A note on the shooting method and its applications in the Stieltjes integral boundary value problems

Huilan Wang; Zigen Ouyang; Hengsheng Tang

doi:10.1186/s13661-015-0359-8

Research
Open Access

A note on the shooting method and its applications in the Stieltjes integral boundary value problems

Huilan Wang¹,
Zigen Ouyang¹Email author and
Hengsheng Tang¹

Boundary Value Problems20152015:102

https://doi.org/10.1186/s13661-015-0359-8

Received: 15 October 2014
Accepted: 25 May 2015
Published: 18 June 2015

Abstract

In this paper, the existence results of positive solutions for three-point Riemann-Stieltjes integral BVPs (boundary value problems) is considered. By applying shooting method and comparison principle, we obtain some new results which extend the known ones. At the same time, the theorems in one of our published articles are corrected by another theorem in this paper.

Keywords

integral boundary value problem
existence
positive solution
shooting method
comparison principle

MSC

34B10
34B18
34C10

1 Introduction

By applying the shooting method, we establish the criteria for the existence of positive solutions to the following Riemann-Stieltjes integral BVPs:

$$\begin{aligned}& u^{\prime\prime}(t)+a(t)f \bigl(u(t) \bigr)=0, \quad 0< t< 1, \end{aligned}$$

(1.1)

$$\begin{aligned}& u(0)=0,\qquad u(1)=\alpha\int_{0}^{\eta}u(s) \,ds, \end{aligned}$$

(1.2)

where $f\in C([0,\infty);[0,\infty))$ and $0<\eta<1$, $\alpha\geq0$ are given constants, and $0<\alpha\eta^{2}<2$.

Set

$$\begin{aligned}& f_{0}=\lim_{u\rightarrow0^{+}} \frac{f(u)}{u}, \qquad f_{\infty}=\lim_{u\rightarrow\infty} \frac{f(u)}{u}, \\& \bar{f}_{x}=\lim_{u\rightarrow x}\sup \frac {f(u)}{u}, \qquad \underline{f}_{x}=\lim_{u\rightarrow x}\inf \frac{f(u)}{u}, \quad x\in\{0,+\infty\}. \end{aligned}$$

By Krasnoselskii’s fixed point theorem in a cone, Tariboon and Sitthiwirattham [1] proved that BVP (1.1)-(1.2) has a positive solution in the case $f_{0}=0$ and $f_{\infty}=\infty$ (super-linear case) or in the case $f_{0}=\infty$ and $f_{\infty}=0$ (sub-linear case) when $0<\alpha\eta^{2}<2$.

Some meaningful results of nonlinear second-order integral BVPs have already been obtained by Kong [2], Webb and Infante [3, 4], etc. The following BVP:

$$ u^{\prime\prime}(t)+f \bigl(u(t) \bigr)=0, \quad 0< t< 1 ; \qquad u(0)=0, \qquad u(1)=\alpha\int_{0}^{\eta }u(s)\,ds, $$

(1.3)

is a special case of Webb and Infante’s [4], where we can deduce the result. Suppose $0<\alpha\eta^{2}<1$; BVP (1.3) has at least one positive solution if one of the following conditions holds:

(i)

$\bar{f}_{0}<\mu$ and $\underline{f}_{\infty}>\mu$;
(ii)

$\bar{f}_{0}>\mu$ and $\underline{f}_{\infty}<\mu$,

where $\mu=1/r(L)$ and $r(L)$ is the spectral radius of the associated linear operator. In [4], the authors used fixed point index theory.

As a numerical method, the shooting method is efficient to find the solution of BVPs [5–7]. Kwong and Wong [7] obtained some results for the Robin boundary condition of the form

$$ \sin\theta u(0)-\cos\theta u'(0) = 0 ,\qquad u(1)- \sum _{i=1}^{m-2} \alpha_{i} u_{i}( \eta_{i})=0, $$

(1.4)

where $\theta\in[0,3\pi/4]$ and $\theta\neq\pi/2$. Kwong and Wong [7] showed that BVP (1.1) with (1.4) has at least one positive solution if $\bar{f}_{0}< L_{\theta}$ and $\underline{f}_{\infty}>L_{\theta}$, where $L_{\theta}$ is a certain but not specified constant related to the associated linear operator.

