Existence and uniqueness of solutions to the nonlinear boundary value problem for fourth-order differential equations with all derivatives

Abstract

This article addresses the existence and uniqueness of solution for fully fourth-order differential equations modeling beams on elastic foundations with nonlinear boundary conditions. The proof will rely on Perov’s fixed point theorem in complete generalized metric spaces to overcome the problems due to the presence of all lower-order derivatives in the nonlinearity. Finally, some illustrating examples of the theory are presented.

1 Introduction

Recently, fourth-order ordinary differential equations for boundary value problems, which frequently appear in various academic and engineering fields, e.g., chemistry, physics, and dynamical system control, have attracted vast attention and have been broadly studied, especially on the existence and multiplicity of solution under different boundary conditions [1–29]. For example, Li [14] considered the existence of positive solutions to fully equations for cantilever beams, which was determined by the fixed point index theory for cones and sublinear or superlinear growth behaviors of the nonlinearity, through

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)),\quad t \in [0,1], \\ u(0)=u'(0)=u''(1)=u'''(1)=0, \end{cases} $$

(1.1)

where f is continuous. In [29], the authors obtained new conclusion on the uniqueness of solutions to problem (1.1) with the order reduction method and the linear operator theory. In [1], Almuthaybiri and Tisdell considered the following nonlinear fourth-order differential equation with linear boundary conditions (LBC) for continuous $f: [0, 1]\times \mathbb{R}^{4}\rightarrow \mathbb{R}$:

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)),\quad t \in [0,1], \\ u(0)=a,\qquad u'(0)=c,\qquad u''(1)=e,\qquad u'''(1)=h, \end{cases} $$

where constants of $a,c,e,h \in \mathbb{R}$ are known. They sharpen traditional uniqueness results by employing Rus’s contraction mapping theorem where it is accepted that the mapping is contraction according to the metric δ when the space $C^{3}[0,1]$ is complete for the other metric d, where

$$ d(y,z)=\sum_{i=0}^{3}\max _{x\in [0,1]} \bigl\vert y^{(i)}(x)-z^{(i)}(x) \bigr\vert , \qquad \delta (y,z)=\sum_{i=0}^{3}L_{i} \biggl( \int _{0}^{1} \bigl\vert y^{(i)}(x)-z^{(i)}(x) \bigr\vert ^{p}\,dx \biggr)^{\frac{1}{p}}. $$

We refer readers to the previous studies on cantilever beam equations [6, 15, 24] and the references therein.

In [18], Ma considered the following beam equation with third-order nonlinear boundary conditions (NLBC):

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t)),\quad 0< t< 1, \\ u(0)=u'(0)=0, \\ u''(1)=0 \quad \text{and} \quad u'''(1)=g(u(1)), \end{cases} $$

(1.2)

where f and g denote continuous functions. Problem (1.2) can model the deformation of an elastic beam on elastic bearings, g expresses the influence of vertical force on the right elastic bearing ($t=1$). Using the contraction principle and some restrictive conditions, the authors proved that the solution of problem (1.2) exists and is unique. The existence and multiplicity for such a problem have also been studied by using critical point theory [9, 10, 19]. For problem (1.2), the nonlinear part does not consist of any derivative terms. Therefore, it is an interesting problem to discuss the existence and uniqueness of solution when the nonlinearity depends on every derivative up to order three and the boundary conditions are nonlinear.

Based on the aforementioned work, the present paper aims to propose an existence and uniqueness theorem for fourth-order differential equations having all derivatives for boundary value problems:

$$ u^{(4)}(t)=f\bigl(t,u(t),u'(t),u''(t),u'''(t) \bigr),\quad 0< t< 1, $$

(1.3)

with either NLBC

$$ u(0)=u'(0)=u''(1)=0,\qquad u'''(1)=g \bigl(u(1)\bigr) $$

(1.4)

or LBC

$$ u(0)=u'(0)=u''(1)=u'''(1)=0, $$

(1.5)

where $f\in C([0,1]\times \mathbb{R}^{4}, \mathbb{R})$ and $g\in C(\mathbb{R}, \mathbb{R})$ are real functions. This paper has two significant features: the nonlinearity function f contains all derivatives of the unknown function with NLBC; the establishment of the existence and uniqueness of solutions for problems (1.3), (1.4) and (1.3), (1.5) is mainly based on Perov’s fixed point theorem (PFPT) and norm $\|\cdot \|_{e}$.

This manuscript is structured as follows. Section 2 provides several essential lemmas associated with this work. Section 3 presents the main results obtained for problems (1.3), (1.4) and (1.3), (1.5).

