We shall reduce problem (1.1) to an integral equation. To this aim, first, by means of the transformation
$$\phi_{m}\bigl(y''(t)\bigr)=-x(t), $$
we convert problem (1.1) into
$$ \left \{ \textstyle\begin{array}{l} x''(t)+f(t,y,y')=0,\quad t\in{J}, \\ x(0)=x(1)=\int^{1}_{0}h(t)x(t)\,dt \end{array}\displaystyle \right . $$
(2.1)
and
$$ \left \{ \textstyle\begin{array}{l} y''(t)=-\phi_{m^{*}}(x(t)),\quad t\in{J}, t\neq t_{k}, \\ \Delta y|_{t=t_{k}}=I_{k}(y(t_{k})),\quad k=1,2,\ldots,m, \\ \Delta y'|_{t=t_{k}}=\bar{I}_{k}(y(t_{k}),y'(t_{k})),\quad k=1,2,\ldots,m, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds, \\ ay(1)+by'(1)=\int^{1}_{0}g(s)y(s)\,ds. \end{array}\displaystyle \right . $$
(2.2)
Lemma 2.1
If (H1), (H2), and (H3) hold, then problem (2.1) has a unique solution
\(x(t)\), which is given by
$$x(t)= \int^{1}_{0}H(t,s)f\bigl(s,y,y'\bigr) \,ds, $$
where
$$\begin{aligned}& H(t,s)=G(t,s)+\frac{1}{1-v} \int^{1}_{0}G(s,\tau)h(\tau)\,d\tau, \end{aligned}$$
(2.3)
$$\begin{aligned}& G(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} t(1-s),& 0\leq t\leq s\leq1, \\ s(1-t),& 0\leq s\leq t\leq1. \end{array}\displaystyle \right . \end{aligned}$$
(2.4)
Proof
Integrating (2.1) from 0 to t we get
$$\begin{aligned}& x'(t)-x'(0)=- \int^{t}_{0}f\bigl(t,y,y'\bigr)\,dt, \\& x'(t)=x'(0)- \int^{t}_{0}f\bigl(t,y,y'\bigr)\,dt. \end{aligned}$$
Integrating it again, we have
$$\begin{aligned}& x(t)-x(0)=x'(0)t- \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds, \\& x(t)=x(0)+x'(0)t- \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
From (2.1) we know that \(x(0)=x(1)=\int^{1}_{0}h(t)x(t)\,dt\). Letting \(t=1\) we then obtain
$$x(1)=x(0)+x'(0)- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds. $$
Hence
$$x'(0)= \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds. $$
Thus we get
$$ x(t)= \int^{1}_{0}h(t)x(t)\,dt+t \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. $$
(2.5)
In order to get the expression of \(x(t)\), different from [3], we multiply both sides of (2.5) with function \(h(t)\) and then integrating it from 0 to 1, we have
$$\begin{aligned} \int^{1}_{0}h(t)x(t)\,dt =& \int^{1}_{0}h(s)\,ds \int^{1}_{0}h(t)x(t)\,dt+ \int ^{1}_{0}th(t)\,dt \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds \end{aligned}$$
and
$$\begin{aligned} (1-\upsilon) \int^{1}_{0}h(t)x(t)\,dt =& \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
Hence,
$$\begin{aligned} \int^{1}_{0}h(t)x(t)\,dt =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ &{}- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds\biggr). \end{aligned}$$
(2.6)
Finally, we obtain