When $\theta= \pi/2$ and $0\leq\sum_{i=1}^{m-2} \alpha_{i} \eta _{i}<1$, Ma [8] has studied BVP (1.1) with (1.4) by using Krasnoselskii’s fixed point theorem in a cone. The sufficient condition for the existence of positive solutions is also the super-linear case or the sub-linear case.

When $a(t)\equiv1$, $m=3$, $\eta=1/2$, as a special case of [8], the BVP

$$ u^{\prime\prime}(t)+f \bigl(u(t) \bigr)=0,\quad 0< t< 1 ; \qquad u(0)=0, \qquad u(1)= \mu u(\eta), $$

(1.5)

was studied by Kwong in [6], where the existence condition is

$$ \bar{f}_{0}< \biggl(2\cos^{-1} \biggl(\frac{\mu}{2} \biggr) \biggr)^{2}< \underline {f}_{\infty}, \quad \mbox{or}\quad \bar{f}_{\infty}< \biggl(2\cos^{-1} \biggl(\frac{\mu }{2} \biggr) \biggr)^{2} < \underline{f}_{0}, $$

(1.6)

which is obtained by the shooting method.

Following the main idea in [6, 7], we considered the generalized multi-point integral BCs [9]

$$ u(0)=0,\qquad u(1)=\sum_{i=1}^{n} \alpha_{i}\int_{0}^{\eta_{i}}u(s) \,ds, $$

(1.7)

where $0<\eta_{1}<\eta_{2}<\cdots<\eta_{n}<1$, $\alpha_{i}\geq0$ for $i= 1,\ldots ,n-1$, and $\alpha_{n}>0$ are given constants.

However, Theorem 1.1 and some proofs in [9] need to be corrected, which is one of the reasons why we write this paper. Furthermore, more general existence criteria are presented in this article as well as the application of the shooting method in the study of BVPs. For simplicity and without loss of generality, we start from BVP (1.1)-(1.2).

2 Preliminaries: some notation and lemmas

The principle of the shooting method is converting the BVP into an IVP (initial value problem) by finding suitable initial slopes $m>0$ such that the solution of (1.1) comes with the initial value condition

$$ u(0)=0,\qquad u'(0)=m. $$

(2.1)

Denote by $u(t,m)$ the solution of the IVP (1.1) with (2.1) provided it exists, and define

$$ k(m)= \frac{\alpha\int_{0}^{\eta}u(s,m)\,ds}{u(1,m)},\qquad \varphi(m)=\alpha\int _{0}^{\eta}u(s,m)\,ds-u(1,m). $$

(2.2)

Then solving the boundary value problem is equivalent to finding a $m^{*}$ such that $k(m^{*})=1$ or $\varphi(m^{*})=0$.

For the sake of convenience, we denote

$$ \max_{0\leq t\leq1} \bigl\{ a(t) \bigr\} =a^{L}, \qquad \min _{0\leq t\leq1} \bigl\{ a(t) \bigr\} =a^{l}. $$

In this paper, we always assume

(H₁):: $f\in C([0,\infty);[0,\infty))$, $a\in C([0,1];[0,\infty))$.

Furthermore, we assume that f is strong continuous enough to guarantee that $u(t, m)$ is uniquely defined and that it depends continuously on both t and m. As for the discussion of this problem, see [6].

Next, we present some comparison theorems which help us to establish the main results.

Lemma 2.1

(Sturm comparison theorem)

Let $\varphi_{1}$ and $\varphi_{2}$ be non-trivial solutions of the equations

$$y''+ q_{1}(x)y = 0,\qquad y''+ q_{2}(x)y = 0, $$

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 2.2

Let $y(t,m)$, $z(t,m)$, $Z(t,m)$ be the positive solution of the initial value problems, respectively,

$$\begin{aligned}& y^{\prime\prime}(t)+f \bigl(y(t) \bigr)=0, \qquad y(0)=0, \qquad y'(0)=m, \\& Z^{\prime\prime}(t)+G(t)Z(t)=0,\qquad Z(0)=0, \qquad Z'(0)=m, \\& z^{\prime\prime}(t)+g(t)z(t)=0,\qquad z(0)=0, \qquad z'(0)=m. \end{aligned}$$

