## Boundary Value Problems

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# Variational approach to a class of impulsive differential equations

Boundary Value Problems20142014:37

DOI: 10.1186/1687-2770-2014-37

Accepted: 14 January 2014

Published: 7 February 2014

## Abstract

In this article, the author discusses the existence of solutions for a class of impulsive differential equations by means of a variational approach different from earlier approaches.

MSC:34B37, 45G10, 47H30, 47J30.

### Keywords

impulsive differential equation integral equation variational method critical point theory

## 1 Introduction

The theory of impulsive differential equations has been emerging as an important area of investigation in recent years [13]. There is a vast literature on the existence of solutions by using topological methods, including fixed point theorems, Leray-Schauder degree theory, and fixed point index theory [415]. But it is quite difficult to apply the variational approach to an impulsive differential equation; therefore, there was no result in this area for a long time. Only in the recent five years, there appeared a few articles which dealt with some impulsive differential equations by using variational methods [1620]. Motivated by [17], in this article we shall use a different variational approach to discuss the existence of solutions for a class of impulsive differential equations and we only deal with classical solutions.

Consider the boundary value problem (BVP) for the second-order nonlinear impulsive differential equation:
$\left\{\begin{array}{l}-{u}^{″}\left(t\right)=f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {J}^{\prime },\\ \mathrm{\Delta }u{|}_{t={t}_{k}}={c}_{k}\phantom{\rule{1em}{0ex}}\left(k=1,2,3,\dots ,m\right),\\ \mathrm{\Delta }{u}^{\prime }{|}_{t={t}_{k}}={d}_{k}\phantom{\rule{1em}{0ex}}\left(k=1,2,3,\dots ,m\right),\\ u\left(0\right)=u\left(1\right)=0,\end{array}$
(1)
where $J=\left[0,1\right]$, $0<{t}_{1}<\cdots <{t}_{k}<\cdots <{t}_{m}<1$, ${J}^{\prime }=J\mathrm{\setminus }\left\{{t}_{1},\dots ,{t}_{k},\dots ,{t}_{m}\right\}$, ${c}_{k}$ and ${d}_{k}$ ($k=1,2,\dots ,m$) are any real numbers, $f\left(t,u\right)$ is a real function defined on $J×R$, where R denotes the set of all real numbers, and $f\left(t,u\right)$ is continuous on ${J}^{\prime }×R$, left continuous at $t={t}_{k}$, i.e.
$\underset{t\to {t}_{k}-0,w\to u}{lim}f\left(t,w\right)=f\left({t}_{k},u\right)$
for any $u\in R$ ($k=1,2,\dots ,m$), and the right limit at $t={t}_{k}$ exists, i.e.
$\underset{t\to {t}_{k}+0,w\to u}{lim}f\left(t,w\right)$
(denoted by $f\left({t}_{k}^{+},u\right)$) exists for any $u\in R$ ($k=1,2,\dots ,m$). $\mathrm{\Delta }u{|}_{t={t}_{k}}$ denotes the jump of $u\left(t\right)$ at $t={t}_{k}$, i.e.
$\mathrm{\Delta }u{|}_{t={t}_{k}}=u\left({t}_{k}^{+}\right)-u\left({t}_{k}^{-}\right),$
where $u\left({t}_{k}^{+}\right)$ and $u\left({t}_{k}^{-}\right)$ represent the right and left limits of $u\left(t\right)$ at $t={t}_{k}$, respectively. Similarly,
$\mathrm{\Delta }{u}^{\prime }{|}_{t={t}_{k}}={u}^{\prime }\left({t}_{k}^{+}\right)-{u}^{\prime }\left({t}_{k}^{-}\right),$

where ${u}^{\prime }\left({t}_{k}^{+}\right)$ and ${u}^{\prime }\left({t}_{k}^{-}\right)$ represent the right and left limits of ${u}^{\prime }\left(t\right)$ at $t={t}_{k}$, respectively. Let $PC\left[J,R\right]$ = {$u:u$ is a real function on J such that $u\left(t\right)$ is continuous at $t\ne {t}_{k}$, left continuous at $t={t}_{k}$, and $u\left({t}_{k}^{+}\right)$ exists, $k=1,2,\dots ,m$} and $P{C}^{1}\left[J,R\right]$ = {$u\in PC\left[J,R\right]:{u}^{\prime }\left(t\right)$ is continuous at $t\ne {t}_{k}$ and ${u}^{\prime }\left({t}_{k}^{+}\right)$, ${u}^{\prime }\left({t}_{k}^{-}\right)$ exist, $k=1,2,\dots ,m$}. A function $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ is called a solution of BVP (1) if $u\left(t\right)$ satisfies (1).

