The method of lower and upper solutions for the cantilever beam equations with fully nonlinear terms

Abstract

In this paper we discuss the existence of solutions of the fully fourth-order boundary value problem

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

which models the deformations of an elastic cantilever beam in equilibrium state, where $f:[0, 1]\times {\mathbb{R}}^{4}\to \mathbb{R}$ is continuous. Using the method of lower and upper solutions and the monotone iterative technique, we obtain some existence results under monotonicity assumptions on nonlinearity.

1 Introduction

In this paper, we are concerned with the existence of the fully fourth-order boundary value problem

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

(1.1)

where $f:[0, 1]\times {\mathbb{R}}^{4}\to \mathbb{R}$ is continuous. This equation models the deformations of an elastic beam in equilibrium state, whose one end-point is fixed and the other is free, and in mechanics it is called cantilever beam equation. In the equation, the physical meaning of the derivatives of the deformation function $u(t)$ is as follows: $u^{(4)}$ is the load density stiffness, $u'''$ is the shear force stiffness, $u''$ is the bending moment stiffness, and $u'$ is the slope [1,2,3,4].

For the special case of BVP (1.1) that f does not contain any derivative terms, namely the simply fourth-order boundary value problem

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

(1.2)

and f only contains first-order derivative term $u'$, namely the fourth-order boundary value problem

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

(1.3)

the existence of positive solutions has been discussed by some authors, see [5,6,7,8,9]. The methods applied in these works are not applicable to BVP (1.1) since they cannot deal with the derivative terms $u''$ and $u'''$.

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

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

(1.4)

the existence of solution has also been discussed by some authors, see [10,11,12,13]. The boundary condition in (1.4) means that the left end of the beam is fixed and the right end of the beam is attached to an elastic bearing device, see [10].

The purpose of this paper is to obtain existence results of solutions to the fully fourth-order nonlinear boundary value problem (1.1). For fully fourth-order nonlinear BVPs with the boundary condition in BVP (1.1) or other boundary conditions, the existence of solution has discussed by several authors, see [14,15,16,17,18,19,20]. In [14], Kaufmann and Kosmatov considered a symmetric fully fourth-order nonlinear boundary value problem. They used a triple fixed point theorem of cone mapping to obtain existence results of triple positive symmetric solutions when f satisfies some range conditions dependent upon three positive parameters a, b and d. Since they did not give the method to determine these parameters, the range conditions are difficult to verify. The authors of [15] used the method of lower and upper solutions to discuss the existence of solution of the fully fourth-order nonlinear boundary value problem

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

(1.5)

where the discussed problem has a pair of ordered lower and upper solutions. But they did not discuss how they found a pair of ordered lower and upper solutions. Under the case that $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is sublinear growth on $x_{0}$, $x_{1}$, $x_{2}$, $x _{3}$, the existence of the following fully fourth-order boundary value problem:

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

(1.6)

is discussed in [16]. In this case, using the method in [16], we can obtain existence results for BVP (1.1). Usually the superlinear problems are more difficult to treat than the sublinear problems. In [17], the present author discussed the case that $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ may be superlinear growth on $x_{0}$, $x_{1}$, $x_{2}$, $x _{3}$ when nonlinearity f is nonnegative by using the fixed point index theory in cones. In recent paper [18], Dang and Ngo dealt with the solvability of BVP (1.1) by using the contraction mapping principle. They showed that if there exists a region

$$ \mathcal{D}_{M}= \biggl\{ (t, x_{0}, x_{1}, x_{2}, x_{3}) \Bigm| t\in [0,1], \vert x_{0} \vert \le \frac{M}{8}, \vert x_{1} \vert \le \frac{M}{6}, \vert x_{2} \vert \le \frac{M}{2}, \vert x_{3} \vert \le M \biggr\} $$

(1.7)

determined by a positive number M such that nonlinearity f satisfies

$$\begin{aligned}& \bigl\vert f(t, x_{0}, x_{1}, x_{2}, x_{3}) \bigr\vert \le M, \end{aligned}$$

(1.8)

$$\begin{aligned}& \bigl\vert f(t, x_{0}, x_{1}, x_{2}, x_{3})-f(t, y_{0}, y_{1}, y_{2}, y_{3}) \bigr\vert \le \sum_{i=0}^{3}c_{i} \vert x_{i}-y_{i} \vert \end{aligned}$$

(1.9)

for any $(t, x_{0}, x_{1}, x_{2}, x_{3})$, $(t, y_{0}, y_{1}, y_{2}, y _{3})\in \mathcal{D}_{M}$, where $c_{0}$, $c_{1}$, $c_{2}$, $c_{3}$ are positive constants and satisfy

$$ q:=\frac{c_{0}}{8}+\frac{c_{1}}{6}+\frac{c_{0}}{2}+c_{3}< 1, $$

(1.10)

then BVP (1.1) has a unique solution u satisfying

$$ \bigl(t, u(t), u'(t), u''(t), u'''(t)\bigr)\in \mathcal{D}_{M}, \quad t \in [0, 1]. $$

(1.11)

See [18, Theorem 2.2]. A similar result is built for BVP (1.6) in [19] and for a fourth-order BVP of Kirchhoff type equation in [20]. Dang and Ngo’s result can be applied to the superlinear equations, and it ensures the uniqueness of solution on $\mathcal{D}_{M}$. However, the key to the application of this result is how to determine the constant M. For the general nonlinearity f, M is not easy to determine and the Lipschitz coefficients condition (1.10) is not easy to satisfy. In this paper we shall discuss the general case that f may be superlinear growth and have negative value.

We will use the method of lower and upper solutions to discuss BVP (1.1). For BVP (1.1), since the boundary conditions are different from BVP (1.5), the definitions of lower and upper solutions are different from those in [16] and the argument methods in [16] are not applicable to BVP (1.1). In Sect. 2, under $f(t, x_{0}, x_{1}, x_{2}, x _{3})$ increasing on $x_{0}$, $x_{1}$, $x_{2}$ and decreasing on $x_{3}$ in the domain surrounded by lower and upper solutions, we use a monotone iterative technique to obtain the existence of a solution between lower and upper solutions. In Sect. 3, under $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ without monotonicity on $x_{3}$, we use a truncating technique to prove the existence of a solution between lower and upper solutions. In Sect. 4, we use the lower and upper theorem built in Sect. 3 to obtain a new existence result of positive solution.

2 Monotone iterative method

The monotone iterative method is an important method for solving nonlinear BVPs. For the special BVP (1.3), a monotone iterative method has been built, see [8]. In this section, we will develop the monotone iterative method of lower and upper solutions for BVP (1.1).

