Multiple solutions to impulsive differential equations

Abstract

In this paper, we study the existence of a second-order impulsive differential equations depending on a parameter λ. By employing a critical point theorem, the existence of at least three solutions is obtained.

1 Introduction

In recent years, the study of the existence of solutions to impulsive differential equation has aroused extensive interest, we refer the reader to [1–5] and the references therein.

In [1], by using some existing critical point theorems, Xie and Luo investigated the existence of multiple solutions of the following Neumann boundary value problem:

$$\begin{aligned}& -\bigl(p(t)u'(t)\bigr)'+q(t)u(t)=\lambda f\bigl(t, u(t)\bigr), \quad t\neq t_{j}, t\in[0, 1], \\& \Delta p(t_{j})u'(t_{j})=I_{j} \bigl(u(t_{j})\bigr), \quad j=1,2,\ldots, m, \\& u'(0)=u'(1)=0. \end{aligned}$$

(1.1)

In [2], Liang and Zhang considered the following boundary value problems:

$$\begin{aligned}& -\bigl(p(t)u'(t)\bigr)'= f\bigl(t, u(t)\bigr),\quad t \neq t_{j}, t\in[0, T], \\& \Delta p(t_{j})u'(t_{j})=I_{j} \bigl(u(t_{j})\bigr),\quad j=1,2,\ldots, m, \\& u(0)=u(T),\qquad p(0)u'(0)=p(T)u'(T). \end{aligned}$$

(1.2)

The authors gave some criteria to guarantee that the problem has at least one solution under some different conditions.

In [3], Li and Shen were concerned with the existence of three solutions for the following boundary value problems:

$$\begin{aligned}& -u''(t)=\lambda f\bigl( u(t)\bigr), \quad t\neq t_{j}, t\in[0, 1], \\& \Delta u'(t_{j})=I_{j}\bigl(u(t_{j}) \bigr), \quad j=1,2,\ldots,m, \\& u(0)=u(1)=0. \end{aligned}$$

(1.3)

Motivated by the previous mentioned paper, in this paper, we will study the existence of at least three solutions for the following boundary value problems:

$$\begin{aligned}& -\bigl(p(t)u'(t)\bigr)'+u(t)=\lambda f\bigl(t, u(t) \bigr), \quad t\neq t_{j}, t\in[0, 1], \\& \Delta p(t_{j})u'(t_{j})=I_{j} \bigl(u(t_{j})\bigr), \quad j=1,2,\ldots, m, \\& u'(0)=u'(1)=0, \end{aligned}$$

(1.4)

where $0=t_{0}< t_{1}<\cdots<t_{m}<t_{m+1}=1$, $p\in PC^{1}([0,1])$, $f\in C([0,1]\times R, R)$, $I_{j}\in C(R, R)$, $j=1,2,\ldots, m$, $\Delta p(t_{j})u'(t_{j})=p(t_{j}^{+})u'(t_{j}^{+})-p(t_{j}^{-})u'(t_{j}^{-})$, $p(t_{j}^{+})u'(t_{j}^{+})$ and $p(t_{j}^{-})u'(t_{j}^{-})$ denote the right and the left limits, respectively, $\lambda\in[0, +\infty)$ is a real parameter.

2 Preliminaries

Let $p_{0}=\min_{t\in[0, 1]}p(t)>0$, $M_{0}=\max\{\frac{1}{p_{0}},1\}$, $X=W^{1,2}[0, 1]$ with the norm

$$\|u\|= \biggl( \int^{1}_{0}\bigl(p(t)\bigl\vert u'(t)\bigr\vert ^{2}+\bigl\vert u(t)\bigr\vert ^{2}\bigr)\,dt \biggr)^{\frac{1}{2}}. $$

Define the norm in $C([0, 1])$ by $\|u\|_{\infty}=\max_{t\in[0, 1]}|u(t)|$.

Lemma 2.1

For any $u\in X$, we have $\|u\|_{\infty}\leq \sqrt{2M_{0}}\|u\|$.

