In this section we state and prove our existence results. We begin with the following lemma.
Lemma 3.1
Suppose the following conditions hold.
- (H1):
-
BVP (1.1)-(1.2) has a pair of lower and upper solutions
α, β
satisfying
$$\begin{aligned} \alpha''(t)\leq\beta''(t),\quad \textit{for }t\in[-\tau,+\infty), \end{aligned}$$
and
f
satisfies Nagumo’s condition with respect to the pair of functions
α, β.
- (H2):
-
There exists a constant
\(\gamma>1\)
such that
$$\begin{aligned} \sup_{0\leq{t}< +\infty}(1+t)^{\gamma}q(t)\varphi (t)< +\infty, \end{aligned}$$
where
φ
is the function in Nagumo’s condition of
f.
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
$$\begin{aligned} \begin{aligned} &\alpha(t)\leq{u(t)}\leq\beta(t),\qquad \alpha'(t) \leq{u'(t)}\leq\beta'(t), \\ &\alpha''(t) \leq{u''(t)}\leq\beta''(t)\quad \textit{for all }t\in[0,+\infty) \end{aligned} \end{aligned}$$
(3.1)
and
$$\begin{aligned} \begin{aligned} &\alpha(t)\leq{u(t)}\leq\beta(t),\qquad \beta'(t) \leq{u'(t)}\leq\alpha'(t),\\ &\alpha''(t) \leq{u''(t)}\leq\beta''(t)\quad \textit{for all }t\in[-\tau,0) \end{aligned} \end{aligned}$$
(3.2)
satisfies
\(\|u\|^{3}_{\infty}<{R}\).
Proof
We can choose \(R>\eta\) such that
$$\begin{aligned} \eta\geq\max \biggl\{ \sup_{t\in[0,+\infty)}\bigl|\beta'''(t)\bigr|, \sup_{t\in[0,+\infty)} \bigl|\alpha'''(t)\bigr|, \frac{\|\beta''-\theta\|_{0}}{a},\frac{\|\alpha''-\theta \|_{0}}{a},C \biggr\} \end{aligned}$$
and
$$\begin{aligned} \int_{\eta}^{R}\frac{s}{h(s)}\,ds>{M} \biggl(\sup _{t\in[0,+\infty)}\frac {\beta''(t)}{(1+t)^{\gamma}}- \inf_{t\in[0,+\infty)} \frac{\alpha''(t)}{(1+t)^{\gamma}}+\frac{\gamma {N}}{\gamma-1} \biggr), \end{aligned}$$
where C is the nonhomogeneous boundary value, and
$$M=\sup_{t\in[0,+\infty)}(1+t)^{\gamma}q(t)\varphi(t),\qquad N=\max\bigl\{ \| \beta\|_{\infty}^{2},\|\alpha\|_{\infty}^{2} \bigr\} . $$
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
$$\bigl|u'''(t)\bigr|< R \quad\mbox{for all } t\geq T. $$
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
$$\begin{aligned} \int_{\eta}^{R}\frac{s}{h(s)}\,ds \leq&\int _{C}^{R}\frac{s}{h(s)}\,ds\\ =& -\int_{t_{1}}^{\infty}\frac{u'''(s)u^{(4)}(s)}{h(u'''(s))}\,ds\\ =& -\int_{t_{1}}^{\infty}\frac {-q(s)f(s,[u(s)],[u'(s)],[u''(s)],u'''(s))u'''(s)}{h(u'''(s))}\,ds\\ \leq&\int_{t_{1}}^{\infty}q(s)\varphi(s)u'''(s)\,ds\\ \leq& M\int_{t_{1}}^{\infty}\frac{u'''(s)}{(1+s)^{\gamma}}\,ds\\ =& M \biggl(\int_{t_{1}}^{\infty} \biggl( \frac{u''(s)}{(1+s)^{\gamma}} \biggr)'\,ds+\int_{t_{1}}^{\infty} \frac{u''(s)}{1+s}\cdot\frac{\gamma }{(1+s)^{\gamma}}\,ds \biggr)\\ \leq&{M} \biggl(\sup_{t\in[0,+\infty)}\frac{\beta''(t)}{(1+t)^{\gamma}}- \inf _{t\in[0,+\infty)}\frac{\alpha''(t)}{(1+t)^{\gamma}}+\frac{\gamma {N}}{\gamma-1} \biggr)\\ < &\int_{\eta}^{R}\frac{s}{h(s)}\,ds, \end{aligned}$$
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
$$-R< {-\eta}\leq\frac{\alpha''(t)-\theta(t)}{a}\leq{u'''(t)}= \frac {u''(t)-\theta(t)}{a}\leq\frac{\beta''(t)- \theta(t)}{a}\leq\eta< {R} $$
for all \(t\in[-\tau,0]\). Therefore, \(|u'''(t)|< R\) for \(t\in[-\tau,0)\). To sum up, we have \(\|u\|^{3}_{\infty}<{R}\). □
Theorem 3.2
Suppose conditions (H1) and (H2) hold. Suppose further that
- (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
\(t\in[0,+\infty)\), \(x_{i},z_{i},w\in{\mathbb{R}}\), \(i=0,\ldots,n\)
when
$$\alpha'\bigl(t-\tau_{1,i}(t)\bigr)\leq{y_{i}} \leq\beta'\bigl(t-\tau_{1,i}(t)\bigr),\quad t- \tau_{1,i}(t)>0, $$
or when
$$\begin{aligned}& \beta'\bigl(t-\tau_{1,i}(t)\bigr)\leq{y_{i}} \leq\alpha'\bigl(t-\tau_{1,i}(t)\bigr), \quad t- \tau_{1,i}(t)\leq0, i=0,1,\ldots,n, \\& f\bigl(t,x_{0},\ldots,x_{n},y_{0},\ldots, \alpha'\bigl(t-\tau_{1,i}(t)\bigr), \ldots,y_{n},z_{0}, \ldots,z_{n},w\bigr) \\& \quad\leq{f}(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{i},\ldots,y_{n},z_{0}, \ldots,z_{n},w) \\& \quad\leq{f}\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,\beta'\bigl(t-\tau _{1,i}(t)\bigr), \ldots,y_{n},z_{0},\ldots,z_{n},w\bigr). \end{aligned}$$
