- Research
- Open Access
Existence of solutions to fourth-order differential equations with deviating arguments
- Received: 20 February 2015
- Accepted: 22 May 2015
- Published: 24 June 2015
Abstract
In this paper, we consider fourth-order differential equations on a half-line with deviating arguments of the form \(u^{(4)}(t)+q(t)f(t,[u(t)],[u'(t)],[u''(t)],u'''(t))=0\), \(0< t<{+\infty}\), with the boundary conditions \(u(0)=A\), \(u'(0)=B\), \(u''(t)-au'''(t)=\theta(t)\), \(-\tau\leq{t}\leq 0\); \(u'''(+\infty)=C\). We present sufficient conditions for the existence of a solution between a pair of lower and upper solutions by using Schäuder’s fixed point theorem. Also, we establish the existence of three solutions between two pairs of lower and upper solutions by using topological degree theory. An important feature of our existence criteria is that the obtained solutions may be unbounded. We illustrate the importance of our results through two simple examples.
Keywords
- fourth-order
- boundary value problem
- half-line
- upper solution
- lower solution
MSC
- 34B15
- 34B40
1 Introduction
In recent years considerable attention has been focused on the existence of solutions to boundary value problems involving differential equations with deviating arguments (DEDA) [1–16]. While most of these works deal with problems on finite intervals and the literature is satisfactory, study of infinite interval problems has been just initiated in [2, 13–17]. This study compare to boundary value problems for second and higher order ordinary differential equations over infinite intervals (and their wide variety of applications to real world problems) [18–24] is far from complete, and needs attention. To fill some of this gap, in this paper we shall provide existence criteria for fourth-order differential equations with deviating arguments of the form where and \(q:(0,+\infty)\rightarrow{(0,+\infty)}\), \(f:[0,+\infty)\times{\mathbb{R}}^{n+1}\times{\mathbb{R}}^{n+1}\times{\mathbb{R}}^{n+1}\times\mathbb{R}\rightarrow{\mathbb{R}}\) are continuous, and \(\tau_{j,i}\): \([0,+\infty)\rightarrow{(0,+\infty)}\) are continuous for all \(j=0,1,2\), \(i=1,2,\ldots,n\). In what follows we shall always assume that \(\lim_{t\rightarrow{+\infty}}(t-\tau_{j,i}(t))=+\infty\), \(j=0,1,2\), \(i=1,2,\ldots,n\). We define the positive real number τ as and seek the solutions of (1.1) which satisfy the boundary conditions where \(\theta\in\mathcal{C}[-\tau,0]\), \(A,B\in\mathbb{R}\), \(a,C\geq 0\), and \(u'''(+\infty)= \lim_{t\rightarrow{+\infty}}u'''(t)\). For this, following as in several above works, and inspired by the contributions in [25–30], we shall employ the method of upper and lower solutions.
$$\begin{aligned} u^{(4)}(t)+q(t)f\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0, \quad 0< t< {+\infty}, \end{aligned}$$
(1.1)
$$\begin{aligned}[b] &\bigl[u(t)\bigr]=\bigl(u(t),u\bigl(t-\tau_{0,1}(t)\bigr),\ldots,u \bigl(t-\tau_{0,n}(t)\bigr)\bigr), \\ &\bigl[u'(t)\bigr]=\bigl(u'(t),u' \bigl(t-\tau_{1,1}(t)\bigr),\ldots,u'\bigl(t- \tau_{1,n}(t)\bigr)\bigr), \\ &\bigl[u''(t)\bigr]=\bigl(u''(t),u'' \bigl(t-\tau_{2,1}(t)\bigr),\ldots,u''\bigl(t- \tau_{2,n}(t)\bigr)\bigr), \end{aligned} $$
$$\begin{aligned} \tau=-\min_{0\leq{j}\leq{2},1\leq{i}\leq{n}}\min_{t\geq0}\bigl(t- \tau_{j,i}(t)\bigr) \end{aligned}$$
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} u(0)=A,\qquad u'(0)=B,\qquad u''(t)-au'''(t)=\theta(t),& {-}\tau\leq{t}\leq 0; \\ u'''(+\infty)=C, \end{array}\displaystyle \right . \end{aligned}$$
(1.2)
The plan of this paper is as follows: In Section 2, we state some definitions and lemmas which are needed to prove the main results. In Section 3, we show that in the presence of a pair of upper and lower solutions the problem (1.1)-(1.2) has at least one solution. Also in this section, we establish that in the presence of two pairs of upper and lower solutions the problem (1.1)-(1.2) has at least three solutions. Finally, in Section 4, we illustrate two examples which show the importance of our results.
2 Preliminaries
In this section we introduce some necessary definitions, lemmas, and preliminary results that will be used in main results which give the existence of solutions of the problem (1.1)-(1.2). First, we construct the Green’s function for the linear boundary value problem
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} u^{(4)}(t)+e(t)=0, & 0< t< {+\infty}; \\ u(0)=A,\qquad u'(0)=B,\qquad u''(t)-au'''(t)=\theta(t), & {-}\tau\leq {t}\leq0; \\ u'''(+\infty)=C. \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned}$$
(2.1)
Lemma 2.1
Let
\(e\in{L^{1}[0,+\infty)}\). Then the solution
\(u\in\mathcal{C}^{3}[-\tau,+\infty) \cap\mathcal {C}^{4}(0,+\infty)\)
of the problem (2.1) can be expressed as
where
and
$$\begin{aligned} u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \phi(t), &{-}\tau\leq{t}\leq0; \\ A+Bt+(aC+\theta(0)) \frac{t^{2}}{2}+C\frac {t^{3}}{3!}+ \int_{0}^{\infty}G(t,s)e(s)\,ds, & 0\leq {t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
(2.2)
$$\begin{aligned} G(t,s)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{a}{2}t^{2}+\frac{st^{2}}{2}-\frac{s^{2}t}{2}+\frac {s^{3}}{3!}, & 0\leq{s}\leq{t}< +\infty; \\ \frac{a}{2}t^{2}+\frac{t^{3}}{3!}, & 0\leq{t}\leq {s}< +\infty, \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned}$$
(2.3)
$$\begin{aligned} \phi(t) =&A+B t+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty}e(s)\,ds \biggr) \bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) \\ &{}+ \int _{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr) \theta(s)\,ds. \end{aligned}$$
Proof
Since \(e\in{L^{1}[0,+\infty)}\), we can integrate (2.1) from t to +∞, and use \(u'''(+\infty)=C\), to get Integrating the above equation on \([0,t]\), and applying Fubini’s theorem and using \(u''(0)-au'''(0)=\theta(0)\), we obtain Again integrating (2.4) twice on \([0,t]\), and applying Fubini’s theorem and using \(u(0)=A\) and \(u'(0)=B\), we find for \(t\in[0,+\infty)\). Now we consider the following third order linear differential equation: If the above equation is rearranged, we have and solving this linear equation on \([t,0]\), we find Next, integrating (2.5) twice on \([t,0]\), and applying Fubini’s theorem and using the following boundary conditions: we obtain for \(t\in[-\tau,0]\). This completes the proof of the lemma. □
$$\begin{aligned} u'''(t)=C+\int_{t}^{\infty}e(s)\,ds,\quad t\geq0. \end{aligned}$$
$$\begin{aligned} u''(t)=aC+a\int_{0}^{\infty}e(s)\,ds+ \theta(0)+Ct+\int_{0}^{t}s e(s)\,ds+\int _{t}^{\infty} t e(s)\,ds. \end{aligned}$$
(2.4)
$$\begin{aligned} u(t) =&A+Bt+\bigl(aC+\theta(0)\bigr) \frac{t^{2}}{2}+C\frac{t^{3}}{3!}+ \int _{0}^{t} \biggl(\frac{a}{2}t^{2}+ \frac {st^{2}}{2}-\frac{s^{2}t}{2}+\frac{s^{3}}{3!} \biggr)e(s)\,ds \\ &{}+ \int _{t}^{\infty} \biggl(\frac {a}{2}t^{2}+ \frac{t^{3}}{3!} \biggr)e(s)\,ds, \end{aligned}$$
$$u''(t)-au'''(t)= \theta(t),\quad t\in[-\tau,0]. $$
$$u'''(t)-\frac{1}{a}u''(t)=- \frac{1}{a}\theta(t),\quad t\in[-\tau,0], $$
$$\begin{aligned} u''(t)=u''(0)e^{\frac{t}{a}}+ \frac{1}{a}\int_{t}^{0}e^{\frac{t-s}{a}} \theta(s)\,ds. \end{aligned}$$
(2.5)
$$u(0)=A,\qquad u'(0)=B \quad\mbox{and}\quad u''(0)= \theta (0)+au'''(0)=\theta(0)+aC+a \int _{0}^{\infty}e(s)\,ds, $$
$$\begin{aligned} u(t) =&A+B t+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty }e(s)\,ds \biggr) \bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) \\ &{}+ \int _{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr) \theta(s)\,ds \end{aligned}$$
Remark 2.2
\(G(t,s)\) defined in (2.3) is the Green’s function of the BVP
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} -u^{(4)}(t)=0, \quad 0< t< {+\infty}; \\ u(0)=u'(0)=0,\qquad u''(0)=au'''(0),\qquad u'''(+\infty)=0. \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned}$$
Lemma 2.3
The Green’s function
\(G(t,s)\)
has the following properties:
- (1)\(G(t,s)\) is twice continuously differentiable on \([0,+\infty )\times[0,+\infty)\) and$$\frac{\partial^{3}G(t,s)}{\partial{t^{3}}} \Big|_{t=s^{+}}-\frac{\partial ^{3}G(t,s)}{\partial{t^{3}}} \Big|_{t=s^{-}}=-1; $$
- (2)
\(\frac{\partial^{i}G(t,s)}{\partial{t^{i}}}\geq 0\), \(\forall(t,s)\in[0,+\infty)\times[0,+\infty)\), for \(i=0,1,2,3\);
- (3)
\(\sup_{t\in[0,+\infty)}\frac{G(t,s)}{1+t^{3}}\leq (\frac{a\sqrt[3]{4}+1}{6} )\), \(\sup_{t\in[0,+\infty)} ( \frac{1}{1+t^{2}}\frac {\partial G(t,s)}{\partial{t}} )\leq (\frac{a+1}{2} )\), \(\sup_{t\in[0,+\infty)} (\frac{1}{1+t}\frac{\partial ^{2}G(t,s)}{\partial{t^{2}}} )\leq(a+1)\), \(\sup_{t\in[0,+\infty)}\frac{\partial^{3}G(t,s)}{\partial {t^{3}}}\leq1\).
