Existence of nontrivial solutions for a nonlinear second order periodic boundary value problem with derivative term

Abstract

In this paper, we study the existence of nontrivial solutions to the following nonlinear differential equation with derivative term:

$$\begin{aligned} {\left\{ \begin{array}{l}u''(t)+a(t)u(t)=f\big (t,u(t),u'(t)\big ),\quad t\in [0,\omega ],\\ u(0)=u(\omega ),\quad u'(0)=u'(\omega ),\end{array}\right. } \end{aligned}$$

where a: $[0,\omega ]\rightarrow \mathbb {R}^{+}\big (\mathbb {R}^{+}=[0,+\infty )\big )$ is a continuous function with $a(t)\not \equiv 0$, f: $[0,\omega ]\times \mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}$ is continuous and may be sign-changing and unbounded from below. Without making any nonnegative assumption on nonlinearity, using the first eigenvalue corresponding to the relevant linear operator and the topological degree, the existence of nontrivial solutions to the above periodic boundary value problem is established in $C^1[0,\omega ]$. Finally, an example is given to demonstrate the validity of our main result.

1 Introduction

Due to wide applications in physics and engineering, second-order periodic boundary value problems (PBVPs) have been extensively studied by many authors, see [1,2,3,4,5,6,7,8,9,10,11,12] and relevant references therein.

In [1], the following problem was discussed by Atici and Guseinov

$$\begin{aligned} \left\{ \begin{array}{l}-\big (p(t)u'(t)\big )'+q(t)u(t)=f\big (t,u(t)\big ),\quad t\in [0,\omega ],\\ u(0)=u(\omega ),\quad p(0)u'(0)=p(\omega )u'(\omega ),\end{array}\right. \end{aligned}$$

where p(x) and q(x) are real-valued measurable functions defined on $[0,\omega ]$ satisfying $p(x)>0, q(x)\ge 0, q(x)\ne 0$ almost everywhere, and

$$\begin{aligned} \int _0^{\omega }\frac{dx}{p(x)}<+\infty , \int _0^{\omega }q(x)dx<+\infty , \end{aligned}$$

f: $[0,\omega ]\times \mathbb {R}\rightarrow \mathbb {R}$ is continuous and $f(t,x)\ge 0$ for $(t,x)\in [0,\omega ]\times [0,\infty ]$. If there exist numbers $0<r<R<+\infty$ such that for all $t\in [0,\omega ]$,

$$\begin{aligned} f(t,x)\le \frac{1}{\omega M}x \, \text {for} \, 0\le x\le r \, \text {and} \, f(t,x)\ge \frac{M}{\omega m^2}x \, \text {for} \, R\le x<+\infty , \end{aligned}$$

where $m=\min _{t,s\in [0,\omega ]}G(t,s), M=\max _{t,s\in [0,\omega ]}G(t,s)$ and G(t, s) is the Green’s function according to its linear problem, the authors established the existence of positive solutions.

Graef et al. [2] investigated the existence of positive solutions, under

$$\begin{aligned}\lim _{u\rightarrow 0}\frac{f(u)}{u}=+\infty , \lim _{u\rightarrow +\infty }\frac{f(u)}{u}=0 \, \text {or} \, \lim _{u\rightarrow 0}\frac{f(u)}{u}=0, \lim _{u\rightarrow +\infty }\frac{f(u)}{u}=+\infty \end{aligned}$$

with f convex and nondecreasing, to

$$\begin{aligned} \left\{ \begin{array}{l} u''(t)+a(t)u(t)=g(t)f(u(t)),\quad t\in [0,2\pi ],\\ u(0)=u(2\pi ),\quad u'(0)=u'(2\pi ),\end{array}\right. \end{aligned}$$

where $f:[0,+\infty )\rightarrow [0,+\infty ), g:[0,2\pi ]\rightarrow [0,+\infty )$ are continuous such that $\min _{t\in [0,2\pi ]}g(t)>0$, and the Green’s function is nonnegative.

Hai [3] proved the existence of positive solutions to

$$\begin{aligned} \left\{ \begin{array}{l} u''(t)+a(t)u(t)=\lambda g(t)f(u(t)),\quad t\in [0,2\pi ],\\ u(0)=u(2\pi ),\quad u'(0)=u'(2\pi )\end{array}\right. \end{aligned}$$

for all $\lambda >0$, where $a:[0,2\pi ]\rightarrow [0,+\infty )$ is continuous with $a(t)\le 1/4$ for all t and $a(t)\not \equiv 0$, $f:[0,+\infty )\rightarrow [0,+\infty )$ is continuous, $g\in L^1(0,2\pi )$ with $g\ge 0$ and $g\not \equiv 0$ on any subinterval of $(0,2\pi )$.

Li and Liang in [4] obtained the existence of positive solutions for

$$\begin{aligned} \left\{ \begin{array}{l} u''(t)+a(t)u(t)=f(t,u(t)),\quad t\in [0,\omega ],\\ u(0)=u(\omega ),\quad u'(0)=u'(\omega ),\end{array}\right. \end{aligned}$$

(1.1)

where $f:[0,\omega ]\times [0,+\infty )\rightarrow [0,+\infty )$ and $a:[0,\omega ]\rightarrow [0,+\infty )$ are continuous with $a(t)\not \equiv 0$, provided that a constant $0<M\le (\pi /\omega )^2$ and

$$\begin{aligned} \limsup _{u\rightarrow 0^+}\max _{t\in [0,\omega ]}\frac{f(t,u)}{u}<\lambda _1<\liminf _{u\rightarrow +\infty }\min _{t\in [0,\omega ]}\frac{f(t,u)}{u} \end{aligned}$$

or

$$\begin{aligned} \liminf _{u\rightarrow 0^+}\min _{t\in [0,\omega ]}\frac{f(t,u)}{u}>\lambda _1>\limsup _{u\rightarrow +\infty }\max _{t\in [0,\omega ]}\frac{f(t,u)}{u}. \end{aligned}$$