- (H5):
-
For any fixed
\(t\in[0,+\infty)\), \(x_{i},y_{i},w\in{\mathbb{R}}\), \(i=0,\ldots,n\)
when
$$\begin{aligned}& \alpha''\bigl(t-\tau_{2,i}(t)\bigr) \leq{z_{i}}\leq\beta''\bigl(t- \tau_{2,i}(t)\bigr), \quad i=0,1,\ldots,n, \\& f\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,\alpha '' \bigl(t-\tau_{2,i}(t)\bigr),\ldots,z_{n},w\bigr)\\& \quad\leq {f}(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,z_{i}, \ldots,z_{n},w)\\& \quad\leq{f}\bigl(t,x_{0},\ldots,x_{n},y_{0}, \ldots,y_{n},z_{0},\ldots,\beta'' \bigl(t-\tau _{2,i}(t)\bigr),\ldots,z_{n},w\bigr), \end{aligned}$$
where
\(\tau_{0,0}=\tau_{1,0}=\tau_{2,0}=0\).
- (H6):
-
$$\int_{0}^{\infty}\max\{s,1\}q(s)\,ds< +\infty,\qquad \int_{0}^{\infty}\max\{s,1\}q(s)\varphi(s)\,ds< +\infty. $$
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
\(\|u\|^{3}_{\infty}<{R}\).
Proof
Let R be a positive number as in Lemma 3.1 and define the auxiliary functions,
$$F_{0},F_{1},F_{2},F_{3}:[0,+\infty) \times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}^{n+1}}\times{{\mathbb{R}}}\rightarrow{{\mathbb{R}}} $$
as follows:
$$\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}$$
where, for \(i=1,2,\ldots,n\),
$$\begin{aligned} \widetilde{x}_{i}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta, & x_{i}>\beta(t-\tau_{0,i}(t)); \\ x_{i}, & \alpha(t-\tau_{0,i}(t))\leq{x_{i}}\leq\beta(t-\tau _{0,i}(t)); \\ \alpha, & x_{i}< \alpha(t-\tau_{0,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
if \(t-\tau_{1,i}(t)>0\),
$$\begin{aligned} \widetilde{y}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta', & {y}_{i}>\beta'(t-\tau_{1,i}(t)); \\ {y}_{i}, & \alpha'(t-\tau_{1,i}(t))\leq{{y}_{i}}\leq\beta '(t-\tau_{1,i}(t)); \\ \alpha', & {y}_{i}< \alpha'(t-\tau_{1,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
if \(t-\tau_{1,i}(t)\leq0\),
$$\begin{aligned} \widetilde{y}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \alpha', & {y}_{i}>\alpha'(t-\tau_{1,i}(t)); \\ {y}_{i}, & \beta'(t-\tau_{1,i}(t))\leq{{y}_{i}}\leq\alpha '(t-\tau_{1,i}(t)); \\ \beta', & {y}_{i}< \beta'(t-\tau_{1,i}(t)), \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
and
$$\begin{aligned} \widetilde{z}_{i}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \beta'', & {z}_{i}>\beta''(t-\tau_{2,i}(t)); \\ {z}_{i}, & \alpha''(t-\tau_{2,i}(t))\leq{{z}_{i}}\leq\beta ''(t-\tau_{2,i}(t)); \\ \alpha'', & {z}_{i}< \alpha''(t-\tau_{2,i}(t)). \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
We consider the modified boundary value problem
$$\begin{aligned} u^{(4)}(t)+q(t)F_{3}\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0,\quad 0< t< {+\infty}, \end{aligned}$$
(3.3)
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
$$\widetilde{T}_{1}u(t)= \int_{0}^{\infty }G(t,s)q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds $$
and
$$\begin{aligned} T_{1}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \psi_{1}(t), &{-}\tau\leq{t}\leq0; \\ l(t)+\widetilde{T}_{1}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
(3.4)
where \(l(t)\) is as in (2.7) and
$$\begin{aligned} \psi_{1}(t) =& A+Bt+ \biggl(\theta(0)+ aC+a\int _{0}^{\infty }q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)\\ &{}\times \bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr)+ \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
We want to show that the operator \(T_{1}\) is completely continuous. We split the proof in the following parts:
(1) \(T_{1}:X\rightarrow{X}\) is well defined. Obviously, for any \(u\in{X}\) by direct calculation, we have
$$\begin{aligned}& (T_{1}u)''(t)-a(T_{1}u)'''(t)= \theta(t) \quad\mbox{for } t\in [-\tau,0] \quad\mbox{and}\\& (T_{1}u)'(0)=B, \qquad(T_{1}u) (0)=A \end{aligned}$$
and for \(t\in(0,+\infty)\),
$$\begin{aligned}& (T_{1}u)'(t)=l'(t)+\int _{0}^{\infty}\frac{\partial{G}(t,s)}{\partial {t}}q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds,\\& (T_{1}u)''(t)=l''(t)+ \int_{0}^{\infty}\frac{\partial ^{2}{G}(t,s)}{\partial{t^{2}}}q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds,\\& (T_{1}u)'''(t)=C+\int _{t}^{\infty }q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds, \end{aligned}$$
