What follows is our main theorem.
Theorem
Suppose that conditions (A1)-(A6) hold. Then all solutions of system (2) are continuable and bounded.
Proof
We define a Lyapunov-Krasovskii functional by
$$ W(t)=W \bigl(t,x(t),y(t) \bigr)=e^{-\frac{\gamma(t)}{\mu}}V_{0} \bigl(t,x(t),y(t) \bigr), $$
(3)
where μ a positive constant,
$$\begin{aligned} \gamma(t) =& \int_{0}^{t} \bigl\vert \theta(s) \bigr\vert \,ds= \int_{0}^{t} \bigl\vert x^{\prime}(s)p^{\prime} \bigl(x(s) \bigr) \bigr\vert \,ds \\ =& \int_{\alpha_{1}(t)}^{\alpha_{2}(t)} \bigl\vert p^{\prime } \bigl(x(u) \bigr) \bigr\vert \,du\leq \int_{-\infty}^{+\infty} \bigl\vert p^{\prime } \bigl(x(u) \bigr) \bigr\vert \,du< \infty, \end{aligned}$$
for \(\theta(t)=x^{\prime}(t)p^{\prime}(x(t))\), \(\alpha_{1}(t)=\min \{ x(0),x(t) \} \), \(\alpha_{2}(t)=\max \{ x(0),x(t) \}\), and
$$\begin{aligned} V_{0}(t) =&V_{0} \bigl(t,x(t),y(t) \bigr)= \frac{1}{2}y^{2}+p(x)\sum_{i=1}^{n}c_{i}(t)\int_{0}^{x}h_{i}(s)\,ds \\ &{}+\sum_{i=1}^{n}\lambda_{i} \int_{-\tau _{i}}^{0} \int_{t+s}^{t}y^{2}(u)\,du\,ds+ \int_{0}^{t} \int_{t}^{\infty} \bigl\vert C ( u,s ) \bigr\vert y^{2}(s)\,du\,ds. \end{aligned}$$
(4)
From the assumptions (A1), (A2), and (A4) it follows that
$$\begin{aligned} V_{0}(t) \geq&\frac{1}{2}y^{2}+p(x)\sum _{i=1}^{n}c_{i}(t) \int _{0}^{x}h_{i}(s)\,ds \\ \geq&\frac{1}{2}y^{2}+\frac{1}{2}\sum _{i=1}^{n}c_{i}\delta_{i}x^{2} \\ \geq&k \bigl( x^{2}+y^{2} \bigr), \end{aligned}$$
(5)
where \(k=\frac{1}{2}\min \{1,\sum_{i=1}^{n}c_{i}\delta_{i} \} \).
Let \(( x(t),y(t) ) \) be a solution of (2). Calculating the time derivative of the functional \(V_{0}(t)\), we obtain
$$\begin{aligned} V_{0}^{\prime}(t) =&y \int_{0}^{t}C ( t,s ) \frac{y ( s ) }{p(x(s))}\,ds-a ( t ) f \biggl( t,x,\frac{y}{p(x)} \biggr) \frac{y^{2}}{p(x)}-b ( t ) g \biggl( t, \frac{y}{p(x)} \biggr) y \\ &{}+y\sum_{i=1}^{n}c_{i} ( t ) \int_{t-\tau _{i}}^{t}h_{i}^{{\prime }} \bigl( x ( s ) \bigr) \frac{y ( s ) }{p(x(s))}\,ds+\theta (t)\sum_{i=1}^{n}c_{i}(t) \int_{0}^{x}h_{i}(s)\,ds \\ &{}+p(x)\sum_{i=1}^{n}c_{i}^{{\prime }}(t) \int_{0}^{x}h_{i}(s)\,ds+\sum _{i=1}^{n} ( \lambda_{i}\tau _{i} ) y^{2}-\sum_{i=1}^{n} \lambda_{i} \int_{t-\tau _{i}}^{t}y^{2} ( s )\,ds \\ &{}+y^{2} \int_{t}^{\infty} \bigl\vert C ( u,t ) \bigr\vert \,du- \int_{0}^{t} \bigl\vert C ( t,s ) \bigr\vert y^{2}(s)\,ds. \end{aligned}$$
By the assumptions (A1)-(A6) and the inequality \(2\vert ab\vert \leq ( a^{2}+b^{2} ) \), the following estimates can be verified:
$$\begin{aligned}& \begin{aligned}[b] y \int_{0}^{t}C ( t,s ) \frac{y ( s ) }{p(x(s))}\,ds &\leq \int_{0}^{t} \bigl\vert C ( t,s ) \bigr\vert \bigl\vert y(t) \bigr\vert \bigl\vert y(s) \bigr\vert \,ds \\ &\leq y^{2} \int_{0}^{t} \bigl\vert C ( t,s ) \bigr\vert \,ds+ \int_{0}^{t} \bigl\vert C ( t,s ) \bigr\vert y^{2} ( s )\,ds, \end{aligned} \\& -a ( t ) f \biggl( t,x,\frac{y}{p(x)} \biggr) \frac{y^{2}}{p(x)}-b ( t ) g \biggl( t,\frac{y}{p(x)} \biggr) y\leq-\frac{m}{p_{1}} \biggl(f \biggl(t,x,\frac{y}{p(x)} \biggr)+g_{0} \biggr)y^{2}, \\& \begin{aligned}[b] y\sum_{i=1}^{n}c_{i} ( t ) \int_{t-\tau _{i}}^{t}h_{i}^{{\prime }} \bigl( x ( s ) \bigr) \frac{y ( s ) }{p(x(s))}\,ds &\leq \sum_{i=1}^{n}C_{i} \gamma_{i} \int_{t-\tau_{i}}^{t} \bigl\vert y ( t ) y ( s ) \bigr\vert \,ds \\ &\leq\sum_{i=1}^{n} ( C_{i} \gamma_{i}\tau_{i} ) y^{2}+\sum _{i=1}^{n}C_{i}\gamma_{i} \int_{t-\tau_{i}}^{t}y^{2} ( s )\,ds, \end{aligned} \\& \theta(t)\sum_{i=1}^{n}c_{i}(t) \int_{0}^{x}h_{i}(s)\,ds\leq \frac{\vert \theta(t)\vert }{2}\sum_{i=1}^{n} ( C_{i}\beta_{i} ) x^{2}, \\& p(x)\sum_{i=1}^{n}c_{i}^{{\prime}}(t) \int_{0}^{x}h_{i}(s)\,ds\leq0. \end{aligned}$$
From these estimates we obtain, quite readily,
$$\begin{aligned} V_{0}^{\prime}(t) \leq& \biggl( R-\frac{m}{p_{1}} \biggl(f \biggl(t,x,\frac{y}{p(x)} \biggr)+g_{0} \biggr) \biggr) y^{2} \\ &{}+\sum_{i=1}^{n} ( C_{i} \gamma_{i}\tau_{i} ) y^{2}+\sum _{i=1}^{n}C_{i}\gamma_{i} \int_{t-\tau_{i}}^{t}y^{2} ( s )\,ds \\ &{}+\sum_{i=1}^{n} ( \lambda_{i} \tau_{i} ) y^{2}-\sum_{i=1}^{n} \lambda_{i} \int_{t-\tau_{i}}^{t}y^{2} ( s )\,ds \\ &{}+\frac{\vert \theta(t)\vert }{2}\sum_{i=1}^{n} ( C_{i}\beta_{i} ) x^{2}. \end{aligned}$$
Let
$$\tau^{\ast}=\max \{ \tau_{1},\tau_{2},\ldots, \tau_{n} \} $$
and
$$\lambda=\sum_{i=1}^{n}\lambda_{i}= \sum_{i=1}^{n}C_{i} \gamma_{i}. $$
Hence, in view of the discussion and (A6), we can conclude that
$$\begin{aligned} V_{0}^{\prime}(t) \leq& \biggl( R+2\lambda \tau^{\ast}- \frac {m}{p_{1}} \biggl(f \biggl(t,x, \frac{y}{p(x)} \biggr)+g_{0} \biggr) \biggr) y^{2}+ \frac{\vert \theta (t)\vert }{2}\sum_{i=1}^{n} ( C_{i} \beta_{i} ) x^{2} \\ \leq&\frac{\vert \theta(t)\vert }{2}\sum_{i=1}^{n} ( C_{i}\beta_{i} ) x^{2}. \end{aligned}$$
(6)
It is now clear that the time derivative of the functional \(W ( t ) \) defined by (3) along any solution of system (2) leads that
$$W^{\prime}(t)=e^{-\frac{\gamma(t)}{\mu}} \biggl( -\frac{\vert \theta (t)\vert }{\mu}V_{0} \bigl(t,x ( t ),y ( t ) \bigr)+\frac {d}{dt}V_{0} \bigl(t,x ( t ) ,y ( t ) \bigr) \biggr) . $$
Therefore, using (5), (6), and taking \(\mu=\frac{2k}{\sum_{i=1}^{n}C_{i}\beta_{i}}\), we obtain
$$W^{\prime}(t)\leq e^{-\frac{\gamma(t)}{\mu}} \Biggl( -\frac{\vert \theta(t)\vert }{2}\sum _{i=1}^{n} ( C_{i}\beta_{i} ) x^{2}+\frac{\vert \theta(t)\vert }{2}\sum_{i=1}^{n} ( C_{i}\beta _{i} ) x^{2} \Biggr) =0. $$
This implies that \(W^{\prime}(t)\leq0\). Since all the functions appearing in equation (1) are continuous, it is obvious that there exists at least a solution of equation (1) defined on \([ t_{0},t_{0}+\delta ) \) for some \(\delta>0\). We need to show that the solution can be extended to the entire interval \([ t_{0},\infty ) \). We assume on the contrary that there is a first time \(T<\infty\) such that the solution exists on \([ t_{0},T ) \) and
$$\lim_{t\rightarrow T^{-}} \bigl( \bigl\vert x(t) \bigr\vert + \bigl\vert y(t) \bigr\vert \bigr) =\infty. $$
