ABSTRACT

In this paper, we prove existence and multiplicities of solutions for asymptotically linear ordinary differential equations satisfying Sturm-Liouville boundary value conditions with resonance. Adding assumption H₃ that is similar to (LL) in Theorem 1.1, by index theory and Morse theory, we obtain more nontrivial solutions.

Keywords:

Asymptotically Linear Ordinary Differential Equations with Resonance, Multiple Solutions, Sturm-Liouville Boundary Value Problem, Index Theory, Morse Theory

1. Introduction

In this paper, we investigate the nontrivial solutions of asymptotically linear ordinary differential equations satisfying Sturm-Liouville BVPs with resonance. Various boundary value problems of asymptotically linear ordinary differential equations have been studied before. Most of them are gotten by the topological degree theory. There are also some papers about resonant problem. But the asymptotically linear ordinary differential equations with resonance aren’t concerned ago. Here, we concern asymptotically linear ordinary differential equations satisfying both Sturm-Liouville boundary value and resonance. We solve the problem to get Theorem 1.1 in the following section.

Now, we consider solutions of the following Sturm-Liouville boundary value problem:

$x^{″}\left(t\right)+f\left(t\mathrm{,}x\left(t\right)\right)=\mathrm{0,}$ (1.1)

$x\left(0\right)\cos \alpha -x^{\prime }\left(0\right)\sin \alpha =\mathrm{0,}$ (1.2)

$x\left(1\right)\cos \beta -x^{\prime }\left(1\right)\sin \beta =\mathrm{0,}$ (1.3)

where $f\in C^1\left(\left[0,1\right]\times R,R\right),0\le \alpha <\text{π},0<\beta \le \text{π}$. In this paper, $f^{\prime }$ denotes the derivative with respect to x. Our main result is the following theorem.

Theorem 1.1 Assume that $a_0\in H_0\left(\mathrm{1,}\left(\mathrm{0,1}\right)\mathrm{,}\frac{\text{π}}{2}\mathrm{,}\frac{\text{π}}{2}\right)$ in [1], i.e. $x^{″}+a_0\left(t\right)x=0$

and (1.2)-(1.3) has one nontrivial solution $x=x_0\left(t\right)>0$,$\forall t\in \left[\mathrm{0,1}\right]$,$\lambda \in L^{\infty }\left[\mathrm{0,1}\right]$. f satisfies the following two conditions:

(H₁) $a_0\left(t\right)\le \frac{f\left(t,x\right)}{x}<\lambda \left(t\right)$ as $\left|x\right|\ge r>0$, r is a constant, $i\left(\lambda \right)=1$,$v\left(\lambda \right)=0$,$i\left(a_0\right)=0$,$v\left(a_0\right)=1$ ;

(H₂) $f\left(t\mathrm{,0}\right)\equiv 0$,$1\bar{\in }\left[i\left({\lambda }_0\right)\mathrm{,}i\left({\lambda }_0\right)+v\left({\lambda }_0\right)\right]$, where ${\lambda }_0\left(t\right):=f^{\prime }\left(t,0\right)$ ;

For the sake of convenience, we denote $f_1\left(t,x\right)=f\left(t,x\right)-a_0\left(t\right)x$,

(H₃)

${\displaystyle {\int }_0^1}{f}_1\left(t,+\infty \right)x_0\text{d}t>0>{\displaystyle {\int }_0^1}{f}_1\left(t,-\infty \right)x_0\text{d}t$,

where $f_1\left(t,+\infty \right)=\underset{x\to +\infty }{\lim \inf }{f}_1\left(t,x\right)$,$f_1\left(t,-\infty \right)=\underset{x\to +\infty }{\lim \sup }{f}_1\left(t,x\right)$, and $x_0\left(t\right)\left(>0\right)$ is a nontrivial solution of $x^{″}+a_0\left(t\right)x=0$ and (1.2)-(1.3), $\forall t\in \left[\mathrm{0,1}\right]$. Then (1.1-1.3) has at least one nontrivial solution. Moreover, if we assume

(H₄) $v\left({\lambda }_0\right)=0,i\left({\lambda }_0\right)\ge 2$.

Then (1.1-1.3) has at least two nontrivial solutions.

In this paper, for any $a\in L^{\infty }\left[\mathrm{0,1}\right]$,$i\left(a\right)$ and $v\left(a\right)$ denote its index and nullity of the associated linear ordinary differential equation (see [2] [3] for reference). In Section 2, we will briefly recall the index and its properties. For the readers’ convenience, we give an example: Assume $\lambda$ is a constant, $\alpha =\frac{\text{π}}{2}$ and $\beta =\frac{\text{π}}{2}$. Then

$i\left(\lambda \right)=\{\begin{array}{l}\begin{array}{l}0\text{as}\lambda \le \mathrm{0,}\\ k+1\text{as}\lambda \in \left(k^2{\text{π}}^2\mathrm{,}{\left(k+1\right)}^2{\text{π}}^2\right]\mathrm{,}\end{array}\hfill \end{array}$

and

$v\left(\lambda \right)=\{\begin{array}{l}\begin{array}{l}0\text{as}\lambda \ne k^2{\text{π}}^2,\\ 1\text{as}\lambda =k^2{\text{π}}^2,k\in N^{+}\cup \left\{0\right\}.\end{array}\hfill \end{array}$

In [2], an index for second order linear Hamiltonian systems was defined. And in [3], an index for more general linear self-adjoint operator equations was developed. In [4] [5] [6] [7], by Conley, Zehnder and Long, an index theory for sympletic paths was defined. More applications about these index theories can be found in [8] - [13]. As in [11], throughout this paper, for $a_1\mathrm{,}{a}_2\in L^{\infty }\left[\mathrm{0,1}\right]$, we write $a_1\le a_2$, if $a_2\left(t\right)-a_1\left(t\right)\ge 0$, for $a\mathrm{.}e\mathrm{.}t\in \left[\mathrm{0,1}\right]$ ; we write $a_1<a_2$, if $a_1\le a_2$, and $a_2\left(t\right)-a_1\left(t\right)>0$ holds on a subset of $\left[\mathrm{0,1}\right]$ with nonzero measure.