Suppose $g(t)\leq G(t)$ be two piecewise continuous functions defined on $[0, 1]$. If

$$ 0\leq g(t)\leq\frac{f(y(t))}{y(t)}\leq G(t) $$

and suppose that $Z(t)$ does not vanish in $(0,1]$, then for any $0\leq s\leq\xi\leq1$, it yields

$$ \frac{z(s,m)}{z(\xi,m)}\leq \frac{y(s ,m)}{y(\xi,m)}\leq \frac{Z(s,m)}{Z(\xi,m)}, $$

(2.3)

and hence, for any $0\leq\eta\leq\xi\leq1$, we have

$$ \frac{\int_{0}^{\eta}z(s,m)\,ds}{z(\xi,m)}\leq \frac{\int_{0}^{\eta}y(s,m)\,ds}{y (\xi,m)}\leq \frac{\int_{0}^{\eta}Z(s,m)\,ds}{Z(\xi,m)}. $$

(2.4)

Proof

Since $0\leq g(t)\leq f(y(t))/y(t)\leq G(t)$ and $Z(t)$ does not vanish in $(0,1]$, from Lemma 2.1, it follows that $y(t)$ and $z(t)$ will not vanish in $(0,1]$. The proof for (2.3) can be seen in [6]. The continuity of the integrands implies the existence of the Riemann integral. In view of the definition of Riemann integral, by using the inequality of the limit, we have (2.4). □

Remark 2.1

Lemma 2.2 is also the correction for Theorem 1.1 in [9].

Lemma 2.3

Consider the BVP

$$\begin{aligned}& y^{\prime\prime}(t)+Ay(t)=0,\quad 0< t< 1, \end{aligned}$$

(2.5)

$$\begin{aligned}& y(0)=0, \qquad y(1)=b. \end{aligned}$$

(2.6)

(i)

If $A=\pi^{2}$, then $y(t)$ vanishes at $t=1$ for the first time on interval $(0,1]$ and $b=0$;
(ii)

if $0< A<\pi^{2}$, then $y(t)$ does not vanish on the interval $(0,1]$ and $b>0$;
(iii)

if $A>\pi^{2}$, then $y(t)$ vanishes before $t=1$ on interval $(0,1]$.

Proof

Obviously, $y(t)=\sin(\pi^{2}t)$ satisfies the conditions $y(0)=0$, $y(1)=0$, and $y(t)>0$ for $t\in(0,1)$, hence (i) is established. According to the Sturm comparison theorem, we can draw the conclusions (ii) and (iii). □

Lemma 2.4

([1])

Assume that (H₁) holds and $\alpha \eta^{2}> 2$, then BVP (1.1)-(1.2) has no positive solution.

In [1] and [9], the proofs are conducted by contradiction to the concavity of solution (also see [4]). In fact, for $m>0$, we compare the solution $u(t,m)$ of the IVP given by (1.1) and (2.1) with the solution $y(t)=mt$ of

$$ y^{\prime\prime}(t)+0y(t)=0,\qquad y(0)=0, \qquad y'(0)=m. $$

(2.7)

If BVP (1.1)-(1.2) has a positive solution $u(t,m)$, then by Lemma 2.2 and the concavity of $u(t,m)$, we have

$$ \frac{1}{\eta}\geq\frac{u(1,m)}{u(\eta,m)}=\frac{\alpha\int_{0}^{\eta }u(s,m)\,ds}{u(\eta,m)}\geq \frac{\alpha\int_{0}^{\eta}y(s,m)\,ds}{y(\eta,m)} =\frac{\alpha\int_{0}^{\eta}ms\,ds}{m\eta}=\frac{\alpha\eta}{2}, $$

(2.8)

that is, $\alpha\eta^{2}\leq2$.

In the following, we always assume that

(H₂):: $0< \alpha\eta^{2} <2 $.

