Solvability of a Third-Order Three-Point Boundary Value Problem on a Half-Line

Abstract

In this paper, we consider the solvability of a third-order three-point boundary value problem on a half-line of the form:

$$\begin{aligned} {\left\{ \begin{array}{ll} x^{\prime \prime \prime }(t)=f\left( t,x(t),{x^{\prime }}(t),{x^{\prime \prime }}(t)\right) ,\quad 0< t<+\infty , \\ x(0)=\alpha x(\eta ),\quad \mathop {\lim }\nolimits _{t\rightarrow +\infty }x^{(i)}(t)=0,\quad i=1,2, \end{array}\right. } \end{aligned}$$

where $\alpha \ne 1$ and $\eta \in (0,+\infty )$, while $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ is $S^2$—Carathéodory function. The existence and uniqueness of solutions for the boundary value problems are obtained by the Leray-Schauder continuation theorem. As an application, an example is given to demonstrate our results.

1 Introduction

In this paper we consider the existence and uniqueness of solutions for third-order three-point boundary value problems on a half-line

$$\begin{aligned} {x^{\prime \prime \prime }}(t)&= f\left( t,x(t),{x^{\prime }}(t),{x^{\prime \prime }}(t)\right) ,\quad 0< t<+\infty , \end{aligned}$$

(1.1)

$$\begin{aligned} x(0)&= \alpha x(\eta ),\quad \displaystyle \lim _{t\rightarrow +\infty }x^{(i)}(t)=0,\quad i=1,2, \end{aligned}$$

(1.2)

where $\alpha \ne 1$ and $\eta \in (0,+\infty )$.

The third-order differential equations arise in many areas, such as the deflection of a curved beam having a constant or a varying cross-section, three layer beam, electromagnetic waves or gravity-driven flows [10]. Meanwhile, the third order boundary value problems in an infinite interval has been widely used to describe the evolution of physical phenomena, for example some draining or coating fluid-flow problems, see [3, 24, 25]. We refer the reader to [2, 4–9, 11–16, 20, 21, 23, 26, 27] for the study of the finite interval problems of third-order differential equations, and to [1, 3, 17, 22, 24, 25] for the study of the infinite interval problems.

Motivated by the above works and [19], in this paper we discuss the solvability of third-order three-point boundary value problems (1.1), (1.2). Based upon the Leray-Schauder continuation theorem, the existence and uniqueness of solutions for BVP (1.1), (1.2) were obtained.

The rest of this paper is organized as follows. In Sect. 2, as the preliminary we give some lemmas which help us to simplify the proofs of our main results. In Sect. 3, firstly we discuss the existence of solutions for BVP (1.1), (1.2) by Leray-Schauder continuation theorem, and then investigate the uniqueness of solutions to BVP (1.1), (1.2). Finally, as an application, we give an example to demonstrate our results.

2 Preliminary

In this section, we present some definitions and lemmas which are useful in the proof of our main results.

Definition 2.1

The function $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ is called an $S^2$—Carathéodory function, if and only if

• (i) for each $(u,v,w )\in {\mathbb {R}}^3,t \mapsto f(t,u,v,w)$ is measurable on $[0,+\infty );$

• (ii) for $\mathrm{a.e.}\ t\in [0,+\infty ),(u,v,w)\mapsto f(t,u,v,w)$ is continuous on ${\mathbb {R}}^3;$

• (iii) for each $r>0,$ there exists $\varphi _{r}(t)\in L^1[0,+\infty )$ with $\varphi _r(t)>0$ on $(0,+\infty )$ and $t\varphi _{r}(t)$, $t^{2}\varphi _{r}(t)\in L^1 [0,+\infty )$, such that $\forall \ u, v, w\in [-r,r]$,

  $$\begin{aligned} |f(t,u,v,w)|\le \varphi _r(t),\quad a.e.\ t\in [0,+\infty ). \end{aligned}$$

Lemma 2.1

For any $h(t)\in L^1[0,+\infty )$ with $th(t),t^{2}h(t)\in L^1 [0,+\infty )$, the BVP

$$\begin{aligned} {\left\{ \begin{array}{ll} x^{\prime \prime \prime }(t)=h(t), \quad 0<t<+\infty ,\\ x(0)=\alpha x(\eta ), \quad \mathop {\lim }\nolimits _{t\rightarrow +\infty }x^{(i)}(t)=0,\quad i=1,2 \end{array}\right. }\end{aligned}$$

(2.1)

has a unique solution

$$\begin{aligned} x(t)=\int _0^{+\infty }G(t,s)h(s)\mathrm{d}s, \end{aligned}$$

where

$$\begin{aligned} G(t,s)= {\left\{ \begin{array}{ll} \frac{1}{2(1-\alpha )}s^{2},&{} s\le \min \left\{ \eta ,t\right\} ;\\ \frac{\alpha }{2(1-\alpha )}s^{2}-\frac{1}{2}t^{2}+ts,&{} t\le s\le \eta ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2},&{} \eta \le s\le t;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) -\frac{1}{2}t^{2}+ts,&{} \max \left\{ \eta ,t\right\} \le s. \end{array}\right. } \end{aligned}$$

Proof

Noticing that $h(t),th(t),t^{2}h(t)\in L^1 [0,+\infty )$, integrate the differential equation in BVP (2.1) from $t$ to $+\infty$, we have

$$\begin{aligned} -{x^{\prime \prime }}(t)=\int _{t}^{+\infty }h(s)\mathrm{d}s. \end{aligned}$$

Integrating this differential equation on $[t,+\infty )$, and applying the Fubini theorem we obtain that

$$\begin{aligned} {x^{\prime }}(t)=\int _{t}^{+\infty }(s-t)h(s)\mathrm{d}s. \end{aligned}$$

Also integrating the above differential on $[0,t]$ one has

$$\begin{aligned} x(t)=x(0)+\frac{1}{2}\int _{0}^{t}s^{2}h(s)\mathrm{d}s+\int _{t}^{+\infty }\left( ts-\frac{1}{2}t^{2}\right) h(s)\mathrm{d}s. \end{aligned}$$