Let us list some conditions.

(H1) There exist $p>2$, $a>0$ and $b>0$ such that
$|f\left(t,u\right)|\le a+b{|u|}^{p-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R.$
(H2) There exist $0 and $d>0$ such that
${\int }_{0}^{u}f\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv\le c{u}^{2}+d,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R.$
Lemma 1 $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ is a solution of BVP (1) if and only if $v\in C\left[J,R\right]$ is a solution of the integral equation
$v\left(t\right)={\int }_{0}^{1}G\left(t,s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,$
(2)
where
$G\left(t,s\right)=\left\{\begin{array}{ll}s\left(1-t\right),& \mathrm{\forall }0\le s\le t\le 1;\\ t\left(1-s\right),& \mathrm{\forall }0\le t
(3)
$g\left(t,v\right)=f\left(t,v+a\left(t\right)-a\left(1\right)t\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,v\in R$
(4)
and
$v\left(t\right)=u\left(t\right)-a\left(t\right)+a\left(1\right)t,\phantom{\rule{2em}{0ex}}a\left(t\right)=\sum _{0<{t}_{k}
(5)
Proof For $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$, we have the formula (see [21], Lemma 1(b))
$\begin{array}{rcl}u\left(t\right)& =& u\left(0\right)+t{u}^{\prime }\left(0\right)+{\int }_{0}^{t}\left(t-s\right){u}^{″}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{0<{t}_{k}
(6)
So, if $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ is a solution of BVP (1), then, by (1) and (6), we have
$\begin{array}{rcl}u\left(t\right)& =& t{u}^{\prime }\left(0\right)-{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{0<{t}_{k}
(7)
It is clear, by (5), that
$a\left(t\right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }0\le t\le {t}_{1};\phantom{\rule{2em}{0ex}}a\left(1\right)=\sum _{k=1}^{m}\left[{c}_{k}+\left(1-{t}_{k}\right){d}_{k}\right],$
(8)
so
${v}^{\prime }\left(0\right)={u}^{\prime }\left(0\right)+a\left(1\right).$
(9)
Substituting (9) into (7), we get
$\begin{array}{rcl}v\left(t\right)& =& t{v}^{\prime }\left(0\right)-{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ =& t{v}^{\prime }\left(0\right)-{\int }_{0}^{t}\left(t-s\right)f\left(s,v\left(s\right)+a\left(s\right)-a\left(1\right)s\right)\phantom{\rule{0.2em}{0ex}}ds\\ =& t{v}^{\prime }\left(0\right)-{\int }_{0}^{t}\left(t-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J.\end{array}$
(10)
By virtue of (5), we see that $v\in C\left[J,R\right]$ (in fact, $v\in {C}^{1}\left[J,R\right]$) and
$v\left(1\right)=u\left(1\right)-a\left(1\right)+a\left(1\right)=u\left(1\right)=0,$
so, letting $t=1$ in (10), we find
${v}^{\prime }\left(0\right)={\int }_{0}^{1}\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$
(11)
Substituting (11) into (10), we get
$\begin{array}{rcl}v\left(t\right)& =& {\int }_{t}^{1}t\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{0}^{t}s\left(1-t\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int }_{0}^{1}G\left(t,s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,\end{array}$

so $v\left(t\right)$ is a solution of the integral equation (2).

