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Nonconstant periodic solutions created by impulses for singular differential equations

Boundary Value Problems20152015:193

https://doi.org/10.1186/s13661-015-0458-6

Received: 8 June 2015

Accepted: 9 October 2015

Published: 23 October 2015

Abstract

In this work we discuss the existence of nonconstant periodic solutions for nonautonomous singular second order differential equations in the presence of impulses. Our approach is variational.

Keywords

nonconstant periodic solutionsimpulsesvariational methodminimization with constraints

MSC

34B3734B16

1 Introduction

This paper is devoted to the study of the existence of nonconstant periodic solutions for non-autonomous singular second order differential equations,
$$ u^{\prime\prime}(t)+f(t,u)=e(t),\quad\mbox{for a.e. }t\in [ 0,T ] , T>0, $$
(1.1)
under impulse conditions
$$ \Delta u^{\prime}(t_{j})=I_{j}\bigl(u(t_{j}) \bigr),\quad j=1,2,\ldots,p-1, $$
(1.2)
where f is a singular negative function and \(\Delta u^{\prime }(t_{j})=u^{\prime}(t_{j}^{+})-u^{\prime}(t_{j}^{-})\), with \(u^{\prime }(t_{j}^{\pm})= \lim_{t\rightarrow t_{j}^{\pm}} u^{\prime }(t)\); \(t_{j}\) for \(j=1,2,\ldots,p-1\), are the instants where the impulses occur with \(0=t_{0}< t_{1}<\cdots<t_{p-1}<t_{p}=T\), \(t_{j+p}=t_{j}+T\). The functions \(I_{j}:\mathbb{R} \rightarrow \mathbb{R} \); \(j=1,2,\ldots,p-1\), are continuous and represent the jump discontinuities of \(u^{\prime}\) at the impulse moments, and \(I_{j+p}\equiv I_{j}\). Applications of impulsive differential equations with or without delays occur in medicine, population dynamics, and chaos theory; see [1, 2]. For the general aspects of impulsive differential equations, we refer the reader to the classical monographs [3, 4]. Due to its significance, a great deal of work has been done in the theory of impulsive differential equations; see for example [57]. It was pointed out in [8] that singular differential equations of the form (1.1) appear in the description of many phenomena in the applied sciences, such as nonlinear elasticity. Singular problems without impulse effects have been investigated extensively in the literature (see [2, 912] and the references therein). Some classical tools have been used to study such problems. These classical techniques include the coincidence degree theory of Mawhin and Willem [13], the method of upper and lower solutions [14], some fixed point theorems [15], and variational methods [16, 17]. For example, the authors in [18] obtained multiple periodic solutions for second-order perturbed Hamiltonian systems with impulse effects via variational methods. We believe that singular problems with impulsive effects have not been sufficiently studied; for some work on the subject, see [19, 20]. Inspired by the above facts, and the following important result (see [12]): if e is an integrable T-periodic function, then (1.1) has a positive T-periodic weak solution if and only if \(\int_{0}^{T}e(t)\,dt<0\), the aim of this paper is to prove a new existence result on a weak nonconstant T-periodic solutions generated by impulses (1.2) for the singular equation (1.1). Here, we say that a solution is generated by impulses if this solution exists when \(I_{j}\neq 0\), for some \(1< j< p-1\), and if it disappears when \(I_{j}\equiv 0\) for all \(1< j< p-1\).

The paper is organized as follows. Section 2 contains the basic preliminaries. An existence result of periodic solutions is given in Section 3. We conclude with an example.