(2.2)

Since $x(0)=\alpha x(\eta )$, we have

$$\begin{aligned} x(0)=\alpha \left[ x(0)+\frac{1}{2}\int _{0}^{\eta }s^{2}h(s)\mathrm{d}s+\int _{\eta }^{+\infty }\left( \eta s-\frac{1}{2}\eta ^{2}\right) h(s)\mathrm{d}s\right] , \end{aligned}$$

and thus

$$\begin{aligned} x(0)=\frac{\alpha }{1-\alpha }\left[ \frac{1}{2}\int _{0}^{\eta }s^{2}h(s)\mathrm{d}s+\int _{\eta }^{+\infty }\left( \eta s-\frac{1}{2}\eta ^{2}\right) h(s)\mathrm{d}s\right] . \end{aligned}$$

Hence from (2.2) it follows that

$$\begin{aligned} x(t)&= \frac{\alpha }{1-\alpha }\left[ \frac{1}{2}\int _{0}^{\eta }s^{2}h(s)\mathrm{d}s + \int _{\eta }^{+\infty }\left( \eta s-\frac{1}{2}\eta ^{2}\right) h(s)\mathrm{d}s\right] \\&\quad {} +\frac{1}{2}\int _{0}^{t}s^{2}h(s)\mathrm{d}s+\int _{t}^{+\infty }\left( ts-\frac{1}{2}t^{2}\right) h(s)\mathrm{d}s. \end{aligned}$$

Therefore, when $0\le t\le \eta$,

$$\begin{aligned} x(t)&= \frac{\alpha }{1-\alpha }\left[ \frac{1}{2}\left( \int _{0}^{t}+\int _{t}^{\eta }\right) s^{2}h(s)\mathrm{d}s +\int _{\eta }^{+\infty }\left( \eta s-\frac{1}{2}\eta ^{2}\right) h(s)\mathrm{d}s\right] \\&\quad {} +\frac{1}{2}\int _{0}^{t}s^{2}h(s)\mathrm{d}s+\left( \int _{t}^{\eta } +\int _{\eta }^{+\infty }\right) \left( ts-\frac{1}{2}t^{2}\right) h(s)\mathrm{d}s, \end{aligned}$$

when $\eta \le t\le +\infty$,

$$\begin{aligned} x(t)&= \frac{\alpha }{1-\alpha }\left[ \frac{1}{2}\int _{0}^{\eta }s^{2}h(s)\mathrm{d}s +\left( \int _{\eta }^{t}+\int _{t}^{+\infty }\right) \left( \eta s-\frac{1}{2}\eta ^{2}\right) h(s)\mathrm{d}s\right] \\&\quad {} +\frac{1}{2}\left( \int _{0}^{\eta }+\int _{\eta }^{t}\right) s^{2}h(s)\mathrm{d}s +\int _{t}^{+\infty }\left( ts-\frac{1}{2}t^{2}\right) h(s)\mathrm{d}s, \end{aligned}$$

that is,

$$\begin{aligned} x(t)=\int _0^{+\infty }G(t,s)h(s)\mathrm{d}s,\quad \forall \ t\in [0,+\infty ). \end{aligned}$$

This completes the proof of the lemma. $\square$

Lemma 2.2

Let the Green function $G(t,s)$ be as in Lemma 2.1. Then $G(t,s)$ has two properties$:$

• (1) For all $t,s\in [0,+\infty )$,

  $$\begin{aligned} |G(t,s)|\le {\left\{ \begin{array}{ll} \frac{1}{2}s^{2},&{} \alpha <0 ;\\ \frac{1}{2(1-\alpha )}s^{2},&{} 0\le \alpha <1;\\ \frac{\alpha }{2(\alpha -1)}s^{2},&{} \alpha >1. \end{array}\right. } \end{aligned}$$
• (2)
  $$\begin{aligned} \lim _{t\rightarrow +\infty }G(t,s)=\overline{G}(s):= {\left\{ \begin{array}{ll} \frac{1}{2(1-\alpha )}s^{2},&{} s\le \eta ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2},&{} \eta \le s. \end{array}\right. } \end{aligned}$$

Proof

(1) It is easy to see that for each fixed $s\in [0,+\infty )$,

$$\begin{aligned} \frac{\partial }{\partial t}G(t,s)\ge 0,\ \forall \ t\in [0,s] \quad \text{ and } \quad \frac{\partial }{\partial t}G(t,s)\equiv 0, \ \forall \ t\in [s,+\infty ), \end{aligned}$$

it follows that

$$\begin{aligned} G(0,s)\le G(t,s)\le G(s,s),\quad \forall \ t, s\in [0,+\infty ). \end{aligned}$$

Since

$$\begin{aligned} G(0,s)&= {\left\{ \begin{array}{ll} \frac{\alpha }{2(1-\alpha )}s^{2},&{} s\le \eta ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) ,&{} \eta \le s, \end{array}\right. } \\ G(s,s)&= {\left\{ \begin{array}{ll} \frac{1}{2(1-\alpha )}s^{2},&{} s\le \eta ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2},&{} \eta \le s, \end{array}\right. } \end{aligned}$$

we have $\forall \ (t,s)\in [0,+\infty )\times [0,\eta ]$,

$$\begin{aligned} \frac{\alpha }{2(1-\alpha )}s^{2}\le G(t,s)\le \frac{1}{2(1-\alpha )}s^{2}, \end{aligned}$$

and $\forall \ (t,s)\in [0,+\infty )\times [\eta ,+\infty )$,

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) \le G(t,s)\le \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2}. \end{aligned}$$

Now we have three cases to consider:

Case 1. $\alpha <0$. In this case, since ${\alpha }/({\alpha -1})>0$ and ${\eta ^{2}}/{2}-\eta s<0$ for $s\ge \eta$, we have

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2}\le \frac{1}{2}s^{2}\quad \hbox {for}\quad s\ge \eta . \end{aligned}$$