Torres [5] gave classical conditions that guarantee the nonnegativity of the Green’s function G(t, s) of the linear problem

$$\begin{aligned} \left\{ \begin{array}{l} u''(t)+a(t)u(t)=0, \, \text {a.e.} \, t\in (0,\omega ),\\ u(0)=u(\omega ),\quad u'(0)=u'(\omega ).\end{array}\right. \end{aligned}$$

(1.2)

The following best Sobolev constants were used:

$$\begin{aligned} K(q) = \left\{ \begin{array}{l} \frac{2\pi }{q\omega ^{1 + \frac{2}{q}}}\left( \frac{2}{2 + q}\right) ^{1-\frac{2}{q}} \left( \frac{\Gamma \left( \frac{1}{q}\right) }{\Gamma \left( \frac{1}{2} + \frac{1}{q}\right) }\right) ^2,\quad 1 \le q < + \infty , \\ \frac{4}{\omega },\quad q =+\infty , \end{array} \right. \end{aligned}$$

(1.3)

where $\Gamma$ is the Gamma function. Let $a\in L^p(0,\omega ) (1\le p\le +\infty )$ and $a\succ 0$ which means that $a(t)\ge 0$ for a.e. $t\in (0,\omega )$ and $a(t)>0$ for t in a subset of positive measure. Then,

$$\begin{aligned} G(t,s)\ge (>)0 \, \text {for} \, (t,s)\in [0,\omega ]\times [0,\omega ] \, \text {when} \, \Vert a\Vert _p\le (<) K(2p^*), \end{aligned}$$

(1.4)

where $1/p+1/p^*=1$. Define

$$\begin{aligned} \Delta =\left\{ a\in L^p(0,\omega ): a\succ 0, \Vert a\Vert _p< K(2p^*) \, \text {for some} \, 1\le p \le +\infty \right\} . \end{aligned}$$

(1.5)

For $a\in \Delta$, the explicit expression of G(t, s) was found by Ma et al. in [6] which shows that it is symmetrical, i.e., $G(t,s)=G(s,t), \forall t,s\in [0,\omega ]$.

For $\omega =1$, PBVP (1.1) was studied in [7] when $a\in L^1[0,1]\bigcap \Delta$ and $f:[0,1]\times (0,+\infty )\rightarrow \mathbb {R}$ with a bound below. Under some other conditions, the existence of positive solutions was obtained.

By means of the fixed point theory, Liu et al. in [8] established the existence of nontrivial solutions for PBVP (1.1) in which $a: [0,\omega ]\rightarrow [0,+\infty )$ is a continuous function with $a(t)\not \equiv 0$, and f: $[0,\omega ]\times \mathbb {R}\rightarrow \mathbb {R}$ is continuous under constraints associated with the first eigenvalue corresponding to the relevant linear operator.

In the works mentioned above, all the nonlinearities are independence of the derivative term $u'$. By Leray–Schauder fixed point theorem, Li and Guo in [9] considered the existence of solution to

$$\begin{aligned} \left\{ \begin{array}{l} u''(t)=f(t,u(t),u'(t)),\quad t\in [0,2\pi ],\\ u(0)=u(2\pi ),\quad u'(0)=u'(2\pi ),\end{array}\right. \end{aligned}$$

where $f: \mathbb {R}^3\rightarrow \mathbb {R}$ is continuous and satisfies the following conditions:

(${\hbox {F}}_1$):

$f(-t,-x,y)=-f(t,x,y), \forall (t,x,y)\in \mathbb {R}^3$;

(${\hbox {F}}_2$):

there exist nonnegative constants a and b satisfying $a+b<1$ and a positive constant $C_0$ such that $f(t,x,y)x\ge -ax^2-by^2-C_0, (t,x,y)\in \mathbb {R}^3$;

(${\hbox {F}}_3$):

the following Nagumo condition is satisfied, that is, for any given $M>0$, there is a positive continuous function $g_M(\rho )$ on $\mathbb {R}^+=[0,+\infty )$ satisfying $\int _0^{+\infty }\frac{\rho d\rho }{g_M(\rho )+1}=+\infty$ such that $|f(t,x,y)|\le g_M(|y|), (t,x,y)\in [0,2\pi ]\times [-M,M]\times \mathbb {R}$.

Inspired by the references cited above and [13,14,15,16,17], we in this paper explore the existence of nontrivial solutions to the following periodic boundary value problem with the nonlinearity dependent on derivative term

$$\begin{aligned} \left\{ \begin{array}{l}u''(t)+a(t)u(t)=f\big (t,u(t),u'(t)\big ),\quad t\in [0,\omega ],\\ u(0)=u(\omega ),\quad u'(0)=u'(\omega ),\end{array}\right. \end{aligned}$$

(1.6)

where $a: [0,\omega ]\rightarrow \mathbb {R}^{+}$ is a continuous function with $a(t)\not \equiv 0$, $f: [0,\omega ]\times \mathbb {R}^2\rightarrow \mathbb {R}$ is continuous and may be sign-changing and unbounded from below. Without making any nonnegative assumption on the nonlinearity, using the first eigenvalue corresponding to the relevant linear operator and the topological degree, the existence of nontrivial solutions is established in $C^1[0,\omega ]$. As far as we know, this kind of PBVP has achieved fewer results.

