## Boundary Value Problems

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# Existence of a solution for a three-point boundary value problem for a second-order differential equation at resonance

Boundary Value Problems20132013:130

https://doi.org/10.1186/1687-2770-2013-130

Accepted: 30 April 2013

Published: 20 May 2013

## Abstract

We present a new existence result for a second-order nonlinear ordinary differential equation with a three-point boundary value problem when the linear part is noninvertible.

MSC:34B10, 34B15.

### Keywords

nonlinear ordinary differential equation three-point boundary value problem problem at resonance existence of solution

## 1 Introduction

The study of multi-point boundary value problems for linear second-order ordinary differential equations goes back to the method of separation of variables [1]. Also, some questions in the theory of elastic stability are related to multi-point problems [2]. In 1987, Il’in and Moiseev [3, 4] studied some nonlocal boundary value problems. Then, for example, Gupta [5] considered a three-point nonlinear boundary value problem. For some recent works on nonlocal boundary value problems, we refer, for example, to [615] and references therein.

As indicated in [16], there has been enormous interest in nonlinear perturbations of linear equations at resonance since the seminal paper of Landesman and Lazer [17]; see [18] for further details.

Here we study the following nonlinear ordinary differential equation of second order subject to the three-point boundary condition:
$\begin{array}{r}-{u}^{″}\left(t\right)=f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left[0,T\right],\\ u\left(0\right)=0,\phantom{\rule{2em}{0ex}}\alpha u\left(\eta \right)=u\left(T\right),\end{array}$
(1)

where $T>0$, $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ is a continuous function $\alpha \in \mathbb{R}$ and $\eta \in \left(0,T\right)$.

In this paper we consider the resonance case $\alpha \eta =T$ to obtain a new existence result. Although this situation has already been considered in the literature [19], we point out that our approach and methodology is different.

## 2 Linear problem

Consider the linear second-order three-point boundary value problem
$\begin{array}{r}-{u}^{″}\left(t\right)=\sigma \left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,T\right],\\ u\left(0\right)=0,\phantom{\rule{2em}{0ex}}\alpha u\left(\eta \right)=u\left(T\right)\end{array}$
(2)

for a given function $\sigma \in C\left[0,T\right]$.

The general solution is
$u\left(t\right)={c}_{1}+{c}_{2}t-{\int }_{0}^{t}\left(t-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds$

with ${c}_{1}$, ${c}_{2}$ arbitrary constants.

From $u\left(0\right)=0$, we get ${c}_{1}=0$. From the second boundary condition, we have
$\left(T-\alpha \eta \right){c}_{2}={\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-\alpha {\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds.$
(3)

### 2.1 Nonresonance case

If $\alpha \eta \ne T$, then
${c}_{2}=\frac{1}{T-\alpha \eta }\left[{\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-\alpha {\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right],$
and the linear problem (2) has a unique solution for any $\sigma \in C\left[0,T\right]$. In this case, we say that (2) is a nonresonant problem since the homogeneous problem has only the trivial solution as a solution, i.e., when $\sigma =0$, ${c}_{1}={c}_{2}=0$ and $u=0$. Note that the solution is given by
$u\left(t\right)={\int }_{0}^{T}g\left(t,s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds$
(4)
with
$g\left(t,s\right)=\left\{\begin{array}{cc}\frac{t\left(T-s\right)}{T-\alpha \eta }-\frac{t\alpha \left(\eta -s\right)}{T-\alpha \eta }-\left(t-s\right),\hfill & 0\le s

For $T=1$ this is precisely the function given in Lemma 2.3 of [20] or in Remark 12 of [21].