It is well known [14] that under non-resonant conditions

${\left(2k\text{π}\right)}^2+\delta \le \frac{f\left(t,x\right)}{x}\le {\left(2\left(k+1\right)\text{π}\right)}^2-\delta ,\text{as}\left|x\right|>r>0,k\in N,$

the existence of solutions of a second order nonlinear ordinary differential equation. Such conditions are called nonresonant. Resonant conditions in [15] [16] :

${\left(2k\text{π}\right)}^2\le \frac{f\left(t,x\right)}{x}\le {\left(2\left(k+1\right)\text{π}\right)}^2,\text{as}\left|x\right|>r>0,k\in N,$

are not enough for existence solutions of (1.1-1.2). An additional condition called the (LL) condition like (H₃) is usually needed. For resonant conditions, we refer to [15] [16] [17]. These three papers [14] [15] [16] are about existence of solutions.

In [18], under resonance conditions, periodic solutions of nonlinear second order ordinary differential equations are considered. Second order Hamiltonian systems satisfying Sturm-Liouville boundary vale with the nonresanonce are considered in [3]. First order asymptotical linear Hamiltonian systems satisfying Sturm-Liouville boundary vale with the nonresanonce are studied in [19]. In [20] [21], $\alpha =0,\beta =0$, the existence of solutions of (1.1-1.3) is investigated.

In this paper, we study the existence of equations with resonance conditions. In order to prove our theorem, we construct the corresponding functional:

$\begin{array}{l}\phi \left(x\right)=\frac{1}{2}\left[{\displaystyle {\int }_0^1}{\left|x^{\prime }\left(t\right)\right|}^2\text{d}t-\left(x\left(1\right)\mathrm{,}x\left(1\right)\right)k\left(\beta \right)+\left(x\left(1\right)\mathrm{,}x\left(1\right)\right)k\left(\alpha \right)\right]\\ -{\displaystyle {\int }_0^{1}F\left(t\mathrm{,}x\left(t\right)\right)\text{d}t}\mathrm{,}\forall x\in E\mathrm{.}\end{array}$ (1.4)

where $F\left(t,u\right)={\displaystyle {\int }_0^u}f\left(t,s\right)\text{d}s,u\in R$,$k\left(t\right)=\cot t=\frac{\cos t}{\sin t},k\left(t\right)=0$ as $t=0$ or

π, and E will be described in Section 2. This functional $\phi \left(x\right)$ is continuous differentiable on E, and any critical point of $\phi$ corresponds to a solution of (1.1)-(1.3).

In Section 3, we will give proofs by the Morse theory following [11] [17].

2. Index Theory for Linear Duffing Equations

For any $a\in L^{\infty }\left[\mathrm{0,1}\right]$, consider the following equation:

$\{\begin{array}{l}\begin{array}{l}{x}^{″}\left(t\right)+a\left(t\right)x=0,\\ x\left(0\right)\cos \alpha -x^{\prime }\left(0\right)\sin \alpha =0,\\ x\left(1\right)\cos \beta -x^{\prime }\left(1\right)\sin \beta =0,\end{array}\hfill \end{array}$ (2.1)

where $0\le \alpha <\text{π},0<\beta \le \text{π}$. Define a Hilbert space $E:=E_{\alpha ,\beta }$. Here $E_{\alpha ,\beta }=H^1\left[0,1\right]$ as $\alpha \mathrm{,}\beta \in \left(\mathrm{0,}\text{π}\right)$ ; $E_{0,\beta }=\left\{x\in H^1\left[0,1\right]|x\left(1\right)=0\right\}$ as $\beta \in \left(\mathrm{0,}\text{π}\right)$ ; $E_{\alpha ,0}=\left\{x\in H^1\left[0,1\right]|x\left(0\right)=0\right\}$ as $\alpha \in \left(\mathrm{0,}\text{π}\right)$ and

$E_{0,\text{π}}=\left\{x\in H^1\left[0,1\right]|x\left(1\right)=0=x\left(0\right)\right\}$. With norm ${\Vert x\Vert }_E\mathrm{<mo>\; :\; </mo>}={\left\{{\displaystyle {\int }_0^1}{\left|x\left(t\right)\right|}^2+{\left|x^{\prime }\left(t\right)\right|}^2\text{d}t\right\}}^{\frac{1}{2}}$ and a bilinear form as $q_a\left(\cdot \mathrm{,}\cdot \right)$ as follows

$\begin{array}{l}{q}_a\left(x\mathrm{,}y\right)={\displaystyle {\int }_0^1\left[{\left|x^{\prime }\left(t\right)\right|}^2\text{d}t-\left(x\left(1\right)\mathrm{,}x\left(1\right)\right)k\left(\beta \right)+\left(x\left(0\right)\mathrm{,}x\left(0\right)\right)k\left(\alpha \right)\right]}\\ -{\displaystyle {\int }_0^{1}a\left(t\right)x\left(t\right)y\left(t\right)\text{d}t}\mathrm{,}\forall x\mathrm{,}y\in E\mathrm{.}\end{array}$ (2.2)

From proposition 2.1.1 and 2.3.3 in [3], we have the following properties.

Proposition 2.1 For any $a\in L^{\infty }\left[\mathrm{0,1}\right]$,

1) The E can be divided into three parts:

$E=E^{+}\left(a\right)\oplus E^0\left(a\right)\oplus E^{-}(a)$

such that $q_a$ is positive definite, null and negative definite on $E^{+}\left(a\right)\mathrm{,}{E}^0\left(a\right)$ and $E^{-}\left(a\right)$ respectively. Furthermore, $E^0\left(A\right)$ and $E^{-}\left(A\right)$ are finitely dimensional. We call $\nu \left(a\right):=\dim E^0\left(a\right)$ and $i\left(a\right):=\dim E^{-}\left(a\right)$ the nullity and index respectively.