3 Main results

Lemma 3.1

Assume that (H₁)-(H₂) holds. Then there exist a solution $x=A_{1}\in(0,\pi)$ such that

$$ g_{1}(x):=\frac{\alpha[1-\cos(\eta x)]}{x\sin x}=1 $$

(3.1)

and a solution $x=A_{2}\in(0,\pi)$ such that

$$ g_{2}(x):=\frac{\alpha\eta\sin(\eta x)}{2\sin x}=1. $$

(3.2)

Proof

It is not difficult to show that

$$\lim_{x\rightarrow0^{+}}g_{1}(x)=\frac{\alpha\eta^{2} }{2}< 1, \qquad \lim _{x\rightarrow\pi^{-}}g_{1}(x)=\infty>1. $$

Since the function $g_{1}(x)$ is continuous on $(0,\pi)$, there must exist a constant $A_{1}\in(0,\pi)$ such that $g_{1}(A_{1})=1$.

Similarly,

$$\lim_{x\rightarrow0^{+}}g_{2}(x)=\frac{\alpha\eta^{2}}{2}< 1, \qquad \lim _{x\rightarrow\pi^{-}}g_{2}(x)=\infty>1. $$

Thus, there exists a positive constant $A_{2}\in(0,\pi)$ such that $g_{2}(A_{2})=1$. □

Theorem 3.1

Assume that (H₁)-(H₂) holds. Suppose one of the following conditions holds:

$$(\mathrm{i}) \quad 0\leq \bar{f}_{0}< \frac{\underline {A}^{2}}{a^{L}} ,\qquad \underline{f}_{\infty}> \frac {\bar{A}^{2}}{a^{l}}; \qquad (\mathrm{ii})\quad 0\leq \bar{f}_{\infty}< \frac{\underline{A}^{2}}{a^{L}}, \qquad\underline {f}_{0}> \frac{\bar{A}^{2}}{a^{l}}. $$

Then problem (1.1)-(1.2) has at least one positive solution, where

$$\underline{A}=\min\{A_{1},A_{2}\},\qquad \bar{A}=\max \{A_{1},A_{2}\}, $$

and $A_{1}$, $A_{2}$ is defined in (3.1) and (3.2), respectively.

Proof

(i) Since $0\leq \bar{f}_{0}< \frac{\underline{A}^{2}}{a^{L}}$, there exists a positive number r such that

$$ \frac{f(u)}{u}< \frac{{\underline{A}}^{2}}{a^{L}}\leq\frac{{A_{1}}^{2}}{a^{L}}, \quad 0< u\leq r. $$

(3.3)

Let $0< m_{1}^{*}< r$, then from the Sturm comparison theorem and the concavity of $u(t,m_{1}^{*})$, it follows that $0\leq u(t,m_{1}^{*})\leq m_{1}^{*}t\leq m_{1}^{*}< r$ for $t\in[0,1]$. Thus

$$ 0\leq a(t)f \bigl(u \bigl(t,m_{1}^{*} \bigr) \bigr)< a^{L} \frac{A_{1}^{2}}{a^{L}}u \bigl(t,m_{1}^{*} \bigr)=A_{1}^{2}u\bigl(t,m_{1}^{*}\bigr)< \pi ^{2}u \bigl(t,m_{1}^{*} \bigr),\quad t\in(0,1]. $$

By Lemma 2.3, it gives $u(t,m_{1}^{*})>0$ for $t\in(0,1]$.

Let $Z(t)=(m_{1}^{*}/A_{1})\sin(A_{1} t)$ for $t\in[0,1]$, then

$$ Z^{\prime\prime}(t)+A_{1}^{2}Z(t)=0,\qquad Z(0)=0,\qquad Z'(0)=m_{1}^{*}. $$

(3.4)

From Lemma 2.2 and Lemma 3.1, we have

$$ k \bigl(m_{1}^{*} \bigr)= \frac{\alpha\int_{0}^{\eta}u(s,m_{1}^{*})\,ds}{u(1,m_{1}^{*})} < \frac{\alpha\int_{0}^{\eta}m_{1}^{*}\sin (A_{1}s)\,ds}{m_{1}^{*}\sin A_{1}}= \frac{\alpha[1-\cos(\eta A_{1})]}{A_{1}\sin A_{1}}=1, $$

(3.5)

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

$$ \frac{f(u)}{u}>\frac{{\bar{A}}^{2}}{a^{l}}\geq\frac{A_{2}^{2}}{a^{l}}, \quad u\geq L, $$

