$ we have

$$\begin{array}{c}\hfill {\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^{2}dx={\left[{\int }_0^{1}q\left(x\right)v\left(x\right)dx\right]}^2-{\int }_0^1\rho \left(x\right)v\left(x\right)w\left(x\right)dx.\end{array}$$

(11)

The Cauchy–Schwarz inequality can be used to estimate the integrals ${\int }_0^{1}q\left(x\right)v\left(x\right)dx$ and ${\int }_0^1\rho \left(x\right)v\left(x\right)w\left(x\right)dx.$

$$\begin{array}{c}\hfill {\left[{\int }_0^{1}q\left(x\right)v\left(x\right)dx\right]}^2\le \left({\int }_0^1{q}^2\left(x\right)dx\right)\left({\int }_0^1{v}^2\left(x\right)dx\right)\end{array}$$

(12)

and

$$\begin{array}{ccc}\hfill \left|{\int }_0^1\rho \left(x\right)v\left(x\right)w\left(x\right)dx\right|& \le & {\left({\int }_0^1\rho \left(x\right)v^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1\rho \left(x\right)w^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(13)

Since ${\displaystyle \underset{[0,1]}{sup}}\rho \left(x\right)={\displaystyle \frac{1}{4}},$ we have

$$\begin{array}{ccc}\hfill \left|{\int }_0^1\rho \left(x\right)v\left(x\right)w\left(x\right)dx\right|& \le & {\displaystyle \frac{1}{4}}{\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(14)

It follows that

$$\begin{array}{cc}\hfill & \left(1-{\int }_0^1{q}^2\left(x\right)dx\right){\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^2\left(x\right)dx\hfill \\ & \le {\displaystyle \frac{1}{4}}{\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

(15)

and so

$${\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^2\left(x\right)dx\le {\displaystyle \frac{1}{4\beta }}{\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}.$$

(16)

But we also have

$$\begin{array}{c}\hfill {\int }_0^1{v}^2\left(x\right)dx\le {\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^2\left(x\right)dx.\end{array}$$

(17)

Thus,

$$\begin{array}{c}\hfill {\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{4\beta }}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\end{array}$$

(18)

and therefore

$${\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(v^{\prime }\right)}^2\left(x\right)dx\le K_1{\int }_0^1{w}^2\left(x\right)dx,$$

(19)

where $K_1={\displaystyle \frac{1}{16{\beta }^2}}$ and $\beta =1-{\int }_0^1{q}^2\left(x\right)dx.$

Now, we multiply both sides of the first equation of (8) by $\rho \left(x\right)u$, and then we repeat the same steps above, considering the nonlocal boundary conditions $u\left(0\right)=u\left(1\right)={\int }_0^{1}p\left(x\right)u\left(x\right)dx.$ This gives

$${\int }_0^1{u}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(u^{\prime }\right)}^2\left(x\right)dx\le K_2{\int }_0^1{v}^2\left(x\right)dx\le K_1{K}_2{\int }_0^1{w}^2\left(x\right)dx,$$

(20)

where $K_2={\displaystyle \frac{1}{16{\alpha }^2}}$ and $\alpha =1-{\int }_0^1{p}^2\left(x\right)dx.$

Furthermore, multiplying both sides of the third equation of (8) by $\rho \left(x\right)w$ repeating the same steps, considering the nonlocal boundary conditions $w\left(0\right)=w\left(1\right)={\int }_0^{1}s\left(x\right)w\left(x\right)dx,$ we obtain

$$\begin{array}{cc}\hfill {\int }_0^1{w}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(w^{\prime }\left(x\right)\right)}^{2}dx& ={\left[{\int }_0^{1}s\left(x\right)w\left(x\right)dx\right]}^2+{\int }_0^{1}A\left(x\right)\rho \left(x\right)w^2\left(x\right)\hfill \\ \hfill & +{\int }_0^{1}B\left(x\right)\rho \left(x\right)v\left(x\right)w\left(x\right)+{\int }_0^{1}C\left(x\right)\rho \left(x\right)u\left(x\right)w\left(x\right)dx\hfill \\ \hfill & -\lambda {\int }_0^{1}f(x,u,v,w)\rho \left(x\right)w\left(x\right)dx.\hfill \end{array}$$

(21)

Note that

$$\begin{array}{ccc}& & \left|{\int }_0^{1}A\left(x\right)\rho \left(x\right)w^2\left(x\right)dx\right|\le {\displaystyle \frac{A_1}{4}}\left({\int }_0^1{w}^2\left(x\right)dx\right),\hfill \end{array}$$

(22)

$$\begin{array}{ccc}& & \left|{\int }_0^{1}B\left(x\right)\rho \left(x\right)v\left(x\right)w\left(x\right)dx\right|\le {\displaystyle \frac{B_1}{4}}{\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

(23)

$$\begin{array}{ccc}& & \left|{\int }_0^{1}C\left(x\right)\rho \left(x\right)u\left(x\right)w\left(x\right)dx\right|\le {\displaystyle \frac{C_1}{4}}{\left({\int }_0^1{u}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

(24)

and

$$\begin{array}{c}\hfill {\left[{\int }_0^{1}s\left(x\right)w\left(x\right)dx\right]}^2\le \left({\int }_0^1{s}^2\left(x\right)dx\right)\left({\int }_0^1{w}^2\left(x\right)dx\right).\end{array}$$

(25)