which show that \(T_{1}u(t)\in{C^{3}[-\tau,+\infty)}\). Further, we have
$$\begin{aligned} &\biggl|\int_{0}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq\int _{0}^{\infty}\max\{s,1\}q(s) \bigl(H\varphi(s)+1 \bigr)\,ds< +\infty, \end{aligned}$$
(3.5)
where \(H= \max_{0\leq{s}\leq\sup_{t\in[0,+\infty )}|u'''(t)|}h(s)\). Now from (3.5) it follows that
$$\begin{aligned} \int_{1}^{\infty}sq(s) \bigl(H\varphi(s)+1\bigr)\,ds\leq \int_{0}^{\infty}\max\{s,1\} q(s) \bigl(H\varphi(s)+1 \bigr)\,ds< +\infty, \end{aligned}$$
which implies
$$\begin{aligned} \lim_{t\rightarrow{+\infty}}tq(t) \bigl(H\varphi(t)+1\bigr)=0. \end{aligned}$$
(3.6)
Next since
$$\begin{aligned} \int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds\leq \int_{t}^{\infty }sq(s) \bigl(H\varphi(s)+1\bigr)\,ds< + \infty,\quad t\geq1, \end{aligned}$$
we also have
$$\begin{aligned} \lim_{t\rightarrow{+\infty}}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0. \end{aligned}$$
(3.7)
By Lebesgue’s dominated convergent theorem, L’Hopital’s rule, and (3.6), (3.7), we obtain
$$\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}$$
that is, \(\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)(t)}{1+t^{3}}=0\), and
$$\begin{aligned} \lim_{t\rightarrow+\infty}\frac {(T_{1}u)(t)}{1+t^{3}} =& \lim_{t\rightarrow+\infty} \frac{l(t)}{1+t^{3}}+\lim_{t\rightarrow+\infty }\frac{(\widetilde{T}_{1}u)(t)}{1+t^{3}} \\ =&\lim_{t\rightarrow+\infty}\frac{A+Bt+(a C+\theta(0))\frac {t^{2}}{2}+C\frac{t^{3}}{3!}}{1+t^{3}}=\frac{C}{6}, \end{aligned}$$
which implies that \(\sup_{t\in[0,+\infty)}\frac {|(T_{1}u)(t)|}{1+t^{3}}<+\infty\). Similarly, we have
$$\begin{aligned} &\biggl|\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}} \biggr|\\ &\quad\leq \lim _{t\rightarrow+\infty} \frac{1}{1+t^{2}}\int_{0}^{\infty} \frac{\partial{G}(t,s)}{\partial {t}}q(s)\bigl|F_{3}\bigl(s,\bigl[u(s)\bigr], \bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\bigr|\,ds \\ &\quad\leq\frac{1}{2}\lim_{t\rightarrow+\infty}\int_{t}^{\infty }q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
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 \(\sup_{t\in[0,+\infty)}\frac {|(T_{1}u)'(t)|}{1+t^{2}}<+\infty\),
$$\begin{aligned} \biggl|\lim_{t\rightarrow+\infty}\frac{(\widetilde {T}_{1}u)''(t)}{1+t} \biggr| \leq&\lim_{t\rightarrow+\infty} \frac{1}{1+t}\int_{0}^{\infty}{ \frac{\partial^{2}{G}(t,s)}{\partial {t^{2}}}}q(s)\bigl|F_{3}\bigl(s,\bigl[u(s)\bigr], \bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\bigr|\,ds \\ \leq&\lim_{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H\varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
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)
$$\begin{aligned} &\biggl|\lim_{t\rightarrow+\infty}\int_{t}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr|\\ &\quad\leq\lim _{t\rightarrow+\infty}\int_{t}^{\infty}q(s) \bigl(H \varphi(s)+1\bigr)\,ds=0, \end{aligned}$$
then
$$\begin{aligned} \lim_{t\rightarrow+\infty}(T_{1}u)'''(t) =& \lim_{t\rightarrow+\infty } \biggl(C+\int_{t}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)\\ =&C< +\infty. \end{aligned}$$
Therefore \(T_{1}u\in{X}\).
(2) \(T_{1}:X\rightarrow{X}\) is continuous. For any convergent sequence \(u_{m}\rightarrow{u}\) in X, we have
$$\begin{aligned}& u_{m}(t)\rightarrow u(t),\qquad u'_{m}(t) \rightarrow u'(t),\qquad u''_{m}(t) \rightarrow{u''(t)},\\& u'''_{m}(t) \rightarrow{u'''(t)},\quad m\rightarrow+\infty, t\in[-\tau,+\infty). \end{aligned}$$
Thus the continuity of \(F_{3}\) implies that
$$\begin{aligned} &\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr)-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\rightarrow 0,\\ &\quad m\rightarrow+\infty. \end{aligned}$$
Since \(u'''_{m}(t)\rightarrow{u'''(t)}\), \(\sigma=\sup\{s_{1}:s_{1}=\sup_{t\in[0,+\infty)}|u_{m}'''(t)|, m\in{N}\}< +\infty\).