2) $\left(i\left(a\right)\mathrm{,}\nu \left(a\right)\right)\in N\times \left\{\mathrm{0,1}\right\}$.

3) $\nu \left(a\right)$ is the dimension of the solution subspace of (2.1-2.3), and $i\left(a\right)={\displaystyle \underset{s<0}{\sum }\nu \left(a+s\right)}$.

4) If $a_1\le a_2$, then $i\left(a_1\right)\le i\left(a_2\right)$ and $i\left(a_1\right)+\nu \left(a_1\right)\le i\left(a_2\right)+\nu \left(a_2\right)$ ; if $a_1<a_2$, then $i\left(a_1\right)+\nu \left(a_1\right)\le i\left(a_2\right)$.

5) There exists $\delta >0$ such that

$q_a\left(u\mathrm{,}u\right)\ge \delta {\Vert u\Vert }_E^2\mathrm{,}\forall u\in E^{+}\left(a\right)\mathrm{.}$

Remarks: 1) The notation $\oplus$ means that the space E is the direct sum of some subspaces.

2) By 4), we can see the index has monotonicity.

1) Let $a\left(t\right)=0$,$\alpha =\frac{\text{π}}{2}$, and $\beta =\frac{\text{π}}{2}$. Then (2.1) has a nontrivial solution $x=c\left(\in R\right)\ne 0$. So $\nu \left(0\right)=1$,$i\left(0\right)=0$. If $a\left(t\right)={\text{π}}^2$,$\alpha =\frac{\text{π}}{2}$, and $\beta =\frac{\text{π}}{2}$, then (2.1) has a nontrivial solution $c_1\sin \text{π}t$,$c_1\ne 0$. So $\nu \left({\text{π}}^2\right)=1$,$i\left({\text{π}}^2\right)=1$. If $a\left(t\right)=k^2{\text{π}}^2$,$\alpha =\frac{\text{π}}{2}$, and $\beta =\frac{\text{π}}{2}$, then (2.1) has a nontrivial solution $c_{k1}\sin k\text{π}t$,$c_{k1}\ne 0$. So by Proposition 2.1 (3), $i\left(k^2{\text{π}}^2\right)={\displaystyle \underset{s<0}{\sum }\nu \left(k^2{\text{π}}^2+s\right)}={\displaystyle \underset{n=1}{\overset{k-1}{\sum }}\nu \left(n^2{\text{π}}^2\right)}=1+\left(k-1\right)=k$,$\nu \left(k^2{\text{π}}^2\right)=1$.

The following lemmas are useful for us to prove the results.

Lemma 2.2 The norm ${\Vert x\Vert }_C:=\underset{0\le t\le 1}{\sup }\left|x\left(t\right)\right|\le C_{\ast }{\Vert x\Vert }_E$, for any $x\in E$, where $C_{\mathrm{<mo>\; *\; </mo>}}\left(\in R\right)$ is a positive constant.

Lemma 2.3 If (H₁) holds, then we have that $E=\text{span}\left\{x_0\right\}\oplus E^{+}\left(\lambda \right)$, where $x_0$ is given in Theorem 1.1.

Proof By Proposition 2.1 (1) and conditions $\nu \left(\lambda \right)=0,i\left(\lambda \right)=1$, we have that

$E^0\left(\lambda \right)=\left\{\theta \right\},\dim E^{-}\left(\lambda \right)=1.$ (2.3)

By Proposition 2.1 (1) and (3), we know that with respect to $\lambda \in L^{\infty }\left[\mathrm{0,1}\right]$, the following decomposition holds,

$E=E^{-}\left(\lambda \right)\oplus E^0\left(\lambda \right)\oplus E^{+}\left(\lambda \right)=E^{-}\left(\lambda \right)\oplus E^{+}\left(\lambda \right).$ (2.4)

Since $E^{-}\left(\lambda \right)$ is one dimensional space, we can assume that $\left\{e^{-}\right\}$ is a base of $E^{-}\left(\lambda \right)$, i.e. $E^{-}\left(\lambda \right)=\text{span}\left\{e^{-}\right\}$. So for any $x\in E$, we have

$x=x^{-}+x^{+}=c_0{e}^{-}+x^{+},$

where $x^{-}\in E^{-}\left(\lambda \right)$,$x^{+}\in E^{+}\left(\lambda \right)$, and $c_0$ is a constant. For $x_0\in E$, we have the decomposition $x_0=c_1{e}^{-}+e^{+}$, where $e^{+}\in E^{+}\left(\lambda \right)$ and $c_1$ is a constant. It is obvious that $c_1\ne 0$. Indeed, if $c_1=0$, then $x_0=e^{+}$. By definition of $q_{\lambda }\left(\cdot \mathrm{,}\cdot \right)$, we will have a contradiction that

$\begin{array}{c}{q}_{\lambda }\left(x_0\mathrm{,}{x}_0\right)={\displaystyle {\int }_0^1\left({x^{\prime }}_0\mathrm{,}{x^{\prime }}_0\right)\text{d}t}-\left(x_0\left(1\right)\mathrm{,}{x}_0\left(1\right)\right)k\left(\beta \right)\\ +\left(x_0\left(1\right)\mathrm{,}{x}_0\left(1\right)\right)k\left(\alpha \right)-{\displaystyle {\int }_0^1\lambda \left(t\right)x_0^2\text{d}t}\\ ={\displaystyle {\int }_0^1\left(a\left(t\right)-\lambda \left(t\right)\right)x_0^2\text{d}t}<0\end{array}$

and $\begin{array}{l}{q}_{\lambda }\left(e^{+},e^{+}\right)={\displaystyle {\int }_0^1}\left(e^{+}{}^{\prime },e^{+}{}^{\prime }\right)\text{d}t-\left(e^{+}\left(1\right),e^{+}\left(1\right)\right)k\left(\beta \right)+\left(e^{+}\left(0\right),e^{+}\left(0\right)\right)k\left(\alpha \right)\\ -{\displaystyle {\int }_0^1}\lambda \left(t\right)\left(e^{+},e^{+}\right)\text{d}t\ge 0\end{array}$. So we obtain