Conversely, suppose that $v\in C\left[J,R\right]$ is a solution of (2), i.e.
$v\left(t\right)=\left(1-t\right){\int }_{0}^{t}sg\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+t{\int }_{t}^{1}\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J.$
(12)
By (4), it is clear that $g\left(t,v\left(t\right)\right)$ is continuous on ${J}^{\prime }$, so differentiation of (12) gives
$\begin{array}{rcl}{v}^{\prime }\left(t\right)& =& -{\int }_{0}^{t}sg\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\left(1-t\right)tg\left(t,v\left(t\right)\right)\\ +{\int }_{t}^{1}\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds-t\left(1-t\right)g\left(t,v\left(t\right)\right)\\ =& -{\int }_{0}^{t}sg\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{t}^{1}\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {J}^{\prime }.\end{array}$
(13)
Differentiating again, we get
${v}^{″}\left(t\right)=-tg\left(t,v\left(t\right)\right)-\left(1-t\right)g\left(t,v\left(t\right)\right)=-g\left(t,v\left(t\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {J}^{\prime }.$
(14)
From (13) we see that ${v}^{\prime }\left({t}_{k}^{+}\right)$ and ${v}^{\prime }\left({t}_{k}^{-}\right)$ ($k=1,2,\dots ,m$) exist and
${v}^{\prime }\left({t}_{k}^{+}\right)={v}^{\prime }\left({t}_{k}^{-}\right)=-{\int }_{0}^{{t}_{k}}sg\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{t}_{k}}^{1}\left(1-s\right)g\left(s,v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$
(15)

It follows from (4), (5), (12), (14), and (15) that $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ and $u\left(t\right)$ satisfies (1). □

Lemma 2 Let condition (H1) be satisfied. If $v\in {L}^{p}\left[J,R\right]$ is a solution of the integral equation (2), then $v\in C\left[J,R\right]$.

Proof It is clear, for function $a\left(t\right)$ defined by (5),
$|a\left(t\right)|\le {a}_{0},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J;\phantom{\rule{2em}{0ex}}{a}_{0}=\sum _{k=1}^{m}\left(|{c}_{k}|+\left(1-{t}_{k}\right)|{d}_{k}|\right).$
(16)
By (4), (5), (16), and condition (H1), we have
$\begin{array}{rcl}|g\left(t,v\right)|& \le & a+b|v+a\left(t\right)-a\left(1\right)t{|}^{p-1}\le a+b{\left(|v|+2{a}_{0}\right)}^{p-1}\\ \le & a+b{\left(2max\left\{|v|,2{a}_{0}\right\}\right)}^{p-1}\le a+b{2}^{p-1}\left({|v|}^{p-1}+{\left(2{a}_{0}\right)}^{p-1}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,v\in R,\end{array}$
so,
$|g\left(t,v\right)|\le {a}_{1}+{b}_{1}{|v|}^{p-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,v\in R,$
(17)
where
${a}_{1}=a+b{2}^{2\left(p-1\right)}{a}_{0}^{p-1},\phantom{\rule{2em}{0ex}}{b}_{1}=b{2}^{p-1}.$
It is clear that $g\left(t,v\right)$ satisfies the Caratheodory condition, i.e. $g\left(t,v\right)$ is measurable with respect to t on J for every $v\in R$ and is continuous with respect to v on R for almost $t\in J$ (in fact, $g\left(t,v\right)$ is discontinuous only at $t={t}_{k}$ ($k=1,2,\dots ,m$)), so (17) implies [22, 23] that the operator g defined by
$\left(gv\right)\left(t\right)=g\left(t,v\left(t\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J$
(18)

is bounded and continuous from ${L}^{p}\left[J,R\right]$ into ${L}^{q}\left[J,R\right]$, where $\frac{1}{p}+\frac{1}{q}=1$ ($q>1$).

Let $v\in {L}^{p}\left[J,R\right]$ be a solution of the integral equation (2). Then by the Hölder inequality,
$|v\left({t}_{1}\right)-v\left({t}_{2}\right)|\le {\left({\int }_{0}^{1}|G\left({t}_{1},s\right)-G\left({t}_{2},s\right){|}^{p}\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}|g\left(s,v\left(s\right)\right){|}^{q}\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{q}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }{t}_{1},{t}_{2}\in J,$

which implies by virtue of the uniform continuity of $G\left(t,s\right)$ on $J×J$ that $v\in C\left[J,R\right]$. □

## 2 Variational approach

Theorem 1 If conditions (H1) and (H2) are satisfied, then BVP (1) has at least one solution $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$.