Let \(H_{1}= \max_{0 \leq s\leq\max\{\sup_{t\in[0,+\infty )}|u'''(t)|, \sigma\}}h(s)\). Then we have
$$\begin{aligned} &\int_{0}^{\infty} q(s)\bigl|\bigl(F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\bigr|\,ds \\ &\quad\leq 2\int_{0}^{\infty}q(s) \bigl(H_{1} \phi(s)+1\bigr)\,ds< +\infty. \end{aligned}$$
Thus, we find
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|_{0}\\ &\quad=\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m}) (t)-(T_{1}u) (t)\bigr|\\ &\quad=\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty} \bigl(-a^{3}-a^{2}t+a^{3}e^{\frac {t}{a}} \bigr)q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\bigl\vert -a^{3}+a^{2}\tau+a^{3}e^{\frac{-\tau}{a}} \bigr\vert \int_{0}^{\infty }q(s)\bigl|F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u \| _{0}\rightarrow0, \end{aligned}$$
(3.8)
as \(m\rightarrow+\infty\).
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|^{0}_{\infty} =& \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})(t)}{1+t^{3}}-\frac{(\widetilde {T}_{1}u)(t)}{1+t^{3}} \biggr| \\ =&\sup_{t\in[0,+\infty)} \biggl|\int_{0}^{\infty} \frac {G(t,s)}{1+t^{3}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&\int_{0}^{\infty} \biggl(\frac{a\sqrt[3]{4}+1}{6} \biggr)q(s)\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{0}_{\infty}\rightarrow0, \end{aligned}$$
(3.9)
as \(m\rightarrow+\infty\).
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|_{1} =&\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m})'(t)-(T_{1}u)'(t)\bigr| \\ =&\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty} \bigl(-a^{2}+a^{2}e^{\frac {t}{a}} \bigr)q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&\bigl\vert -a^{2}+a^{2}e^{\frac{-\tau}{a}}\bigr\vert \int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|_{1}\rightarrow0, \end{aligned}$$
(3.10)
as \(m\rightarrow+\infty\).
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|^{1}_{\infty}\\ &\quad= \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})'(t)}{1+t^{2}}-\frac{(\widetilde {T}_{1}u)'(t)}{1+t^{2}} \biggr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\frac{1}{1+t^{2}}\int_{0}^{\infty} \frac {\partial G(t,s)}{\partial {t}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty} \biggl(\frac{a+1}{2} \biggr)q(s)\bigl|F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{1}_{\infty}\rightarrow0, \end{aligned}$$
(3.11)
as \(m\rightarrow+\infty\).
$$\begin{aligned} \|T_{1}u_{m}-T_{1}u\|_{2} =&\max _{t\in[-\tau ,0]}\bigl|(T_{1}u_{m})''(t)-(T_{1}u)''(t)\bigr| \\ =&\max_{t\in[-\tau,0]} \biggl|\int_{0}^{\infty}ae^{\frac {t}{a}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ \leq&a\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|_{2}\rightarrow0, \end{aligned}$$
(3.12)
as \(m\rightarrow +\infty\).
$$\begin{aligned} &\|T_{1}u_{m}-T_{1}u\|^{2}_{\infty}\\ &\quad= \sup_{t\in[0,+\infty)} \biggl|\frac {(\widetilde{T}_{1}u_{m})''(t)}{1+t}-\frac{(\widetilde {T}_{1}u)''(t)}{1+t} \biggr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\frac{1}{1+t}\int_{0}^{\infty} \frac {\partial^{2}G(t,s)}{\partial {t^{2}}}q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s) \bigr],\bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s)\bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty }(a+1)q(s)\bigl|F_{3} \bigl(s,\bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s) \bigr],\bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{2}_{\infty}\rightarrow0, \end{aligned}$$
(3.13)
as \(m\rightarrow+\infty\).
To show \(\|T_{1}u_{m}-T_{1}u\|^{3}_{\infty}\rightarrow0\), as \(m\rightarrow+\infty\), we need the following:
$$\begin{aligned} &\sup_{t\in[0,+\infty)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad= \sup_{t\in[0,+\infty)}\bigl|{(\widetilde {T}_{1}u_{m})'''(t)}-( \widetilde{T}_{1}u)'''(t)\bigr| \\ &\quad=\sup_{t\in[0,+\infty)} \biggl|\int_{t}^{\infty }q(s) \bigl(F_{3}\bigl(s,\bigl[u_{m}(s)\bigr], \bigl[u_{m}'(s)\bigr],\bigl[u_{m}''(s) \bigr],u_{m}'''(s) \bigr)\\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr)\,ds \biggr| \\ &\quad\leq\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds \end{aligned}$$
and
$$\begin{aligned} &\sup_{t\in[-\tau,0)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad= \sup_{t\in[-\tau,0)}\frac {1}{a}\bigl|{(T_{1}u_{m})''(t)}-(T_{1}u)''(t)\bigr|\\ &\quad\leq\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds. \end{aligned}$$
Hence, it follows that
$$\begin{aligned} &\| T_{1}u_{m}-T_{1}u\|^{3}_{\infty}\\ &\quad=\sup_{t\in[-\tau,+\infty)}\bigl|{(T_{1}u_{m})'''(t)}-(T_{1}u)'''(t)\bigr|\\ &\quad\leq\sup_{t\in[0,+\infty)}\bigl|{(T_{1}u_{m})'''(t)}-T_{1}u)'''(t)\bigr|+ \sup_{t\in[-\tau,0)}\bigl| (T_{1}u_{m})'''(t)-(T_{1}u)'''(t)\bigr|\\ &\quad\leq 2\int_{0}^{\infty }q(s)\bigl|F_{3}\bigl(s, \bigl[u_{m}(s)\bigr],\bigl[u_{m}'(s)\bigr], \bigl[u_{m}''(s)\bigr],u_{m}'''(s) \bigr) \\ &\qquad{}-F_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr], \bigl[u''(s)\bigr],u'''(s) \bigr)\bigr|\,ds, \end{aligned}$$
that is,
$$\begin{aligned} \| T_{1}u_{m}-T_{1}u \|^{3}_{\infty}\rightarrow0, \end{aligned}$$
(3.14)
as \(m\rightarrow+\infty\).