But from ${1}/({1-\alpha })<1$, it follows that

$$\begin{aligned} \frac{1}{2(1-\alpha )}s^{2}\le \frac{1}{2}s^{2}\quad \hbox {for}\quad s\in [0,\eta ]. \end{aligned}$$

Therefore

$$\begin{aligned} G(t,s)\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

(2.3)

Also $\forall \ s\in [0,+\infty )$,

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) -\frac{\alpha }{2(1-\alpha )}s^{2}=\frac{\alpha }{2(\alpha -1)}(\eta -s)^{2}\ge 0, \end{aligned}$$

then

$$\begin{aligned} G(t,s)\ge \frac{\alpha }{2(1-\alpha )}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

This together with (2.3) implies that

$$\begin{aligned} \frac{\alpha }{2(1-\alpha )}s^{2}\le G(t,s)\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

Hence from the fact $0<{\alpha }/({\alpha -1})<1$, it follows that

$$\begin{aligned} |G(t,s)|\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

Case 2. $0\le \alpha <1$. In this case, obviously $G(t,s)\ge 0$ on $[0,+\infty )\times [0,+\infty )$. Since $\forall \ s\in [0,+\infty )$,

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2}-\frac{1}{2(1-\alpha )}s^{2} =\frac{\alpha }{2(\alpha -1)}(\eta -s)^{2}\le 0, \end{aligned}$$

it follows that

$$\begin{aligned} G(t,s)\le \frac{1}{2(1-\alpha )}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

Hence

$$\begin{aligned} 0\le G(t,s)\le \frac{1}{2(1-\alpha )}s^{2},\quad \forall \ t,s\in [0,+\infty ), \end{aligned}$$

which implies that

$$\begin{aligned} |G(t,s)|\le \frac{1}{2(1-\alpha )}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

Case 3. $\alpha >1$. In this case, obviously $s^{2}/(2(1-\alpha ))\le 0$ for $s\in [0,+\infty )$. On the other hand,

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) <0\quad \hbox {for}\quad s\ge \eta , \end{aligned}$$

and thus

$$\begin{aligned} \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2}<\frac{1}{2}s^{2}\quad \hbox {for}\quad s\ge \eta . \end{aligned}$$

Therefore

$$\begin{aligned} G(t,s)\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

(2.4)

Also since

$$\begin{aligned} \frac{\alpha }{2(1-\alpha )}s^{2}-\frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) =\frac{\alpha }{2(1-\alpha )}(s-\eta )^{2}\le 0,\quad \forall \ s\in [0,+\infty ), \end{aligned}$$

it follows from (2.4) that

$$\begin{aligned} \frac{\alpha }{2(1-\alpha )}s^{2}\le G(t,s)\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

This together with the fact ${\alpha }/({1-\alpha })<-1$ implies that

$$\begin{aligned} |G(t,s)|\le \frac{\alpha }{2(\alpha -1)}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

(2) From the definition of $G(t,s)$, it follows that

$$\begin{aligned} G(t,s)= {\left\{ \begin{array}{ll} \frac{1}{2(1-\alpha )}s^{2},&{} s\le \eta \le t ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2},&{} \eta \le s\le t, \end{array}\right. } \end{aligned}$$

and thus

$$\begin{aligned} \lim _{t\rightarrow +\infty }G(t,s)= {\left\{ \begin{array}{ll} \frac{1}{2(1-\alpha )}s^{2},&{} s\le \eta ;\\ \frac{\alpha }{\alpha -1}\left( \frac{1}{2}\eta ^{2}-\eta s\right) +\frac{1}{2}s^{2},&{} \eta \le s. \end{array}\right. } \end{aligned}$$

This completes the proof of the lemma. $\square$

Consider the space

$$\begin{aligned} C_{\infty }^2[0,+\infty )=\left\{ x \in C^2[0,+\infty ):\lim _{t\rightarrow +\infty }x^{(i)}(t)\ \hbox {exists},\ i=0,1,2 \right\} \end{aligned}$$

with the norm $\Vert x\Vert :=\Vert x\Vert _{\infty }+\Vert x^{\prime }\Vert _{\infty }+\Vert x^{\prime \prime }\Vert _{\infty },\ \forall \ x\in C_{\infty }^2[0,+\infty ).$ Then by the standard arguments, we can prove that $(C_{\infty }^2[0,+\infty ), \Vert \cdot \Vert )$ is a Banach space. Now, we introduce the subspace $X$ of $C_{\infty }^2[0,+\infty )$ as follows:

$$\begin{aligned} X=\left\{ x \in C_{\infty }^2[0,+\infty ):x(0)=\alpha x(\eta ),\lim _{t\rightarrow +\infty }x^{(i)}(t)\hbox {=0},\ i=1,2\right\} . \end{aligned}$$

Then it is clear that $X$ is closed in $C_{\infty }^2[0,+\infty )$, and hence is itself a Banach space.

Lemma 2.3

[1] Let $M \subset C_{\infty }=\{x\in C[0,+\infty ):\lim _{t\rightarrow +\infty }x(t)\ \hbox {exists}\}.$ Then $M$ is relatively compact if the following conditions hold$:$

• (i) all functions from $M$ are uniformly bounded$;$

• (ii) all functions from $M$ are equicontinuous on any compact interval of $[0,+\infty );$

• (iii) all functions from $M$ are equiconvergent at infinity, that is, for any given $\varepsilon >0,$ there exists a $T(\varepsilon )>0,$ such that $|f(t)-f(+\infty )|<\varepsilon$ for all $t>T$ and $f\in M$.

Let us denote the operator $T$ as

$$\begin{aligned} (Tx)(t)= \int _0^{+\infty }G(t,s) f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s,\quad 0<t<+\infty . \end{aligned}$$

It is easy to check from Lemma 2.1, 2.2 and Lebesgue’s dominated convergence theorem that if $f$ satisfies the $S^2$—Carathéodory condition, then $T:X\rightarrow X$ is well defined.

Lemma 2.4

Let $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ be an $S^2$—Carathéodory function, then $T:X\rightarrow X$ is compact.