2 Preliminaries

Let $E=C^1[0,\omega ]$ be the Banach space of all continuously differentiable functions on $[0,\omega ]$ with the norm $\Vert {u}\Vert _{C^{1}}=\max \{\Vert {u}\Vert _{C} , \Vert {u'\Vert _{C}}\}$ for all $u \in E$. Set $P=\{u\in C[0,\omega ]: u(t) \ge 0, t\in [0,\omega ] \}$, it is clear that P is a solid cone in $C[0,\omega ]$, that is, the interior point set of P is nonempty. Thus P is a total cone in $C[0,\omega ]$, i.e. $C[0,\omega ]=\overline{P-P}$, which means that the set $P-P=\{u-v: u, v \in P\}$ is dense in $C[0,\omega ]$ (see [18, 19]).

To state our main theorem in this paper, we make the following hypotheses:

$(C_1)$:

$a: [0,\omega ]\rightarrow \mathbb {R}^{+}$ is a continuous function with $a(t)\not \equiv 0$ and $\Vert a\Vert _C<\big (\frac{\pi }{\omega }\big )^2=K(2)$, where K is defined by (1.3);

$(C_2)$:

f: $[0,\omega ]\times \mathbb {R}^2\rightarrow \mathbb {R}$ is continuous.

Lemma 2.1

If $(C_1)$ and $(C_2)$ hold, then PBVP (1.6) is equivalent to

$$\begin{aligned} u\left( t\right) =\int _0^\omega G(t,s)f(s,u(s),u'(s))ds, \end{aligned}$$

where G(t, s) is the Green’s function of (1.2).

Lemma 2.2

[5, 6] If $(C_1)$ holds, then G(t, s) has the following properties:

• (1) $G(t,s)=G(s,t)>0$ for all $(t,s)\in [0,\omega ] \times [0,\omega ]$;

• (2) Let $l_{1}=\min _{0\le t,s \le \omega }G(t,s)$ and $l_{2}=\max _{0\le t,s \le \omega }G(t,s)$, then $l_{2}>l_{1}>0$;

• (3) Let $l_{3}=\sup _{0\le t,s \le \omega }\left| \frac{\partial G(t,s)}{\partial t}\right|$, then $l_{3}>0;$

• (4) For $c=\frac{l_1}{l_2}\in (0,1)$, $G(t,s) \ge cG(\tau ,s), \quad \forall ~t,s,\tau \in [0,\omega ]$.

Remark 2.3

By $(C_1)$, $a\in L^{\infty }[0,\omega ]$ and thus $a\in \Delta$ defined by (1.4) for $p=+\infty$ and $p^*=1$. Lemma 2.1 follows from (1.4). If $a(t) \equiv m^{2}$ , $m \in (0,\frac{1}{2})$ and $\omega =2\pi$, we can obtain that G(t, s) defined as follows satisfies Lemma 2.2.

$$\begin{aligned}G(t,s) = \left\{ \begin{array}{l} \frac{\sin m(t - s) + \sin m(2\pi - t + s)}{2m(1 - \cos 2m\pi )}, 0 \le s \le t \le 2\pi , \\ \frac{\sin m(s - t) + \sin m(2\pi - s + t)}{2m(1 - \cos 2m\pi )}, 0 \le t \le s \le 2\pi , \\ \end{array} \right. \\ \frac{\partial G(t,s)}{\partial t}= \left\{ \begin{array}{l} \frac{\cos m(t - s) - \cos m(2\pi - t + s)}{2(1 - \cos 2m\pi )}, 0 \le s \le t \le 2\pi , \\ \frac{- \cos m(s - t) + \cos m(2\pi - s + t)}{2(1 - \cos 2m\pi )}, 0 \le t \le s \le 2\pi , \\ \end{array} \right. \\ l_{1} = \frac{\sin 2m\pi }{2m(1 - \cos 2m\pi )}, \quad l_{2} = \frac{\sin m\pi }{m(1 - \cos 2m\pi )}, \quad l_{3}=\frac{1}{2}. \end{aligned}$$

Define a linear operator $L: C[0,\omega ]\rightarrow C[0,\omega ]$ by

$$\begin{aligned} (Lu)(t)=\int _0^\omega {G(t,s)u(s)ds} \end{aligned}$$

(2.1)

and an operator $A: E \rightarrow E$ as follows:

$$\begin{aligned} (Au)(t)=\int _0^\omega G(t,s)f(t,u(t),u'(t))ds, \quad u \in E. \end{aligned}$$

(2.2)

Clearly, $A: E \rightarrow E$ is a completely continuous operator, and the existence of solutions of (1.6) is equivalent to the fixed points of A. Moreover, $L: C[0,\omega ]\rightarrow C[0,\omega ]$ is a completely continuous linear operator, satisfying $L(P) \subset P$. Since $G(t,s) > 0$ for $(t,s)\in [0,\omega ] \times [0,\omega ]$, the spectral radius r(L) of L is positive by Lemma 2.2. The Krein-Rutman theorem [20] then asserts that there exists $\varphi \in P \backslash \left\{ 0 \right\}$ corresponding to the first eigenvalue $\lambda _{1}=\frac{1}{r(L)}$ of L such that

$$\begin{aligned} \lambda _{1}L\varphi =\varphi . \end{aligned}$$

(2.3)

It is easy to see that $\varphi \in E$. According to (2.1), (2.3) and Lemma 2.2 (1), one has

$$\begin{aligned} \int _0^\omega \varphi (t)(Lu)(t )dt = \frac{1}{\lambda _1}\int _0^\omega \varphi (t)u(t)dt\quad \, \text {for all} \, u\in E. \end{aligned}$$

(2.4)

Let $\delta = c\int _0^\omega \varphi (t )dt$, so $\delta >0$.