Proof
(1) and (2) are obvious. Here we shall prove the first inequality of (3). We note that for all integers k and l
For \(s\leq{t}\), we have and for \(s\geq{t}\)
The other parts can be proved similarly. □
$$\begin{aligned} \sup_{t\in[0,+\infty)}\frac{t^{k}}{1+t^{l}} =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{l-k}{l}(\frac{k}{l-k})^{\frac{k}{l}}, & k< l; \\ 1, & k=l; \\ +\infty, & k>l. \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} \sup_{t\in[0,+\infty)}\frac{G(t,s)}{1+t^{3}} =&\sup_{t\in[0,+\infty )} \biggl(\frac{\frac{a}{2}t^{2}+\frac{st^{2}}{2}-\frac{s^{2}t}{2}+\frac {s^{3}}{6}}{1+t^{3}} \biggr) \leq\sup_{t\in[0,+\infty)} \biggl( \frac{\frac{at^{2}}{2}}{1+t^{3}}+\frac {\frac{t^{3}}{6}}{1+t^{3}} \biggr)\\ \leq&\frac{a}{2}\sup_{t\in[0,+\infty)}\frac{t^{2}}{1+t^{3}}+ \frac {1}{6}\sup_{t\in[0,+\infty)}\frac{t^{3}}{1+t^{3}}\leq \frac{a\sqrt[3]{4}+1}{6} \end{aligned}$$
$$\begin{aligned} \sup_{t\in[0,+\infty)}\frac{G(t,s)}{1+t^{3}} =&\sup_{t\in[0,+\infty )} \biggl(\frac{\frac{a}{2}t^{2}+\frac{t^{3}}{6}}{1+t^{3}} \biggr) \leq\sup_{t\in[0,+\infty)} \biggl( \frac{\frac{at^{2}}{2}}{1+t^{3}}+\frac{\frac {t^{3}}{6}}{1+t^{3}} \biggr)\\ \leq&\frac{a}{2}\sup_{t\in[0,+\infty)}\frac{t^{2}}{1+t^{3}}+ \frac {1}{6}\sup_{t\in[0,+\infty)}\frac{t^{3}}{1+t^{3}}\leq \frac{a\sqrt[3]{4}+1}{6}. \end{aligned}$$
We consider the space X defined by with the norm where
$$\begin{aligned} X =& \biggl\{ u\in{C^{3}[-\tau,+\infty)} : \sup_{t\in[0,+\infty)} \frac{|u(t)|}{1+t^{3}}< +\infty, \sup_{t\in[0,+\infty)}\frac{|u'(t)|}{1+t^{2}}< + \infty,\\ &{} \sup_{t\in[0,+\infty)}\frac{|u''(t)|}{1+t}< +\infty, \lim _{t\rightarrow{+\infty}}u'''(t) \mbox{ exists} \biggr\} \end{aligned}$$
$$\begin{aligned} \|u\|=\max\bigl\{ \|u\|_{0},\|u\|_{1},\|u\|_{2}, \|u\|^{0}_{\infty},\|u\| ^{1}_{\infty},\|u \|^{2}_{\infty},\|u\|^{3}_{\infty}\bigr\} , \end{aligned}$$
$$\begin{aligned}& \|u\|_{0}=\max_{t\in[-\tau,0]}\bigl|u(t)\bigr|, \qquad \|u \|^{0}_{\infty}=\sup_{t\in[0,+\infty)}\frac{|u(t)|}{1+t^{3}},\\& \|u\|_{1}=\max_{t\in[-\tau,0]}\bigl|u'(t)\bigr|,\qquad \|u \|^{1}_{\infty}=\sup_{t\in[0,+\infty)}\frac{|u'(t)|}{1+t^{2}},\\& \|u\|_{2}=\max_{t\in[-\tau,0]}\bigl|u''(t)\bigr|,\qquad \|u\|^{2}_{\infty}=\sup_{t\in[0,+\infty)} \frac{|u''(t)|}{1+t}, \qquad \|u\|^{3}_{\infty}=\sup_{t\in[-\tau,+\infty)}\bigl|{u'''(t)}\bigr|. \end{aligned}$$
It is clear that \((X,\|\cdot\|)\) is a Banach space. Next we define the mapping \(T:X\rightarrow\mathcal{C}^{3}[-\tau,+\infty )\cap\mathcal{C}^{4}(0,+\infty)\) by where and
$$\begin{aligned} Tu(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \psi(t), &{-}\tau\leq{t}\leq0; \\ l(t)+ \int_{0}^{\infty }G(t,s)q(s)f(s,[u(s)],[u'(s)],[u''(s)],u'''(s))\,ds, & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
(2.6)
$$\begin{aligned} \psi(t) =&A+B t+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty }q(s)f \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)\\ &{}\times \bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr)+ \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds \end{aligned}$$
$$\begin{aligned} l(t)=A+Bt+\bigl(aC+\theta(0)\bigr) \frac{t^{2}}{2}+C \frac{t^{3}}{3!}. \end{aligned}$$
(2.7)
Lemma 2.4
The mapping
\(T:X\rightarrow\mathcal{C}^{3}[-\tau,+\infty)\cap\mathcal {C}^{4}(0,+\infty)\)
in (2.6) has the following properties:
- (1)
\(Tu(0)=A\), \((Tu)'(0)=B\), \((Tu)''(t)-a(Tu)'''(t)=\theta(t)\) for \(t\in [-\tau,0]\),
- (2)
\(Tu(t)\) is three-times continuously differentiable on \(t\in[-\tau ,+\infty)\),
- (3)
\((Tu)^{(4)}(t)=-q(t)f(t,[u(t)],[u'(t)],[u''(t)],u'''(t))\), \(t\in (0,+\infty)\),
- (4)
When applying the Schäuder fixed point theorem to prove the existence result, it is necessary to show that the operator \(T_{1}\) (defined later) is completely continuous. For this, we need the following modified version of the Arzela-Ascoli lemma (see [18, 20]).
Lemma 2.5
\(M\subset{X}\)
is relatively compact if the following conditions hold:
- (1)
all functions belonging to M are uniformly bounded,
- (2)
all functions belonging to M are equi-continuous on any compact sub-interval of \([-\tau,+\infty)\),
- (3)all functions from M are equi-convergent at infinity, that is, for any \(\epsilon>0\), there exists a \(T=T(\epsilon)>0\) such that, for all \(t\geq T\) and any \(u\in{M}\),$$\begin{aligned}& \biggl|\frac{u(t)}{1+t^{3}}-\lim_{t\rightarrow+\infty}\frac{u(t)}{1+t^{3}} \biggr| < \epsilon,\qquad \biggl|\frac{u'(t)}{1+t^{2}}-\lim_{t\rightarrow+\infty }\frac{u'(t)}{1+t^{2}} \biggr|< \epsilon,\\& \biggl|\frac{u''(t)}{1+t}-\lim_{t\rightarrow+\infty}\frac {u''(t)}{1+t} \biggr|< \epsilon \quad\textit{and} \quad \bigl|u'''(t)-\lim_{t\rightarrow+\infty}u'''(t)\bigr|< \epsilon. \end{aligned}$$
Definition 2.6
A function \(\alpha\in X\cap{\mathcal {C}^{4}(0,+\infty)}\) is called a lower solution of (1.1)-(1.2) provided
Similarly, a function \(\beta\in X\cap{\mathcal{C}^{4}(0,+\infty)}\) is called an upper solution of (1.1)-(1.2) provided
Also, we say \(\alpha(\beta)\) is a strict lower solution (strict upper solution) for problem (1.1)-(1.2) if all the above inequalities are strict.
$$\begin{aligned}& \alpha^{(4)}(t)+{q(t)f\bigl(t,\bigl[\alpha(t)\bigr],\bigl[ \alpha'(t)\bigr],\bigl[\alpha''(t)\bigr], \alpha '''(t)\bigr)}\geq0,\quad 0< t< +\infty; \end{aligned}$$
(2.8)
$$\begin{aligned}& \begin{aligned} & \alpha(0)\leq A, \qquad \alpha'(0)= B,\qquad \alpha''(t)-a\alpha '''(t)\leq\theta(t), \quad -\tau\leq{t}\leq0; \\ &\alpha'''(+\infty)\leq C. \end{aligned} \end{aligned}$$
(2.9)
$$\begin{aligned}& \beta^{(4)}(t)+{q(t)f\bigl(t,\bigl[\beta(t)\bigr],\bigl[ \beta'(t)\bigr],\bigl[\beta''(t)\bigr], \beta '''(t)\bigr)}\leq0, \quad 0< t< {+ \infty}; \end{aligned}$$
(2.10)
$$\begin{aligned}& \begin{aligned} &\beta(0)\geq A, \qquad\beta'(0)= B, \qquad \beta''(t)-a\beta '''(t)\geq\theta(t),\quad -\tau\leq{t}\leq0; \\ &\beta'''(+\infty)\geq C. \end{aligned} \end{aligned}$$
(2.11)
Remark 2.7
If then on integrating (2.12) and using the boundary restrictions in Definition 2.6, we find that \(\alpha'(t)\leq\beta'(t)\), \(\alpha (t)\leq\beta(t)\) for all \(t\in[0,+\infty)\) and \(\beta'(t)\leq\alpha '(t)\), \(\alpha(t)\leq\beta(t)\) for all \(t\in[-\tau,0)\).
$$\begin{aligned} \alpha''(t)\leq\beta''(t), \quad\mbox{for every } t\in [-\tau,+\infty), \end{aligned}$$
(2.12)
Definition 2.8
Let α, β be lower and upper solutions for the problem (1.1)-(1.2) satisfying A continuous function f is said to satisfy Nagumo’s condition with respect to the pair of functions α, β if there exist positive functions φ and \(h\in{\mathcal{C}[0,+\infty)}\) such that for all \(t\in[0,+\infty)\), and \((x_{0},\ldots,x_{n})\in{[[\alpha(t)],[\beta(t)]]}\), \(y_{i}(t-\tau_{1,i}(t))\in[\alpha'(t-\tau_{1,i}(t)),\beta'(t-\tau _{1,i}(t))]\) if \(t-\tau_{1,i}(t)>0\), \(y_{i}(t-\tau_{1,i}(t))\in[\beta'(t-\tau_{1,i}(t)),\alpha'(t-\tau _{1,i}(t))]\) if \(t-\tau_{1,i}(t)\leq0\), \(0\leq i\leq n\), \(\tau_{1,0}=0\), \((z_{0},\ldots,z_{n})\in[[\alpha''(t)], [\beta''(t)]] \), \(w\in\mathbb{R}\), and
$$\alpha''(t)\leq\beta''(t), \quad\mbox{for all } t\in[-\tau,+\infty). $$
$$\begin{aligned} \bigl|f(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w)\bigr|\leq \varphi(t)h\bigl(|w|\bigr) \end{aligned}$$
$$\begin{aligned} \int_{0}^{\infty}q(s)\varphi(s)\,ds< +\infty,\qquad \int _{0}^{\infty}\frac{s}{h(s)}\,ds=+\infty. \end{aligned}$$
3 Main results
In this section we state and prove our existence results. We begin with the following lemma.
Lemma 3.1
Suppose the following conditions hold.
Then there exists a constant
\(R>0\) (depending on
α, β, h, and
C) such that every solution
u
of (1.1)-(1.2) with
and
satisfies
\(\|u\|^{3}_{\infty}<{R}\).
- (H1):
- (H2):
-
There exists a constant \(\gamma>1\) such thatwhere φ is the function in Nagumo’s condition of f.$$\begin{aligned} \sup_{0\leq{t}< +\infty}(1+t)^{\gamma}q(t)\varphi (t)< +\infty, \end{aligned}$$
$$\begin{aligned} \begin{aligned} &\alpha(t)\leq{u(t)}\leq\beta(t),\qquad \alpha'(t) \leq{u'(t)}\leq\beta'(t), \\ &\alpha''(t) \leq{u''(t)}\leq\beta''(t)\quad \textit{for all }t\in[0,+\infty) \end{aligned} \end{aligned}$$
(3.1)
$$\begin{aligned} \begin{aligned} &\alpha(t)\leq{u(t)}\leq\beta(t),\qquad \beta'(t) \leq{u'(t)}\leq\alpha'(t),\\ &\alpha''(t) \leq{u''(t)}\leq\beta''(t)\quad \textit{for all }t\in[-\tau,0) \end{aligned} \end{aligned}$$
(3.2)
Proof
We can choose \(R>\eta\) such that and where C is the nonhomogeneous boundary value, and Let u be a solution of the differential equation (1.1) satisfying (3.1) and (3.2). If \(|u'''(t)|< R\), for all \(t\in[0,+\infty)\), there is nothing to prove. If this is not true, there exists a \(t_{0}\in[0,+\infty)\) such that \(|u'''(t_{0})|\geq R\). Since \(\lim_{t\rightarrow+\infty} u'''(t)=C < R\), there exists a \(T>0\) such that Let \(t_{1}=\inf\{t\leq T: |u'''(s)|< R\mbox{ for all }s\in[t,+\infty)\}\). Then \(|u'''(t_{1})|=R\) and \(|u'''(t)|< R\) for all \(t>t_{1}\) and there exists a \(t_{2}\) such that \(|u'''(t)|\geq R\) for \(t\in[t_{2},t_{1}]\). So we have two cases to consider \(u'''(t_{1})=R\) and \(u'''(t)\geq R\) for \(t\in [t_{2},t_{1}]\), or \(u'''(t_{1})=-R\) and \(u'''(t)\leq-R\) for \(t\in[t_{2},t_{1}]\). We assume that \(u'''(t_{1})=R\) and \(u'''(t)\geq R\) for \(t\in[t_{2},t_{1}]\), then we have which is a contradiction. In the case \(u'''(t_{1})=-R\) and \(u'''(t)\leq -R\) for \(t\in[t_{2},t_{1}]\), we obtain a similar contradiction. Thus, \(|u'''(t)|< R\) for all \(t\in[0,+\infty)\). From the boundary condition (1.2) we also have for all \(t\in[-\tau,0]\). Therefore, \(|u'''(t)|< R\) for \(t\in[-\tau,0)\). To sum up, we have \(\|u\|^{3}_{\infty}<{R}\). □
$$\begin{aligned} \eta\geq\max \biggl\{ \sup_{t\in[0,+\infty)}\bigl|\beta'''(t)\bigr|, \sup_{t\in[0,+\infty)} \bigl|\alpha'''(t)\bigr|, \frac{\|\beta''-\theta\|_{0}}{a},\frac{\|\alpha''-\theta \|_{0}}{a},C \biggr\} \end{aligned}$$
$$\begin{aligned} \int_{\eta}^{R}\frac{s}{h(s)}\,ds>{M} \biggl(\sup _{t\in[0,+\infty)}\frac {\beta''(t)}{(1+t)^{\gamma}}- \inf_{t\in[0,+\infty)} \frac{\alpha''(t)}{(1+t)^{\gamma}}+\frac{\gamma {N}}{\gamma-1} \biggr), \end{aligned}$$
$$M=\sup_{t\in[0,+\infty)}(1+t)^{\gamma}q(t)\varphi(t),\qquad N=\max\bigl\{ \| \beta\|_{\infty}^{2},\|\alpha\|_{\infty}^{2} \bigr\} . $$
$$\bigl|u'''(t)\bigr|< R \quad\mbox{for all } t\geq T. $$
$$\begin{aligned} \int_{\eta}^{R}\frac{s}{h(s)}\,ds \leq&\int _{C}^{R}\frac{s}{h(s)}\,ds\\ =& -\int_{t_{1}}^{\infty}\frac{u'''(s)u^{(4)}(s)}{h(u'''(s))}\,ds\\ =& -\int_{t_{1}}^{\infty}\frac {-q(s)f(s,[u(s)],[u'(s)],[u''(s)],u'''(s))u'''(s)}{h(u'''(s))}\,ds\\ \leq&\int_{t_{1}}^{\infty}q(s)\varphi(s)u'''(s)\,ds\\ \leq& M\int_{t_{1}}^{\infty}\frac{u'''(s)}{(1+s)^{\gamma}}\,ds\\ =& M \biggl(\int_{t_{1}}^{\infty} \biggl( \frac{u''(s)}{(1+s)^{\gamma}} \biggr)'\,ds+\int_{t_{1}}^{\infty} \frac{u''(s)}{1+s}\cdot\frac{\gamma }{(1+s)^{\gamma}}\,ds \biggr)\\ \leq&{M} \biggl(\sup_{t\in[0,+\infty)}\frac{\beta''(t)}{(1+t)^{\gamma}}- \inf _{t\in[0,+\infty)}\frac{\alpha''(t)}{(1+t)^{\gamma}}+\frac{\gamma {N}}{\gamma-1} \biggr)\\ < &\int_{\eta}^{R}\frac{s}{h(s)}\,ds, \end{aligned}$$
$$-R< {-\eta}\leq\frac{\alpha''(t)-\theta(t)}{a}\leq{u'''(t)}= \frac {u''(t)-\theta(t)}{a}\leq\frac{\beta''(t)- \theta(t)}{a}\leq\eta< {R} $$
Theorem 3.2
Suppose conditions (H1) and (H2) hold. Suppose further that
Then BVP (1.1)-(1.2) has at least one solution
\(u\in X \cap{\mathcal{C}^{4}(0,+\infty)}\)
satisfying (3.1)-(3.2) and
\(\|u\|^{3}_{\infty}<{R}\).