Let $I=[0, 1]$ and $C(I)$ denote the Banach space of all continuous functions $u(t)$ on I with norm $\|u\|_{C}=\max_{t\in I}|u(t)|$. Generally, for $n\in \mathbb{N}$, we use $C^{n}(I)$ to denote the Banach space of all nth-order continuous differentiable functions on I with the norm $\|u\|_{C^{n}}=\max \{ \|u\|_{C}, \|u'\|_{C}, \ldots , \|u^{(n)}\|_{C}\}$. Let $C^{+}(I)$ denote the cone of all nonnegative functions in $C(I)$.

Let $f: I\times \mathbb{R}^{4}\to \mathbb{R}$ be continuous and consider BVP (1.1). If a function $v\in C^{4}(I)$ satisfies

$$ \textstyle\begin{cases} v^{(4)}(t)\le f(t, v(t), v'(t), v''(t), v'''(t)), \quad t\in I, \\ v(0)\le 0,\qquad v'(0)\le 0,\qquad v''(1)\le 0,\qquad v'''(1)\ge 0, \end{cases} $$

(2.1)

we call it a lower solution of BVP (1.1), and if a function $w\in C ^{4}(I)$ satisfies

$$ \textstyle\begin{cases} w^{(4)}(t)\ge f(t, w(t), w'(t), w''(t), w'''(t)), \quad t\in I, \\ w(0)\ge 0,\qquad w'(0)\ge 0,\qquad w''(1)\ge 0,\qquad w'''(1)\le 0, \end{cases} $$

(2.2)

we call it an upper solution of BVP (1.1).

Lemma 2.1

Let $v_{0}\in C^{4}(I)$ be a lower solution of BVP (1.1) and $w_{0}$ be an upper solution, and ${v_{0}}'''\ge {w_{0}}'''$. Then

$$ v_{0}\le w_{0},\qquad {v_{0}}'\le {w_{0}}',\qquad {v_{0}}'' \le {w_{0}}''. $$

(2.3)

Proof

Let $u=w_{0}-v_{0}$, then $u'''(t)\le 0$ for every $t\in I$. By the definitions of lower and upper solutions, we have

$$\begin{aligned}& u''(t) =u''(1)- \int _{t}^{1} u'''(s)\,ds \ge 0,\quad t\in I, \\& u'(t) =u'(0)+ \int _{0}^{t} u''(s)\,ds\ge 0, \quad t\in I, \\& u(t) =u(0)+ \int _{0}^{t} u'(s)\,ds\ge 0, \quad t\in I. \end{aligned}$$

Hence, (2.3) holds. □

Given $h\in C(I)$, consider the linear boundary value problem (LBVP)

$$ \textstyle\begin{cases} u^{(4)}(t)=h(t) , \quad t\in I, \\ u(0)=u'(0)=u''(1)=u'''(1)=0 . \end{cases} $$

(2.4)

Lemma 2.2

For every $h\in C(I)$, LBVP (2.4) has a unique solution $u:=S h\in C ^{4}(I)$. Moreover, the solution operator $S: C(I)\to C^{3}(I)$ is a completely continuous linear operator.

Proof

For given any $h\in C(I)$, it is easy to verify that

$$ u(t)= \int _{0}^{t}(t-\tau ) \int _{\tau }^{1}(s-\tau )h(s)\,d s \,d\tau :=Sh(t), \quad t\in I, $$

(2.5)

is a unique solution of LBVP (2.4). From expression (2.5), we easily see that $S: C(I)\to C^{3}(I)$ is a completely continuous linear operator. □

Lemma 2.3

If $u\in C^{4}(I)$ and satisfies

$$ \textstyle\begin{cases} u^{(4)}(t)\ge 0, \quad t\in I, \\ u(0)\ge 0,\qquad u'(0)\ge 0,\qquad u''(1)\ge 0,\qquad u'''(1)\le 0, \end{cases} $$

(2.6)

then $u\ge 0$, $u'\ge 0$, $u''\ge 0$, and $u'''\le 0$.

Proof

Similar to the proof of Lemma 2.1, we have

$$\begin{aligned}& u'''(t) =u'''(1)- \int _{t}^{1} u^{(4)}(s)\,ds\le 0,\quad t\in I, \\& u''(t) =u''(1)- \int _{t}^{1} u''(s)\,ds\ge 0,\quad t\in I, \\& u'(t) =u'(0)+ \int _{0}^{t} u''(s)\,ds\ge 0, \quad t\in I, \\& u(t) =u(0)+ \int _{0}^{t} u'(s)\,ds\ge 0, \quad t\in I. \end{aligned}$$

Hence, the conclusion of Lemma 2.3 holds. □

We introduce a semi-ordering ⪯ in $C^{3}(I)$ by

$$ v\preceq w\quad \Longleftrightarrow\quad v\le w,\qquad v'\le w',\qquad v''\le w'',\quad \text{and}\quad v'''\ge w'''. $$

(2.7)

Then $C^{3}(I)$ is an ordered Banach space by this semi-ordering. We also use $w\succeq v$ to denote $v\preceq w$. Letting $v, w\in C^{3}(I)$ and $v\preceq w$, we denote the order-interval in $C^{3}(I)$ by

$$ [v, w]_{C^{3}}=\bigl\{ u\in C^{3}(I): v\preceq u\preceq w \bigr\} . $$

(2.8)

Theorem 2.1

Assume that $f: I\times \mathbb{R}^{4}\to \mathbb{R}$ is continuous, BVP (1.1) has a lower solution $v_{0}$ and an upper solution $w_{0}$ with ${v_{0}}''' \ge {w_{0}}'''$, and f satisfies the following monotone conditions:

(F1):

for every $t\in I$ and $x_{3}\in [w_{0}'''(t), v_{0}'''(t)]$, $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is increasing on $x_{0}$, $x _{1}$, and $x_{2}$ in $[v_{0}(t), w_{0}(t)]\times [v_{0}'(t), w_{0}'(t)] \times [v_{0}''(t), w_{0}''(t)]$;

(F2):

for every $t\in I$ and $(x_{0},x_{1},x_{2})\in [v_{0}(t),w_{0}(t)] \times [v_{0}'(t),w_{0}'(t)]\times [v_{0}''(t),w_{0}''(t)]$, $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is decreasing on $x_{3}$ in $[w_{0}'''(t), v_{0}'''(t)]$.