Choose a subcone $P_{1}$ of P given by

$$\begin{aligned} P_{1} = \left\{ u \in P: \int _0^\omega \varphi (t)u(t) dt\ge \delta \Vert u\Vert _C \right\} . \end{aligned}$$

Lemma 2.4

$L(P) \subset P_{1}$.

Proof

Suppose $u \in P$, the following equation can be obtained in light of Lemma 2.2:

$$\begin{aligned} \int _0^\omega \varphi (t)(Lu)(t)dt&=\int _0^\omega \varphi (t)\int _0^\omega G(t,s)u(s)dsdt\\&\ge c\int _0^\omega \varphi (t)dt\int _0^\omega G(\tau ,s)u(s)ds=\delta (Lu)(\tau ), \end{aligned}$$

where $\tau \in [0,\omega ]$. Obviously, $\int _0^\omega \varphi (t)(Lu)(t)dt\ge \delta \Vert Lu\Vert _{C}$.

Lemma 2.5

[18, 19] Let E be a Banach space and $\Omega$ be a bounded open set in E with $0\in \Omega$. Suppose that $A: \Omega \rightarrow E$ is a completely continuous operator. If

$$\begin{aligned} Au \ne \tau u, \quad \forall u \in \partial \Omega , \quad \tau \ge 1, \end{aligned}$$

then the topological degree $\deg (I-A,\Omega ,0)=1$.

Lemma 2.6

[18, 19] Let E be a Banach space and $\Omega$ be a bounded open set in E. Suppose that $A: \Omega \rightarrow E$ is a completely continuous operator. If there exists $u_{0} \ne 0$ such that

$$\begin{aligned} u-Au\ne \mu u_{0}, \quad \forall u \in \partial \Omega , \quad \mu \ge 0, \end{aligned}$$

then the topological degree $\deg (I-A,\Omega ,0)=0$.

Three sets are given for the sake of simplicity for later writing as follows:

$$\begin{aligned} B_{r}= & {} \left\{ u\in E: \Vert u\Vert _{C^{1}} < r \right\} , \quad \partial B_{r} = \left\{ u\in E: \Vert u\Vert _{C^{1}} = r \right\} ,\\ \overline{B}_r= & {} \left\{ u\in E: \Vert u\Vert _{C^{1}} \le r \right\} \, \text {for} \, r>0. \end{aligned}$$

3 Main result

Theorem 3.1

Under the hypotheses $(C_1)$-$(C_2)$ suppose that

$(C_{3})$:

there exists a constant $\alpha >1$ such that $\liminf _{x +\alpha |y| \rightarrow +\infty }\frac{f(t,x,y)}{x+\alpha |y|} > \lambda _{1} \, \text {uniformly on} \, t \in [0,\omega ];$

$(C_{4})$:

$\limsup _{x + |y| \rightarrow -\infty }\frac{f(t,x,y)}{x+|y|} < \lambda _{1} \, \text {uniformly on} \, t \in [0,\omega ];$

$(C_{5})$:

there exist nonnegative constants $a \ge 0, b \ge 0$ and $r > 0$ satisfying

$$\begin{aligned} \omega (a+b) \max \left\{ l_{2}, l_{3}\right\} < 1, \end{aligned}$$

(3.1)

such that

$$\begin{aligned} |f(t,x,y)| \le a|x|+b|y|, \end{aligned}$$

(3.2)

for all $(t, x, y) \in [0, \omega ] \times [-r, r]^2$, where $\lambda _{1}$ is the first eigenvalue of the operator L defined by (2.1) and $l_2, l_3$ are as Lemma 2.2. If the following Nagumo condition is fulfilled, i.e.

$(C_6)$:

for any $M>0$ there is a positive continuous function $H_M(\rho )$ on $\mathbb {R}^+$ satisfying

$$\begin{aligned} \int _0^{+\infty }\frac{\rho d\rho }{H_M(\rho )+1}=+\infty , \end{aligned}$$

(3.3)

such that

$$\begin{aligned} |f(t,x,y)|\le H_M(|y|), \forall (t,x,y)\in [0,\omega ]\times [-M,M]\times \mathbb {R}, \end{aligned}$$

(3.4)

then PBVP (1.6) has at least one nontrivial solution.

Proof

(i) First, we prove that $Au \ne \tau u$ for $u \in \partial B_{r}$ and $\tau \ge 1.$ In fact, if there exist $u_{1} \in \partial B_{r}$ and $\tau _{0} \ge 1$ such that $Au_{1} = \tau _{0} u_{1},$ then we deduce from Lemma 2.1, (3.1), (3.7) and $-r \le u_{1}(t) \le r, -r \le u'_{1}(t) \le r, \forall t \in [0, \omega ]$ that

$$\begin{aligned} \Vert u_{1}\Vert _{C}&=\frac{1}{\tau _{0}} \max _{0 \le t\le \omega } \Big |\int _0^\omega G(t,s)f(s, u_{1}(s), u'_{1}(s))ds\Big |\\&\le \int _0^\omega l_{2} |f\big (s, u_{1}(s), u'_{1}(s)\big )|ds \le \int _0^\omega l_{2} \big (a |u_{1} (s)| + b|u'_{1} (s)|\big )ds\\&\le \omega (a+b) l_{2}\Vert u_{1}\Vert _{C^{1}} < \Vert u_{1}\Vert _{C^{1}}=r \end{aligned}$$

and

$$\begin{aligned} \Vert u'_{1}\Vert _{C}= & {} \frac{1}{\tau }_{0}\max _{0 \le t\le \omega } \Big |\int _0^\omega \frac{\partial G(t,s)}{\partial t} f\big (s, u_{1}(s), u'_{1}(s)\big )ds\Big | \\\le & {} \frac{1}{\tau }_{0}\max _{0 \le t\le \omega } \int _0^\omega \Big |\frac{\partial G(t,s)}{\partial t}\Big | |f\big (s, u_{1}(s), u'_{1}(s)\big )| ds\\\le & {} \int _0^\omega l_{3} \big (a |u_{1} (s)| + b|u'_{1} (s)|\big )ds\le \omega (a+b) l_{3}\Vert u_{1}\Vert _{C^{1}} < \Vert u_{1}\Vert _{C^{1}}=r. \end{aligned}$$

