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The shooting method and integral boundary value problems of thirdorder differential equation
Advances in Difference Equations volumeÂ 2016, ArticleÂ number:Â 138 (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.
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 [2â€“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 [6â€“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 the existing literature, there are very few papers dealing with thirdorder differential equations with RiemannStieltjes integral boundary conditions. We found that Graef and Webb [6] studied the following problem:
where \(p > \frac{1}{2}\), and \(\alpha[u]\), \(\beta[u]\), and \(\lambda[v]\) are linear functional on \(C[0, 1]\) given by a RiemannStieltjes integral. The existence of multiple positive solutions is obtained by the application of the fixed point index theory.
In 2012, Jankowski [7] used a fixed point theorem to establish the existence of at least three nonnegative solutions of some nonlocal BVPs to the thirdorder differential equation
where Î» denotes a linear functional on \(C(J)\) given by \(\lambda [x]=\int^{1}_{0}x(t)\,d\Lambda(t)\) involving a Stieltjes integral with a suitable function Î› of bounded variation.
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 [18â€“24] and the references therein for details.
Using the shooting method, Henderson [23] obtained solutions of the three point BVP for the secondorder equation
where \(f : (a,b) \times\mathbb{R}^{2}\to\mathbb{R}\) is continuous, \(a< x_{1}< x_{2}< x_{3}< b\), and \(y_{1},y_{2}\in\mathbb{R}\).
In [24], by applying the shooting method and the comparison principle, Wang investigated the existence results of positive solutions for the RiemannStieltjes integrals BVPs
where \(f\in C([0,\infty);[0,\infty))\) and \(0<\eta<1\), \(\alpha\geq0\) are given constants, \(0<\alpha\eta^{2}<2\).
However, to the best of our knowledge, no paper has considered the existence of positive solutions for thirdorder differential equation with the shooting method till now. Motivated by the excellent work mentioned above, in this paper, we try to employ the shooting method to establish the criteria for the existence of positive solutions to the following thirdorder differential equation with integral boundary condition:
where \(\alpha[u] = \int_{0}^{1}u(s)\,dA(s)\), \(\beta[u] = \int _{0}^{1}u(s)\,dB(s)\), and \(\alpha[u]\), \(\beta[u]\) are linear functions on \(C[0,1]\) given by the RiemannStieltjes integral, \(A(t)\), \(B(t)\) are suitable functions of a bounded variation.
Set
In this paper, we always assume
 (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
Define an operator \(A : C[0,1] \to C^{1}[0,1]\) as
for \(t \in[0,1] \), where
is the Green function for the following firstorder differential equation:
Let \(y=u'\), then BVP (1.1) is equivalent to the following secondorder BVP:
Lemma 2.1
If y is a positive solution of (2.2), then u is a positive solution of (1.1).
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 \(t\in(0,1)\).â€ƒâ–¡
The principle of the shooting method converts the BVP into an IVP (initial value problem) by finding suitable initial values m such that equation (2.2) comes with the initial value condition as
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.
Denote
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]
Let \(\varphi_{1}\) and \(\varphi_{2}\) be nontrivial solutions of the equations
respectively, on an interval I; here \(q_{1}\) and \(q_{2}\) are continuous functions such that \(q_{1}(x) \leq q_{2}(x)\) on I. Then between any two consecutive zeros \(x_{1}\) and \(x_{2}\) of \(\varphi_{1}\), there exists at least one zero of \(\varphi_{2}\) unless \(q_{1}(x) \equiv q_{2}(x)\) on \((x_{1},x_{2})\).
Lemma 2.3
Let \(y(t,m)\), \(z(t,m)\), \(Z(t,m)\) be the solution of the IVPs, respectively,
and suppose that \(F(t)\), \(g(t)\), and \(G(t)\) are continuous functions defined on \([0, 1]\) such that
If \(Z(t,m)\) does not vanish in \((0,1]\), then for any \(0 \leq\xi\leq s \leq1\), we have
and hence, for any \(0 \leq s \leq1\), we have
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).â€ƒâ–¡
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
If BVP (2.2) has a positive solution \(y(t)\), then \(y(t,m)\) is the positive solution of IVP (2.3). For \(m > 0\), we compare the solution \(y(t, m)\) of IVP given by (2.3) with the solution \(z(t)=m\) of
By LemmaÂ 2.3, we have
In fact, \(y(0,m)=\int^{1}_{0}y(s,m)\,dA(s)\). That is, \(\int ^{1}_{0}\,dA(s)\geq1\).
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).â€ƒâ–¡
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 \(\int^{1}_{0}\,dA(t)=\int^{1}_{0}\alpha(t)\,dt > 1\).
For the sake of convenience, we denote
It is obvious that \(\alpha^{L}\geq\alpha^{l} > 1\).
Lemma 3.1
Assume that (H_{1})(H_{3}) hold. Then there exist a solution \(x=A_{1}\in (0,\frac{\pi}{2})\) such that
and a solution \(x=A_{2} \in(0,\frac{\pi}{2})\) such that
Proof
From (H_{3}) and the FigureÂ 1, we can easily get LemmaÂ 3.1.
â€ƒâ–¡
Theorem 3.2
Assume that (H_{1})(H_{3}) hold. Suppose one of the following conditions holds:

(i)
\(0\leq f^{0} < \frac{\underline{A}^{2}}{h^{L}}\), \(f_{\infty} > \frac{\bar{A}^{2}}{h^{l}}\);

(ii)
\(0\leq f^{\infty} < \frac{\underline{A}^{2}}{h^{L}}\), \(f_{0} > \frac{\bar{A}^{2}}{h^{l}}\).