Make iterative sequences $\{v_{n}\}$ and $\{w_{n}\}$ starting from $v_{0}$ and $w_{0}$ respectively by using the iterative equation

$$ \textstyle\begin{cases} {u_{n}}^{(4)}(t)=f(t, u_{n-1}(t), u_{n-1}'(t), u_{n-1}''(t), u _{n-1}'''(t)),\quad t\in I, \\ u_{n}(0)=u_{n}'(0)=u_{n}''(1)=u_{n}'''(1)=0, \end{cases}\displaystyle n=1, 2,\ldots . $$

(2.9)

Then $\{v_{n}\}$ and $\{w_{n}\}$ satisfy the monotone condition

$$ v_{0}\preceq v_{n}\preceq v_{n+1}\preceq w_{n+1}\preceq w_{n}\preceq w_{0},\quad n=1, 2, \ldots , $$

(2.10)

and converge in $C^{3}(I)$. Moreover, $\underline{u}=\lim_{n\to \infty }v_{n}$ and $\overline{u}=\lim_{n\to \infty }w_{n}$ are minimal and maximal solutions of BVP (1.1) in $[v_{0}, w_{0}]_{C^{3}}$.

Proof

By Lemma 2.1, $v_{0}\preceq w_{0}$. Define a mapping $F: C^{3}(I)\to C(I)$ by

$$ F(u) (t):=f\bigl(t, u(t), u'(t), u''(t), u'''(t)\bigr),\quad t\in I, u \in C^{3}(I). $$

(2.11)

Then $F: C^{3}(I)\to C(I)$ is continuous and by Assumptions (F1) and (F2) we can verify that

$$ v_{0}\preceq u_{1} \preceq u_{2}\preceq w_{0}\quad \Longrightarrow\quad F(u _{1})\le F(u_{2}). $$

(2.12)

By Lemma 2.2, $A=S\circ F: C^{3}(I)\to C^{3}(I)$ is completely continuous and the solution of BVP(1) is equivalent to the fixed point of A. By the definition of S, the iterative sequences $\{v_{n}\}$ and $\{w_{n}\}$ satisfy

$$ v_{n}=A v_{n-1},\qquad w_{n}=A w_{n-1}, \quad n=1, 2, \ldots . $$

(2.13)

We show that

$$ v_{0}\preceq v_{1},\qquad w_{1} \preceq w_{0}. $$

(2.14)

Let $u=v_{1}-v_{0}$. Then by the definition of the lower solution $v_{0}$, u satisfies (2.6). By Lemma 2.3, $u\succeq 0$, and hence $v_{0}\preceq v_{1}$. Similarly, $w_{1} \preceq w_{0}$ can be showed. By Lemma 2.3 and (2.12), we can prove that

$$ v_{0}\preceq u_{1} \preceq u_{2}\preceq w_{0}\quad \Longrightarrow\quad A u _{1}\preceq Au_{2}. $$

(2.15)

By (2.14) and (2.15), we see that (2.10) holds. Note that $\{v_{n}\}= \{S(F(v_{n-1}))\}$ and $\{w_{n}\}=\{S(F(w_{n-1}))\}$ are relatively compact in $C^{3}(I)$ by the complete continuity of S. Combining this fact with (2.10), we conclude that

$$ v_{n}\to \underline{u}, \qquad w_{n}\to \overline{u}\quad \text{in } C^{3}(I). $$

(2.16)

By (2.13) and (2.15) we can prove that $\underline{u}$ and u̅ are minimal and maximal fixed points of A in $[v_{0}, w_{0}]_{C^{3}}$. Hence, they are minimal and maximal solutions of BVP (1.1) in $[v_{0}, w_{0}]_{C^{3}}$. □

Example 2.1

Consider the following fourth-order boundary value problem with superlinear terms:

$$ \textstyle\begin{cases} u^{(4)}(t)=\frac{1}{3}\sqrt[3]{u}+(u')^{2}+(u'')^{3}-\frac{1}{2}(u''')^{3}+t ^{2}(1-t)^{2}, \quad t\in I, \\ u(0)=u'(0)=u''(1)=u'''(1)=0 . \end{cases} $$

(2.17)

Corresponding to BVP (1.1), the nonlinearity is

$$ f(t, x_{0}, x_{1}, x_{2}, x_{3})= \frac{1}{3}\sqrt[3]{x_{0}}+x_{1} ^{2}+x_{2}^{3}- \frac{1}{2}x_{3}^{3}+t^{2}(1-t)^{2}, $$

(2.18)

which is cubic growth on $x_{2}$ and $x_{3}$. We use Theorem 2.1 to show that BVP (2.17) has a positive solution. Clearly, $v_{0}(t)\equiv 0$ is a lower solution of BVP (2.17). We verify that

$$ w_{0}(t)=\frac{1}{24} t^{4}+\frac{1}{12} t^{3}+\frac{1}{4} t^{2}(1-t), \quad t\in I $$

is an upper solution of BVP (2.17). Since

$$\begin{aligned}& w_{0}'(t) =\frac{1}{6} t^{3}+ \frac{1}{2} t(1-t)\ge 0,\quad t\in I, \\& w_{0}''(t) =\frac{1}{2} (1-t)^{2}\ge 0,\quad t\in I, \\& w_{0}'''(t) =t-1\le 0,\quad t \in I, \end{aligned}$$

we obtain that

$$ \Vert w_{0} \Vert _{C}=\frac{1}{8},\qquad \bigl\Vert w_{0}' \bigr\Vert _{C}= \frac{1}{6},\qquad \bigl\Vert w _{0}'' \bigr\Vert _{C}=\frac{1}{2},\qquad \bigl\Vert w_{0}''' \bigr\Vert _{C}=1. $$

(2.19)

By (2.18) and (2.19), we have

$$\begin{aligned} f\bigl(t, w_{0}, w_{0}', w_{0}'', w_{0}''' \bigr) &=\frac{1}{3}\sqrt[3]{w _{0}(t)}+\bigl(w_{0}'(t) \bigr)^{2}+\bigl(w_{0}''(t) \bigr)^{3}-\frac{1}{2}\bigl(w_{0}'''(t) \bigr)^{3}+t ^{2}(1-t)^{2} \\ &\le \frac{1}{3}\sqrt[3]{ \Vert w_{0} \Vert _{C}}+ \bigl\Vert w_{0}' \bigr\Vert _{C}^{2}+ \bigl\Vert w_{0}'' \bigr\Vert _{C}^{3}+\frac{1}{2} \bigl\Vert w_{0}''' \bigr\Vert _{C}^{3}+t^{2}(1-t)^{2} \\ &\le \frac{1}{6}+\frac{1}{36}+\frac{1}{8}+ \frac{1}{2}+\frac{1}{16} \\ &< 1=w_{0}^{(4)}(t),\quad t\in I. \end{aligned}$$