- (H3):
-
For any fixed \(t\in[0,+\infty)\), \(y_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,\ldots,n\), when$$\begin{aligned} &\alpha\bigl(t-\tau_{0,i}(t)\bigr)\leq{x_{i}}\leq\beta \bigl(t-\tau_{0,i}(t)\bigr), \quad i=0,1,\ldots,n,\\ &f\bigl(t,x_{0},x_{1},\ldots,\alpha\bigl(t- \tau_{0,i}(t)\bigr),\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w\bigr) \\ &\quad\leq {f}(t,x_{0},x_{1},\ldots,x_{i}, \ldots,x_{n},y_{0},\ldots,y_{n},z_{0}, \ldots,z_{n},w) \\ &\quad\leq{f}\bigl(t,x_{0},x_{1},\ldots,\beta\bigl(t-\tau _{0,i}(t)\bigr),\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w\bigr). \end{aligned}$$
- (H4):
-
For any fixed \(t\in[0,+\infty)\), \(x_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,\ldots,n\) whenor when$$\alpha'\bigl(t-\tau_{1,i}(t)\bigr)\leq{y_{i}} \leq\beta'\bigl(t-\tau_{1,i}(t)\bigr),\quad t- \tau_{1,i}(t)>0, $$$$\begin{aligned}& \beta'\bigl(t-\tau_{1,i}(t)\bigr)\leq{y_{i}} \leq\alpha'\bigl(t-\tau_{1,i}(t)\bigr), \quad t- \tau_{1,i}(t)\leq0, i=0,1,\ldots,n, \\& f\bigl(t,x_{0},\ldots,x_{n},y_{0},\ldots, \alpha'\bigl(t-\tau_{1,i}(t)\bigr), \ldots,y_{n},z_{0}, \ldots,z_{n},w\bigr) \\& \quad\leq{f}(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{i},\ldots,y_{n},z_{0}, \ldots,z_{n},w) \\& \quad\leq{f}\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,\beta'\bigl(t-\tau _{1,i}(t)\bigr), \ldots,y_{n},z_{0},\ldots,z_{n},w\bigr). \end{aligned}$$
- (H5):
-
For any fixed \(t\in[0,+\infty)\), \(x_{i},y_{i},w\in{\mathbb{R}}\), \(i=0,\ldots,n\) whenwhere \(\tau_{0,0}=\tau_{1,0}=\tau_{2,0}=0\).$$\begin{aligned}& \alpha''\bigl(t-\tau_{2,i}(t)\bigr) \leq{z_{i}}\leq\beta''\bigl(t- \tau_{2,i}(t)\bigr), \quad i=0,1,\ldots,n, \\& f\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,\alpha '' \bigl(t-\tau_{2,i}(t)\bigr),\ldots,z_{n},w\bigr)\\& \quad\leq {f}(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{i}, \ldots,z_{n},w)\\& \quad\leq{f}\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,\beta'' \bigl(t-\tau _{2,i}(t)\bigr),\ldots,z_{n},w\bigr), \end{aligned}$$
- (H6):
-
$$\int_{0}^{\infty}\max\{s,1\}q(s)\,ds< +\infty,\qquad \int_{0}^{\infty}\max\{s,1\}q(s)\varphi(s)\,ds< +\infty. $$
Proof
Let R be a positive number as in Lemma 3.1 and define the auxiliary functions, as follows: where, for \(i=1,2,\ldots,n\), if \(t-\tau_{1,i}(t)>0\), if \(t-\tau_{1,i}(t)\leq0\), and We consider the modified boundary value problem with the boundary conditions (1.2). We will show that the problem (3.3)-(1.2) has at least one solution u in X. Now for \(u\in X\), we define two operators \(\widetilde{T}_{1}\), \(T_{1}\) by and where \(l(t)\) is as in (2.7) and We want to show that the operator \(T_{1}\) is completely continuous. We split the proof in the following parts:
$$F_{0},F_{1},F_{2},F_{3}:[0,+\infty) \times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}}\rightarrow{{\mathbb{R}}} $$
$$\begin{aligned}& F_{0} (t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w)\\& \quad=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} f(t,\beta,\widetilde{x}_{1},\ldots,\widetilde {x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},w), & x_{0}>\beta(t), \\ f(t,x_{0},\widetilde{x}_{1},\ldots,\widetilde {x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},w), & \alpha(t)\leq{x_{0}}\leq\beta(t), \\ f(t,\alpha,\widetilde{x}_{1},\ldots,\widetilde {x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},w), & x_{0}< {\alpha(t)}, \end{array}\displaystyle \displaystyle \right . \\& F_{1} (t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w) \\& \quad=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} F_{0}(t,x_{0},\ldots,{x}_{n},\beta',\widetilde{y}_{1},\ldots,\widetilde {y}_{n},z_{0},\ldots,z_{n},w), & y_{0}>\beta'(t), \\ F_{0}(t,x_{0},\ldots,{x}_{n},y_{0},\widetilde{y}_{1},\ldots,\widetilde {y}_{n},z_{0},\ldots,z_{n},w), & \alpha'(t)\leq{y_{0}}\leq\beta'(t), \\ F_{0}(t,x_{0},\ldots,{x}_{n},\alpha',\widetilde{y}_{1},\ldots,\widetilde {y}_{n},z_{0},\ldots,z_{n},w), & y_{0}< {\alpha'(t)}, \end{array}\displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \right . \\& F_{2}(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{n},w) \\& \quad=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} F_{1}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},\beta'',\widetilde {z}_{1},\ldots,\widetilde{z}_{n},w)- \frac{z_{0}-\beta''}{1+|z_{0}-\beta''|}, & z_{0}>\beta''(t), \\ F_{1}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},z_{0},\widetilde {z}_{1},\ldots,\widetilde{z}_{n},w), & \alpha''(t)\leq {z_{0}}\leq\beta''(t), \\ F_{1}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},\alpha'',\widetilde {z}_{1},\ldots,\widetilde{z}_{n},w)+\frac{z_{0}-\alpha''}{1+|z_{0}-\alpha ''|},& { z_{0}}< {\alpha''(t)}, \end{array}\displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \right . \\& F_{3}(t,x_{0},x_{1},\ldots,x_{n},y_{0},y_{1}, \ldots,y_{n},z_{0},\ldots,z_{n},w) \\& \quad=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} F_{2}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},R), & w>R, \\ F_{2}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},w),& -R\leq {w}\leq{R}, \\ F_{2}(t,x_{0},\ldots,{x}_{n},y_{0},\ldots,y_{n},z_{0},\ldots,z_{n},-R),& w< {R}, \end{array}\displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} \widetilde{x}_{i}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta, & x_{i}>\beta(t-\tau_{0,i}(t)); \\ x_{i}, & \alpha(t-\tau_{0,i}(t))\leq{x_{i}}\leq\beta(t-\tau _{0,i}(t)); \\ \alpha, & x_{i}< \alpha(t-\tau_{0,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} \widetilde{y}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta', & {y}_{i}>\beta'(t-\tau_{1,i}(t)); \\ {y}_{i}, & \alpha'(t-\tau_{1,i}(t))\leq{{y}_{i}}\leq\beta '(t-\tau_{1,i}(t)); \\ \alpha', & {y}_{i}< \alpha'(t-\tau_{1,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} \widetilde{y}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \alpha', & {y}_{i}>\alpha'(t-\tau_{1,i}(t)); \\ {y}_{i}, & \beta'(t-\tau_{1,i}(t))\leq{{y}_{i}}\leq\alpha '(t-\tau_{1,i}(t)); \\ \beta', & {y}_{i}< \beta'(t-\tau_{1,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} \widetilde{z}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta'', & {z}_{i}>\beta''(t-\tau_{2,i}(t)); \\ {z}_{i}, & \alpha''(t-\tau_{2,i}(t))\leq{{z}_{i}}\leq\beta ''(t-\tau_{2,i}(t)); \\ \alpha'', & {z}_{i}< \alpha''(t-\tau_{2,i}(t)). \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} u^{(4)}(t)+q(t)F_{3}\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0,\quad 0< t< {+\infty}, \end{aligned}$$
(3.3)
$$\widetilde{T}_{1}u(t)= \int_{0}^{\infty }G(t,s)q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds $$
$$\begin{aligned} T_{1}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \psi_{1}(t), &{-}\tau\leq{t}\leq0; \\ l(t)+\widetilde{T}_{1}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
(3.4)
$$\begin{aligned} \psi_{1}(t) =& A+Bt+ \biggl(\theta(0)+ aC+a\int _{0}^{\infty }q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)\\ &{}\times \bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr)+ \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
(1) \(T_{1}:X\rightarrow{X}\) is well defined. Obviously, for any \(u\in{X}\) by direct calculation, we have and for \(t\in(0,+\infty)\), which show that \(T_{1}u(t)\in{C^{3}[-\tau,+\infty)}\). Further, we have where \(H= \max_{0\leq{s}\leq\sup_{t\in[0,+\infty )}|u'''(t)|}h(s)\). Now from (3.5) it follows that which implies Next since we also have By Lebesgue’s dominated convergent theorem, L’Hopital’s rule, and (3.6), (3.7), we obtain that is, \(\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)(t)}{1+t^{3}}=0\), and which implies that \(\sup_{t\in[0,+\infty)}\frac {|(T_{1}u)(t)|}{1+t^{3}}<+\infty\). Similarly, we have that is, \(\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}}=0\) and \(\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}}= \lim_{t\rightarrow+\infty}\frac{l'(t)}{1+t^{2}}+\lim_{t\rightarrow +\infty}\frac{(T_{1}u)'(t)}{1+t^{2}}=\frac{C}{2}\), which implies that \(\sup_{t\in[0,+\infty)}\frac {|(T_{1}u)'(t)|}{1+t^{2}}<+\infty\), that is, \(\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)''(t)}{1+t}=0\) and \(\lim_{t\rightarrow+\infty}\frac{(T_{1}u)''(t)}{1+t}= \lim_{t\rightarrow+\infty}\frac{l''(t)}{1+t}+\lim_{t\rightarrow+\infty }\frac{(\widetilde{T}_{1}u)''(t)}{1+t}=C\), which implies that \(\sup_{t\in[0,+\infty)}\frac {|(T_{1}u)''(t)|}{1+t}<+\infty\), and by (3.5) then Therefore \(T_{1}u\in{X}\).