Proof By Lemma 1 and Lemma 2, we need only to show that the integral equation (2) has a solution $v\in {L}^{p}\left[J,R\right]$. The integral equation (2) can be written in the form
$v=Ggv,$
(19)
where G is the linear integral operator defined by
$\left(Gv\right)\left(t\right)={\int }_{0}^{1}G\left(t,s\right)v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,$
(20)
and the nonlinear operator g is defined by (18), which is bounded and continuous from ${L}^{p}\left[J,R\right]$ into ${L}^{q}\left[J,R\right]$ ($\frac{1}{p}+\frac{1}{q}=1$). It is well known that $G\left(t,s\right)$ is a ${L}^{2}$ positive-definite kernel with eigenvalues $\left\{\frac{1}{{n}^{2}{\pi }^{2}}\right\}$ ($n=1,2,3,\dots$) and, by the continuity of $G\left(t,s\right)$, we have
${\int }_{0}^{1}{\int }_{0}^{1}{\left[G\left(t,s\right)\right]}^{p}\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty },$
(21)
so [22, 23] the linear operator G defined by (20) is completely continuous from ${L}^{2}\left[J,R\right]$ into ${L}^{2}\left[J,R\right]$ and also from ${L}^{q}\left[J,R\right]$ into ${L}^{p}\left[J,R\right]$, and $G=H{H}^{\ast }$, where $H={G}^{\frac{1}{2}}$ (the positive square-root operator of G) is completely continuous from ${L}^{2}\left[J,R\right]$ into ${L}^{p}\left[J,R\right]$ and ${H}^{\ast }$ denotes the adjoint operator of H, which is completely continuous from ${L}^{q}\left[J,R\right]$ into ${L}^{2}\left[J,R\right]$. We now show that (19) has a solution $v\in {L}^{p}\left[J,R\right]$ is equivalent to the equation
$u={H}^{\ast }gHu$
(22)
has a solution $u\in {L}^{2}\left[J,R\right]$. In fact, if $v\in {L}^{p}\left[J,R\right]$ is a solution of (19), i.e. $v=H{H}^{\ast }gv$, then ${H}^{\ast }gv={H}^{\ast }gH{H}^{\ast }gv$, so, $u={H}^{\ast }gv\in {L}^{2}\left[J,R\right]$ and u is a solution of (22). Conversely, if $u\in {L}^{2}\left[J,R\right]$ is a solution of (22), then $Hu=H{H}^{\ast }gHu=GgHu$, so, $v=Hu\in {L}^{p}\left[J,R\right]$ and v is a solution of (19). Consequently, we need only to show that (22) has a solution $u\in {L}^{2}\left[J,R\right]$. It is well known [22, 23] that the functional Φ defined by
$\mathrm{\Phi }\left(u\right)=\frac{1}{2}\left(u,u\right)-{\int }_{0}^{1}dt{\int }_{0}^{\left(Hu\right)\left(t\right)}g\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {L}^{2}\left[J,R\right]$
(23)
is a ${C}^{1}$ functional on ${L}^{2}\left[J,R\right]$ and its Fréchet derivative is
${\mathrm{\Phi }}^{\prime }\left(u\right)=u-{H}^{\ast }gHu,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {L}^{2}\left[J,R\right].$
(24)

Hence we need only to show that there exists a $u\in {L}^{2}\left[J,R\right]$ such that ${\mathrm{\Phi }}^{\prime }\left(u\right)=\theta$ (θ denotes the zero element of ${L}^{2}\left[J,R\right]$), i.e. u is a critical point of functional Φ.