Hence, $w_{0}$ is an upper solution of BVP (2.17). By (2.18), $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is increasing on $x_{0}$, $x _{1}$, $x_{2}$ in $[0, +\infty )^{3}$ and decreasing on $x_{3}$ in all $\mathbb{R}$. Hence f satisfies Assumptions (F1) and (F2). By Theorem 2.1, BVP (2.17) has at least one solution $u_{0}\in [v_{0}, w_{0}]_{C ^{3}}$, which is a positive solution. Since f does not satisfy the Nagumo condition on $x_{2}$ and $x_{3}$ in [17], this result cannot be obtained from [17]. This result also cannot be obtained from [18]. In fact, for any $M>0$, the first term $\frac{1}{3}\sqrt[3]{x_{0}}$ of expression (2.18) of f does not satisfy the Lipschitz condition on $\mathcal{D}_{M}$. Hence the Lipschitz condition (1.9) does not hold on $\mathcal{D}_{M}$. So [18, Theorem 2.2] is not applicable for BVP (2.17), and our existence result for BVP (2.17) cannot be obtained from [18].

3 A theorem of lower and upper solutions

In this section, we discuss the existence of a solution between a lower solution and an upper solution for BVP (1.1) under the case of nonlinearity $f(t,x_{0},x_{1},x_{2},x_{3})$ without monotonicity on $x_{3}$. In [15], an existence result between a lower solution and an upper solution was established for BVP (1.5), in which the authors requested nonlinearity $f(t,x_{0},x_{1},x_{2},x_{3})$ to satisfy a Nagumo-type condition on $x_{3}$, see [15, Theorem 3.1]. Since the boundary conditions and the definitions of lower and upper solutions of BVP (1.5) are different from those of BVP (1.1), the results presented in [15] are not applicable to BVP (1.1). We will use a directly truncating function technique to establish a similar existence result. A remarkable difference is that our existence result does not need the Nagumo-type condition. Our result is as follows:

Theorem 3.1

Let $f: I\times \mathbb{R}^{4}\to \mathbb{R}$ be continuous and BVP (1.1) have a lower solution $v_{0}$ and an upper solution $w_{0}$ with ${v_{0}}''' \ge {w_{0}}'''$. If f satisfies the following condition:

(F3):

for any $t\in I$ and $(x_{0}, x_{1}, x_{2})\in [v_{0}(t), w _{0}(t)]\times [v_{0}'(t), w_{0}'(t)]\times [v_{0}''(t), w_{0}''(t)]$,

$$\begin{aligned}& f\bigl(t, x_{0}, x_{1}, x_{2}, v_{0}'''(t)\bigr) \ge f \bigl(t, v_{0}(t), v _{0}'(t), v_{0}''(t), v_{0}'''(t) \bigr), \\& f\bigl(t, x_{0}, x_{1}, x_{2}, w_{0}'''(t)\bigr) \le f \bigl(t, w_{0}(t), w _{0}'(t), w_{0}''(t), w_{0}'''(t) \bigr); \end{aligned}$$

then BVP (1.1) has at least one solution in $[v_{0}, w_{0}]_{C^{3}}$.

In Theorem 3.1, condition (F3) is weaker than condition (F1) of Theorem 2.1, and Theorem 3.1 does not need the monotonicity of $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ on $x_{3}$. For the existence, Theorem 3.1 is more applicable than Theorem 2.1, but it has no monotone iterative procedure of seeking solutions. The proof of Theorem 3.1 needs the following lemma.

Lemma 3.1

Let $f: I\times \mathbb{R}^{4}\to \mathbb{R}$ be continuous and bounded. Then BVP (1.1) has at least one solution $u\in C^{4}(I)$.

Proof

Let $F: C^{3}(I)\to C(I)$ be the mapping defined by (2.10). Then, by Lemma 2.2, $A=S\circ F: C^{3}(I)\to C^{3}(I)$ is completely continuous and the solutions of BVP (1.1) are equivalent to the fixed points of A. We show that A has a fixed point in $C^{3}(I)$. By the boundedness of f, there exists a positive constant $M>0$ such that

$$ \bigl\vert f(t, x_{0}, x_{1}, x_{2}, x_{3}) \bigr\vert \le M,\quad (t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times \mathbb{R}^{4}. $$

(3.1)

By (2.10) and (3.1), $F: C^{3}(I)\to C(I)$ satisfies

$$ \bigl\Vert F(u) \bigr\Vert _{C}\le M,\quad u\in C^{3}(I). $$

(3.2)

Choose $R\ge M\|S\|$ and set $\varOmega =\{u\in C^{3}(I): \|u\|_{C^{3}} \le R\}$, where $\|S\|$ denotes the norm of linear bounded operator $S: C(I)\to C^{3}(I)$. Then Ω is a bounded and convex closed set in $C^{3}(I)$. For every $u\in \varOmega$, by (3.2), we have

$$ \Vert Au \Vert _{C^{3}}= \bigl\Vert S\bigl(F(u)\bigr) \bigr\Vert _{C^{3}}\le \Vert S \Vert \cdot \bigl\Vert F(u) \bigr\Vert _{C}\le M \Vert S \Vert \le R. $$

Hence $Au\in \varOmega$. This means that $A(\varOmega )\subset \varOmega$. By the Schauder fixed point theorem, A has a fixed point in Ω, which is a solution of BVP (1.1). □

Proof of Theorem 3.1

By Lemma 2.3, $v_{0}\le w_{0}$, $v_{0}'\le w_{0}'$, $v_{0}''\le w_{0}''$. Define functions $\eta _{0}, \eta _{1}, \eta _{2}, \eta _{3}: T\times \mathbb{R}\to \mathbb{R}$ by

$$\begin{aligned} \begin{gathered} \eta _{0}(t,y) =\min \bigl\{ \max \bigl\{ v_{0}(t), y\bigr\} , w_{0}(t)\bigr\} , \\ \eta _{1}(t,y) =\min \bigl\{ \max \bigl\{ v_{0}'(t), y\bigr\} , w_{0}'(t)\bigr\} , \\ \eta _{2}(t,y) =\min \bigl\{ \max \bigl\{ v_{0}''(t), y\bigr\} , w_{0}''(t)\bigr\} , \\ \eta _{3}(t,y) =\min \bigl\{ \max \bigl\{ w_{0}'''(t), y\bigr\} , v_{0}'''(t) \bigr\} . \end{gathered} \end{aligned}$$

(3.3)