2 Preliminaries

In this section we introduce some basic notions that will be used in the rest of the paper. \(P_{T}\) denotes the set of T-periodic functions \(u: \mathbb{R} \rightarrow \mathbb{R} \) satisfying \(u(t+T)=u(t)\) for all \(t\in \mathbb{R} \); \(C_{T}=\{u\in P_{T}; u\mbox{ is continuous}\}\). For \(u\in C_{T}\) we denote its norm by \(\Vert u\Vert _{\infty}=\sup \{ \vert u(t)\vert ; t\in [ 0,T ] \} \). Then \(( C_{T}, \Vert \cdot \Vert _{\infty} ) \) is a Banach space. For \(p\geq1\), \(L^{p}:=L^{p}(0,T;\mathbb{R} )\) is the classical Lebesgue space of measurable functions \(u:[0;T]\rightarrow \mathbb{R} \) such that \(\vert u(\cdot)\vert ^{p}\) is integrable, and for \(u\in L^{p}\) we define its norm by
$$ \Vert u\Vert _{L^{p}}= \biggl( \int_{0}^{T} \bigl\vert u(t)\bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}}. $$
We consider the Sobolev space \(H_{T}^{1}=\{u:\mathbb{R} \rightarrow \mathbb{R} ; u \mbox{ is absolutely continuous}, u^{\prime}\in L^{2}, \mbox{ and}\mbox{ }u(t)=u(t+T) \mbox{ for }t\in \mathbb{R} \}\). \(H_{T}^{1}\), equipped with the inner product
$$ (u,v)=\int_{0}^{T}u^{\prime}(t)v^{\prime }(t)\,dt+ \int_{0}^{T}u(t)v(t)\,dt $$
and the norm
$$ \Vert u\Vert _{H_{T}^{1}}:= \bigl( \Vert u\Vert _{L^{2}}^{2}+ \bigl\Vert u^{\prime}\bigr\Vert _{L^{2}}^{2} \bigr) ^{\frac{1}{2}} $$
is a reflexive Banach space. Also, \(H_{T}^{1}\) admits the orthogonal decomposition, \(H_{T}^{1}=E+F\), where F is the subspace of constant functions in \(H_{T}^{1}\) and E denotes the subspace of functions in \(H_{T}^{1}\) with zero mean value. E is a weakly closed subspace of \(H_{T}^{1}\). If \(u\in E\), then the Wirtinger inequality
$$ \int_{0}^{T}\bigl\vert u(t)\bigr\vert ^{2}\,dt\leq\frac {T^{2}}{4\pi ^{2}}\int_{0}^{T} \bigl\vert u^{\prime}(t)\bigr\vert ^{2}\,dt $$
(2.1)
implies that, on E, we can obtain the equivalent norm
$$ \| u\|:=\bigl\Vert u^{\prime}\bigr\Vert _{L^{2}}. $$
Also, for \(u\in E\) we have
$$ \Vert u\Vert _{\infty}\leq\sqrt{T}\Vert u\Vert . $$
(2.2)
It is easy to see that a T-periodic solution of (1.1), (1.2) with zero mean value must be a nonconstant T-periodic solution of (1.1), (1.2).

Definition 1

\(u\in H_{T}^{1}\) is solution of (1.1), (1.2) if \(u\in C_{T}\) such that for every j, \(u_{j}=u|_{[t_{j},t_{j+1}]} \in H^{2}(t_{j},t_{j+1})\), and it satisfies (1.1) for a.e. \(t\in[0,T]\), \(t\neq t_{j}\), the limits \(u^{\prime}(t_{j}^{-})\), \(u^{\prime}(t_{j}^{+})\) exist and the impulsive conditions (1.2) are satisfied.

3 Main result

We consider the impulsive second-order periodic boundary value problem,
$$ \left \{ \textstyle\begin{array}{@{}l} u^{\prime\prime}(t)+f(t,u)=e(t),\quad\mbox{for }t\in(0,T), t\neq t_{j}, \\ \Delta u^{\prime}(t_{j})=I_{j}(u(t_{j})),\quad j=1,2,\ldots,p-1, \\ u(0)-u(T)=0, \qquad u^{\prime}(0)-u^{\prime}(T)=0,\end{array}\displaystyle \right . $$
(3.1)
under the following assumptions:
  1. (H1)
    1. (i)

      \(f: \mathbb{R} \times(0,+\infty)\rightarrow \mathbb{R} \), is a negative Carathéodory function which is T-periodic in its first argument,

       
    2. (ii)

      \(\lim_{u\rightarrow0^{+}}f(t,u)=-\infty\), for a.e. \(t \in [0,T]\),

       
    3. (iii)

      \(\lim_{u\rightarrow +\infty}f(t,u)=0\), for a.e. \(t \in[0,T]\),

       
     
  2. (H2)
    1. (i)

      e is a locally integrable T-periodic function and \(\bar{e}:=\frac{1}{T}\int_{0}^{T}e(t)\,dt>0\),

       
    2. (ii)

      \(I_{j}:\mathbb{R} \rightarrow \mathbb{R} \), is a continuous bounded function for all \(j=1,\ldots,p-1\), such that \(m=\inf I_{j}<\sup I_{j}=M<-\frac{T}{p-1}\bar{e}\).

       
     

Remark 1

(H1)(iii) implies that \(\lim_{u\rightarrow+\infty}\frac {F(t,u)}{u^{2}}=0\), for a.e. \(t \in [0,T]\) where \(F(t,u):=\int_{1}^{u}f(t,s)\,ds\).

Remark 2

Consider (1.1) and suppose that \(I_{j}\equiv0\) for all \(1< j< p-1\). In this case f verifies the conditions of the second result in [12]. Then (1.1) has a positive T-periodic solution if and only if \(\bar{e}<0\). This means that (1.1) under (H1) and (H2)(i) does not have a T-periodic weak solution. However, if the impulses happen, i.e. if (H2)(ii) is fulfilled for this singular equation (1.1), there may exist a positive T-periodic weak solution. Such a solution is called a periodic solution generated by impulses as pointed out in [20].