$\begin{array}{c}x=c_0{e}^{-}+x^{+}\\ =\frac{c_0}{c_1}\left(c_1{e}^{-}+e^{+}\right)-\frac{c_0}{c_1}{e}^{+}+x^{+}\\ =\frac{c_0}{c_1}{x}_0+\left(x^{+}-\frac{c_0}{c_1}{e}^{+}\right).\end{array}$

We have proved that $E=\text{span}\left\{x_0\right\}\cup E^{+}\left(\lambda \right)$. It is also obvious that $\text{span}\left\{x_0\right\}\cap E^{+}\left(\lambda \right)=\left\{\theta \right\}$. In fact that if $x\left(\ne \theta \right)\in \text{span}\left\{x_0\right\}\cap E^{+}\left(\lambda \right)$, we have that on the one hand for $x\in \text{span}\left\{x_0\right\}$. Then there exists a $c\left(\in R\right)\ne 0$ such that $x=c{x}_0$ and

$\begin{array}{c}{q}_{\lambda }\left(x\mathrm{,}x\right)={\displaystyle {\int }_0^1{c}^2{\left|{x^{\prime }}_0\right|}^2}-c^2\left(x_0\left(1\right)\mathrm{,}{x}_0\left(1\right)\right)k\left(\beta \right)\\ +c^2\left(x_0\left(1\right)\mathrm{,}{x}_0\left(1\right)\right)k\left(\alpha \right)-{\displaystyle {\int }_0^1{c}^2\lambda \left(t\right)x_0^2\text{d}t}\\ ={\displaystyle {\int }_0^1{c}^2\left(a_0\left(t\right)-\lambda \left(t\right)\right)x_0^2\text{d}t}<\mathrm{0\; <mo>\; ;\; </mo>}\end{array}$

on the other hand for $x\in E^{+}\left(\lambda \right)$,

$\begin{array}{l}{q}_{\lambda }\left(x\mathrm{,}x\right)={\displaystyle {\int }_0^1}{\left|x^{\prime }\left(t\right)\right|}^2\text{d}t-\left(x\left(1\right)\mathrm{,}x\left(1\right)\right)k\left(\beta \right)+\left(x\left(1\right)\mathrm{,}x\left(1\right)\right)k\left(\alpha \right)\\ -{\displaystyle {\int }_0^1}\lambda \left(t\right)x^2\left(t\right)\text{d}t\ge 0.\end{array}$

By Proposition 2.1 (1) and (2.2), we have $x=\theta$. This is a contradiction. So the proof is completed.

Remark: For $x=c{x}_0\in \text{span}\left\{x_0\right\}$, we can define ${\Vert x\Vert }_E=\left|c\right|$.

In order to prove Theorem 1.1, we need some lemmas. Let X be a Hilbert space and $\psi \in C^1\left(X\mathrm{,}R\right)$. As in [17], let $K=\left\{x\in X|{\psi }^{\prime }\left(x\right)=\theta \right\}$,${\psi }_m=\left\{x\in X|\psi \left(x\right)\le m\right\}$. For an isolated critical point $x_0$, the critical group is defined by $C_q\left(\psi ,x_0\right)=H_q\left({\psi }_c\cap U,\left({\psi }_c\backslash \left\{x_0\right\}\right)\cap U;R\right)$ for $q=0,1,2,3,\cdots$, where U is a neighborhood of $\left\{x_0\right\}$ such that $K\cap \left({\psi }_c\cap U\right)=\left\{x_0\right\}$ and $c=\psi \left(x_0\right)$.

When $\psi \in C^2\left(X\mathrm{,}R\right)$ and $p\in K$, we have ${\psi }^{″}\left(p\right)$ is a self-adjoint operator. We call the dimension of negative space corresponding to the spectral decomposing the Morse index of p and denote it by $m^{-}\left({\psi }^{″}\left(p\right)\right)$, and denote by $m^0\left({\psi }^{″}\left(p\right)\right)=\dim \left(\ker {\psi }^{″}\left(p\right)\right)$. If ${\psi }^{″}\left(p\right)$ has a bounded inverse we say that p is nondegenerate.

From Theorem 3.1 in Chapter 3, Theorem 5.1, 5.2, Corollary 5.2 in Chapter 5 in [17], one can prove the following lemma.

Lemma 2.4. Assume $\psi \in C^2\left(X\mathrm{,}R\right)$ satisfies the (PS) condition, ${\psi }^{\prime }\left(\theta \right)={\theta }_1$, where $\theta$ is the zero vector in X and ${\theta }_1$ is the zero vector in $X^{\mathrm{<mo>\; *\; </mo>}}$ which is the dual space of X, and there is a positive integer $\gamma$ such that

$\gamma \bar{\in }\left[m^{-}\left({\psi }^{″}\left(\theta \right)\right)\mathrm{,}{m}^{-}\left({\psi }^{″}\left(\theta \right)\right)+m^0\left({\psi }^{″}\left(\theta \right)\right)\right]$ and $H_q\left(X\mathrm{,}{\psi }_m\mathrm{<mo>\; ;\; </mo>}R\right)={\delta }_{q\gamma }R$ for

some regular $m<\psi \left(\theta \right)$, here ${\delta }_{q\gamma }=\{\begin{array}{l}\begin{array}{l}1q=\gamma \hfill \end{array}\\ 0q\ne \gamma \end{array}$. Then $\psi$ has a critical point

$p_0\ne \theta$ with $C_{\gamma }\left(\psi \mathrm{,}{p}_0\right)\ne 0$. Moreover, if $\theta$ is a nondegenerate critical point, and $m^0\left({\psi }^{″}\left(p_0\right)\right)\le \left|\gamma -m^{-}\left({\psi }^{″}\left(\theta \right)\right)\right|$, then $\psi$ has another critical point $p_1\ne p_0\mathrm{,}\theta$.