Then

$$\begin{array}{cc}\hfill \left|{\int }_0^{1}f(x,u,v,w)\rho \left(x\right)w\left(x\right)dx\right|& \le {\displaystyle \frac{a}{4}}{\int }_0^1\left|u\left(x\right)w\left(x\right)\right|dx+{\displaystyle \frac{b}{4}}{\int }_0^1\left|v\left(x\right)w\left(x\right)\right|dx\hfill \\ \hfill & +{\displaystyle \frac{c}{4}}{\int }_0^1\left|w^2\left(x\right)\right|dx+{\displaystyle \frac{1}{4}}{\int }_0^1\left|d\left(x\right)w\left(x\right)\right|dx.\hfill \end{array}$$

(26)

By means of the $ϵ-$ inequality, we estimate the integral ${\int }_0^1\left|d\left(x\right)w\left(x\right)\right|dx$ to obtain

$$\begin{array}{c}\hfill {\int }_0^1\left|d\left(x\right)w\left(x\right)\right|dx\le {\displaystyle \frac{1}{2ϵ}}{d}^2+{\displaystyle \frac{ϵ}{2}}{\int }_0^1{w}^2\left(x\right)dx,ϵ>0.\end{array}$$

(27)

Thus

$$\begin{array}{cc}\hfill \left|{\int }_0^{1}f(x,u,v,w)\rho \left(x\right)w\left(x\right)dx\right|& \le {\displaystyle \frac{a}{4}}{\left({\int }_0^1{u}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{b}{4}}{\left({\int }_0^1{v}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{c}{4}}\left({\int }_0^1{w}^2\left(x\right)dx\right)\hfill \\ & +\left({\displaystyle \frac{d^2}{8ϵ}}+{\displaystyle \frac{ϵ}{8}}{\int }_0^1{w}^2\left(x\right)dx\right),ϵ>0.\hfill \end{array}$$

(28)

Since

$$\begin{array}{ccc}& & {\int }_0^1{v}^2\left(x\right)dx\le {\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right)v^{\prime 2}\left(x\right)dx\le K_1{\int }_0^1{w}^2\left(x\right)dx,\hfill \end{array}$$

(29)

$$\begin{array}{ccc}& & {\int }_0^1{u}^2\left(x\right)dx\le {\int }_0^1{u}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(u^{\prime }\right)}^2\left(x\right)dx\le K_1{K}_2{\int }_0^1{w}^2\left(x\right)dx,\hfill \end{array}$$

(30)

we substitute (29) and (30) into (23) and (24)

$$\begin{array}{ccc}& & \left|{\int }_0^{1}B\left(x\right)\rho \left(x\right)v\left(x\right)w\left(x\right)dx\right|\le {\displaystyle \frac{B_1}{4}}\sqrt{K_1}{\int }_0^1{w}^2\left(x\right)dx,\hfill \end{array}$$

(31)

$$\begin{array}{ccc}& & \left|{\int }_0^{1}C\left(x\right)\rho \left(x\right)u\left(x\right)w\left(x\right)dx\right|\le {\displaystyle \frac{C_1}{4}}\sqrt{K_1{K}_2}{\int }_0^1{w}^2\left(x\right)dx.\hfill \end{array}$$

(32)

Also, substituting (29) and (30) into (28)

$$\begin{array}{ccc}& & \left|{\int }_0^{1}f(x,u,v,w)\rho \left(x\right)w\left(x\right)dx\right|\hfill \\ & & \le \left[{\displaystyle \frac{a}{4}}\sqrt{K_1{K}_2}+{\displaystyle \frac{b}{4}}\sqrt{K_1}+{\displaystyle \frac{c}{4}}+{\displaystyle \frac{ϵ}{8}}\right]{\int }_0^1{w}^2\left(x\right)dx+{\displaystyle \frac{d^2}{8ϵ}},ϵ>0.\hfill \end{array}$$

(33)

From (25), (31)–(33), we obtain

$$\begin{array}{cc}\hfill & \left[1-\left({\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{\lambda c}{4}}+{\displaystyle \frac{\lambda ϵ}{8}}+\sqrt{K_1}\left({\displaystyle \frac{B_1}{4}}+{\displaystyle \frac{\lambda b}{4}}+\sqrt{K_2}({\displaystyle \frac{C_1}{4}}+{\displaystyle \frac{\lambda a}{4}})\right)\right)\right]{\int }_0^1{w}^2\left(x\right)dx\hfill \\ \hfill & +{\int }_0^1\rho \left(x\right)w^{\prime 2}\left(x\right)dx\le {\displaystyle \frac{\lambda d^2}{8ϵ}}.\hfill \end{array}$$

(34)

Define

$$\delta =1-\left({\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{\lambda c}{4}}+{\displaystyle \frac{\lambda ϵ}{8}}+\sqrt{K_1}\left({\displaystyle \frac{B_1}{4}}+{\displaystyle \frac{\lambda b}{4}}+\sqrt{K_2}({\displaystyle \frac{C_1}{4}}+{\displaystyle \frac{\lambda a}{4}})\right)\right)$$