Then $\eta _{0}, \eta _{1}, \eta _{2}, \eta _{3}: T\times \mathbb{R} \to \mathbb{R}$ are continuous and satisfy

$$\begin{aligned} \begin{gathered} v_{0}(t) \le \eta _{0}(t,y)\le w_{0}(t),\quad (t,y)\in I\times \mathbb{R}, \\ v_{0}'(t) \le \eta _{1}(t,y)\le w_{0}'(t),\quad (t,y)\in I\times \mathbb{R}, \\ v_{0}''(t) \le \eta _{2}(t,y) \le w_{0}''(t),\quad (t,y)\in I\times \mathbb{R}, \\ w_{0}'''(t) \le \eta _{3}(t,y)\le v_{0}'''(t), \quad (t,y)\in I\times \mathbb{R}. \end{gathered} \end{aligned}$$

(3.4)

Make a truncating function $f^{*}$ of f by

$$\begin{aligned} f^{*}(t, x_{0}, x_{1}, x_{2}, x_{3})&= f\bigl(t, \eta _{0}(t,x_{0}), \eta _{1}(t,x_{1}), \eta _{2}(t,x_{2}), \eta _{3}(t,x_{2})\bigr) \\ &\quad {} +\frac{x_{3}-\eta _{3}(t,x_{3})}{{x_{3}}^{2}+1}, \quad (t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times \mathbb{R}^{4}. \end{aligned}$$

(3.5)

Then by (3.3) and (3.4), $f^{*}:I\times \mathbb{R}^{4}\to \mathbb{R}$ is continuous and bounded. By Lemma 3.1, the boundary value problem

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

(3.6)

has a solution $u_{0}\in C^{4}(I)$. We show that

$$ w_{0}'''\le u_{0}'''\le v_{0}'''. $$

(3.7)

In fact, if $w_{0}'''\not \le u_{0}'''$, then for the function

$$ \phi (t)=u_{0}'''(t)-w_{0}'''(t), \quad t\in I, $$

(3.8)

$\min_{t\in I} \phi (t)<0$. Since $\phi (1)\ge 0$, there exists $t_{0}\in [0, 1)$ such that

$$ \phi (t_{0})=\min_{t\in I} \phi (t)< 0, \qquad \phi '(t_{0})\ge 0, $$

from which and (3.8) it follows that

$$ u_{0}'''(t_{0})< w_{0}'''(t_{0}), \qquad u_{0}^{(4)}(t_{0})\ge w_{0}^{(4)}(t_{0}). $$

(3.9)

Hence from definition (3.3), we see that

$$ \eta _{3}\bigl(t_{0},u_{0}'''(t_{0}) \bigr)= w_{0}'''(t_{0}). $$

(3.10)

By Eq. (3.6), (3.10), (3.4), condition (F3) and the definition of the upper solution $w_{0}$, we have

$$\begin{aligned} u_{0}^{(4)}(t_{0}) &=f^{*} \bigl(t_{0}, u_{0}(t_{0}), u_{0}'(t_{0}), u _{0}''(t_{0}), u_{0}'''(t_{0}) \bigr) \\ & =f\bigl(t_{0},\eta _{0}\bigl(t_{0},u_{0}(t_{0}) \bigr), \eta _{1}\bigl(t_{0},u_{0}'(t _{0})\bigr), \eta _{2}\bigl(t_{0},u_{0}''(t_{0}) \bigr), \eta _{3}\bigl(t_{0},u_{0}'''(t _{0})\bigr)\bigr) \\ & \quad {}+\frac{u_{0}'''(t_{0})-\eta _{3}(t_{0},u_{0}'''(t_{0}))}{ {[u_{0}'''(t_{0})]}^{2}+1} \\ & =f\bigl(t_{0},\eta _{0}\bigl(t_{0},u_{0}(t_{0}) \bigr), \eta _{1}\bigl(t_{0},u_{0}'(t _{0})\bigr), \eta _{2}\bigl(t_{0},u_{0}''(t_{0}) \bigr),w_{0}'''(t_{0}) \bigr) \\ & \quad {}+\frac{u_{0}'''(t_{0})-w_{0}'''(t_{0})}{{[u_{0}'''(t_{0})]} ^{2}+1} \\ & \le f\bigl(t_{0}, w_{0}(t_{0}), w_{0}'(t_{0}), w_{0}''(t_{0}), w _{0}'''(t_{0}) \bigr)+\frac{u_{0}'''(t_{0})-w_{0}'''(t_{0}))}{{[u_{0}'''(t _{0})]}^{2}+1} \\ & < f\bigl(t_{0}, w_{0}(t_{0}), w_{0}'(t_{0}), w_{0}''(t_{0}), w_{0}'''(t _{0}) \bigr) \\ & \le w_{0}^{(4)}(t_{0}), \end{aligned}$$

that is, $u_{0}^{(4)}(t_{0})< w_{0}^{(4)}(t_{0})$, which contradicts (3.9). Hence, $w_{0}'''\le u_{0}'''$.

With a similar argument, we can show that $u_{0}'''\le v_{0}'''$, so (3.7) holds. Now by Lemma 2.1,

$$ v_{0}\le u_{0}\le w_{0},\qquad {v_{0}}'\le {u_{0}}'\le {w_{0}}',\qquad {v_{0}}'' \le {u_{0}}''\le {w_{0}}''. $$

(3.11)

From (3.7), (3.11), and the definition (3.3) of $\eta _{i}$ ($i=0, 1, 2, 3$), it follows that

$$ \eta _{i}\bigl(t, u^{(i)}(t)\bigr)=u^{(i)}(t), \quad t\in I, i=0, 1, 2, 3. $$

Hence by Eq. (3.6) we have

$$\begin{aligned} u_{0}^{(4)}(t) &=f^{*}\bigl(t, u_{0}(t), u_{0}'(t), u_{0}''(t), u_{0}'''(t)\bigr) \\ & =f\bigl(t,\eta _{0}\bigl(t,u_{0}(t)\bigr), \eta _{1}\bigl(t,u_{0}'(t)\bigr), \eta _{2}\bigl(t,u _{0}''(t)\bigr), \eta _{3}\bigl(t,u_{0}'''(t) \bigr)\bigr) \\ & \quad{} +\frac{u_{0}'''(t)-\eta _{3}(t,u_{0}'''(t))}{{[u_{0}'''(t)]} ^{2}+1} \\ & = f\bigl(t, u_{0}(t), u_{0}'(t), u_{0}''(t), u_{0}'''(t) \bigr),\quad t \in I. \end{aligned}$$

