Solvability of a nonlinear second order m-point boundary value problem with p-Laplacian at resonance

Abstract

We study the existence of solutions of the nonlinear second order m-point boundary value problem with p-Laplacian at resonance

$$ \textstyle\begin{cases} (\phi _{p}(x'))'=f(t,x,x'),\quad t\in [0,1],\\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}a_{i}x(\xi _{i}), \end{cases} $$

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, $f:[0,1]\times \mathbb{R}^{2}\to \mathbb{R}$ is a continuous function, $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$. Based on the topological transversality method together with the barrier strip technique and the cut-off technique, we obtain new existence results of solutions of the above problem. Meanwhile some examples are also given to illustrate our main results.

1 Introduction

In this paper, we consider the existence of solutions of the following m-point boundary value problem with p-Laplacian at resonance:

$$ \textstyle\begin{cases} (\phi _{p}(x'))'=f(t,x,x'), \quad t\in [0,1], \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}a_{i}x(\xi _{i}), \end{cases} $$

(1.1)

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, $f:[0,1]\times \mathbb{R}^{2}\rightarrow \mathbb{R}$ is a continuous function, $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, and $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$. By a solution to problem (1.1), we mean a function $x\in C^{1}[0,1]$ with $\phi _{p}(x')\in C^{1}[0,1]$ that satisfies problem (1.1).

It is well known that the research of m-point boundary value problems is significant in the theory of ordinary differential equations and practical applications, see [4, 13].

We specifically mention that the existence of solutions of problem (1.1) with $p=2$ has been studied by Gupta [12], Feng and Webb [3], Infante and Webb [14], Ma [20], Infante and Zima [15], Liang and Lin [18], in which the Leray–Schauder continuation theorem, the Mawhin’s continuation theorem, the theory of fixed point index, the nonlinear alternative of Leray–Schauder, the Leggett–Williams norm-type theorem due to O’Regan and Zima [21], and the fixed point theorem in cones are used, respectively. In the case that $p>1$, problem (1.1) has been studied by García-Huidobro et al. [5–7] via some continuation lemma of Leray–Schauder type due to themselves, Zhu and Wang [24] via Mawhin’s continuation theorem. For other types of works regarding the second order nonlocal boundary value problems involving p-Laplacian, we refer the reader to [1, 2, 8–10, 16, 23].

Inspired by the above works and [17, 22], the aim of this paper is to establish the new existence results of solutions to problem (1.1) by using the topological transversality method together with the barrier strip technique and the cut-off technique.

We would like to emphasize that the results of this paper are new even when $p=2$, and the main research tool used is the topological transversality theorem, which is different from those in [5–7, 20, 24]. The advantage of the topological transversality theorem is that it can transform the resonance problem into a nonresonance problem by appropriately selecting a convex subset U of X. Besides, in [5–7], the order of growth of f with respect to the derivative term is less than or equal to p. However, in our results, we just impose some local sign conditions on the nonlinear term f and do not require the growth constraints, so the degree of the derivative term in f can exceed p.

This work is organized as follows: In Sect. 2, we first briefly introduce the topological transversality theory. And then, a modified boundary value problem is constructed by the cut-off technique. Finally, the barrier strip technique is used to estimate a prior bound of solutions of the modified boundary value problems in $C^{1}[0,1]$. In Sect. 3, the topological transversality method is used to establish some existence theorems of solutions of problem (1.1). As applications of our main results, some examples are given in the last section.

Throughout this paper, the following local conditions on f will be used.

(H₁):

There exists $r_{1}\leq 0\leq r_{2}$ with $r_{1}^{2}+r_{2}^{2}>0$ such that

$$ f(t,r_{1},0) \leq 0, \qquad f(t,r_{2},0) \geq 0, \quad \forall t\in [0,1]. $$

(H₂):

There exists $R_{1}\leq 0\leq R_{2}$ with $R_{1}^{2}+R_{2}^{2}>0$ such that

$$ f(t,x,R_{1})\geq 0, \qquad f(t,x,R_{2})\leq 0, \quad \forall (t,x)\in [0,1] \times [r_{1},r_{2}]. $$

(H₃):

There exists $R_{1}\leq 0\leq R_{2}$ with $R_{1}^{2}+R_{2}^{2}>0$ such that

$$ f(t,x,R_{1})\geq 0,\qquad f(t,x,R_{2})\ge 0, \quad \forall (t,x)\in [0,1] \times [r_{1},r_{2}], $$

and

$$ f(t,x,y)\leq \frac{\phi _{p}(R_{2})}{1-\xi _{1}},\quad \forall (t,x,y) \in [0,1]\times [r_{1},r_{2}]\times [R_{1},R_{2}]. $$

(H₄):

There exists $R_{1}\leq 0 \leq R_{2}$ with $R_{1}^{2}+R_{2}^{2}>0$ such that

$$ f(t,x,R_{1})\leq 0, \qquad f(t,x,R_{2})\leq 0, \quad \forall (t,x)\in [0,1] \times [r_{1},r_{2}], $$

and

$$ f(t,x,y)\geq \frac{\phi _{p}(R_{1})}{1-\xi _{1}}, \quad \forall (t,x,y) \in [0,1]\times [r_{1},r_{2}]\times [R_{1},R_{2}]. $$

(H₅):

There exists $R_{1}\leq 0\leq R_{2}$ with $R_{1}^{2}+R_{2}^{2}>0$ such that

$$ f(t,x,R_{1})\leq 0, \qquad f(t,x,R_{2})\geq 0, \quad \forall (t,x)\in [0,1] \times [r_{1},r_{2}], $$

and

$$ \frac{\phi _{p}(R_{1})}{1-\xi _{1}}\leq f(t,x,y)\leq \frac{\phi _{p}(R_{2})}{1-\xi _{1}},\quad \forall (t,x,y)\in [0,1] \times [r_{1},r_{2}]\times [R_{1},R_{2}]. $$

Here, the constants $r_{1}$ and $r_{2}$ in (H₂)–(H₅) are as in (H₁).

2 Preliminaries

Firstly, we briefly review some concepts and results of topological transversality theory. Let U be a convex subset of a Banach space X and $\mathcal{D}\subset U$ be an open set. Denote by $H_{\partial \mathcal{D}}(\overline{\mathcal{D}},U)$ the set of compact operators $F:\overline{\mathcal{D}}\rightarrow U$ that are fixed point free on $\partial \mathcal{D}$. We say that $F\in H_{\partial \mathcal{D}}(\overline{ \mathcal{D}},U)$ is essential if every operator in $H_{\partial \mathcal{D}}(\overline{\mathcal{D}},U)$ that agrees with F on $\partial \mathcal{D}$ has a fixed point in $\mathcal{D}$.