The following lemma is also useful for us to prove the main result.

Lemma 2.5 (Fatou’s lemma). Given a measure space $\left(\Omega \mathrm{,}\Sigma \mathrm{,}\mu \right)$ and a set $X\in \Sigma$, let $f_n$ be a sequence of $\left(\Sigma \mathrm{,}{\mathcal{B}}_{R_{\ge 0}}\right)$ -measurable non-negative functions $f_n\mathrm{<mo>\; :\; </mo>}X\to \left[\mathrm{0,}+\infty \right]$, where ${\mathcal{B}}_{R_{\ge 0}}$ denotes the σ-algebra of Borel sets on $\left[\mathrm{0,}+\infty \right]$. Define the function $f\mathrm{<mo>\; :\; </mo>}X\to \left[\mathrm{0,}+\infty \right]$ by setting

$f\left(x\right)=\underset{n\to \infty }{\lim \inf }{f}_n\left(x\right),f\left(x\right)=\underset{n\to \infty }{\lim \inf }{f}_n\left(x\right),$

for every $x\in X$. Then f is $\left(\Sigma \mathrm{,}{\mathcal{B}}_{R_{\ge 0}}\right)$ -measurable, and

${\displaystyle {\int }_X}f\text{d}\mu \le \underset{n\to \infty }{\lim \inf }{\displaystyle {\int }_X}{f}_n\text{d}\mu \mathrm{.}$

Remark The integrals may be finite or infinite.

3. Proof of the Main Result

The proof of Theorem 1.1 will depend on the following lemma.

Lemma 3.1 Under (H₁),(H₂), and (H₃), the functional $\phi$ satisfies the (PS) condition.

Proof For $\left\{x_n\right\}\subset E\mathrm{,}{\phi }^{\prime }\left(x_n\right)\to \theta$, and $\phi \left(x_n\right)$ is bounded, we shall find a convergent subsequence in E. By (1.3), for $u\in E$, we have

$\begin{array}{c}\langle {\phi }^{\prime }\left(x_n\right),u\rangle ={\displaystyle {\int }_0^1{x^{\prime }}_n\left(t\right)u^{\prime }\left(t\right)\text{d}t}-\left(x_n\left(1\right),u\left(1\right)\right)k\left(\beta \right)+\left(x_n\left(0\right),u\left(0\right)\right)k\left(\alpha \right)\\ -{\displaystyle {\int }_0^{1}f\left(t,x_n\left(t\right)\right)u\left(t\right)\text{d}t}.\end{array}$ (3.1)

Next, we will prove ${\left\{{\Vert x_n\Vert }_E\right\}}_1^{\infty }$ is bounded. Indeed, it suffices to prove that ${\Vert x_n\Vert }_C$ is bounded. By a contradiction, we assume that ${\Vert x_n\Vert }_C\to +\infty$, as $n\to \infty$.

Defining

$q_n\left(t\right)=\{\begin{array}{l}\begin{array}{l}\frac{f\left(t,x_n\left(t\right)\right)}{x_n\left(t\right)}\left|x_n\left(t\right)\right|\ge r\\ \lambda \left(t\right)\left|x_n\left(t\right)\right|<r\end{array}\hfill \end{array},\text{and}{h}_n\left(t\right)=f\left(t,x_n\left(t\right)\right)-q_n\left(t\right)x_n\left(t\right),$ (3.2)

from (H₁), (H₂) and $f\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,1}\right]\times R\to R$ is continuous, we have

$a_0\le q_n\left(t\right)\le \lambda \left(t\right)\text{and}\left|h_n\left(t\right)\right|<r\left|\lambda \left(t\right)\right|+C_0\mathrm{,}$ (3.3)

where $C_0$ is a constant. Then we get

$f\left(t\mathrm{,}x\left(t\right)\right)=q_n\left(t\right)x_n\left(t\right)+h_n\left(t\right)\mathrm{.}$ (3.4)

By (3.1), it follows that

$\begin{array}{c}{\displaystyle {\int }_0^1}\left({x^{\prime }}_n\left(t\right)\mathrm{,}{u}^{\prime }\left(t\right)\right)\text{d}t={\displaystyle {\int }_0^1}\left(f\left(t\mathrm{,}{x}_n\left(t\right)\right)\mathrm{,}u\left(t\right)\right)\text{d}t+\left(x_n\left(1\right)\mathrm{,}u\left(1\right)\right)k\left(\beta \right)\\ -\left(x_n\left(0\right)\mathrm{,}u\left(0\right)\right)k\left(\alpha \right)+\langle {\phi }^{\prime }\left(x_n\right)\mathrm{,}u\rangle \mathrm{.}\end{array}$ (3.5)

Assuming $y_n=\frac{x_n}{{\Vert x_n\Vert }_C}$, by (3.4), and multiplying ${\Vert x_n\Vert }_C^{-1}$ on both sides of (3.5), we can get that

