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

*Proof* By Lemma 1 and Lemma 2, we need only to show that the integral equation (

2) has a solution

$v\in {L}^{p}[J,R]$. The integral equation (

2) can be written in the form

where

*G* is the linear integral operator defined by

$(Gv)(t)={\int}_{0}^{1}G(t,s)v(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,$

(20)

and the nonlinear operator

*g* is defined by (18), which is bounded and continuous from

${L}^{p}[J,R]$ into

${L}^{q}[J,R]$ (

$\frac{1}{p}+\frac{1}{q}=1$). It is well known that

$G(t,s)$ is a

${L}^{2}$ positive-definite kernel with eigenvalues

$\{\frac{1}{{n}^{2}{\pi}^{2}}\}$ (

$n=1,2,3,\dots $) and, by the continuity of

$G(t,s)$, we have

${\int}_{0}^{1}{\int}_{0}^{1}{[G(t,s)]}^{p}\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty},$

(21)

so [

22,

23] the linear operator

*G* defined by (20) is completely continuous from

${L}^{2}[J,R]$ into

${L}^{2}[J,R]$ and also from

${L}^{q}[J,R]$ into

${L}^{p}[J,R]$, and

$G=H{H}^{\ast}$, where

$H={G}^{\frac{1}{2}}$ (the positive square-root operator of

*G*) is completely continuous from

${L}^{2}[J,R]$ into

${L}^{p}[J,R]$ and

${H}^{\ast}$ denotes the adjoint operator of

*H*, which is completely continuous from

${L}^{q}[J,R]$ into

${L}^{2}[J,R]$. We now show that (19) has a solution

$v\in {L}^{p}[J,R]$ is equivalent to the equation

has a solution

$u\in {L}^{2}[J,R]$. In fact, if

$v\in {L}^{p}[J,R]$ is a solution of (19),

*i.e.* $v=H{H}^{\ast}gv$, then

${H}^{\ast}gv={H}^{\ast}gH{H}^{\ast}gv$, so,

$u={H}^{\ast}gv\in {L}^{2}[J,R]$ and

*u* is a solution of (22). Conversely, if

$u\in {L}^{2}[J,R]$ is a solution of (22), then

$Hu=H{H}^{\ast}gHu=GgHu$, so,

$v=Hu\in {L}^{p}[J,R]$ and

*v* is a solution of (19). Consequently, we need only to show that (22) has a solution

$u\in {L}^{2}[J,R]$. It is well known [

22,

23] that the functional Φ defined by

$\mathrm{\Phi}(u)=\frac{1}{2}(u,u)-{\int}_{0}^{1}dt{\int}_{0}^{(Hu)(t)}g(t,v)\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in {L}^{2}[J,R]$

(23)

is a

${C}^{1}$ functional on

${L}^{2}[J,R]$ and its Fréchet derivative is

${\mathrm{\Phi}}^{\prime}(u)=u-{H}^{\ast}gHu,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in {L}^{2}[J,R].$

(24)

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

By (4), (5), (16), and condition (H

_{1}), we have

${\int}_{0}^{u}g(t,v)\phantom{\rule{0.2em}{0ex}}dv={\int}_{0}^{u+a(t)-a(1)t}f(t,w)\phantom{\rule{0.2em}{0ex}}dw-{\int}_{0}^{a(t)-a(1)t}f(t,w)\phantom{\rule{0.2em}{0ex}}dw,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R$

(25)

and

$\begin{array}{rcl}|{\int}_{0}^{a(t)-a(1)t}f(t,w)\phantom{\rule{0.2em}{0ex}}dw|& \le & |a(t)-a(1)t|(a+b|a(t)-a(1)t{|}^{p-1})\\ \le & 2{a}_{0}(a+b{2}^{p-1}{a}_{0}^{p-1})={a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J.\end{array}$

(26)

So, (25), (26), and condition (H

_{2}) imply

$\begin{array}{rcl}{\int}_{0}^{(Hu)(t)}g(t,v)\phantom{\rule{0.2em}{0ex}}dv& \le & {\int}_{0}^{(Hu)(t)+a(t)-a(1)t}f(t,w)\phantom{\rule{0.2em}{0ex}}dw+{a}_{2}\\ \le & c{\{(Hu)(t)+a(t)-a(1)t\}}^{2}+d+{a}_{2}\\ \le & 2c\{{[(Hu)(t)]}^{2}+{[a(t)-a(1)t]}^{2}\}+d+{a}_{2}\\ \le & 2c{[(Hu)(t)]}^{2}+8c{a}_{0}^{2}+d+{a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in {L}^{2}[J,R],t\in J.\end{array}$

