# Existence and uniqueness of solutions for periodic-integrable boundary value problem of second order differential equation

- Hongtu Hua
^{1, 2}Email author, - Fuzhong Cong
^{1}and - Yi Cheng
^{1, 2}

**2012**:89

https://doi.org/10.1186/1687-2770-2012-89

© Hua et al.; licensee Springer 2012

**Received: **26 March 2012

**Accepted: **16 July 2012

**Published: **7 August 2012

## Abstract

In this paper we deal with one kind of second order periodic-integrable boundary value problem. Using the lemma on bilinear form and Schauder’s fixed point theorem, we give the existence and uniqueness of solutions for the problem under Lazer type nonresonant condition.

**MSC:**34B15, 34B16, 37J40.

### Keywords

lemma on bilinear forms Schauder’s fixed point theorem existence and uniqueness periodic-integrable boundary value problems## 1 Introduction and main results

where $p(t)\in {C}^{1}(R,R)$ is a given *T*-periodic function in $t\in R$, and $p(t)>0$; $f\in {C}^{1}(R\times R,R)$ is *T*-periodic in *t*.

Throughout this paper, we assume

*m*and

*M*such that

for all $t\in R$ and $x\in R$;

Motivated by the above works, we will consider periodic-integrable boundary value problem (1.1). The main result obtained by us is the following theorem.

**Theorem 1** *Assume that* (*A* 1) *and* (*A* 2) *are satisfied*. *Then PIBVP* (1.1) *has a unique solution*.

This paper is organized as follows. Section 2 deals with a linear problem. There, using the bilinear lemma developed by Lazer, one proves the uniqueness of solutions for linear equations. In Section 3, applying the result in Section 2 and Schauder’s fixed point theorem, we complete the proof of Theorem 1.

## 2 Linear equation

here $p(t)\in {C}^{1}(R,R)$ is a given *T*-periodic function in $t\in R$, and $p(t)>0$; $q(t)\in C(R,R)$ is a *T*-periodic function. Assume that

*m*and

*M*such that

for all $t\in R$. Moreover, *m* and *M* suit (A2).

**Theorem 2** *Assume that* (*L* 1) *and* (*A* 2) *are satisfied*, *then PIBVP* (2.1) *has only a trivial solution*.

In order to prove Theorem 2, let us give some following concepts.

where *N* suits assumption (L1), and ${a}_{m}$, ${b}_{m}$, ${c}_{0}$, ${c}_{k}$ and ${d}_{k}$ are some constants. Then ${\mathcal{O}}_{\alpha ,\beta}={\mathcal{X}}_{\alpha ,\beta}\oplus {\mathcal{Y}}_{\alpha ,\beta}$.

for all $y\in {\mathcal{Y}}_{\alpha ,\beta}$. Thus, ${H}_{\alpha ,\beta}$ is positive definite on ${\mathcal{X}}_{\alpha ,\beta}$ and negative definite on ${\mathcal{Y}}_{\alpha ,\beta}$. By the lemma in [1], we assert that if ${H}_{\alpha ,\beta}(u,v)=0$ for all $u\in {\mathcal{O}}_{\alpha ,\beta}$, then $v\equiv 0$.

*x*on $[\alpha ,\beta ]$ with $x(\alpha )=x(\beta )=0$, we introduce an auxiliary function

The following lemma is very useful in our proofs.

**Lemma 1**

*If*$p(t)$, $q(t)$

*are continuous and satisfy*(

*L*1)

*and*(

*A*2),

*then the following two points boundary value problem*

*has only a trivial solution*.

*Proof*It is clear that 0 is a solution of two points boundary value problem (2.2). If $v(t)$ is a solution of problem (2.2), then ${v}^{\alpha ,\beta}(t)\in {\mathcal{O}}_{\alpha ,\beta}$. For any $u\in {\mathcal{O}}_{\alpha ,\beta}$, we have

By assumption (L1), ${H}_{\alpha ,\beta}$ is positive definite on ${\mathcal{X}}_{\alpha ,\beta}$ and negative definite on ${\mathcal{Y}}_{\alpha ,\beta}$. These show ${v}^{\alpha ,\beta}(t)\equiv 0$ for $t\in [0,T]$, *that is*, $v(t)\equiv 0$ for $t\in [\alpha ,\beta ]$. The proof of Lemma 1 is ended. □

*Proof of Theorem 2* It is clear that PIBVP (2.1) has at least one solution, for example, ${x}_{\ast}\equiv 0$. Assume that PIBVP (2.1) possesses a nontrivial solution ${x}^{\ast}\ne 0$. The proof is divided into three parts.

