Variational approach to a class of impulsive differential equations

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.

1 Introduction

The theory of impulsive differential equations has been emerging as an important area of investigation in recent years [1–3]. 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 [4–15]. 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 [16–20]. 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:

$$\{\begin{array}{l}-u^{″}(t)=f(t,u(t)),\mathrm{\forall }t\in J^{\prime },\\ \mathrm{\Delta }u{|}_{t=t_k}=c_k(k=1,2,3,\dots ,m),\\ \mathrm{\Delta }{u}^{\prime }{|}_{t=t_k}=d_k(k=1,2,3,\dots ,m),\\ u(0)=u(1)=0,\end{array}$$

(1)

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, $f(t,u)$ is a real function defined on $J\times R$, where R denotes the set of all real numbers, and $f(t,u)$ is continuous on $J^{\prime }\times R$, left continuous at $t=t_k$, i.e.

$$\underset{t\to t_k-0,w\to u}{lim}f(t,w)=f(t_k,u)$$

for any $u\in R$ ($k=1,2,\dots ,m$), and the right limit at $t=t_k$ exists, i.e.

$$\underset{t\to t_k+0,w\to u}{lim}f(t,w)$$

(denoted by $f(t_k^{+},u)$) exists for any $u\in R$ ($k=1,2,\dots ,m$). $\mathrm{\Delta }u{|}_{t=t_k}$ denotes the jump of $u(t)$ at $t=t_k$, i.e.

$$\mathrm{\Delta }u{|}_{t=t_k}=u\left(t_k^{+}\right)-u\left(t_k^{-}\right),$$

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

Let us list some conditions.

(H₁) There exist $p>2$, $a>0$ and $b>0$ such that

$$|f(t,u)|\le a+b{|u|}^{p-1},\mathrm{\forall }t\in J,u\in R.$$

(H₂) There exist $0<c<\frac{{\pi }^2}{4}$ and $d>0$ such that

$${\int }_0^{u}f(t,v)dv\le c{u}^2+d,\mathrm{\forall }t\in J,u\in R.$$

Lemma 1 $u\in P{C}^1[J,R]\cap C^2[J^{\prime },R]$ is a solution of BVP (1) if and only if $v\in C[J,R]$ is a solution of the integral equation

$$v(t)={\int }_0^{1}G(t,s)g(s,v(s))ds,\mathrm{\forall }t\in J,$$

(2)

where

$$G(t,s)=\{\begin{array}{ll}s(1-t),& \mathrm{\forall }0\le s\le t\le 1;\\ t(1-s),& \mathrm{\forall }0\le t<s\le 1,\end{array}$$

(3)

$$g(t,v)=f(t,v+a(t)-a(1)t),\mathrm{\forall }t\in J,v\in R$$

(4)

and

$$v(t)=u(t)-a(t)+a(1)t,a(t)=\sum _{0<t_k<t}[c_k+(t-t_k)d_k],\mathrm{\forall }t\in J.$$

(5)

Proof For $u\in P{C}^1[J,R]\cap C^2[J^{\prime },R]$, we have the formula (see [21], Lemma 1(b))

$$\begin{array}{rcl}u(t)& =& u(0)+t{u}^{\prime }(0)+{\int }_0^t(t-s)u^{″}(s)ds\\ +\sum _{0<t_k<t}\{[u\left(t_k^{+}\right)-u(t_k)]+(t-t_k)[u^{\prime }\left(t_k^{+}\right)-u^{\prime }\left(t_k^{-}\right)]\},\mathrm{\forall }t\in J.\end{array}$$

(6)

So, if $u\in P{C}^1[J,R]\cap C^2[J^{\prime },R]$ is a solution of BVP (1), then, by (1) and (6), we have

$$\begin{array}{rcl}u(t)& =& t{u}^{\prime }(0)-{\int }_0^t(t-s)f(s,u(s))ds+\sum _{0<t_k<t}[c_k+(t-t_k)d_k]\\ =& t{u}^{\prime }(0)-{\int }_0^t(t-s)f(s,u(s))ds+a(t),\mathrm{\forall }t\in J.\end{array}$$

(7)

It is clear, by (5), that

$$a(t)=0,\mathrm{\forall }0\le t\le t_1;a(1)=\sum _{k=1}^m[c_k+(1-t_k)d_k],$$

(8)

so

$$v^{\prime }(0)=u^{\prime }(0)+a(1).$$

(9)

Substituting (9) into (7), we get

$$\begin{array}{rcl}v(t)& =& t{v}^{\prime }(0)-{\int }_0^t(t-s)f(s,u(s))ds\\ =& t{v}^{\prime }(0)-{\int }_0^t(t-s)f(s,v(s)+a(s)-a(1)s)ds\\ =& t{v}^{\prime }(0)-{\int }_0^t(t-s)g(s,v(s))ds,\mathrm{\forall }t\in J.\end{array}$$