Hence ${\Vert u_{1}\Vert }_{C^{1}} < r$, which contradicts $u_{1}\in \partial B_{r}.$

Therefore, it follows from Lemma 2.5 that

$$\begin{aligned} \deg (I-A, B_{r}, 0)=1. \end{aligned}$$

(3.5)

(ii) It follows from $(C_{3})$ and $(C_{4})$ that there exist $\varepsilon _{0} \in (0, \lambda _{1})$ and $X_{0} > 0$ such that

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} + \varepsilon _{0}) (x + \alpha |y|) \, \text {for} \, x+\alpha |y|>X_{0} \, \text {and} \, t \in [0, \omega ] \end{aligned}$$

and

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} - \varepsilon _{0}) (x + |y|) \, \text {for} \, x+|y|<-X_{0} \, \text {and} \, t \in [0, \omega ]. \end{aligned}$$

Let

$$\begin{aligned} \Omega =\mathbb {R}^{2}\backslash \left( \{(x,y): x+\alpha |y|>X_0\}\cup \{(x,y): x+|y|<-X_0\} \right) =\Omega _1 \cup \Omega _2, \end{aligned}$$

where $\Omega _1 = \Omega \cap \{(x,y):x+|y|\ge 0 \}, \Omega _2= \Omega \cap \{(x,y):x+|y|\le 0 \}$, it is easy to see that $\Omega$, $\Omega _1$ and $\Omega _2$ are closed bounded sets in $\mathbb {R}^{2}$.

Take $C_1>0$ such that

$$\begin{aligned} -C_{1}\le \min _{t \in [0,\omega ], (x,y)\in \Omega _1}\{f(t, x, y)-\left( \lambda _{1}+\varepsilon _{0}\right) (x+\alpha |y|)\}, \end{aligned}$$

hence

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} + \varepsilon _{0}) (x + \alpha |y|) - C_{1}\ge (\lambda _{1} + \varepsilon _{0})x - C_{1} \end{aligned}$$

(3.6)

and

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} + \varepsilon _{0}) (x + \alpha |y|) - C_{1}\ge (\lambda _{1}-\varepsilon _{0})x - C_{1} \end{aligned}$$

(3.7)

on $\{(x,y): x+\alpha |y|>X_0\}\cup \Omega _1$ and $t \in [0,\omega ]$.

Take $C_2>0$ such that

$$\begin{aligned} -C_{2}\le \min _{t \in [0, \omega ], (x,y)\in \Omega _2}\left\{ f(t, x, y)-\left( \lambda _{1}-\varepsilon _{0}\right) (x+|y|)\right\} , \end{aligned}$$

hence

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} - \varepsilon _{0}) (x + |y|) - C_{2}\ge (\lambda _{1} - \varepsilon _{0})x - C_{2} \end{aligned}$$

(3.8)

and

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} - \varepsilon _{0}) (x + |y|) - C_{2}\ge (\lambda _{1}+\varepsilon _{0})x - C_{2} \end{aligned}$$

(3.9)

on $\{(x,y): x+|y|<-X_0\}\cup \Omega _2$ and $t \in [0,\omega ]$.

Let $C= \max \left\{ C_{1} , C_{2}\right\}$. We can derive from (3.7) and (3.8) that

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} - \varepsilon _{0}) x - C, \quad \forall (t, x, y) \in [0, \omega ] \times \mathbb {R}^{2}, \end{aligned}$$

(3.10)

from (3.6) and (3.9) that

$$\begin{aligned} f(t, x, y) \ge (\lambda _{1} + \varepsilon _{0})x- C, \quad \forall (t, x, y) \in [0, \omega ] \times \mathbb {R}^{2}. \end{aligned}$$

(3.11)

(iii) Let $\varphi _0(t)\equiv 1, \forall t\in [0,\omega ]$, then

$$\begin{aligned} \int _0^\omega \varphi (t)\varphi _0(t)dt\ge c\int _0^\omega \varphi (t)dt=\delta \Vert \varphi _0\Vert _C, \end{aligned}$$

and hence $\varphi _0\in P_{1}$. Let

$$\begin{aligned} D= \left\{ u\in C^{1} [0, \omega ]: \,\text {there exists some} \, \mu \ge 0 \, \text {such that} \, u = Au + \mu \varphi _0\right\} . \end{aligned}$$

We claim that there is $M > 0$ such that $\Vert u\Vert _C \le M, \forall u \in D$. Indeed, if $u_{0} \in D$, then there is $\mu _{0}\ge 0$ such that by (3.11)

$$\begin{aligned} u_0 (t)&=(Au_0) (t) + \mu _0 \varphi _0(t)\\&= \int _0^\omega G(t,s) f(s, u_{0}(s), u'_0(s)) ds + \mu _{0}\varphi _0(t)\\&\ge \int _0^\omega G(t,s) f(s, u_{0}(s), u'_0(s)) ds\\&\ge \int _0^\omega G(t,s) \left( (\lambda _{1} + \varepsilon _0 )u_0 (s) - C \right) ds \\&= (\lambda _1 + \varepsilon _0 ) (Lu_0)(t) - C(L\varphi _0)(t), \end{aligned}$$

i.e.,

$$\begin{aligned} u_0 (t) \ge (\lambda _1 + \varepsilon _0 ) (Lu_0)(t) - C(L\varphi _0)(t). \end{aligned}$$