$$\begin{aligned}& (T_{1}u)''(t)-a(T_{1}u)'''(t)= \theta(t) \quad\mbox{for } t\in [-\tau,0] \quad\mbox{and}\\& (T_{1}u)'(0)=B, \qquad(T_{1}u) (0)=A \end{aligned}$$
$$\begin{aligned}& (T_{1}u)'(t)=l'(t)+\int _{0}^{\infty}\frac{\partial{G}(t,s)}{\partial {t}}q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds,\\& (T_{1}u)''(t)=l''(t)+ \int_{0}^{\infty}\frac{\partial ^{2}{G}(t,s)}{\partial{t^{2}}}q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds,\\& (T_{1}u)'''(t)=C+\int _{t}^{\infty }q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds, \end{aligned}$$
$$\begin{aligned} &\biggl|\int_{0}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq\int _{0}^{\infty}\max\{s,1\}q(s) \bigl(H\varphi(s)+1 \bigr)\,ds< +\infty, \end{aligned}$$
(3.5)
$$\begin{aligned} \int_{1}^{\infty}sq(s) \bigl(H\varphi(s)+1\bigr)\,ds\leq \int_{0}^{\infty}\max\{s,1\} q(s) \bigl(H\varphi(s)+1 \bigr)\,ds< +\infty, \end{aligned}$$
$$\begin{aligned} \lim_{t\rightarrow{+\infty}}tq(t) \bigl(H\varphi(t)+1\bigr)=0. \end{aligned}$$
(3.6)
$$\begin{aligned} \int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds\leq \int_{t}^{\infty }sq(s) \bigl(H\varphi(s)+1\bigr)\,ds< + \infty,\quad t\geq1, \end{aligned}$$
$$\begin{aligned} \lim_{t\rightarrow{+\infty}}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0. \end{aligned}$$
(3.7)
$$\begin{aligned} & \biggl| \lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)(t)}{1+t^{3}} \biggr| \\ &\quad\leq\lim _{t\rightarrow+\infty}\int_{0}^{\infty} \frac{|G(t,s)|}{1+t^{3}}q(s)\bigl|F_{3}\bigl(s,\bigl[u(s)\bigr], \bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\bigr|\,ds\\ &\quad\leq\lim_{t\rightarrow+\infty} \biggl( \int_{0}^{\infty} \frac{|G(t,s)|}{1+t^{3}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds \biggr) \\ &\quad=\lim_{t\rightarrow+\infty} \biggl[\int_{0}^{t} \frac{(\frac {a}{2}t^{2}+\frac{st^{2}}{2}-\frac{s^{2}t}{2}+\frac {s^{3}}{6})}{1+t^{3}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds \\ &\qquad{}+\int_{t}^{\infty} \frac{(\frac{a}{2}t^{2}+\frac {t^{3}}{6})}{1+t^{3}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds \biggr] \\ &\quad=\lim_{t\rightarrow+\infty} \biggl[\int_{0}^{t} \frac{(at+st-\frac {s^{2}}{2})}{3t^{2}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds+\frac{(\frac{a}{2}t^{2}+\frac {t^{3}}{6})}{3t^{2}}q(t) \bigl(H \varphi(t)+1\bigr) \biggr] \\ &\qquad{}+\lim_{t\rightarrow+\infty} \biggl[\int_{t}^{\infty} \frac{(at+\frac {t^{2}}{2})}{3t^{2}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds-\frac{(\frac{a}{2}t^{2}+\frac {t^{3}}{6})}{3t^{2}}q(t) \bigl(H \varphi(t)+1\bigr) \biggr] \\ &\quad=\lim_{t\rightarrow+\infty}\int_{0}^{t} \frac{(at+st-\frac {s^{2}}{2})}{3t^{2}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds +\lim_{t\rightarrow+\infty} \frac{(\frac{a}{2}t+\frac {t^{2}}{6})}{3t^{2}}{t}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\qquad{}+\lim_{t\rightarrow+\infty}\int_{t}^{\infty} \frac{(at+\frac {t^{2}}{2})}{3t^{2}}q(s) \bigl(H\varphi(s)+1\bigr)\,ds- \lim_{t\rightarrow+\infty} \frac{(\frac{a}{2}t+\frac {t^{2}}{6})}{3t^{2}}{t}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\quad=\lim_{t\rightarrow+\infty}\int_{0}^{t} \biggl[\frac {(a+s)}{6t}q(s) \bigl(H\varphi(s)+1\bigr)\,ds+\frac{(at+\frac {t^{2}}{2})}{6t}q(t) \bigl(H\varphi(t)+1\bigr) \biggr] \\ &\qquad{}+\lim_{t\rightarrow+\infty} \biggl[\int_{t}^{\infty} \frac {a+t}{6t}q(s) \bigl(H\varphi(s)+1\bigr)\,ds-\frac{(at+\frac {t^{2}}{2})}{6t}q(t) \bigl(H\varphi(t)+1\bigr) \biggr] \\ &\quad=\lim_{t\rightarrow+\infty}\int_{0}^{t} \frac{(a+s)}{6t}q(s) \bigl(H\varphi (s)+1\bigr)\,ds+\lim_{t\rightarrow+\infty} \frac{(a+\frac {t}{2})}{6t}{t}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\qquad{}+\lim_{t\rightarrow+\infty}\int_{t}^{\infty} \frac {a+t}{6t}q(s) \bigl(H\varphi(s)+1\bigr)\,ds-\lim_{t\rightarrow+\infty} \frac {(a+\frac{t}{2})}{6t}{t}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\quad=\lim_{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds -\lim_{t\rightarrow+\infty}\frac{(a+t)}{6{}}{}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\quad=\lim_{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds -\lim_{t\rightarrow+\infty}\frac{(a+t)}{6{t}}{t}q(t) \bigl(H\varphi(t)+1\bigr) \\ &\quad=\frac{1}{6}\lim_{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds =0, \end{aligned}$$
$$\begin{aligned} \lim_{t\rightarrow+\infty}\frac {(T_{1}u)(t)}{1+t^{3}} =& \lim_{t\rightarrow+\infty} \frac{l(t)}{1+t^{3}}+\lim_{t\rightarrow+\infty }\frac{(\widetilde{T}_{1}u)(t)}{1+t^{3}} \\ =&\lim_{t\rightarrow+\infty}\frac{A+Bt+(a C+\theta(0))\frac {t^{2}}{2}+C\frac{t^{3}}{3!}}{1+t^{3}}=\frac{C}{6}, \end{aligned}$$
$$\begin{aligned} &\biggl|\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}} \biggr|\\ &\quad\leq \lim _{t\rightarrow+\infty} \frac{1}{1+t^{2}}\int_{0}^{\infty} \frac{\partial{G}(t,s)}{\partial {t}}q(s)\bigl|F_{3}\bigl(s,\bigl[u(s)\bigr], \bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\bigr|\,ds \\ &\quad\leq\frac{1}{2}\lim_{t\rightarrow+\infty}\int_{t}^{\infty }q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
$$\begin{aligned} \biggl|\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)''(t)}{1+t} \biggr| \leq&\lim_{t\rightarrow+\infty} \frac{1}{1+t}\int_{0}^{\infty}{ \frac{\partial^{2}{G}(t,s)}{\partial {t^{2}}}}q(s)\bigl|F_{3}\bigl(s,\bigl[u(s)\bigr], \bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\bigr|\,ds \\ \leq&\lim_{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
$$\begin{aligned} &\biggl|\lim_{t\rightarrow+\infty}\int_{t}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr|\\ &\quad\leq\lim _{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H \varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
$$\begin{aligned} \lim_{t\rightarrow+\infty}(T_{1}u)'''(t) =& \lim_{t\rightarrow+\infty } \biggl(C+\int_{t}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)\\ =&C< +\infty. \end{aligned}$$
(2) \(T_{1}:X\rightarrow{X}\) is continuous. For any convergent sequence \(u_{m}\rightarrow{u}\) in X, we have Thus the continuity of \(F_{3}\) implies that Since \(u'''_{m}(t)\rightarrow{u'''(t)}\), \(\sigma=\sup\{s_{1}:s_{1}=\sup_{t\in[0,+\infty)}|u_{m}'''(t)|, m\in{N}\}< +\infty\).
$$\begin{aligned}& u_{m}(t)\rightarrow u(t),\qquad u'_{m}(t) \rightarrow u'(t),\qquad u''_{m}(t) \rightarrow{u''(t)},\\& u'''_{m}(t) \rightarrow{u'''(t)},\quad m\rightarrow+\infty, t\in[-\tau,+\infty). \end{aligned}$$
$$\begin{aligned} &\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr)-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\rightarrow 0,\\ &\quad m\rightarrow+\infty. \end{aligned}$$
Let \(H_{1}= \max_{0 \leq s\leq\max\{\sup_{t\in[0,+\infty )}|u'''(t)|, \sigma\}}h(s)\). Then we have Thus, we find that is, as \(m\rightarrow+\infty\). that is, as \(m\rightarrow+\infty\). that is, as \(m\rightarrow+\infty\). that is, as \(m\rightarrow+\infty\). that is, as \(m\rightarrow +\infty\). that is, as \(m\rightarrow+\infty\).
$$\begin{aligned} &\int_{0}^{\infty} q(s)\bigl|\bigl(F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\bigr|\,ds \\ &\quad\leq 2\int_{0}^{\infty}q(s) \bigl(H_{1} \phi(s)+1\bigr)\,ds< +\infty. \end{aligned}$$
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|_{0}\\ &\quad=\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m}) (t)-(T_{1}u) (t)\bigr|\\ &\quad=\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty} \bigl(-a^{3}-a^{2}t+a^{3}e^{\frac {t}{a}} \bigr)q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\bigl\vert -a^{3}+a^{2}\tau+a^{3}e^{\frac{-\tau}{a}} \bigr\vert \int_{0}^{\infty }q(s)\bigl|F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u \| _{0}\rightarrow0, \end{aligned}$$
(3.8)
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|^{0}_{\infty} =& \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})(t)}{1+t^{3}}-\frac{(\widetilde {T}_{1}u)(t)}{1+t^{3}} \biggr| \\ =&\sup_{t\in[0,+\infty)} \biggl|\int_{0}^{\infty} \frac {G(t,s)}{1+t^{3}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&\int_{0}^{\infty} \biggl(\frac{a\sqrt[3]{4}+1}{6} \biggr)q(s)\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{0}_{\infty}\rightarrow0, \end{aligned}$$
(3.9)
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|_{1} =&\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m})'(t)-(T_{1}u)'(t)\bigr| \\ =&\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty} \bigl(-a^{2}+a^{2}e^{\frac {t}{a}} \bigr)q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&\bigl\vert -a^{2}+a^{2}e^{\frac{-\tau}{a}}\bigr\vert \int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|_{1}\rightarrow0, \end{aligned}$$
(3.10)
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|^{1}_{\infty}\\ &\quad= \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})'(t)}{1+t^{2}}-\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}} \biggr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\frac{1}{1+t^{2}}\int_{0}^{\infty} \frac {\partial G(t,s)}{\partial {t}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty} \biggl(\frac{a+1}{2} \biggr)q(s)\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{1}_{\infty}\rightarrow0, \end{aligned}$$
(3.11)
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|_{2} =&\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m})''(t)-(T_{1}u)''(t)\bigr| \\ =&\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty}ae^{\frac {t}{a}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&a\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|_{2}\rightarrow0, \end{aligned}$$
(3.12)
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|^{2}_{\infty}\\ &\quad= \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})''(t)}{1+t}-\frac{(\widetilde {T}_{1}u)''(t)}{1+t} \biggr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\frac{1}{1+t}\int_{0}^{\infty} \frac {\partial^{2}G(t,s)}{\partial {t^{2}}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty }(a+1)q(s)\bigl|F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{2}_{\infty}\rightarrow0, \end{aligned}$$
(3.13)
To show \(\|T_{1}u_{m}-T_{1}u\|^{3}_{\infty}\rightarrow0\), as \(m\rightarrow+\infty\), we need the following: and Hence, it follows that that is, as \(m\rightarrow+\infty\).
$$\begin{aligned} &\sup_{t\in[0,+\infty)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad= \sup_{t\in[0,+\infty)}\bigl|{(\widetilde {T}_{1}u_{m})'''(t)}-( \widetilde{T}_{1}u)'''(t)\bigr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\int_{t}^{\infty }q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s) \bigr)\\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds \end{aligned}$$
$$\begin{aligned} &\sup_{t\in[-\tau,0)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad= \sup_{t\in[-\tau,0)}\frac {1}{a}\bigl|{(T_{1}u_{m})''(t)}-(T_{1}u)''(t)\bigr|\\ &\quad\leq\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds. \end{aligned}$$
$$\begin{aligned} &\| T_{1}u_{m}-T_{1}u\|^{3}_{\infty}\\ &\quad=\sup_{t\in[-\tau,+\infty)}\bigl|{(T_{1}u_{m})'''(t)}-(T_{1}u)'''(t)\bigr|\\ &\quad\leq\sup_{t\in[0,+\infty)}\bigl|{(T_{1}u_{m})'''(t)}-T_{1}u)'''(t)\bigr|+ \sup_{t\in[-\tau,0)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad\leq 2\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{3}_{\infty}\rightarrow0, \end{aligned}$$
(3.14)
Combining (3.8)-(3.14), we find \(\|(T_{1}u_{m})-(T_{1}u)\| \rightarrow0\), as \(m\rightarrow+\infty\); so \(T_{1}:X\rightarrow{X}\) is continuous.