Proof

First we show that $T$ is continuous. To do this, let $x_{n},x_{0}\in X(n=1,2,\cdot \cdot \cdot )$ and $x_{n}\rightarrow x_{0}(n\rightarrow +\infty )$. Then there exists $r_{0}>0$, such that

$$\begin{aligned} \Vert x_{n}\Vert \le r_{0},\quad n=0,1,2,\cdots . \end{aligned}$$

Since $f$ is an $S^2$—Carathéodory function, then for the above $r_{0}>0$, there exists a positive function $\varphi _{r_{0}}\in L^{1}[0,+\infty )$ with $t\varphi _{r_{0}}(t),t^{2}\varphi _{r_{0}}(t)\in L^{1}[0,+\infty )$ such that for each $n=0,1,2,\cdots ,$

$$\begin{aligned} \left| f\left( t,x_{n}(t),{x_{n}^{\prime }}(t),{x_{n}^{\prime \prime }}(t)\right) \right| \le \varphi _{r_{0}}(t),\quad \mathrm{a.e}.\ t\in [0,+\infty ). \end{aligned}$$

Thus

$$\begin{aligned}&\int _0^{+\infty }\left| \overline{G}(s)\left( f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_{0}(s),{x_{0}^{\prime }}(s),{x_{0}^{\prime \prime }}(s)\right) \right) \right| \mathrm{d}s\\&\le 2\int _0^{+\infty }\left| \overline{G}(s)\right| \varphi _{r_{0}}(s)\mathrm{d}s<+\infty ,\quad n=1,2,\cdots . \end{aligned}$$

Consequently from the Lebesgue’s dominated convergence theorem, it follows that

$$\begin{aligned}&|(Tx_{n})(t)-(Tx_{0})(t)|\\&\le \int _0^{+\infty }|G(t,s)|\left| f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_0(s),{x_0^{\prime }}(s),{x_0^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\&\le \frac{1}{2}\left( 1+\frac{1+|\alpha |}{|1-\alpha |}\right) \int _0^{+\infty }s^{2} \left| f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) \right. \\&\quad \quad \left. -f\left( s,x_0(s),{x_0^{\prime }}(s),{x_0^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\&\rightrightarrows 0\quad \hbox {on}\ [0,+\infty )\ (n\rightarrow +\infty ),\\&|{(Tx_{n})^{\prime }}(t)-{(Tx_{0})^{\prime }}(t)|\\&\le \int _{t}^{+\infty }(s-t)\left| f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_{0}(s),{x_{0}^{\prime }}(s),{x_{0}^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\&\le \int _{0}^{+\infty }s\left| f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_{0}(s),{x_{0}^{\prime }}(s),{x_{0}^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\&\rightrightarrows 0\quad \hbox {on}\ [0,+\infty )\ (n\rightarrow +\infty ) \end{aligned}$$

and

$$\begin{aligned}&|{(Tx_{n})^{\prime \prime }}(t)-{(Tx_0)^{\prime \prime }}(t)|\\&= \left| -\int _{t}^{+\infty }\left( f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_0(s),{x_0^{\prime }}(s),{x_0^{\prime \prime }}(s)\right) \right) \mathrm{d}s\right| \\&\le \int _{0}^{+\infty }\left| f\left( s,x_{n}(s),{x_{n}^{\prime }}(s),{x_{n}^{\prime \prime }}(s)\right) -f\left( s,x_0(s),{x_0^{\prime }}(s),{x_0^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\&\rightrightarrows 0\quad \hbox {on}\ [0,+\infty )\ (n\rightarrow +\infty ). \end{aligned}$$

Therefore, $T:X\rightarrow X$ is continuous.

Next we show that $T$ maps bounded sets into relatively compact set. Let $B\subset X$ be a bounded set. Then there exists $r_{1}>0$, such that $\forall \ x\in B$,

$$\begin{aligned} \Vert x\Vert =\Vert x\Vert _{\infty }+\Vert {x^{\prime }}\Vert _{\infty }+\Vert {x^{\prime \prime }}\Vert _{\infty }\le r_{1}. \end{aligned}$$

Since $f$ is an $S^2$—Carathéodory function, there exists a positive function $\varphi _{r_{1}}\in L^{1}[0,+\infty )$ with $t\varphi _{r_{1}}(t),t^{2}\varphi _{r_{1}}(t)\in L^{1}[0,+\infty ),$ such that $\forall \ x\in B$,

$$\begin{aligned} \left| f\left( t,x(t),{x^{\prime }}(t),{x^{\prime \prime }}(t)\right) \right| \le \varphi _{r_{1}}(t),\quad \mathrm{a.e}.\ t\in [0,+\infty ). \end{aligned}$$

Therefore $\forall \ x\in B$, we have

$$\begin{aligned} |(Tx)(t)|&\le \int _0^{+\infty }|G(t,s)|\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s\nonumber \\&\le \frac{1}{2}\left( 1+\frac{1+|\alpha |}{|1-\alpha |}\right) \int _0^{+\infty }s^{2}\varphi _{r_{1}}(s)\mathrm{d}s<+\infty ,\quad t\in [0,+\infty ),\nonumber \\ |(Tx)'(t)|&\le \int _{t}^{+\infty }s\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s\nonumber \\&\le \int _0^{+\infty }s\varphi _{r_{1}}(s)\mathrm{d}s<+\infty ,\quad t\in [0,+\infty ) \end{aligned}$$

(2.5)

and

$$\begin{aligned} |{(Tx)^{\prime \prime }}(t)|&\le \int _{t}^{+\infty }\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s\nonumber \\&\le \int _0^{+\infty }\varphi _{r_{1}}(s)\mathrm{d}s<+\infty ,\quad t\in [0,+\infty ). \end{aligned}$$

(2.6)

Thus $\{(Tx)(t):x\in B\}, \{{(Tx)^{\prime }}(t):x\in B\}$ and $\left\{ {(Tx)^{\prime \prime }}(t):x\in B\right\}$ are uniformly bounded. Also, from (2.5) and (2.6) it follows that, $\left\{ (Tx)(t): x\in B\right\}$ and $\left\{ {(Tx)^{\prime }}(t):x\in B\right\}$ are equi-continuous on any compact interval of $[0,+\infty )$. Meanwhile, $\forall \ t_{1},t_{2}\in [0,+\infty )$ and $\forall \ x\in B$, we have

$$\begin{aligned} \left| {(Tx)^{\prime \prime }}(t_{2})-{(Tx)^{\prime \prime }}(t_{1})\right| = \left| \int _{t_{1}}^{t_{2}}f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \le \left| \int _{t_{1}}^{t_{2}}\varphi _{r_{1}}(s)\mathrm{d}s\right| , \end{aligned}$$

and so by the absolute continuity of Lebesgue integral, $\left\{ {(Tx)^{\prime \prime }}(t):x\in B\right\}$ is equicontinuous on any compact interval of $[0,+\infty )$.

