The shooting method and integral boundary value problems of thirdorder differential equation
 Wenyu Xie^{1} and
 Huihui Pang^{1}Email author
https://doi.org/10.1186/s1366201608244
© Xie and Pang 2016
Received: 12 January 2016
Accepted: 30 March 2016
Published: 20 May 2016
Abstract
In this paper, the existence of at least one positive solution for thirdorder differential equation boundary value problems with RiemannStieltjes integral boundary conditions is discussed. By applying the shooting method and the comparison principle, we obtain some new results which extend the known ones. Meanwhile, an example is worked out to demonstrate the main results.
Keywords
1 Introduction
It is well known that thirdorder equations arise from many branches of applied mathematics and physics. For example, in the deflection of a curved beam having a constant or varying cross section, a three layer beam, electromagnetic waves or gravity driven flows [1]. There have been extensive studies on thirdorder differential equation BVPs (boundary value problems), for example [2–5]. Most of these results are obtained via applying the topological degree theory, the fixed point theorems on cones, the lower and upper solution method, the critical point theory and monotone technique. We refer the reader to [6–17] and the references therein.
Recently, the attention has shifted to BVPs with Stieltjes integral boundary condition since this kind of conditions has been considered a single framework of multipoint and integral type boundary conditions. For more comments on the RiemannStieltjes integral boundary condition and its importance, we refer the reader to [4, 5] and other related work such as [6, 7].
In [8], the author applied the method of lower and upper solutions to generate an iterative technique and discussed the existence of solutions of nonlinear thirdorder ordinary differential equations with integral boundary conditions. Pang and Xie [9] investigated the existence of concave positive solutions and established corresponding iterative schemes for a thirdorder differential equation with RiemannStieltjes integral boundary conditions using the monotone iterative technique.
It is well known that the classical shooting method could be effectively used to establish the existence and multiplicity results for differential equation BVPs. To some extent, this approach has an advantage over the traditional methods. Readers can see [18–24] and the references therein for details.
 (H_{1}):

\(f \in C([0,\infty)\times[0,\infty); [0,\infty))\), \(f(u, v)\not \equiv0\);
 (H_{2}):

\(h\in C([0,1];[0,\infty))\);
 (H_{3}):

\(\int^{1}_{0}\,dA(t)> 1\), \(0< \int^{1}_{0}\,dB(t)<1\).
2 Preliminaries
Proof
Assume y is a positive solution of (2.2), then \(y(t)> 0\) for \(t\in (0,1)\) and it follows from \(u(t)=Ay(t)\) that \(u(t)\) satisfies (1.1). Assume on the contrary that there is a \(t_{0} \in(0,1)\) such that \(u(t_{0})= \min_{t\in(0,1)}u(t)\leq0\), then \(u'(t_{0})=0\) and \(u''(t_{0}) \geq0\), which yields \(y(t_{0})=u'(t_{0})=0\). This contradicts the assumption that y is a positive solution of (2.2). Hence, \(u(t)>0\) for all \(t\in(0,1)\). □
Under the assumptions (H_{1})(H_{3}), denote by \(y(t,m)\) the solution of the IVP (2.3). We assume that f is strong continuous enough to guarantee that \(y(t,m)\) is uniquely defined and that it depends continuously on both t and m. The discussion of this problem can be found in [18]. Therefore the solution of IVP (2.3) exists.
Then solving (2.2) is equivalent to finding a \(m^{*}\) such that \(k(m^{*}) = 1 \) or \(\varphi(m^{*})=0\).
Lemma 2.2
(Sturm comparison theorem) [25]
Lemma 2.3
Proof
The proof for (2.4) can be found in [18]. The continuity of the integrands implies the existence of the Riemann integral. In view of the definition of Stieltjes integral, by using the inequality of the limit, we have (2.5). □
Lemma 2.4
Assume that (H_{1})(H_{2}) hold and \(0 < \int^{1}_{0}\,dA(t)< 1\), then BVP (2.2) has no positive solution.
Proof
Hence, we need \(\int^{1}_{0}\,dA(s)\geq1\), and we assume \(\int ^{1}_{0}\,dA(s)> 1\) in (H_{3}) in order to satisfy (3.1). □
3 Main results
In the following, we assume that \(A(t)\) has continuous derivative function \(\alpha(t)\) and \(\alpha(t)> 1\) for \(t\in[0,1]\) such that \(\int^{1}_{0}\,dA(t)=\int^{1}_{0}\alpha(t)\,dt > 1\).
It is obvious that \(\alpha^{L}\geq\alpha^{l} > 1\).
Lemma 3.1
Theorem 3.2
 (i)
\(0\leq f^{0} < \frac{\underline{A}^{2}}{h^{L}}\), \(f_{\infty} > \frac{\bar{A}^{2}}{h^{l}}\);
 (ii)
\(0\leq f^{\infty} < \frac{\underline{A}^{2}}{h^{L}}\), \(f_{0} > \frac{\bar{A}^{2}}{h^{l}}\).
Proof
As we mentioned above, BVP (1.1) having a positive solution is equivalent to BVP (2.2) having a positive solution.
Next, we will find a positive number \(m^{*}_{2}\) such that \(\varphi (m^{*}_{2})\geq0\).
Since the solution \(y(t,m)\) is concave and \(y'(0,m)=0\), it hits the line \(y=L\) at most one time for the constant L defined in (3.7) and \(t\in(0,1]\). We denote the intersecting time by \(\bar{\delta}_{m}\) provided it exists. Henceforth, denote \(I_{m}=(0,\bar{\delta}_{m}]\subseteq(0,1]\). If \(y(1,m)\geq L\), then \(\bar{\delta}_{m}=1\).
The discussion is divided into three steps.
Step 1. We claim that there exists a value \(m_{0}\) large enough such that \(0\leq y(t,m_{0})\leq L\) for \(t\in[\bar{\delta}_{m_{0}},1]\) and \(y(t,m_{0})\geq L\) for \(t\in I_{m_{0}}\).
Since \(y(t,m)\) is continuous and concave, there exists a number \(m_{0}\) large enough such that \(y(t,m_{0})\geq L\) for \(t\in I_{m_{0}}\).
Step 3. Seek a value \(m^{*}_{2}\) and a positive number σ such that \(0< \frac{A_{2}}{A_{2}+\epsilon}\leq\sigma\leq1\) and \(y(t,m^{*}_{2})\geq L\) for \(t\in(0,\sigma]\).
In the following, we prove that \(k(m^{*}_{2})\geq1\) for the selected \(m^{*}_{2}\) and σ.
From (3.6) and (3.14), we can find a \(m^{*}\) between \(m^{*}_{1}\) and \(m^{*}_{2}\) such that \(y(t,m^{*})\) is the solution of (2.2). So that \(u(t,m^{*})=Ay(t,m^{*})\) is the solution of (1.1).
Now, we prove for (ii).
From (3.17) and (3.18), we can find a \(m^{*}\) between \(m^{*}_{3}\) and \(m^{*}_{4}\) such that \(y(t,m^{*})\) is the solution of (2.2). So \(u(t,m^{*})=Ay(t,m^{*})\) is the solution of (1.1). The proof of the theorem is complete. □
Example
Declarations
Acknowledgements
The work is supported by Chinese Universities Scientific Fund (Project No.2016LX002)
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
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