(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

${L}^{2}[J,R]$ into

${L}^{2}[J,R]$, and

${\lambda}_{1}$ denotes the largest eigenvalue of

*G*. It follows from (23), (27), and (28) that

$\begin{array}{rcl}\mathrm{\Phi}(u)& \ge & \frac{1}{2}(u,u)-2c(Hu,Hu)-8c{a}_{0}^{2}-d-{a}_{2}\\ =& \frac{1}{2}(u,u)-2c(Gu,u)-8c{a}_{0}^{2}-d-{a}_{2}\ge \frac{1}{2}(u,u)-\frac{2c}{{\pi}^{2}}(u,u)-8c{a}_{0}^{2}-d-{a}_{2}\\ =& (\frac{1}{2}-\frac{2c}{{\pi}^{2}}){\parallel u\parallel}^{2}-8c{a}_{0}^{2}-d-{a}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in {L}^{2}[J,R],\end{array}$

(29)

which implies by virtue of

$0<c<\frac{{\pi}^{2}}{4}$ (see condition (H

_{2})) that

$\underset{\parallel u\parallel \to \mathrm{\infty}}{lim}\mathrm{\Phi}(u)=\mathrm{\infty}.$

(30)

So, there exists a

$r>0$ such that

$\mathrm{\Phi}(u)>\mathrm{\Phi}(\theta )=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in {L}^{2}[J,R],\parallel u\parallel >r.$

(31)

It is well known [

22,

23] that the ball

$T(\theta ,r)=\{u\in {L}^{2}[J,R]:\parallel u\parallel \le r\}$ is weakly closed and weakly compact and the functional

$\mathrm{\Phi}(u)$ is weakly lower semicontinuous, so, there exists

${u}^{\ast}\in T(\theta ,r)$ such that

$\mathrm{\Phi}\left({u}^{\ast}\right)=\underset{u\in T(\theta ,r)}{inf}\mathrm{\Phi}(u)\le \mathrm{\Phi}(\theta ).$

(32)

It follows from (31) and (32) that

$\mathrm{\Phi}\left({u}^{\ast}\right)=\underset{u\in {L}^{2}[J,R]}{inf}\mathrm{\Phi}(u).$

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

**Example 1** Consider the BVP

$\{\begin{array}{l}-{u}^{\u2033}(t)=\frac{9}{2}u(t)sin(t-u(t))-{t}^{3},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in {J}^{\prime},\\ \mathrm{\Delta}u{|}_{t={t}_{k}}={c}_{k}\phantom{\rule{1em}{0ex}}(k=1,2,\dots ,m),\\ \mathrm{\Delta}{u}^{\prime}{|}_{t={t}_{k}}={d}_{k}\phantom{\rule{1em}{0ex}}(k=1,2,\dots ,m),\\ u(0)=u(1)=0,\end{array}$

(33)

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

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

*Proof* Evidently, (33) is a BVP of the form (1) with

$f(t,u)=\frac{9}{2}usin(t-u)-{t}^{3}.$

(34)

It is clear that

$f\in C[J\times R,R]$. By (34), we have

$|f(t,u)|\le \frac{9}{2}|u|+1,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R.$

(35)

Moreover, it is well known that

$|u|\le \frac{1}{2}(1+{u}^{2}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in R.$

(36)

So, (35) and (36) imply that

$|f(t,u)|\le \frac{9}{4}{u}^{2}+\frac{13}{4},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R,$

and consequently, condition (H

_{1}) is satisfied for

$p=3$,

$a=\frac{13}{4}$ and

$b=\frac{9}{4}$. On the other hand, choose

${\u03f5}_{0}$ such that

$0<{\u03f5}_{0}<\frac{1}{4}({\pi}^{2}-9).$

(37)

For

$|u|\ge \frac{1}{{\u03f5}_{0}}$, we have

$|u|\le {\u03f5}_{0}{u}^{2}$, so,

$|u|\le {\u03f5}_{0}{u}^{2}+\frac{1}{{\u03f5}_{0}},\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in R.$

(38)

By (35), we have

${\int}_{0}^{u}f(t,v)\phantom{\rule{0.2em}{0ex}}dv\le \frac{9}{4}{u}^{2}+|u|,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R.$

(39)

It follows from (38) and (39) that

${\int}_{0}^{u}f(t,v)\phantom{\rule{0.2em}{0ex}}dv\le (\frac{9}{4}+{\u03f5}_{0}){u}^{2}+\frac{1}{{\u03f5}_{0}},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R.$

(40)

Since, by virtue of (37),

$0<\frac{9}{4}+{\u03f5}_{0}<\frac{{\pi}^{2}}{4},$

we see that (40) implies that condition (H_{2}) is satisfied for $c=\frac{9}{4}+{\u03f5}_{0}$ and $d=\frac{1}{{\u03f5}_{0}}$. 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*.