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:
$$\begin{aligned}& \|T_{1}u\|_{0}\leq |A|+\tau|B|+ \bigl(\bigl|a^{2}C+a \theta(0)\bigr|+\tau\|\theta\| _{0} \bigr) \bigl(-a+\tau+ae^{\frac{-\tau}{a}} \bigr) \\& \hphantom{\|T_{1}u\|_{0}\leq}{}+\bigl(-a^{3}+a^{2}\tau+a^{3}e^{\frac{- \tau}{a}}\bigr) \int_{0}^{\infty }q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \|T_{1}u\|^{0}_{\infty}\leq |A|+|B|+\bigl|aC+ \theta(0)\bigr|+C+ \biggl(\frac{a \sqrt [3]{4}+1}{6} \biggr)\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| _{1}\leq |B|+\bigl(\bigl|a^{2}C+a \theta(0)\bigr|+\tau\|\theta\| _{0}\bigr) \bigl(1-e^{\frac{-\tau}{a}} \bigr) \\& \hphantom{\bigl\| (T_{1}u)\bigr\| _{1}}{}+\bigl(a^{2}-a^{2}e^{\frac{-\tau}{a}}\bigr) \int_{0}^{\infty }q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{1}_{\infty}\leq |B|+\bigl|aC+\theta(0)\bigr|+C+ \biggl(\frac{a+2 }{2} \biggr)\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| _{2}\leq \bigl|aC+\theta(0)\bigr|+\|\theta \|_{0} +a\int_{0}^{\infty }q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{2}_{\infty}\leq \bigl|aC+\theta(0)\bigr|+C+ (a+1 )\int_{0}^{\infty}q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds, \\& \bigl\| (T_{1}u)\bigr\| ^{3}_{\infty}\leq \frac{1}{a}\bigl|aC+ \theta(0)\bigr|+ \frac{2}{a}\| \theta\|_{0}+C +2\int _{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1 \bigr)\,ds, \end{aligned}$$
which implies that
$$\begin{aligned} \|T_{1}u\| \leq&|A|+\xi|B|+\upsilon\bigl|aC+\theta (0)\bigr|+C+\gamma\|\theta \|_{0}+\chi\int_{0}^{\infty}q(s) \bigl(r_{3}\varphi(s)+1\bigr)\,ds, \end{aligned}$$
where
$$\begin{aligned}& \chi=\max \biggl\{ \bigl(-a^{3}+a^{2}\tau+a^{3}e^{\frac{- \tau }{a}}\bigr), \biggl(\frac{a \sqrt[3]{4}+1}{6} \biggr),\bigl(a^{2}-a^{2}e^{\frac{- \tau}{a}}\bigr), (a+1),2 \biggr\} , \\& \upsilon=\max \biggl\{ \bigl(-a^{2}+a\tau+a^{2}e^{\frac{- \tau}{a}}\bigr),\bigl(a-ae^{\frac {-\tau}{a}}\bigr),\frac{1}{a},1 \biggr\} ,\qquad \xi=\max \{\tau,1 \}, \\& \gamma=\max \biggl\{ \bigl(-a\tau+\tau^{2}+a\tau e^{\frac{-\tau}{a}} \bigr),\bigl(\tau -\tau e^{\frac{-\tau}{a}}\bigr),1,\frac{2}{a} \biggr\} . \end{aligned}$$
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 \(t_{1},t_{2}\in [-\tau,0]\),
$$\begin{aligned}& \bigl|T_{1}u{(t_{1})}-T_{1}u(t_{2}) \bigr|=\bigl| \psi_{1}(t_{1})-\psi_{1}(t_{2})\bigr| \rightarrow 0 \quad\mbox{as } t_{1}\rightarrow t_{2},\\& \bigl|(T_{1}u)'{(t_{1})}-(T_{1}u)'(t_{2})\bigr|=\bigl| \psi'_{1}(t_{1})-\psi '_{1}(t_{2})\bigr| \rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}, \end{aligned}$$
further since
$$\begin{aligned} (T_{1}u)''{(t)} =&\psi''_{1}(t)\\ =& \biggl(\theta(0)+ aC+a\int_{0}^{\infty }q(s)F_{3} \bigl(s,\bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr)e^{\frac{t}{a}}\\ &{}+ \frac{1}{a}\int_{t}^{0}e^{\frac{t-s}{a}} \theta(s)\,ds \end{aligned}$$
is continuous on \([-\tau,0]\), we find
$$\bigl|(T_{1}u)''{(t_{1})}-(T_{1}u)''(t_{2})\bigr|=\bigl| \psi''_{1}(t_{1})-\psi ''_{1}(t_{2})\bigr|\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}. $$
Next, for \(t_{1},{t_{2}}\in[0,\varepsilon]\) with \(\varepsilon>0\) a constant, we have
$$\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}$$
and
$$\begin{aligned} &\bigl| {(T_{1}u)'''{(t_{1})}}-{(T_{1}u)'''(t_{2})}\bigr|\\ &\quad= \biggl|\int_{t_{1}}^{\infty} q(s)F_{3}\bigl(s, \bigl[u(s)\bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds\\ &\qquad{}-\int _{t_{2}}^{\infty} q(s)F_{3}\bigl(s,\bigl[u(s) \bigr],\bigl[u'(s)\bigr],\bigl[u''(s) \bigr],u'''(s)\bigr)\,ds \biggr| \\ &\quad\leq\int_{t_{1}}^{t_{2}}q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds \\ &\quad\rightarrow0 \quad\mbox{as } t_{1}\rightarrow{t_{2}}. \end{aligned}$$
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