Theorem 1

Suppose (H1) and (H2) hold. Then (3.1) admits at least one weak nonconstant T-periodic solution.

Proof

To prove this result, we rely on a variational method. In order to study problem (3.1), we consider the following modified problem:
$$ \left \{ \textstyle\begin{array}{@{}l} u^{\prime\prime}(t)+f_{r}(t,u(t))=e(t),\quad \mbox{for a.e. } t\in (0,T), t\neq t_{j}, \\ \Delta u^{\prime}(t_{j})=I_{j}(u(t_{j})),\quad j=1,2,\ldots,p-1, \\ u(0)-u(T)=0, \qquad u^{\prime}(0)-u^{\prime}(T)=0, \end{array}\displaystyle \right . $$
(3.2)
where \(f_{r}:[0,T]\times \mathbb{R} \rightarrow \mathbb{R} \) is the truncation function defined for \(r\in(0,1]\) by
$$ f_{r}(t,u)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} f(t,r), & u\leq r, \\ f(t,u), & u>r. \end{array}\displaystyle \right . $$
\(f_{r}\) is a negative, continuous, and T-periodic function in t, which satisfies (H1)(iii). Let \(F_{r}(t,u):=\int_{1}^{u}f_{r}(t,s)\,ds\).
Take \(v\in H_{T}^{1}\) and multiply the two sides of the equality \(-u^{\prime \prime}-f_{r}(t;u)+e(t)=0\) by v and integrate from 0 to T
$$ \int_{0}^{T}\bigl[-u^{\prime\prime}(t)-f_{r} \bigl(t;u(t)\bigr)+e(t)\bigr]v(t)\,dt=0. $$
(3.3)
Due to the jump discontinuities of \(u^{\prime}\) at each \(t_{j}\), \(j=1,2,\ldots,p-1\), and since v is T-periodic and \(u^{\prime }(0)-u^{\prime }(T)=0\), the first term of (3.3) becomes
$$\begin{aligned} \int_{0}^{T}u^{\prime\prime}(t)v(t)\,dt =&\sum _{j=0}^{p}\int_{t_{j}}^{t_{j+1}}u^{\prime\prime }(t)v(t)\,dt \\ =&u^{\prime}(T)v(T)-u^{\prime}(0)v(0)-\sum _{j=1}^{p-1}\Delta u^{\prime}(t_{j})v(t_{j})- \int_{0}^{T}u^{\prime}(t)v^{\prime}(t)\,dt \\ =&-\sum_{j=1}^{p-1}\Delta u^{\prime }(t_{j})v(t_{j})-\int_{0}^{T}u^{\prime}(t)v^{\prime}(t)\,dt. \end{aligned}$$
Combining the above with (3.3) we obtain
$$ \sum_{j=1}^{p-1}\Delta u^{\prime }(t_{j})v(t_{j})+ \int_{0}^{T}u^{\prime}(t)v^{\prime }(t)\,dt- \int_{0}^{T}f_{r}\bigl(t;u(t)\bigr)\,dt+ \int_{0}^{T}e(t)]v(t)\,dt=0. $$
(3.4)

As a result, we introduce the concept of a weak solution for problem (3.2). We say that a function u is a weak solution of problem (3.2) if (3.4) holds for any \(v\in H_{T}^{1}\).