- (a)
*There exist* $p>2$ *and* $a>0$,

$b>0$ *such that* $|g(t,v)|\le a+b{|v|}^{p-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,v\in R.$

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

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

*Then the integral equation* (2)

*has at least one nontrivial solution in* ${L}^{p}[J,R]$.

*If*,

*in addition*,

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

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

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

(H

_{3}) There exist

$0\le \tau <\frac{1}{2}$ and

$M>0$ such that

${\int}_{0}^{u}f(t,v+a(t)-a(1)t)\phantom{\rule{0.2em}{0ex}}dv\le \tau uf(t,u+a(t)-a(1)t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,|u|\ge M.$

(H_{4}) $\frac{f(t,u+a(t)-a(1)t)}{u}\to 0$ as $u\to 0$ uniformly for $t\in J$, and $\frac{f(t,u+a(t)-a(1)t)}{u}\to \mathrm{\infty}$ as $|u|\to \mathrm{\infty}$ uniformly for $t\in J$.

(H_{5}) $f(t,-u+a(t)-a(1)t)=-f(t,u+a(t)-a(1)t)$, $\mathrm{\forall}t\in J$, $u\in R$.

**Theorem 2** *Suppose that conditions* (H_{1}), (H_{3}), *and* (H_{4}) *are satisfied*. *Then BVP* (1) *has at least one solution* $u\in P{C}^{1}[J,R]\cap {C}^{2}[{J}^{\prime},R]$. *If*, *in addition*, *condition* (H_{5}) *is satisfied*, *then BVP* (1) *has infinitely many solutions* ${u}_{n}\in P{C}^{1}[J,R]\cap {C}^{2}[{J}^{\prime},R]$ ($n=1,2,3,\dots $).

*Proof* In the proof of Lemma 2, we see that condition (H_{1}) implies condition (a) of Lemma 3 (see (17)). On the other hand, it is clear that conditions (H_{3}), (H_{4}), (H_{5}) 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

$\{\begin{array}{l}-{u}^{\u2033}(t)=\{\begin{array}{ll}{[u(t)-t]}^{3},& \mathrm{\forall}0\le t<\frac{1}{2};\\ {[u(t)+3t-3]}^{3},& \mathrm{\forall}\frac{1}{2}<t\le 1,\end{array}\\ \mathrm{\Delta}u{|}_{t=\frac{1}{2}}=1,\\ \mathrm{\Delta}{u}^{\prime}{|}_{t=\frac{1}{2}}=-4,\\ u(0)=u(1)=0.\end{array}$

(41)

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

*Proof* Obviously, (41) is a BVP of form (1). In this situation,

$J=[0,1]$,

$m=1$,

${t}_{1}=\frac{1}{2}$,

${J}^{\prime}=[0,1]\mathrm{\setminus}\{\frac{1}{2}\}$,

${c}_{1}=1$,

${d}_{1}=-4$, and

$f(t,u)=\{\begin{array}{ll}{(u-t)}^{3},& \mathrm{\forall}0\le t\le \frac{1}{2};\\ {(u+3t-3)}^{3},& \mathrm{\forall}\frac{1}{2}<t\le 1.\end{array}$

(42)

It is clear that

$f(t,u)$ is continuous on

${J}^{\prime}\times R$, left continuous at

$t={t}_{1}$, and the right limit

$f({t}_{1}^{+},u)$ exists. By (42), we have

$\begin{array}{rcl}|f(t,u)|& \le & {(|u|+\frac{3}{2})}^{3}\le {(2max\{|u|,\frac{3}{2}\})}^{3}\\ \le & {2}^{3}(|u{|}^{3}+{\left(\frac{3}{2}\right)}^{3})=8{|u|}^{3}+27,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R,\end{array}$

so, condition (H

_{1}) is satisfied for

$p=4$,

$a=27$ and

$b=8$. By (5), we have

$a(t)=\{\begin{array}{ll}0,& \mathrm{\forall}0\le t\le \frac{1}{2};\\ 3-4t,& \mathrm{\forall}\frac{1}{2}<t\le 1,\end{array}$

(43)

so,

$a(1)=-1$ and (42) and (43) imply

$f(t,u+a(t)-a(1)t)={u}^{3},\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,u\in R,$

(44)

and, consequently, (H_{3}) is satisfied for $\tau =\frac{1}{4}$ and any $M>0$. On the other hand, from (44) we see that conditions (H_{4}) and (H_{5}) are all satisfied. Hence, our conclusion follows from Theorem 2. □