The following assumptions are made in this paper:

$(H_{1})$: There exist four constants $p_{i}>0$ such that, for all $u_{i}, v_{i} \in \mathbb{R}$, $i=1,2,3,4$,

$$\begin{aligned}& \bigl\vert f(t,u_{1},u_{2},u_{3},u_{4})-f(t,v_{1},v_{2},v_{3},v_{4}) \bigr\vert \\& \quad \leq p_{1} \vert u_{1}-v_{1} \vert +p_{2} \vert u_{2}-v_{2} \vert +p_{3} \vert u_{3}-v_{3} \vert +p_{4} \vert u_{4}-v_{4} \vert . \end{aligned}$$

$(H_{2})$: By $(H_{1})$, we obtain the following matrix:

$$M=\left(\begin{array}{cccc}\frac{p_1}{8}+\frac{L_g}{3}& \frac{p_2}{6}& \frac{p_3}{4}& \frac{p_4}{4}\\ \frac{p_1}{4}+\frac{L_g}{2}& \frac{p_2}{3}& \frac{p_3}{2}& \frac{p_4}{2}\\ \frac{p_1}{4}+L_g& \frac{p_2}{3}& \frac{p_3}{2}& \frac{p_4}{2}\\ \frac{p_1}{3}+L_g& \frac{p_2}{2}& p_3& p_4\end{array}\right),$$

where $L_{g}$ is a constant and g is a Lipschitz function, and we suppose that its spectral radius $\rho (M)<1$.

$(H_{3})$: By $(H_{1})$, we obtain the matrix

$$M_1=\left(\begin{array}{cccc}\frac{p_1}{8}+L_g& \frac{p_2}{8}& \frac{p_3}{8}& \frac{p_4}{8}\\ \frac{p_1}{6}+L_g& \frac{p_2}{6}& \frac{p_3}{6}& \frac{p_4}{6}\\ \frac{p_1}{2}+L_g& \frac{p_2}{2}& \frac{p_3}{2}& \frac{p_4}{2}\\ p_1+L_g& p_2& p_3& p_4\end{array}\right),$$

and we suppose that its spectral radius $\rho (M_{1})<1$.

2 Preliminaries

Let the norm of Banach space $E=C[0,1]$ be $\|u(t)\|_{\infty}=\max_{0 \leq t \leq 1}|u(t)|$.

Lemma 2.1

([18])

Given $h\in E$, there exists a unique solution for the linear problem

$$ \textstyle\begin{cases} u^{(4)}(t)=h(t), \quad 0< t< 1, \\ u(0)=u'(0)=0, \\ u''(1)=0 \quad \textit{and} \quad u'''(1)=\gamma \in \mathbb{R}, \end{cases} $$

and the solution can be expressed by

$$ u(t)= \int _{0}^{1}G_{1}(t,s)h(s)\,ds+\gamma \frac{t^{2}}{6}(t-3) $$

with Green’s function G given by

$$ G_{1}(t,s)=\textstyle\begin{cases} \frac{1}{6}t^{2}(3s-t),& 0 \leq t \leq s \leq 1, \\ \frac{1}{6}s^{2}(3t-s),& 0 \leq s \leq t \leq 1. \end{cases} $$

Based on Lemma 2.1, we see that u is the solution of problem (1.3), (1.4) if and only if u is the solution of the formula

$$ u(t)= \int _{0}^{1}G_{1}(t,s)f \bigl(s,u(s),u'(s),u''(s),u'''(s) \bigr)\,ds+g\bigl(u(1)\bigr) \frac{t^{2}}{6}(t-3). $$

(2.1)

Differentiating equation (2.1) yields

$$ u'(t)= \int _{0}^{1}G_{2}(t,s)f \bigl(s,u(s),u'(s),u''(s),u'''(s) \bigr)\,ds+g\bigl(u(1)\bigr) \frac{t}{2}(t-2), $$

where

$$ G_{2}(t,s)=\textstyle\begin{cases} \frac{1}{2}t(2s-t), & 0 \leq t \leq s \leq 1, \\ \frac{1}{2}s^{2}, & 0 \leq s < t \leq 1. \end{cases} $$

Similarly, we can obtain

$$ u''(t)= \int _{0}^{1}G_{3}(t,s)f \bigl(s,u(s),u'(s),u''(s),u'''(s) \bigr)\,ds+g\bigl(u(1)\bigr) (t-1) $$

and

$$ u'''(t)= \int _{0}^{1}G_{4}(t,s)f \bigl(s,u(s),u'(s),u''(s),u'''(s) \bigr)\,ds+g\bigl(u(1)\bigr), $$

where

$$\begin{aligned}& G_{3}(t,s)=\textstyle\begin{cases} s-t, & 0 \leq t \leq s \leq 1, \\ 0, & 0 \leq s < t \leq 1, \end{cases}\displaystyle \\& G_{4}(t,s)=\textstyle\begin{cases} -1, & 0 \leq t \leq s \leq 1, \\ 0, & 0 \leq s < t \leq 1. \end{cases}\displaystyle \end{aligned}$$