(3.12)

Multiplying (3.12) by $\varphi (t)$ on both sides and integrating over $[0,\omega ]$, from (2.4) we obtain

$$\begin{aligned} \int _0^\omega u_0(t) \varphi (t)dt&\ge (\lambda _1 + \varepsilon _0) \int _0^\omega (Lu_0)(t)\varphi (t)dt- C \int _0^\omega (L\varphi _0) (t) \varphi (t)dt\\&= \Big (1 + \frac{\varepsilon _0}{\lambda _1}\Big ) \int _0^\omega u_0 (t) \varphi (t) dt- C \int _0^\omega (L\varphi _0) (t)\varphi (t) dt. \end{aligned}$$

Thus by (2.4) we have

$$\begin{aligned} \int _0^\omega u_0(t) \varphi (t) dt\le \frac{\lambda _1 C}{\varepsilon _0} \int _0^\omega (L\varphi _0)(t) \varphi (t) dt= \frac{C}{\varepsilon _0} \int _0^\omega \varphi (t) dt. \end{aligned}$$

(3.13)

By the definition of D,

$$\begin{aligned}&u_0(t) - (\lambda _1 - \varepsilon _0) (Lu_0)(t) +C(L\varphi _0)(t)\\&\quad = (Au_0)(t) + \mu _{0}\varphi _0(t)-(\lambda _1- \varepsilon _0)(Lu_0)(t)+C(L\varphi _0)(t)\\&\quad = L[(Fu_0)-(\lambda _1- \varepsilon _0)u_0 +C \varphi _0](t) + \mu _{0}\varphi _0(t), \end{aligned}$$

where $(Fu_0)(t)=f(t,u_0(t),u'_0(t))$. In light of (3.10), one has

$$\begin{aligned} (Fu_0)-(\lambda _1- \varepsilon _0)u_0 +C\varphi _0\in P, \end{aligned}$$

which implies that $L[(Fu_0)-(\lambda _1- \varepsilon _0)u_0 +C\varphi _0] \in P_1$ from Lemma 2.4. So we have

$$\begin{aligned} u_0 - (\lambda _1 - \varepsilon _0)Lu_0+CL\varphi _0\in P_1. \end{aligned}$$

Therefore, by (2.4) and (3.13), we have

$$\begin{aligned}&\Vert u_0 - (\lambda _1 - \varepsilon _0)Lu_0+CL\varphi _0\Vert _C \\&\quad \le \frac{1}{\delta } \int _0^\omega \varphi (t) [u_0 - (\lambda _1 - \varepsilon _0)Lu_0+ CL\varphi _0](t) dt\\&\quad = \frac{1}{\delta \lambda _1} \left( \varepsilon _0 \int _0^\omega \varphi (t) u_0(t) dt +C \int _0^\omega \varphi (t) dt\right) \\&\quad \le \frac{1}{\delta \lambda _1} \left( C \int _0^\omega \varphi (t) dt+C \int _0^\omega \varphi (t) dt\right) \le \frac{2C \omega }{\delta \lambda _1}\Vert \varphi \Vert _C. \end{aligned}$$

Since $(\lambda _1- \varepsilon _0) r(L) < 1$, the operator $I - (\lambda _1- \varepsilon _0)L$ has the bounded inverse operator $(I - (\lambda _1- \varepsilon _0)L)^{-1}$ in $C[0,\omega ]$. Therefore, there is $M>0$ such that $\Vert u\Vert _C \le M, \forall u \in D$.

(iv) By (3.3), it is easy to see that there exists $M_1>2M$ such that

$$\begin{aligned} \int _{0}^{M_1}\frac{\rho d\rho }{H_M(\rho )+B}>2M, \end{aligned}$$

(3.14)

where $B=\frac{\pi ^2}{\omega ^2} M$.

Let $R>\max \{r, M_1\}$ and we will show that

$$\begin{aligned} u-A u \ne \mu \varphi _0, \quad \forall u \in \partial B_R, \quad \mu \ge 0. \end{aligned}$$

(3.15)

If it does not hold, there exist $u_2\in \partial B_R$ and $\mu _1\ge 0$ such that

$$\begin{aligned} u_2 - A u_2 =\mu _1\varphi _0, \end{aligned}$$

(3.16)

thus $u_2 \in D,$$\Vert u_2\Vert _C \le M.$ We can derive from (3.4) and (3.16) that

$$\begin{aligned} u_2''(t)&=-a(t)u_2(t)+f\big (t,u_2(t),u_2'(t)\big )\nonumber \\&\le \Vert a\Vert _C \Vert u_2\Vert _C + |f\big (t,u_2(t),u_2'(t)\big )|\nonumber \\&\le \frac{\pi ^2}{\omega ^2} M + |f\big (t,u_2(t),u_2'(t)\big ) |\le H_M(|u'_2(t)|)+B. \end{aligned}$$

(3.17)

In the same way,

$$\begin{aligned} -u_2''(t)&=a(t)u_2(t)-f\big (t,u_2(t),u_2'(t)\big )\nonumber \\&\le \Vert a\Vert _C \Vert u_2\Vert _C + |f\big (t,u_2(t),u_2'(t)\big ) |\nonumber \\&\le \frac{\pi ^2}{\omega ^2} M + |f\big (t,u_2(t),u_2'(t)\big )|\le H_M(|u'_2(t)|)+B. \end{aligned}$$

(3.18)

We will use (3.14), (3.17) and (3.18) to show that $\Vert u'_2\Vert _C \le M_1$.