$$\begin{aligned}& \biggl|\frac{(T_{1}u){(t)}}{1+t^{3}}-\lim_{t\rightarrow{+\infty}} \frac{(T_{1}u)(t)}{1+t^{3}} \biggr| = \biggl| \frac{(T_{1}u){(t)}}{1+t^{3}}-\frac{C}{6} \biggr|\rightarrow0, \quad \mbox{as } t\rightarrow{+ \infty}, \\& \biggl|\frac{(T_{1}u)'{(t)}}{1+t^{2}}-\lim_{t\rightarrow {+\infty}}\frac{(T_{1}u)'(t)}{1+t^{2}} \biggr| = \biggl| \frac{(T_{1}u)'{(t)}}{1+t^{2}}-\frac{C}{2} \biggr|\rightarrow0, \quad\mbox{as } t\rightarrow{+ \infty}, \\& \biggl|\frac{(T_{1}u)''{(t)}}{1+t}-\lim_{t\rightarrow {+\infty}}\frac{(T_{1}u)''(t)}{1+t} \biggr| = \biggl| \frac{(T_{1}u)''{(t)}}{1+t}-C \biggr|\rightarrow0, \quad\mbox{as } t\rightarrow{+\infty}, \end{aligned}$$
and
$$\begin{aligned} &\Bigl|(T_{1}u)'''{(t)}-\lim _{t\rightarrow{+\infty }}(T_{1}u)'''(t)\Bigr|=\bigl|(T_{1}u)'''{(t)}-C\bigr| \leq \biggl|\int_{t}^{\infty }q(s) \bigl(r_{3} \varphi(s)+1\bigr)\,ds \biggr|\rightarrow0 \\ &\quad\mbox{as } t\rightarrow{+\infty}. \end{aligned}$$
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.
(4) \(T_{1}:X\rightarrow{X}\) has at least one fixed point. Let \(\Omega=\{u\in{X},\|u\|\leq N\}\) where
$$\begin{aligned} N=|A|+\xi|B|+\upsilon\bigl|aC+\theta(0)\bigr|+C+\gamma\|\theta \|_{0}+\chi\int_{0}^{\infty}q(s) \bigl(H \varphi(s)+1\bigr)\,ds. \end{aligned}$$
(3.15)
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
$$\begin{aligned} \omega\bigl(t^{*}\bigr)>0,\qquad \omega'\bigl(t^{*} \bigr)=0,\qquad \omega''\bigl(t^{*}\bigr)\leq0. \end{aligned}$$
(3.16)
By the definition of auxiliary functions and \(R>\sup_{t\in[0,+\infty )}|\beta'''(t)|\), we have
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) =& -q\bigl(t^{*} \bigr)F_{3}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \bigl[u''\bigl(t^{*}\bigr) \bigr],u'''\bigl(t^{*}\bigr) \bigr) \\ =& -q\bigl(t^{*}\bigr)F_{2}\bigl(t^{*},\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[u'\bigl(t^{*} \bigr)\bigr],\bigl[u''\bigl(t^{*}\bigr) \bigr],\beta '''\bigl(t^{*} \bigr)\bigr) \\ =& -q\bigl(t^{*}\bigr) \biggl[F_{1}\bigl(t^{*}, \bigl[u\bigl(t^{*}\bigr)\bigr],\bigl[u' \bigl(t^{*}\bigr)\bigr],\beta '' \bigl({t^{*}}\bigr),u''\bigl(t^{*}- \tau_{2,1}\bigl(t^{*}\bigr)\bigr),\ldots,\beta''' \bigl(t^{*}\bigr)\bigr) \\ &{}-\frac{u''(t^{*}) -\beta''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|} \biggr]. \end{aligned}$$
Now if \(u''(t^{*}-\tau_{2,1}(t^{*}))>\beta''(t^{*}-\tau_{2,1}(t^{*}))\) from the definition of \(\widetilde{z}_{1}\) it follows that
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) =&-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \beta''\bigl({t^{*}}\bigr), \beta''\bigl(t^{*}-\tau_{2,1} \bigl(t^{*}\bigr)\bigr),u'' \bigl(t^{*}-\tau_{2,2}\bigl(t^{*}\bigr)\bigr), \ldots,\\ &{}\beta '''\bigl(t^{*} \bigr)\bigr)+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|} \end{aligned}$$
and if \(u''(t^{*}-\tau_{2,1}(t^{*}))\leq\beta''(t^{*}-\tau_{2,1}(t^{*}))\) from the condition (H5), we have
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr], \beta''\bigl({t^{*}}\bigr), \beta''\bigl(t^{*}-\tau_{2,1} \bigl(t^{*}\bigr)\bigr),u'' \bigl(t^{*}-\tau _{2,2}\bigl(t^{*}\bigr)\bigr), \ldots,\\ &{}\beta'''\bigl(t^{*}\bigr) \bigr) +q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
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
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*} \bigr)F_{1}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[u'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
Next, if \(u'(t^{*})>\beta'(t^{*})\) from the definition of \(F_{1}\) it follows that
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta '\bigl({t^{*}}\bigr),u' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr),\ldots, \bigl[\beta''\bigl(t^{*}\bigr)\bigr],\beta '''\bigl(t^{*}\bigr)\bigr)\\ &{}+q \bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|} \end{aligned}$$
and if \(u'(t^{*})\leq\beta'(t^{*})\) from the definition of \(F_{1}\) and the condition (H4) we have
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta '\bigl({t^{*}}\bigr),u' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr),\ldots, \bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
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