The next two lemmas can be found in [11].

Lemma 2.1

If $q\in \mathcal{D}$ and $F\in H_{\partial \mathcal{D}}(\overline{\mathcal{D}},U)$ is a constant operator, $F(x)=q$ for $x\in \overline{\mathcal{D}}$, then F is essential.

Lemma 2.2

Assume that

• (i) $F\in H_{\partial \mathcal{D}}(\overline{\mathcal{D}},U)$ is essential;

• (ii) $H:\overline{\mathcal{D}}\times [0,1]\rightarrow U$ is a compact homotopy, $H(\cdot ,0)=F$ and $H(x,\lambda )\neq x$ for $x\in \partial \mathcal{D}$ and $\lambda \in [0,1]$.

Then $H(\cdot ,1)$ is essential and therefore it has a fixed point in $\mathcal{D}$.

According to the intermediate value property of continuous functions, it is not difficult to obtain the following.

Lemma 2.3

Assume that

• (i) $x \in C^{1}[0,1]$;

• (ii) $x(1)=\sum_{i=1}^{m-2}a_{i} x(\xi _{i})$, where $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$.

Then there exists $\eta \in [\xi _{1},\xi _{m-2}]$ such that $x(1)=x(\eta )$.

Let the constants $r_{1}$, $r_{2}$, $R_{1}$, $R_{2}$ be such that

$$ r_{1}\leq 0\leq r_{2} \quad (r_{1}\neq r_{2}), \qquad R_{1}\leq 0\leq R_{2} \quad (R_{1} \neq R_{2}). $$

We define the modification g of f as follows:

$$ g(t,x,y) = \textstyle\begin{cases} f(t,r_{2},y)+x-r_{2}, \quad x>r_{2}; \\ f(t,x,y), \quad r_{1} \leq x\leq r_{2}; \\ f(t,r_{1},y)+x-r_{1}, \quad x< r_{1}, \end{cases} $$

and define the modification h of g by setting

$$ h(t,x,y)= \textstyle\begin{cases} g(t,x,R_{2}),\quad y>R_{2}; \\ g(t,x,y), \quad R_{1}\leq y\leq R_{2}; \\ g(t,x,R_{1}), \quad y< R_{1}. \end{cases} $$

Obviously, both $g(t,x,y)$ and $h(t,x,y)$ are continuous on $[0,1]\times \mathbb{R}^{2}$.

We note here that in the following discussion, we agree that when (H₁) is true, the constants $r_{1}$ and $r_{2}$ in function g are the ones in (H₁), while for $i=2,3,4,5$, when (H_i) is true, the constants $R_{1}$ and $R_{2}$ in function h are the ones in (H_i).

Consider the family of the following modified boundary value problem:

$$\begin{aligned}& \bigl(\phi _{p}\bigl(x'(t)\bigr)\bigr)'= \lambda h\bigl(t,x(t),x'(t)\bigr), \quad t\in [0,1], \end{aligned}$$

(2.1)

$$\begin{aligned}& x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}a_{i}x( \xi _{i}), \end{aligned}$$

(2.2)

where $\lambda \in (0,1]$.

The following lemma is a prior estimate of the possible solutions of the modified problem (2.1), (2.2) in $C^{1}[0,1]$.

Lemma 2.4

Assume that (H₁) and (H₂) hold. Let x be a solution of the modified problem (2.1), (2.2) for some $\lambda \in (0,1]$. Then

$$\begin{aligned}& r_{1}\leq x(t)\leq r_{2}, \quad \forall t\in [0,1], \end{aligned}$$

(2.3)

$$\begin{aligned}& R_{1}\leq x'(t)\leq R_{2}, \quad \forall t\in [0,1]. \end{aligned}$$

(2.4)

Proof

We divided the proof into two steps.

Step 1. We prove that (2.3) holds.

Suppose on the contrary that there exists $t_{0}\in [0,1]$ such that $x(t_{0})< r_{1}$ or $x(t_{0})>r_{2}$. Without loss of generality, we may assume that $x(t_{0})< r_{1}$. Let $t_{1}\in [0,1]$ be such that

$$ x(t_{1})=\min_{t\in [0,1]}x(t)< r_{1}. $$

(2.5)

By Lemma 2.3, we can assume that $t_{1}\in [0,1)$, and then from (2.2) we have $x'(t_{1})=0$. It follows from (H₁) and the definition of h that

$$ \begin{aligned} \bigl(\phi _{p}\bigl(x'(t) \bigr)\bigr)'|_{t=t_{1}} &=\lambda h\bigl(t_{1},x(t_{1}),0 \bigr) \\ &=\lambda g\bigl(t_{1},x(t_{1}),0\bigr) \\ &=\lambda \bigl(f(t_{1},r_{1},0)+x(t_{1})-r_{1} \bigr)< 0, \end{aligned} $$

and thus there exists $\delta >0$ such that $\phi _{p}(x'(t))$ is decreasing on $[t_{1},t_{1}+\delta )\subset [0,1)$. This together with the monotonicity of $\phi _{p}^{-1}(\cdot )$ implies that

$$ x'(t)< x'(t_{1})=0,\quad \forall t\in (t_{1},t_{1}+\delta ), $$

which contradicts (2.5). Therefore $x(t)\geq r_{1}$, $t\in [0,1]$. Similarly, we can obtain $x(t)\leq r_{2}$, $t\in [0,1]$. This implies that (2.3) holds.

Step 2. We use the barrier strip technique to show that (2.4) holds.