Therefore, we suppose that there exists a solution $u_{1}(t)$ for problem (1.3), (1.4) that can be rewritten as follows:

$$ \textstyle\begin{cases} u_{1}(t)=\int _{0}^{1}G_{1}(t,s)f(s,u_{1}(s),u_{2}(s),u_{3}(s),u_{4}(s))\,ds+g(u_{1}(1)) \frac{t^{2}}{6}(t-3), \\ u_{2}(t)=\int _{0}^{1}G_{2}(t,s)f(s,u_{1}(s),u_{2}(s),u_{3}(s),u_{4}(s))\,ds+g(u_{1}(1)) \frac{t}{2}(t-2), \\ u_{3}(t)=\int _{0}^{1}G_{3}(t,s)f(s,u_{1}(s),u_{2}(s),u_{3}(s),u_{4}(s))\,ds+g(u_{1}(1))(t-1), \\ u_{4}(t)=\int _{0}^{1}G_{4}(t,s)f(s,u_{1}(s),u_{2}(s),u_{3}(s),u_{4}(s))\,ds+g(u_{1}(1)), \end{cases} $$

where $u_{2}$, $u_{3}$, $u_{4}$ are the first-, second-, and third-order derivatives of $u_{1}$, respectively. This is a fixed point problem in $E^{4}$ for a completely continuous operator

$$ T=(T_{1},T_{2},T_{3},T_{4}),\quad T: E^{4}\rightarrow E^{4}. $$

(2.2)

$T_{i}v\in E$ ($i=1,2,3,4$) with $v(t)=(v_{1}(t),v_{2}(t),v_{3}(t),v_{4}(t))$ are defined by the following integrals:

$$\begin{aligned}& (T_{1}v) (t)= \int _{0}^{1}G_{1}(t,s)f\bigl(s,v(s)\bigr) \,ds+g\bigl(v_{1}(1)\bigr) \frac{t^{2}}{6}(t-3), \end{aligned}$$

(2.3)

$$\begin{aligned}& (T_{2}v) (t)= \int _{0}^{1}G_{2}(t,s)f\bigl(s,v(s)\bigr) \,ds+g\bigl(v_{1}(1)\bigr)\frac{t}{2}(t-2), \\& (T_{3}v) (t)= \int _{0}^{1}G_{3}(t,s)f\bigl(s,v(s)\bigr) \,ds+g\bigl(v_{1}(1)\bigr) (t-1), \\& (T_{4}v) (t)= \int _{0}^{1}G_{4}(t,s)f\bigl(s,v(s)\bigr) \,ds+g\bigl(v_{1}(1)\bigr), \end{aligned}$$

(2.4)

respectively.

Thus, it is straightforward to conclude that, for $t,s\in [0,1]$,

$$ 0 \leq G_{1}(t,s) \leq \frac{1}{2}t^{2},\quad 0 \leq G_{2}(t,s) \leq t, $$

(2.5)

and

$$ 0 \leq G_{3}(t,s) \leq 1,\quad -1 \leq G_{4}(t,s) \leq 0. $$

Moreover, by a simple calculation, we obtain

$$\begin{aligned}& \begin{aligned} \int _{0}^{1}G_{1}(t,s)s^{2}\,ds &= \int _{0}^{t} \frac{1}{6}s^{2}(3t-s)s^{2} \,ds+ \int _{t}^{1}\frac{1}{6}t^{2}(3s-t)s^{2} \,ds \\ &= \frac{1}{360}\bigl(t^{4}-20t+45\bigr)t^{2} \leq \frac{t^{2}}{8}, \end{aligned} \end{aligned}$$

(2.6)

$$\begin{aligned}& \begin{aligned} \int _{0}^{1}G_{1}(t,s)s\,ds &= \int _{0}^{t} \frac{1}{6}s^{2}(3t-s)s \,ds+ \int _{t}^{1}\frac{1}{6}t^{2}(3s-t)s \,ds \\ &= \frac{1}{120}\bigl(t^{3}-10t+20\bigr)t^{2} \leq \frac{t^{2}}{6}, \end{aligned} \end{aligned}$$

(2.7)

$$\begin{aligned}& \begin{aligned} \int _{0}^{1}G_{1}(t,s)\,ds &= \int _{0}^{t} \frac{1}{6}s^{2}(3t-s) \,ds+ \int _{t}^{1}\frac{1}{6}t^{2}(3s-t) \,ds \\ &= \frac{1}{24}\bigl(t^{2}-4t+6\bigr)t^{2} \leq \frac{t^{2}}{4}. \end{aligned} \end{aligned}$$

(2.8)

Similarly, we obtain

$$\begin{aligned}& \int _{0}^{1}G_{2}(t,s)s^{2}\,ds \leq \frac{t}{4},\qquad \int _{0}^{1}G_{2}(t,s)s\,ds \leq \frac{t}{3},\qquad \int _{0}^{1}G_{2}(t,s)\,ds \leq \frac{t}{2};\\& \int _{0}^{1}G_{3}(t,s)s^{2}\,ds \leq \frac{1}{4},\qquad \int _{0}^{1}G_{3}(t,s)s\,ds \leq \frac{1}{3},\qquad \int _{0}^{1}G_{3}(t,s)\,ds \leq \frac{1}{2}; \end{aligned}$$

and

$$ \int _{0}^{1} \bigl\vert G_{4}(t,s) \bigr\vert s^{2}\,ds \leq \frac{1}{3},\qquad \int _{0}^{1} \bigl\vert G_{4}(t,s) \bigr\vert s\,ds \leq \frac{1}{2},\qquad \int _{0}^{1} \bigl\vert G_{4}(t,s) \bigr\vert \,ds \leq 1. $$