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta'\bigl(t^{*}\bigr),\beta ' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr), \ldots,\bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta '''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
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
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\beta'\bigl(t^{*}\bigr),\beta ' \bigl(t^{*}-\tau_{1,1}\bigl(t^{*}\bigr)\bigr), \ldots,\bigl[\beta''\bigl(t^{*}\bigr)\bigr], \beta '''\bigl(t^{*}\bigr) \bigr)\\ &{}+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
Following exactly as above, using the definition of \(\widetilde{y}_{i}\) and (H4), we consider the cases \(i=2,\ldots,n\) to finally obtain
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*} \bigr)F_{0}\bigl(t^{*},\bigl[u\bigl(t^{*}\bigr) \bigr],\bigl[\beta'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}. \end{aligned}$$
Next, if \(u(t^{*})>\beta(t^{*})\) from the definition of \(F_{0}\) it follows that
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*}\bigr)f \bigl(t^{*},\beta\bigl({t^{*}}\bigr),u\bigl(t^{*}- \tau _{0,1}\bigl(t^{*}\bigr)\bigr),\ldots,\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[\beta'' \bigl(t^{*}\bigr)\bigr],\beta ''' \bigl(t^{*}\bigr)\bigr)\\ &{}+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|} \end{aligned}$$
and if \(u(t^{*})\leq\beta(t^{*})\) from the definition of \(F_{0}\) and the condition (H3) we have
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr) \geq&-q\bigl(t^{*}\bigr)f \bigl(t^{*},\beta\bigl({t^{*}}\bigr),u\bigl(t^{*}- \tau _{0,1}\bigl(t^{*}\bigr)\bigr),\ldots,\bigl[u \bigl(t^{*}\bigr)\bigr],\bigl[\beta'' \bigl(t^{*}\bigr)\bigr],\beta ''' \bigl(t^{*}\bigr)\bigr)\\ &{}+q\bigl(t^{*}\bigr) \frac{u''(t^{*})-\beta''(t^{*})}{1+|u''(t^{*})-\beta ''(t^{*})|}. \end{aligned}$$
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
$$\begin{aligned} u^{(4)}\bigl(t^{*}\bigr)\geq-q\bigl(t^{*}\bigr)f \bigl(t^{*},\bigl[\beta\bigl(t^{*}\bigr)\bigr],\bigl[ \beta'\bigl(t^{*}\bigr)\bigr],\bigl[\beta ''\bigl(t^{*}\bigr)\bigr], \beta'''\bigl(t^{*}\bigr) \bigr)+q\bigl(t^{*}\bigr)\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}, \end{aligned}$$
which implies that
$$\begin{aligned} \omega''\bigl(t^{*}\bigr)\geq{q \bigl(t^{*}\bigr)}\frac{u''(t^{*})-\beta ''(t^{*})}{1+|u''(t^{*})-\beta''(t^{*})|}>0, \end{aligned}$$
which is a contractions.
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
$$\alpha''(t)\leq u''(t)\leq \beta''(t), \quad t\in[-\tau,+\infty), $$
which on integration and using boundary conditions gives
$$\beta'(t)\leq{u'(t)}\leq\alpha'(t),\quad t \in[-\tau,0),\quad \mbox{and} \quad \alpha'(t)\leq{u'(t)}\leq \beta'(t), \quad t\in[0,+\infty) $$
and now a further integration leads to
$$\alpha(t)\leq{u(t)}\leq\beta(t),\quad t\in[-\tau,+\infty). $$
Further, since all conditions of the Lemma 3.1 are satisfied, \(\|u\|^{3}_{\infty}< R\). Consequently, we have
$$\begin{aligned} u^{(4)}(t) =&-q(t)F_{3}\bigl(t,\bigl[u(t)\bigr], \bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)\\ =&-q(t)f\bigl(t, \bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr) \end{aligned}$$
and hence, u is a solution of (1.1)-(1.2). □
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
$$\begin{aligned} \begin{aligned} &\alpha^{(i)}_{2}(t) \nleq\beta^{(i)}_{1}(t),\quad i=0,1,2, \\ &\alpha^{(i)}_{1}(t)\leq\alpha^{(i)}_{2}(t)\leq\beta^{(i)}_{2}(t),\qquad \alpha^{(i)}_{1}(t)\leq\beta^{(i)}_{1}(t)\leq\beta^{(i)}_{2}(t), \quad t\in[-\tau,+\infty), i=0,2, \\ &\beta^{\prime}_{2}(t)\leq\alpha^{\prime}_{2}(t)\leq\alpha^{\prime}_{1}(t), \qquad\beta ^{\prime}_{2}(t)\leq\beta^{\prime}_{1}(t)\leq\alpha^{\prime}_{1}(t), \quad t\in[-\tau ,0), \\ &\alpha^{\prime}_{1}(t)\leq\alpha^{\prime}_{2}(t)\leq\beta^{\prime}_{2}(t), \qquad\alpha ^{\prime}_{1}(t)\leq\beta^{\prime}_{1}(t)\leq\beta^{\prime}_{2}(t),\quad t\in[0,+\infty), \end{aligned} \end{aligned}$$
(3.17)
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