(3) \(T_{1}:X\rightarrow{X}\) is compact. The operator \(T_{1}\) is compact if \(T_{1}\) maps bounded subsets of X into relatively compact sets. Let K be any bounded subset of X, then \(r_{3}= \sup_{0\leq s\leq\{\sup_{t\in[0,+\infty )}|u'''(t)|,u\in K\}} h(s)<+\infty\). For any \(u\in{K}\), we have the following: which implies that where Therefore, \(T_{1}K\) is uniformly bounded. We also know that \(\psi_{1}(t)\) and \(\psi'_{1}(t)\) are continuous on \([-\tau ,0]\). Thus in view of \([-\tau,0]\) compact, \(\psi_{1}(t)\) and \(\psi'_{1}(t)\) are also uniformly continuous. Thus it follows that for \(t_{1},t_{2}\in [-\tau,0]\), further since is continuous on \([-\tau,0]\), we find Next, for \(t_{1},{t_{2}}\in[0,\varepsilon]\) with \(\varepsilon>0\) a constant, we have and Thus, \(T_{1}K\) is equi-continuous. Finally, we will show that \(T_{1}K\) is equi-convergent at infinity. In fact, when \(t>0\) we have and Hence all conditions of Lemma 2.5 are fulfilled, so \(T_{1}K\) is relatively compact. Therefore, \(T_{1}:X\rightarrow{X}\) is completely continuous.
$$\begin{aligned}& \|T_{1}u\|_{0}\leq |A|+\tau|B|+ \bigl(\bigl|a^{2}C+a \theta(0)\bigr|+\tau\|\theta\| _{0} \bigr) \bigl(-a+\tau+ae^{\frac{-\tau}{a}} \bigr) \\& \hphantom{\|T_{1}u\|_{0}\leq}{}+\bigl(-a^{3}+a^{2}\tau+a^{3}e^{\frac{- \tau}{a}}\bigr) \int_{0}^{\infty }q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \|T_{1}u\|^{0}_{\infty}\leq |A|+|B|+\bigl|aC+ \theta(0)\bigr|+C+ \biggl(\frac{a \sqrt [3]{4}+1}{6} \biggr)\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| _{1}\leq |B|+\bigl(\bigl|a^{2}C+a \theta(0)\bigr|+\tau\|\theta\| _{0}\bigr) \bigl(1-e^{\frac{-\tau}{a}} \bigr) \\& \hphantom{\bigl\| (T_{1}u)\bigr\| _{1}}{}+\bigl(a^{2}-a^{2}e^{\frac{-\tau}{a}}\bigr) \int_{0}^{\infty }q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{1}_{\infty}\leq |B|+\bigl|aC+\theta(0)\bigr|+C+ \biggl(\frac{a+2 }{2} \biggr)\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| _{2}\leq \bigl|aC+\theta(0)\bigr|+\|\theta \|_{0} +a\int_{0}^{\infty }q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{2}_{\infty}\leq \bigl|aC+\theta(0)\bigr|+C+ (a+1 )\int_{0}^{\infty}q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{3}_{\infty}\leq \frac{1}{a}\bigl|aC+ \theta(0)\bigr|+ \frac{2}{a}\| \theta\|_{0}+C +2\int _{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1 \bigr)\,ds, \end{aligned}$$
$$\begin{aligned} \|T_{1}u\| \leq&|A|+\xi|B|+\upsilon\bigl|aC+\theta (0)\bigr|+C+\gamma\|\theta \|_{0}+\chi\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \end{aligned}$$
$$\begin{aligned}& \chi=\max \biggl\{ \bigl(-a^{3}+a^{2}\tau+a^{3}e^{\frac{- \tau }{a}}\bigr), \biggl(\frac{a \sqrt[3]{4}+1}{6} \biggr),\bigl(a^{2}-a^{2}e^{\frac{- \tau}{a}}\bigr), (a+1),2 \biggr\} , \\& \upsilon=\max \biggl\{ \bigl(-a^{2}+a\tau+a^{2}e^{\frac{- \tau}{a}}\bigr),\bigl(a-ae^{\frac {-\tau}{a}}\bigr),\frac{1}{a},1 \biggr\} ,\qquad \xi=\max \{\tau,1 \}, \\& \gamma=\max \biggl\{ \bigl(-a\tau+\tau^{2}+a\tau e^{\frac{-\tau}{a}} \bigr),\bigl(\tau -\tau e^{\frac{-\tau}{a}}\bigr),1,\frac{2}{a} \biggr\} . \end{aligned}$$
$$\begin{aligned}& \bigl|T_{1}u{(t_{1})}-T_{1}u(t_{2}) \bigr|=\bigl| \psi_{1}(t_{1})-\psi_{1}(t_{2})\bigr| \rightarrow 0 \quad\mbox{as } t_{1}\rightarrow t_{2},\\& \bigl|(T_{1}u)'{(t_{1})}-(T_{1}u)'(t_{2})\bigr|=\bigl| \psi'_{1}(t_{1})-\psi '_{1}(t_{2})\bigr| \rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}, \end{aligned}$$
$$\begin{aligned} (T_{1}u)''{(t)} =&\psi''_{1}(t)\\ =& \biggl(\theta(0)+ aC+a\int_{0}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)e^{\frac{t}{a}}\\ &{}+ \frac{1}{a}\int_{t}^{0}e^{\frac{t-s}{a}} \theta(s)\,ds \end{aligned}$$
$$\bigl|(T_{1}u)''{(t_{1})}-(T_{1}u)''(t_{2})\bigr|=\bigl| \psi''_{1}(t_{1})-\psi ''_{1}(t_{2})\bigr|\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}. $$
$$\begin{aligned} & \biggl|\frac{(T_{1}u){(t_{1})}}{1+t_{1}^{3}}-\frac {(T_{1}u)(t_{2})}{1+t_{2}^{3}} \biggr|\\ &\quad= \biggl|\frac{l(t_{1})}{1+t_{1}^{3}}-\frac {l(t_{2})}{1+t_{2}^{3}}+\int_{0}^{\infty} \biggl(\frac {G(t_{1},s)}{1+t_{1}^{3}}-\frac{G(t_{2},s)}{1+t_{2}^{3}} \biggr)q(s)\\ &\qquad{}\times F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq \biggl|\frac{l(t_{1})}{1+t_{1}^{3}}-\frac {l(t_{2})}{1+t_{2}^{3}} \biggr|+\int_{0}^{\infty} \biggl|\frac {{G}(t_{1},s)}{1+t_{1}^{3}}-\frac{{G}(t_{2},s)}{1+t_{2}^{3}} \biggr| q(s) \bigl(r_{3}\varphi(s)+1 \bigr)\,ds \\ &\quad\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}, \\ &\biggl|\frac{(T_{1}u)'{(t_{1})}}{1+t_{1}^{2}}-\frac {(T_{1}u)'(t_{2})}{1+t_{2}^{2}} \biggr| \\ &\quad= \biggl|\frac{l'(t_{1})}{1+t_{1}^{2}}-\frac {l'(t_{2})}{1+t_{2}^{2}}+\int_{0}^{\infty} \biggl(\frac{\frac{\partial G(t_{1},s)}{\partial{t}}}{1+t_{1}^{2}}-\frac {\frac{\partial G(t_{2},s)}{\partial{t}}}{1+t_{2}^{2}} \biggr)q(s)\\ &\qquad{}\times F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq \biggl|\frac{l'(t_{1})}{1+t_{1}^{2}}-\frac {l'(t_{2})}{1+t_{2}^{2}} \biggr| +\int_{0}^{\infty} \biggl|\frac{\frac{\partial{G}(t_{1},s)}{\partial {t}}}{1+t_{1}^{2}}-\frac{\frac{\partial{G}(t_{2},s)}{\partial {t}}}{1+t_{2}^{2}} \biggr|q(s) \bigl(r_{3}\varphi(s)+1 \bigr)\,ds\\ &\quad\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}, \\ &\biggl|\frac{(T_{1}u)''{(t_{1})}}{1+t_{1}}-\frac {(T_{1}u)''(t_{2})}{1+t_{2}} \biggr| \\ &\quad= \biggl|\frac{l''(t_{1})}{1+t_{1}}-\frac {l''(t_{2})}{1+t_{2}}+\int_{0}^{\infty} \biggl(\frac{\frac{\partial^{2}G(t_{1},s)}{\partial {t^{2}}}}{1+t_{1}^{2}}-\frac{\frac{\partial^{2}G(t_{2},s)}{\partial }{t^{2}}}{1+t_{2}^{2}} \biggr)q(s)\\ &\qquad{}\times F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr|\\ &\quad\leq \biggl|\frac{l''(t_{1})}{1+t_{1}}-\frac {l''(t_{2})}{1+t_{2}} \biggr| +\int_{0}^{\infty} \biggl|\frac{\frac{\partial^{2}{G}(t_{1},s)}{\partial {t^{2}}}}{1+t_{1}}-\frac{\frac{\partial^{2}{G}(t_{2},s)}{\partial {t^{2}}}}{1+t_{2}} \biggr|q(s) \bigl(r_{3}\varphi(s)+1 \bigr)\,ds\\ &\quad\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}, \end{aligned}$$
$$\begin{aligned} &\bigl| {(T_{1}u)'''{(t_{1})}}-{(T_{1}u)'''(t_{2})}\bigr|\\ &\quad= \biggl|\int_{t_{1}}^{\infty} q(s)F_{3}\bigl(s, \bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds\\ &\qquad{}-\int _{t_{2}}^{\infty} q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq\int_{t_{1}}^{t_{2}}q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds \\ &\quad\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}. \end{aligned}$$
$$\begin{aligned}& \biggl|\frac{(T_{1}u){(t)}}{1+t^{3}}-\lim_{t\rightarrow{+\infty}} \frac{(T_{1}u)(t)}{1+t^{3}} \biggr| = \biggl| \frac{(T_{1}u){(t)}}{1+t^{3}}-\frac{C}{6} \biggr|\rightarrow0, \quad \mbox{as } t\rightarrow{+ \infty}, \\& \biggl|\frac{(T_{1}u)'{(t)}}{1+t^{2}}-\lim_{t\rightarrow {+\infty}}\frac{(T_{1}u)'(t)}{1+t^{2}} \biggr| = \biggl| \frac{(T_{1}u)'{(t)}}{1+t^{2}}-\frac{C}{2} \biggr|\rightarrow0, \quad\mbox{as } t\rightarrow{+ \infty}, \\& \biggl|\frac{(T_{1}u)''{(t)}}{1+t}-\lim_{t\rightarrow {+\infty}}\frac{(T_{1}u)''(t)}{1+t} \biggr| = \biggl| \frac{(T_{1}u)''{(t)}}{1+t}-C \biggr|\rightarrow0, \quad\mbox{as } t\rightarrow{+\infty}, \end{aligned}$$
$$\begin{aligned} &\Bigl|(T_{1}u)'''{(t)}-\lim _{t\rightarrow{+\infty }}(T_{1}u)'''(t)\Bigr|=\bigl|(T_{1}u)'''{(t)}-C\bigr| \leq \biggl|\int_{t}^{\infty }q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds \biggr|\rightarrow0 \\ &\quad\mbox{as } t\rightarrow{+\infty}. \end{aligned}$$
(4) \(T_{1}:X\rightarrow{X}\) has at least one fixed point. Let \(\Omega=\{u\in{X},\|u\|\leq N\}\) where For any \(u\in\Omega\), it is easy to see that \(\|T_{1}u\|\leq\Omega\), and thus \(T_{1}\Omega\subset\Omega\). The Schäuder fixed point theorem now guarantees that the operator \(T_{1}\) has at least one fixed point in Ω, which is a solution of BVP (3.3)-(1.2). Now we shall show that this solution u satisfies the inequalities (3.1) and (3.2) which in view of the definitions of \(F_{3}\), \(F_{2}\), \(F_{1}\), and \(F_{0}\) will imply that u is in fact a solution of (1.1)-(1.2). For this, we only prove that \({u''(t)}\leq\beta''(t)\), \(t\in[-\tau ,+\infty)\). A similar argument can be used to prove \({\alpha''(t)\leq u''(t)}\), \(t\in[-\tau,+\infty)\). If not true, we set \(\omega(t)=u''(t)-\beta''(t)\), then there exists \(t^{*}\in[-\tau,+\infty)\) such that \(\omega(t^{*})=\sup_{-\tau\leq{t}<+\infty}\omega(t)>0\). Obviously, if \(t^{*}=-\tau\) then \(\omega'(t^{*})\leq0\), and if \(t^{*}\in(-\tau,0]\) then \(\omega'(t^{*})=0\). However, from the boundary condition, we have \(\omega'(t^{*})= \frac{1}{a}\omega(t^{*})>0\), which gives a contraction. If \(t^{*}\in(0,+\infty)\), then we have By the definition of auxiliary functions and \(R>\sup_{t\in[0,+\infty )}|\beta'''(t)|\), we have Now if \(u''(t^{*}-\tau_{2,1}(t^{*}))>\beta''(t^{*}-\tau_{2,1}(t^{*}))\) from the definition of \(\widetilde{z}_{1}\) it follows that and if \(u''(t^{*}-\tau_{2,1}(t^{*}))\leq\beta''(t^{*}-\tau_{2,1}(t^{*}))\) from the condition (H5), we have Similarly, we consider the cases \(u''(t^{*}-\tau_{2,i}(t^{*}))>\beta''(t^{*}-\tau_{2,i}(t^{*}))\) or \(u''(t^{*}-\tau_{2,i}(t^{*}))\leq\beta''(t^{*}-\tau_{2,i}(t^{*}))\), \(i=2,3,\ldots,n\), and obtain the inequality Next, if \(u'(t^{*})>\beta'(t^{*})\) from the definition of \(F_{1}\) it follows that and if \(u'(t^{*})\leq\beta'(t^{*})\) from the definition of \(F_{1}\) and the condition (H4) we have If \(t^{*}-\tau_{1,1}(t^{*})>0\), while discussing the cases \(u'(t^{*}-\tau_{1,1}(t^{*}))>\beta'(t^{*}-\tau_{1,1}(t^{*}))\) we use the definition of \(\widetilde{y}_{1}\), and when discussing the cases \(u'(t^{*}-\tau _{1,1}(t^{*}))\leq\beta'(t^{*}-\tau_{1,1}(t^{*}))\) we use (H4), and obtain Similarly, if \(t^{*}-\tau_{1,1}(t^{*})\leq0\), while discussing the cases \(u'(t^{*}-\tau_{1,1}(t^{*}))\geq\beta'(t^{*}-\tau_{1,1}(t^{*}))\) we use (H4), while when discussing the cases \(u'(t^{*}-\tau _{1,1}(t^{*}))<\beta'(t^{*}-\tau_{1,1}(t^{*}))\) we use the definition of \(\widetilde{y}_{1}\), to again find Following exactly as above, using the definition of \(\widetilde{y}_{i}\) and (H4), we consider the cases \(i=2,\ldots,n\) to finally obtain Next, if \(u(t^{*})>\beta(t^{*})\) from the definition of \(F_{0}\) it follows that and if \(u(t^{*})\leq\beta(t^{*})\) from the definition of \(F_{0}\) and the condition (H3) we have Similarly, we use the definition of \(\widetilde{x}_{i}\) and (H3) while discussing the cases \(u(t^{*}-\tau_{0,i}(t^{*}))>\beta(t^{*}-\tau _{0,i}(t^{*}))\) or \(u(t^{*}-\tau_{0,i}(t^{*}))\leq\beta(t^{*}-\tau_{0,i}(t^{*}))\), \(i=1,\ldots,n\), to get which implies that which is a contractions.