Hence, we define the energy functional \(\Phi_{r}\): \(H_{T}^{1}\rightarrow \mathbb{R} \), associated to (3.2) by
$$ \Phi_{r}(u):=\frac{1}{2}\int_{0}^{T}\bigl|u^{\prime }(t)\bigr|^{2}\,dt+ \sum_{j=1}^{p-1}\int_{0}^{u(t_{j})}I_{j}(s)\,ds-\int_{0}^{T}F_{r}\bigl(t,u(t)\bigr)\,dt+ \int_{0}^{T}e(t)u(t)\,dt. $$
(3.5)
Clearly, \(\Phi_{r}\) is well defined on \(H_{T}^{1}\). Combining the weak lower semicontinuity of the \(L^{2}\)-norm and Fatou’s lemma we infer that \(\Phi_{r}\) is weakly lower semi continuous, by means of the assumptions (H1)(i), (H2). Also, it is a differentiable functional whose derivative is the functional \(\Phi _{r}^{\prime}(u)\), given by
$$ \Phi_{r}^{\prime}(u)v=\int_{0}^{T}u^{\prime}(t)v^{\prime }(t)\,dt+ \sum_{j=1}^{p-1}I_{j} \bigl(u(t_{j})\bigr)v(t_{j})-\int _{0}^{T}f_{r}\bigl(t,u(t)\bigr)v(t)\,dt+\int _{0}^{T}e(t)v(t)\,dt. $$
Obviously, from (3.4), if \(u\in H_{T}^{1}\) is a critical point of the functional \(\Phi_{r}\), then u is a weak solution of problem (3.2). So, to obtain nonconstant weak solutions, it is sufficient to prove the existence of critical points of \(\Phi_{r}\), on the weakly closed subspace E of \(H_{T}^{1}\).
Now, we claim that \(\Phi_{r}\) is coercive on E. Indeed, the assumption (H1)(iii), implies that, for all \(\varepsilon\in(0,\min(1,\frac{\pi^{2}}{T^{2}}))\), there exists \(\delta _{\varepsilon}>0\) such that, for almost every \(t\in[0,T]\), we have
$$ \bigl\vert f_{r}(t,u)\bigr\vert \leq2\varepsilon u $$
(3.6)
whenever \(\vert u\vert >\delta _{\varepsilon }\). Using (3.2), (3.6), and (H1)(iii) we obtain for all \(u\in \mathbb{R} \) and a.e \(t\in[0,T]\)
$$ \bigl\vert F_{r}(t,u)\bigr\vert \leq\varepsilon u^{2}+ \max_{\vert u\vert \leq\delta_{\varepsilon}} \bigl\vert F_{r}(t;u)\bigr\vert - \varepsilon, $$
(3.7)
so that
$$ \int_{0}^{T}F_{r}\bigl(t,u(t)\bigr)\,dt \leq\varepsilon \int_{0}^{T}\bigl\vert u(t)\bigr\vert ^{2}\,dt+C_{\varepsilon}, $$
(3.8)
where \(C_{\varepsilon}=\int_{0}^{T}\max_{\vert u\vert \leq\delta} \vert F_{r}(t;u)\vert \,dt-T\varepsilon<+\infty\). Also, one can easily see that
$$ \int_{0}^{T}e(t)u(t)\,dt+\sum _{j=1}^{p-1}\int _{0}^{u(t_{j})}I_{j}(s)\,ds \geq \bigl( m(p-1)-\Vert e\Vert _{L^{1}} \bigr) \Vert u\Vert _{\infty}. $$
Thus, for \(u\in E\), by (H2) and the previous inequalities we obtain
$$\begin{aligned} \Phi_{r}(u) =&\frac{1}{2}\int_{0}^{T}\bigl|u^{\prime }(t)\bigr|^{2}\,dt- \int_{0}^{T}F_{r}\bigl(t,u(t)\bigr)\,dt+ \int _{0}^{T}e(t)u(t)\,dt+\sum _{j=1}^{p-1}\int _{0}^{u(t_{j})}I_{j}(s)\,ds \\ \geq&\frac{1}{2}\int_{0}^{T}\bigl|u^{\prime}(t)\bigr|^{2}\,dt- \varepsilon \int_{0}^{T}u(t)^{2}\,dt-C_{\varepsilon}T+ \bigl( m(p-1)-\Vert e\Vert _{L^{1}} \bigr) \Vert u\Vert _{\infty} \\ \geq&\frac{1}{2}\biggl(1-\frac{T^{2}}{2\pi^{2}}\varepsilon\biggr)\Vert u \Vert ^{2}+\sqrt{T} \bigl( m(p-1)-\Vert e\Vert _{L^{1}} \bigr) \Vert u\Vert -C\varepsilon T. \end{aligned}$$
So, \(\Phi_{r}(u)\rightarrow+\infty\) as \(\Vert u\Vert \rightarrow+\infty\), which shows that \(\Phi_{r}\) is coercive on E. Since E is a weakly closed subspace of \(H_{T}^{1}\), using the direct method of the calculus of variations, we see that there exists \(u^{\ast}\in E\) such that
$$ \Phi_{r}\bigl(u^{\ast}\bigr)=\inf_{E} \Phi_{r}. $$

Notice that by (H1)(i) and (H2)(i), we have \(\int _{0}^{T}f_{r}(t,0)\,dt=\int_{0}^{T}f(t,r)\,dt\leq 0<\int_{0}^{T}e(t)\,dt\), so that the function \(u\equiv0\) cannot be a solution of (3.2). Hence \(u^{\ast}\) is a nontrivial solution of (3.2). □

In the following, we shall show that \(u^{\ast}\) is a solution of (3.1). For this purpose we introduce the following auxiliary result.

Lemma 1

There exist \(r_{0}\in(0,1)\) and a constant \(\beta_{0}>0\) such that each solution u of (3.2) satisfies \(r_{0}\leq u(t)\leq\beta_{0}\), for all t. In particular, any T-periodic solution of (3.2) is a solution of (3.1).