(35)

and let $\sqrt{K_1}={\displaystyle \frac{1}{4\beta }}$ and $\sqrt{K_2}={\displaystyle \frac{1}{4\alpha }}.$ Given the fact that $\alpha <1, \beta <1$ and $\gamma =1-{\int }_0^1{s}^2\left(x\right)dx,$ we have

$$\begin{array}{ccc}& & {\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{\lambda c}{4}}+{\displaystyle \frac{1}{4\beta }}\left({\displaystyle \frac{B_1}{4}}+{\displaystyle \frac{\lambda b}{4}}+{\displaystyle \frac{1}{4\alpha }}({\displaystyle \frac{C_1}{4}}+{\displaystyle \frac{\lambda a}{4}})\right)\hfill \\ & & ={\displaystyle \frac{1}{64\alpha \beta }}\left[16\alpha \beta A_1+16\alpha \beta c\lambda +4\alpha B_1+4\alpha b\lambda +C_1+\lambda a\right]\hfill \\ & & <{\displaystyle \frac{1}{64\alpha \beta }}\left[16A_1+4B_1+C_1+21k\lambda \right]\hfill \\ \\ & & <1-\underset{0}{\overset{1}{\int }}{s}^2\left(x\right)dx,\hfill \end{array}$$

which implies

$$\begin{array}{c}\hfill 1-\left({\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{\lambda c}{4}}+\sqrt{K_1}\left({\displaystyle \frac{B_1}{4}}+{\displaystyle \frac{\lambda b}{4}}+\sqrt{K_2}({\displaystyle \frac{C_1}{4}}+{\displaystyle \frac{\lambda a}{4}})\right)\right)>0.\end{array}$$

(36)

We choose $ϵ$ small enough to guarantee that $\delta >0.$ It follows that

$$\begin{array}{c}\hfill {\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{\lambda c}{4}}+{\displaystyle \frac{\lambda ϵ}{8}}+\sqrt{K_1}\left({\displaystyle \frac{B_1}{4}}+{\displaystyle \frac{\lambda b}{4}}+\sqrt{K_2}({\displaystyle \frac{C_1}{4}}+{\displaystyle \frac{\lambda a}{4}})\right)<1.\end{array}$$

(37)

Hence,

$$\begin{array}{c}\hfill {\int }_0^1{w}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right)w^{\prime 2}\left(x\right)dx\le M_1,\end{array}$$

(38)

where $M_1={\displaystyle \frac{\frac{\lambda d^2}{8ϵ}}{\delta }}.$ Now, we combine (38) with (19) to obtain

$${\int }_0^1{v}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right)v^{\prime 2}\left(x\right)dx\le M_2,$$

(39)

where $M_2=K_1{M}_1.$

In a similar fashion, we use (20) to obtain

$${\int }_0^1{u}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right)u^{\prime 2}\left(x\right)dx\le M_3,$$

(40)

where $M_3=K_1{K}_2{M}_1.$

But, on the other hand we have

$$\rho \left(x\right)u\left(x\right)={\int }_0^x{\left(\rho \left(x\right)u\left(x\right)\right)}^{\prime }dx+\rho \left(0\right)u\left(0\right)={\int }_0^x{\left(\rho \left(x\right)u\left(x\right)\right)}^{\prime }dx.$$

(41)

Hence,

$$\left|\rho \left(x\right)u\left(x\right)\right|\le {\int }_0^1\left|{\left(\rho \left(x\right)u\left(x\right)\right)}^{\prime }\right|dx={\int }_0^1\left|{\rho }^{\prime }\left(x\right)u\left(x\right)+\rho \left(x\right)u^{\prime }\left(x\right)\right|dx.$$

(42)

Using ${\displaystyle \underset{[0,1]}{sup}}\mid {\rho }^{\prime }\left(x\right)\mid =1,$ employing Hölder’s inequality and using ${\rho }^2\left(x\right)\le \rho \left(x\right),x\in [0,1],$ we obtain

$$\begin{array}{cc}\hfill \left|\rho \left(x\right)u\left(x\right)\right|& \le {\left[{\int }_0^1{\left(\left|u\left(x\right)+\rho \left(x\right)u^{\prime }\left(x\right)\right|\right)}^2\right]}^{{\displaystyle \frac{1}{2}}}dx\hfill \\ \hfill & \le \sqrt{2}{\left[{\int }_0^1\left(u^2\left(x\right)+{\left(\rho \left(x\right)u^{\prime }\left(x\right)\right)}^2\right)dx\right]}^{{\displaystyle \frac{1}{2}}}\le \sqrt{2M_3}.\hfill \end{array}$$

(43)

Likewise,

$$\left|\rho \left(x\right)v\left(x\right)\right|\le \sqrt{2}{\left[{\int }_0^1\left(v^2\left(x\right)+\rho \left(x\right)v^{\prime 2}\left(x\right)\right)dx\right]}^{{\displaystyle \frac{1}{2}}}\le \sqrt{2M_2}$$

(44)

and

$$\left|\rho \left(x\right)w\left(x\right)\right|\le \sqrt{2}{\left[{\int }_0^1\left(w^2\left(x\right)+\rho \left(x\right)w^{\prime 2}\left(x\right)\right)dx\right]}^{{\displaystyle \frac{1}{2}}}\le \sqrt{2M_1}.$$

(45)

The proof of the proposition follows now from these three inequalities.

$$M=\sqrt{2}\left(M_1^{{\displaystyle \frac{1}{2}}}+M_2^{{\displaystyle \frac{1}{2}}}+M_3^{{\displaystyle \frac{1}{2}}}\right).$$

□

Remark 1. 