To compute the fixed point T defined in equation (2.2), we review several conclusions and concepts from vector-valued metric spaces and PFPT. Let $d:E\times E\rightarrow R^{n}$ be a function, where E is a nonempty set. Then, if the following features are satisfied for all $u,v,w\in E$:

• (1) $d(u,v)=0$ if and only if $u=v$, and $d(u,v) \geq 0$ otherwise;

• (2) $d(u,v)=d(v,u)$;

• (3) $d(u,v)\leq d(u,w)+d(w,v)$,

d can be regarded as a vector metric on E. The meaning of “≤” is the natural componentwise order relation of $\mathbb{R}^{n}$. Let $\alpha =(\alpha _{1},\alpha _{2},\ldots ,\alpha _{n})^{T}$, $\beta =(\beta _{1},\beta _{2},\ldots ,\beta _{n})^{T}$, and $\alpha ,\beta \in \mathbb{R}^{n}$, “$\alpha \leq \beta$” means $\alpha _{i} \leq \beta _{i}$ for $i=1,2,\ldots ,n$.

Under these circumstances, $(E,d)$ is a generalized metric space, which has similar features, e.g., the completeness of the space, and the Cauchyness and convergence of a sequence, with usual metric spaces.

Theorem 2.2

((PFPT) [21, 22])

Given a vector metric d and a complete generalized metric space $(E,d)$, let $N:E\rightarrow E$, then

$$ d\bigl(N(u),N(v)\bigr)\leq Md(u,v) $$

for every $u,v\in E$, and square matrix M belongs to the set of all n-order square matrices with positive elements, $\mathcal{M}_{n\times n}(\mathbb{R}_{+})$. If $\rho (M)<1$, then there is a unique fixed point $u^{*}$ for N and

$$ d\bigl(N^{k}(v),u^{*}\bigr)\leq M^{k}(I-M)^{-1}d \bigl(N(v),v\bigr) $$

for every $v\in E$ and $k\geq 1$.

Theorem 2.3

([30])

Let δ be another vector metric on E, $(E,d)$ be a complete generalized metric space. If $N:E\rightarrow E$ is continuous with respect to d and

• (i) There exists $U\in \mathcal{M}_{n\times n}(\mathbb{R}_{+})$ such that $d(Nu,Nv)\leq U\cdot \delta (u,v)$ for all $u,v\in E$;

• (ii) There exists $M\in \mathcal{M}_{n\times n}(\mathbb{R}_{+})$ with $\rho (M)<1$ such that $\delta (N(u),N(v))\leq M\delta (u,v)$ for all $u,v\in E$.

Then N has a unique fixed point in E.

Remark 2.4

We refer readers to [22] for the properties of a nonnegative matrix with a spectral radius smaller than one.

3 Results

Theorem 3.1

Problem (1.3), (1.4) has a unique solution if $(H_{1})$ and $(H_{2})$ hold.

Proof

By (2.3) and (2.5), for every $v=(v_{1},v_{2},v_{3},v_{4})\in E^{4}$, we have

$$ \bigl\vert T_{1}v(t) \bigr\vert \leq \frac{1}{2} \biggl( \biggl\vert \int _{0}^{1}f\bigl(s,v(s)\bigr)\,ds \biggr\vert + \bigl\vert g\bigl(v_{1}(1)\bigr) \bigr\vert \biggr)t^{2}, $$

which shows that $T_{1}$ maps $E^{4}$ into $E_{1}$, a vector subspace of E. It is a Banach space given by

$$ E_{1}=\bigl\{ u(t)\in E,\exists \lambda >0, \text{s.t. } \bigl\vert u(t) \bigr\vert \leq \lambda t^{2}, t\in [0,1]\bigr\} $$

and the norm of $E_{1}$ $\|u\|_{e_{1}}=\inf \{\lambda :|u(t)|\leq \lambda e_{1}(t),t\in [0,1] \}$ with $e_{1}(t)=t^{2}$.

Similarly, by (2.4) and (2.5), for every $v\in E^{4}$, we obtain

$$ \bigl\vert T_{2}v(t) \bigr\vert \leq \biggl( \biggl\vert \int _{0}^{1}f\bigl(s,v(t)\bigr)\,ds \biggr\vert + \bigl\vert g\bigl(v_{1}(1)\bigr) \bigr\vert \biggr)t, $$

and $T_{2}$ maps $E^{4}$ into the Banach space $(E_{2},\|\cdot \|_{e_{2}})$, where

$$ E_{2}=\bigl\{ u(t)\in E,\exists \lambda >0, \text{s.t. }\bigl\vert u(t) \bigr\vert \leq \lambda t,t \in [0,1]\bigr\} $$

and the norm $\|u\|_{e_{2}}=\inf \{\lambda :|u(t)|\leq \lambda e_{2}(t),t\in [0,1] \}$ with $e_{2}(t)=t$.

Hence, we conclude that $T(E^{4})\subset X:=E_{1}\times E_{2}\times E^{2}$. Therefore, only the existence of a unique fixed point v for T in X has to be demonstrated to get the uniqueness solutions for problem (1.3), (1.4).

