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Existence and uniqueness of solutions for the second order periodic-integrable boundary value problem
- Xue Feng^{1, 2}Email author and
- Fuzhong Cong^{1, 2}
- Received: 3 January 2017
- Accepted: 5 July 2017
- Published: 26 July 2017
Abstract
This paper is mainly devoted to studying one kind of the second order differential equation. Under periodic-integrable boundary value condition, the existence of the solutions of this equation is discussed by the method of the operator theory and the Schauder fixed point theorem.
Keywords
- periodic-integrable boundary value
- existence and uniqueness
- Schauder’s fixed point theorem
1 Introduction and the main results
Assumption A _{1}
Assumption A _{2}
Assumption A _{3}
Assumption A _{4}
We can now state our two main results by the following theorems.
Theorem 1
If Assumptions \(A_{1}\) and \(A_{2}\) hold, then the PIBVP (1) has at least one solution.
Theorem 2
If Assumptions \(A_{3}\) and \(A_{4}\) hold, then the PIBVP (1) has a unique solution.
In Section 2, we introduce two lemmas which will be used in later sections. In Section 3, the linear problem will be discussed by the theory of ordinary differential equation, thus the uniqueness of solutions of linear equations is proved. In Sections 4 and 5, we apply the conclusions in Sections 2 and 3 and Schauder’s fixed point theorem to proving Theorems 1 and 2. In Section 6, as applications of the main results, we introduce two examples.
2 Preliminary
Let us first state some lemmas which will be used in the proof of the main results.
Lemma 1
Proof
Expand \(x(t)\) as a Fourier series and substitute the expressions into the integrals. Thus, the proof is completed. □
From the above steps, we can deduce the following lemma.
Lemma 2
The function f is denoted by \(f(t,x,y) = h_{1}(t,x,y)+xg_{0}(t,x,y)+yg_{1}(t,x,y)\), whereas \(\vert h_{1}(t,x,y) \vert \leq4\sup\limits_{0\leq t\leq2\pi, \vert x \vert \leq M_{1}, \vert y \vert \leq M_{3}} \vert f(t,x,y) \vert \), \(a\leq g_{0}\leq b\) and \(\vert g_{1} \vert \leq2M_{2}\).
3 Linear equation
Lemma 3
Proof
Now we prove it. There are two cases:
Case 1. \(x(0)=\eta<0\). Let \(t_{0} =\operatorname{inf}\{{t| t\in[0,{2\pi}]}\mbox{ and } x(t)=0\}\), which implies that \(x(t_{0})=0 \mbox{ and } x'(t_{0})>0\). Define \(t_{\star} =\operatorname{inf}\{{t | t\in(t_{0},{2\pi}]}\mbox{ and }x(t)=0\}\). If \(t_{\star}= t_{0}\), then there will exist the sequences \(\{t^{i}\}\), \(x(t^{i})= 0\) as \(t^{i}\rightarrow t_{0}\) \((i\rightarrow\infty)\). By Rolle’s theorem, there is a number \(\xi^{i}\) in \([t^{i-1},t^{i}]\), such that \(x'(\xi^{i})= 0\), meanwhile \(\xi^{i}\rightarrow t_{0} \), so \(x'(t_{0})= 0\), a contradiction. Therefore \(t_{0}\) is the first zero point and \(t_{\star}\) is the next zero point. By the periodic-integral boundary conditions, there exists \(t_{1}\in[t_{0},t_{\star}]\), such that \(x(t)>0\), for \(t\in(t_{0},t_{1})\), \(x'(t_{1})= 0\).
Case 2. \(x(0)=\eta>0\). By the linear property of the problem, \(-x(t)\) is also a solution. Thus, the case is translated into Case 1.
Lemma 4
Problem (13) has a unique solution.
Proof
4 The proof of Theorem 1
Step 2: P̅ is compact. For any bounded set \(S\subset{\mathcal{C}}\), we assert that \(\overline{P}(S)\) is the bounded set in \({\mathcal{C}}\). If not, similar to the proof of step 1, we will be led to a contradiction. For any \({x}\in{S}\), \(y=\overline{P}x\) is defined by (17). Because \(\vert {y'} \vert \), \(\vert {y} \vert \), \(\vert {f}_{x} \vert \) and \(\vert {f}_{x'} \vert \) are all bounded, proceeding as the proof of step 1, we show that \(\{y_{k}\}\) and \(\{y'_{k}\}\) are both uniformly family bounded degree of equicontinuous. By the Ascoli-Arzela theorem, P̅ is a compact operator.
Let \(E=\{{x}\in{\mathcal{C}}\vert \Vert x \Vert \leq{K}\}\). By the fixed point theorem, \(\overline{P}:E\rightarrow E\) has at least one fixed point and thus the PIBVP (1) has at least one solution. The proof of Theorem 1 is completed.
5 The proof of Theorem 2
Hence, by Lemma 3, \(u(t)\equiv0\) on \([0,2\pi]\), that is, \(x_{1}(t)=x_{2}(t)\).
Define the linear operator \({T}:{\mathcal{C}}\rightarrow{\mathcal{C}}\). For each \({x}\in {\mathcal{C}}\), \({T}[x](t)=y(t)\) is the unique solution of (25). Thus, the existence of the solution of Problem (1) is equivalent to the existence of the fixed point of T in Banach space \(\mathcal{C}\). We will prove that T is continuous and compact, and \({T({\mathcal{C}})}\) is a bounded subset in \({\mathcal{C}}\).
Hence, by the uniqueness we know \({y}_{0}={T}{x}_{0}\). Thus, operator T is continuous.
Step 2: T is compact. For any bounded set \(S\subset{\mathcal{C}}\), we assert that \({T}(S)\) is the bounded set in \({\mathcal{C}}\). If not, similar to the proof of step 1, we are led to a contradiction. For any \({x}\in{S}\), \(y={T}x\) is defined by (25). Because \(\vert {y'} \vert \), \(\vert {y} \vert \), \(\vert {f}_{x} \vert \), and \(\vert {f}_{x'} \vert \) are all bounded, and then \(\Vert {y'} \Vert <\infty\). Proceeding as in the proof of step 1, we show that \(\{y_{j}\}\) and \(\{y'_{j}\}\) are both uniformly family bounded degree of equicontinuous. By the Ascoli-Arzela theorem, T is a compact operator.
Let \(\overline{E}=\{{x}\in{\mathcal{C}} \vert \Vert x \Vert \leq{K}\}\). By the fixed point theorem, \({T}:\overline{E}\rightarrow\overline{E}\) has one fixed point. The proof of Theorem 2 is completed.
6 Examples
In this section, to illustrate significance and effectiveness of the results, we introduce two examples.
Example 1
According to (35) and (36), we derive that f satisfies Assumptions A _{1} and A _{2} . By Theorem 1, the PIBVP (34) has at least one solution.
Example 2
Declarations
Acknowledgements
The authors express their thanks to the referee for their useful comments. The research of F. Cong was partially supported by NFSC Grant (11171350).
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Authors’ Affiliations
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