In Proposition 1, we can replace the weight function $\rho \left(x\right)=x(1-x)$ with any positive function $\rho \in {\mathbb{C}}^2[0,1]$ that satisfies the following conditions: ${\rho }^{″}<0,$ $\rho \left(0\right)=\rho \left(1\right)=0$ with ${\rho }^{\prime }\left(0\right)>0$ and ${\rho }^{\prime }\left(1\right)<0.$

Furthermore, if ${\displaystyle \underset{[0,1]}{sup}}\rho \left(x\right)<1,$ ${\displaystyle \underset{[0,1]}{sup}}{\rho }^{\prime }\left(x\right)\le 1.$ Then, in this case, we write ${\displaystyle \underset{[0,1]}{sup}}\rho \left(x\right)={\displaystyle \frac{1}{\sigma }}$ for some $\sigma >1,$ $\sqrt{K_1}={\displaystyle \frac{1}{\sigma \beta }}$ and $\sqrt{K_1}={\displaystyle \frac{1}{\sigma \alpha }},$ and so the value of δ becomes

$$\delta =1-\left({\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{\sigma }}+{\displaystyle \frac{\lambda c}{\sigma }}+{\displaystyle \frac{\lambda ϵ}{2\sigma }}+{\displaystyle \frac{B_1}{{\sigma }^2\beta }}+{\displaystyle \frac{\lambda b}{{\sigma }^2\beta }}+{\displaystyle \frac{C_1}{{\sigma }^3\alpha \beta }}+{\displaystyle \frac{\lambda a}{{\sigma }^3\alpha \beta }}\right).$$

Therefore, in order to obtain the desired result, we can write estimate (7) as

$$\lambda <{\displaystyle \frac{\sigma \alpha \beta -A_1-B_1-C_1}{3k}}.$$

The Schauder’s fixed-point theorem plays a crucial role in proving the existence of the solution. To make use of this theorem, we need the following lemmas.

Lemma 1 

([25]). Let $g:[0,1]\to \mathbb{R}$ and $h:[0,1]\to \mathbb{R}$ be continuous functions. If $1-{\int }_0^{1}h\left(x\right)dx\ne 0,$ then the unique solution u of the problem

$$\begin{array}{ccc}& & u^{″}=g\left(x\right)\hfill \end{array}$$

(46)

subject to the nonlocal boundary conditions $u\left(0\right)=u\left(1\right)={\int }_0^{1}h\left(x\right)u\left(x\right)dx$ is given by

$$\begin{array}{ccc}& & u\left(x\right)={\int }_0^{1}G(x,y;h\left(y\right))g\left(y\right)dy,\hfill \end{array}$$

(47)

where $G(x;y;h)$ is the Green function of this BVP and is given explicitly in [25].

From (8) and this lemma, we therefore obtain an equivalent integral system:

$$\begin{array}{c}\hfill \left\{\begin{array}{ccc}\hfill u& =& {\int }_0^1{G}_1(x,s)v\left(s\right)ds,x\in [0,1],\hfill \\ \hfill v& =& {\int }_0^1{G}_2(x,s)w\left(s\right)ds,x\in [0,1],\hfill \\ \hfill w& =& -{\int }_0^1{G}_3(x,s)A\left(s\right)w\left(s\right)ds-{\int }_0^1{G}_3(x,s)B\left(s\right)v\left(s\right)ds\hfill \\ & & -{\int }_0^1{G}_3(x,s)C\left(s\right)u\left(s\right)ds+\lambda {\int }_0^1{G}_3(x,s)f(s,u\left(s\right),v\left(s\right),w\left(s\right))ds,\hfill \end{array}\right.\end{array}$$

(48)

where $G_1(x,y)=G(x,y;p),$ $G_2(x,y)=G(x,y;q)$ and $G_3(x,y)=G(x,y;s)$ such that $1-{\int }_0^{1}p\left(x\right)dx\ne 0,$ $1-{\int }_0^{1}q\left(x\right)dx\ne 0$ and $1-{\int }_0^{1}s\left(x\right)dx\ne 0$. Since $G_i(x,y),i=1,2,3$ are continuous on $[0,1],$ we let ${\displaystyle \underset{[0,1]}{sup}}\mid G_i(x,y)\mid \le L,i=1,2,3,$ and L will be chosen such that

$$KL\le \sqrt{2},$$

(49)

where K is given by

$$K=max\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}(A_1+\lambda c+\lambda \sqrt{{\displaystyle \frac{8ϵ\delta }{\lambda }}}),{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}(B_1+\lambda b),{\displaystyle \frac{1}{4}}(C_1+\lambda a)\right).$$

(50)

Now, let us define the following Banach space

$$\begin{array}{c}\hfill {\mathbb{Y}}_{\rho }=\left\{u\in {\mathbb{C}}^4[0,1]:u\mathrm{satisfies}\left(48\right)\mathrm{and}\left(49\right)\right\}\end{array}$$

(51)

with norm ${\Vert u\Vert }_{\rho ,3}={\Vert u\Vert }_{\rho ,0}+{\Vert u\Vert }_{\rho ,1}+{\Vert u\Vert }_{\rho ,2},$ where ${\Vert u\Vert }_{\rho ,1}={\Vert u^{″}\Vert }_{\rho ,0}$, ${\Vert u\Vert }_{\rho ,2}={\Vert u^{\left(4\right)}\Vert }_{\rho ,0}.$ We also define the operator $T:\mathbb{X}⟶\mathbb{X}$ by

$$\begin{array}{c}\hfill T(u,v,w)=\left(T_1(u,v,w),T_2(u,v,w),T_3(u,v,w)\right),\end{array}$$

(52)

where $\mathbb{X}={\mathbb{Y}}_{\rho }\times {\mathbb{Y}}_{\rho }\times {\mathbb{Y}}_{\rho }$ with norm