For $w=(w_{1},w_{2},w_{3}, w_{4})$, $v=(v_{1},v_{2},v_{3},v_{4}) \in X$, we define

$$ d(w,v)=\bigl( \Vert w_{1}-v_{1} \Vert _{e_{1}}, \Vert w_{2}-v_{2} \Vert _{e_{2}}, \Vert w_{3}-v_{3} \Vert _{\infty}, \Vert w_{4}-v_{4} \Vert _{\infty}\bigr)^{T}. $$

Obviously, X is a complete generalized Banach metric space equipped with a vector-valued metric d.

To apply Theorem 2.2, one should prove that, for all $v, w \in X$, T satisfies

$$ d(Tw, Tv)\leq Md(w,v) $$

and some nonnegative matrix M with $\rho (M)<1$. To this end, let $w, v \in X$ be any elements of X. By (2.6), (2.7), (2.8), and $(H_{1})$, we get

$$\begin{aligned}& \bigl\vert (T_{1}w) (t)-(T_{1}v) (t) \bigr\vert \\& \quad \leq \int _{0}^{1}G_{1}(t,s) \bigl\vert f \bigl(s,w(s)\bigr)-f\bigl(s,v(s)\bigr) \bigr\vert \,ds + \frac{t^{2}}{6}(3-t) \bigl\vert g\bigl(w_{1}(1)\bigr)-g\bigl(v_{1}(1)\bigr) \bigr\vert \\& \quad \leq \int _{0}^{1}|G_{1}(t,s)\bigl[ p_{1} \bigl\vert w_{1}(s)-v_{1}(s) \bigr\vert +p_{2} \bigl\vert w_{2}(s)-v_{2}(s) \bigr\vert +p_{3} \bigl\vert w_{3}(s)-v_{3}(s) \bigr\vert \\& \qquad {}+p_{4} \bigl\vert w_{4}(s)-v_{4}(s) \bigr\vert \bigr]\,ds+ \frac{1}{3} \bigl\vert g\bigl(w_{1}(1) \bigr)-g\bigl(v_{1}(1)\bigr) \bigr\vert \\& \quad \leq p_{1} \Vert w_{1}-v_{1} \Vert _{e_{1}} \int _{0}^{1}G_{1}(t,s)s^{2} \,ds+p_{2} \Vert w_{2}-v_{2} \Vert _{e_{2}} \int _{0}^{1}G_{1}(t,s)s\,ds \\& \qquad {}+p_{3} \Vert w_{3}-v_{3} \Vert _{\infty} \int _{0}^{1}G_{1}(t,s)\,ds+ p_{4} \Vert w_{4}-v_{4} \Vert _{\infty} \int _{0}^{1}G_{1}(t,s)\,ds+ \frac{t^{2}}{3}L_{g} \Vert w_{1}-v_{1} \Vert _{e_{1}} \\& \quad \leq \biggl[\frac{t^{2}}{8}p_{1}+\frac{t^{2}}{3}L_{g} \biggr] \Vert w_{1}-v_{1} \Vert _{e_{1}}+ \frac{t^{2}}{6}p_{2} \Vert w_{2}-v_{2} \Vert _{e_{2}} \\& \qquad {} +\frac{t^{2}}{4}p_{3} \Vert w_{3}-v_{3} \Vert _{\infty}+\frac{t^{2}}{4}p_{4} \Vert w_{4}-v_{4} \Vert _{\infty}. \end{aligned}$$

This result yields that

$$|(T_{1}w)(t)-(T_{1}v)(t)|\le t^2(\frac{p_1}{8}+\frac{L_g}{3}\frac{p_2}{6}\frac{p_3}{4}\frac{p_4}{4})\left(\begin{array}{c}{\parallel w_1-v_1\parallel }_{e_1}\\ {\parallel w_2-v_2\parallel }_{e_2}\\ {\parallel w_3-v_3\parallel }_{\mathrm{\infty }}\\ {\parallel w_4-v_4\parallel }_{\mathrm{\infty }}\end{array}\right).$$

Now, the definition of norm $\|\cdot \|_{e_{1}}$ implies that

$${\parallel T_{1}w-T_{1}v\parallel }_{e_1}\le (\frac{p_1}{8}+\frac{L_g}{3}\frac{p_2}{6}\frac{p_3}{4}\frac{p_4}{4})\left(\begin{array}{c}{\parallel w_1-v_1\parallel }_{e_1}\\ {\parallel w_2-v_2\parallel }_{e_2}\\ {\parallel w_3-v_3\parallel }_{\mathrm{\infty }}\\ {\parallel w_4-v_4\parallel }_{\mathrm{\infty }}\end{array}\right).$$