Case 1: ${x}^{\ast}(0)={x}^{\ast}(T)=0$. By Lemma 1 ($\alpha =0$ and $\beta =T$), PIBVP (2.1) has only a trivial solution. This contradicts ${x}^{\ast}\ne 0$.

*S*, which implies that $0<a<b<2\pi $ and ${x}^{\ast}(a)={x}^{\ast}(b)=0$. By Lemma 1 ($\alpha =a$ and $\beta =b$) the two points boundary value problem

*a*and

*b*, one has

This contradicts ${\int}_{0}^{a}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>0$ and ${\int}_{b}^{T}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>0$.

Case 3: ${x}^{\ast}(0)={x}^{\ast}(T)=\eta <0$. This case is similar to Case 2.

Thus, we complete the proof of Theorem 2. □

**Theorem 3**

*If*$p(t)$, $q(t)$

*are continuous and satisfy*(

*L*1)

*and*(

*A*2),

*then the following PIBVP*

*has a unique solution*.

*Proof*Let ${x}_{\ast 1}(t)$ and ${x}_{\ast 2}(t)$ be two linear independent solutions of the following linear equation:

From (2.5) constants ${c}_{3}$, ${c}_{4}$ are unique. Thus, PIBVP (2.4) has only one solution. □

## 3 Nonlinear equations

To prove the main result, we need the following Lemma 2.

**Lemma 2** *If* *f* *satisfies* (*A* 1) *and* (*A* 2), *then for any given* $y\in {\mathcal{O}}^{\ast}$, *PIBVP* (3.2) *has only one solution*, *denoted as* ${x}_{y}(t)$ *and* $\parallel {x}_{y}\parallel \le M$.

*Proof*From condition (A2), it follows that

By Theorem 3, PIBVP (3.2) has only one solution ${x}_{y}(t)$. If $\parallel {x}_{y}\parallel \le M$ does not hold, there would exist a sequence $\{{y}_{m}(t)\}$ such that $\parallel {x}_{{y}_{m}}\parallel \to \mathrm{\infty}$, $m\to \mathrm{\infty}$. Choose a subsequence of ${\{h(t,{y}_{m})\}}_{m=1}^{\mathrm{\infty}}$, without loss of generality, express as itself, such that the sequences are weakly convergent in ${L}^{2}(0,T)$. Denote the limit as ${h}_{0}(t)$. It is obvious that ${h}_{0}(t)\in {L}^{2}(0,T)$.

and ${x}_{m}^{\prime}\to z(t)$ in $C([0,T],R)$. Thus, ${x}_{0}(0)={x}_{0}(T)$ and ${\int}_{0}^{T}{x}_{0}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0$.

which implies $z(t)={x}_{0}^{\prime}(t)$, for any $t\in [0,T]$. Hence, $\parallel {x}_{0}\parallel =1$.

On the other hand, by Theorem 2, PIBVP (3.4) has only zero, which leads to a contradiction. The proof of Lemma 2 is completed. □

Define an operator $F:{\mathcal{O}}^{\ast}\to {\mathcal{O}}^{\ast}$ by $Fy={x}_{y}$. Applying Lemma 2, $F:{B}_{{M}^{\ast}}\to {B}_{{M}^{\ast}}$.

**Lemma 3** *Operator* *F* *is completely continuous on* ${\mathcal{O}}^{\ast}$.

*Proof*We first prove that

*F*is continuous. Given any $\{{y}_{m}\}\subset {\mathcal{O}}^{\ast}$ such that ${y}_{m}\to {y}_{0}\in {\mathcal{O}}^{\ast}$. Put ${u}_{m}={x}_{{y}_{m}}-{x}_{{y}_{0}}$. From the definition

Hence, from Theorem 2, ${u}_{0}(t)\equiv 0$. This implies *F* is continuous. By Lemma 2, for any bounded subset $D\subset {\mathcal{O}}^{\ast}$, $F(D)$ is also bounded. Hence, applying the continuity of *F* and Arzela-Ascoli theorem, $F(D)$ is relatively compact. This shows *F* is completely continuous on ${\mathcal{O}}^{\ast}$. The proof of Lemma 3 is completed. □

*Proof of Theorem 1* By Lemma 2, Lemma 3 and Schauder’s fixed point theorem, *F* has a fixed point in ${\mathcal{O}}^{\ast}$, *that is*, PIBVP (1.1) has a solution $x(t)$.

Hence by Theorem 3, $x(t)\equiv 0$. The uniqueness is proved. □

## Declarations

### Acknowledgements

The authors are grateful to the referees for their useful comments. The research of F. Cong was partially supported by NSFC Grant (11171350) and Natural Science Foundation of Jilin Province of China (201115133).

## Authors’ Affiliations

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