By (4), (5), (16), and condition (H1), we have
${\int }_{0}^{u}g\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv={\int }_{0}^{u+a\left(t\right)-a\left(1\right)t}f\left(t,w\right)\phantom{\rule{0.2em}{0ex}}dw-{\int }_{0}^{a\left(t\right)-a\left(1\right)t}f\left(t,w\right)\phantom{\rule{0.2em}{0ex}}dw,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R$
(25)
and
$\begin{array}{rcl}|{\int }_{0}^{a\left(t\right)-a\left(1\right)t}f\left(t,w\right)\phantom{\rule{0.2em}{0ex}}dw|& \le & |a\left(t\right)-a\left(1\right)t|\left(a+b|a\left(t\right)-a\left(1\right)t{|}^{p-1}\right)\\ \le & 2{a}_{0}\left(a+b{2}^{p-1}{a}_{0}^{p-1}\right)={a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J.\end{array}$
(26)
So, (25), (26), and condition (H2) imply
$\begin{array}{rcl}{\int }_{0}^{\left(Hu\right)\left(t\right)}g\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv& \le & {\int }_{0}^{\left(Hu\right)\left(t\right)+a\left(t\right)-a\left(1\right)t}f\left(t,w\right)\phantom{\rule{0.2em}{0ex}}dw+{a}_{2}\\ \le & c{\left\{\left(Hu\right)\left(t\right)+a\left(t\right)-a\left(1\right)t\right\}}^{2}+d+{a}_{2}\\ \le & 2c\left\{{\left[\left(Hu\right)\left(t\right)\right]}^{2}+{\left[a\left(t\right)-a\left(1\right)t\right]}^{2}\right\}+d+{a}_{2}\\ \le & 2c{\left[\left(Hu\right)\left(t\right)\right]}^{2}+8c{a}_{0}^{2}+d+{a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {L}^{2}\left[J,R\right],t\in J.\end{array}$
(27)
It is well known [24],
$\parallel G\parallel ={\lambda }_{1}=\frac{1}{{\pi }^{2}},$
(28)
where G is defined by (20) and is regarded as a positive-definite operator from ${L}^{2}\left[J,R\right]$ into ${L}^{2}\left[J,R\right]$, and ${\lambda }_{1}$ denotes the largest eigenvalue of G. It follows from (23), (27), and (28) that
$\begin{array}{rcl}\mathrm{\Phi }\left(u\right)& \ge & \frac{1}{2}\left(u,u\right)-2c\left(Hu,Hu\right)-8c{a}_{0}^{2}-d-{a}_{2}\\ =& \frac{1}{2}\left(u,u\right)-2c\left(Gu,u\right)-8c{a}_{0}^{2}-d-{a}_{2}\ge \frac{1}{2}\left(u,u\right)-\frac{2c}{{\pi }^{2}}\left(u,u\right)-8c{a}_{0}^{2}-d-{a}_{2}\\ =& \left(\frac{1}{2}-\frac{2c}{{\pi }^{2}}\right){\parallel u\parallel }^{2}-8c{a}_{0}^{2}-d-{a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {L}^{2}\left[J,R\right],\end{array}$
(29)
which implies by virtue of $0 (see condition (H2)) that
$\underset{\parallel u\parallel \to \mathrm{\infty }}{lim}\mathrm{\Phi }\left(u\right)=\mathrm{\infty }.$
(30)
So, there exists a $r>0$ such that
$\mathrm{\Phi }\left(u\right)>\mathrm{\Phi }\left(\theta \right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {L}^{2}\left[J,R\right],\parallel u\parallel >r.$
(31)
It is well known [22, 23] that the ball $T\left(\theta ,r\right)=\left\{u\in {L}^{2}\left[J,R\right]:\parallel u\parallel \le r\right\}$ is weakly closed and weakly compact and the functional $\mathrm{\Phi }\left(u\right)$ is weakly lower semicontinuous, so, there exists ${u}^{\ast }\in T\left(\theta ,r\right)$ such that
$\mathrm{\Phi }\left({u}^{\ast }\right)=\underset{u\in T\left(\theta ,r\right)}{inf}\mathrm{\Phi }\left(u\right)\le \mathrm{\Phi }\left(\theta \right).$
(32)
It follows from (31) and (32) that
$\mathrm{\Phi }\left({u}^{\ast }\right)=\underset{u\in {L}^{2}\left[J,R\right]}{inf}\mathrm{\Phi }\left(u\right).$

Hence ${\mathrm{\Phi }}^{\prime }\left({u}^{\ast }\right)=\theta$ and the theorem is proved. □

Example 1 Consider the BVP
$\left\{\begin{array}{l}-{u}^{″}\left(t\right)=\frac{9}{2}u\left(t\right)sin\left(t-u\left(t\right)\right)-{t}^{3},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {J}^{\prime },\\ \mathrm{\Delta }u{|}_{t={t}_{k}}={c}_{k}\phantom{\rule{1em}{0ex}}\left(k=1,2,\dots ,m\right),\\ \mathrm{\Delta }{u}^{\prime }{|}_{t={t}_{k}}={d}_{k}\phantom{\rule{1em}{0ex}}\left(k=1,2,\dots ,m\right),\\ u\left(0\right)=u\left(1\right)=0,\end{array}$
(33)

where $J=\left[0,1\right]$, $0<{t}_{1}<\cdots <{t}_{k}<\cdots <{t}_{m}<1$, ${J}^{\prime }=J\mathrm{\setminus }\left\{{t}_{1},\dots ,{t}_{k},\dots ,{t}_{m}\right\}$, ${c}_{k}$ and ${d}_{k}$ ($k=1,2,\dots ,m$) are any real numbers.