${\Vert (u,v,w)\Vert }_{\rho ,3}={\Vert u\Vert }_{\rho ,0}+{\Vert v\Vert }_{\rho ,0}+{\Vert w\Vert }_{\rho ,0}$ such that

$$\begin{array}{ccc}& & T_1(u,v,w)={\int }_0^1{G}_1(x,s)v\left(s\right)ds,\hfill \end{array}$$

(53)

$$\begin{array}{ccc}& & T_2(u,v,w)={\int }_0^1{G}_2(x,s)w\left(s\right)ds\hfill \end{array}$$

(54)

and

$$\begin{array}{ccc}\hfill T_3(u,v,w)=& & -{\int }_0^1{G}_3(x,s)A\left(s\right)w\left(s\right)ds\hfill \\ & & -{\int }_0^1{G}_3(x,s)B\left(s\right)v\left(s\right)ds-{\int }_0^1{G}_3(x,s)C\left(s\right)u\left(s\right)ds\hfill \\ & & +\lambda {\int }_0^1{G}_3(x,s)f(s,u\left(s\right),v\left(s\right),w\left(s\right))ds.\hfill \end{array}$$

(55)

Consider now the closed convex set

$$\begin{array}{c}\hfill \mathbb{S}=\left\{(u,v,w)\in {\mathbb{X}:\parallel (u,v,w)\parallel }_{\rho ,3}\le M\right\}.\end{array}$$

(56)

Lemma 2. 

For any $(u,v,w)\in \mathbb{S},$ $T(u,v,w)$ is contained in $\mathbb{S}.$

Proof. 

From the definition of $T(u,v,w),$ we have

$$\begin{array}{ccc}\hfill \left|\rho \left(x\right)T_1(u,v,w)\right|& & \le \rho \left(x\right){\int }_0^1\left|G_1(x,s)\right|\left|v\left(s\right)\right|ds\hfill \end{array}$$

$$\begin{array}{ccc}& & \le {\displaystyle \frac{1}{4}}{\int }_0^1\left|G_1(x,s)\right|\left|v\left(s\right)\right|ds,\hfill \\ \hfill \left|\rho \left(x\right)T_2(u,v,w)\right|& & \le \rho \left(x\right){\int }_0^1\left|G_2(x,s)\right|\left|w\left(s\right)\right|ds\hfill \end{array}$$

(57)

$$\begin{array}{ccc}& & \le {\displaystyle \frac{1}{4}}{\int }_0^1\left|G_2(x,s)\right|\left|w\left(s\right)\right|ds.\hfill \end{array}$$

(58)

Hence,

$$\begin{array}{c}\hfill \left|\rho \left(x\right)T_1(u,v,w)\right|\le {\displaystyle \frac{1}{4}}{\left({\int }_0^1{\left|G_1(x,s)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{\left|v\left(s\right)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}},\end{array}$$

(59)

$$\begin{array}{c}\hfill \left|\rho \left(x\right)T_2(u,v,w)\right|\le {\displaystyle \frac{1}{4}}{\left({\int }_0^1{\left|G_2(x,s)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{\left|w\left(s\right)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}.\end{array}$$

(60)

From (38) and (39), we have ${\int }_0^1{v}^2\left(x\right)dx\le M_2,$ and ${\int }_0^1{w}^2\left(x\right)dx\le M_1.$ Therefore,

$$\begin{array}{ccc}& & \left|\rho \left(x\right)T_1(u,v,w)\right|\le {\displaystyle \frac{1}{4}}L{M}_2^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

(61)

$$\begin{array}{ccc}& & \left|\rho \left(x\right)T_2(u,v,w)\right|\le {\displaystyle \frac{1}{4}}L{M}_1^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(62)

Similarly,

$$\begin{array}{cc}\hfill & \left|\rho \left(x\right)T_3(u,v,w)\right|\le \rho \left(x\right){\int }_0^1\left|G_3(x,s)\right|\left|A\left(s\right)\right|\left|w\left(s\right)\right|ds\hfill \\ & +\rho \left(x\right){\int }_0^1\left|G_3(x,s)\right|\left|B\left(s\right)\right|\left|v\left(s\right)\right|ds\hfill \\ \hfill & +\rho \left(x\right){\int }_0^1\left|G_3(x,s)\right|\left|C\left(s\right)\right|\left|u\left(s\right)\right|ds\hfill \\ \hfill & +\lambda \rho \left(x\right){\int }_0^1\left|G_3(x,s)\right|\left|f(s,u(s),v(s),w(s\left)\right)\right|ds.\hfill \end{array}$$

(63)

Consequently,

$$\begin{array}{cc}\hfill \left|\rho \left(x\right)T_3(u,v,w)\right|& \le {\displaystyle \frac{1}{4}}L\left(A_1{M}_1^{{\displaystyle \frac{1}{2}}}+B_1{M}_2^{{\displaystyle \frac{1}{2}}}+C_1{M}_3^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ & +{\displaystyle \frac{\lambda L}{4}}\left(a{M}_3^{{\displaystyle \frac{1}{2}}}+b{M}_2^{{\displaystyle \frac{1}{2}}}+c{M}_1^{{\displaystyle \frac{1}{2}}}+d\right).\hfill \end{array}$$

(64)

Thus

$$\left|\rho \left(x\right)T_3(u,v,w)\right|\le {\displaystyle \frac{1}{4}}L\left[(A_1+\lambda c)M_1^{{\displaystyle \frac{1}{2}}}+(B_1+\lambda b)M_2^{{\displaystyle \frac{1}{2}}}+(C_1+\lambda a)M_3^{{\displaystyle \frac{1}{2}}}+d\right].$$

