\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 210, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/210\hfil Asymptotic behavior of solutions] {Asymptotic behavior of solutions to mixed type differential equations} \author[S. Pinelas \hfil EJDE-2014/210\hfilneg] {Sandra Pinelas} % in alphabetical order \address{Sandra Pinelas \newline Academia Militar, Departamento de Ci\^encias Exatas e Naturais, Conde Castro Guimar\~aes, Amadora, Portugal} \email{sandra.pinelas@gmail.com} \thanks{Submitted May 5, 2014. Published October 10, 2014.} \subjclass[2000]{34K06, 34D04, 47H10} \keywords{Differential equations; fixed point; asymptotic behavior} \begin{abstract} This work concerns the asymptotic behavior of solutions to the differential equation $$ \dot{x}(t)+\sum_{i=1}^{m}a_i(t)x(r_i(t))+\sum_{j=1}^{n}b_j(t)x(\tau_j(t))=0, $$ where $a_j(t)$ and $b_j(t)$ are real-valued continuous functions and $r_j(t)$ and $\tau_j(t)$ are non-negative functions such that \begin{gather*} r_i(t)\leq t,\; t\geq t_0,\quad\lim_{t\to \infty}r_i(t)=\infty,\; i=1,\dots,m;\\ \tau_j(t)\geq t,\; t\geq t_0,\quad\lim_{t\to \infty}\tau_j(t)=\infty,\; j=1,\dots,n. \end{gather*} \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In recent years, the theory of delay differential equations with advanced and retarded arguments (mixed type) has provided a natural framework for mathematical modeling of many real world phenomena, namely optimal control problems \cite{p1}, nerve conduction theory \cite{c1}, the slowing down of neutrons in nuclear reactors \cite{s1}, models for economic dynamics \cite{p1,r1} and the description of traveling waves in a spatial lattice \cite{m1,m2}. See Bellman and Cooke \cite{b1} for more applications of differential equations of mixed type. The concept of delay is related to the memory of systems, where past events influence the current behavior. The concept of advance is related to a potential future events which are known at the current time, and which could be useful for decision making. It is well known that the solutions of these types of equations cannot be obtained in closed form. In the absence of a closed form, a viable alternative is studying the qualitative behavior of solutions. As a first step, we need existence and uniqueness of solutions which can be a complicated issue for mixed type equations. In this article we study the asymptotic behavior of the advanced and retarded differential equation \begin{equation} x'(t)+\sum_{i=1}^{m}a_i(t)x(r_i(t))+\sum_{j=1}^{n}b_j(t)x(\tau_j(t))=0, \quad t\geq t_0,\label{e1} \end{equation} where $a_i(t)$ and $b_j(t)$ are real continuous functions and $r_j(t)$ and $\tau_j(t)$ are non-negative functions such that \begin{gather*} r_i(t)\leq t,\; t\geq t_0,\quad\lim_{t\to \infty}r_i(t)=\infty,\; i=1,\dots,m;\\ \tau_j(t)\geq t,\; t\geq t_0,\quad\lim_{t\to \infty}\tau_j(t)=\infty,\; j=1,\dots,n. \end{gather*} About fifteen years ago, a new technique of fixed points was developed for studying stability of delay