(10)

By virtue of (5), we see that $v\in C[J,R]$ (in fact, $v\in C^1[J,R]$) and

$$v(1)=u(1)-a(1)+a(1)=u(1)=0,$$

so, letting $t=1$ in (10), we find

$$v^{\prime }(0)={\int }_0^1(1-s)g(s,v(s))ds.$$

(11)

Substituting (11) into (10), we get

$$\begin{array}{rcl}v(t)& =& {\int }_t^{1}t(1-s)g(s,v(s))ds+{\int }_0^{t}s(1-t)g(s,v(s))ds\\ =& {\int }_0^{1}G(t,s)g(s,v(s))ds,\mathrm{\forall }t\in J,\end{array}$$

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

Conversely, suppose that $v\in C[J,R]$ is a solution of (2), i.e.

$$v(t)=(1-t){\int }_0^{t}sg(s,v(s))ds+t{\int }_t^1(1-s)g(s,v(s))ds,\mathrm{\forall }t\in J.$$

(12)

By (4), it is clear that $g(t,v(t))$ is continuous on $J^{\prime }$, so differentiation of (12) gives

$$\begin{array}{rcl}{v}^{\prime }(t)& =& -{\int }_0^{t}sg(s,v(s))ds+(1-t)tg(t,v(t))\\ +{\int }_t^1(1-s)g(s,v(s))ds-t(1-t)g(t,v(t))\\ =& -{\int }_0^{t}sg(s,v(s))ds+{\int }_t^1(1-s)g(s,v(s))ds,\mathrm{\forall }t\in J^{\prime }.\end{array}$$

(13)

Differentiating again, we get

$$v^{″}(t)=-tg(t,v(t))-(1-t)g(t,v(t))=-g(t,v(t)),\mathrm{\forall }t\in J^{\prime }.$$

(14)

From (13) we see that $v^{\prime }(t_k^{+})$ and $v^{\prime }(t_k^{-})$ ($k=1,2,\dots ,m$) exist and

$$v^{\prime }\left(t_k^{+}\right)=v^{\prime }\left(t_k^{-}\right)=-{\int }_0^{t_k}sg(s,v(s))ds+{\int }_{t_k}^1(1-s)g(s,v(s))ds.$$

(15)

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

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

Proof It is clear, for function $a(t)$ defined by (5),

$$|a(t)|\le a_0,\mathrm{\forall }t\in J;a_0=\sum _{k=1}^m(|c_k|+(1-t_k)|d_k|).$$

(16)

By (4), (5), (16), and condition (H₁), we have

$$\begin{array}{rcl}|g(t,v)|& \le & a+b|v+a(t)-a(1)t{|}^{p-1}\le a+b{(|v|+2a_0)}^{p-1}\\ \le & a+b{(2max\{|v|,2a_0\})}^{p-1}\le a+b{2}^{p-1}({|v|}^{p-1}+{(2a_0)}^{p-1}),\mathrm{\forall }t\in J,v\in R,\end{array}$$

so,

$$|g(t,v)|\le a_1+b_1{|v|}^{p-1},\mathrm{\forall }t\in J,v\in R,$$

(17)

where

$$a_1=a+b{2}^{2(p-1)}{a}_0^{p-1},b_1=b{2}^{p-1}.$$

It is clear that $g(t,v)$ satisfies the Caratheodory condition, i.e. $g(t,v)$ 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, $g(t,v)$ is discontinuous only at $t=t_k$ ($k=1,2,\dots ,m$)), so (17) implies [22, 23] that the operator g defined by

$$(gv)(t)=g(t,v(t)),\mathrm{\forall }t\in J$$

(18)

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

Let $v\in L^p[J,R]$ be a solution of the integral equation (2). Then by the Hölder inequality,

$$|v(t_1)-v(t_2)|\le {\left({\int }_0^1\right|G(t_1,s)-G(t_2,s){|}^{p}ds)}^{\frac{1}{p}}{\left({\int }_0^1\right|g(s,v(s)){|}^{q}ds)}^{\frac{1}{q}},\mathrm{\forall }{t}_1,t_2\in J,$$

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

2 Variational approach

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

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

$$v=Ggv,$$

(19)

where G is the linear integral operator defined by

$$(Gv)(t)={\int }_0^{1}G(t,s)v(s)ds,\mathrm{\forall }t\in J,$$

(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

$${\int }_0^1{\int }_0^1{[G(t,s)]}^{p}dsdt<\mathrm{\infty },$$

(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

$$u=H^{\ast }gHu$$

(22)

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

$$\mathrm{\Phi }(u)=\frac{1}{2}(u,u)-{\int }_0^{1}dt{\int }_0^{(Hu)(t)}g(t,v)dv,\mathrm{\forall }u\in L^2[J,R]$$