Similarly, we obtain

$$\begin{array}{c}{\parallel T_{2}w-T_{2}v\parallel }_{e_2}\le (\frac{p_1}{4}+\frac{L_g}{2}\frac{p_2}{3}\frac{p_3}{2}\frac{p_4}{2})\left(\begin{array}{c}{\parallel w_1-v_1\parallel }_{e_1}\\ {\parallel w_2-v_2\parallel }_{e_2}\\ {\parallel w_3-v_3\parallel }_{\mathrm{\infty }}\\ {\parallel w_4-v_4\parallel }_{\mathrm{\infty }}\end{array}\right),\hfill \\ {\parallel T_{3}w-T_{3}v\parallel }_{\mathrm{\infty }}\le (\frac{p_1}{4}+L_g\frac{p_2}{3}\frac{p_3}{2}\frac{p_4}{2})\left(\begin{array}{c}{\parallel w_1-v_1\parallel }_{e_1}\\ {\parallel w_2-v_2\parallel }_{e_2}\\ {\parallel w_3-v_3\parallel }_{\mathrm{\infty }}\\ {\parallel w_4-v_4\parallel }_{\mathrm{\infty }}\end{array}\right),\hfill \end{array}$$

and

$${\parallel T_{4}w-T_{4}v\parallel }_{\mathrm{\infty }}\le (\frac{p_1}{3}+L_g\frac{p_2}{2}{p}_3{p}_4)\left(\begin{array}{c}{\parallel w_1-v_1\parallel }_{e_1}\\ {\parallel w_2-v_2\parallel }_{e_2}\\ {\parallel w_3-v_3\parallel }_{\mathrm{\infty }}\\ {\parallel w_4-v_4\parallel }_{\mathrm{\infty }}\end{array}\right).$$

We can put the above four inequalities under the following vector inequality by using vector-valued metric on X:

$$d(Tw,Tv)\le \left(\begin{array}{cccc}\frac{p_1}{8}+\frac{L_g}{3}& \frac{p_2}{6}& \frac{p_3}{4}& \frac{p_4}{4}\\ \frac{p_1}{4}+\frac{L_g}{2}& \frac{p_2}{3}& \frac{p_3}{2}& \frac{p_4}{2}\\ \frac{p_1}{4}+L_g& \frac{p_2}{3}& \frac{p_3}{2}& \frac{p_4}{2}\\ \frac{p_1}{3}+L_g& \frac{p_2}{2}& p_3& p_4\end{array}\right)d(w,v).$$

By $(H_{2})$, there is a unique solution to problem (1.3), (1.4), which follows from Theorem 2.2. □

From the previous argument, we know that the basic complete generalized Banach metric space used in obtaining Theorem 3.1 is X, not $E^{4}$. If we consider problem (1.3), (1.4) in $E^{4}$ by using an appropriate vector-valued metric on $E^{4}$ and PFPT, the result of Theorem 3.1 remains true, except that $(H_{2})$ is replaced with $(H_{3})$.

Theorem 3.2

Problem (1.3), (1.4) has a unique solution if $(H_{1})$ and $(H_{3})$ hold.

Proof

PFPT is applied to the complete generalized Banach metric space $E^{4}$ having the following vector-valued metric for $w,v \in E^{4}$:

$$ d_{1}(w,v)=\bigl( \Vert w_{1}-v_{1} \Vert _{\infty}, \Vert w_{2}-v_{2} \Vert _{\infty}, \Vert w_{3}-v_{3} \Vert _{\infty}, \Vert w_{4}-v_{4} \Vert _{\infty}\bigr)^{T}. $$

Note that

$$ \max_{t\in [0,1]} \int _{0}^{1}G_{1}(t,s)\,ds = \frac{1}{8},\qquad \max_{t \in [0,1]} \int _{0}^{1}G_{2}(t,s)\,ds = \frac{1}{6}, $$

and

$$ \max_{t\in [0,1]} \int _{0}^{1}G_{3}(t,s)\,ds = \frac{1}{2}, \qquad \max_{t \in [0,1]} \int _{0}^{1} \bigl\vert G_{4}(t,s) \bigr\vert \,ds = 1. $$

Similar to proving Theorem 3.1, we obtain

$$d_1(Tw,Tv)\le \left(\begin{array}{cccc}\frac{p_1}{8}+L_g& \frac{p_2}{8}& \frac{p_3}{8}& \frac{p_4}{8}\\ \frac{p_1}{6}+L_g& \frac{p_2}{6}& \frac{p_3}{6}& \frac{p_4}{6}\\ \frac{p_1}{2}+L_g& \frac{p_2}{2}& \frac{p_3}{2}& \frac{p_4}{2}\\ p_1+L_g& p_2& p_3& p_4\end{array}\right)d_1(w,v)$$

for $\forall w,v\in E^{4}$. Thus, Perov’s fixed point theorem can be applied. □

Example 3.3

We consider problem (1.3), (1.4) with

$$ f(t,u_{1},u_{2},u_{3},u_{4})=a\sin u_{1}+a\cos u_{2}+au_{3}+a\arctan u_{4}+t^{2}+1 $$

and

$$ g(u)=b\ln \bigl(1+u^{2}\bigr). $$

According to the mean value theorem, $|b|$ is a constant and g is a Lipschitz function. Taking $a=\frac{1}{3}$ and $b=1$, we deduce that $\rho (M)=0.9629$ and $\rho (M_{1})=1.2538$. The hypothesis of Theorem 3.1 can therefore be satisfied. But, if we take $a=\frac{1}{2}$ and $b=\frac{1}{4}$, then we deduce that $\rho (M)=1.0291$ and $\rho (M_{1})=0.9813$. The hypotheses of Theorem 3.2 are satisfied in this case.