Let

$$ \begin{aligned} &S_{1}=\bigl\{ t\in [0,1]: R_{1}-1 \leq x'(t)< R_{1}\bigr\} , \\ &S_{2}=\bigl\{ t\in [0,1]: R_{2}< x'(t)\leq R_{2}+1\bigr\} . \end{aligned} $$

(2.6)

We now assert that the sets $S_{1}$ and $S_{2}$ are empty. We only prove that $S_{1}=\emptyset$. Similarly, it can be shown that $S_{2}=\emptyset$. Suppose on the contrary that $S_{1}\neq \emptyset$. Choose $t_{0}\in S_{1}$. Then $R_{1}-1\leq x'(t_{0})< R_{1}$, and so it follows from (2.2) that $0< t_{0}\leq 1$. By the continuity of $x'(t)$ on $[0,1]$ and the fact of $x'(0)=0$, there exist $0< t_{1}< t_{2}\leq t_{0}$ such that

$$ R_{1}-1\leq x'(t_{0})\leq x'(t_{2})< x'(t_{1})< R_{1} $$

(2.7)

and

$$ x'(t_{2})\leq x'(t)\leq x'(t_{1}), \quad \forall t\in [t_{1},t_{2}]. $$

Consequently, $[t_{1},t_{2}]\subset S_{1}$; whereas from (2.3) and $({\mathrm{H}}_{2})$ we obtain

$$ \begin{aligned} \bigl(\phi _{p}\bigl(x'(t) \bigr)\bigr)' &=\lambda h\bigl(t,x(t),x'(t)\bigr) \\ &=\lambda g\bigl(t,x(t),R_{1}\bigr) \\ &=\lambda f\bigl(t,x(t),R_{1}\bigr)\geq 0, \quad \forall t\in [t_{1},t_{2}]. \end{aligned} $$

Hence $\phi _{p}(x'(t))$ is nondecreasing on $[t_{1},t_{2}]$. Furthermore, by the monotonicity of $\phi _{p}^{-1}(\cdot )$, we have that $x'(t)$ is nondecreasing on $[t_{1},t_{2}]$. Consequently,

$$ x'(t_{1})\leq x'(t_{2}), $$

which contradicts (2.7). This implies that $S_{1}=\emptyset$. Notice that $x'(0)=0$ and $x'\in C[0,1]$, it follows that (2.4) holds. This completes the proof of the lemma. □

Lemma 2.5

Assume that (H₁) and $({\mathrm{H}}_{3})$ hold. Let x be a solution of the modified problem (2.1), (2.2) for some $\lambda \in (0,1]$. Then $x(t)$ satisfies (2.3) and (2.4).

Proof

From the proof of Step 1 of Lemma 2.4, it is easy to see that $x(t)$ satisfies (2.3). We now show that (2.4) holds. To do this, we let $S_{i}$ ($i=1,2$) be as in (2.6). Then $S_{1}=\emptyset$ by the proof of Step 2 of Lemma 2.4. Therefore, it follows from (2.2) that

$$ x'(t)\geq R_{1},\quad \forall t\in [0,1]. $$

(2.8)

We now prove that $S_{2}=\emptyset$. Indeed, by contradiction, we assume that there exists $t_{0}\in S_{2}$ such that $R_{2}< x'(t_{0})\leq R_{2}+1$. According to (2.2), we have $0< t_{0}\leq 1$. In view of Lemma 2.3 and the mean value theorem of differentials, there exists $\zeta \in (\eta ,1)\subset [\xi _{1},1]$ such that $x'(\zeta )=0$. Integrating (2.1) from ζ to 1, applying (2.3) and $({\mathrm{H}}_{3})$, we obtain

$$ \begin{aligned} \phi _{p}\bigl(x'(1)\bigr) & = \lambda \int _{\zeta}^{1} h\bigl(t,x(t),x'(t) \bigr){\,\mathrm{d}}t \\ & = \lambda \int _{\zeta}^{1} g\bigl(t,x(t),\theta \bigl(x'(t)\bigr)\bigr){\,\mathrm{d}}t \\ & = \lambda \int _{\zeta }^{1} f\bigl(t,x(t),\theta \bigl(x'(t)\bigr)\bigr){\,\mathrm{d}}t \\ &\leq \lambda \int _{\zeta}^{1} \frac{\phi _{p}(R_{2})}{1-\xi _{1}}{ \,\mathrm{d}}t \\ & < \phi _{p}(R_{2}), \end{aligned} $$

where $\theta (y):=\max \{R_{1},\min \{y,R_{2}\}\}$ for $y\in \mathbb{R}$. Consequently, $x'(1)< R_{2}$, and so $0< t_{0}<1$. From the continuity of $x'(t)$ on $[0,1]$, there exist $t_{0}\leq t_{1}< t_{2}<1$ such that

$$ R_{2}< x'(t_{2})< x'(t_{1}) \leq x'(t_{0})\leq R_{2}+1, $$

(2.9)

and

$$ x'(t_{2})\leq x'(t)\leq x'(t_{1}), \quad \forall t\in [t_{1},t_{2}]. $$

Consequently, $[t_{1},t_{2}]\subset S_{2}$. It follows from the definition of h, (2.3), and $({\mathrm{H}}_{3})$ that

$$ \begin{aligned} \bigl(\phi _{p}\bigl(x'(t) \bigr)\bigr)' & = \lambda h\bigl(t,x(t),x'(t)\bigr) \\ & = \lambda g\bigl(t,x(t),R_{2}\bigr) \\ & = \lambda f\bigl(t,x(t),R_{2}\bigr)\geq 0, \quad \forall t\in [t_{1},t_{2}]. \end{aligned} $$

This implies that

$$ \phi _{p}\bigl(x'(t_{1})\bigr)\leq \phi _{p}\bigl(x'(t_{2})\bigr), $$

and thus, by the monotonicity of $\phi _{p}^{-1}(\cdot )$, we obtain

$$ x'(t_{1})\leq x'(t_{2}), $$

which contradicts (2.9). This shows that $S_{2}=\emptyset$. Notice that $x'(0)=0$ and $x'\in C[0,1]$, we have

$$ x'(t)\leq R_{2}, \quad \forall t\in [0,1]. $$

This together with (2.8) implies that (2.4) holds. This completes the proof of the lemma. □

Similar to the proof of Lemma 2.5, we can easily show the following two results.

Lemma 2.6

Assume that (H₁) and (H₄) hold. Let x be a solution of the modified problem (2.1), (2.2) for some $\lambda \in (0,1]$. Then $x(t)$ satisfies (2.3) and (2.4).

Lemma 2.7

Assume that (H₁) and (H₅) hold. Let x be a solution of the modified problem (2.1), (2.2) for some $\lambda \in (0,1]$. Then $x(t)$ satisfies (2.3) and (2.4).