On the other hand, from Lebesgue’s dominated convergence theorem, it follows that

$$\begin{aligned} \begin{array}{ll} |(Tx)(t)-(Tx)(+\infty )| &{} \le \displaystyle \int _0^{+\infty }\left| G(t,s)-\overline{G}(s)\right| \left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\ &{} \le \displaystyle \int _0^{+\infty }\left| G(t,s)-\overline{G}(s)\right| \varphi _{r_{1}}(s)\mathrm{d}s\\ &{} \rightarrow 0\quad (t\rightarrow +\infty ),\quad \forall \ x\in B,\\ \left| {(Tx)^{\prime }}(t)-{(Tx)^{\prime }}(+\infty )\right| &{} \le \displaystyle \int _{t}^{+\infty }(s-t)\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s\\ &{}\le \displaystyle \int _{t}^{+\infty }s\varphi _{r_{1}}(s)\mathrm{d}s\\ &{} \rightarrow 0\quad (t\rightarrow +\infty ),\quad \forall \ x\in B \end{array} \end{aligned}$$

and

$$\begin{aligned} \left| {(Tx)^{\prime \prime }}(t)-{(Tx)^{\prime \prime }}(+\infty )\right|&\le \int _{t}^{+\infty }\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \le \int _{t}^{+\infty }\varphi _{r_{1}}(s)\mathrm{d}s\\&\rightarrow 0\quad (t\rightarrow +\infty ),\quad \forall \ x\in B, \end{aligned}$$

we have that $\left\{ (Tx)(t):x\in B\right\} ,\left\{ {(Tx)^{\prime }}(t):x\in B\right\}$ and $\left\{ {(Tx)^{\prime }}(t):x\in B\right\}$ are equiconvergent at infinity. Hence from Lemma 2.4, $TB$ is relatively compact in $X$.

In summary, $T:X\rightarrow X$ is compact. This completes the proof of the lemma. $\square$

Lemma 2.5

[18] (Leray-Schauder continuation theorem) Let $X$ be a real Banach space and let $\Omega$ be a bounded open neighborhood of $0$ in $X$. Let $T: \bar{\Omega }\rightarrow X$ be a completely continuous operator such that for all $\lambda \in (0,1)$ and $x\in \partial \Omega$, $x\ne \lambda Tx$. Then the operator equation

$$\begin{aligned} x=Tx \end{aligned}$$

has a solution $x\in \bar{\Omega }$.

3 Main Results

Now we apply the Leray-Schauder continuation theorem to establish the existence theorems for BVP (1.1), (1.2).

Theorem 3.1

Assume that $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ is $S^2-$ Carathéodory function. Suppose also that there exist nonnegative functions $p(t),q(t),r(t),e(t)\in L^{1}[0,+\infty )$ with $t^ip(t),t^iq(t),$ $t^ir(t),t^ie(t)\in L^{1}[0,+\infty )(i=1,2)$, such that for all $(u,v,w)\in {\mathbb {R}}^3$,

$$\begin{aligned} |f(t,u,v,w)|\le p(t)|u|+q(t)|v|+r(t)|w|+e(t),\quad \mathrm{a.e}.\ t\in [0,+\infty ). \end{aligned}$$

Then BVP (1.1), (1.2) has at least one solution, provided

$$\begin{aligned} \max \left\{ P, Q, R\right\} <1, \end{aligned}$$

where

$$\begin{aligned} P&= {\left\{ \begin{array}{ll} P_{0}+P_{1}+\frac{1}{2}P_{2}, &{} \alpha <0; \\ P_{0}+P_{1}+\frac{1}{2(1-\alpha )}P_{2}, &{} 0\le \alpha <1; \\ P_{0}+P_{1}+\frac{\alpha }{2(\alpha -1)}P_{2}, &{} \alpha >1, \end{array}\right. } \quad P_{i}=\displaystyle \int _0^{+\infty }s^{i}p(s)\mathrm{d}s,\ i=0,1,2, \\ Q&= {\left\{ \begin{array}{ll} Q_{0}+Q_{1}+\frac{1}{2}Q_{2}, &{} \alpha <0; \\ Q_{0}+Q_{1}+\frac{1}{2(1-\alpha )}Q_{2}, &{} 0\le \alpha <1; \\ Q_{0}+Q_{1}+\frac{\alpha }{2(\alpha -1)}Q_{2}, &{} \alpha >1, \end{array}\right. } \quad Q_{i}=\displaystyle \int _0^{+\infty }s^{i}q(s)\mathrm{d}s,\ i=0,1,2, \\ R&= {\left\{ \begin{array}{ll} R_{0}+R_{1}+\frac{1}{2}R_{2}, &{} \alpha <0; \\ R_{0}+R_{1}+\frac{1}{2(1-\alpha )}R_{2}, &{} 0\le \alpha <1; \\ R_{0}+R_{1}+\frac{\alpha }{2(\alpha -1)}R_{2}, &{} \alpha >1, \end{array}\right. } \quad R_{i}=\displaystyle \int _0^{+\infty }s^{i}r(s)\mathrm{d}s,\ i=0,1,2. \end{aligned}$$