If the function g in the nonlinear boundary condition is equal to zero, problem (1.3), (1.4) reduces to problem (1.3), (1.5).

Theorem 3.4

If $(H_{1})$ holds, then problem (1.3), (1.5) has a unique solution providing $\rho (M)<1$ with $L_{g}=0$, or $\rho (M_{1})<1$ with $L_{g}=0$.

The above theorems present a method for the existence and uniqueness of solution for fourth-order differential equations with all derivatives. For the nonlinear boundary, there are few studies at present, and the theories need to be further studied. For the linear boundary, by contrast to [1], the condition of Theorem 3.4 is not optimal. The results in [1] offer an advancement over traditional approaches based on the Rus fixed point theorem and two metrics. It is notable that the method used in [1] seems to be invalid for (1.3), (1.4) because the nonlinear boundary conditions are dominated by the function of $u(1)$. Now we present the same existence and uniqueness results as [1] by utilizing Perov’s fixed point theorem with two metrics. Let p, q be constants. They satisfy $\frac{1}{p}+\frac{1}{q}=1$ with $p>1$ and $q<1$. Define positive constants $\beta _{i}$, $c_{i}$, and $\gamma _{i}$ for $i=1,2,3,4$ by

$$ \beta _{i}\geq \max_{t\in [0,1]} \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert \,ds, \qquad c_{i}= \biggl( \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert ^{q}\,ds \biggr)^{\frac{1}{q}}, $$

and

$$ \gamma _{i}= \biggl( \int _{0}^{1} \biggl( \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert ^{q}\,ds \biggr)^{\frac{p}{q}}\,dt \biggr)^{\frac{1}{q}}. $$

Theorem 3.5

If $(H_{1})$ holds, then problem (1.3), (1.5) has a unique solution providing that $\sum_{i=1}^{4}\gamma _{i}p_{i}<1$.

Proof

The main idea of this result is adopted from [1]. For $w,v\in E^{4}$, we let

$$ \delta (w,v)=\bigl( \Vert w_{1}-v_{1} \Vert _{p}, \Vert w_{2}-v_{2} \Vert _{p}, \Vert w_{3}-v_{3} \Vert _{p}, \Vert w_{4}-v_{4} \Vert _{p}\bigr)^{T}, $$

where $\|w_{1}\|_{p}= ( \int _{0}^{1} |w_{1}(t)|^{p}\,dt )^{ \frac{1}{p}}$. Clearly, $\delta (\cdot ,\cdot )$ is a vector-valued metric in E; however, it is not complete. From the definition of two metrics $d_{1}$ and δ on $E^{4}$, we have

$$\delta (w,v)\le \left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)d_1(w,v).$$

(3.1)

Note that problem (1.3), (1.5) having a unique solution means that the operator $S:E^{4}\rightarrow E^{4}$ having a unique fixed point is proved, where S is given by

$$ S=(S_{1},S_{2},S_{3},S_{4}), $$

and for $v(t)=(v_{1}(t),v_{2}(t),v_{3}(t),v_{4}(t))$, we define $S_{i}v\in E$ ($i=1,2,3,4$) by

$$ (S_{i}v) (t)= \int _{0}^{1}G_{i}(t,s)f\bigl(s,v(s) \bigr)\,ds, $$

respectively. For $w,v\in E^{4}$, by $(H_{1})$ and Hölder’s inequality, we obtain

$$\begin{aligned}& \bigl\vert (S_{i}w) (t)-(S_{i}v) (t) \bigr\vert \\& \quad \leq \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert \cdot \bigl\vert f\bigl(s,w(s)\bigr)-f\bigl(s,v(s)\bigr) \bigr\vert \,ds \\& \quad \leq \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert \sum_{j=1}^{4}p_{j} \bigl\vert w_{j}(s)-v_{j}(s) \bigr\vert \,ds \\& \quad \leq \biggl( \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert ^{q}\,ds \biggr)^{ \frac{1}{q}}\sum _{j=1}^{4}p_{j} \biggl( \int _{0}^{1} \bigl\vert w_{j}(s)-v_{j}(s) \bigr\vert ^{p}\,ds \biggr)^{\frac{1}{p}} \\& \quad \leq c_{i}\sum_{j=1}^{4}p_{j} \Vert w_{j}-v_{j} \Vert _{p},\quad i=1,2,3,4. \end{aligned}$$

(3.2)

With the help of norm $\|\cdot \|_{\infty}$, we obtain

$$ \Vert S_{i}w-S_{i}v \Vert _{\infty}\leq c_{i}\sum_{j=1}^{4}p_{j} \Vert w_{j}-v_{j} \Vert _{p}. $$

Combining the above inequalities, we have

$$d_1(Sw,Sv)\le \left(\begin{array}{cccc}{c}_1{p}_1& c_1{p}_2& c_1{p}_3& c_1{p}_4\\ c_2{p}_1& c_2{p}_2& c_2{p}_3& c_2{p}_4\\ c_3{p}_1& c_3{p}_2& c_3{p}_3& c_3{p}_4\\ c_4{p}_1& c_4{p}_2& c_4{p}_3& c_4{p}_4\end{array}\right)\delta (w,v).$$

This together with (3.1) indicates that condition (1) of Theorem 2.3 can be met and S is continuous on $E^{4}$ with respect to metric $d_{1}$.