Now, we denote by $X=C^{1}[0,1]\times \mathbb{R}$ the Banach space equipped with the norm

$$ \bigl\Vert (x,r) \bigr\Vert = \Vert x \Vert _{\infty}+ \bigl\Vert x' \bigr\Vert _{\infty}+ \vert r \vert , \quad (x,r)\in X. $$

Set

$$ U=\bigl\{ (x,r)\in X: x(0)=0, r\in \mathbb{R}\bigr\} $$

and

$$ \mathcal{D}= \left \{(x,r)\in U\left | \begin{gathered} r_{1}-r_{2}-1< x(t)< r_{2}-r_{1}+1 \text{ on } [0,1], \\ R_{1}-1< x'(t)< R_{2}+1 \text{ on } [0,1], \\ r_{1}-1< r< r_{2}+1 \end{gathered} \right . \right \}. $$

Obviously, U is a convex subset of X and $\mathcal{D}$ is an open subset of U.

Lemma 2.8

Let the operator $H:\overline{\mathcal{D}}\times [0,1]\rightarrow U$ be defined by

$$ H(x,r,\lambda )= \Biggl(0,\lambda r-\lambda \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1} \phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \Biggr). $$

(2.10)

Then $H(x,r,\lambda )$ is a compact operator.

Proof

Notice that to prove $H(x,r,\lambda )$ is a compact operator, it is sufficient to show the second component of $H(x,r,\lambda )$, that is,

$$ H_{2}(x,r,\lambda ):=\lambda r-\lambda \sum _{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1} \phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s $$

is compact since the first component $H_{1}(x,r,\lambda )\equiv 0$ of $H(x,r,\lambda )$ is compact.

We now divided the proof into two steps.

Step 1. We prove that $H_{2}(x,r,\lambda )$ is continuous. To do this, we let $\{(x_{n},r_{n},\lambda _{n})\}_{n=1}^{\infty} \subset \overline{\mathcal{D}}\times [0,1]$ be such that $(x_{n},r_{n},\lambda _{n})\to (x_{0},r_{0},\lambda _{0})\in \overline{\mathcal{D}}\times [0,1](n\to \infty )$. Then

$$ \Vert x_{n}-x_{0} \Vert _{\infty}\to 0, \qquad \bigl\Vert x'_{n}-x'_{0} \bigr\Vert _{\infty}\to 0,\qquad \vert r_{n}-r_{0} \vert \to 0,\qquad \vert \lambda _{n}-\lambda _{0} \vert \to 0,\quad n \to \infty . $$

From the continuity of h and $\phi _{p}^{-1}$, we have

$$ \begin{aligned} H_{2}(x_{n},r_{n}, \lambda _{n})& = \lambda _{n} r_{n}-\lambda _{n} \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{n}(\tau )+r_{n},x'_{n}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){ \,\mathrm{d}}s \\ &\to \lambda _{0} r_{0}-\lambda _{0}\sum _{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau )+r_{0},x'_{0}( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \\ & = H_{2}(x_{0},r_{0},\lambda _{0})\quad (n\to \infty ). \end{aligned} $$

Hence $H_{2}(x,r,\lambda )$ is continuous on $\overline{\mathcal{D}}\times [0,1]$.

Step 2. We show that $H_{2}(\overline{\mathcal{D}}\times [0,1])$ is a relatively compact set in $\mathbb{R}$. Notice that for all $(x,r,\lambda )\in \overline{\mathcal{D}}\times [0,1]$, we have

$$ \begin{aligned} \bigl\vert H_{2}(x,r,\lambda ) \bigr\vert = {}& \Biggl\vert \lambda r-\lambda \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \Biggr\vert \\ \leq {}& \vert r \vert +\sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{1} \bigl\vert h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr) \bigr\vert {\,\mathrm{d}}\tau \biggr){ \,\mathrm{d}}s \\ \leq {}& \max \{-r_{1}+1,r_{2}+1\}+\phi _{p}^{-1}(M)=:C>0, \end{aligned} $$

where

$$ M=\max \left\{ \bigl\vert h(t,x,y) \bigr\vert \left| \begin{gathered} 0\leq t \leq 1, 2r_{1}-r_{2}-2\leq x\leq 2r_{2}-r_{1}+2, \\ R_{1}-1\leq y\leq R_{2}+1 \end{gathered} \right . \right\}. $$

(2.11)

Thus, $H_{2}(\overline{\mathcal{D}}\times [0,1])$ is a relatively compact set in $\mathbb{R}$. This completes the proof of the lemma. □

Lemma 2.9

Assume that (H₁) holds. Let the operator $F:\overline{\mathcal{D}}\rightarrow U$ be defined by

$$ F(x,r)= \Biggl(0,r-\sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1} \phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr){\,\mathrm{d}} \tau \biggr){\,\mathrm{d}}s \Biggr). $$

Then F is essential.

Proof

It follows from (2.10) that $H(\cdot ,\cdot ,1)=F(\cdot ,\cdot )$, $H(x,r,0)=(0,0)\in \mathcal{D}$ for $(x,r)\in \overline{\mathcal{D}}$, and thus from Lemma 2.1 we know that $H(x,r,0)$ is essential. Besides, $H(x,r,\lambda )$ is compact by Lemma 2.8.

We now show that

$$ H(x,r,\lambda )\neq (x,r),\quad \forall (x,r)\in \partial \mathcal{D}, \lambda \in [0,1]. $$

(2.12)

Obviously, $H(x,r,0)\neq (x,r)$ for all $(x,r)\in \partial \mathcal{D}$. Suppose that $H(x_{0},r_{0},\lambda _{0})=(x_{0},r_{0})$ for some $(x_{0},r_{0})\in \partial \mathcal{D}$ and $\lambda _{0}\in (0,1]$. Then $x_{0}=0$ and

$$ \lambda _{0}r_{0}-\lambda _{0}\sum _{i=1}^{m-2}a_{i} \int _{ \xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau )+r_{0},x'_{0}( \tau )\bigr)\,\text{d}\tau \biggr)\,\text{d}s=r_{0}. $$

Consequently,

$$ \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h(\tau ,r_{0},0)\,\text{d} \tau \biggr)\,\text{d}s=r_{0} \biggl(1- \frac{1}{\lambda _{0}} \biggr). $$

If $r_{0}>r_{2}$, from (H₁) we have

$$ h(\tau ,r_{0},0)=g(\tau ,r_{0},0)=f(\tau ,r_{2},0)+r_{0}-r_{2}>0,\quad \forall \tau \in [0,1]. $$