### 2.2 Resonance case

If $T=\alpha \eta$, then (3) is solvable if and only if
${\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds=\alpha {\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$
(5)
and then (2) has a solution if and only if (5) holds. In such a case, (2) has an infinite number of solutions given by
$u\left(t\right)=ct-{\int }_{0}^{t}\left(t-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}c\in \mathbb{R}.$
In particular ct, $c\in \mathbb{R}$ is a solution of the homogeneous linear equation
$-{u}^{″}\left(t\right)=0,\phantom{\rule{1em}{0ex}}t\in \left[0,T\right]$
satisfying the boundary conditions
$u\left(0\right)=0,\phantom{\rule{2em}{0ex}}\alpha u\left(\eta \right)=u\left(T\right).$
Note that
$u\left(T\right)-u\left(\eta \right)={c}_{2}T-{\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{c}_{2}\eta +{\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$
and then
${c}_{2}=\frac{1}{T-\eta }\left[u\left(T\right)-u\left(\eta \right)+{\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right].$
We now use that $u\left(T\right)=\frac{T}{\eta }u\left(\eta \right)$ to get
$\frac{1}{T-\eta }\left[u\left(T\right)-u\left(\eta \right)\right]=\frac{1}{T}u\left(T\right)$
and
${c}_{2}=\frac{1}{T-\eta }\left[{\int }_{0}^{T}\left(T-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{\eta }\left(\eta -s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right]+\frac{1}{T}u\left(T\right).$
Hence the solution of (2) is given, implicitly, as
$u\left(t\right)={\int }_{0}^{T}\frac{t\left(T-s\right)}{T-\eta }\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{\eta }\frac{t\left(\eta -s\right)}{T-\eta }\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{t}\left(t-s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{t}{T}u\left(T\right)$
or, equivalently,
$u\left(t\right)={\int }_{0}^{T}k\left(t,s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{t}{T}u\left(T\right),$
(6)
where
$k\left(t,s\right)=\left\{\begin{array}{cc}s,\hfill & 0\le s

We note that $k\in C\left(\left[0,T\right]×\left[0,T\right],\mathbb{R}\right)$ and $k\left(t,s\right)\ge 0$ for every $\left(t,s\right)\in \left[0,T\right]×\left[0,T\right]$.

## 3 Nonlinear problem

Defining the operators:
$\begin{array}{c}\begin{array}{r}F:C\left[0,T\right]\to C\left[0,T\right],\\ \left[Fu\right]\left(t\right)=f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}u\in C\left[0,T\right],t\in \left[0,T\right],\end{array}\hfill \\ \begin{array}{r}K:C\left[0,T\right]\to C\left[0,T\right],\\ \left[K\sigma \right]\left(t\right)={\int }_{0}^{T}k\left(t,s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\sigma \in C\left[0,T\right],t\in \left[0,T\right],\end{array}\hfill \\ \begin{array}{r}L:C\left[0,T\right]\to C\left[0,T\right],\\ \left[Lu\right]\left(t\right)=\frac{t}{T}u\left(T\right),\phantom{\rule{1em}{0ex}}u\in C\left[0,T\right],t\in \left[0,T\right],\end{array}\hfill \end{array}$
the nonlinear problem is equivalent to
$u=Nu,$

where $N=K\circ F+L$.

We note that (6) can be written as
$u\left(t\right)-\frac{t}{T}u\left(T\right)={\int }_{0}^{T}k\left(t,s\right)\sigma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds$
and the nonlinear problem (1) as
$u\left(t\right)-\frac{t}{T}u\left(T\right)={\int }_{0}^{T}k\left(t,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

This suggests to introduce the new function $v\left(t\right)=u\left(t\right)-\frac{t}{T}u\left(T\right)$. To find a solution u, we have to find v and $u\left(T\right)$.

For every constant $c\in \mathbb{R}$, we solve
$v\left(t\right)={\int }_{0}^{T}k\left(t,s\right)f\left(s,v\left(s\right)+\frac{s}{T}c\right)\phantom{\rule{0.2em}{0ex}}ds$
(7)
and let $\phi \left(c\right)$ be the set of solutions of (7). This set may be empty (no solution), a singleton (unique solution) or with more than one element (multiple solutions). For every ${v}_{c}\in \phi \left(c\right)$, we consider
${u}_{c}\left(t\right)={v}_{c}\left(t\right)+\frac{t}{T}c,$
and hence
${u}_{c}\left(t\right)={\int }_{0}^{T}k\left(t,s\right)f\left(s,{u}_{c}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{t}{T}c.$
If $c={u}_{c}\left(T\right)$, then ${u}_{c}$ is a solution of the nonlinear problem (1). We then look for fixed points of the map
$c\in \mathbb{R}⟶{u}_{c}\left(T\right)\in \mathbb{R}.$

For $c\in \mathbb{R}$ fixed, we try to solve the integral equation (7).

Assume that there exist $a,b\in C\left[0,T\right]$ and $\alpha \in \left[0,1\right)$ such that
$|f\left(t,u\right)|\le a\left(t\right)+b\left(t\right){|u|}^{\alpha }$
(8)

for every $t\in \left[0,T\right]$, $u\in \mathbb{R}$.

For $v\in C\left[0,T\right]$, define ${F}_{c}v\in C\left[0,T\right]$ as
$\left[{F}_{c}v\right]\left(t\right)=f\left(t,v\left(t\right)+\frac{t}{T}c\right).$

Thus, a solution of (7) is precisely a fixed point of $K\circ {F}_{c}={K}_{c}$. Note that ${K}_{c}$ is a compact operator. For $v\in C\left[0,T\right]$, let $\parallel v\parallel ={sup}_{t\in \left[0,T\right]}|v\left(t\right)|$.