Conclusion BVP (33) has at least one solution $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$.

Proof Evidently, (33) is a BVP of the form (1) with
$f\left(t,u\right)=\frac{9}{2}usin\left(t-u\right)-{t}^{3}.$
(34)
It is clear that $f\in C\left[J×R,R\right]$. By (34), we have
$|f\left(t,u\right)|\le \frac{9}{2}|u|+1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R.$
(35)
Moreover, it is well known that
$|u|\le \frac{1}{2}\left(1+{u}^{2}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in R.$
(36)
So, (35) and (36) imply that
$|f\left(t,u\right)|\le \frac{9}{4}{u}^{2}+\frac{13}{4},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R,$
and consequently, condition (H1) is satisfied for $p=3$, $a=\frac{13}{4}$ and $b=\frac{9}{4}$. On the other hand, choose ${ϵ}_{0}$ such that
$0<{ϵ}_{0}<\frac{1}{4}\left({\pi }^{2}-9\right).$
(37)
For $|u|\ge \frac{1}{{ϵ}_{0}}$, we have $|u|\le {ϵ}_{0}{u}^{2}$, so,
$|u|\le {ϵ}_{0}{u}^{2}+\frac{1}{{ϵ}_{0}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in R.$
(38)
By (35), we have
${\int }_{0}^{u}f\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv\le \frac{9}{4}{u}^{2}+|u|,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R.$
(39)
It follows from (38) and (39) that
${\int }_{0}^{u}f\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv\le \left(\frac{9}{4}+{ϵ}_{0}\right){u}^{2}+\frac{1}{{ϵ}_{0}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R.$
(40)
Since, by virtue of (37),
$0<\frac{9}{4}+{ϵ}_{0}<\frac{{\pi }^{2}}{4},$

we see that (40) implies that condition (H2) is satisfied for $c=\frac{9}{4}+{ϵ}_{0}$ and $d=\frac{1}{{ϵ}_{0}}$. Hence, our conclusion follows from Theorem 1. □

By using the Mountain Pass Lemma and the Minimax Principle established by Ambrosetti and Rabinowitz [25, 26], we have obtained in [23] the existence of a nontrivial solution and the existence of infinitely many nontrivial solutions for a class of nonlinear integral equations. Since (2) is a special case of such nonlinear integral equations, we get the following result for (2).

Lemma 3 (Special case of Theorem 1 and Theorem 2 in [23])

Suppose the following.
1. (a)
There exist $p>2$ and $a>0$, $b>0$ such that
$|g\left(t,v\right)|\le a+b{|v|}^{p-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,v\in R.$

2. (b)
There exist $0\le \tau <\frac{1}{2}$ and $M>0$ such that
${\int }_{0}^{v}g\left(t,w\right)\phantom{\rule{0.2em}{0ex}}dw\le \tau vg\left(t,v\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,|v|\ge M.$

3. (c)

$\frac{g\left(t,v\right)}{v}\to 0$ as $v\to 0$ uniformly for $t\in J$ and $\frac{g\left(t,v\right)}{v}\to \mathrm{\infty }$ as $|v|\to \mathrm{\infty }$ uniformly for $t\in J$.

Then the integral equation (2) has at least one nontrivial solution in ${L}^{p}\left[J,R\right]$. If, in addition,
1. (d)

$g\left(t,-v\right)=-g\left(t,v\right)$, $\mathrm{\forall }t\in J$, $v\in R$.

Then the integral equation (2) has infinite many nontrivial solutions in ${L}^{p}\left[J,R\right]$.

Let us list more conditions for the function $f\left(t,u\right)$.

