# The shooting method and integral boundary value problems of third-order differential equation

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## Abstract

In this paper, the existence of at least one positive solution for third-order differential equation boundary value problems with Riemann-Stieltjes 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

shooting method positive solution third-order boundary conditions including Stieltjes integrals## 1 Introduction

It is well known that third-order 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 third-order differential equation BVPs (boundary value problems), for example [2, 3, 4, 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, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 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 Riemann-Stieltjes integral boundary condition and its importance, we refer the reader to [4, 5] and other related work such as [6, 7].

*λ*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 Λ 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 third-order 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 third-order differential equation with Riemann-Stieltjes 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, 19, 20, 21, 22, 23, 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 \(t\in(0,1)\). □

*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.

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 non*-

*trivial 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). □

### 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

*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*

### 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.

*r*such that

*L*large enough such that

Next, we will find a positive number \(m^{*}_{2}\) such that \(\varphi (m^{*}_{2})\geq0\).

*σ*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 \(t\in I_{m_{0}}\).

*t*, we have

*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,

*f*guarantees that \(y(t,m)\) is uniquely defined, the solution \(y(t,m_{k-1})\) and \(y(t,m_{k})\) have no intersection in the interval \([\bar{\delta}_{m_{k-1}},1)\). It follows from

*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 \(t\in(0,\sigma]\).

*n*large enough such that

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

_{1})-(H

_{3}) and the condition (ii) of Theorem 3.2 are satisfied. It implies that (3.19) has at least one positive solution \(u(t)\).

## Notes

### Acknowledgements

The work is supported by Chinese Universities Scientific Fund (Project No.2016LX002)

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