That is, $u_{0}$ is a solution of BVP (1.1) in $[v_{0}, w_{0}]_{C^{3}}$. □

Example 3.1

Consider the fourth-order boundary value problem

$$ \textstyle\begin{cases} u^{(4)}(t)=\frac{1}{8} \sqrt[3]{u}+(u')^{3}+(u'')^{3}+\frac{1}{2}(u''')^{3}+t(1-t), \quad t\in I, \\ u(0)=u'(0)=u''(1)=u'''(1)=0 . \end{cases} $$

(3.12)

Similar to Example 2.1, one can verify that $v_{0}=0$ is a lower solution and $w_{0}$ given by

$$ w_{0}(t)=\frac{1}{24} t^{4}+\frac{1}{12} t^{3}+\frac{1}{4} t^{2}(1-t), \quad t\in I, $$

(3.13)

is an upper solution of BVP (3.12). Since $w_{0}'''(t)=t-1\le 0=v_{0}'''(t)$ and the corresponding nonlinearity

$$ f(t, x_{0}, x_{1}, x_{2}, x_{3})= \frac{1}{8} \sqrt[3]{x_{0}}+x _{1}^{3}+x_{2}^{3}+ \frac{1}{2}x_{3}^{3}+t(1-t) $$

(3.14)

is increasing on $x_{0}$, $x_{1}$, $x_{2}$, all the conditions of Theorem 3.1 are satisfied. By Theorem 3.1, BVP (3.12) has at least one solution $u_{0}\in [v_{0}, w_{0}]_{C^{3}}$; clearly this solution is a positive solution. Since f is increasing on $x_{3}$ and it does not satisfy condition (F2), this result cannot be obtained by Theorem 2.1. For any $M>0$, by expression (3.14) of f, the Lipschitz condition (1.9) does not hold on $\mathcal{D}_{M}$, and hence the result of [18] is not applicable for BVP (3.12).

4 Existence of positive solutions

In [17], the present first author have discussed the existence of positive solution of BVP (1.1) by using fixed point theory in cones. In this section we present a different existence result of positive for BVP (1.1) by Theorem 3.1.

Theorem 4.1

Let $f: I\times \mathbb{R}^{4}\to \mathbb{R}$ be continuous and satisfy the following conditions:

(F4):

for every $t\in I$ and $x_{3}\in (-\infty , 0]$, $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is increasing on $x_{0}$, $x_{1}$, and $x_{2}$ in $[0, +\infty )$;

(F5):

there exists a positive constant $\delta >0$ such that

$$ f(t, x_{0}, x_{1}, x_{2}, x_{3}) \ge 21 x_{0} \quad \textit{for all }(t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times [0, \delta ]^{3}\times [- \delta , 0]; $$

(F6):

there exist nonnegative constants $a_{0}$, $a_{1}$, $a_{2}$, $a _{3}$ satisfying $a_{0}+a_{1}+a_{2}+a_{3}<1$ and a positive constant $C_{0}>0$ such that

$$ f(t, x_{0}, x_{1}, x_{2}, x_{3}) \le a_{0} x_{0}+a_{1} x_{1}+a _{2} x_{2}+a_{3} \vert x_{3} \vert +C_{0} $$

for all $(t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times [0, + \infty )^{3}\times (-\infty , 0]$.

Then BVP (1.1) has at least one positive solution.

The proof of Theorem 4.1 needs the following existence and uniqueness result of a general fourth-order linear boundary value problem.

Lemma 4.1

Let $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$ be nonnegative constants and satisfy $a_{0}+a_{1}+a_{2}+a_{3}<1$. Then, for every $h\in C(I)$, the fourth-order linear boundary value problem

$$ \textstyle\begin{cases} u^{(4)}(t)=a_{0}u(t)+a_{1}u'(t)+a_{2}u''(t)-a_{3}u'''(t)+h(t) , \quad t\in I, \\ u(0)=u'(0)=u''(1)=u'''(1)=0 \end{cases} $$

(4.1)

has a unique solution $u\in C^{4}(I)$, and when $h\in C^{+}(I)$, the solution u satisfies

$$ u\ge 0,\qquad u'\ge 0,\qquad u''\ge 0,\qquad u'''\le 0. $$

(4.2)

Proof

Choose a closed subset space of $C^{3}(I)$ by

$$ E=\bigl\{ u\in C^{3}(I): u(0)=u'(0)=u''(1)=u'''(1)=0 \bigr\} . $$

(4.3)

For every $u\in E$, we show that

$$ \Vert u \Vert _{C} \le \bigl\Vert u' \bigr\Vert _{C}\le \bigl\Vert u'' \bigr\Vert _{C}\le \bigl\Vert u''' \bigr\Vert _{C}. $$

(4.4)

For every $t\in I$, by the boundary condition of E, we have

$$\begin{aligned}& \bigl\vert u(t) \bigr\vert = \biggl\vert \int _{0}^{t} u'(s) \,ds \biggr\vert \le \int _{0}^{t} \bigl\vert u'(s) \bigr\vert \,ds\le t \bigl\Vert u' \bigr\Vert _{C}\le \bigl\Vert u' \bigr\Vert _{C} , \\& \bigl\vert u'(t) \bigr\vert = \biggl\vert \int _{0}^{t} u''(s) \,ds \biggr\vert \le \int _{0}^{t} \bigl\vert u''(s) \bigr\vert \,ds\le t \bigl\Vert u'' \bigr\Vert _{C}\le \bigl\Vert u'' \bigr\Vert _{C} , \\& \bigl\vert u''(t) \bigr\vert = \biggl\vert - \int _{t}^{1} u''(s) \,ds \biggr\vert \le \int _{t}^{1} \bigl\vert u'''(s) \bigr\vert \,ds\le (1-t) \bigl\Vert u''' \bigr\Vert _{C}\le \bigl\Vert u''' \bigr\Vert _{C} . \end{aligned}$$

From these inequalities we conclude that

$$ \Vert u \Vert _{C}\le \bigl\Vert u' \bigr\Vert _{C},\qquad \bigl\Vert u' \bigr\Vert _{C} \le \bigl\Vert u'' \bigr\Vert _{C},\qquad \bigl\Vert u'' \bigr\Vert _{C}\le \bigl\Vert u''' \bigr\Vert _{C}. $$

Hence, (4.4) holds. By (4.4), we have

$$ \Vert u \Vert _{C^{3}}= \bigl\Vert u''' \bigr\Vert _{C},\quad u\in E. $$

(4.5)