Proof

By Lemma 2.1, it is easy to see that $x\in X$ is a solution of BVP (1.1), (1.2) if and only if $x$ is a fixed point of $T$. Now, we apply Leray-Schauder continuation theorem, to prove that $T$ has a fixed point in $X$. To do this, it is sufficient to show that operator equations

$$\begin{aligned} x=\lambda Tx,\quad \lambda \in (0,1) \end{aligned}$$

(3.1)

has a priori bound $M$ independently of $\lambda$, that is we need only to show that boundary value problems

$$\begin{aligned} {\left\{ \begin{array}{ll} {x^{\prime \prime \prime }}(t)=\lambda f\left( t,x(t),{x^{\prime }}(t),{x^{\prime \prime }}(t)\right) ,\quad 0<t<+\infty ,\ \lambda \in (0,1),\\ x(0)=\alpha x(\eta ),\quad \mathop {\lim }\nolimits _{t\rightarrow +\infty }{x^{\prime }}(t)=0,\quad \mathop {\lim }\nolimits _{t\rightarrow +\infty }{x^{\prime \prime }}(t)=0 \end{array}\right. } \end{aligned}$$

(3.2)

has a priori bound $M$ independently of $\lambda \in (0,1)$.

Suppose that $x=x(t)$ is a possible solution of (3.2), and let

$$\begin{aligned} E_{0}=\int _0^{+\infty }e(s)\mathrm{d}s,\quad E_{1}=\int _0^{+\infty }se(s)\mathrm{d}s,\quad E_{2}=\int _0^{+\infty }s^{2}e(s)\mathrm{d}s. \end{aligned}$$

We will now divide the proof into three cases.

Case 1. $\alpha <0$. In this case, by Lemma 2.2 we have

$$\begin{aligned} |G(t,s)|\le \frac{1}{2}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

From (3.2) and Lemma 2.1 it follows that

$$\begin{aligned} \begin{array}{lll} |x(t)| &{} = \left| \lambda \displaystyle \int _0^{+\infty }G(t,s)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\ &{} \le \displaystyle \int _0^{+\infty }|G(t,s)|\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \\ &{} \le \displaystyle \int _0^{+\infty }\frac{1}{2}s^{2}\left( p(s)|x(s)|+q(s)\left| {x^{\prime }}(s)\right| +r(s)\left| {x^{\prime \prime }}(s)\right| +e(s)\right) \mathrm{d}s \\ &{} \le \frac{1}{2}\left( P_{2}\Vert x\Vert _{\infty }+Q_{2}\left\| {x^{\prime }}\right\| _{\infty }+R_{2}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{2}\right) ,\quad \forall \ t\in [0,+\infty ), \\ |{x^{\prime }}(t)| &{} = \left| \lambda \displaystyle \int _t^{+\infty }(s-t)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\ &{} \le \displaystyle \int _0^{+\infty }s\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \\ &{} \le \displaystyle \int _0^{+\infty }s\left( p(s)|x(s)|+q(s)\left| {x^{\prime }}(s)\right| +r(s)\left| {x^{\prime \prime }}(s)\right| +e(s)\right) \mathrm{d}s \\ &{} \le P_{1}\Vert x\Vert _{\infty }+Q_{1}\left\| {x^{\prime }}\right\| _{\infty }+R_{1}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{1},\quad \forall \ t\in [0,+\infty ) \\ \end{array} \end{aligned}$$

and

$$\begin{aligned} |{x^{\prime \prime }}(t)|&= \left| -\lambda \int _t^{+\infty }f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\&\le \int _0^{+\infty }\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \\&\le \int _0^{+\infty }\left( p(s)|x(s)|+q(s)\left| {x^{\prime }}(s)\right| +r(s)\left| {x^{\prime \prime }}(s)\right| +e(s)\right) \mathrm{d}s \\&\le P_{0}\Vert x\Vert _{\infty }+Q_{0}\left\| {x^{\prime }}\right\| _{\infty }+R_{0}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{0},\quad \forall \ t\in [0,+\infty ). \end{aligned}$$

Thus

$$\begin{aligned} \Vert x\Vert _{\infty }&\le \frac{1}{2}\left( P_{2}\Vert x\Vert _{\infty }+Q_{2}\left\| {x^{\prime }}\right\| _{\infty }+R_{2}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{2}\right) , \\ \Vert {x^{\prime }}\Vert _{\infty }&\le P_{1}\Vert x\Vert _{\infty }+Q_{1}\left\| {x^{\prime }}\right\| _{\infty }+R_{1}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{1},\\ \Vert {x^{\prime \prime }}\Vert _{\infty }&\le P_{0}\Vert x\Vert _{\infty }+Q_{0}\left\| {x^{\prime }}\right\| _{\infty }+R_{0}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{0}. \end{aligned}$$

Consequently

$$\begin{aligned} \Vert x\Vert&\le P\Vert x\Vert _{\infty }+Q\left\| {x^{\prime }}\right\| _{\infty }+R\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{0}+E_{1}+\frac{1}{2}E_{2}\\&\le \max \left\{ P,Q,R\right\} \Vert x\Vert +E_{0}+E_{1}+\frac{1}{2}E_{2}. \end{aligned}$$

Therefore

$$\begin{aligned} \Vert x\Vert \le \frac{E_{0}+E_{1}+\frac{1}{2}E_{2}}{1-\max \left\{ P,Q,R\right\} } =: M_{1}. \end{aligned}$$

Case 2. $0\le \alpha <1$. In this case, by Lemma 2.2 we have

$$\begin{aligned} |G(t,s)|\le \frac{1}{2(1-\alpha )}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