Proof

Here, we shall use some ideas from [10].

We proceed by contradiction. Suppose, on the contrary, that, for each \(r\in (0,1)\) and for each constant \(\beta>0\), there exists a T-periodic solution u of (3.2) which satisfies
$$ u(t)< r\quad \mbox{or}\quad u(t)>\beta\quad\mbox{for some }t\in [ 0,T ] . $$
(3.9)
In particular, if for each integer \(n>1\) we consider \(r_{n}=\frac{1}{n}\) and \(\beta=n\), the above assumption implies that there exists a solution \(u_{n}\) of (3.2) for \(r=r_{n}\) such that
$$ \bigl\{ u_{n}(t);t\in \mathbb{R} \bigr\} \nsubseteqq [ r_{n},n ] . $$
(3.10)
We will show that this assumption leads to a contradiction.
First, we claim that for every \(n>1\) there must exist \(\tau _{n}\in[ 0,T ] \) such that
$$ u_{n}(\tau_{n})\in\biggl[\frac{1}{n},n\biggr]. $$
Indeed, suppose on the contrary that there exists a subsequence of \((u_{n})_{n}\), which we label the same, for which \(\min u_{n}(t)>n\). It follows from (H1)(iii) and the Fatou lemma that
$$\begin{aligned} (p-1)M \geq&\liminf_{n\rightarrow+\infty}\sum _{j=1}^{p-1} I_{j}\bigl(u_{n}(t_{j})\bigr) =\liminf _{n\rightarrow +\infty}\int_{0}^{T}\bigl(f_{r_{n}} \bigl(t,u_{n}(t)\bigr)-e(t)\bigr)\,dt\\ \geq&\int_{0}^{T}\liminf_{n\rightarrow+\infty} \bigl(f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t)\bigr)\,dt \end{aligned}$$
and
$$\begin{aligned} \int_{0}^{T}\liminf_{n\rightarrow+\infty} \bigl(f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t)\bigr)\,dt =&\int _{0}^{T} \liminf_{x\rightarrow+\infty} \bigl(f(t,x)-e(t)\bigr)\,dt\\ =&\int_{0}^{T}\lim_{x\rightarrow+\infty} \bigl(f(t,x)-e(t)\bigr)\,dt=-T\bar{e}, \end{aligned}$$
which leads to
$$ (p-1)M\geq-T\bar{e}. $$
This is a contradiction to (H2)(ii). Similarly, we will arrive at a contradiction with (H2), if we assume that \(\max u_{n}<\frac{1}{n}\). In fact, by the Fatou lemma we have
$$\begin{aligned} \limsup_{n\rightarrow+\infty}\sum_{j=1}^{p-1} I_{j}\bigl(u_{n}(t_{j})\bigr) =&\limsup _{n\rightarrow+\infty} \int_{0}^{T} \bigl( f_{r_{n}}\bigl(t,u_{n}(t) \bigr) -e(t)\bigr)\,dt\\ \leq&\int_{0}^{T}\limsup_{n\rightarrow+\infty} \bigl(f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t)\bigr)\,dt\\ \leq&\int_{0}^{T}\limsup_{x\rightarrow0^{+}} \bigl[ f(t,x)-e(t) \bigr]\,dt. \end{aligned}$$
Hence
$$ \limsup_{n\rightarrow+\infty}\sum _{j=1}^{p-1}I_{j} \bigl(u_{n}(t_{j})\bigr)\leq\int_{0}^{T} \lim_{x\rightarrow0^{+}} \bigl[ f(t,x)-e(t) \bigr]\,dt=-\infty. $$
This contradicts the assumption that \(I_{j}\) is bounded.
Next, we show that \(u_{n}\) is bounded from above. Since for all \(n>1\), \(u_{n} \) is a T-periodic solution of (3.2), \(\Phi_{r}^{\prime }(u_{n})=0\). Hence for all \(v\in H_{T}^{1}\) and for all \(n>1\) we have, for all ε, \(0<\varepsilon<1\),
$$ \Biggl\vert \int_{0}^{T} \bigl[ u_{n}^{{\prime }}(t)v^{{\prime}}(t)-f_{r_{n}} \bigl(t,u_{n}(t)\bigr)v(t)+e(t)v(t) \bigr] +\sum _{j=1}^{p-1}I_{j}\bigl(u_{n}(t_{j+sp}) \bigr)v(t_{j+sp})\Biggr\vert \leq \varepsilon \Vert v\Vert . $$
(3.11)
Taking \(v(t)\equiv-1\) in the above inequality, we obtain
$$\begin{aligned} &\biggl\vert \int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr]\,dt\biggr\vert - \Biggl\vert \sum_{j=1}^{p-1}I_{j}(u_{n}(t_{j}) \Biggr\vert \\ &\quad\leq\biggl\vert \int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr]\,dt-\sum _{j=1}^{p-1}I_{j}(u_{n}(t_{j}) \biggr\vert \leq\varepsilon\sqrt{T}. \end{aligned}$$
Then
$$\begin{aligned} \biggl\vert \int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr]\,dt\biggr\vert \leq&\Biggl\vert \sum_{j=1}^{p-1}I_{j}(u_{n}(t_{j}) \Biggr\vert +\varepsilon\sqrt {T}\ \\ \leq&(p-1)\vert m\vert +\varepsilon\sqrt{T}. \end{aligned}$$
Now, from the above inequality, we get for all \(n>1\)