By Lemma 2.2, the solution operator of LBVP (2.3) $S:C(I)\to E$ is a completely linear operator. For every $h\in C(I)$ and $t\in I$, setting $u=Sh$, by Eq. (2.4), we have

$$ \bigl\vert u'''(t) \bigr\vert = \biggl\vert - \int _{t}^{1}u^{(4)}(s) \,ds \biggr\vert = \biggl\vert \int _{t} ^{1} h(s) \,ds \biggr\vert \le \Vert h \Vert _{C}. $$

Hence

$$ \Vert Sh \Vert _{C^{3}}= \Vert u \Vert _{C^{3}}= \bigl\Vert u''' \bigr\Vert _{C} \le \Vert h \Vert _{C}. $$

This means that the norm of the linear bounded operator $S:C(I)\to E$ satisfies

$$ \Vert S \Vert _{\mathcal{B}(C(I), E)}\le 1. $$

(4.6)

Define a linear operator $B: E\to C(I)$ by

$$ Bu(t):=a_{0} u(t)+a_{1} u'(t)+a_{2} u''(t)-a_{3} u'''(t), \quad u \in E, t\in I. $$

(4.7)

Then, by the definition of the operator $S: C(I)\to E$, LBVP (4.1) is rewritten to the form of the operator equation in Banach space E:

$$ (\mathrm{I}-S B) u=Sh , $$

(4.8)

where I is the identity operator in E. We prove that the norm of the composite operator SB in $\mathcal{B}(E, E)$ satisfies $\|T B\|_{\mathcal{B}(E, E)}< 1$.

For every $u\in E$, by the definition of B and (4.4), we have

$$\begin{aligned} \Vert Bu \Vert _{C} &\le a_{0} \Vert u \Vert _{C}+a_{1} \bigl\Vert u' \bigr\Vert _{C}+a_{2} \bigl\Vert u'' \bigr\Vert _{C}+ a _{3} \bigl\Vert u''' \bigr\Vert _{C} \\ &\le (a_{0}+a_{1}+a_{2}+a_{3}) \bigl\Vert u''' \bigr\Vert _{C}, \\ &=(a_{0}+a_{1}+a_{2}+a_{3}) \Vert u \Vert _{C^{3}}. \end{aligned}$$

(4.9)

From (4.6) and (4.9) it follows that

$$\begin{aligned} \Vert SB u \Vert _{C^{3}} &= \bigl\Vert S(Bu) \bigr\Vert _{C^{3}} \le \Vert S \Vert _{\mathcal{B}(C(I), E)} \cdot \Vert Bu \Vert _{C} \\ &\le (a_{0}+a_{1}+a_{2}+a_{3}) \Vert u \Vert _{C^{3}}. \end{aligned}$$

This means that $\|S B\|_{\mathcal{B}(E, E)}\le a_{0}+a_{1}+a_{2}+a _{3}<1$.

Since $\|S B\|_{\mathcal{B}(E, E)}< 1$, it follows that $\mathrm{I}-S B$ has a bounded inverse operator given by the series

$$ ( \mathrm{I}-S B )^{-1}=\sum_{n=0}^{\infty }(S B)^{n}. $$

Hence, Eq. (4.8), equivalently, LBVP (4.1), has the unique solution

$$ u=( \mathrm{I}-S B )^{-1}S h=\sum_{n=0}^{\infty } (S B)^{n} S h . $$

(4.10)

Set $K_{3}=\{ u\in C^{3}: u\succeq 0\}$. Then $K_{3}$ is a closed convex cone in $C^{3}(I)$. For every $v\in K_{3}$, by the definition (4.7) of B, $Bv\in C^{+}(I)$. By Lemma 2.3, $SBv=S(Bv)\in K_{3}$. Hence, $SB(K_{3})\subset K_{3}$. Let $h\in C^{+}(I)$. By Lemma 2.3, $v=Sh\in K_{3}$. Hence, for every $n\in \mathbb{N}$, $(S B)^{n} S h=(SB)^{n} v\in K_{3}$. By (4.10) and the completeness of $K_{3}$, $u\in K_{3}$, that is, u satisfies (4.2). □

Proof of Theorem 4.1

By [17, Lemma 2.3 and Lemma 2.4], the fourth-order linear eigenvalue problem(EVP)

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

(4.11)

has a minimum positive real eigenvalue $\lambda _{1}\in [8, 21)$, and $\lambda _{1}$ has a positive unit eigenfunction, namely there exists $\phi _{1}\in C^{4}(I)\cap C^{+}(I)$ with $\|\phi _{1}\|_{C}=1$ which satisfies the equation

$$ \textstyle\begin{cases} \phi _{1}^{(4)}(t)=\lambda _{1} {\phi _{1}}(t) , \quad t\in I, \\ {\phi _{1}}(0)=\phi _{1}'(0)=\phi _{1}''(1)=\phi _{1}'''(1)=0 . \end{cases} $$

(4.12)

By Lemma 2.3, $\phi _{1}\in K_{3}$. Choose a positive constant

$$ \varepsilon =\min \bigl\{ \delta /\max \bigl\{ 1, \bigl\Vert \phi _{1}' \bigr\Vert _{C}, \bigl\Vert \phi _{1}'' \bigr\Vert _{C}, \bigl\Vert \phi _{1}''' \bigr\Vert _{C}\bigr\} , C_{0}/21\bigr\} , $$

(4.13)

and let $v_{0}=\varepsilon \phi _{1}(t)$. Then, for every $t\in I$,

$$\begin{aligned}& 0\le v_{0}(t)\le \varepsilon \Vert \phi _{1} \Vert _{C}\le \delta , \qquad 0\le v_{0}'(t)\le \varepsilon \bigl\Vert \phi _{1}' \bigr\Vert _{C} \le \delta \\& 0\le v_{0}''(t)\le \varepsilon \bigl\Vert \phi _{1}'' \bigr\Vert _{C}\le \delta , \qquad 0\ge v_{0}'''(t) \ge - \varepsilon \bigl\Vert \phi _{1}''' \bigr\Vert _{C} \ge -\delta . \end{aligned}$$

By Assumption (F5), we have

$$ f\bigl(t, v_{0}(t), v_{0}'(t), v_{0}''(t), v_{0}'''(t) \bigr) \ge 21v_{0}(t) \ge \lambda _{1}v_{0}(t)={v_{0}}^{(4)}(t), \quad t\in I. $$

Hence $v_{0}$ is a lower solution of BVP (1.1).