$\begin{array}{c}{\displaystyle {\int }_0^1}\left({y^{\prime }}_n\left(t\right)\mathrm{,}{u}^{\prime }\left(t\right)\right)\text{d}t={\displaystyle {\int }_0^1{q}_n\left(t\right)y_{n}u\text{d}t}+\left(y_n\left(1\right)\mathrm{,}u\left(1\right)\right)k\left(\beta \right)-\left(y_n\left(0\right)\mathrm{,}u\left(0\right)\right)k\left(\alpha \right)\\ +{\Vert x_n\Vert }_C^{-1}\left({\displaystyle {\int }_0^1{h}_n\left(t\right)u\left(t\right)\text{d}t}+\langle \phi \left(x_n\right)\mathrm{,}u\rangle \right)\mathrm{.}\end{array}$ (3.6)

Furthermore, we add ${\displaystyle {\int }_0^1{y}_n\left(t\right)u\left(t\right)\text{d}t}$ on two sides of (3.6) to obtain that

$\begin{array}{c}{\left(y_n\mathrm{,}u\right)}_E={\displaystyle {\int }_0^1{y}^{\prime }{u}^{\prime }\text{d}t}+{\displaystyle {\int }_0^1{y}_{n}u\text{d}t}\\ ={\displaystyle {\int }_0^1{q}_n\left(t\right)y_{n}u\text{d}t}+{\displaystyle {\int }_0^1{y}_{n}u\text{d}t}+\left(y_n\left(1\right)\mathrm{,}u\left(1\right)\right)k\left(\beta \right)\\ -\left(y_n\left(0\right)\mathrm{,}u\left(0\right)\right)k\left(\alpha \right)+{\Vert x_n\Vert }_C^{-1}\left({\displaystyle {\int }_0^1{h}_n\left(t\right)u\left(t\right)\text{d}t}+\langle \phi \left(x_n\right)\mathrm{,}u\rangle \right)\mathrm{.}\end{array}$

So, by ${\Vert y_n\Vert }_{L^2}={\left({\displaystyle {\int }_0^1{y}_n^2\left(t\right)\text{d}t}\right)}^{\frac{1}{2}}\le {\Vert y_n\Vert }_C=1$ and (3.3) we have

$\begin{array}{c}{\Vert y_n\Vert }_E=\underset{{\Vert u\Vert }_E\le 1}{\sup }{\left(y_n,u\right)}_E\\ \le {\displaystyle {\int }_0^1}\left(q_n\left(t\right)y_n\left(t\right)u\right)\text{d}t+{\displaystyle {\int }_0^1}\left(y_n\left(t\right)u\right)\text{d}t+\left|k\left(\beta \right)\right|+\left|k\left(\alpha \right)\right|+C_2\\ \le C_3{\Vert y_n\Vert }_{L^2}{\Vert u\Vert }_{L^2}+\left|k\left(\beta \right)\right|+\left|k\left(\alpha \right)\right|+C_2\\ \le C^{*},\end{array}$

where $C_2$,$C_3$ and $C^{\mathrm{<mo>\; *\; </mo>}}$ are constants. So ${\left\{{\Vert y_n\Vert }_E\right\}}_1^{\infty }$ is bounded. Then $\left\{y_n\right\}$ has a convergent subsequence. Without loss of generality, we also denoted by $\left\{y_n\right\}$. Then $y_n\rightharpoonup y_0$ in E and $y_n\to y_0$ in $C\left[\mathrm{0,1}\right]$. By inequality $a_0\left(t\right)\le q_n\left(t\right)\le \lambda \left(t\right)$, we have $q_n\rightharpoonup q_0$ in $L^2\left[\mathrm{0,1}\right]$. Then taking the limits on both sides of (3.6), we have, for any $u\in E$,

$\begin{array}{l}{\displaystyle {\int }_0^1{y^{\prime }}_0{u}^{\prime }\text{d}t}-\left(y_0\left(1\right)\mathrm{,}u\left(1\right)\right)k\left(\beta \right)+\left(y_0\left(0\right)\mathrm{,}u\left(0\right)\right)k\left(\alpha \right)-{\displaystyle {\int }_0^1{q}_0\left(t\right)y_{0}u\text{d}t}=\mathrm{0,}\\ \forall u\in E\mathrm{.}\end{array}$ (3.7)

From (3.7) and [3], we have that $y_0$ is a solution of the following problem:

$\{\begin{array}{l}\begin{array}{l}{y}^{″}\left(t\right)+q_0\left(t\right)y=\mathrm{0,}a\mathrm{.}e\mathrm{.}t\in \left[\mathrm{0,1}\right]\\ y\left(0\right)\cos \alpha -y^{\prime }\left(0\right)\sin \alpha =\mathrm{0,}\\ y\left(1\right)\cos \beta -y^{\prime }\left(1\right)\sin \beta =\mathrm{0,}\end{array}\hfill \end{array}$ (3.8)

What’s more, since $a_0\left(t\right)\le q_n\left(t\right)\le \lambda \left(t\right)$, we have $q_0=0$. In fact, by the meaning of the notation “<” and “ $\le$ ”, on the one hand, if $q_0\left(t\right)=\lambda \left(t\right)$, then $\nu \left(q_0\right)=\nu \left(\lambda \right)=0$. Therefore, by the definition $\nu (\cdot )$, this means that (3.8) only has a trivial solution. In fact, by ${\Vert y_n\Vert }_C=1$, we obtain ${\Vert y_0\Vert }_C=1$. So (3.8) has a nontrivial solution. This is a contradiction. On the other hand, if $a_0\left(t\right)<q_0\left(t\right)<\lambda \left(t\right)$, then $1=i\left(a_0\right)+\nu \left(a_0\right)\le i\left(q_0\right)\le i\left(\lambda \right)=1$ holds. While $y_0$ is a nontrivial solution of (3.8), this leads $\nu \left(q_0\right)=1$. So by Proposition 2.1 (4), we get $2=i\left(q_0\right)+\nu \left(q_0\right)\le i\left(\lambda \right)=1$. This is also a contradiction. From discussion above, we obtain the conclusion that $q_0=a_0$. So we immediately get $y_0\in \text{span}\left\{x_0\right\}$.