$$\begin{aligned} N=|A|+\xi|B|+\upsilon\bigl|aC+\theta(0)\bigr|+C+\gamma\|\theta \|_{0}+\chi\int_{0}^{\infty}q(s) \bigl(H \varphi(s)+1\bigr)\,ds. \end{aligned}$$
(3.15)
$$\begin{aligned} \omega\bigl(t^{*}\bigr)>0,\qquad \omega'\bigl(t^{*} \bigr)=0,\qquad \omega''\bigl(t^{*}\bigr)\leq0. \end{aligned}$$
(3.16)
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) =& -q\bigl(t^{*} \bigr)F_{3}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \bigl[u''\bigl(t^{*}\bigr) \bigr],u'''\bigl(t^{*}\bigr) \bigr) \\ =& -q\bigl(t^{*}\bigr)F_{2}\bigl(t^{*},\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[u'\bigl(t^{*} \bigr)\bigr],\bigl[u''\bigl(t^{*}\bigr) \bigr],\beta '''\bigl(t^{*} \bigr)\bigr) \\ =& -q\bigl(t^{*}\bigr) \biggl[F_{1}\bigl(t^{*}, \bigl[u\bigl(t^{*}\bigr)\bigr],\bigl[u' \bigl(t^{*}\bigr)\bigr],\beta '' \bigl({t^{*}}\bigr),u''\bigl(t^{*}- \tau_{2,1}\bigl(t^{*}\bigr)\bigr),\ldots,\beta''' \bigl(t^{*}\bigr)\bigr) \\ &{}-\frac{u''(t^{*}) -\beta''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|} \biggr]. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) =&-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \beta''\bigl({t^{*}}\bigr), \beta''\bigl(t^{*}-\tau_{2,1} \bigl(t^{*}\bigr)\bigr),u'' \bigl(t^{*}-\tau_{2,2}\bigl(t^{*}\bigr)\bigr), \ldots,\\ &{}\beta '''\bigl(t^{*} \bigr)\bigr)+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|} \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \beta''\bigl({t^{*}}\bigr), \beta''\bigl(t^{*}-\tau_{2,1} \bigl(t^{*}\bigr)\bigr),u'' \bigl(t^{*}-\tau _{2,2}\bigl(t^{*}\bigr)\bigr), \ldots,\\ &{}\beta'''\bigl(t^{*}\bigr) \bigr) +q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta '\bigl({t^{*}}\bigr),u' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr),\ldots, \bigl[\beta''\bigl(t^{*}\bigr)\bigr],\beta '''\bigl(t^{*}\bigr)\bigr)\\ &{}+q \bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|} \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta '\bigl({t^{*}}\bigr),u' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr),\ldots, \bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta'\bigl(t^{*}\bigr),\beta ' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr), \ldots,\bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta '''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta'\bigl(t^{*}\bigr),\beta ' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr), \ldots,\bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta '''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[\beta'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*}\bigr)f \bigl(t^{*},\beta\bigl({t^{*}}\bigr),u\bigl(t^{*}- \tau _{0,1}\bigl(t^{*}\bigr)\bigr),\ldots,\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[\beta'' \bigl(t^{*}\bigr)\bigr],\beta ''' \bigl(t^{*}\bigr)\bigr)\\ &{}+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|} \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*}\bigr)f \bigl(t^{*},\beta\bigl({t^{*}}\bigr),u\bigl(t^{*}- \tau _{0,1}\bigl(t^{*}\bigr)\bigr),\ldots,\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[\beta'' \bigl(t^{*}\bigr)\bigr],\beta ''' \bigl(t^{*}\bigr)\bigr)\\ &{}+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*}\bigr)f \bigl(t^{*},\bigl[\beta\bigl(t^{*}\bigr)\bigr],\bigl[ \beta'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}, \end{aligned}$$
$$\begin{aligned} \omega''\bigl(t^{*}\bigr)\geq{q \bigl(t^{*}\bigr)}\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}>0, \end{aligned}$$
If \(t^{*}=+\infty\) then \(\omega(+\infty)= \sup_{t\in[-\tau ,+\infty)}\omega(t)>0\). From the boundary conditions, we also have \(\omega'(+\infty)=u'''(+\infty)-\beta'''(+\infty)\leq0\). But this implies that \(\omega''(+\infty)\leq0\) and \(\omega'(+\infty )=0\). However, now following as above, we find \(\omega''(+\infty)>0\), which is a contradiction. Thus, \(u''(t)\leq\beta''(t)\), \(t\in[-\tau ,+\infty)\).
Consequently, we have which on integration and using boundary conditions gives and now a further integration leads to Further, since all conditions of the Lemma 3.1 are satisfied, \(\|u\|^{3}_{\infty}< R\). Consequently, we have and hence, u is a solution of (1.1)-(1.2). □
$$\alpha''(t)\leq u''(t)\leq \beta''(t), \quad t\in[-\tau,+\infty), $$
$$\beta'(t)\leq{u'(t)}\leq\alpha'(t),\quad t \in[-\tau,0),\quad \mbox{and} \quad \alpha'(t)\leq{u'(t)}\leq \beta'(t), \quad t\in[0,+\infty) $$
$$\alpha(t)\leq{u(t)}\leq\beta(t),\quad t\in[-\tau,+\infty). $$
$$\begin{aligned} u^{(4)}(t) =&-q(t)F_{3}\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)\\ =&-q(t)f\bigl(t, \bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr) \end{aligned}$$
Theorem 3.3
Assume that there exist two pairs of upper and lower solutions
\(\beta _{k}\), \(\alpha_{k}\), \(k=1,2\)
of BVP (1.1)-(1.2), where
\(\alpha_{2}\), \(\beta_{1}\)
are strict and
and
f
satisfies Nagumo’s condition with respect to
\(\alpha_{1}\), \(\beta_{2}\). Suppose further that conditions (H2)-(H6) hold with
α
and
β
replaced by
\(\alpha_{1}\)
and
\(\beta_{2}\), respectively. Then the problem (1.1)-(1.2) has at least three solutions
\(u_{1}\), \(u_{2}\), and
\(u_{3}\)
such that
$$\begin{aligned} \begin{aligned} &\alpha^{(i)}_{2}(t) \nleq\beta^{(i)}_{1}(t),\quad i=0,1,2, \\ &\alpha^{(i)}_{1}(t)\leq\alpha^{(i)}_{2}(t)\leq\beta^{(i)}_{2}(t),\qquad \alpha^{(i)}_{1}(t)\leq\beta^{(i)}_{1}(t)\leq\beta^{(i)}_{2}(t), \quad t\in[-\tau,+\infty), i=0,2, \\ &\beta^{\prime}_{2}(t)\leq\alpha^{\prime}_{2}(t)\leq\alpha^{\prime}_{1}(t), \qquad\beta ^{\prime}_{2}(t)\leq\beta^{\prime}_{1}(t)\leq\alpha^{\prime}_{1}(t), \quad t\in[-\tau ,0), \\ &\alpha^{\prime}_{1}(t)\leq\alpha^{\prime}_{2}(t)\leq\beta^{\prime}_{2}(t), \qquad\alpha ^{\prime}_{1}(t)\leq\beta^{\prime}_{1}(t)\leq\beta^{\prime}_{2}(t),\quad t\in[0,+\infty), \end{aligned} \end{aligned}$$
(3.17)
$$\begin{aligned}& \alpha^{\prime\prime}_{k}(t)\leq{u^{\prime\prime}_{k}(t)} \leq\beta^{\prime\prime}_{k}(t),\qquad \alpha _{k}(t) \leq{u_{k}(t)}\leq\beta_{k}(t), \quad t\in[-\tau,+\infty), k=1,2, \\& \beta^{\prime}_{k}(t)\leq{u^{\prime}_{k}(t)} \leq\alpha^{\prime}_{k}(t), \quad t\in[-\tau ,0), \qquad\alpha^{\prime}_{k}(t) \leq{u^{\prime}_{k}(t)}\leq\beta^{\prime}_{k}(t),\quad t\in [0,+\infty), k=1,2, \\& {u^{(i)}_{3}(t)}\nleq\beta^{(i)}_{1}(t),\qquad {u^{(i)}_{3}(t)} \ngeq\alpha ^{(i)}_{2}(t), \quad t\in[-\tau,+\infty), i=0,1,2. \end{aligned}$$
Proof
First we define the truncated functions \(\widetilde {F}_{0}\), \(\widetilde{F}_{1}\), \(\widetilde{F}_{2}\), \(\widetilde{F}_{3}\) the same as \(F_{0}\), \(F_{1}\), \(F_{2}\), \(F_{3}\) in Theorem 3.2 with α, β replaced by \(\alpha_{1}\) and \(\beta_{2}\), respectively. Consider the modified differential equation To show that (3.18)-(1.2) has at least three solutions, we define operators \(\widetilde{T}_{2}\), \(T_{2} \) as and where \(l(t)\) is as in (2.7) and
$$\begin{aligned} u^{(4)}(t)+q(t)\widetilde {F}_{3}\bigl(t, \bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0,\quad 0\leq{t}< +\infty. \end{aligned}$$
(3.18)
$$\widetilde{T}_{2}u(t)=\int_{0}^{\infty}{G}(t,s)q(s) \widetilde {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds $$
$$\begin{aligned} T_{2}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} {\psi}_{2}(t), &-\tau\leq{t}\leq 0; \\ l(t)+\widetilde{T}_{2}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} {\psi}_{2}(t) =&A+Bt \\ &{}+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty}q(s) \widetilde {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds \biggr) \\ &{}\times\bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) + \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
As in Theorem 3.2, \(T_{2}\) is completely continuous. By using the degree theory, we will show that \(T_{2}\) has at least three fixed points which are solutions of (3.18)-(1.2). We note that R in Lemma 3.1 instead of α, β now depends on \(\alpha_{1}\), \(\beta_{2}\). Set \(\Omega_{2}=\{u\in{X},\|u\|< N\}\) where N is as in (3.15) then for any \(u\in\overline{\Omega}_{2}\), it follows that \(\|T_{2}u\|< N\), thus \(T_{2}\overline{\Omega}_{2}\subset{\Omega_{2}}\), and so we have \(\deg(I-T_{2},\Omega_{2},0)=1\). Set Since \(\alpha_{2}''\nleq\beta_{1}''\), \(\alpha^{\prime\prime}_{1}(t)\leq{\alpha ^{\prime\prime}_{2}(t)}\leq\beta^{\prime\prime}_{2}(t)\), \(\alpha^{\prime\prime}_{1}(t)\leq{\beta ^{\prime\prime}_{1}(t)}\leq\beta^{\prime\prime}_{2}(t)\), we find \(\Omega_{\alpha_{2}}\neq\emptyset\neq\Omega^{\beta_{1}}\), \(\overline{\Omega}_{\alpha_{2}}\cap\overline{\Omega^{\beta _{1}}}=\emptyset\), \(\Omega_{2}\backslash{\overline{\Omega_{\alpha_{2}}\cup\Omega^{\beta _{1}}}}\neq\emptyset\). Now since \(\alpha_{2}\), \(\beta_{1}\) are strict lower and upper solutions there is no solution in \(\partial\Omega_{\alpha_{2}}\cup\partial\Omega^{\beta_{1}}\). The additivity of degree implies that We will show that \(\deg(I-T_{2},\Omega_{\alpha_{2}},0)=\deg(I-T_{2},\Omega^{\beta _{1}},0)=1\). For this, we define new operators \(\widetilde{T}_{3}:\overline{\Omega }_{{2}}\rightarrow\overline{\Omega}_{{2}}\) and \(T_{3}:\overline{\Omega }_{{2}}\rightarrow\overline{\Omega}_{{2}}\) as and where \(l(t)\) is as in (2.7) and Here the functions \(\widehat{F}_{0}\), \(\widehat{F}_{1}\), \(\widehat{F}_{2}\), \(\widehat {F}_{3}\) are same as \(\widetilde{F}_{0}\), \(\widetilde{F}_{1}\), \(\widetilde {F}_{2}\), \(\widetilde{F}_{3}\) except that \(\alpha_{1}\) is replaced by \(\alpha_{2}\). Now similar to the proof of Theorem 3.2 we find that u is a fixed point of \(T_{3}\) only when \(\alpha''_{2}(t)\leq{u''(t)}\leq\beta''_{2}(t)\), \(t\in[-\tau,+\infty)\). Since the lower solution \(\alpha_{2}\) is strict, \(\alpha''_{2}(t)\neq u''(t)\), \(t\in(-\tau,+\infty)\). Therefore, \(u\in\Omega_{\alpha_{2}}\). Hence, it follows that Also, \(T_{3}\overline{\Omega}_{2}\subset\Omega_{2}\), so that we have Therefore, Similarly, we have and this leads to Finally, using the properties of the degree, we conclude that \(T_{2}\) has at least three fixed points which are the claimed solutions of the BVP (1.1)-(1.2). □