(3.6)

and there exists a positive number $\epsilon< A_{2}(1-\eta)/\eta$ small enough that

$$ \frac{f(u)}{u}\geq\frac{(A_{2}+\epsilon)^{2}}{a^{l}},\quad u\geq L. $$

(3.7)

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

Claim. There exist a slope $m_{2}^{*}$ and two positive numbers ρ and σ such that

$$0< \rho\leq\eta\leq\frac{A_{2}}{A_{2}+\epsilon} \leq\sigma\leq1 \quad \mbox{and}\quad u \bigl(t,m_{2}^{*} \bigr)\geq L \quad\mbox{for } t\in[\rho, \sigma]. $$

Since the solution $u(t,m)$ is concave, it hits the line $u=L$ at most two times for the constant L defined in (3.6) and $t\in(0,1]$. We denote the left intersecting time by $\underline{\delta}_{m}$ and the right one by $\overline{\delta}_{m}$ provided they exist. Henceforth, denote $I_{m}=[\underline{\delta}_{m},\overline{\delta}_{m}]\subseteq (0,1]$. If $u(1,m)\geq L$, then $\overline{\delta}_{m}=1$.

The discussion is divided into three steps.

Step 1. We claim that there exists a slope $m_{0}$ large enough such that $0\leq u(t,m_{0})\leq L$ for $t\in[0,\underline{\delta }_{m_{0}}]$ and $u(t,m_{0})\geq L$ for $t\in I_{m_{0}}$.

Otherwise, provided $u(t,m)\leq L$ for all $t\in[0,1]$ as $m\rightarrow\infty$, then by integrating both sides of (1.1) from 0 to t, we have

$$ u(t,m)= mt-\int_{0}^{t} (t-s)a(s)f \bigl(u(s,m) \bigr)\,ds. $$

(3.8)

Hence, from (3.3) and the continuity of $f(u)$, we have

$$ m= u(1,m)+\int_{0}^{1}(1-s) a(s)f \bigl(u(s,m) \bigr)\,ds\leq L+L_{f} a^{L}, $$

(3.9)

where $L_{f}=\max_{u\in[0,L]}{f(u)}$. If we choose $m>L+L_{f} a^{L}$, (3.9) will lead to a contradiction.

Since $u(t,m)$ is continuous and concave, there exists a number $m_{0}$ large enough such that $u(t,m_{0})\geq L$ for $t\in I_{m_{0}}$.

Step 2. There exists a monotonically increasing sequence $\{m_{k}\} $ such that the sequence ${\underline{\delta}_{m_{k}}}$ is decreasing on $m_{k}$ and ${\overline{\delta}_{m_{k}}}$ is increasing on $m_{k}$. That is,

$$I_{m_{0}}\subset I_{m_{1}}\subset\cdots\subset I_{m_{k}} \subset\cdots \subseteq(0,1] $$

and $u(t,m_{k})\geq L$ for $t\in I_{m_{k}}$.

First, we prove that

$$ \underline{\delta}_{m_{k}}< \underline{\delta}_{m_{k-1}},\quad k=1,2 \ldots \mbox{ for } m_{k}>m_{k-1}. $$

(3.10)

When $k=1$, we have

$$u(\underline{\delta}_{m_{0}},m_{1})>u(\underline{ \delta}_{m_{0}},m_{0}) $$

in the case

$$ m_{1}>m_{0}+2a^{L}L_{f}\underline{ \delta}_{m_{0}}. $$

(3.11)

Otherwise, provided

$$ u(\underline{\delta}_{m_{0}},m_{1})\leq u(\underline{\delta }_{m_{0}},m_{0})=L, $$

(3.12)

then from (3.8) and (3.11), we have

$$\begin{aligned} & u(\underline{\delta}_{m_{0}},m_{1})-u(\underline{\delta }_{m_{0}},m_{0}) \\ &\quad= (m_{1}-m_{0})\underline{\delta}_{m_{0}}-\int _{0}^{\underline {\delta}_{m_{0}}} (\underline{\delta }_{m_{0}}-s)a(s) \bigl[f \bigl(u(s,m_{1}) \bigr)-f \bigl(u(s,m_{0}) \bigr) \bigr] \,ds \\ &\quad> (m_{1}-m_{0})\underline{\delta}_{m_{0}}-2a^{L}L_{f} \underline {\delta}_{m_{0}}^{2} \\ &\quad= \underline{\delta}_{m_{0}} \bigl[(m_{1}-m_{0})-2a^{L}L_{f} \underline {\delta}_{m_{0}} \bigr]>0, \end{aligned}$$

which contradicts (3.12).