differential equations \cite{b2}. In the present article we apply this technique to mixed type differential equation. It is possible to find in the literature some conditions to ensure the stability of a solution of a delay differential equation, but it is not easy to find conditions for stability for mixed type differential equations. We will establish necessary and sufficient conditions for all solutions of \eqref{e1} to converge to zero. In the second section we establish the main results, and in the third section we illustrate the previous results with an example. \section{Main results} Let $r_0=\inf\{r_i(s):s\geq t_0, i=1,\dots ,m\}$. Then $r_0\leq t_0$, and the initial condition for \eqref{e1} is determined by a function $\phi$, continuous on $[r_0,t_0]$, \begin{equation} \label{e2} x(t)=\phi(t)\quad \text{for } r_0\leq t\leq t_0\,. \end{equation} For short notation, we write $x_0=x(t_0)$ and $y_0=y(t_0)$. By a solution to \eqref{e1} we mean a continuous function $x:[r_0,\infty)\to\mathbb{R}$ satisfying \eqref{e2}, and differentiable on $[t_0,\infty)$ and satisfies \eqref{e1}. \begin{theorem} \label{thm1} Let $a_i(t)$ and $b_j(t)$ non-positive functions. Suppose that the inequality \begin{equation} \label{e3} y(t)\geq-\sum_{i=1}^{m}a_i(t)e^{-\int_{r_i(t)}^{t}y(s)\,ds} -\sum_{j=1}^{n}b_j(t)e^{\int_{t}^{\tau_j(t)}y(s)\,ds} \quad \text{for }t\geq t_0 \end{equation} has a nonnegative solution which is integrable on each interval $[t_0,b]$. Then \eqref{e1} has a positive solution. \end{theorem} \begin{proof} Let $y_0(t)$ be a nonnegative solution of \eqref{e3}. Define the iteration \[ y_{k+1}(t)=\begin{cases} y_k(t) & r_0\leq t\leq t_0,\\ -\sum_{i=1}^{m}a_i(t)e^{-\int_{r_i(t)}^{t}y_{k}(s)\,ds} -\sum_{j=1}^{n}b_j(t)e^{\int_{t}^{\tau_j(t)}y_{k}(s)\,ds},& t_0\leq t\\ \end{cases} \] for $k=0,1,\dots$. Then, by \eqref{e3}, we have \[ y_1(t)=-\sum_{i=1}^{m}a_i(t)e^{-\int_{r_i(t)}^{t}y_0(s)\,ds} -\sum_{j=1}^{n}b_j(t)e^{\int_{t}^{\tau_j(t)}y_0(s)\,ds}\leq y_0(t)\,. \] By induction we have $0\leq y_{k+1}(t)\leq y_{k}(t)\leq\dots\leq y_0(t)$. Hence, there exists a pointwise limit $y(t)=\lim_{k\to \infty} y_{k}(t)$. By the Lesbesgue convergence theorem, we have \[ y(t)=-\sum_{i=1}^{m}a_i(t)e^{-\int_{r_i(t)}^{t}y(s)\,ds} -\sum_{j=1}^{n}b_j(t)e^{\int_{t}^{\tau_j(t)}y(s)\,ds}. \] Then, the function \[ x(t)=\begin{cases} y(t_0)e^{\int_{t_0}^{t}y(s)\,ds} & t\geq t_0,\\ y(t) & r_0\leq t\leq t_0 \end{cases} \] is a positive solution of \eqref{e1}. \end{proof} \begin{theorem} \label{thm2} Let $a_i(t)$ and $b_j(t)$ be non-positive functions. If \[ \int_{t_0}^{\infty}\sum_{j=1}^n b_j(s)\,ds=-\infty \] and $x(t)$ is a non-oscillatory solution of \eqref{e1}, then $\lim_{t\to \infty}x(t)=\infty$. \end{theorem} \begin{proof} Since $x$ is non-oscillatory, it must be eventually positive or eventually negative. We consider only the positive case, because in the negative case we can consider $-x$ which is also a solution. Suppose that $x(t)>0$ for $t\geq t_1'$. Select $t_1\geq t_1'$ such that $t_1'\leq \inf\{r_i(s):s\geq t_1,\, i=1,\dots,m\}$. Then $x'(t)\geq0$ for $t\geq t_1$, and \begin{align*} x'(t) &= -\sum_{i=1}^{m}a_i(t)x(r_i(t))-\sum_{j=1}^{n}b_j(t)x(\tau_j(t))\\ &\geq -\sum_{j=1}^{n}b_j(t)x(\tau_j(t))\\ &\geq -x(t_1)\sum_{j=1}^{n}b_j(t), \end{align*} which implies \[ x(t)\geq-x(t_1)\int_{t_0}^{t}\sum_{j=1}^{n}b_j(s)\,ds. \] Thus, $\lim_{t\to \infty}x(t)=\infty$. \end{proof} \begin{theorem} \label{thm3} Let $a_i(t)$ and $b_j(t)$ non-negative functions. If, either \[ \int_{t_0}^{\infty}\sum_{i=1}^n a_i(s)\,ds=\infty \quad\text{or}\quad \int_{t_0}^{\infty}\sum_{j=1}^n b_j(s)\,ds=\infty\,, \] and $x(t)$ is a non-oscillatory solution of \eqref{e1}, then $\lim_{t\to \infty}x(t)=0$. \end{theorem} \begin{proof} Suppose that $x(t)>0$ for $t\geq t_1$. Select $t_1\geq t_1'$ such that $t_1'\leq \inf\{r_i(s):s\geq t_1,\, i=1,\dots,m\}$. Then $x'(t)\leq0$ for $t\geq t_1$. thus, $x(t)$ is non-increasing and positive. It must have a finite limit. If $\lim_{t\to \infty}x(t)=d>0$, then $x(t)>d$ for $t\geq t_1$, and \[ x'(t) \leq -d\Big(\sum_{i=1}^{m}a_i(s)+\sum_{j=1}^{n}b_j(t)\Big) \] which implies $\lim_{t\to \infty}x(t)=-\infty$. This contradicts to the assumption that $x(t)$ is positive, and therefore $\lim_{t\to \infty}x(t)=0$. \end{proof} Next we study the asymptotic behavior of \eqref{e1}, independently of the sign of the coefficients. In the next lemma we establish an equivalence between the differential equation \eqref{e1} and an integral equation. \begin{lemma}\label{lem1} A function $x(t)$ is a solution of \eqref{e1} and \eqref{e2} if and only if $x(t)$ is a solution of \begin{equation} \begin{aligned} \label{e4} x(t) & =x_0e^{-\int_{t_0}^{t}(A(s)+B(s))\,ds} -\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds}\\ &\quad \times \Big(\sum_{i=1}^{m}a_i(u)\int_{r_i(u)}^{u}E_x(s)\,ds -\sum_{j=1}^{n}b_j(u) \int_{u}^{\tau_j(u)}E_x(s)\,ds\Big)\,du \end{aligned} \end{equation} for $t\geq t_0$, and \eqref{e2} is satisfied. Here we use the notation \begin{gather*} A(t)=\sum_{i=1}^{m}a_i(t), \quad B(t)=\sum_{j=1}^{m}b_j(t), \\ E_x(t)=\sum_{i=1}^{m}a_i(t)x(r_i(t))+\sum_{j=1}^{n}b_j(t)x(\tau_j(t)). \end{gather*} \end{lemma} \begin{proof} Note that by \eqref{e1}, $x'(t)=-E_x(t)$ and that \begin{gather*} x(r_i(t))=x(t)-\int_{r_i(t)}^{t}x'(u)\,du =x(t)+\int_{r_i(t)}^{t} E_x(u)\,du,\\ x(\tau_j(t))=x(t)+\int_t^{\tau_j(t)} x'(u)\,du =x(t)-\int_t^{\tau_j(t)} E_x(u)\,du\,. \end{gather*} Then \eqref{e1} can be re-written as \[ x'(t) +\sum_{i=1}^{m}a_i(t)\Big(x(t)-\int_{r_i(t)}^{t}\dot{x}(s)\,ds\Big) +\sum_{j=1}^{n}b_j(t)\Big(x(t)-\int_{\tau_j(t)}^{t}\dot{x}(s)\,ds\Big)=0 \] which is equivalent to \begin{equation} \label{e6} x'(t)+(A(t)+B(t))x(t) =-\sum_{i=1}^{m}a_i(t)\int_{r_i(t)}^{t}E_x(s)\,ds +\sum_{j=1}^{n}b_j(t)\int_{t}^{\tau_j(t)}E_x(s)\,ds. \end{equation} Multiplying both sides by the integrating factor $\exp(\int_{t_0}^{t}(A(s)+B(s))\,ds)$, we obtain differential equation equivalent to the one above: \begin{align*} &\frac{d}{dt}\Big(e^{\int_{t_0}^{t}(A(s)+B(s))\,ds}x(t)\Big)\\ &=-e^{\int_{t_0}^{t}(A(s)+B(s))\,ds} \Big(\sum_{i=1}^{m}a_i(t)\int_{r_i(t)}^{t}E_x(s)\,ds -\sum_{j=1}^{n}b_j(t)\int_{t}^{\tau_j(t)}E_x(s)\,ds\Big). \end{align*} Integrating from $t_0$ to $t$, we obtain \eqref{e4}. Now starting from \eqref{e4} differentiate with respect to $t$, and retrace the steps above to obtain \eqref{e1}. The proof is complete. \end{proof} \begin{theorem}\label{thm4} Assume that there exists a constant $c$ such that \begin{equation} \label{e10} \begin{aligned} &\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \Big( \sum_{i=1}^{m}|a_i(u)| \int_{r_i(u)}^{u}\Big(\sum_{k=1}^{m}|a_{k}(s)| +\sum_{\ell=1}^{n}|b_{\ell}(s)|\Big)\,ds \\ &+\sum_{j=1}^{n}|b_j(u)|\int_{u}^{\tau_j(u)} \Big(\sum_{k=1}^{m}|a_{k}(s)|+\sum_{\ell=1}^{n}|b_{\ell}(s)|\Big)\,ds\Big)\,du \leq c<1\,. \end{aligned} \end{equation} Then for each initial condition \eqref{e2}, there exists a unique solution to \eqref{e1}. Furthermore, if \begin{equation} \lim_{t\to \infty} \int_{t_0}^{t}(A(s)+B(s))\,ds=\infty, \label{e12} \end{equation} then, every solution of \eqref{e1} converges to zero. \end{theorem} \begin{proof} Let $C([t_0,\infty))$ be the set of real-valued functions, continuous on $[t_0,\infty)$. Then this is a Banach space with the norm \[ \| x\| =\sup_{t\geq t_0} |x(t)|. \] Based on \eqref{e4}, for a function $x\in C([r_0,\infty))$, we define the operator \[ (Tx)(t)= \begin{cases} x(t) \quad \text{if } r_0\leq t \leq t_0,\\[4pt] e^{-\int_{t_0}^{t}(A(s)+B(s))\,ds}x(t_0) -\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds}\\ \times \Big(\sum_{i=1}^{m}a_i(u)\int_{r_i(u)}^{u}E_x(s)\,ds -\sum_{j=1}^{n}b_j(u)\int_{u}^{\tau_j(u)}E_x(s)\,ds\Big)\,du\\ \quad\text{if } t_0\leq t. \end{cases} \] It is clear that $T$ maps $C([t_0,\infty))$ into $C([t_0,\infty))$ and preserves the values of $x(t)$ for $t\in [r_0,t_0]$. We will proof that $T$ is a contraction. Let $x$, $y$ be two continuous function on $[t_0,\infty)$, and satisfying the same initial conditions \eqref{e2}. Then for $t\geq t_0$, we have \begin{align*} &|(Tx)(t)-(Ty)(t)|\\ &\leq e^{-\int_{t_0}^{t}(A(s)+B(s))\,ds}|(x(t_0)-y(t_0))| +\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds}\\ &\quad\times\Big( \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)-E_{y}(s)|\,ds\,du\\ &\quad +\sum_{j=1}^{n}|b_j(u)|\int_{u}^{\tau_j(u)}|E_x(s)-E_{y}(s)|\,ds\,du \Big). \end{align*} Since $x(t)=y(t)$ for $r_0\leq t\leq t_0$, and \begin{align*} |E_x(s)-E_{y}(s)| & \leq |\sum_{k=1}^{m}a_{k}(t)x(r_{k}(t))-\sum_{k=1}^{m}a_{k}(t)y(r_{k}(t))|\\ & \quad +|\sum_{\ell=1}^{n}b_{\ell}(t)x(\tau_{\ell}(t)) -\sum_{\ell=1}^{n}b_{\ell}(t)y(\tau_{\ell}(t))|\\ & \leq \sum_{k=1}^{m}|a_{k}(t)||x(r_{k}(t))-y(r_{k}(t))| +\sum_{\ell=1}^{n}|b_{\ell}(t)||x(\tau_{\ell}(t))-y(\tau_{\ell}(t))|, \end{align*} by \eqref{e10}, we obtain \[ |(Tx)(t)-(Ty)(t)| \leq c\| x(t)-y(t)\| \,. \] Consequently, the operator $T$ has a unique fixed point in $C([t_0,\infty))$, and this fixed point satisfies \eqref{e2}. Let \[ L=\big\{ x\in C([t_0,\infty)):\underset{t\to\infty}{\lim}x(t)=0\big\}, \] which is a closed subspace of $C([t_0,\infty))$. Now, we claim that $T(L)\subset L$ and that $T$ preserves the initial conditions \eqref{e2}. Indeed, for $x\in L$, we have \begin{equation} \label{e14} \begin{aligned} |(Tx)(t)| & \leq |x_0|e^{-\int_{t_0}^{t}(A(s)+B(s))\,ds} +\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds}\\ &\quad\times \Big(\sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds +\sum_{j=1}^{n}|b_j(u)|\int_{u}^{\tau_j(u)}|E_x(s)|\,ds\Big)\,du\,. \end{aligned} \end{equation} Note that by \eqref{e12}, \[ \lim_{t\to\infty} |x_0| e^{-\int_{t_0}^{t}(A(s)+B(s))\,ds}=0\,. \] On the other hand, since $x\in L$, for each $\epsilon>0$ there exists $t_1'\geq t_0$ such that $| x(t)| <\epsilon/2$ for all $t\geq t_1'$. Select $t_1\geq t_1'$ such that $t_1'\leq \inf\{r_i(s):s\geq t_1,\, i=1,\dots,m\}$. Then for $t\geq t_1$, \begin{align*} &\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds\,du\\ &\leq\int_{t_0}^{t_1}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds\,du\\ &\quad +\frac{\epsilon}{2}\int_{t_1}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_j(u)}^{u} \Big(\sum_{k=1}^{m}|a_{k}(s)|+\sum_{\ell=1}^{n}|b_{\ell}(s)|\Big)\,ds\,du \end{align*} We observe that the first term on the right-hand side approaches zero as $t\to\infty$. Then there exists $t_2\geq t_1$ such that \[ \int_{t_0}^{t_1}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds\,du<\frac{\epsilon}{2}\,. \] Then by \eqref{e10} \begin{align*} &\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds\,du \\ &\leq\frac{\epsilon}{2}+\frac{\epsilon}{2} \int_{t_1}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds}\sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u} \Big(\sum_{k=1}^{m}|a_{k}(s)|+\sum_{\ell=1}^{n}|b_{\ell}(s)|\Big)\,ds\,du <\epsilon. \end{align*} Since $\epsilon$ is arbitrarily small, \begin{equation} \label{e16} \lim_{t\to\infty} \int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{i=1}^{m}|a_i(u)|\int_{r_i(u)}^{u}|E_x(s)|\,ds\,du=0. \end{equation} By a similar process, we prove that \begin{equation} \label{e18} \lim_{t\to\infty}\int_{t_0}^{t}e^{-\int_{u}^{t}(A(s)+B(s))\,ds} \sum_{j=1}^{n}|b_j(u)|\int_{u}^{\tau_j(u)}|E_x(s)|\,ds= 0\,. \end{equation} Therefore $(Tx)(t)\to 0$; i.e., the fixed point $x=Tx$, satisfies $\lim_{t\to\infty} x(t)=0$. \end{proof} The above theorem provides sufficient conditions for the convergence of solutions to zero. The next theorem provides necessary conditions. \begin{theorem} \label{thm5} Assume that \eqref{e12} holds, and \begin{equation} \liminf_{t\to\infty} \int_{t_0}^{t}(A(s)+B(s))\,ds>-\infty.