The monotonicity of $\phi _{p}^{-1}(\cdot )$ and $\phi _{p}^{-1}(0)=0$ imply that

$$ \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h(\tau ,r_{0},0)\,\text{d} \tau \biggr) \,\text{d}s>\sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1}(0) \,\text{d}s=0, $$

which contradicts $r_{0}(1-1/\lambda _{0})\leq 0$. Hence, $r_{0}\leq r_{2}$. Similarly, we can obtain $r_{0}\geq r_{1}$. Consequently, $(0,r_{0})\in \mathcal{D}$, which contradicts $(0,r_{0})\in \partial \mathcal{D}$. This shows that (2.12) holds. Therefore, according to Lemma 2.2, $F(\cdot ,\cdot )=H(\cdot ,\cdot ,1)$ is essential. This completes the proof of the lemma. □

Lemma 2.10

Let the operator $G:\overline{\mathcal{D}}\times [0,1]\rightarrow U$ be defined by

$$G(x,r,\lambda )={\left(\begin{array}{c}{\int }_0^t{\varphi }_p^{-1}\left(\lambda {\int }_0^{s}h(\tau ,x(\tau )+r,x^{\prime }(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\\ r-{\sum }_{i=1}^{m-2}{a}_i{\int }_{{\xi }_i}^1{\varphi }_p^{-1}\left({\int }_0^{s}h(\tau ,x(\tau )+r,x^{\prime }(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\end{array}\right)}^{\ast },$$

where the symbol “∗” denotes the transpose of vector. Then G is a compact operator.

Proof

Firstly, we define operators $G_{1}:\overline{\mathcal{D}}\times [0,1]\rightarrow C^{1}[0,1]$ and $G_{2}:\overline{\mathcal{D}}\times [0,1]\rightarrow \mathbb{R}$ by

$$ G_{1}(x,r,\lambda )= \int _{0}^{t}\phi _{p}^{-1} \biggl(\lambda \int _{0}^{s} h\bigl( \tau ,x(\tau )+r,x'(\tau )\bigr)\,\text{d}\tau \biggr)\,\text{d}s $$

and

$$ G_{2}(x,r,\lambda )=r-\sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1} \phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr)\,\text{d} \tau \biggr)\,\text{d}s. $$

Next, we divided the proof into three steps.

Step 1. We prove that $G_{1}:\overline{\mathcal{D}}\times [0,1]\rightarrow C^{1}[0,1]$ is continuous.

Let $\{(x_{n},r_{n},\lambda _{n})\}_{n=1}^{\infty}\subset \overline{\mathcal{D}}\times [0,1]$ be such that $(x_{n},r_{n},\lambda _{n})\to (x_{0},r_{0},\lambda _{0})\in \overline{\mathcal{D}}\times [0,1](n\to \infty )$. Then

$$\begin{aligned}& \Vert x_{n}-x_{0} \Vert _{\infty}\to 0, \qquad \bigl\Vert x'_{n}-x'_{0} \bigr\Vert _{\infty}\to 0, \\& \vert r_{n}-r_{0} \vert \to 0,\qquad \vert \lambda _{n}-\lambda _{0} \vert \to 0\quad (n\to \infty ). \end{aligned}$$

The continuity of h and $\phi _{p}^{-1}$ imply that

$$ \phi _{p}^{-1}\biggl(\lambda _{n} \int _{0}^{t} h\bigl(\tau ,x_{n}(\tau )+r_{n},x'_{n}( \tau )\bigr){\,\mathrm{d}}\tau \biggr) \to \phi _{p}^{-1}\biggl(\lambda _{0} \int _{0}^{t} h\bigl( \tau ,x_{0}(\tau )+r_{0},x'_{0}(\tau )\bigr){\,\mathrm{d}}\tau \biggr) $$

uniformly on $[0,1]$ as $n\to \infty$, and

$$\begin{aligned}& \int _{0}^{t}\phi _{p}^{-1} \biggl(\lambda _{n} \int _{0}^{s} h\bigl(\tau ,x_{n}( \tau ) +r_{n},x'_{n}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \\& \quad \to \int _{0}^{t} \phi _{p}^{-1} \biggl(\lambda _{0} \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau ) +r_{0},x'_{0}( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \end{aligned}$$

uniformly on $[0,1]$ as $n\to \infty$. That is,

$$ \biggl\Vert \frac{\mathrm{d}}{{\mathrm{d}}t}G_{1}(x_{n},r_{n}, \lambda _{n})- \frac{\mathrm{d}}{{\mathrm{d}}t}G_{1}(x_{0},r_{0}, \lambda _{0}) \biggr\Vert _{ \infty}\to 0\quad (n\to \infty ) $$

and

$$ \bigl\Vert G_{1}(x_{n},r_{n},\lambda _{n})-G_{1}(x_{0},r_{0},\lambda _{0}) \bigr\Vert _{\infty} \to 0 \quad (n\to \infty ). $$

This shows that $G_{1}:\overline{\mathcal{D}}\times [0,1]\rightarrow C^{1}[0,1]$ is continuous.

Step 2. We show that the set $G_{1}(\overline{\mathcal{D}}\times [0,1])$ is relatively compact in $C^{1}[0,1]$. Obviously, for all $(x,r)\in \overline{\mathcal{D}}$, $\lambda \in [0,1]$, we have

$$\begin{aligned}& \bigl\vert G_{1}(x,r,\lambda ) \bigr\vert = \biggl\vert \int _{0}^{t}\phi _{p}^{-1} \biggl(\lambda \int _{0}^{s} h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr) {\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \biggr\vert \leq \phi _{p}^{-1}(M),\\& \biggl\vert \frac{\mathrm{d}}{{\mathrm{d}}t}G_{1}(x,r,\lambda ) \biggr\vert = \biggl\vert \phi _{p}^{-1}\biggl( \lambda \int _{0}^{t} h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr){\,\mathrm{d}}\tau \biggr) \biggr\vert \leq \phi _{p}^{-1}(M), \end{aligned}$$

where M is defined by (2.11). Since $\phi _{p}^{-1}(\cdot )$ is uniformly continuous on $[-M,M]$, for any $\varepsilon >0$, there exists $\mu >0$ such that when $|s_{2}-s_{1}|<\mu$ ($\forall s_{1},s_{2}\in [-M,M]$),

$$ \bigl\vert \phi _{p}^{-1}(s_{2})-\phi _{p}^{-1}(s_{1}) \bigr\vert < \varepsilon . $$