Then problem (1.1) has at least one positive solution, where
and \(A_{1}\), \(A_{2}\) are defined in (3.1) and (3.2), respectively.
Proof
As we mentioned above, BVP (1.1) having a positive solution is equivalent to BVP (2.2) having a positive solution.
(i) Since \(0\leq f^{0} < \frac{\underline{A}^{2}}{h^{L}}\), there exists a positive number r such that
Let \(0< m^{*}_{1}< r\), from the Sturm comparison theorem and the concavity of \(y(t,m^{*}_{1})\) for \(t\in[0,1]\), we have
Let \(Z(t)= m^{*}_{1}\cos(A_{1}t)\) for \(t\in[0,1]\), then \(Z(t)\) satisfies the following IVP:
From (3.4), LemmaÂ 2.3, and LemmaÂ 3.1, we have
that is, \(\varphi(m^{*}_{1})\leq0\).
On the other hand, the second inequality in (i) implies that there exists a number L large enough such that
and there exists a positive number \(\epsilon< A_{2}\) small enough that
Next, we will find a positive number \(m^{*}_{2}\) such that \(\varphi (m^{*}_{2})\geq0\).
There exist a value \(m^{*}_{2}\) and a positive number Ïƒ such that
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 \(t\in I_{m_{0}}\).
Otherwise, provided \(y(t,m)\leq L\) for all \(t\in[0,1]\) as \(m\to\infty \), then by integrating both sides of equation (2.3) from 0 to t, we have
Hence, from (3.3) and the continuity of \(f(Ay,y)\), we have
Since A is defined in (2.1) as a continuous operator that depends on y, for \(f(Ay,y)\) there exists a maximum for \(y \in[0,L]\). Denote \(L_{f}=\max_{y\in[0,L]}f(Ay,y)\). If we choose \(m>L+L_{f}h^{L}\), (3.10) will lead to a contradiction.
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 2. There exists a monotonically increasing sequence \(\{m_{k}\} \) such that the sequence \(\bar{\delta}_{m_{k}}\) is increasing on \(m_{k}\). That is,
and \(y(t,m_{k})\geq L\) for \(t\in I_{m_{k}}\).
We prove that
Since f guarantees that \(y(t,m)\) is uniquely defined, the solution \(y(t,m_{k1})\) and \(y(t,m_{k})\) have no intersection in the interval \([\bar{\delta}_{m_{k1}},1)\). It follows from
that
Thus we have (3.11).
When \(k=1\), see the relationship of m and \(I_{m}\) in FigureÂ 2.
Step 3. Seek a value \(m^{*}_{2}\) and a positive number Ïƒ such that \(0< \frac{A_{2}}{A_{2}+\epsilon}\leq\sigma\leq1\) and \(y(t,m^{*}_{2})\geq L\) for \(t\in(0,\sigma]\).
Following step 1, step 2, and the extension principle of solutions, there exists a positive integer n large enough such that
If we taken \(m^{*}_{2}=m_{n}\), \(\sigma=\bar{\delta}_{m_{n}}\), then
In the following, we prove that \(k(m^{*}_{2})\geq1\) for the selected \(m^{*}_{2}\) and Ïƒ.
Set \(z(t)= m^{*}_{2} \cos\sigma(A_{2}+\epsilon)t\), then \(z(t)\) satisfies the following IVP:
where \(\sigma\leq1\). From (3.8), we have
Further, noting that \(y(1,m^{*}_{2})>L\) (this time \(\sigma=1\)) or \(y(1,m^{*}_{2})\leq y(\sigma, m^{*}_{2})=L\), then by LemmaÂ 2.3 and LemmaÂ 3.1 we have
which implies \(\varphi(m^{*}_{2})\geq0\).
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).
Set \(z(t)= m^{*}_{3} \cos\sigma(A_{1}+\epsilon)t\) and \(Z(t)= m^{*}_{4}\cos (A_{2}t)\) for \(t\in[0,1]\), then \(z(t)\) and \(Z(t)\) satisfy the following IVPs, respectively:
where \(\sigma\leq1\).
Similar to (3.6) and (3.14), it follows from (2.3) and (3.1)(3.2) that
where \(0< \sigma\leq\frac{A_{1}}{A_{1}+\epsilon}\leq1\), and
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.â€ƒâ–¡
Example
Consider the BVP
where
Simple calculation shows that
We can find the proper \(A_{1}=\frac{1}{2}\) and \(A_{2}=\frac{3}{2}\) such that \(\frac{\alpha^{l} \sin A_{1}}{A_{1}}\geq1\) and \(\frac{\alpha^{L} \sin A_{2}}{A_{2}}\leq1\). Therefore (H_{1})(H_{3}) and the condition (ii) of TheoremÂ 3.2 are satisfied. It implies that (3.19) has at least one positive solution \(u(t)\).
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The work is supported by Chinese Universities Scientific Fund (Project No.2016LX002)
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Xie, W., Pang, H. The shooting method and integral boundary value problems of thirdorder differential equation. Adv Differ Equ 2016, 138 (2016). https://doi.org/10.1186/s1366201608244
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DOI: https://doi.org/10.1186/s1366201608244