(65)

Since $M_1={\displaystyle \frac{\lambda d^2}{8ϵ\delta }}$ (from (38)), we have

$$\begin{array}{cc}\hfill \left|\rho \left(x\right)T_3(u,v,w)\right|& \le {\displaystyle \frac{1}{4}}L\left[(A_1+\lambda c+\lambda \sqrt{{\displaystyle \frac{8ϵ\delta }{\lambda }}})M_1^{{\displaystyle \frac{1}{2}}}+(B_1+\lambda b)M_2^{{\displaystyle \frac{1}{2}}}+(C_1+\lambda a)M_3^{{\displaystyle \frac{1}{2}}}\right].\hfill \end{array}$$

(66)

Therefore,

$$\left|\rho \left(x\right)T_1(u,v,w)\right|+\left|\rho \left(x\right)T_2(u,v,w)\right|+\left|\rho \left(x\right)T_3(u,v,w)\right|\le M^{*},$$

(67)

where

$$M^{*}={\displaystyle \frac{1}{4}}L\left[(1+A_1+\lambda c+\lambda \sqrt{{\displaystyle \frac{8ϵ\delta }{\lambda }}})M_1^{{\displaystyle \frac{1}{2}}}+(1+B_1+\lambda b)M_2^{{\displaystyle \frac{1}{2}}}+(C_1+\lambda a)M_3^{{\displaystyle \frac{1}{2}}}\right].$$

(68)

Using (50), we obtain

$$M^{*}\le KL\left(M_1^{{\displaystyle \frac{1}{2}}}+M_2^{{\displaystyle \frac{1}{2}}}+M_3^{{\displaystyle \frac{1}{2}}}\right).$$

(69)

From (49), we have

$$M^{*}\le \sqrt{2}\left(M_1^{{\displaystyle \frac{1}{2}}}+M_2^{{\displaystyle \frac{1}{2}}}+M_3^{{\displaystyle \frac{1}{2}}}\right).$$

(70)

Noting that $M=\sqrt{2}\left(M_1^{{\displaystyle \frac{1}{2}}}+M_2^{{\displaystyle \frac{1}{2}}}+M_3^{{\displaystyle \frac{1}{2}}}\right)$ is indeed the same constant of Proposition 1.

We therefore have ${\Vert T(u,v,w)\Vert }_{\rho ,3}\le M.$ By continuity of $f(x,u,v,w),$u and $v,$ we see that $T(u,v,w)$ is continuous. This proves that $T(u,v,w)$ is also contained in $\mathbb{S}.$ □

Hence $T:\mathbb{S}\to \mathbb{S}$ maps the closed, bounded and convex set $\mathbb{S}$ into itself. A similar argument can be applied to ${\left(T(u,v,w)\right)}^{\prime }\left(x\right),$ where

$$\begin{array}{ccc}& & {\left(T_1(u,v,w)\right)}^{\prime }\left(x\right)={\int }_0^1{G}_{1,x}(x,s)v\left(s\right)ds,\hfill \end{array}$$

(71)

$$\begin{array}{ccc}& & {\left(T_2(u,v,w)\right)}^{\prime }\left(x\right)={\int }_0^1{G}_{2,x}(x,s)w\left(s\right)ds\hfill \end{array}$$

(72)

and

$$\begin{array}{ccc}\hfill {\left(T_3(u,v,w)\right)}^{\prime }\left(x\right)=& & -{\int }_0^1{G}_{3,x}(x,s)A\left(s\right)w\left(s\right)ds\hfill \\ & & -{\int }_0^1{G}_{3,x}(x,s)B\left(s\right)v\left(s\right)ds-{\int }_0^1{G}_{3,x}(x,s)C\left(s\right)u\left(s\right)ds\hfill \\ & & +\lambda {\int }_0^1{G}_{3,x}(x,s)f(s,u\left(s\right),v\left(s\right),w\left(s\right))ds,\hfill \end{array}$$

(73)

where $G_{i,x}(x,s)={\displaystyle \frac{\partial G_i(x,s)}{\partial x}},i=1,2,3$.

To prove that $T(u,v)$ is compact, we use the Arzela–Ascoli lemma; that is, $T\left(\mathbb{S}\right)$ must be closed, bounded and equicontinuous.

In order to prove that $T\left(\mathbb{S}\right)$ is equicontinuous, take x and y in the interval $[0,1],$ then there exists $\xi$ between x and y such that

$$G_1(x,s)-G_1(y,s)=(y-x)G_{1,x}(\xi ,s).$$

(74)

It follows by the definition of $T(u,v,w)$ that

$$\left|\rho \left(x\right)\left(T_1(u\left(x\right),v\left(x\right),w\left(x\right))-T_1(u\left(y\right),v\left(y\right),w\left(y\right))\right)\right|=\rho \left(x\right)\left|{\int }_0^1\left[G_1(x,s)-G_1(y,s)\right]v\left(s\right)ds\right|.$$

(75)

Hence,

$$\begin{array}{cc}\hfill \left|\rho \left(x\right)\left(T_1(u\left(x\right),v\left(x\right),w\left(x\right))-T_1(u\left(y\right),v\left(y\right),w\left(y\right))\right)\right|& =\rho \left(x\right)\left|{\int }_0^1\left[G_1(x,s)-G_1(y,s)\right]v\left(s\right)\right|ds\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}{\int }_0^1\left|(y-x)G_{1,x}(\xi ,s)v\left(s\right)\right|ds\hfill \\ \hfill & \le L_1^{\prime }{M}_2^{{\displaystyle \frac{1}{2}}}\left|x-y\right|,\hfill \end{array}$$