$$\begin{aligned} x(t) =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}th(t)\,dt \int ^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}h(t)\,dt \int ^{t}_{0}(t-s)f\bigl(s,y,y'\bigr) \,ds\biggr) \\ &+t \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =&\frac{1}{1-\upsilon}\biggl( \int^{1}_{0} \int ^{1}_{0}th(t) (1-s)f\bigl(s,y,y' \bigr)\,dt\,ds- \int^{1}_{0} \int ^{t}_{0}h(t) (t-s)f\bigl(s,y,y' \bigr)\,ds\,dt\biggr) \\ &{}+ \int^{1}_{0}t(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =& \int^{1}_{0}\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}t(1-s)h(t)\,dt- \int ^{t}_{0}(t-s)h(t)\,dt\biggr)f\bigl(s,y,y' \bigr)\,ds \\ &{}+ \int^{1}_{0}t(1-s)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{0}(1-s)f\bigl(s,y,y'\bigr) \,ds \\ =& \int^{1}_{0}\frac{1}{1-\upsilon}\biggl( \int^{1}_{0}t(1-s)h(t)\,dt- \int ^{t}_{0}(t-s)h(t)\,dt\biggr)f\bigl(s,y,y' \bigr)\,ds \\ &{}+ \int^{t}_{0}s(1-t)f\bigl(s,y,y'\bigr) \,ds- \int^{1}_{t}t(1-s)f\bigl(s,y,y'\bigr) \,ds. \end{aligned}$$
Thus
$$x(t)= \int^{1}_{0}H(t,s)f\bigl(s,y,y'\bigr) \,ds, $$
where
$$\begin{aligned}& H(t,s)=G(t,s)+\frac{1}{1-v} \int^{1}_{0}G(s,\tau)h(\tau)\,d\tau, \\& G(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} t(1-s),& 0\leq t\leq s\leq1, \\ s(1-t),& 0\leq s\leq t\leq1. \end{array}\displaystyle \right . \end{aligned}$$
This completes the proof. □
Let \(e(t)=t(1-t)\). Then from (2.3) and (2.4) we can prove that \(H(t,s)\) and \(G(t,s)\) have the following properties.
Lemma 2.2
Let
\(G(t,s)\)
and
\(H(t,s)\)
be given as in Lemma
2.1. Assume that (H3) holds. Then we have
$$\begin{aligned}& H(t,s)>0, \qquad G(t,s)>0,\quad \forall t,s\in(0,1), \\& H(t,s)\geq0,\qquad G(t,s)\geq0,\quad \forall t,s\in{J}, \\& e(t)e(s)\leq G(t,s)\leq G(t,t)=t(1-t)=e(t)\leq\bar{e}= \max_{t\in {J}}e(t)= \frac{1}{4},\quad \forall t,s\in{J}, \\& \rho e(s)\leq H(t,s)\leq\gamma s(1-s)=\gamma e(s)\leq\frac{1}{4}\gamma , \quad \forall t,s\in{J}, \end{aligned}$$
where
$$\gamma=\frac{1}{1-\upsilon}, \qquad \rho=\frac{\int^{1}_{0}e(\tau)h(\tau)\,d\tau }{1-\upsilon}. $$
Lemma 2.3
If (H1), (H2), and (H3) hold, then problem (2.2) has a unique solution
\(y(t)\)
expressed in the form
$$\begin{aligned} y(t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \bigl(x(s)\bigr)\,ds-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr), \end{aligned}$$
where
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau, \end{aligned}$$
(2.7)
$$\begin{aligned}& G_{1}(t,s)=\frac{1}{d} \left \{ \textstyle\begin{array}{l@{\quad}l} (b+as)(b+a(1-t)), &0\leq s\leq t\leq1, \\ (b+at)(b+a(1-s)), &0\leq t\leq s\leq1, \end{array}\displaystyle \right . \end{aligned}$$
(2.8)
$$\begin{aligned}& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t) \,dt, \end{aligned}$$
(2.9)
and
\(d=a(a+2b)\).