Since $y_0\in \text{span}\left\{x_0\right\}$, there are two cases about $y_0$. One is that $y_0>0$, the other is that $y_0<0$. Without loss of generality, if $y_0>0$, we assume $y_0=x_0$, and if $y_0<0$, we assume $y_0=-x_0$. Firstly, we discuss the situation that $y_0>0$. If $y_0>0$, i.e. $y_0=x_0$, then for $\forall \epsilon >0,\exists N=N\left(\epsilon \right)$ such that for $n>N$,$\left|y_n-x_0\right|\mathrm{<mo>\; <\; </mo>}\epsilon$ holds. Here, we take the $\epsilon$ such that $x_0-\epsilon >0$, i.e. when $n>N$,$\left\{y_n\left(t\right)\right\}$ belong to the neighborhood of $y_0$,$\left(x_0-\epsilon \mathrm{,}{x}_0+\epsilon \right)$, for all $t\in \left[\mathrm{0,1}\right]$. This means $y_n\left(t\right)>x_0-\epsilon$, as $n>N$, for all $t\in \left[\mathrm{0,1}\right]$.

So by $y_n=\frac{x_n\left(t\right)}{{\Vert x_n\Vert }_C}$, we can get that for any $t\in \left[\mathrm{0,1}\right]$,$x_n\left(t\right)=y_n\left(t\right){\Vert x_n\Vert }_C>0$

for $n>N$. Then $x_n\left(t\right)\to +\infty$ for all $t\in \left[\mathrm{0,1}\right]$, as ${\Vert x_n\Vert }_C\to \infty$. By the assumption that ${\Vert x_n\Vert }_C\to \infty$, as $n\to \infty$, taking the limits on both sides of (3.1) and letting $u=x_0$, we can obtain

$\begin{array}{l}\langle {\phi }^{\prime }\left(x_n\right),x_0\rangle \\ ={\displaystyle {\int }_0^1{x^{\prime }}_n{x^{\prime }}_0\text{d}t}-\left(x_n\left(1\right),x_0\left(1\right)\right)k\left(\beta \right)+\left(x_n\left(0\right),x_0\left(0\right)\right)k\left(\alpha \right)-{\displaystyle {\int }_0^{1}f\left(t,x_n\left(t\right)\right)x_0\text{d}t}\\ ={\displaystyle {\int }_0^1\left(q_0\left(t\right)x_n-f\left(t,x_n\right)\right)x_0\text{d}t}\\ =-{\displaystyle {\int }_0^1{f}_1\left(t,x_n\right)x_0\text{d}t}.\end{array}$ (3.9)

So, by (H₃), and (3.9), the following holds

${\displaystyle {\int }_0^1{f}_1\left(t\mathrm{,}{x}_n\left(t\right)\right)x_0\text{d}t}\to \mathrm{0,}\text{as}n\to \infty \mathrm{.}$ (3.10)

Furthermore, by the Fatou’s Lemma and (3.10), we have

$\begin{array}{c}0=\underset{n\to \infty }{\underset{\_}{\lim }}{\displaystyle {\int }_0^1{f}_1\left(t\mathrm{,}{x}_n\left(t\right)\right)x_0\text{d}t}\ge {\displaystyle {\int }_0^1\underset{n\to \infty }{\underset{\_}{\lim }}{f}_1\left(t\mathrm{,}{x}_n\left(t\right)\right)x_0\text{d}t}\\ ={\displaystyle {\int }_0^1{f}_1\left(t\mathrm{,}+\infty \right)x_0\text{d}t}\mathrm{,}\end{array}$

a contradiction to assumption (H₃). Hence, if $y_0>0$, this leads to a contradiction. Secondly, in a similar way, we can show that if $y_0<0$, there also be a contradiction. Therefore, the sequence ${\left\{{\Vert x_n\Vert }_C\right\}}_1^{\infty }$ is a bounded sequence. By the equality $x_n=y_n{\Vert x_n\Vert }_C$ and the fact that ${\Vert y_n\Vert }_E$ is bounded, we can get that ${\left\{{\Vert x_n\Vert }_E\right\}}_1^{\infty }$ is bounded in E. Furthermore, ${\left\{x_n\right\}}_1^{\infty }$ has a weak convergent subsequence in E, without loss of generality, still denoted by ${\left\{x_n\right\}}_1^{\infty }$. So we have $x_n\rightharpoonup x_{\mathrm{<mo>\; *\; </mo>}}$ in E and $x_n\to x_{\mathrm{<mo>\; *\; </mo>}}$ in $C\left[\mathrm{0,1}\right]$. In addition, by (3.1), we also have

$\begin{array}{l}{\displaystyle {\int }_0^1{x^{\prime }}_{\mathrm{<mo>\; *\; </mo>}}\left(t\right)u^{\prime }\left(t\right)\text{d}t}-\left(x_{\mathrm{<mo>\; *\; </mo>}}\left(1\right)\mathrm{,}u\left(1\right)\right)k\left(\beta \right)+\left(x_{\mathrm{<mo>\; *\; </mo>}}\left(0\right)\mathrm{,}u\left(0\right)\right)k\left(\alpha \right)\\ -{\displaystyle {\int }_0^1\left(f\left(t\mathrm{,}{x}_{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\right)\mathrm{,}u\left(t\right)\right)\text{d}t}=0.\end{array}$ (3.11)

At last, we only need to finish the mission that $x_n\to x_{\mathrm{<mo>\; *\; </mo>}}$ in E. Indeed, by (3.5), (3.11) and $x_n\rightharpoonup x_{\mathrm{<mo>\; *\; </mo>}}$, we obtain the fact that