Now, we choose $\delta =\mu /M$. Then, for all $t_{1},t_{2}\in [0,1]$ with $|t_{2}-t_{1}|<\delta$ and for all $(x,r)\in \overline{\mathcal{D}}$, we have

$$ \biggl\vert \int _{t_{1}}^{t_{2}}h\bigl(\tau ,x(\tau )+r,x'(\tau )\bigr){\,\mathrm{d}}\tau \biggr\vert \leq M \vert t_{2}-t_{1} \vert < \mu . $$

Thus, when $|t_{2}-t_{1}|<\delta$ ($t_{1},t_{2}\in [0,1]$), we have for all $(x,r,\lambda ) \in \overline{\mathcal{D}}\times [0,1]$,

$$ \biggl\vert \phi _{p}^{-1}\biggl(\lambda \int _{0}^{t_{2}}h\bigl(\tau ,x(\tau )+r,x'( \tau )\bigr)\,\text{d}\tau \biggr) -\phi _{p}^{-1} \biggl(\lambda \int _{0}^{t_{1}}h\bigl(\tau ,x( \tau )+r,x'(\tau )\bigr)\,\text{d}\tau \biggr) \biggr\vert < \varepsilon . $$

This shows that $\{\frac{\mathrm{d}}{{\mathrm{d}}t}G_{1} (x,r,\lambda ): (x,r,\lambda ) \in \overline{\mathcal{D}} \times [0,1] \}$ is equicontinuous on $[0,1]$. Therefore, $G_{1}(\overline{\mathcal{D}}\times [0,1])$ is a relatively compact set in $C^{1}[0,1]$ by Arzelà-Ascoli theorem.

Step 3. We prove that $G_{2}:\overline{\mathcal{D}}\times [0,1]\rightarrow \mathbb{R}$ is a compact operator.

The compactness of the operator $G_{2}$ is clear, the proof is similar to the one of $H_{2}$ in Lemma 2.8.

In summary, $G:\overline{\mathcal{D}}\times [0,1]\rightarrow U$ is a compact operator. This completes the proof of the lemma. □

3 Main results

With the preparatory work in Sect. 2, we can now establish the existence results of solutions of problem (1.1).

Theorem 3.1

Assume that (H₁) and (H₂) hold. Then boundary value problem (1.1) has at least one solution $x=x(t)$ satisfying (2.3) and (2.4).

Proof

We note that to obtain the existence of solutions satisfying (2.3) and (2.4) of problem (1.1), it is sufficient to prove that the modified problem (2.1), (2.2) has a solution $x=x(t)$ satisfying (2.3) and (2.4).

Below, we will prove in two steps.

Step 1. We prove that if the operator $G(\cdot ,\cdot ,1)$, which is defined in Lemma 2.10, has a fixed point, then the modified problem (2.1), (2.2) has a solution satisfying (2.3) and (2.4).

Suppose that $(x_{1},r_{1})$ is a fixed point of $G(\cdot ,\cdot ,1)$. It follows from the definition of the operator G that

$$ x_{1}(t)= \int _{0}^{t}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{1}( \tau )+r_{1},x'_{1}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s, \quad \forall t\in [0,1] $$

and

$$ \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{1}(\tau )+r_{1},x'_{1}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s=0. $$

Furthermore, we have

$$ x'_{1}(t)=\phi _{p}^{-1}\biggl( \int _{0}^{t} h\bigl(\tau ,x_{1}(\tau )+r_{1},x'_{1}( \tau )\bigr){\,\mathrm{d}}\tau \biggr),\quad \forall t\in [0,1], $$

and thus

$$ x_{1}'(0)=0,\qquad x_{1}(1)=\sum _{i=1}^{m-2}a_{i}x_{1}(\xi _{i}). $$

Setting $x_{2}(t)=x_{1}(t)+r_{1}$ for $t\in [0,1]$, it is easy to see that $x_{2}$ is a solution of the modified problem (2.1), (2.2) with $\lambda =1$, and the validity of (2.3) and (2.4) follows from Lemma 2.4.

Step 2. We show that the operator $G(\cdot ,\cdot ,1)$ has a fixed point.

Notice that $G(\cdot ,\cdot ,0)=F(\cdot ,\cdot )$, F is essential by Lemma 2.9, and G is a compact operator by Lemma 2.10. For the existence of a fixed point of $G(\cdot ,\cdot ,1)$, it is sufficient to verify that

$$ G(x,r,\lambda )\neq (x,r), \quad \forall (x,r)\in \partial \mathcal{D}, \lambda \in [0,1]. $$

Suppose on the contrary that $G(x_{0},r_{0},\lambda _{0})=(x_{0},r_{0})$ for some $(x_{0},r_{0})\in \partial \mathcal{D}$ and $\lambda _{0}\in [0,1]$. If $\lambda _{0}=0$, then from the proof of Lemma 2.10, we have $(x_{0},r_{0})\notin \partial \mathcal{D}$, which contradicts $(x_{0},r_{0})\in \partial \mathcal{D}$. If $\lambda _{0}\in (0,1]$, then we have

$$ x_{0}(t)= \int _{0}^{t}\phi _{p}^{-1} \biggl(\lambda _{0} \int _{0}^{s} h\bigl( \tau ,x_{0}(\tau ) +r_{0},x'_{0}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s,\quad \forall t\in [0,1] $$

and

$$ \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau )+r_{0},x'_{0}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){ \,\mathrm{d}}s=0. $$

(3.1)

Hence

$$ x'_{0}(t)=\phi _{p}^{-1}\biggl( \lambda _{0} \int _{0}^{t} h\bigl(\tau ,x_{0}( \tau ) +r_{0},x'_{0}(t)\bigr){\,\mathrm{d}}\tau \biggr),\quad \forall t\in [0,1]. $$