\label{e19} \end{equation} If all solutions of \eqref{e1} converge to zero, then \eqref{e14} holds. \end{theorem} \begin{proof} For the shake of contradiction suppose that \eqref{e14} does not hold; i. e., \[ \liminf_{t\to\infty} \int_{t_0}^{t_n}(A(s)+B(s))\,ds=\alpha <\infty. \] Then by \eqref{e19}, $\alpha>-\infty$. Then there exists a sequence $\{t_n\}$ approaching $\infty$, such that \[ \lim_{n\to\infty} \int_{t_0}^{t_n} (A(s)+B(s))\,ds=\alpha\,. \] Let $x$ be a solution such that $x(t_0)=x_0\neq 0$. then \begin{equation} \label{e20} \lim_{n\to\infty} x_0e^{-\int_{t_0}^{t_n} (A(s)+B(s))\,ds} =x_0e^\alpha\neq 0\,. \end{equation} By Lemma \ref{lem1}, $x(t_n)$ satisfies \eqref{e4} with $t_n$ instead of $t$. By \eqref{e16} and \eqref{e18}, \begin{equation} \label{e22} \begin{aligned} &\lim_{n\to\infty} \int_{t_0}^{t_n} e^{-\int_{u}^{t_n}(A(s)+B(s))\,ds}\\ &\times \Big(\sum_{i=1}^{m}a_i(u)\int_{r_i(u)}^{u}E_x(s)\,ds -\sum_{j=1}^{n}b_j(u) \int_{u}^{\tau_j(u)}E_x(s)\,ds\Big)\,du=0\,. \end{aligned} \end{equation} Since all solutions approach zero, by \eqref{e4}, \eqref{e20} and \eqref{e22}, it follows that \[ 0=\lim_{n\to\infty} x(t_n) = x_0e^\alpha+0 \neq 0\,. \] This contradiction competes the proof. \end{proof} \section{An example} In this section we provide an example to illustrate our results. \begin{example} \label{examp8} \rm Consider the equation \begin{equation} \dot{x}(t)+\sin(t)e^{-(a-1)t/a}x(t/a)+\big(1-\sin(t)\big)e^{(b-1)t}x(bt)=0, \quad t>0\label{eq:4.1} \end{equation} where $a>b>1$ and $b-1<(a-1)/a$. Note that $r_0=t_0=0$ in this example, so the initial condition \eqref{e2} reduces to $x(t_0)=x_0$. We want to check that all the assumptions of Theorem \ref{thm4} are satisfied. \begin{align*} &\int_0^{t}(A(s)+B(s))\,ds\\ &=\int_0^{t}(\sin(t)e^{-(a-1)t/a}+\big(1-\sin(t)\big)e^{(b-1)t})\,ds\\ &=\frac{e^{-\alpha t}}{\alpha^2 +1}\big(\alpha\sin(t)-\cos(t)\big) +e^{\beta t}\big(\frac{1}{\beta}-\frac{\beta}{\beta^2 +1}\sin(t) +\frac{1}{\beta^2 +1}\cos(t)\big)+k, \end{align*} where $\alpha=(a-1)/a$, $\beta=b-1$ and \[ k=\frac{1}{\alpha^2 +1}+\frac{1}{\beta}+\frac{1}{\beta^2 +1}. \] Since \begin{gather*} \frac{e^{-\alpha t}}{\alpha^2 +1}(\alpha\sin(t)-\cos(t)) \to 0,\quad\text{and}\\ e^{\beta t}(\frac{1}{\beta}-\frac{\beta}{\beta^2 +1}\sin(t) +\frac{1}{\beta^2 +1}\cos(t)) \to \infty, \end{gather*} as $t\to \infty$, it follows that \eqref{e12} holds. Also note that \begin{align*} &\int_0^{t}e^{-\int_{u}^{t}\big(\sin(s)e^{-\alpha s}+(1-\sin(s))e^{\beta s}\big)\,ds} \Big\{ |\sin(u)|e^{-\alpha u}\int_{u/a}^{u}\big(|\sin(s)| e^{-\alpha s} \\ &\quad +|1-\sin(s)|e^{\beta s}\big)\,ds +|1-\sin(u)|e^{\beta u}\int_{u}^{bu}(|\sin(s)|e^{-\alpha s} +|1-\sin(s)|e^{\beta s})\,ds\Big\} \,du \\ &<\int_0^{t}e^{-\int_{u}^{t}(\sin(s)e^{-\alpha s}+(1-\sin(s))e^{\beta s})\,ds} \Big\{ e^{-\alpha u}\int_{u/a}^{u}(e^{-\alpha s} \\ &\quad +2e^{\beta s})\,ds+2e^{\beta u} \int_{u}^{bu}(e^{-\alpha s}+2e^{\beta s})\,ds\Big\} \,du \\ &=\int_0^{t}e^{-\int_{u}^{t}(\sin se^{-\alpha s} +(1-\sin s)e^{\beta s})\,ds} \Big\{ e^{-\alpha u}\big(\frac{e^{-\alpha u/a}-e^{-\alpha u}}{\alpha} +2\frac{e^{\beta u}-e^{\beta u/a}}{\beta}\big) \\ &\quad +2e^{\beta u}\big(\frac{e^{-\alpha u}-e^{-\alpha bu}}{\alpha} +2\frac{e^{\beta bu}-e^{\beta u}}{\beta}\big)\Big\} \,du \\ &b>1$ and $b-1<(a-1)/a$. So, there exists an $\epsilon$, $0<\epsilon