(76)

where ${\displaystyle \underset{[0,1]}{sup}}\mid G_1(x,s)\mid \le L_1^{\prime }.$ Let $ϵ>0.$ If $\left|x-y\right|<\delta ={\displaystyle \frac{ϵ}{3L_1^{\prime }{M}_2^{\frac{1}{2}}}},$ we have

$$\left|\rho \left(x\right)\left(T_1(u\left(x\right),v\left(x\right),w\left(x\right))-T_1(u\left(y\right),v\left(y\right),w\left(y\right))\right)\right|\le {\displaystyle \frac{ϵ}{3}}$$

(77)

for all x and y in the interval $[0,1]$ satisfying $\left|x-y\right|<\delta .$

A similar argument can be applied to $T_2(u\left(x\right),v\left(x\right),w\left(x\right))$ and $T_3(u\left(x\right),v\left(x\right),w\left(x\right)),$ we have

$$\left|\rho \left(x\right)\left(T_2(u\left(x\right),v\left(x\right),w\left(x\right))-T_2(u\left(y\right),v\left(y\right),w\left(y\right))\right)\right|\le {\displaystyle \frac{ϵ}{3}}$$

(78)

and

$$\left|\rho \left(x\right)\left(T_3(u\left(x\right),v\left(x\right),w\left(x\right))-T_3(u\left(y\right),v\left(y\right),w\left(y\right))\right)\right|\le {\displaystyle \frac{ϵ}{3}}$$

(79)

for all x and y in the interval $[0,1]$ satisfying $\left|x-y\right|<\delta .$ This proves the equicontinuity of $T\left(\mathbb{S}\right)$.

Therefore, $T(u,v,w)$ has a fixed point by the Schauder’s fixed point theorem.

Theorem 1. 

Under the hypothesis of Proposition 1, there exists a continuous solution $(u,v,w)$ which satisfies system (8).

3. A Uniqueness Theorem

A uniqueness theorem can also be obtained by assuming that $f(x,u,u^{\left(2\right)},u^{\left(4\right)})$ satisfies a Lipschitz condition in $u,u^{\left(2\right)}$ and $u^{\left(4\right)}$ with constants $t_i>0,i=0,1,2,$ that is

$$\left|f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})\right|\le t_0\left|u-v\right|+t_1\left|u^{″}-v^{″}\right|+t_2\left|u^{\left(4\right)}-v^{\left(4\right)}\right|.$$

(80)

If we assume that

$${\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{B_1\sqrt{K_1}}{4}}+{\displaystyle \frac{\sqrt{K_1{K}_2}{C}_1}{4}}+{\displaystyle \frac{\lambda t_0\sqrt{K_1{K}_2}}{4}}+{\displaystyle \frac{\lambda \sqrt{K_1}{t}_1}{4}}+{\displaystyle \frac{\lambda t_2}{4}}<1,$$

(81)

where $K_1={\displaystyle \frac{1}{16{(1-{\int }_0^1{q}^2\left(x\right)dx)}^2}},$ $K_2={\displaystyle \frac{1}{16{(1-{\int }_0^1{p}^2\left(x\right)dx)}^2}},$

Then, we have

Theorem 2. 

If f is Lipschitz in $u,u^{″}$ and $u^{\left(4\right)}$ such that the constants $K_i>0,i=0,1$ satisfy (81) and if (6) and (7) hold with (5), then the system (8) has a unique solution $(u,v,w).$

Proof. 

Let u and v be two solutions of Pr. (1)–(4) such that $u\ne v$. Assuming $w=u-v,$ we have

$$w^{\left(6\right)}+A\left(x\right)w^{\left(4\right)}+B\left(x\right)w^{″}+C\left(x\right)w=\lambda \left[f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})\right]$$

(82)

subject to

$$\begin{array}{c}\hfill w\left(0\right)=w\left(1\right)={\int }_0^{1}p\left(x\right)w\left(x\right)dx,\end{array}$$

(83)

$$\begin{array}{c}\hfill w^{″}\left(0\right)=w^{″}\left(1\right)={\int }_0^{1}q\left(x\right)w^{″}\left(x\right)dx\end{array}$$

(84)

and

$$\begin{array}{c}\hfill w^{\left(4\right)}\left(0\right)=w^{\left(4\right)}\left(1\right)={\int }_0^{1}s\left(x\right)w^{\left(4\right)}\left(x\right)dx.\end{array}$$

(85)

Thus, we have

$$\left\{\begin{array}{ccc}\hfill w^{″}& =& y,w\left(0\right)=w\left(1\right)={\int }_0^{1}p\left(x\right)w\left(x\right)dx,\hfill \\ \hfill y^{″}& =& z,y\left(0\right)=y\left(1\right)={\int }_0^{1}q\left(x\right)y\left(x\right)dx,\hfill \\ \hfill z^{″}& =& -A\left(x\right)z-B\left(x\right)y-C\left(x\right)w+\lambda \left[f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})\right],\hfill \\ \hfill & & z\left(0\right)=z\left(1\right)={\int }_0^{1}s\left(x\right)z\left(x\right)dx.\hfill \end{array}\right.$$

(86)