In particular,

$$ x'_{0}(0)=0. $$

Notice that $\phi _{p}^{-1}(s_{1}s_{2})=\phi _{p}^{-1}(s_{1})\phi _{p}^{-1}(s_{2})$ for all $s_{1},s_{2}\in \mathbb{R}$ and (3.1), we have

$$ \begin{aligned} x_{0}(1)-\sum _{i=1}^{m-2}{a_{i}}x_{0}(\xi _{i}) = {}& \sum_{i=1}^{m-2}a_{i} \int _{\xi _{i}}^{1}\phi _{p}^{-1} \biggl(\lambda _{0} \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau )+r_{0},x'_{0}(\tau )\bigr){\,\mathrm{d}}\tau \biggr){ \,\mathrm{d}}s \\ = {}& \phi _{p}^{-1}(\lambda _{0})\sum _{i=1}^{m-2}a_{i} \int _{ \xi _{i}}^{1}\phi _{p}^{-1} \biggl( \int _{0}^{s} h\bigl(\tau ,x_{0}(\tau )+r_{0},x'_{0}( \tau )\bigr){\,\mathrm{d}}\tau \biggr){\,\mathrm{d}}s \\ = {}& 0, \end{aligned} $$

that is,

$$ x_{0}(1)=\sum_{i=1}^{m-2}a_{i}x_{0}( \xi _{i}). $$

Setting $x(t)=x_{0}(t)+r_{0}$ for $t\in [0,1]$. Then we can see that x is a solution of the modified problem (2.1), (2.2) with $\lambda =\lambda _{0}$. It follows from Lemma 2.4 that

$$ r_{1}\leq x(t)=x_{0}(t)+r_{0}\leq r_{2}, \quad \forall t\in [0,1] $$

(3.2)

and

$$ R_{1}-1< R_{1}\leq x'(t)=x'_{0}(t) \leq R_{2}< R_{2}+1, \quad \forall t \in [0,1]. $$

Notice that $x_{0}(0)=0$, and so (3.2) yields $r_{1}-1< r_{1}\leq r_{0}\leq r_{2}< r_{2}+1$. Thus

$$ r_{1}-r_{2}-1< r_{1}-r_{0}\leq x_{0}(t)\leq r_{2}-r_{0}< r_{2}-r_{1}+1,\quad \forall t\in [0,1]. $$

Consequently, $(x_{0},r_{0})\in \mathcal{D}$, which contradicts $(x_{0},r_{0})\in \partial \mathcal{D}$.

In summary, the conclusion of Theorem 3.1 holds. This completes the proof of the theorem. □

The following conclusion can be obtained by Theorem 3.1 immediately.

Corollary 3.1

Assume that all the conditions of Theorem 3.1with $r_{1}=0$ hold. Then problem (1.1) has at least one nonnegative solution $x=x(t)$ satisfying (2.3) and (2.4). Furthermore, the solution $x=x(t)$ is nondecreasing on $[0,1]$ provided $R_{1}=0$.

The following theorems can be proved by using similar arguments to those of Theorem 3.1.

Theorem 3.2

Assume that $({\mathrm{H}}_{1})$ and $({\mathrm{H}}_{3})$ hold. Then the boundary value problem (1.1) has at least one solution $x=x(t)$ satisfying (2.3) and (2.4).

Corollary 3.2

Assume that all the conditions of Theorem 3.2with $r_{1}=0$ hold. Then problem (1.1) has at least one nonnegative solution $x=x(t)$ satisfying (2.3) and (2.4). Furthermore, the solution $x=x(t)$ is nondecreasing on $[0,1]$ provided $R_{1}=0$.

Theorem 3.3

Assume that $({\mathrm{H}}_{1})$ and $({\mathrm{H}}_{4})$ hold. Then the boundary value problem (1.1) has at least one solution $x=x(t)$ satisfying (2.3) and (2.4).

Corollary 3.3

Assume that all the conditions of Theorem 3.3with $r_{1}=0$ hold. Then problem (1.1) has at least one nonnegative solution $x=x(t)$ satisfying (2.3) and (2.4). Furthermore, the solution $x=x(t)$ is nondecreasing on $[0,1]$ provided $R_{1}=0$.

Theorem 3.4

Assume that $({\mathrm{H}}_{1})$ and $({\mathrm{H}}_{5})$ hold. Then the boundary value problem (1.1) has at least one solution $x=x(t)$ satisfying (2.3) and (2.4).

Remark 3.1

Theorem 3.4 takes Theorem 2.1 of [19, 20] as a special case.

Corollary 3.4

Assume that all the conditions of Theorem 3.4with $r_{1}=0$ hold. Then problem (1.1) has at least one nonnegative solution $x=x(t)$ satisfying (2.3) and (2.4). Furthermore, the solution $x=x(t)$ is nondecreasing on $[0,1]$ provided $R_{1}=0$.

Remark 3.2

If the $\phi _{p}$ in problem (1.1) is replaced by ϕ, which is an increasing homeomorphism from $\mathbb{R}$ onto $\mathbb{R}$ and satisfies $\phi (s_{1}s_{2})=\phi (s_{1})\phi (s_{2})$ for all $s_{1},s_{2}\in \mathbb{R}$, Theorems $3.1-3.4$ and Corollaries $3.1-3.4$ are still true.

4 Some examples

As applications of our results, this section will provide four illustrative examples.

Example 4.1

Consider the m-point boundary value problem with p-Laplacian at resonance

$$ \textstyle\begin{cases} (\phi _{p}(x'(t)))'=\sum_{i=0}^{n} c_{i}(x(t))^{i}+P_{l}(x'(t)), \quad t\in [0,1], \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}a_{i} x(\xi _{i}), \end{cases} $$

(4.1)

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$; n is an even number, $c_{i}\in \mathbb{R}$ ($i=0,1,\ldots ,n$); the polynomial $\sum_{i=0}^{n} c_{i} x^{i}$ has at least two distinct real roots with different signs; $P_{l}(y)$ is a polynomial of degree $l\in \mathbb{N}$, $P_{l}(0)=0$; and $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$.

Let

$$ f(t,x,y)=\sum_{i=0}^{n} c_{i}x^{i}+P_{l}(y), \quad (t,x,y) \in [0,1]\times{ \mathbb{R}^{2}}. $$

Obviously, $f\in C([0,1]\times \mathbb{R}^{2})$. Let $r_{1}$ and $r_{2}$ be the minimum and maximum real root of the polynomial $\sum_{i=0}^{n} c_{i}x^{i}$, respectively. Then from the assumption it follows that $r_{1}\leq 0\leq r_{2}$, $r_{1}^{2}+r_{2}^{2}>0$ and

$$ f(t,r_{1},0)=0,\qquad f(t,r_{2},0)=0,\quad \forall t\in [0,1]. $$

Hence, condition $({\mathrm{H}}_{1})$ of Theorem 3.1 is satisfied.