From (3.2) and Lemma 2.1 it follows that

$$\begin{aligned} |x(t)|&\le \int _0^{+\infty }|G(t,s)|\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \\&\le \int _0^{+\infty }\frac{1}{2(1-\alpha )}s^{2}\left( p(s)|x(s)|+q(s)\left| {x^{\prime }}(s)\right| +r(s)\left| {x^{\prime \prime }}(s)\right| +e(s)\right) \mathrm{d}s \\&\le \frac{1}{2(1-\alpha )}\left( P_{2}\Vert x\Vert _{\infty }+Q_{2}\left\| {x^{\prime }}\right\| _{\infty }+R_{2}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{2}\right) ,\quad \forall \ t\in [0,+\infty ), \\ |{x^{\prime }}(t)|&= \left| \lambda \int _t^{+\infty }(s-t)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\&\le P_{1}\Vert x\Vert _{\infty }+Q_{1}\left\| {x^{\prime }}\right\| _{\infty }+R_{1}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{1},\quad \forall \ t\in [0,+\infty ) \end{aligned}$$

and

$$\begin{aligned} |{x^{\prime \prime }}(t)|&= \left| -\lambda \int _t^{+\infty }f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\&\le P_{0}\Vert x\Vert _{\infty }+Q_{0}\left\| {x^{\prime }}\right\| _{\infty }+R_{0}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{0},\quad \forall \ t\in [0,+\infty ). \end{aligned}$$

Thus

$$\begin{aligned} \Vert x\Vert \le \max \left\{ P,Q,R\right\} \Vert x\Vert +E_{0}+E_{1}+\frac{1}{2(1-\alpha )}E_{2}, \end{aligned}$$

and hence

$$\begin{aligned} \Vert x\Vert \le \frac{E_{0}+E_{1}+\frac{1}{2(1-\alpha )}E_{2}}{1-\max \left\{ P,Q,R\right\} } =: M_{2}. \end{aligned}$$

Case 3. $\alpha >1$. In this case, by Lemma 2.2 we have

$$\begin{aligned} |G(t,s)|\le \frac{\alpha }{2(\alpha -1)}s^{2},\quad \forall \ t,s\in [0,+\infty ). \end{aligned}$$

From (3.2) and Lemma 2.1 it follows that

$$\begin{aligned} |x(t)|&\le \int _0^{+\infty }|G(t,s)|\left| f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \right| \mathrm{d}s \\&\le \int _0^{+\infty }\frac{\alpha }{2(\alpha -1)}s^{2}\left( p(s)|x(s)|+q(s)\left| {x^{\prime }}(s)\right| +r(s)\left| {x^{\prime \prime }}(s)\right| +e(s)\right) \mathrm{d}s \\&\le \frac{\alpha }{2(\alpha -1)}\left( P_{2}\Vert x\Vert _{\infty }+Q_{2}\left\| {x^{\prime }}\right\| _{\infty }+R_{2}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{2}\right) ,\quad \forall \ t\in [0,+\infty ), \\ |{x^{\prime }}(t)|&= \left| \lambda \int _t^{+\infty }(s-t)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\&\le P_{1}\Vert x\Vert _{\infty }+Q_{1}\left\| {x^{\prime }}\right\| _{\infty }+R_{1}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{1},\quad \forall \ t\in [0,+\infty ) \end{aligned}$$

and

$$\begin{aligned} |{x^{\prime \prime }}(t)|&= \left| -\lambda \int _t^{+\infty }f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s\right| \\&\le P_{0}\Vert x\Vert _{\infty }+Q_{0}\left\| {x^{\prime }}\right\| _{\infty }+R_{0}\left\| {x^{\prime \prime }}\right\| _{\infty }+E_{0},\quad \forall \ t\in [0,+\infty ). \end{aligned}$$

Thus

$$\begin{aligned} \Vert x\Vert \le \max \left\{ P,Q,R\right\} \Vert x\Vert +E_{0}+E_{1}+\frac{\alpha }{2(\alpha -1)}E_{2}, \end{aligned}$$

and hence

$$\begin{aligned} \Vert x\Vert \le \frac{E_{0}+E_{1}+\frac{\alpha }{2(\alpha -1)}E_{2}}{1-\max \left\{ P,Q,R\right\} } =: M_{3}. \end{aligned}$$

In summary, the operator equations (3.1) has a priori bound $M:=M_{1}+M_{2}+M_{3}$ which is independent of $\lambda \in (0,1)$. Hence by Lemma 2.5 (Leray-Schauder continuation theorem), BVP (1.1), (1.2) has at least one solution. This completes the proof of the theorem. $\square$

Next, we give a result on the uniqueness of solutions for BVP (1.1), (1.2).

Theorem 3.2

Assume that $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ is $S^2-$ Carathéodory function. Suppose also that there exist nonnegative functions $p(t),q(t),r(t)\in L^{1}[0,+\infty )$ with $t^ip(t),t^iq(t),$ $t^ir(t)\in L^{1}[0,+\infty )(i=1,2)$, such that

$$\begin{aligned} |f(t,u_1,v_1,w_1)-f(t,u_2,v_2,w_2)| \le p(t)|u_1-u_2|+q(t)|v_1-v_2|+r(t)|w_1-w_2| \end{aligned}$$

(3.3)

for a.e. $t\in [0,+\infty )$ and all $(u_i,v_i,w_i)\in {\mathbb {R}}^3, i=1,2$.

Then BVP (1.1), (1.2) has a unique solution, provided

$$\begin{aligned} \max \left\{ P, Q, R\right\} <1, \end{aligned}$$

where $P, Q, R$ as in Theorem 3.1.

Proof

We note that assumption (3.3) implies

$$\begin{aligned} |f(t,u,v,w)| \le p(t)|u|+q(t)|v|+r(t)|w|+|f(t,0,0,0)| \end{aligned}$$

for a.e. $x \in [0,+\infty )$ and all $(u,v,w) \in {\mathbb {R}}^3$. Accordingly from Theorem 3.1, BVP (1.1), (1.2) has at least one solution.