Proof
First, we assume that \(t\in I_{i}\), \(I_{i}=(t_{i},t_{i+1})\) (\(i=0,1,2,\ldots,m\)). Integrating both sides of (2.2) from \(t_{i}\) to \(t^{-}_{i+1}\), we get
$$\begin{aligned}& y'\bigl(t^{-}_{1}\bigr)-y'(0)=- \int^{t_{1}}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'\bigl(t^{-}_{2}\bigr)-y' \bigl(t^{+}_{1}\bigr)=- \int^{t_{2}}_{t_{1}}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& \ldots, \\& y'(t)-y'\bigl(t^{+}_{i}\bigr)=- \int^{t}_{t_{i}}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt. \end{aligned}$$
Adding the above equations, we find
$$\begin{aligned}& y'(t)-y'(0)-\sum_{k:t_{k}< t} \bigl(y'\bigl(t^{+}_{k}\bigr)-y' \bigl(t^{-}_{k}\bigr)\bigr)=- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'(t)=y'(0)+\sum_{k:t_{k}< t} \bigl(y'\bigl(t^{+}_{k}\bigr)-y' \bigl(t^{-}_{k}\bigr)\bigr)- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt, \\& y'(t)=y'(0)+\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)- \int ^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr) \,dt. \end{aligned}$$
(2.10)
Similarly, we can get
$$\begin{aligned} y(t) =&y(0)+y'(0)t- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds \\ &{}+\sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
(2.11)
Let \(t=1\) in (2.10) and (2.11). We obtain
$$\left \{ \textstyle\begin{array}{l} by'(1)=by'(0)+b\sum_{k:t_{k}< t}\bar{I}_{k}(y(t_{k}),y'(t_{k}))-b\int ^{t}_{0}\phi_{m^{*}}(x(t))\,dt, \\ ay(1)=ay(0)+ay'(0)-a\int^{1}_{0}(1-s)\phi_{m^{*}}(x(s))\,ds \\ \hphantom{ay(1)={}}{}+a \sum_{k:t_{k}< t}(1-t_{k})\bar{I}_{k}(y(t_{k}),y'(t_{k}))+a\sum_{k:t_{k}< t}I_{k}(y(t_{k})). \end{array}\displaystyle \right . $$
It is easy to get
$$\begin{aligned} ay(1)+by'(1) =&ay(0)+(a+b)y'(0)-a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}+a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}+b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \\ =& \int^{1}_{0}g(s)y(s)\,ds, \end{aligned}$$
which implies
$$\begin{aligned} ay(0)+(a+b)y'(0) =&a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a\sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt+ \int^{1}_{0}g(s)y(s)\,ds. \end{aligned}$$
Then we have the following equations:
$$\left \{ \textstyle\begin{array}{l} ay(0)+(a+b)y'(0) \\ \quad =a\int^{1}_{0}(1-s)\phi_{m^{*}}(x(s))\,ds-a\sum_{k:t_{k}< t}(1-t_{k})\bar{I}_{k}(y(t_{k}),y'(t_{k})) \\ \qquad {}-a \sum_{k:t_{k}< t}I_{k}(y(t_{k}))-b\sum_{k:t_{k}< t}\bar{I}_{k}(y(t_{k}),y'(t_{k})) \\ \qquad {}-b\int^{t}_{0}\phi_{m^{*}}(x(t))\,dt+\int^{1}_{0}g(s)y(s)\,ds, \\ ay(0)-by'(0)=\int^{1}_{0}g(s)y(s)\,ds. \end{array}\displaystyle \right . $$
Obviously,
$$\begin{aligned}& y'(0) =\frac{1}{a+2b}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\& \hphantom{y'(0) ={}}{} -a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr], \\& y(0) =\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\& \hphantom{y(0) ={}}{} -a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\& \hphantom{y(0) ={}}{} +\frac{1}{a} \int^{1}_{0}g(s)y(s)\,ds. \end{aligned}$$
Substituting (2.10), (2.11) in (2.9), we have
$$\begin{aligned} y(t) =&\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{1}{a} \int^{1}_{0}g(s)y(s)\,ds+\frac{t}{a+2b}\biggl[a \int ^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
In a similar way as the proof of (2.6) in Lemma 2.1, we get
$$\begin{aligned} \int^{1}_{0}g(s)y(s)\,ds =&\frac{a}{a-\xi}\biggl\{ \frac{b}{a(a+2b)}\biggl[ \int ^{1}_{0}g(t)\, dt a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(t)\bigr) \,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{t}{a+2b} \int^{1}_{0}g(t)\,dt\biggl[a \int^{1}_{0}(1-s)\phi _{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{1}_{0}g(t)\,dt \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds \\ &{}+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \int^{1}_{0}g(t)\,dt + \int^{1}_{0}g(t) \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)\,dt\biggr\} . \end{aligned}$$