Proof

For $u\in X$ by the mean-value theorem, there exists $\tau\in(0, 1)$ such that $\int^{1}_{0}u(s)\,ds=u(\tau)$. Hence, for $t\in[0, 1]$, we have

$$\begin{aligned} \bigl\vert u(t)\bigr\vert =&\biggl\vert u(\tau)+ \int^{t}_{\tau}u'(s)\,ds\biggr\vert \leq\bigl\vert u(\tau)\bigr\vert + \int ^{1}_{0}\bigl\vert u'(s)\bigr\vert \,ds \\ \leq& \int^{1}_{0}\bigl\vert u(s)\bigr\vert \,ds+ \int^{1}_{0}\bigl\vert u'(s)\bigr\vert \,ds \\ \leq& \biggl( \int^{1}_{0}\bigl\vert u(s)\bigr\vert ^{2}\,ds\biggr)^{\frac{1}{2}}+\sqrt{1/p_{0}}\biggl( \int ^{1}_{0}p(t)\bigl\vert u'(t) \bigr\vert ^{2}\,dt\biggr)^{\frac{1}{2}} \\ \leq& \sqrt{2M_{0}}\|u\|. \end{aligned}$$

For every $u\in X$, we define the functional $\varphi(u): X\to R$ by

$$\varphi(u)=\Phi(u)-\lambda\Psi(u); $$

here

$$\Phi(u)=\frac{1}{2}\|u\|^{2}+\sum^{m}_{j=1} \int^{u(t_{j})}_{0} I_{j}(s)\,ds $$

and

$$\Psi(u)= \int^{1}_{0} F(t, u)\,dt, $$

where $F(t, u)=\int^{u(t)}_{0}f(t, s)\,ds$.

We easily show that φ is differentiable at any $u\in X$ and

$$\varphi'(u)v= \int^{1}_{0}\bigl(p(t)u'(t)v'(t)+u(t)v(t) \bigr)\,dt +\sum^{m}_{j=1}I_{j} \bigl(u(t_{j})\bigr)v(t_{j})-\lambda \int^{1}_{0}f\bigl(t, u(t)\bigr)v(t)\,dt. $$

Obviously, Φ is a nonnegative continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on $X^{*}$, and Ψ is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. □

Lemma 2.2

[1]

If $u\in X$ is a critical point of the functional φ, then u is a classical solution of problem (1.4).

Suppose that $E\subset X$. We denote $\overline{E}^{\omega}$ as the weak closure of E, that is, $u\in \overline{E}^{\omega}$ if there exists a sequence $\{u_{n}\}\subset E$ such that $g(u_{n})\to g(u)$ for every $g\in X^{*}$. Our main tool is the following three critical points theorem obtained in [6].

Lemma 2.3

[6], Theorem 2.1

Let X be separable and reflexive real Banach space. $\Phi: X\to R$ a nonnegative continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on $X^{*}$. $J: X\to R$ a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Assume that there exists $x_{0}\in X$ such that $\Phi(x_{0})=J(x_{0})=0$ and that

• (i) $\lim_{\|x\|\to+\infty}(\Phi(x)-\lambda J(x))=+\infty$ for all $\lambda \in[0, +\infty)$.

Further, assume that there are $r>0$, $x_{1}\in X$ such that

• (ii) $r<\Phi(x_{1})$.

• (iii) $\sup_{x\in\overline{\Phi^{-1}((-\infty, r))}^{\omega}} J(x)<\frac{r}{r+\Phi(x_{1})}J(x_{1})$.

Then, for each

$$\lambda\in\Lambda_{1}= \biggl( \frac{\Phi(x_{1})}{J(x_{1})-\sup_{x\in\overline{\Phi^{-1}((-\infty, r))}^{\omega}} J(x)}, \frac{r}{\sup_{x\in \overline{\Phi^{-1}((-\infty, r))}^{\omega}}J(x)} \biggr), $$

the equation

$$ \Phi'(x)-\lambda J'(x)=0 $$

(2.1)

has at least three solutions in X and, moreover, for each $h>1$, there exist an open interval

$$\Lambda_{2}\subseteq \biggl[0, \frac{hr}{r(J(x_{1})/\Phi(x_{1}))-\sup_{x\in \overline{\Phi^{-1}((-\infty, r))}^{\omega}}J(x)} \biggr) $$

and a positive real number σ such that, for each $\lambda\in \Lambda_{2}$, (1.2) has at least three solutions in X whose norms are less than σ.

3 Main results

Theorem 3.1

The following conditions are given.

