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

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.

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 [6–15] 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:

$$\begin{array}{r}-u^{″}(t)=f(t,u(t)),t\in [0,T],\\ u(0)=0,\alpha u(\eta )=u(T),\end{array}$$

(1)

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

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

$$\begin{array}{r}-u^{″}(t)=\sigma (t),t\in [0,T],\\ u(0)=0,\alpha u(\eta )=u(T)\end{array}$$

(2)

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

The general solution is

$$u(t)=c_1+c_{2}t-{\int }_0^t(t-s)\sigma (s)ds$$

with $c_1$, $c_2$ arbitrary constants.

From $u(0)=0$, we get $c_1=0$. From the second boundary condition, we have

$$(T-\alpha \eta )c_2={\int }_0^T(T-s)\sigma (s)ds-\alpha {\int }_0^{\eta }(\eta -s)\sigma (s)ds.$$

(3)

2.1 Nonresonance case

If $\alpha \eta \ne T$, then

$$c_2=\frac{1}{T-\alpha \eta }[{\int }_0^T(T-s)\sigma (s)ds-\alpha {\int }_0^{\eta }(\eta -s)\sigma (s)ds],$$

and the linear problem (2) has a unique solution for any $\sigma \in C[0,T]$. 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

$$u(t)={\int }_0^{T}g(t,s)\sigma (s)ds$$

(4)

with

$$g(t,s)=\{\begin{array}{cc}\frac{t(T-s)}{T-\alpha \eta }-\frac{t\alpha (\eta -s)}{T-\alpha \eta }-(t-s),\hfill & 0\le s<min(\eta ,t),\hfill \\ \frac{t(T-s)}{T-\alpha \eta }-\frac{t\alpha (\eta -s)}{T-\alpha \eta },\hfill & 0\le t<s<\eta <T,\hfill \\ \frac{t(T-s)}{T-\alpha \eta }-(t-s),\hfill & 0\le \eta <s<t\le T,\hfill \\ \frac{t(T-s)}{T-\alpha \eta },\hfill & max(\eta ,t)<s\le T.\hfill \end{array}$$

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

$${\int }_0^T(T-s)\sigma (s)ds=\alpha {\int }_0^{\eta }(\eta -s)\sigma (s)ds,$$

(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

$$u(t)=ct-{\int }_0^t(t-s)\sigma (s)ds,c\in \mathbb{R}.$$

In particular ct, $c\in \mathbb{R}$ is a solution of the homogeneous linear equation

$$-u^{″}(t)=0,t\in [0,T]$$

satisfying the boundary conditions

$$u(0)=0,\alpha u(\eta )=u(T).$$

Note that

$$u(T)-u(\eta )=c_{2}T-{\int }_0^T(T-s)\sigma (s)ds-c_2\eta +{\int }_0^{\eta }(\eta -s)\sigma (s)ds,$$

and then

$$c_2=\frac{1}{T-\eta }[u(T)-u(\eta )+{\int }_0^T(T-s)\sigma (s)ds-{\int }_0^{\eta }(\eta -s)\sigma (s)ds].$$

We now use that $u(T)=\frac{T}{\eta }u(\eta )$ to get

$$\frac{1}{T-\eta }[u(T)-u(\eta )]=\frac{1}{T}u(T)$$

and

$$c_2=\frac{1}{T-\eta }[{\int }_0^T(T-s)\sigma (s)ds-{\int }_0^{\eta }(\eta -s)\sigma (s)ds]+\frac{1}{T}u(T).$$

Hence the solution of (2) is given, implicitly, as

$$u(t)={\int }_0^T\frac{t(T-s)}{T-\eta }\sigma (s)ds-{\int }_0^{\eta }\frac{t(\eta -s)}{T-\eta }\sigma (s)ds-{\int }_0^t(t-s)\sigma (s)ds+\frac{t}{T}u(T)$$

or, equivalently,

$$u(t)={\int }_0^{T}k(t,s)\sigma (s)ds+\frac{t}{T}u(T),$$

(6)

where

$$k(t,s)=\{\begin{array}{cc}s,\hfill & 0\le s<min(\eta ,t),\hfill \\ t,\hfill & 0\le t<s<\eta \le T,\hfill \\ \frac{t(T-s)}{T-\eta }-(t-s),\hfill & 0\le \eta <s<t\le T,\hfill \\ \frac{t(T-s)}{T-\eta },\hfill & max(\eta ,t)<s\le T.\hfill \end{array}$$