Hence, we finally get
$$\begin{aligned} y(t) =&\frac{b}{a(a+2b)}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{a}{a-\xi}\biggl\{ \frac{b}{a(a+2b)}\biggl[ \int^{1}_{0}g(t)\, dt a \int ^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s) \bigr)\,ds \\ &{}- a\sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr) \\ &{}-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}+\frac{t}{a+2b} \int^{1}_{0}g(t)\,dt\biggl[a \int^{1}_{0}(1-s)\phi _{m^{*}}\bigl(x(s) \bigr)\,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{1}_{0}g(t)\,dt \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \int^{1}_{0}g(t)\,dt \\ &{}+ \int^{1}_{0}g(t)\,dt \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)\biggr\} \\ &{}+\frac{t}{a+2b}\biggl[a \int^{1}_{0}(1-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds-a \sum_{k:t_{k}< t}(1-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr) \\ &{}-a \sum_{k:t_{k}< t}I_{k} \bigl(y(t_{k})\bigr)-b\sum_{k:t_{k}< t} \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+b \int^{t}_{0}\phi_{m^{*}}\bigl(x(t)\bigr)\,dt \biggr] \\ &{}- \int^{t}_{0}(t-s)\phi_{m^{*}}\bigl(x(s)\bigr) \,ds+ \sum_{k:t_{k}< t}(t-t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)+\sum_{k:t_{k}< t}I_{k}\bigl(y(t_{k}) \bigr). \end{aligned}$$
Hence
$$y(t)= \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \bigl(x(s)\bigr)\,ds-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr), $$
where
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau, \\& G_{1}(t,s)=\frac{1}{d} \left \{ \textstyle\begin{array}{l@{\quad}l} (b+as)(b+a(1-t)),& 0\leq s\leq t\leq1, \\ (b+at)(b+a(1-s)),& 0\leq t\leq s\leq1, \end{array}\displaystyle \right . \\& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t)\,dt. \end{aligned}$$
Then the proof is completed. □
It follows from (2.7), (2.8), and (2.9) that \(H_{1}(t,s)\), \(G_{1}(t,s)\), and \(H_{2}(t)\) have the following properties.
Lemma 2.4
Suppose (H3) holds and assume that
\(G_{1}(t,s)\)
and
\(H_{1}(t,s)\)
are given as in Lemma
2.3. Then we have
$$\begin{aligned}& \frac{1}{d}b^{2}\leq G_{1}(t,s)\leq G_{1}(s,s)\leq\frac{(a+b)^{2}}{d},\quad \forall t,s\in{J}, \end{aligned}$$
(2.12)
$$\begin{aligned}& \rho_{1}\leq H_{1}(t,s)\leq\frac{a}{a-\xi}G_{1}(s,s) \leq\rho_{2},\quad \forall t,s\in{J}, \end{aligned}$$
(2.13)
$$\begin{aligned}& H_{2}(t)\leq\rho_{2}, \end{aligned}$$
(2.14)
where
$$\delta=\frac{b}{a+b}, \qquad \rho_{1}=\frac{b^{2}\gamma^{1}}{a+2b},\qquad \rho _{2}=\frac{a(a+b)^{2}}{(a-\xi)d}. $$
Proof
Clearly, it follows from the definition of \(G_{1}(t,s)\) that (2.12) holds. Now we show that (2.13) and (2.14) are true.
In fact, for \(t\in[\zeta,1]\) and \(s\in J\), we have
$$\begin{aligned}& H_{1}(t,s)=G_{1}(t,s)+\frac{1}{a-\xi} \int^{1}_{0}G_{1}(s,\tau)g(\tau )\,d\tau \\& \hphantom{H_{1}(t,s)}\leq G_{1}(s,s)+\frac{\xi}{a-\xi}G_{1}(s,s)\leq \frac {a(a+b)^{2}}{(a-\xi)d}=\rho_{2}, \\& H_{2}(t)=\frac{at-a-b}{a+2b}+\frac{1}{a-\xi} \int^{1}_{0}\frac {at-a-b}{a+2b}g(t)\,dt\leq \frac{at}{a+2b}+\frac{\xi}{a-\xi}\frac{at}{a+2b}, \\& \frac{at}{a+2b}\leq\frac{a}{a+2b}\leq\frac{(a+b)^{2}}{a+2b}. \end{aligned}$$
Consequently,
$$ H_{2}(t)\leq\frac{(a+b)^{2}}{a+2b}+\frac{\xi}{a-\xi}\frac {(a+b)^{2}}{a+2b}= \frac{a(a+b)^{2}}{(a-\xi)d}=\rho_{2}. $$
(2.15)
This completes the proof. □
Combining Lemma 2.1 with Lemma 2.3, we can get directly the following result.