For $\lambda \in \left(0,1\right)$, if $v=\lambda {K}_{c}\left(v\right)$ we have
$v\left(t\right)=\lambda {\int }_{0}^{T}k\left(t,s\right)f\left(s,v\left(s\right)+\frac{s}{T}c\right)\phantom{\rule{0.2em}{0ex}}ds,$
and
$|v\left(t\right)|\le \parallel k\parallel {\int }_{0}^{T}f\left(s,v\left(s\right)+\frac{s}{T}c\right)\phantom{\rule{0.2em}{0ex}}ds\le \parallel k\parallel \cdot T\left[\parallel a\parallel +\parallel b\parallel {\left(\parallel v\parallel +c\right)}^{\alpha }\right].$
Hence there exist constants ${a}_{0}$, ${b}_{0}$ such that
$\parallel v\parallel \le {a}_{0}+{b}_{0}{\left(\parallel v\parallel +c\right)}^{\alpha }$
(9)

for any $v\in C\left[0,T\right]$ and $\lambda \in \left(0,1\right)$ solution of $v=\lambda {K}_{c}\left(v\right)$. This implies that v is bounded independently of $\lambda \in \left(0,1\right)$, and hence by Schaefer’s fixed point theorem (Theorem 4.3.2 of [22]), ${K}_{c}$ has at least a fixed point, i.e., for given c, equation (7) is solvable.

Now suppose f is Lipschitz continuous.

Then there exists $l>0$ such that
$|f\left(t,x\right)-f\left(t,y\right)|\le l|x-y|$
(10)

for every $t\in \left[0,T\right]$ and $x,y\in \mathbb{R}$.

Then, for $v,w\in C\left[0,T\right]$, we have
$|\left[{K}_{c}v\right]\left(t\right)-\left[{K}_{c}w\right]\left(t\right)|\le {\int }_{0}^{T}k\left(t,s\right)l|v\left(s\right)-w\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds$
and
$\parallel {K}_{c}v-{K}_{c}w\parallel \le \parallel k\parallel \cdot l\cdot T\parallel v-w\parallel .$

Thus, for $l>0$ small, equation (7) has a unique solution in view of the classical Banach contraction fixed point theorem.

Now, under conditions (8) and (10), set
$c\in \mathbb{R}⟶{v}_{c}\in C\left[0,T\right],$

where ${v}_{c}$ is the unique solution of (7), and as a consequence of the contraction principle, this map is continuous.

Define the map
$\begin{array}{c}\phi :\mathbb{R}⟶\mathbb{R},\hfill \\ \phi \left(c\right)={v}_{c}\left(T\right).\hfill \end{array}$
If there exists $c\in \mathbb{R}$ such that $\phi \left(c\right)=0$, then for that c we have ${v}_{c}\left(T\right)$, and the function
${u}_{c}\left(t\right)={v}_{c}\left(t\right)+\frac{t}{T}c$

is such that ${u}_{c}\left(T\right)=c$, and therefore ${u}_{c}$ is a solution of the original nonlinear problem (1).

Now, assume that
$\underset{u\to ±\mathrm{\infty }}{lim}f\left(t,u\right)=±\mathrm{\infty }$
(11)

uniformly on $t\in \left[0,T\right]$.

Then the growth of $\parallel v\parallel$ is sublinear in view of estimate (9). However, c growths linearly. Hence the norm of the function
${v}_{c}\left(s\right)+\frac{s}{T}c$

growths asymptotically as c.

This implies that ${lim}_{c\to ±\mathrm{\infty }}\phi \left(c\right)=±\mathrm{\infty }$, and there exists $c\in \mathbb{R}$ with $\phi \left(c\right)=0$.

We have the following result.

Theorem 3.1 Suppose that f satisfies the growth conditions (8) and (10). If (11) holds, then (1) is solvable for l sufficiently small.

Note that condition (11) is crucial since for $f\left(t,u\right)=\sigma \left(t\right)$ and, in view of (5), the problem (1) may have no solution.

## Declarations

### Acknowledgements

This research has been partially supported by Ministerio de Economía y Competitividad (Spain), project MTM2010-15314, and co-financed by the European Community fund FEDER. The author is thankful to the referees for their useful suggestions.

## Authors’ Affiliations

(1)
Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela
(2)
Department of Mathematics, Faculty of Science, King Abdulaziz University

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