Let \(( x(t),y(t) ) \) be such a solution of system (2) with initial condition \(( x_{0},y_{0} ) \). Since the Lyapunov-Krasovskii type functional \(W(t)\) is positive definite and decreasing, \(W^{\prime }(t)\leq0\), along the trajectories of system (2), we can say that \(W(t)\) is bounded \([ t_{0},T ] \). We have
$$W\bigl(T,x(T),y(T)\bigr)\leq W(t_{0},x_{0},y_{0})=W_{0}. $$
Hence, it follows from (3) and (5) that
$$x^{2}(T)+y^{2}(T)\leq\frac{W_{0}}{D}, $$
where \(D=k\exp ( -\gamma ( T ) \mu^{-1} ) \). This inequality implies that \(\vert x(t)\vert \) and \(\vert y(t)\vert \) are bounded as \(t\rightarrow T^{-}\). Thus, we can conclude that \(T<\infty\) is not possible, we must have \(T=\infty\). This completes the proof of the theorem. □
Example
We consider the following nonlinear integro-differential equation of second order with two constants delays, \(\tau_{1}>0\) and \(\tau_{2}>0\):
$$\begin{aligned} & \biggl( \biggl( 2+\frac{\sin x}{1+x^{2}} \biggr) x^{\prime} \biggr) ^{{\prime }}+ \biggl( 1+\frac{2}{1+t^{2}} \biggr) \bigl( e^{-t}+\sin x+\cos x^{{\prime }}+6 \bigr) x^{\prime} \\ &\quad{}+ \biggl( 1+\frac{1}{1+t^{2}} \biggr) \biggl( 3x^{\prime}+ \frac{x^{\prime }}{1+x^{\prime^{2}}} \biggr) +2 \bigl( e+e^{-t} \bigr) x ( t-\tau _{1} ) \\ &\quad{}+2 \bigl( e^{2}+e^{-t} \bigr) x ( t- \tau_{2} ) = \int_{0}^{t}\frac{s}{ ( 1+2t ) ^{2}}x^{\prime}(s)\,ds. \end{aligned}$$
(7)
When we compare equation (7) with equation (1), the existence can be seen of the following estimates:
$$\begin{aligned}& p ( x ) =2+\frac{\sin x}{1+x^{2}},\quad 1\leq p ( x ) \leq3, \\& \int_{-\infty}^{\infty} \bigl\vert p^{\prime}(u) \bigr\vert \,du\leq \int_{-\infty}^{\infty} \biggl( \biggl\vert \frac{\cos u}{1+u^{2}} \biggr\vert + \biggl\vert \frac{2u\sin u}{ ( 1+u^{2} ) ^{2}} \biggr\vert \biggr)\,du\leq\pi, \\& a ( t ) =1+\frac{2}{1+t^{2}},\qquad b ( t ) =1+\frac{1}{1+t^{2}}, \\& m=1\leq b(t)\leq a(t)\leq3=M, \\& c_{1} ( t ) =e+e^{-t},\qquad c_{2} ( t ) =e^{2}+e^{-t}, \\& c_{1}=e\leq c_{1}(t)\leq e+1=C_{1}, \\& c_{2}=e^{2}\leq c_{2}(t)\leq e^{2}+1=C_{2}, \\& c_{1}^{{\prime}}(t)\leq0,\qquad c_{2}^{{\prime}} (t) \leq0,\quad t \geq0, \\& f \bigl( t,x,x^{{\prime}} \bigr) =e^{-t}+\sin x+\cos x^{{\prime }}+6,\qquad 4\leq f \bigl( t,x,x^{{\prime}} \bigr) \leq9, \\& g \bigl( t,x^{{\prime}} \bigr) =3x^{\prime}+\frac{x^{\prime }}{1+x^{\prime ^{2}}}, \\& g ( t,0 ) =0,\qquad g_{0}=3\leq\frac{g ( t,x^{{\prime }} ) }{x^{{\prime}}} \leq4=g_{1} \quad( y \neq0 ), \\& h_{1} ( x ) = h_{2} ( x ) =2x, \\& h_{1} ( 0 ) =h_{2} ( 0 ) =0,\qquad\frac{h_{1} ( x ) }{x}= \frac{h_{2} ( x ) }{x}=2 \quad( x\neq0 ) , \\& \bigl\vert h_{1}^{{\prime}} ( x ) \bigr\vert = \bigl\vert h_{2}^{{\prime}} ( x ) \bigr\vert =2, \\& \int_{0}^{t} \bigl\vert C ( t,s ) \bigr\vert \,ds+ \int_{t}^{\infty} \bigl\vert C ( u,t ) \bigr\vert \,du \\& \quad= \int_{0}^{t} \biggl\vert \frac {s}{ ( 1+2t ) ^{2}} \biggr\vert \,ds+ \int_{t}^{\infty} \biggl\vert \frac{t}{ ( 1+2u ) ^{2}} \biggr\vert \,du\leq\frac{3}{8}=R. \end{aligned}$$
Thus, all the assumptions of the theorem hold. So we can conclude that all solutions of (7) are continuable and bounded.