Let $u'_2(t) \not \equiv 0$. Since $u_2(0)=u_2(\omega )$, there exist $t_0 \in (0, \omega )$ and $t_1 \in [0,\omega ]$ with $t_1 \ne t_0$ such that

$$\begin{aligned} u'_2(t_0) = 0, \Vert u'_2\Vert _C =|u'_2(t_1)| >0. \end{aligned}$$

(3.19)

Case 1. $u'_2(t_1) >0, t_0 < t_1$. Set

$$\begin{aligned} s_1=\sup \{ s\in [t_0, t_1) : u'_2 (s) = 0\}. \end{aligned}$$

(3.20)

Then $s_1 < t_1$ and by the definition of supremum,

$$\begin{aligned} u'_2(t) > 0, t \in (s_1, t_1]; \quad u'_2(s_1) = 0. \end{aligned}$$

(3.21)

Hence, for every $t\in [s_1, t_1]$, by (3.17) we have

$$\begin{aligned} \frac{u'_2(t)u''_2(t)}{H_M(u'_2(t))+B} \le u'_2(t), t\in [s_1,t_1]. \end{aligned}$$

(3.22)

Then integrating the inequality (3.22) over $[s_1,t_1]$ and making the variable transformation $\rho =u'_2(t)$, we have

$$\begin{aligned} \int _{0}^{u'_2(t_1)}\frac{\rho d\rho }{H_M(\rho )+B}=\int _{u'_2(s_1)}^{u'_2(t_1)}\frac{\rho d\rho }{H_M(\rho )+B} \le u_2(t_1)-u_2(s_1) \le 2M. \end{aligned}$$

From this inequality and (3.14) it follows that $u'_2(t_1) \le M_1$. Hence, $\Vert u'_2\Vert _C =u'_2(t_1) \le M_1.$

Case 2. $u'_2(t_1)>0 , t_0 > t_1$. Set

$$\begin{aligned} s_2=\inf \{s\in (t_1, t_0] : u'_2 (s) = 0\}. \end{aligned}$$

(3.23)

Then $t_1 < s_2$ and by the definition of infimum,

$$\begin{aligned} u'_2(t) > 0, t\in [t_1, s_2); \quad u'_2(s_2) = 0. \end{aligned}$$

(3.24)

Hence, for every $t\in [t_1, s_2]$, by (3.18) we have

$$\begin{aligned} \frac{-u'_2(t)u''_2(t)}{H_M(u'_2(t))+B} \le u'_2(t), t\in [t_1,s_2]. \end{aligned}$$

(3.25)

Then integrating the inequality (3.25) over $[t_1,s_2]$ and making the variable transformation $\tilde{\rho }=u'_2(t)$, we have

$$\begin{aligned} \int _{0}^{u'_2(t_1)}\frac{\tilde{\rho } d\tilde{\rho }}{H_M(\tilde{\rho })+B}=- \int _{u'_2(t_1)}^{u'_2(s_2)}\frac{\tilde{\rho } d\tilde{\rho }}{H_M(\tilde{\rho })+B} \le u_2(s_2)-u_2(t_1) \le 2M. \end{aligned}$$

From this inequality and (3.14) it follows that $u'_2(t_1) \le M_1$. Hence, ${\Vert u'_2\Vert }_C =u'_2(t_1) \le M_1.$

Case 3. $u'_2(t_1)<0 , t_0 < t_1$. Set

$$\begin{aligned} s_3=\sup \{s\in [t_0, t_1) : u'_2 (s) = 0\}. \end{aligned}$$

(3.26)

Then $s_3 < t_1$ and by the definition of supremum,

$$\begin{aligned} u'_2(t) < 0, t \in (s_3, t_1]; \quad u'_2(s_3) = 0. \end{aligned}$$

(3.27)

Hence, for every $t\in [s_3, t_1]$, by (3.18) we have

$$\begin{aligned} \frac{-u'_2(t)u''_2(t)}{H_M(-u'_2(t))+B} \ge u'_2(t), t\in [s_3,t_1]. \end{aligned}$$

(3.28)

Then integrating the inequality (3.28) over $[s_3,t_1]$ and making the variable transformation $\tau =-u'_2(t)$, we have

$$\begin{aligned} -\int _{0}^{-u'_2(t_1)}\frac{\tau d\tau }{H_M(\tau )+B}=- \int _{-u'_2(s_3)}^{-u'_2(t_1)}\frac{\tau d\tau }{H_M(\tau )+B} \ge u_2(t_1)-u_2(s_3) \ge -2M. \end{aligned}$$

From this inequality and (3.14), it follows that $-u'_2(t_1) \le M_1$. Hence, ${\Vert u'_2\Vert }_C =-u'_2(t_1) \le M_1.$

Case 4. $u'_2(t_1) <0 , t_0 > t_1$. Set

$$\begin{aligned} s_4=\inf \{ s\in (t_1, t_0] : u'_2 (s) = 0\}. \end{aligned}$$

(3.29)

Then $t_1 < s_4$ and by the definition of infimum,

$$\begin{aligned} u'_2(t) < 0, t \in [t_1, s_4); \quad u'_2(s_4) = 0. \end{aligned}$$

(3.30)

Hence, for every $t\in [t_1, s_4]$, by (3.17) we have

$$\begin{aligned} \frac{-u'_2(t)u''_2(t)}{H_M(-u'_2(t))+B} \le -u'_2(t),\ t\in [t_1,s_4]. \end{aligned}$$

(3.31)

Then integrating the inequality (3.31) over $[t_1,s_4]$ and making the variable transformation $\tilde{\tau }=-u'_2(t)$, we have

$$\begin{aligned} \int _{0}^{-u'_2(t_1)}\frac{\tilde{\tau } d\tilde{\tau }}{H_M(\tilde{\tau })+B}=- \int _{-u'_2(t_1)}^{-u'_2(s_4)}\frac{\tilde{\tau } d\tilde{\tau }}{H_M(\tilde{\tau })+B} \le u_2(t_1)-u_2(s_4) \le 2M, \end{aligned}$$

From this inequality and (3.14) it follows that $-u'_2(t_1) \le M_1$. Hence, ${\Vert u'_2\Vert }_C =-u'_2(t_1) \le M_1.$

In summary, $\Vert u'_2\Vert _C\le M_1$ and thus $\Vert u_2\Vert _{C^1}\le M_1$, which is a contradiction to $\Vert u_2\Vert _{C^1}=R>M_1$. Therefore, (3.15) holds.