$$\begin{aligned}& \alpha^{\prime\prime}_{k}(t)\leq{u^{\prime\prime}_{k}(t)} \leq\beta^{\prime\prime}_{k}(t),\qquad \alpha _{k}(t) \leq{u_{k}(t)}\leq\beta_{k}(t), \quad t\in[-\tau,+\infty), k=1,2, \\& \beta^{\prime}_{k}(t)\leq{u^{\prime}_{k}(t)} \leq\alpha^{\prime}_{k}(t), \quad t\in[-\tau ,0), \qquad\alpha^{\prime}_{k}(t) \leq{u^{\prime}_{k}(t)}\leq\beta^{\prime}_{k}(t),\quad t\in [0,+\infty), k=1,2, \\& {u^{(i)}_{3}(t)}\nleq\beta^{(i)}_{1}(t),\qquad {u^{(i)}_{3}(t)} \ngeq\alpha ^{(i)}_{2}(t), \quad t\in[-\tau,+\infty), i=0,1,2. \end{aligned}$$
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
$$\begin{aligned} u^{(4)}(t)+q(t)\widetilde {F}_{3}\bigl(t, \bigl[u(t)\bigr],\bigl[u'(t)\bigr],\bigl[u''(t) \bigr],u'''(t)\bigr)=0,\quad 0\leq{t}< +\infty. \end{aligned}$$
(3.18)
To show that (3.18)-(1.2) has at least three solutions, we define operators \(\widetilde{T}_{2}\), \(T_{2} \) as
$$\widetilde{T}_{2}u(t)=\int_{0}^{\infty}{G}(t,s)q(s) \widetilde {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds $$
and
$$\begin{aligned} T_{2}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} {\psi}_{2}(t), &-\tau\leq{t}\leq 0; \\ l(t)+\widetilde{T}_{2}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
where \(l(t)\) is as in (2.7) and
$$\begin{aligned} {\psi}_{2}(t) =&A+Bt \\ &{}+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty}q(s) \widetilde {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds \biggr) \\ &{}\times\bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) + \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
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
$$\begin{aligned}& \Omega_{\alpha_{2}}=\bigl\{ u\in{\Omega_{2}}:u''(t)> \alpha_{2}''(t), t\in[-\tau,+\infty)\bigr\} , \\& \Omega^{\beta_{1}}=\bigl\{ u\in{\Omega_{2}}:u''(t)< \beta_{1}''(t), t\in[-\tau,+\infty)\bigr\} . \end{aligned}$$
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
$$\begin{aligned} \deg(I-T_{2},\Omega_{{2}},0)={}&\deg\bigl(I-T_{2}, \Omega_{2}\backslash{\overline {\Omega_{\alpha_{2}}\cup \Omega^{\beta_{1}}}},0\bigr) \\ &{}+\deg(I-T_{2},\Omega_{\alpha_{2}},0)+ \deg\bigl(I-T_{2},\Omega^{\beta_{1}},0\bigr). \end{aligned}$$
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
$$\widetilde{T}_{3}u(t)=\int_{0}^{\infty}q(s) \widehat {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds $$
and
$$\begin{aligned} T_{3}u(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} {\psi}_{3}(t), &{-}\tau\leq{t}\leq 0; \\ l(t)+\widetilde{T}_{3}u(t), & 0\leq{t}< +\infty, \end{array}\displaystyle \displaystyle \right . \end{aligned}$$
where \(l(t)\) is as in (2.7) and
$$\begin{aligned} {\psi}_{3}(t) =&A+Bt \\ &{}+ \biggl(\theta(0)+ aC+a\int_{0}^{\infty}q(s) \widehat {F}_{3}\bigl(s,\bigl[u(s)\bigr],\bigl[u'(s) \bigr],\bigl[u''(s)\bigr],u'''(s) \bigr)\,ds \biggr) \\ &{}\times\bigl(-at-a^{2}+a^{2}e^{\frac{t}{a}} \bigr) + \int_{t}^{0} \bigl(s-a-t+ae^{\frac{t-s}{a}} \bigr)\theta(s)\,ds. \end{aligned}$$
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
$$\deg(I-T_{3},\Omega_{2}\backslash\overline{ \Omega}_{\alpha_{2}},0)=0. $$
Also, \(T_{3}\overline{\Omega}_{2}\subset\Omega_{2}\), so that we have
$$\deg(I-T_{3},\Omega_{2},0)=1. $$
Therefore,
$$\begin{aligned} \deg(I-T_{2},\Omega_{\alpha_{2}},0) =&\deg (I-T_{3}, \Omega_{\alpha_{2}},0) \\ =&\deg(I-T_{3},\Omega_{\alpha_{2}},0)+\deg(I-T_{3}, \Omega_{2} \backslash\overline{\Omega}_{\alpha_{2}},0) \\ =&\deg(I-T_{3},\Omega_{2},0)=1. \end{aligned}$$
Similarly, we have
$$\deg\bigl(I-T_{2},\Omega^{\beta_{1}},0\bigr)=1, $$
and this leads to
$$\deg\bigl(I-T_{2},\Omega_{2}\backslash{\overline{ \Omega_{\alpha_{2}}\cup \Omega^{\beta_{1}}}},0\bigr)=-1. $$
Finally, using the properties of the degree, we conclude that \(T_{2}\) has at least three fixed points
$$u_{1}\in\Omega_{\alpha_{2}},\qquad u_{2}\in \Omega^{\beta_{1}},\qquad u_{3}\in\Omega_{2}\backslash \overline{\Omega_{\alpha_{2}}\cup\Omega ^{\beta_{1}}} $$
which are the claimed solutions of the BVP (1.1)-(1.2). □