$$\begin{aligned} \int_{0}^{T}\bigl\vert f_{r_{n}}\bigl(t,u_{n}(t)\bigr)\bigr\vert \,dt \leq&\int _{0}^{T}\bigl\vert f_{r_{n}} \bigl(t,u_{n}(t)\bigr)-e(t)\bigr\vert \,dt+\int_{0}^{T} e(t)\,dt \\ =& \biggl\vert \int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr]\,dt\biggr\vert + \bar{e}T \\ \leq&(p-1)\vert m\vert +\varepsilon\sqrt{T}+T\bar{e}. \end{aligned}$$
(3.12)
Also, taking \(v=u_{n}\) in (3.11), we obtain
$$ \varepsilon \Vert u_{n}\Vert \geq\bigl\Vert u^{\prime }\bigr\Vert _{{L_{2}}}^{2}-\int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr] u_{n}(t)\,dt+\sum _{j=1}^{p-1}I_{j} \bigl(u_{n}(t_{j})\bigr)u_{n}(t_{j}). $$
(3.13)
Using (3.12) we get for all \(n >1\)
$$\begin{aligned} \biggl\vert \int_{0}^{T} \bigl[ f_{r_{n}}\bigl(t,u_{n}(t)\bigr)-e(t) \bigr] u_{n}(t)\,dt\biggr\vert \leq&\Vert u_{n}\Vert _{\infty } \biggl( \int_{0}^{T}\bigl\vert f_{r_{n}}\bigl(t,u_{n}(t)\bigr)\bigr\vert \,dt+\int _{0}^{T}\bigl\vert e(t)\bigr\vert \,dt \biggr) \\ \leq&\Vert u_{n}\Vert _{\infty} \bigl( \varepsilon\sqrt {T}+T\bar{e}+(p-1)|m|+\Vert e\Vert _{L^{1}} \bigr) . \end{aligned}$$
Thus (3.13) implies that
$$ \varepsilon \Vert u_{n}\Vert \geq \Vert u_{n}\Vert ^{2} -\Vert u_{n}\Vert _{\infty} \bigl( \varepsilon \sqrt{T}+T\bar{e}+\Vert e\Vert _{L^{1}}+(p-1)|m| \bigr) . $$
(3.14)
Wirtinger’s inequality (2.1) combined with (3.14) gives, for all \(n>1\),
$$ \varepsilon \Vert u_{n}\Vert \geq \Vert u_{n}\Vert ^{2}-\sqrt{\frac{T}{12}}\Vert u_{n}\Vert \bigl( \varepsilon \sqrt{T}+T\bar{e}+\Vert e\Vert _{L^{1}}+(p-1)|m| \bigr) . $$
We deduce that, for \(n >1\),
$$ \Vert u_{n}\Vert \leq\beta_{1}, $$
where
$$ \beta_{1}= \biggl( 1+\frac{T}{2\sqrt{3}} \biggr) +\sqrt{ \frac {T}{12}} \bigl( \Vert e\Vert _{L^{1}}+T\bar{e}+(p-1)|m| \bigr) . $$
Notice that \(\beta_{1}\) is independent of n. Hence \((u_{n})_{n}\) is bounded in \(H_{T}^{1}\). Since \(\Vert u_{n}\Vert _{\infty }\leq \sqrt{T}\Vert u_{n}\Vert \), we deduce that
$$ u_{n}(t)\leq\beta:=\beta_{1}\sqrt{T}. $$
Consequently, for n sufficiently large (\(n>\beta\)), for all \(t\in [ 0,T ] \), we have \(u_{n}(t)\leq n\). Furthermore, we cannot have \(u_{n}(t)\geq\frac{1}{n}\) for all \(t\in [ 0,T ] \); otherwise we would get \(\frac{1}{n}\leq u_{n}(t)\leq n\) for all \(t\in [ 0,T ] \) and this contradicts the assumption (3.10). Therefore, for n sufficiently large (\(n>\beta\)), there must exist a \(t_{n}^{\ast}\in [ 0,T ] \) such that \(u_{n}(t_{n}^{\ast})<\frac{1}{n}\). This means that \(t_{n}^{\ast}\in I_{\frac{1}{n}}\), where \(I_{\frac{1}{n}}\) is the set defined by
$$ I_{\frac{1}{n}}=\bigl\{ t\in [ 0,T ] ; u_{n}(t)< r_{n} \bigr\} . $$
(3.15)
Hence the set \(I_{\frac{1}{n}}\) is not empty. The continuity of the solution \(u_{n}\) at \(t=t_{n}^{\ast}\) implies that \(\operatorname {meas}(I_{\frac {1}{n}})>0\), which implies
$$ \int_{I_{\frac{1}{n}}}\bigl[f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)-e(t)\bigr]\,dt\neq0. $$
Now, consider the sets
$$\begin{aligned}& I_{1,\beta}= \bigl\{ t\in [ 0,T ] ; 1\leq u_{n}(t)\leq \beta \bigr\} , \end{aligned}$$
(3.16)
$$\begin{aligned}& I_{\frac{1}{n},1}= \bigl\{ t\in [ 0,T ] ; r_{n}\leq u_{n}(t)< 1 \bigr\} , \end{aligned}$$
(3.17)
so that we can write
$$ [ 0,T ] =I_{\frac{1}{n}}\cup I_{\frac{1}{n},1}\cup I_{1,\beta}. $$
Then integrating the differential equation in (3.2) from 0 to T we obtain
$$\begin{aligned} \Upsilon_{n} :=&\int_{0}^{T}-u_{n}^{\prime\prime }(t)\,dt= \int_{0}^{T}\bigl(f_{r_{n}} \bigl(t,u_{n}(t)\bigr)-e(t)\bigr)\,dt \\ =&\int_{I_{\frac{1}{n}}}\bigl[f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)-e(t)\bigr]\,dt \\ &{}+\int_{I_{\frac{1}{n},1}}\bigl[f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)-e(t)\bigr]\,dt+\int_{I_{1,\beta}}\bigl[f_{{r_{n}}} \bigl(t,u_{n}(t)\bigr)-e(t)\bigr]\,dt. \end{aligned}$$
(3.18)