$$\begin{aligned}& \Omega_{\alpha_{2}}=\bigl\{ u\in{\Omega_{2}}:u''(t)> \alpha_{2}''(t), t\in[-\tau,+\infty)\bigr\} , \\& \Omega^{\beta_{1}}=\bigl\{ u\in{\Omega_{2}}:u''(t)< \beta_{1}''(t), t\in[-\tau,+\infty)\bigr\} . \end{aligned}$$
$$\begin{aligned} \deg(I-T_{2},\Omega_{{2}},0)={}&\deg\bigl(I-T_{2}, \Omega_{2}\backslash{\overline {\Omega_{\alpha_{2}}\cup \Omega^{\beta_{1}}}},0\bigr) \\ &{}+\deg(I-T_{2},\Omega_{\alpha_{2}},0)+ \deg\bigl(I-T_{2},\Omega^{\beta_{1}},0\bigr). \end{aligned}$$
$$\widetilde{T}_{3}u(t)=\int_{0}^{\infty}q(s) \widehat {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds $$
$$\begin{aligned} T_{3}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} {\psi}_{3}(t), &{-}\tau\leq{t}\leq 0; \\ l(t)+\widetilde{T}_{3}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
$$\begin{aligned} {\psi}_{3}(t) =&A+Bt \\ &{}+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty}q(s) \widehat {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds \biggr) \\ &{}\times\bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) + \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
$$\deg(I-T_{3},\Omega_{2}\backslash\overline{ \Omega}_{\alpha_{2}},0)=0. $$
$$\deg(I-T_{3},\Omega_{2},0)=1. $$
$$\begin{aligned} \deg(I-T_{2},\Omega_{\alpha_{2}},0) =&\deg (I-T_{3}, \Omega_{\alpha_{2}},0) \\ =&\deg(I-T_{3},\Omega_{\alpha_{2}},0)+\deg(I-T_{3}, \Omega_{2} \backslash\overline{\Omega}_{\alpha_{2}},0) \\ =&\deg(I-T_{3},\Omega_{2},0)=1. \end{aligned}$$
$$\deg\bigl(I-T_{2},\Omega^{\beta_{1}},0\bigr)=1, $$
$$\deg\bigl(I-T_{2},\Omega_{2}\backslash{\overline{ \Omega_{\alpha_{2}}\cup \Omega^{\beta_{1}}}},0\bigr)=-1. $$
$$u_{1}\in\Omega_{\alpha_{2}},\qquad u_{2}\in \Omega^{\beta_{1}},\qquad u_{3}\in\Omega_{2}\backslash \overline{\Omega_{\alpha_{2}}\cup\Omega ^{\beta_{1}}} $$
4 Examples
Example 4.1
Consider the fourth-order nonlinear differential equation on the half-line with deviating arguments where Clearly, (4.1) is a particular case of (1.1) with \(q(t)= \frac{1}{(1+t)^{3}}\), It follows that We consider (4.1) together with the following boundary conditions: Comparing this with (1.2), we find \(\theta(t)= \frac{4}{3}\), \(a=\frac{1}{3}\), \(A=2\), \(B=0\), \(C=0\).
$$\begin{aligned} u^{(4)}(t)+\frac {1}{(1+t)^{3}}f\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0,\quad t\in(0,+\infty), \end{aligned}$$
(4.1)
$$\begin{aligned} &f\bigl(t,\bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t)\bigr],u'''(t)\bigr)\\ &\quad=\frac{(u'''(t)-1)^{2}}{(1+t)^{4}} \bigl[\bigl(2t+u''(t)\bigr)+ \bigl(t+u''(t-1)\bigr)+ \bigl(u' ({t}/{3}-{1}/{3} )-1 \bigr)^{2}\\ &\qquad{}+ \bigl(t^{3}+u ({t}/{2} -{1}/{2} ) \bigr) \bigr]. \end{aligned}$$
$$\begin{aligned}& \bigl[u(t)\bigr]= \bigl(u(t),u (t-{t}/{2}-{1}/{2} ) \bigr),\qquad \bigl[u'(t)\bigr]= \bigl(u'(t),u' (t-{2t}/{3}-{1}/{3} ) \bigr),\\& \bigl[u''(t)\bigr]=\bigl(u''(t),u''(t-1)\bigr),\\& \tau_{2,1}(t)=1,\qquad \tau_{1,1}(t)=\frac{2t}{3}+ \frac{1}{3},\qquad \tau _{0,1}(t)=\frac{t}{2}+ \frac{1}{2}. \end{aligned}$$
$$\begin{aligned} \tau=-\min_{0\leq{j}\leq{2}}\min_{t\geq0}\bigl(t- \tau_{j,1}(t)\bigr)=1. \end{aligned}$$
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l } u(0)=2,\qquad u'(0)=0,\qquad u''(t)- \frac{1}{3}u'''(t)=\frac{4}{3}\quad\mbox{for }t\in [-1,0]; \\ u'''(+\infty)=0. \end{array}\displaystyle \right . \end{aligned}$$
(4.2)
For (4.1)-(4.2) a direct substitution shows that are upper and lower solutions such that \(\beta,\alpha\in X\cap{\mathcal {C}^{4}(0,+\infty)}\). Further, for these functions we have We also note that when \(t\in[0,+\infty)\) and the function f is continuous and satisfies Nagumo’s condition with respect to α and β, that is, Hence we can take \(\varphi(t)=12\) and \(h(w)=(w+1)^{2}\). Now if \(1<\gamma \leq3\), then and and these imply that conditions (H1), (H2), and (H6) are fulfilled. Now we will show that f satisfies conditions (H3)-(H5) of Theorem 3.2. For \(t\in[0,+\infty)\), \(y_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,1\), when since f is increasing with respect to \(x_{1}\), for \(x_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,1\), when or since f is decreasing on \([\beta'(t-\tau_{1,1}),\alpha'(t-\tau _{1,1}(t))]\) for \(t\in[0,1)\) and increasing on \([\alpha'(t-\tau _{1,1}(t)),\beta'(t-\tau_{1,1})]\) for \(t\in[1,+\infty)\) with respect to \(y_{1}\), and for \(t\in[0,+\infty)\), \(x_{i},y_{i},w\in{\mathbb{R}}\), \(i=0,1\), when since f is increasing with respect to \(z_{0}\), also when since f is increasing respect to \(z_{1}\), Theorem 3.2 now ensures that the BVP (4.1)-(4.2) has at least one solution \(u(t)\) such that and also \(\|u\|^{3}_{\infty}<{R}\) where \(R>\sqrt{\exp(192)(1+\eta^{2})}\), \(\eta \geq4\), and \(\gamma=2\).
$$\begin{aligned} \beta(t) = \frac{t^{3}}{6}+\frac{4t^{2}}{3}+\frac{11}{3},\qquad \alpha(t) = -\frac{t^{3}}{6} \end{aligned}$$
$$\begin{aligned} \alpha''(t)=-t\leq\beta''(t)=t+ \frac{8}{3},\quad t\in[-1,+\infty). \end{aligned}$$
$$\begin{aligned}& \alpha ({t}/{2}-{1}/{2} )=\frac{1}{48}-\frac{t}{16}+ \frac {t^{2}}{16} - \frac{t^{3}}{48}\leq{x_{1}}\leq\beta ({t}/{2}-{1}/{2} ) =\frac{191}{48} - \frac{29t}{48} + \frac{13t^{2}}{48} + \frac{t^{3}}{48}, \\& \beta' ({t}/{3}-{1}/{3} )=-\frac{5}{6} + \frac{7t}{9} + \frac {t^{2}}{18}\leq{y_{1}}\leq \alpha' ({t}/{3}-{1}/{3} ) =-\frac{1}{18} + \frac{t}{9} - \frac{t^{2}}{18} ,\quad t\in[0,1), \\& \alpha' ({t}/{3}-{1}/{3} )=-\frac{1}{18} + \frac{t}{9} - \frac{t^{2}}{18}\leq{y_{1}}\leq \beta' ({t}/{3}-{1}/{3} )= -\frac{5}{6} + \frac{7t}{9} + \frac{t^{2}}{18},\quad t\in[1,+\infty), \\& \alpha''(t)=-t\leq{z_{0}}\leq \beta''(t)={t+\frac{8}{3}}, \qquad \alpha''(t-1)=-t+1\leq{z_{1}}\leq \beta''(t-1)={t+\frac{5}{3}}, \end{aligned}$$
$$\begin{aligned} &\bigl\vert f(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w) \bigr\vert \\ &\quad=\biggl\vert (w-1)^{2}\frac{(2t+z_{0})+(t+z_{1})+(y_{1}-1)^{2}+(t^{3} +x_{1})}{(1+t)^{4}}\biggr\vert \\ &\quad\leq \biggl(\sup_{t\in[0,+\infty)}\frac{\frac{t^{4}}{324}+\frac {1{,}435t^{3}}{1{,}296}+\frac{871t^{2}}{1{,}296}+\frac{5{,}393t}{1{,}296}+\frac {1{,}681}{144}}{(1+t)^{4}} \biggr) \bigl(|w|+1\bigr)^{2} \\ &\quad\leq 12\bigl(|w|+1\bigr)^{2}. \end{aligned}$$
$$\sup_{t\in[0,+\infty)}(1+t)^{\gamma}\frac{12}{(1+t)^{3}}=\sup _{t\in [0,+\infty)}\frac{12}{(1+t)^{3-\gamma}}\leq12< +\infty\ $$
$$\begin{aligned}& \int_{0}^{\infty}\frac{1}{(1+s)^{3}}\,ds< +\infty, \qquad \int_{0}^{\infty}\frac{s}{(1+s)^{3}}\,ds< +\infty, \\& \int _{0}^{\infty }\frac{s}{h(s)}\,ds=\int _{0}^{\infty}\frac{s}{(s+1)^{2}}\,ds=+\infty, \end{aligned}$$
$$\alpha\bigl(t-\tau_{0,1}(t)\bigr)=\alpha ({t}/{2}-{1}/{2} )\leq {x_{1}}\leq\beta ({t}/{2}-{1}/{2} )=\beta\bigl(t-\tau_{0,1}(t) \bigr) $$
$$\begin{aligned} f\bigl(t,x_{0},\alpha({t}/{2}-{1}/{2}),y_{0},y_{1},z_{0},z_{1},w \bigr) \leq&{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w)\\ \leq&{f}\bigl(t,x_{0},\beta({t}/{2}-{1}/{2}),y_{0},y_{1},z_{0},z_{1},w \bigr), \end{aligned}$$
$$\begin{aligned} &\beta'\bigl(t-\tau_{1,1}(t)\bigr)=\beta' ({t}/{3}-{1}/{3} )\leq {y_{1}}\leq\alpha'\bigl(t- \tau_{1,1}(t)\bigr)=\alpha' ({t}/{3}-{1}/{3} ),\\ &\quad \mbox{if } t-\tau_{1,1}\leq0, t\in[0,1), \end{aligned}$$
$$\begin{aligned} &\alpha'\bigl(t-\tau_{1,1}(t)\bigr)=\alpha' ({t}/{3}-{1}/{3} )\leq {y_{1}}\leq\beta'\bigl(t- \tau_{0,1}(t)\bigr)=\beta' ({t}/{3}-{1}/{3} ),\\ &\quad \mbox{if } t-\tau_{1,1}> 0, t\in[1,+\infty), \end{aligned}$$
$$\begin{aligned} f\bigl(t,x_{0},x_{1},y_{0},\alpha'({t}/{3}-{1}/{3}),z_{0},z_{1},w \bigr) \leq&{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w) \\ \leq&{f}\bigl(t,x_{0},x_{1},y_{0}, \beta'({t}/{3}-{1}/{3}),z_{0},z_{1},w\bigr), \end{aligned}$$
$$\alpha''(t)\leq{z_{0}}\leq \beta''(t), $$
$$\begin{aligned} f\bigl(t,x_{0},x_{1},y_{0},y_{1}, \alpha''(t),z_{1},w\bigr) \leq{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w) \leq{f}\bigl(t,x_{0},x_{1},y_{0},y_{1}, \beta''(t),z_{1},w\bigr), \end{aligned}$$
$$\alpha''\bigl(t-\tau_{2,1}(t)\bigr)= \alpha''(t-1)\leq{z_{1}}\leq \beta'\bigl(t-\tau _{2,1}(t)\bigr)=\beta''(t-1), $$
$$\begin{aligned} f\bigl(t,x_{0},x_{1},y_{0},y_{1},z_{0}, \alpha''(t-1),w\bigr) \leq&{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w)\\ \leq&{f}\bigl(t,x_{0},x_{1},y_{0},y_{1},z_{0}, \beta''(t-1),w\bigr). \end{aligned}$$
$$\begin{aligned} -\frac{t^{3}}{6}\leq{u(t)}\leq\frac{t^{3}}{6}+\frac{4t^{2}}{3}+ \frac {11}{3},\quad -t\leq{u''(t)}\leq t, \mbox{for all } t\in[-1,+\infty), \end{aligned}$$
$$\begin{aligned}& \frac{t^{2}}{2}+\frac{8t}{3}\leq{u'(t)}\leq- \frac{t^{2}}{2}, \quad\mbox{for all } t\in[-1,0),\\& -\frac{t^{2}}{2} \leq{u'(t)}\leq\frac{t^{2}}{2}+\frac{8t}{3},\quad \mbox{for all } t\in[0,+\infty), \end{aligned}$$
Example 4.2
Consider the fourth-order nonlinear differential equation on the half-line with deviating arguments where Clearly, (4.3) is a particular case of (1.1) with \(q(t)= \frac{1}{(1+t)^{3}}\), It follows that We consider (4.3) together with the following boundary conditions: Comparing this with (1.2), we find \(\theta(t)= t-1\), \(a= \frac{1}{3}\), \(A= \frac{13}{2}\), \(B=4\), \(C= \frac{23}{28}\).