$\begin{array}{l}{\Vert x_n-x_{\mathrm{<mo>\; *\; </mo>}}\Vert }_E=\underset{{\Vert u\Vert }_E\le 1}{\sup }{\left(x_n-x_{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}u\right)}_E=\underset{{\Vert u\Vert }_E\le 1}{\sup }\left[{\displaystyle {\int }_0^1}\left({x^{\prime }}_n-{x^{\prime }}_{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}{u}^{\prime }\right)\text{d}t+{\displaystyle {\int }_0^1}\left(x_n-x_{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}u\right)\text{d}t\right]\\ =\underset{{\Vert u\Vert }_E\le 1}{\sup }\{{\displaystyle {\int }_0^1}\left(f\left(t\mathrm{,}{x}_n\right)-f\left(t\mathrm{,}{x}_{\mathrm{<mo>\; *\; </mo>}}\right)\mathrm{,}u\left(t\right)\right)\text{d}t+\langle {\phi }^{\prime }\left(x_n\right)-{\phi }^{\prime }\left(x_{\mathrm{<mo>\; *\; </mo>}}\right)\mathrm{,}u\rangle \\ +\left(x_n\left(1\right)-x_{\mathrm{<mo>\; *\; </mo>}}\left(1\right)\mathrm{,}u\right)k\left(\beta \right)-\left(x_{\mathrm{<mo>\; *\; </mo>}}\left(0\right)-x_n\left(0\right)\mathrm{,}u\right)k\left(\alpha \right)\\ +{\displaystyle {\int }_0^1}\left(x_n-x_{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}u\right)\text{d}t\}\to \mathrm{0,}\text{as}n\to \infty \mathrm{.}\end{array}$

The (PS) condition is verified.

After the preliminary work, we can prove Theorem 1.1.

Proof of Theorem 1.1. Since $\nu \left(a\right)\le 1$ for any $a\in L^{\infty }\left[\mathrm{0,1}\right]$, by Lemma 2.4, we only need to prove

$H_q\left(E\mathrm{,}{\phi }_z\mathrm{<mo>\; ;\; </mo>}R\right)\cong {\delta }_{q\gamma }R$ (3.12)

for $-z>-\phi \left(\theta \right)$ large enough, where $\gamma =\nu \left(0\right)=i\left(\lambda \right)=1$. By Lemma 2.3, we know that E can be split into two subspaces $\text{span}\left\{x_0\right\}$ and $E^{+}\left(\lambda \right)$, i.e.

$E=\text{span}\left\{x_0\right\}\oplus E^{+}\left(\lambda \right)\mathrm{.}$

Next, we will take two steps to obtain the proof of (3.12).

First step: For $-z>-\phi \left(\theta \right)$ large enough, we have

$H_q\left(E\mathrm{,}{\phi }_z\mathrm{<mo>\; ;\; </mo>}R\right)\cong H_q\left(\mathcal{M}\mathrm{,}\mathcal{M}\cap {\phi }_z\mathrm{<mo>\; ;\; </mo>}R\right)\mathrm{,}q=\mathrm{0,1,2,}\cdots \mathrm{,}$ (3.13)

where $\mathcal{M}\subset H$ will be defined later. By assumption, for any $y\in E$, we have

$\begin{array}{l}\langle {\phi }^{\prime }\left(x\right)\mathrm{,}y\rangle \\ ={\displaystyle {\int }_0^1{x}^{\prime }{y}^{\prime }\text{d}t}-\left(x\left(1\right)\mathrm{,}y\left(1\right)\right)k\left(\beta \right)+\left(x\left(0\right)\mathrm{,}y\left(0\right)\right)k\left(\alpha \right)-{\displaystyle {\int }_0^{1}f\left(t\mathrm{,}x\right)y\text{d}t}\mathrm{.}\end{array}$ (3.14)

We will consider the behavior of f in two subintervals of $\left[\mathrm{0,1}\right]$. One is $\left\{t\mathrm{<mo>\; |\; </mo>}x\left(t\right)\ge r\right\}\subset \left[\mathrm{0,1}\right]$, the other is $\left\{t|\left|x\left(t\right)\right|<r\right\}\subset \left[0,1\right]$. Since f is continuous on $\left[\mathrm{0,1}\right]\times \left(-\infty \mathrm{,}+\infty \right)$, it is obvious that $f\left(t\mathrm{,}x\left(t\right)\right)$ is bounded on $\left\{t|\left|x\left(t\right)\right|<r\right\}$. So there exists a constant $M_1\in R$ such that $\left|f\left(t,x\left(t\right)\right)\right|<M_1$ when $\left|x\left(t\right)\right|<r$.

By Lemma 2.3, we have a decomposition with respect to $x\left(t\right)\in E$, i.e. there exist $x_{+}\left(t\right)\in E^{+}\left(\lambda \right)$, and $c\in R$ such that $x=x_{+}+c{x}_0$. When $\left|x_{+}+c{x}_0\right|<r$, we have $\left|c{x}_0\right|<r+\left|x_{+}\right|<r+{\Vert x_{+}\Vert }_C$. Furthermore, we get

$\left|c{x}_0{\displaystyle \underset{\left|x_{+}+c{x}_0\right|<r}{\int }}f\left(t,x_{+}+c{x}_0\right)\text{d}t\right|<M_1\left(r+{\Vert x_{+}\Vert }_C\right).$ (3.15)

So by (15), we have

$\begin{array}{l}{\displaystyle \underset{\left|x_{+}+c{x}_0\right|<r}{\int }}f\left(t,x_{+}+c{x}_0\right)\left(x_{+}-c{x}_0\right)\text{d}t\\ ={\displaystyle \underset{\left|x_{+}+c{x}_0\right|<r}{\int }}f\left(t,x_{+}+c{x}_0\right)x_{+}\text{d}t-c{x}_0{\displaystyle \underset{\left|x_{+}+c{x}_0\right|<r}{\int }}f\left(t,x\right)\text{d}t\\ \le M_1{\Vert x_{+}\Vert }_C+M_1\left(r+{\Vert x_{+}\Vert }_C\right).\end{array}$ (3.16)