(1) Assume we are integrating positively on all subintervals of \([ 0,T ] \).

If \(t\in I_{\frac{1}{n}}\) then \(u_{n}(t)< r_{n}\). It follows from (3.15) and (H1)(ii) that
$$ \int_{I_{\frac{1}{n}}}\bigl[f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)-e(t)\bigr]\,dt= \int_{I_{\frac{1}{n}}}\bigl[f(t,r_{n})-e(t) \bigr]\,dt< 0, $$
which yields
$$ \Upsilon_{n}< \int_{I_{\frac{1}{n},1}} \bigl[f_{{r_{n}}} \bigl(t,u_{n}(t)\bigr)-e(t)\bigr]\,dt+\int_{I_{1,\beta}} \bigl[f_{{r_{n}}}\bigl(t,u_{n}(t)\bigr)-e(t)\bigr]\,dt. $$
(3.19)
If \(t\in I_{1,\beta}\) then \(u_{n}(t)\in [1,\beta]\). This means that \(u_{n}(t)\) is bounded on \(I_{1,\beta}\), since \(f_{{r_{n}}}\) is continuous in u, then \(f_{{r_{n}}}\) is bounded almost everywhere in \(I_{1,\beta}\). Let
$$ C=C(\beta)=\max \bigl\{ \bigl\vert f_{{r_{n}}}(t,x)\bigr\vert ; t\in [ 0,T ] ,1\leq x\leq\beta \bigr\} . $$
(3.20)
Then
$$ \biggl\vert \int_{I_{1,\beta}} \bigl[ f_{{r_{n}}} \bigl(t,u_{n}(t)\bigr)-e(t) \bigr]\,dt\biggr\vert \leq\int _{I_{1,\beta}}\bigl\vert f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)\bigr\vert +\bigl\vert e(t)\bigr\vert \,dt\leq T\bigl(C+\Vert e\Vert _{L^{1}}\bigr), $$
(3.21)
and (3.19) leads to
$$ \Upsilon_{n}\leq\int_{I_{\frac{1}{n},1}} \bigl[ f_{{r_{n}}} \bigl(t,u_{n}(t)\bigr)-\bar{e} \bigr]\,dt+T(C+\bar{e}). $$
(3.22)
By (H1)(ii), we see that, for every \(\sigma>0\), there exists \(\gamma_{\sigma}>0\) such that \(f(t,x)-\bar{e} <-\sigma\), for all \(x\in I_{\gamma_{\sigma}}:= ( 0,\gamma _{\sigma } ) \) and for every \(t\in [ 0,T ] \). Then, for n large enough (\(n>\beta\)), we have \(J:=I_{\frac{1}{n},1}\cap I_{\gamma _{\sigma }}\neq\emptyset\). Hence, (H1)(i) implies
$$ \int_{I_{\frac{1}{n},1}} \bigl[ f_{{r_{n}}}\bigl(t,u_{n}(t) \bigr)-\bar {e} \bigr]\,dt< \int_{J} \bigl[ f_{{r_{n}}}\bigl(t,u_{n}(t)\bigr)-\bar {e} \bigr]\,dt< -\sigma \operatorname {meas}(J). $$
(3.23)
Thus, for \(\sigma=\frac{1}{\operatorname {meas}(J)}n^{2}T(C+\bar{e})\), we obtain
$$\begin{aligned} \Upsilon_{n} < &\int_{I_{{\frac{1}{n},1}}} \bigl[ f_{{r_{n}}}\bigl(t,u_{n}(t)\bigr)-\bar{e} \bigr]\,dt+T(C+\bar{e}) \\ < &T(C+\bar{e}) \bigl(1-n^{2}\bigr)\mathop{\rightarrow}\limits_{n\rightarrow+\infty} - \infty. \end{aligned}$$
(3.24)
Then \(\Upsilon_{n}\) is not bounded.
(2) If we integrate negatively on all subintervals of \([ 0,T ] \) then, instead of (3.23), we get