(23)

is a $C^1$ functional on $L^2[J,R]$ and its Fréchet derivative is

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

(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 (H₁), we have

$${\int }_0^{u}g(t,v)dv={\int }_0^{u+a(t)-a(1)t}f(t,w)dw-{\int }_0^{a(t)-a(1)t}f(t,w)dw,\mathrm{\forall }t\in J,u\in R$$

(25)

and

$$\begin{array}{rcl}|{\int }_0^{a(t)-a(1)t}f(t,w)dw|& \le & |a(t)-a(1)t|(a+b|a(t)-a(1)t{|}^{p-1})\\ \le & 2a_0(a+b{2}^{p-1}{a}_0^{p-1})=a_2,\mathrm{\forall }t\in J.\end{array}$$

(26)

So, (25), (26), and condition (H₂) imply

$$\begin{array}{rcl}{\int }_0^{(Hu)(t)}g(t,v)dv& \le & {\int }_0^{(Hu)(t)+a(t)-a(1)t}f(t,w)dw+a_2\\ \le & c{\{(Hu)(t)+a(t)-a(1)t\}}^2+d+a_2\\ \le & 2c\{{[(Hu)(t)]}^2+{[a(t)-a(1)t]}^2\}+d+a_2\\ \le & 2c{[(Hu)(t)]}^2+8c{a}_0^2+d+a_2,\mathrm{\forall }u\in L^2[J,R],t\in J.\end{array}$$

(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

$$\begin{array}{rcl}\mathrm{\Phi }(u)& \ge & \frac{1}{2}(u,u)-2c(Hu,Hu)-8c{a}_0^2-d-a_2\\ =& \frac{1}{2}(u,u)-2c(Gu,u)-8c{a}_0^2-d-a_2\ge \frac{1}{2}(u,u)-\frac{2c}{{\pi }^2}(u,u)-8c{a}_0^2-d-a_2\\ =& (\frac{1}{2}-\frac{2c}{{\pi }^2}){\parallel u\parallel }^2-8c{a}_0^2-d-a_2,\mathrm{\forall }u\in L^2[J,R],\end{array}$$

(29)

which implies by virtue of $0<c<\frac{{\pi }^2}{4}$ (see condition (H₂)) that

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

(30)

So, there exists a $r>0$ such that

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

(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

$$\{\begin{array}{l}-u^{″}(t)=\frac{9}{2}u(t)sin(t-u(t))-t^3,\mathrm{\forall }t\in J^{\prime },\\ \mathrm{\Delta }u{|}_{t=t_k}=c_k(k=1,2,\dots ,m),\\ \mathrm{\Delta }{u}^{\prime }{|}_{t=t_k}=d_k(k=1,2,\dots ,m),\\ u(0)=u(1)=0,\end{array}$$

(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

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

(35)

Moreover, it is well known that

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

(36)

So, (35) and (36) imply that

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

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

$$0<{ϵ}_0<\frac{1}{4}({\pi }^2-9).$$

(37)

For $|u|\ge \frac{1}{{ϵ}_0}$, we have $|u|\le {ϵ}_0{u}^2$, so,

$$|u|\le {ϵ}_0{u}^2+\frac{1}{{ϵ}_0},\mathrm{\forall }u\in R.$$

(38)

By (35), we have

$${\int }_0^{u}f(t,v)dv\le \frac{9}{4}{u}^2+|u|,\mathrm{\forall }t\in J,u\in R.$$

(39)

It follows from (38) and (39) that

$${\int }_0^{u}f(t,v)dv\le (\frac{9}{4}+{ϵ}_0)u^2+\frac{1}{{ϵ}_0},\mathrm{\forall }t\in J,u\in R.$$

(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.

• (a) There exist $p>2$ and $a>0$, $b>0$ such that

  $$|g(t,v)|\le a+b{|v|}^{p-1},\mathrm{\forall }t\in J,v\in R.$$

• (b) There exist $0\le \tau <\frac{1}{2}$ and $M>0$ such that

  $${\int }_0^{v}g(t,w)dw\le \tau vg(t,v),\mathrm{\forall }t\in J,|v|\ge M.$$

• (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 $f(t,u)$.

(H₃) There exist $0\le \tau <\frac{1}{2}$ and $M>0$ such that

$${\int }_0^{u}f(t,v+a(t)-a(1)t)dv\le \tau uf(t,u+a(t)-a(1)t),\mathrm{\forall }t\in J,|u|\ge M.$$

(H₄) $\frac{f(t,u+a(t)-a(1)t)}{u}\to 0$ 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 $t\in J$.

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

Theorem 2 Suppose that conditions (H₁), (H₃), and (H₄) 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 (H₅) 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 (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

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

(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

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

so, condition (H₁) is satisfied for $p=4$, $a=27$ and $b=8$. By (5), we have

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

(43)

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

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

(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. □