Hence, for a slope $m_{1}>m_{0}+2a^{L}L_{f}\underline{\delta}_{m_{0}}$, there exists a number $0<\underline{\delta}_{m_{1}}<\underline{\delta}_{m_{0}}$ such that

$$u(\underline{\delta}_{m_{1}},m_{1})=L, \quad \mbox{and} \quad u(t,m_{1})\leq L \quad \mbox{for } t\in (0,\underline{ \delta}_{m_{1}}]. $$

See Figure 1.

Figure 1
**The relationship of** m **and** $\pmb{I_{m}}$ . $m_{1}>m_{0}$, $I_{m_{0}}\subset I_{m_{1}}$.

By mathematical induction, it is not difficult to show that $\underline{\delta}_{m_{k}}<\underline{\delta}_{m_{k-1}}$, $k=1,2,\ldots$ .

Further, we turn to the right hand of the interval $I_{m_{k}}$. Since f guarantees that $u(t, m)$ is uniquely defined, the solutions $u(t,m_{k-1})$ and $u(t,m_{k})$ have no intersection in the interval $[\underline{\delta}_{m_{k-1}},1)$. It follows from

$$u(\underline{\delta}_{m_{k-1}},m_{k})>u(\underline{ \delta}_{m_{k-1}},m_{k-1}) $$

that

$$u(\overline{\delta}_{m_{k-1}},m_{k})>u(\overline{ \delta}_{m_{k-1}},m_{k-1}). $$

Thus we have

$$ \overline{\delta}_{m_{k}}>\overline{\delta}_{m_{k-1}},\quad k=1,2, \ldots\mbox{ for } m_{k}>m_{k-1}. $$

(3.13)

When $k=1$, also see Figure 1.

Step 3. Seek out a slope $m_{2}^{*}$ and two positive numbers ρ and σ such that $0<\rho\leq\eta\leq\frac{A_{2}}{A_{2}+\epsilon} \leq\sigma\leq1$ and $u(t,m_{2}^{*})\geq L$ for $t\in[\rho,\sigma]$.

Subcase 1. $\eta\in[\underline{\delta}_{m_{0}},\overline{\delta }_{m_{0}}]$ and $u(1,m_{0})\geq L$. In this case, we take $m_{2}^{*}=m_{0}$ and $\rho=\underline{\delta}_{m_{0}}$, $\sigma=\overline{\delta}_{m_{0}}=1$.

Subcase 2. $\eta\nsubseteq[\underline{\delta}_{m_{0}},\overline {\delta}_{m_{0}}]$ or $u(1,m_{0})< L$. Following the step 1, step 2, and the extension principle of solutions, there exists a positive integer n large enough such that

$$ \underline{\delta}_{m_{n}}< \eta, \qquad \overline{\delta}_{m_{n}} \geq\frac {A_{2}}{A_{2}+\epsilon}. $$

(3.14)

If we take $m_{2}^{*}=m_{n}$ and $\rho=\underline{\delta}_{m_{n}}$, $\sigma =\overline{\delta}_{m_{n}}$, then

$$ \sigma(A_{2}+\epsilon)\geq A_{2}. $$

(3.15)

Two of the possible cases of $I_{m_{0}}$ can be seen in Figure 2.

Figure 2
**Two of the possible cases of** $\pmb{I_{m_{0}}}$ . Subcase 1: $\eta\in[\underline{\delta}_{m_{0}},\overline{\delta }_{m_{0}}]$ and $u(1,m_{0})\geq L$; Subcase 2: $\eta\nsubseteq[\underline{\delta }_{m_{0}},\overline{\delta}_{m_{0}}]$ or $u(1,m_{0})< L$.

In the following, we prove that $k(m_{2}^{*})\geq1$ or $\varphi(m_{2}^{*})>0$ for the selected $m_{2}^{*}$ and ρ, σ.