$$ \int_{I_{{\frac{1}{n},1}}} \biggl[ f\biggl(t,u_{0}+ \frac {1}{n}\biggr)-\bar{e} \biggr]\,dt>\sigma \operatorname {meas}(I_{J}). $$
This, together with (3.20), leads to
$$ \Upsilon_{n}\rightarrow+\infty,\quad\mbox{as }n \rightarrow+\infty. $$
(3.25)
On the other hand, integrating the differential equation in (3.2) from 0 to T and using T-periodicity of \(u_{n}^{\prime}\), we obtain
$$\begin{aligned} \Upsilon_{n} =&-\int_{0}^{T}u_{n}^{\prime\prime }(t)\,dt=- \sum_{j=0}^{p}\int_{t_{j}^{+}}^{t_{j+1}^{-}}u_{n}^{\prime\prime}(t)\,dt \\ =&\sum_{j=1}^{p-1}\Delta u_{n}^{\prime }(t_{j})=\sum _{j=1}^{p-1}I_{j}\bigl(u_{n}(t_{j}) \bigr)\leq(p-1)M. \end{aligned}$$
Thus by (H2)
$$ \mbox{for each }n\in \mathbb{N} ^{\ast}, \Upsilon_{n}< 0 \mbox{ and }\Upsilon_{n}\mbox{ is bounded}. $$
(3.26)
We see that (3.26) contradicts (3.24) and (3.25). This contradiction shows that Lemma 1 is proved. In particular, Lemma 1 shows that there exists \(r\in(0,1)\) such that every T-periodic solution u of (3.2) is a solution of (3.1), since it satisfies \(u(t)\geq r\) for all \(t\in \mathbb{R} \) and \(f_{r}(t,u(t))=f(t,u(t))\), if \(u(t)\geq r\). Therefore \(u^{\ast}\) is a nonconstant T-periodic solution of (1.1), (1.2). This completes the proof of our main result. □

4 Example

Consider the impulsive singular problem
$$ \left \{ \textstyle\begin{array}{@{}l} u^{\prime\prime}(t)-\frac{e^{t}}{u^{\alpha}}=e(t),\quad \mbox{for } t\in(0,T), t\neq t_{j}, \\ \Delta u^{\prime}(t_{j})=I_{j}(u(t_{j})),\quad j=1, \\ u(0)-u(T)=0, \end{array}\displaystyle \right . $$
(4.1)
where \(\alpha>1\) and \(T>0\). Take \(I_{j}(s)=\cos s-2\), and \(e\in L^{2}([0;T],\mathbb{R} )\) such that \(\bar{e}<\frac{1}{T}\). In this case \(m=-3\) and \(M=-1\). Then (H1)-(H2) hold. Therefore, by Theorem 1, problem (4.1) has at least one nonconstant T-periodic solution.

Declarations

Acknowledgements

The authors would like to thank an anonymous referee and Professor Ravi Agarwal for comments that led to the improvement of the presentation of the manuscript. Also, A Boucherif is grateful to King Fahd University of Petroleum and Minerals for its constant support.

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)
Department of Mathematics, University of Tlemcen, Tlemcen, Algeria
(2)
Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran, Saudi Arabia

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© Daoudi-Merzagui and Boucherif 2015

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