$$\begin{aligned} u^{(4)}(t)+\frac{1}{(1+t)^{3}}f\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0, \quad t\in(0,+\infty), \end{aligned}$$
(4.3)
$$\begin{aligned} &f\bigl(t,\bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t)\bigr],u'''(t)\bigr)\\ &\quad= \biggl(\frac {23}{28}-u'''(t) \biggr)\\ &\qquad{}+\frac{(1-(u'''(t))^{2})^{2}(\frac{9}{16} -(u'''(t))^{2})^{2}(\frac{6}{7} -u'''(t))^{2}[(1+u''(t-1))+(u'(t-\frac{1}{2})-\frac{1}{2})^{2} +(u(t-\frac{1}{3})+1)]}{(u'''(t)^{2}+1)^{4}(1+t)^{4}}. \end{aligned}$$
$$\begin{aligned}& \bigl[u(t)\bigr]= \bigl(u(t),u (t-{1}/{3} ) \bigr),\qquad \bigl[u'(t) \bigr]= \bigl(u'(t),u' (t-{1}/{2} ) \bigr),\\& \bigl[u''(t)\bigr]=\bigl(u''(t),u''(t-1) \bigr), \\& \tau_{2,1}(t)=1, \qquad\tau_{1,1}(t)={1}/{2}, \qquad\tau_{0,1}(t)={1}/{3}. \end{aligned}$$
$$\begin{aligned} \tau=-\min_{0\leq{j}\leq{2}}\min_{t\geq0}(t- \tau_{j,1})=1. \end{aligned}$$
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} u(0)= \frac{13}{2},\qquad u'(0)=4, \qquad u''(t)-2u'''(t)=t-1\quad t\in[-1,0]; \\ u'''(+\infty)= \frac{23}{28}. \end{array}\displaystyle \right . \end{aligned}$$
(4.4)
For (4.3)-(4.4) we take \(\alpha_{1}\), \(\alpha_{2}\) and \(\beta _{1}\), \(\beta_{2}\) as follows: and and by direct substitution verify that \(\alpha_{2},\beta_{1}\in X\cap {\mathcal{C}^{4}(0,+\infty)}\) are its strict lower and upper solutions, and \(\alpha_{1},\beta_{2}\in X\cap{\mathcal{C}^{4}(0,+\infty)}\) are lower and upper solutions, and satisfy the assumption (3.17).
$$\alpha_{1}(t)=-\frac{t^{3}}{6}-2t^{2}+4t,\qquad \alpha_{2}(t)=\frac{t^{3}}{8}+t^{2}+4t+6 $$
$$\beta_{1}(t)=\frac{t^{3}}{7}+4t+7, \qquad \beta_{2}(t)= \frac{t^{3}}{6}+2t^{2}+4t+7, $$
We also verify that for every \(t\in[0,+\infty)\), \(w\in\mathbb{R}\), we have Hence the function f satisfies Nagumo’s condition with \(h(w)=(w+1)^{2}\) and \(\varphi(t)=41\). Now if \(1<\gamma\leq3\), then and and these imply that conditions (H1), (H2), and (H6) are fulfilled. Now we shall show that f satisfies conditions (H3)-(H5) of Theorem 3.2. For \(t\in[0,+\infty)\), \(y_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,1\), when since f is increasing with respect to \(x_{1}\), for \(x_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,1\), when or Since f is decreasing on \([\beta'(t-\tau_{1,1}),\alpha'(t-\tau _{1,1}(t))]\) for \(t\in[0,{1}/{2})\) and increasing on \([\alpha'(t-\tau _{1,1}(t)),\beta'(t-\tau_{1,1})]\) for \(t\in[{1}/{2},+\infty)\) with respect to \(y_{1}\), and for \(t\in[0,+\infty)\), \(x_{i},y_{i},w\in{\mathbb{R}}\), \(i=0,1\), when since f is increasing with respect to \(z_{1}\), This ensures that in Theorem 3.3 all assumptions (H1)-(H6) are fulfilled. Therefore, we conclude that the problem (4.3)-(4.4) has at least three solutions.
$$\begin{aligned}& -t-3\leq z_{1}\leq t+3, \qquad {-}\frac{t^{3}}{6}-\frac{11t^{2}}{6}+ \frac{95t}{18}-\frac{251}{162}\leq x_{1}\leq \frac{t^{3}}{6}+\frac{11t^{2}}{6}+\frac{49t}{18}+\frac{953}{162}, \\& \frac{t^{2}}{2}+\frac{7t}{2}+\frac{17}{8} \leq y_{1} \leq-\frac {t^{2}}{2}-\frac{7t}{2}+\frac{47}{8},\quad t\in[0,{1}/{2}) \quad\mbox{and}\\& -\frac{t^{2}}{2}-\frac{7t}{2}+\frac{47}{8}\leq y_{1}\leq\frac {t^{2}}{2}+\frac{7t}{2}+\frac{17}{8},\quad t \in[{1}/{2},+\infty), \end{aligned}$$
$$\begin{aligned} &\bigl|f(t,x_{0},x_{1},y_{0},y_{1},z_{0}, z_{1},w)\bigr|\\ &\quad=\biggl\vert ({23}/{28}-w) +\frac{(1-w^{2})^{2}(\frac{9}{16} -w^{2})^{2}(\frac{6}{7} -w)^{2}[(1+z_{1})+(y_{1}-\frac{1}{2})^{2} +(x_{1}+1)]}{(w^{2}+1)^{4}(1+t)^{4}}\biggr\vert \\ &\quad\leq\bigl(1+|w|\bigr)+\bigl(1+|w|\bigr)^{2}\frac {\frac{206{,}185}{5{,}184} + \frac{1{,}087 t}{72} + \frac{377 t^{2}}{24} + \frac {11 t^{3}}{3} + \frac{t^{4}}{4}}{(1+t)^{4}}\\ &\quad\leq\bigl(1+|w|\bigr)^{2} \biggl[1+\sup_{t\in[0,+\infty)} \frac{\frac{206{,}185}{5{,}184} + \frac{1{,}087 t}{72} + \frac{377 t^{2}}{24} + \frac{11 t^{3}}{3} + \frac {t^{4}}{4}}{(1+t)^{4}} \biggr]\\ &\quad\leq41\bigl(1+|w|\bigr)^{2}=\varphi(t)h\bigl(|w|\bigr). \end{aligned}$$
$$\sup_{t\in[0,+\infty)}(1+t)^{\gamma}\frac{41}{(1+t)^{3}}=\sup _{t\in [0,+\infty)}\frac{41}{(1+t)^{3-\gamma}}\leq41< +\infty, $$
$$\begin{aligned}& \int_{0}^{\infty}\frac{1}{(1+s)^{3}}\,ds< +\infty, \qquad \int_{0}^{\infty}\frac{s}{(1+s)^{3}}\,ds< +\infty,\\& \int _{0}^{\infty }\frac{s}{h(s)}\,ds=\int _{0}^{\infty}\frac{s}{(s+1)^{2}}\,ds=+\infty, \end{aligned}$$
$$\alpha\bigl(t-\tau_{0,1}(t)\bigr)=\alpha(t-{1}/{3}) \leq{x_{1}}\leq\beta (t-{1}/{3})=\beta\bigl(t-\tau_{0,1}(t) \bigr), $$
$$\begin{aligned} f\bigl(t,x_{0},\alpha(t-{1}/{3}),y_{0},y_{1},z_{0},z_{1},w \bigr) \leq& f(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w) \\ \leq& f\bigl(t,x_{0},\beta(t-{1}/{3}),y_{0},y_{1},z_{0},z_{1},w \bigr), \end{aligned}$$
$$\begin{aligned} &\beta'\bigl(t-\tau_{1,1}(t)\bigr)=\beta'(t-{1}/{2}) \leq {y_{1}}\leq\alpha'\bigl(t-\tau_{1,1}(t) \bigr)=\alpha'(t-{1}/{2}), \\ &\quad\mbox{if } t-\tau_{1,1}\leq0, t \in[0,{1}/{2}), \end{aligned}$$
$$\begin{aligned} &\alpha'\bigl(t-\tau_{1,1}(t)\bigr)=\alpha'(t-{1}/{2}) \leq{y_{1}}\leq\beta'\bigl(t-\tau _{0,1}(t)\bigr)= \beta'(t-{1}/{2}), \\ &\quad\mbox{if } t-\tau_{1,1}> 0, t \in[{1}/{2},+\infty). \end{aligned}$$
$$\begin{aligned} f\bigl(t,x_{0},x_{1},y_{0},\alpha'(t-{1}/{2}),z_{0},z_{1},w \bigr) \leq&{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w)\\ \leq&{f}\bigl(t,x_{0},x_{1},y_{0}, \beta'(t-{1}/{2}),z_{0},z_{1},w\bigr), \end{aligned}$$
$$\alpha''\bigl(t-\tau_{2,1}(t)\bigr)= \alpha''(t-1)\leq{z_{1}}\leq \beta'\bigl(t-\tau _{2,1}(t)\bigr)=\beta''(t-1), $$
$$\begin{aligned} f\bigl(t,x_{0},x_{1},y_{0},y_{1},z_{0}, \alpha''(t-1),w\bigr) \leq&{f}(t,x_{0},x_{1},y_{0},y_{1},z_{0},z_{1},w)\\ \leq&{f}\bigl(t,x_{0},x_{1},y_{0},y_{1},z_{0}, \beta''(t-1),w\bigr). \end{aligned}$$
Declarations
Acknowledgements
The authors are grateful for the referees’ careful reading and comments on this paper that led to the improvement of the original manuscript. This work was done when the third author was on academic leave, visiting Texas A&M University Kingsville, Department of Mathematics. He gratefully acknowledges the financial support of The Scientific and Technological Research Council of Turkey (TUBITAK).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
Economics, Commercial and Management Sciences, Preparatory School of Oran, BP 65 CH 2 Achaba Hnifi, Technople de l’USTO, Bir El Djir, Algeria
(2)
Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, TX 78363-8202, USA
(3)
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
(4)
Department of Mathematics, Faculty of Science, Ege University, Bornova, Izmir, 35100, Turkey
(5)
Mathematics Faculty of Science, Oran University, Es-Senia, BP1524, Algeria
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