By (3.2), for $w,v\in E^{4}$ and $i=1,2,3,4$, we have

$$\begin{aligned} \Vert S_{i}w-S_{i}v \Vert _{p} =& \biggl( \int _{0}^{1} \bigl\vert (S_{i}w) (t)-(S_{i}v) (t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \\ \leq& \biggl( \int _{0}^{1} \biggl( \int _{0}^{1} \bigl\vert G_{i}(t,s) \bigr\vert ^{q}\,ds \biggr)^{\frac{p}{q}}\,dt \biggr)^{\frac{1}{q}} \sum_{j=1}^{4}p_{j} \Vert w_{j}-v_{j} \Vert _{p} \\ =& \gamma _{i} \sum_{j=1}^{4}p_{j} \Vert w_{j}-v_{j} \Vert _{p}. \end{aligned}$$

Putting the above inequality over i under the following vector inequality:

$$\delta (Sw,Sv)\le \left(\begin{array}{cccc}{\gamma }_1{p}_1& {\gamma }_1{p}_2& {\gamma }_1{p}_3& {\gamma }_1{p}_4\\ {\gamma }_2{p}_1& {\gamma }_2{p}_2& {\gamma }_2{p}_3& {\gamma }_2{p}_4\\ {\gamma }_3{p}_1& {\gamma }_3{p}_2& {\gamma }_3{p}_3& {\gamma }_3{p}_4\\ {\gamma }_4{p}_1& {\gamma }_4{p}_2& {\gamma }_4{p}_3& {\gamma }_4{p}_4\end{array}\right)\delta (w,v):=M_2\delta (w,v).$$

Take $P=(p_{1}, p_{2}, p_{3}, p_{4})$, $\Gamma =(\gamma _{1},\gamma _{2},\gamma _{3},\gamma _{4})\in \mathbb{R}^{4}$. Then $M_{2}=P^{T}\Gamma$. So, the rank of matrix $M_{2}$ is equal to 1 and $\rho (M_{2})=P\Gamma ^{T}=\sum_{i=1}^{4}\gamma _{i}p_{i}$. Thus, S is contractive on $E^{4}$ with respect to the metric δ. Consequently, problem (1.3), (1.5) has a unique solution following Theorem 2.3. □

We all know that the Banach fixed point theorem and its generalization play a crucial role in the theory with unique fixed point properties. Surprisingly, the Banach fixed point theorem is equivalent to some of its generalization such as Perov’s fixed point theorem [31]. More surprisingly, all the uniqueness results can be obtained by the Banach fixed point theorem from Theorem 17.5 in [32]. However, in some cases, the introduction of the generalizations of the Banach fixed point will make the study more effective and convenient in a way which may overcome some difficulties encountered in the differential equation and differential system. Those difficulties contain the choice of a complete metric such that an operator T or $T^{m}$ for some m is a strict contraction. For example, we consider problem (1.3), (1.5), as shown in [31], take

$$ D(w,v)=\max \bigl\{ \Vert w_{1}-v_{1} \Vert _{\infty}, \Vert w_{2}-v_{2} \Vert _{\infty}, \Vert w_{3}-v_{3} \Vert _{\infty}, \Vert w_{4}-v_{4} \Vert _{\infty}\bigr\} . $$

Then $(E^{4}, D)$ is a metric space. Note that $\sum_{i=1}^{4}\gamma _{i}p_{i}<1$. The operator S may not be contractive with respect to metric D since the operator S such that

$$ D(Sw,Sv)\leq \Biggl(\max \{c_{i}\}\sum_{i=1}^{4}p_{i} \Biggr) D(w,v),\quad w,v\in E^{4}. $$

For problem (1.3), (1.4), if we take

$$ D_{1}(w,v)=\max \bigl\{ \Vert w_{1}-v_{1} \Vert _{e_{1}}, \Vert w_{2}-v_{2} \Vert _{e_{2}}, \Vert w_{3}-v_{3} \Vert _{\infty}, \Vert w_{4}-v_{4} \Vert _{\infty}\bigr\} , $$

then we obtain that

$$ D_{1}(Tw,Tv)\leq \biggl(\frac{p_{1}}{3}+L_{g}+ \frac{p_{2}}{2}+p_{3}+p_{4}\biggr) D_{1}(w,v), $$

which shows that $T:X\rightarrow X$ may not be contractive with respective to metric $D_{1}$ even if $\rho (M)<1$ holds. But, as shown in [31], there must be an integer $n_{0} > 0$ so that for $n\geq n_{0}$, operator $S^{n}$ (or $T^{n}$) is contractive and the constant $n_{0}$ depends heavily on the spectral radius of matrix $M_{2}$ (or M).