According to Lemma 2.6, one has

$$\begin{aligned} \deg (I-A, B_R,0)=0. \end{aligned}$$

(3.32)

From (3.7) and (3.32) we have

$$\begin{aligned} \deg (I-A, B_R\backslash \overline{B}_r, 0) =\deg (I-A, B_R, 0)-\deg (I-A, B_r, 0)=-1. \end{aligned}$$

As a result, A has at least one fixed point on $B_R \backslash \overline{B}_r$, which means that PBVP (1.6) has at least one nontrivial solution.

Example

Consider the following PBVP

$$\begin{aligned} \left\{ \begin{array}{l}u''(t)+\frac{1}{16}u(t)=f(u(t),u'(t)),\quad t\in [0,2\pi ],\\ u(0)=u(2\pi ),\quad u'(0)=u'(2\pi ),\end{array}\right. \end{aligned}$$

(3.33)

where

$$\begin{aligned} f(x,y)=\left\{ \begin{array}{ll} \frac{1}{72}(x+2|y|), &{} x+2|y|\le 0,\\ (x+2|y|)^2, &{} x+2|y|>0. \end{array} \right. \end{aligned}$$

Obviously, $a(t)=\frac{1}{16}$ and f(x, y) satisfy $(C_1)$ and $(C_2)$, respectively. It is easy to see that $l_2 = 2\sqrt{2}$ and $l_3 = \frac{1}{2}$.

It follows from (2.1) and Lemma 2.2 that for $u\in C[0,\omega ]$,

$$\begin{aligned} (Lu)(t)|\le \int _0^\omega |G(t,s)u(s)|ds\le 2\pi l_2\Vert u\Vert _C \end{aligned}$$

and the spectral radius $r(L)\le \Vert L\Vert \le 2\pi l_2$, thus $\lambda _1\ge \frac{1}{2\pi l_2}=\frac{1}{4\sqrt{2}\pi }$.

Obviously, for $\alpha =2$, we have

$$\begin{aligned} \liminf _{x + 2|y| \rightarrow +\infty }\frac{f(x,y)}{x+2|y|}=\liminf _{x + 2|y| \rightarrow +\infty }\frac{(x+2|y|)^2}{x+2|y|}=+\infty >\lambda _{1}. \end{aligned}$$

If $x+2|y|\le 0$, then $x+|y|\le x+2|y|\le 0$ and $|x+2|y||\le |x+|y||$, hence

$$\begin{aligned} \limsup _{x + |y| \rightarrow -\infty }\frac{f(x,y)}{x+|y|}=\limsup _{x + |y| \rightarrow -\infty }\frac{x+2|y|}{72(x+|y|)}\le \frac{1}{72} < \lambda _{1}; \end{aligned}$$

if $x+2|y|>0$, then

$$\begin{aligned} \limsup _{x + |y| \rightarrow -\infty }\frac{f(x,y)}{x+|y|}=\limsup _{x + |y| \rightarrow -\infty }\frac{(x+2|y|)^2}{x+|y|}\le 0< \lambda _{1}. \end{aligned}$$

Therefore, $(C_3)$ and $(C_4)$ are satisfied.

Take $a=b=\frac{1}{36}, r=\frac{1}{180}$, and thus $2 \pi (a+b) \max \{l_2, l_3\} = \frac{2\sqrt{2}\pi }{9} <1.$ If $x+2|y|\le 0$,

$$\begin{aligned} |f(x,y)|\le \frac{|x|}{72}+\frac{|y|}{36}\le a|x|+b|y|; \end{aligned}$$

if $x+2|y|>0$ and $(x,y)\in [-r,r]^2$,

$$\begin{aligned} |f(x,y)|\le (|x|+2|y|)^2\le 5r|x|+4r|y|\le a|x|+b|y|. \end{aligned}$$

So $(C_5)$ holds.

For any $M>0$ define $H_M(\rho )=(M+2\rho )^2+1$ on $\mathbb {R}^+$, it is clear that (3.3) holds. If $-1\le x+2|y|\le 0$,

$$\begin{aligned} |f(x,y)|=\frac{1}{72}|x+2|y||\le 1\le H_M(|y|); \end{aligned}$$

if $x+2|y|<-1$ and $(x,y)\in [-M,M]\times \mathbb {R}$,

$$\begin{aligned} |f(x,y)|=\frac{1}{72}|x+2|y||\le |x+2|y||^2\le (|x|+2|y|)^2\le H_M(|y|); \end{aligned}$$

if $x+2|y|>0$ and $(x,y)\in [-M,M]\times \mathbb {R}$,

$$\begin{aligned} |f(x,y)|=|x+2|y||^2\le (|x|+2|y|)^2\le H_M(|y|). \end{aligned}$$

Therefore, (3.4) also holds. By Theorem 3.1, we know that PBVP (3.4) has at least one nontrivial solution.