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

3 Nonlinear problem

Defining the operators:

$$\begin{array}{c}\begin{array}{r}F:C[0,T]\to C[0,T],\\ [Fu](t)=f(t,u(t)),u\in C[0,T],t\in [0,T],\end{array}\hfill \\ \begin{array}{r}K:C[0,T]\to C[0,T],\\ [K\sigma ](t)={\int }_0^{T}k(t,s)\sigma (s)ds,\sigma \in C[0,T],t\in [0,T],\end{array}\hfill \\ \begin{array}{r}L:C[0,T]\to C[0,T],\\ [Lu](t)=\frac{t}{T}u(T),u\in C[0,T],t\in [0,T],\end{array}\hfill \end{array}$$

the nonlinear problem is equivalent to

$$u=Nu,$$

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

We note that (6) can be written as

$$u(t)-\frac{t}{T}u(T)={\int }_0^{T}k(t,s)\sigma (s)ds$$

and the nonlinear problem (1) as

$$u(t)-\frac{t}{T}u(T)={\int }_0^{T}k(t,s)f(s,u(s))ds.$$

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

For every constant $c\in \mathbb{R}$, we solve

$$v(t)={\int }_0^{T}k(t,s)f(s,v(s)+\frac{s}{T}c)ds$$

(7)

and let $\phi (c)$ 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 $v_c\in \phi (c)$, we consider

$$u_c(t)=v_c(t)+\frac{t}{T}c,$$

and hence

$$u_c(t)={\int }_0^{T}k(t,s)f(s,u_c(s))ds+\frac{t}{T}c.$$

If $c=u_c(T)$, then $u_c$ is a solution of the nonlinear problem (1). We then look for fixed points of the map

$$c\in \mathbb{R}⟶u_c(T)\in \mathbb{R}.$$

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

Assume that there exist $a,b\in C[0,T]$ and $\alpha \in [0,1)$ such that

$$|f(t,u)|\le a(t)+b(t){|u|}^{\alpha }$$

(8)

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

For $v\in C[0,T]$, define $F_{c}v\in C[0,T]$ as

$$[F_{c}v](t)=f(t,v(t)+\frac{t}{T}c).$$

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 $v\in C[0,T]$, let $\parallel v\parallel ={sup}_{t\in [0,T]}|v(t)|$.

For $\lambda \in (0,1)$, if $v=\lambda K_c(v)$ we have

$$v(t)=\lambda {\int }_0^{T}k(t,s)f(s,v(s)+\frac{s}{T}c)ds,$$

and

$$|v(t)|\le \parallel k\parallel {\int }_0^{T}f(s,v(s)+\frac{s}{T}c)ds\le \parallel k\parallel \cdot T[\parallel a\parallel +\parallel b\parallel {(\parallel v\parallel +c)}^{\alpha }].$$

Hence there exist constants $a_0$, $b_0$ such that

$$\parallel v\parallel \le a_0+b_0{(\parallel v\parallel +c)}^{\alpha }$$

(9)

for any $v\in C[0,T]$ and $\lambda \in (0,1)$ solution of $v=\lambda K_c(v)$. This implies that v is bounded independently of $\lambda \in (0,1)$, 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

$$|f(t,x)-f(t,y)|\le l|x-y|$$

(10)

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

Then, for $v,w\in C[0,T]$, we have

$$|[K_{c}v](t)-[K_{c}w](t)|\le {\int }_0^{T}k(t,s)l|v(s)-w(s)|ds$$

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

$$c\in \mathbb{R}⟶v_c\in C[0,T],$$

where $v_c$ is the unique solution of (7), and as a consequence of the contraction principle, this map is continuous.

Define the map

$$\begin{array}{c}\phi :\mathbb{R}⟶\mathbb{R},\hfill \\ \phi (c)=v_c(T).\hfill \end{array}$$

If there exists $c\in \mathbb{R}$ such that $\phi (c)=0$, then for that c we have $v_c(T)$, and the function

$$u_c(t)=v_c(t)+\frac{t}{T}c$$

is such that $u_c(T)=c$, and therefore $u_c$ is a solution of the original nonlinear problem (1).

Now, assume that

$$\underset{u\to \pm \mathrm{\infty }}{lim}f(t,u)=\pm \mathrm{\infty }$$

(11)

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

Then the growth of $\parallel v\parallel$ is sublinear in view of estimate (9). However, c growths linearly. Hence the norm of the function

$$v_c(s)+\frac{s}{T}c$$

growths asymptotically as c.

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

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 $f(t,u)=\sigma (t)$ and, in view of (5), the problem (1) may have no solution.