(H₁):

$\sum^{m}_{j=1}\int^{u(t_{j})}_{0}I_{j}(t)\,dt\geq0$.

(H₂):

Let $a_{i}>0$ ($i=1,2$), $M>0$, and $0<\mu<2$ such that

$$F(t, u)\leq a_{1}|u|^{\mu}- a_{2}, \quad \textit{for } |u|\geq M, t\in[0, 1]. $$

(H₃):

There exist two positive constants c, $c_{1}$ with $c_{1}>\frac{c}{\sqrt{2M_{0}}}$, such that

$$4M_{0} \int^{1}_{0}\max_{|u|\leq c}F(t, u) \,dt< c^{2} \Biggl(\frac{c^{2}}{4M_{0}}+\frac{c_{1}^{2}}{2}+\sum ^{m}_{j=1} \int ^{c_{1}}_{0}I_{j}(t)\,dt \Biggr)^{-1} \int^{1}_{0}F(t, c_{1})\,dt. $$

Furthermore, put

$$ \begin{aligned} &\lambda_{1}=\frac{4M_{0}\int^{1}_{0}\max_{ |u|\leq c}F(t, u)\,dt}{c^{2}}, \\ &\lambda_{2}=\frac{\int^{1}_{0}F(t, c_{1})\,dt-\int^{1}_{0}\max_{ |u|\leq c} F(t, u)\,dt}{\frac{c_{1}^{2}}{2}+\sum^{m}_{j=1}\int^{c_{1}}_{0}I_{j}(s)\,ds}. \end{aligned} $$

(3.1)

Then, for each $\lambda\in(\frac{1}{\lambda_{2}}, \frac{1}{\lambda_{1}})$, problem (1.4) has at least three solutions in X.

Proof

Now we show the conditions (i)-(iii) of Lemma 2.3 are satisfied.

For any $u\in X$, $|u|\geq M$, and $\lambda\geq0$, and the assumptions (H₁)-(H₂) we have

$$\begin{aligned} \Phi(u)-\lambda \Psi(u) =&\frac{1}{2}\|u\|^{2}+\sum ^{m}_{j=1} \int ^{u(t_{j})}_{0}I_{j}(s)\,ds-\lambda \int^{1}_{0}F\bigl(t, u(t)\bigr)\,dt \\ \geq&\frac{1}{2}\|u\|^{2}-\lambda\bigl[a_{1}|u|^{\mu}-a_{2} \bigr] \\ \geq& \frac{1}{2}\|u\|^{2}-\lambda\bigl[a_{1}(2M_{0})^{\mu/2} \|u\|^{\mu}-a_{2}\bigr], \end{aligned}$$

$0<\mu<2$ implies that

$$\lim_{\|u\|\to\infty}\bigl(\Phi(u)-\lambda J(u)\bigr)=+\infty, $$

which shows the condition (i) of Lemma 2.3 is satisfied.

Let $u_{1}=c_{1}\in X$ and $c_{1}>\frac{c}{\sqrt{2M_{0}}}$. Then

$$\begin{aligned} \Phi (u_{1}) =&\frac{1}{2}\|u_{1}\|^{2}+\sum ^{m}_{j=1} \int ^{u_{1}(t_{j})}_{0}I_{j}(s)\,ds \\ =& \frac{1}{2}c_{1}^{2}+\sum^{m}_{j=1} \int ^{c_{1}}_{0}I_{j}(s)\,ds\geq \frac{1}{2}c_{1}^{2}>\frac{c^{2}}{4M_{0}}=r, \end{aligned}$$

so the condition (ii) of Lemma 2.3 is obtained.