By Lemma 4.1, the linear boundary value

$$ \textstyle\begin{cases} u^{(4)}(t)=a_{0}u(t)+a_{1}u'(t)+a_{2}u''(t)-a_{3}u'''(t)+C_{0} , \quad t\in I, \\ u(0)=u'(0)=u''(1)=u'''(1)=0 \end{cases} $$

(4.14)

has a unique solution $w_{0}\in K_{3}$, where $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, and $C_{0}$ are the constants in Assumption (F6). By Assumption (F6), $w_{0}$ is an upper solution of BVP (1.1). We show that $v_{0}\preceq w_{0}$. Set $u_{0}=w_{0}-v_{0}$, since $v_{0}, w_{0} \in K_{3}$, by the definitions of $v_{0}$ and $w_{0}$ and (4.13), we have

$$\begin{aligned} u_{0}^{(4)}(t) &= w_{0}^{(4)}(t)-v_{0}^{(4)}(t) \\ &=a_{0} w_{0}(t)+a_{1}w_{0}'(t)+a_{2}w_{0}''(t)-a_{3}w_{0}'''(t)+C _{0}-\varepsilon \lambda _{1}\phi _{1}(t) \\ &\ge a_{0} w_{0}(t)+a_{1}w_{0}'(t)+a_{2}w_{0}''(t)-a_{3}w_{0}'''(t) \ge 0, \quad t\in I. \end{aligned}$$

(4.15)

By this inequality and Lemma 2.3, $u_{0}\succeq 0$. Hence $v_{0} \preceq w_{0}$. Now by Assumption (F4), condition (F3) of Theorem 3.1 holds. By Theorem 3.1, BVP (1.1) has a solution between $v_{0}$ and $w_{0}$, which is a positive solution of BVP (1.1). □

Example 4.1

Consider the following fourth-order nonlinear boundary value problem:

$$ \textstyle\begin{cases} u^{(4)}(t)=a\sqrt{ \vert u \vert }+b\sqrt[3]{ (u')^{2}}+c\sqrt[5]{(u'')^{2}}-d u'''-e (u''')^{4},\quad t\in [0, 1], \\ u(0)=u'(0)=u''(1)=u'''(1)=0 , \end{cases} $$

(4.16)

where a, b, c, d, e are positive constants. We verify that the nonlinearity term of BVP (4.16)

$$ f(t, x_{0}, x_{1}, x_{2}, x_{3})=a \sqrt{ \vert x_{0} \vert }+b \sqrt[3]{x _{1}^{2}}+c \sqrt[5]{x_{2}^{2}} -d x_{3}-e x_{3}^{4} $$

(4.17)

satisfies conditions (F4)–(F6).

By expression (4.17), for every $t\in I$ and $x_{3}\in (-\infty , 0]$, $f(t, x_{0}, x_{1}, x_{2}, x_{3})$ is increasing on $x_{0}$, $x _{1}$, and $x_{2}$ on $[0, +\infty )$. Hence (F4) holds. Choose $\delta =\min \{\frac{a^{2}}{441}, \sqrt[3]{\frac{d}{e}}\}$, then for any $(t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times [0, \delta ]^{3} \times [-\delta , 0]$, by (4.17) we have

$$\begin{aligned} f(t, x_{0}, x_{1}, x_{2}, x_{3}) \ge & a\sqrt{x_{0}}-x_{3}\bigl(d+e x_{3}^{3} \bigr)=\frac{a}{\sqrt{x_{0}} } x_{0}+ \vert x_{3} \vert \bigl(d-e \vert x_{3} \vert ^{3}\bigr) \\ \ge & \frac{a}{\sqrt{\delta } } x_{0} + \vert x_{3} \vert \bigl(d-e \delta ^{3}\bigr) \ge \frac{a}{\sqrt{\delta } } x_{0} \ge 21x_{0}. \end{aligned}$$

Hence (F5) holds. For any $(t, x_{0}, x_{1}, x_{2}, x_{3})\in I\times [0, +\infty )^{3}\times (-\infty , 0]$, using the Young inequality

$$ rs\le \frac{1}{\alpha }s^{\alpha }+\frac{1}{\beta }s^{\beta }, \qquad \alpha , \beta >0,\qquad \frac{1}{\alpha }+\frac{1}{\beta }=1;\qquad r, s \ge 0, $$

we have

$$\begin{aligned} &a \sqrt{ \vert x_{0} \vert } =(2a) \bigl(x_{0}^{1/2}/2 \bigr)\le 2a^{2}+\frac{1}{8} x _{0}\quad (\alpha =\beta =2), \\ &b \sqrt[3]{x_{1}^{2}} =\bigl(2^{2/3}b\bigr) \bigl(x_{1}^{2/3}/2^{2/3}\bigr)\le \frac{4}{3} b^{3}+\frac{1}{3} x_{1}\quad (\alpha =3, \beta =3/2), \\ &c \sqrt[5]{x_{2}^{2}} =\bigl(8^{2/5}c\bigr) \bigl(x_{1}^{2/5}/8^{2/5}\bigr)\le \frac{12}{5} c^{5/3}+\frac{1}{20} x_{2}\quad (\alpha =5/3, \beta =5/2). \end{aligned}$$

(4.18)

By these inequalities and (4.17), we obtain that

$$\begin{aligned} f(t, x_{0}, x_{1}, x_{2}, x_{3}) &\le \frac{1}{8} x_{0}+\frac{1}{3} x_{1}+ \frac{1}{20} x_{2}+C_{0}+ \vert x_{3} \vert \bigl(d-e \vert x_{3} \vert ^{3}\bigr) \\ &\le \frac{1}{8} x_{0}+\frac{1}{3} x_{1}+ \frac{1}{20} x_{2}+C_{0}+ \max_{x_{3}\in \mathbb{R}} \vert x_{3} \vert \bigl(d-e \vert x_{3} \vert ^{3}\bigr) \\ &= \frac{1}{8} x_{0}+\frac{1}{3} x_{1}+ \frac{1}{20} x_{2}+C_{0}+ \frac{3}{4} \,d \biggl(\frac{d}{4e} \biggr)^{1/3}, \end{aligned}$$

(4.19)

where $C_{0}=2a^{2}+\frac{4}{3} b^{3}+\frac{12}{5} c^{5/3}$. Choose $a_{0}=\frac{1}{8}$, $a_{1}=\frac{1}{3}$, $a_{2}=\frac{1}{20}$, $a_{3}=0$, and $C=C_{0}+\frac{3}{4} \,d (\frac{d}{4e} )^{1/3}$, then $a_{0}+a_{1}+a_{2}+a_{3}=\frac{61}{120}<1$. From (4.19) it follows that (F6) holds.

Consequently, the nonlinearity f of BVP (4.16) satisfies conditions (F4)–(F6). By Theorem 4.1, BVP (4.16) has at least one positive solution.