(H3) There exist $0\le \tau <\frac{1}{2}$ and $M>0$ such that
${\int }_{0}^{u}f\left(t,v+a\left(t\right)-a\left(1\right)t\right)\phantom{\rule{0.2em}{0ex}}dv\le \tau uf\left(t,u+a\left(t\right)-a\left(1\right)t\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,|u|\ge M.$

(H4) $\frac{f\left(t,u+a\left(t\right)-a\left(1\right)t\right)}{u}\to 0$ as $u\to 0$ uniformly for $t\in J$, and $\frac{f\left(t,u+a\left(t\right)-a\left(1\right)t\right)}{u}\to \mathrm{\infty }$ as $|u|\to \mathrm{\infty }$ uniformly for $t\in J$.

(H5) $f\left(t,-u+a\left(t\right)-a\left(1\right)t\right)=-f\left(t,u+a\left(t\right)-a\left(1\right)t\right)$, $\mathrm{\forall }t\in J$, $u\in R$.

Theorem 2 Suppose that conditions (H1), (H3), and (H4) are satisfied. Then BVP (1) has at least one solution $u\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$. If, in addition, condition (H5) is satisfied, then BVP (1) has infinitely many solutions ${u}_{n}\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ ($n=1,2,3,\dots$).

Proof In the proof of Lemma 2, we see that condition (H1) implies condition (a) of Lemma 3 (see (17)). On the other hand, it is clear that conditions (H3), (H4), (H5) are the same as conditions (b), (c), (d) in Lemma 3, respectively. Hence the conclusion of Theorem 2 follows from Lemma 3, Lemma 2, and Lemma 1. □

Example 2 Consider the BVP
$\left\{\begin{array}{l}-{u}^{″}\left(t\right)=\left\{\begin{array}{ll}{\left[u\left(t\right)-t\right]}^{3},& \mathrm{\forall }0\le t<\frac{1}{2};\\ {\left[u\left(t\right)+3t-3\right]}^{3},& \mathrm{\forall }\frac{1}{2}
(41)

Conclusion BVP (41) has infinite many solutions ${u}_{n}\in P{C}^{1}\left[J,R\right]\cap {C}^{2}\left[{J}^{\prime },R\right]$ ($n=1,2,3,\dots$).

Proof Obviously, (41) is a BVP of form (1). In this situation, $J=\left[0,1\right]$, $m=1$, ${t}_{1}=\frac{1}{2}$, ${J}^{\prime }=\left[0,1\right]\mathrm{\setminus }\left\{\frac{1}{2}\right\}$, ${c}_{1}=1$, ${d}_{1}=-4$, and
$f\left(t,u\right)=\left\{\begin{array}{ll}{\left(u-t\right)}^{3},& \mathrm{\forall }0\le t\le \frac{1}{2};\\ {\left(u+3t-3\right)}^{3},& \mathrm{\forall }\frac{1}{2}
(42)
It is clear that $f\left(t,u\right)$ is continuous on ${J}^{\prime }×R$, left continuous at $t={t}_{1}$, and the right limit $f\left({t}_{1}^{+},u\right)$ exists. By (42), we have
$\begin{array}{rcl}|f\left(t,u\right)|& \le & {\left(|u|+\frac{3}{2}\right)}^{3}\le {\left(2max\left\{|u|,\frac{3}{2}\right\}\right)}^{3}\\ \le & {2}^{3}\left(|u{|}^{3}+{\left(\frac{3}{2}\right)}^{3}\right)=8{|u|}^{3}+27,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R,\end{array}$
so, condition (H1) is satisfied for $p=4$, $a=27$ and $b=8$. By (5), we have
$a\left(t\right)=\left\{\begin{array}{ll}0,& \mathrm{\forall }0\le t\le \frac{1}{2};\\ 3-4t,& \mathrm{\forall }\frac{1}{2}
(43)
so, $a\left(1\right)=-1$ and (42) and (43) imply
$f\left(t,u+a\left(t\right)-a\left(1\right)t\right)={u}^{3},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in J,u\in R,$
(44)

and, consequently, (H3) is satisfied for $\tau =\frac{1}{4}$ and any $M>0$. On the other hand, from (44) we see that conditions (H4) and (H5) are all satisfied. Hence, our conclusion follows from Theorem 2. □

## Declarations

### Acknowledgements

Research was supported by the National Nature Science Foundation of China (No. 10671167).

## Authors’ Affiliations

(1)
Department of Mathematics, Shandong University

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