By Lemma 2.1, if $\Phi(u)\leq r$, then

$$\bigl\vert u(t)\bigr\vert ^{2}\leq2M_{0}\|u \|^{2}\leq4M_{0}\Phi(u)\leq4M_{0}r=c^{2}, \quad \mbox{for }t\in [0, 1], $$

which implies that

$$\Phi^{-1}(-\infty, r)\subseteq\bigl\{ u\in X, \bigl\vert u(t)\bigr\vert \leq c, t\in[0, 1]\bigr\} . $$

So for any $u\in X$, we have

$$\sup_{u\in\overline{\Phi^{-1}(-\infty, r)}^{\omega}} \Psi(u) =\sup_{u\in\Phi^{-1}(-\infty, r)} \Psi(u)\leq \int^{1}_{0}\max_{ |u|\leq c}F(t, u)\,dt. $$

On the other hand, we obtain

$$ \frac{r}{r+\Phi(u_{1})}\Psi(u_{1}) =c^{2} \Biggl[4M_{0} \Biggl(\frac{c^{2}}{4M_{0}}+\frac{c_{1}^{2}}{2}+\sum^{m}_{j=1} \int ^{c_{1}}_{0}I_{j}(t)\,dt\Biggr) \Biggr]^{-1} \int^{1}_{0}F(t, c_{1})\,dt. $$

From the assumption (H₃) we have

$$\sup_{u\in\overline{\Phi^{-1}(-\infty, r)}^{\omega}} \Psi(u)< \frac{r}{r+\Phi(u_{1})}\Psi(u_{1}), $$

which shows the condition (iii) of Lemma 2.3 is satisfied.

Note that

$$\begin{aligned}& \frac{\Phi(u_{1})}{\Psi(u_{1})-\sup_{u\in\overline{\Phi^{-1}(-\infty , r)}^{\omega}} \Psi(u)}\leq \frac{\frac{1}{2}c_{1}^{2}+\sum^{m}_{j=1}\int^{c_{1}}_{0}I_{j}(s)\,ds}{\int ^{1}_{0}F(t, c_{1})\,dt-\int^{1}_{0}\max_{ |u|\leq c}F(t, u)\,dt}=\frac{1}{\lambda_{2}}, \\& \frac{r}{\sup_{u\in\overline{\Phi^{-1}(-\infty, r)}^{\omega}} \Psi(u)}\geq\frac{c^{2}}{4M_{0}\int^{1}_{0}\max_{ |u|\leq c}F(t, u)}=\frac{1}{\lambda_{1}}. \end{aligned}$$

The condition (H₃) implies $\lambda_{2}>\lambda_{1}$. In the light of Lemma 2.3, the problem (1.4) has at least three solutions in X for each $\lambda\in (1/\lambda_{2}, 1/\lambda_{1})$.

The proof is complete. □

4 Examples

Consider the following problem:

$$\begin{aligned}& -\bigl(e^{t}u'(t)\bigr)'+u(t)=\lambda f(t, u), \quad t\in[0, 1], t\neq t_{1}, \\& \Delta\bigl(e^{t_{1}}u'(t_{1}) \bigr)=u(t_{1}),\quad t_{1}=\frac{1}{2}, \\& u'(0)=u'(1)=0, \end{aligned}$$

(4.1)

where

$$f(t, u)=\left \{ \textstyle\begin{array}{l@{\quad}l} e^{2u},& u\leq4, \\ u^{1/2}+e^{8}-4,& u>4, \end{array}\displaystyle \right . $$

then

$$F(t, u)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{2}(e^{2u}-1),& u\leq4, \\ \frac{2}{3}u^{3/2}+(e^{8}-4)u+\frac{61}{6}-\frac{7}{2}e^{8}, & u>4. \end{array}\displaystyle \right . $$

Clearly $M_{0}=1$. Let $c=1$, $c_{1}=4$, it follows that

$$\begin{aligned}& 4M_{0} \int^{1}_{0}\max_{|u|\leq c}F(t, u) \,dt \\& \quad =2\bigl(e^{2}-1\bigr) < c^{2} \Biggl(\frac{c^{2}}{4M_{0}}+\frac{c_{1}^{2}}{2}+\sum ^{m}_{j=1} \int ^{c_{1}}_{0}I_{j}(s)\,ds \Biggr)^{-1} \int^{1}_{0}F(t, c_{1})\,dt= \frac{2(e^{8}-1)}{65}, \\& \frac{1}{\lambda_{1} }=\frac{1}{2(e^{2}-1)}, \qquad \frac{1}{\lambda_{2} }= \frac {32}{e^{8}-e^{2}}, \end{aligned}$$

which shows that all conditions of Theorem 3.1 are satisfied, so the problem (4.1) admits at least three solutions for $\lambda\in(\frac{32}{e^{8}-e^{2}}, \frac{1}{2(e^{2}-1)})$.