Open Access

Multiple solutions to impulsive differential equations

Advances in Difference Equations20172017:20

https://doi.org/10.1186/s13662-017-1079-4

Received: 4 August 2016

Accepted: 6 January 2017

Published: 18 January 2017

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.

Keywords

multiple solutionscritical point theoremimpulses

MSC

34A3734K10

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 [15] 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
  1. (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
  1. (ii)

    \(r<\Phi(x_{1})\).

     
  2. (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 \(\lambda\in(\frac{32}{e^{8}-e^{2}}, \frac{1}{2(e^{2}-1)})\).

Declarations

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huaian, P.R. China

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