Set $z(t)= (m_{2}^{*}/\sigma(A_{2}+\epsilon) )\sin (\sigma (A_{2}+\epsilon)t )$, then

$$ z''(t)+\sigma^{2}(A_{2}+ \epsilon)^{2}z(t)=0, \qquad z(0)=0, \qquad z'(0)=m_{2}^{*}, \quad t\in[\rho,\sigma], $$

(3.16)

where $\rho\leq\eta<\sigma\leq1$. From (3.7), we have

$$\frac{f(u)}{u}\geq\frac{\sigma^{2}(A_{2}+\epsilon)^{2}}{a^{l}},\quad u\geq L. $$

Further, noting that $u(1,m_{2}^{*})>L$ (this time $\sigma=1$) or $u(1,m_{2}^{*})\leq u(\sigma,m_{2}^{*})=L$ and the function

$$S(x)=\frac{\sin\eta x}{\sin x} $$

is increasing for $x\in(0,\pi)$, then by Lemma 2.2, Lemma 3.1, and inequality (3.15), we have

$$\begin{aligned} k \bigl(m_{2}^{*} \bigr)&= \frac{\alpha\int_{0}^{\eta}u(s,m_{2}^{*})\,ds}{u(1,m_{2}^{*})} \geq \frac{\alpha\eta u(\eta,m_{2}^{*})}{2u(1,m_{2}^{*})}\geq \frac{\alpha\eta u(\eta,m_{2}^{*})}{2u(\sigma,m_{2}^{*})} \\ &\geq \frac{\alpha\eta\sin\eta\sigma(A_{2}+\epsilon)}{2\sin \sigma(A_{2}+\epsilon)}\geq \frac{\alpha\eta\sin(\eta A_{2})}{2\sin A_{2}}=1, \end{aligned}$$

(3.17)

which implies $\varphi(m_{2}^{*})\geq0$.

From (3.5) and (3.17), we can find a $m^{*}$ between $m_{1}^{*}$ and $m_{2}^{*}$ such that $u(t,m^{*})$ is the solution of (1.1)-(1.2). The theorem is complete.

The proof for (ii) is similar, so we omit it. □

Now, we present the result for BVP (1.1) with (1.7), which is also the correction of Theorem 3.1 and Theorem 3.2 in [9].

Theorem 3.2

Assume that (H₁)-(H₂) hold. Suppose one of the following conditions holds:

$$(\mathrm{i}) \quad 0\leq \bar{f}_{0}< \frac{\underline {A}^{2}}{a^{L}} , \qquad \underline{f}_{\infty}> \frac {\bar{A}^{2}}{a^{l}}; \qquad (\mathrm{ii}) \quad 0\leq \bar{f}_{\infty}< \frac{\underline{A}^{2}}{a^{L}}, \qquad \underline {f}_{0}> \frac{\bar{A}^{2}}{a^{l}}. $$

Then problem (1.1) with (1.7) has at least one positive solution, where

$$\underline{A}=\min\{A_{1},A_{2}\},\qquad \bar{A}=\max \{A_{1},A_{2}\} $$

and $A_{1}$, $A_{2}$ is defined by

$$ \frac{\sum_{i=1}^{n}\alpha_{i}[1-\cos(A_{1}\eta_{i} )]}{A_{1}\sin A_{1}}=1 $$

(3.18)

and

$$ \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(A_{2}\eta_{i} )}{2\sin A_{2}}=1. $$

(3.19)

Proof

Similar to (3.5) and (3.17), it follows from (1.7) and (3.18)-(3.19) that

$$\begin{aligned} k \bigl(m_{1}^{*} \bigr)&= \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}u(s,m_{1}^{*})\,ds}{u(1,m_{1}^{*})} < \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}m_{1}^{*}\sin(A_{1}s)\,ds}{m_{1}^{*}\sin A_{1}} \\ &= \frac{\sum_{i=1}^{n}\alpha_{i}[1-\cos(A_{1}\eta_{i} )]}{A_{1}\sin A_{1}}=1 \end{aligned}$$