Now, suppose that $x_1(t)$, $x_2(t)$ are two solutions of BVP (1.1), (1.2). Let $y(t)=x_1(t)-x_2(t)$. Then $y(t)$ satisfies the boundary condition (1.2) and

$$\begin{aligned} {y^{\prime \prime \prime }}(t)= f\left( t,x_1(t),{x_{1}^{\prime }}(t),{x_1^{\prime \prime }}(t)\right) -f\left( t,x_2(t),{x_2^{\prime }}(t),{x_2^{\prime \prime }}(t)\right) ,\quad 0<t<+\infty . \end{aligned}$$

Hence from Lemma 2.1 we have

$$\begin{aligned} y(t)=\int _0^{+\infty }G(t,s)\left[ f\left( s,x_1(s),{x_{1}^{\prime }}(s),{x_1^{\prime \prime }}(s)\right) -f\left( t,x_2(s),{x_2^{\prime }}(s),{x_2^{\prime \prime }}(s)\right) \right] \mathrm{d}s. \end{aligned}$$

Similar to the proof of Theorem 3.1, we can show easily that

$$\begin{aligned} \Vert y\Vert \le \max \left\{ P,Q,R\right\} \Vert y\Vert , \end{aligned}$$

that is

$$\begin{aligned} \left( 1-\max \left\{ P,Q,R\right\} \right) \Vert y\Vert \le 0. \end{aligned}$$

Since $\max \left\{ P,Q,R\right\} <1$, it follows that $\Vert y\Vert =0$, and hence $y(t) \equiv 0$ on $[0,+\infty )$, i.e., $x_1(t) \equiv x_2(t)$ on $[0,+\infty )$. This completes the proof of the theorem. $\square$

Finally, as an application, we give an example to demonstrate our results.

Example 3.1

Consider the third-order boundary value problem

$$\begin{aligned} {x^{\prime \prime \prime }}&= \frac{1}{4}\min \left\{ 1,t^{-4}\right\} \frac{1}{1+x^2}+\frac{1}{3}\min \left\{ t,t^{-5}\right\} \sqrt{({x^{\prime }})^{2}+1} + \frac{1}{2}e^{-2t}|{x^{\prime \prime }}|,\end{aligned}$$

(3.4)

$$\begin{aligned} x(0)&= \frac{1}{2}x(\eta ),\quad \lim _{t\rightarrow +\infty }{x^{\prime }}(t)=0,\quad \lim _{t\rightarrow +\infty }{x^{\prime \prime }}(t)=0, \end{aligned}$$

(3.5)

where $0<\eta <+\infty$.

Let

$$\begin{aligned} f(t,u,v,w)&=\frac{1}{4}\min \left\{ 1,t^{-4}\right\} \frac{1}{1+u^2}+\frac{1}{3}\min \left\{ t,t^{-5}\right\} \sqrt{v^{2}+1}+\frac{1}{2}e^{-2t}|w|,\\ p(t)&=\frac{1}{4}\min \left\{ 1,t^{-4}\right\} ,\quad q(t)=\frac{1}{3}\min \left\{ t,t^{-5}\right\} ,\quad r(t)=\frac{1}{2}e^{-2t}. \end{aligned}$$

Then it is easy to check that $f:[0,+\infty )\times {\mathbb {R}}^3 \rightarrow {\mathbb {R}}$ is an $S^2-$ Carathéodory function, and

$$\begin{aligned} |f(t,u_1,v_1,w_1){-}f(t,u_2,v_2,w_2)| \le p(t)|u_1{-}u_2|+q(t)|v_1-v_2|+r(t)|w_1{-}w_2| \end{aligned}$$

for all $t\in [0,+\infty )$ and all $(u_i,v_i,w_i)\in {\mathbb {R}}^3, i=1,2$. Meanwhile, obviously $t^ip(t)$, $t^iq(t)$, $t^ir(t)\in L^{1}[0,+\infty )$, $i=0,1,2$.

It is easy to compute that

$$\begin{aligned} P_{0}&=\frac{1}{3}, \quad P_{1}=\frac{1}{4}, \quad P_{2}=\frac{1}{3}, \\ Q_{0}&=\frac{1}{4}, \quad Q_{1}=\frac{2}{9}, \quad Q_{2}=\frac{1}{4}, \\ R_{0}&=\frac{1}{4}, \quad R_{1}=\frac{1}{8}, \quad R_{2}=\frac{1}{8}. \end{aligned}$$

It follows that

$$\begin{aligned} \max \big \{P,Q,R\big \}&= \max \left\{ P_{0}+P_{1}+P_{2},Q_{0}+Q_{1}+Q_{2},R_{0}+R_{1}+R_{2}\right\} \\&= \max \left\{ \frac{11}{12},\frac{13}{18},\frac{1}{2}\right\} =\frac{11}{12}<1. \end{aligned}$$

In summary, all conditions of Theorem 3.2 are satisfied for BVP (3.4), (3.5), and hence BVP (3.4), (3.5) has a unique solution $x=x(t)$.

Noticing that the Green’s function corresponding to BVP (3.4), (3.5) satisfy

$$\begin{aligned} G(t,s)\ge G(0,s) = {\left\{ \begin{array}{ll} \frac{1}{2}s^{2}\ge 0, &{} s\le \eta ; \\ \eta s-\frac{1}{2}\eta ^{2}>0, &{} s\ge \eta , \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} f(t,u,v,w)> 0,\quad \forall \ (t,u,v,w)\in [0,+\infty )\times {\mathbb {R}}^{3}. \end{aligned}$$

It follows that

$$\begin{aligned} x(t)&= \int _0^{+\infty }G(t,s)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s \\&\ge \int _0^{+\infty }G(0,s)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s >0,\quad t\in [0,+\infty ). \end{aligned}$$

Also since

$$\begin{aligned} {x^{\prime }}(t)&= \int _{t}^{+\infty }(s-t)f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s>0,\quad t\in [0,+\infty ), \\ {x^{\prime \prime }}(t)&= -\int _{t}^{+\infty }f\left( s,x(s),{x^{\prime }}(s),{x^{\prime \prime }}(s)\right) \mathrm{d}s<0,\quad t\in [0,+\infty ), \end{aligned}$$

we have that the unique solution $x=x(t)$ is strictly monotone increasing convex positive on $[0,+\infty )$, that is, BVP (3.4), (3.5) has a unique solution which is strictly monotone increasing, convex and positive on $[0,+\infty )$.