Proceeding as in Proposition 1, we obtain

$$\begin{array}{cc}\hfill {\int }_0^1{z}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right){\left(z^{\prime }\left(x\right)\right)}^{2}dx& ={\left[{\int }_0^{1}s\left(x\right)z\left(x\right)dx\right]}^2+{\int }_0^{1}A\left(x\right)\rho \left(x\right)z^2\left(x\right)\hfill \\ \hfill & +{\int }_0^{1}B\left(x\right)\rho \left(x\right)y\left(x\right)z\left(x\right)+{\int }_0^{1}C\left(x\right)\rho \left(x\right)w\left(x\right)z\left(x\right)dx\hfill \\ \hfill & -\lambda {\int }_0^1\left[f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})\right]\rho \left(x\right)z\left(x\right)dx,\hfill \\ \hfill \end{array}$$

(87)

where

$$\begin{array}{ccc}& & \left|{\int }_0^{1}A\left(x\right)\rho \left(x\right)z^2\left(x\right)dx\right|\le {\displaystyle \frac{A_1}{4}}\left({\int }_0^1{z}^2\left(x\right)dx\right),\hfill \\ & & \left|{\int }_0^{1}B\left(x\right)\rho \left(x\right)y\left(x\right)z\left(x\right)dx\right|\le {\displaystyle \frac{B_1}{4}}{\left({\int }_0^1{y}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{z}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}},\hfill \\ & & \left|{\int }_0^{1}C\left(x\right)\rho \left(x\right)w\left(x\right)z\left(x\right)dx\right|\le {\displaystyle \frac{C_1}{4}}{\left({\int }_0^1{w}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{z}^2\left(x\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}$$

(88)

and

$$\begin{array}{c}\hfill {\left[{\int }_0^{1}s\left(x\right)z\left(x\right)dx\right]}^2\le \left({\int }_0^1{s}^2\left(x\right)dx\right)\left({\int }_0^1{z}^2\left(x\right)dx\right).\end{array}$$

(89)

Applying the Lipschitz condition to $f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})$ to obtain

$$\begin{array}{cc}\hfill \left|{\int }_0^{1}f(x,u,u^{″},u^{\left(4\right)})-f(x,v,v^{″},v^{\left(4\right)})\rho \left(x\right)z\left(x\right)dx\right|\le & {\displaystyle \frac{t_0}{4}}{\int }_0^1\left|w\left(x\right)z\left(x\right)\right|dx\hfill \\ \hfill & +{\displaystyle \frac{t_1}{4}}{\int }_0^1\left|y\left(x\right)z\left(x\right)\right|dx\hfill \\ \hfill & +{\displaystyle \frac{t_2}{4}}{\int }_0^1\left|z^2\left(x\right)\right|dx.\hfill \end{array}$$

(90)

Combining these inequalities, we obtain

$${\delta }_1{\int }_0^1{z}^2\left(x\right)dx+{\int }_0^1\rho \left(x\right)z^{\prime 2}\left(x\right)dx\le 0,$$

(91)

where

$${\delta }_1=1-\left({\int }_0^1{s}^2\left(x\right)dx+{\displaystyle \frac{A_1}{4}}+{\displaystyle \frac{B_1\sqrt{K_1}}{4}}+{\displaystyle \frac{\sqrt{K_1{K}_2}{C}_1}{4}}+{\displaystyle \frac{\lambda t_0\sqrt{K_1{K}_2}}{4}}+{\displaystyle \frac{\lambda \sqrt{K_1}{t}_1}{4}}+{\displaystyle \frac{\lambda t_2}{4}}\right)>0.$$

(92)

This is a contradiction.

This completes the proof. □

4. Conclusions

The paper investigates the existence and uniqueness of the solution of sixth-order nonlinear beam differential equations with four parameters, subject to the conditions (2)–(7), where (2)–(4) give nonlocal boundary conditions rather than local, (6) implies the loaded force f satisfies a growth condition rather being bounded, and (7) implies that an upper estimate is imposed on the eigenvalues which refer to the reciprocal of the flexural rigidity that measures the resistance to bend. So, by imposing an upper bound on $\lambda ,$ we maintain a small deflection to the beam. The study establishes the existence and uniqueness of deflection u as a solution to (1). This facilitates the idea of modeling the system in concrete applications and enables researchers to study the behavior of the system by studying the qualitative properties of the unique solution and look for situations to optimize the solution. Furthermore, as suggested by the equation and its conditions, keeping $\lambda$ small can be guaranteed if either the loaded force f is small, or the material of the beam has high flexural rigidity, and this might be advantageous for the car and aircraft industries and construction engineering regarding safety standards since large deflections could cause large vibrations which could annoy drivers and passengers, or cause cracks in the beams which could be dangerous in construction engineering.

The objective of this research is to determine whether small deflections occur in the beams of the form of circular ring segment (which are modeled by the sixth-order beam equations). The result shows that the solution to the problem exists, which implies that small deflections continue to exist on the beam whether the edges of the beam (end points) are fixed, slightly relaxed, or slightly supported. This result should help the manufacturers to reduce noise and vibrations and thus improve their designs.

Finally, the study opens the doors to further mathematical research (e.g., finding an explicit form of the solution or approximating it using semi-analytic or numerical methods). Numerical solutions of sixth-order differential equations with local boundary conditions have been studied in the literature, see [12,13,14,26,27,28,29,30]. However, the nonlocal conditions requirement is new to the literature, and the results of this paper can indeed motivate researchers to investigate the numerical solution of Pr. (1)–(4), subject to (6) and (7). It is worth noting that the assumption of small deflections and having small values of the parameter $\lambda$ suggests using the homotopy perturbation method since it works perfectly for nonlinear equations containing small parameters and can provide the solution in a rapid convergent series.