(3.20)

and

$$\begin{aligned} k \bigl(m_{2}^{*} \bigr)&= \frac{\sum_{i=1}^{n}\alpha_{i}\int_{0}^{\eta _{i}}u(s,m_{2}^{*})\,ds}{u(1,m_{2}^{*})} \geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i} u(\eta _{i},m_{2}^{*})}{2u(1,m_{2}^{*})}\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i} u(\eta_{i},m_{2}^{*})}{2u(\sigma,m_{2}^{*})} \\ &\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(\eta _{i}\sigma(A_{2}+\epsilon))}{2\sin\sigma(A_{2}+\epsilon)}\geq \frac{\sum_{i=1}^{n}\alpha_{i}\eta_{i}\sin(A_{2}\eta_{i} )}{2\sin A_{2}}=1, \end{aligned}$$

(3.21)

where $\eta_{n}<\sigma\leq1$ and (3.15) holds.

The remainder of the proof is similar, so we omit it. □

4 Conclusion and discussion

The conditions in [8] and [1] are easy to verify; however, they are not as general as ours, because the sup-linear case or the sub-linear case is sufficient for the conditions in Theorem 3.1. As an example of [4], where the constant μ is related to the Green’s function and the spectral radius of associated linear operator, our calculation is more direct. The idea of this paper was illuminated by [6, 7]; however, the certain constant $L_{\theta}$ could not be given explicitly in [7] and η only equals $1/2$ in [6]. From this point of view, this paper extends the work of [6, 7] and presents another way to find the ‘eigenvalue’ by numerical calculation, though it is related to a transcendental equation which has at least one numerical solution.

In fact, we can extent our results to [8]. The proof is fit, where

$$k(m)= \frac{\sum_{i=1}^{m-2} \alpha_{i} u(\eta_{i},m)}{u(1,m)} $$

and the constant $A=A_{1}=A_{2}\in(0,\pi)$ is explicitly determined by

$$ \frac{\sum_{i=1}^{m-2} \alpha_{i} \sin(A\eta_{i} )}{\sin A}=1. $$

(4.1)

In other words, we can substitute the condition

(i):: $f_{0} = 0$ and $f_{\infty}= \infty$, or

(ii):: $f_{0}= \infty$ and $f_{\infty}= 0$,

with

(i′):: $0\leq\bar{f}_{0}< A^{2}<\underline{f}_{\infty}$; or

(ii′):: $0\leq\bar{f}_{\infty}< A^{2}< \underline{f}_{0}$,

where A is defined in (4.1).

Next, we apply the result to the special case BVP (1.5), where $a^{L}=a^{l}=1$, $m=3$, $\alpha=\mu$, $\eta=1/2$. From (4.1), we have

$$A=2\cos^{-1} \biggl(\frac{\mu}{2} \biggr). $$

By plugging it into (i′) and (ii′), we have the same result as (1.6).

Further, when $\alpha\eta^{2}=2$, BVP (1.1)-(1.2) is at resonant. There may not exist a solution $x=A_{1}\in(0,\pi)$ and $x=A_{2}\in(0,\pi)$ to (3.1) and (3.2), respectively. If (3.1) and (3.2) has a solution $x=A_{1}\in(0,\pi)$ and $x=A_{2}\in(0,\pi)$, respectively, then we can also obtain the existence result for (1.1)-(1.2), similarly for (1.1) with (1.7).

When $\theta= \pi/2$ and $\sum_{i=1}^{m-2} \alpha_{i} \eta_{i}=1$, BVP (1.1) with (1.4) is resonant. If there exists a number $A\in (0,\pi)$ such that (4.1), then the existence result for BVP (1.1) with (1.4) can be obtained, similarly for BVP (1.5).

Declarations

Acknowledgements

It was remarked to the authors by Professor Webb that the result is not correct in [9]. The authors would like to express their sincere gratitude to Professor Webb for his helpful comments and suggestion on the manuscript, as well as the anonymous reviewers’ comments. Moreover, the first author HW is sorry for having cited the comparison theorem by mistake in [9]. This project was supported by the Scientific Research Fund of Hunan Provincial Educational Department (No. 13A088), the Scientific Research Foundation of Hengyang City (No. 2012KJ2) and the Construct Program in USC.

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

(1)

School of Mathematics and Physics, University of South China, Hengyang, 421001, P.R. China

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