Lemma 2.5
Assume that (H1)-(H3) hold. Then
\(y(t)\)
has the following form:
$$\begin{aligned} y(t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\tau)f\bigl(\tau ,y( \tau),y'(\tau)\bigr)\,d\tau\biggr)\,ds \\ &{}-\sum ^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
Proof
The conclusion is so straightforward that we omit it here. □
We next give some notations and a fixed point theorem which will be used to prove our main results. Let
$$\begin{aligned} \mathit{PC}^{1}\bigl[J,R^{n}\bigr] =&\bigl\{ y:J\rightarrow R^{n}\mid y'(t) \mbox{ is continuous when } t\neq t_{k},y\bigl(t^{+}_{k}\bigr),y \bigl(t^{+}_{-}\bigr),y'\bigl(t^{+}_{k} \bigr),y'\bigl(t^{-}_{k}\bigr) \\ &{}\mbox{exist and } y\bigl(t^{-}_{k}\bigr)=y(t_{k}),k=1,2, \ldots,m \bigr\} . \end{aligned}$$
Clearly, \(\mathit{PC}^{1}[J,R^{n}]\) is a Banach space with the norm \(\|y\|_{\mathit{PC}^{1}}=\max\{\|y\|_{\mathit{PC}^{1}},\|y'\|_{\mathit{PC}^{1}}\}\).
Lemma 2.6
[19]
\(H\subset \mathit{PC}^{1}[J,R^{n}]\)
is a relatively compact set if and only if
\(\forall y\in H\), y
and
\(y'\)
are uniformly bounded in
J
and equi-continuous on
\(J_{k}\) (\(k=0, 1, 2, \ldots, m\)).
Definition 2.1
A function \(y\in \mathit{PC}^{1}\) is said to be a solution of (1.1) if it satisfies every equation in system (1.1).
Lemma 2.7
(Schauder fixed point theorem)
If
K
is a nonempty convex subset of a Banach space
V
and
T
is a continuous mapping of
K
into itself such that
\(T(K)\)
is contained in a compact subset of
K, then
T
has a fixed point.
Definition 2.2
Define an operator \(A:\mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\) by
$$\begin{aligned} (Ay) (t) =& \int^{1}_{0}H_{1}(t,s)\phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\alpha )f\bigl(\alpha,y,y' \bigr)\, d\alpha \biggr)\,ds \\ &{}-\sum^{m}_{k=1}H_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
(2.16)
Lemma 2.8
Assume that (H1)-(H3) hold. Then
\(y(t)\in{J}\)
is a fixed point of
A
if and only if
\(y(t)\)
is a solution of problem (1.1).
Lemma 2.9
The operator
\(A: \mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\)
is completely continuous.
Proof
According to (2.16) we have
$$\begin{aligned} (Ay)'(t) =& \int^{1}_{0}H'_{1t}(t,s) \phi_{m^{*}} \biggl( \int^{1}_{0}H(s,\tau )f\bigl(\tau,y,y' \bigr)\,d\tau \biggr)\,ds \\ &{}-\sum^{m}_{k=1}H'_{1}(t,t_{k}) \bar{I}_{k}\bigl(y(t_{k}),y'(t_{k}) \bigr)-\sum^{m}_{k=1}H'_{2}(t)I_{k} \bigl(y(t_{k})\bigr). \end{aligned}$$
(2.17)
From (2.16) and (2.17) we know that \(A:\mathit{PC}^{1}[J,R^{n}]\rightarrow \mathit{PC}^{1}[J,R^{n}]\) is continuous. For any bounded set \(S\in \mathit{PC}^{1}[J,R^{n}]\), and any function \(y(t)\in S\), we see that \((Ay)(t)\) and \((Ay)'(t)\) are uniformly bounded and equi-continuous on \(J_{k}\) (\(k=0,1,2,\ldots,m\)). Hence, according to Lemma 2.6 we see that \(A(S)\) is a relatively compact set, therefore A is a completely continuous operator. □