Let

$$ F=\mathop{\max} _{{r_{1}}\leq x\leq r_{2}} \Biggl\vert \sum_{i=0}^{n} c_{i}x^{i} \Biggr\vert . $$

Then, for $x\in [r_{1},r_{2}]$, $y\in \mathbb{R}$, we have

$$ P_{l}(y)-F\leq f(t,x,y)\leq P_{l}(y)+F. $$

Thus, if there exists $y_{1}<0$ such that $P_{l}(y_{1})-F>0$ and there exists $y_{2}>0$ such that $P_{l}(y_{2})+F<0$, then condition $({\mathrm{H}}_{2})$ of Theorem 3.1 is satisfied.

In summary, problem (4.1) has at least one solution $x=x(t)$ provided

$$ \inf_{y>0}P_{l}(y)+F< 0< \sup_{y< 0}P_{l}(y)-F. $$

Note that if l is an odd number and the coefficient of $y^{l}$ in $P_{l}(y)$ is negative, then the above inequality holds.

Example 4.2

Consider the m-point boundary value problem with p-Laplacian at resonance

$$ \textstyle\begin{cases} (\phi _{p}(x'(t)))'=x^{2}(t)-\frac{1}{4}- \frac{1}{2^{n}}(x'(t))^{n}\cos (\frac{\pi}{2}x'(t)), \quad t\in [0,1], \hfill \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}a_{i}x(\xi _{i}), \end{cases} $$

(4.2)

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, n is an even number, $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, and $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$.

Let

$$ f(t,x,y)=x^{2}-\frac{1}{4}-\frac{1}{2^{n}}y^{n} \cos{\biggl(\frac{\pi}{2}y\biggr)}, \quad (t,x,y)\in [0,1]\times{ \mathbb{R}^{2}}. $$

Obviously, $f\in C([0,1]\times{\mathbb{R}^{2}})$. We choose $r_{1}=-1/2$, $r_{2}=1/2$, $R_{1}=-2$, $R_{2}=2$. Then

$$ f(t,r_{1},0)=f(t,r_{2},0)=0,\quad \forall t\in [0,1]. $$

This implies that (H₁) holds. On the other hand, we have

$$ f(t,x,R_{1})=f(t,x,R_{2})>0,\quad \forall (t,x)\in [0,1]\times [r_{1},r_{2}]. $$

Moreover, we have

$$ f(t,x,y)\leq 1< \frac{\phi _{p}(R_{2})}{1-\xi _{1}}, \quad \forall (t,x,y) \in [0,1]\times \biggl[- \frac{1}{2},\frac{1}{2}\biggr]\times [-2,2]. $$

Hence, (H₃) holds. From Theorem 3.2, problem (4.2) has at least one solution.

Example 4.3

Consider the m-point boundary value problem with p-Laplacian at resonance

$$ \textstyle\begin{cases} (\phi _{p}(x'(t)))'=x^{2}(t)-\frac{1}{4}+(x'(t))^{3} \sin (\frac{\pi}{2}x'(t)), \quad t \in [0,1], \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}{{a_{i}}x({\xi _{i}})}, \end{cases} $$

(4.3)

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, and $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$.

Let

$$ f(t,x,y)=x^{2}-{\frac{1}{4}}+y^{3}\sin \biggl( \frac{\pi}{2}y \biggr),\quad (t,x,y)\in [0,1]\times{\mathbb{R}^{2}}. $$

Clearly, $f\in C([0,1]\times{\mathbb{R}^{2}})$. We select $r_{1}=-1/2$, $r_{2}=1/2$, $R_{1}=-2$, $R_{2}=2$. Then

$$ f(t,r_{1},0)=f(t,r_{2},0)=0, \quad \forall t\in [0,1]. $$

This implies that (H₁) holds. On the other hand, we have

$$ f(t,x,R_{1})=f(t,x,R_{2})\leq 0,\quad \forall t\in [0,1]\times [r_{1},r_{2}] $$

and

$$ f(t,x,y)>-1>\frac{\phi _{p}(R_{1})}{1-\xi _{1}}, \quad \forall (t,x,y) \in [0,1]\times \biggl[- \frac{1}{2},\frac{1}{2}\biggr]\times [-2,2]. $$

Hence, (H₄) is also true. From Theorem 3.3, problem (4.3) has at least one solution.

Example 4.4

Consider the m-point boundary value problem with p-Laplacian at resonance

$$ \textstyle\begin{cases} (\phi _{p}(x'(t)))'=\frac{1}{4}-\frac{1}{2}x(t)\cos (2 \pi x(t))+\frac{1}{2^{n+1}}(x'(t))^{n} \sin (\frac{\pi}{4}x'(t)), \quad t \in [0,1], \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}{{a_{i}}x({\xi _{i}})}, \end{cases} $$

(4.4)

where $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, n is an even number, $a_{i}>0$ ($i=1,2,\ldots ,m-2$) with $\sum_{i=1}^{m-2}a_{i}=1$, and $0<\xi _{1}<\xi _{2}<\cdots <\xi _{m-2}<1$.

Let

$$ f(t,x,y)=\frac{1}{4}-\frac{1}{2}x\cos (2\pi x)+ \frac{1}{{{2^{n+1}}}}{y^{n}} \sin \biggl(\frac{\pi}{4}y\biggr),\quad (t,x,y)\in [0,1] \times {\mathbb{R}^{2}}. $$

Evidently, $f\in C([0,1]\times {\mathbb{R}^{2}})$. We choose $r_{1}=-1/2$, $r_{2}=1/2$, $R_{1}=-2$, $R_{2}=2$. Then, for all $t\in [0,1]$, we have

$$ f(t,r_{1},0)=0,\qquad f(t,r_{2},0)>0. $$

This implies that (H₁) holds. On the other hand, for all $(t,x)\in [0,1]\times [r_{1},r_{2}]$, it follows that

$$ f(t,x,R_{1})\leq 0,\qquad f(t,x,R_{2})>0. $$

Moreover, when $(t,x,y)\in [0,1]\times [-1/2,1/2]\times [-2,2]$, we have

$$ \frac{\phi _{p}(R_{1})}{1-\xi _{1}}< f(t,x,y)< \frac{\phi _{p}(R_{2})}{1-\xi _{1}}. $$

This shows that (H₅) is satisfied. From Theorem 3.4, problem (4.4) has at least one solution.