Singular and Nonsingular Third-Order Periodic Boundary Value Problems with Indefinite Weight

Abstract

We are concerned with the existence of positive periodic solutions of third-order periodic boundary value problems

$$\begin{aligned} \begin{array}{c} u'''(t)+k_1u''(t)+k_2u'(t)+k_3u(t)=\lambda a(t)f(u),~~t\in (0,\omega ),\\ u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega ), \end{array} \end{aligned}$$

where $k_1,k_3\in (0,\infty )$ and $k_2\in (0,(\frac{\pi }{\omega })^2)$ are constants. $\lambda$ is a positive parameter. The weight function $a\in C([0,\omega ],\mathbb {R})$ may change sign. $f\in C([0,\infty ),\mathbb {R})$ with $f(0):=\lim \nolimits _{u\rightarrow 0^+}f(u)>0$. We show that there exists a constant $\lambda ^*>0$, such that the above problem has at least one positive periodic solution for $\lambda \in (0,\lambda ^*)$ where f(0) is bounded. This result is based upon bifurcation theory and Leray–Schauder fixed point theorem. On the other hand, by using Krasnoselskii’s fixed point theorem in a cone, we show that there exists a positive constant $\lambda _0$ such that for all $\lambda \in (0,\lambda _0)$, the above problem has at least one positive periodic solution where f(0) is unbounded, namely that f has a singularity at $u=0$. And this result is applicable to weak as well as strong singularities.

1 Introduction

In this paper, we proceed with the investigation of positive solutions for a class of third-order differential equations

$$\begin{aligned} u'''(t)+k_1u''(t)+k_2u'(t)+k_3u(t)=\lambda a(t)f(u(t)),~~~t\in (0,\omega ) \end{aligned}$$

(1.1)

subjects to the periodic boundary conditions

$$\begin{aligned} u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega ), \end{aligned}$$

(1.2)

where $k_1,k_3\in (0,\infty )$ and $k_2\in (0,(\frac{\pi }{\omega })^2)$ are constants and $\lambda$ is a positive parameter. $a:[0,\omega ]\rightarrow \mathbb {R}$ is a continuous function which may change sign. Define that $a^{+}(t):=\max \nolimits _{t\in [0,\omega ]}\{a(t),0\}$ and $a^{-}(t):=\max \nolimits _{t\in [0,\omega ]}\{-a(t),0\}$. $f:[0,\infty )\rightarrow \mathbb {R}$ is a continuous function with $f(0):=\lim \nolimits _{u\rightarrow 0^+}f(u)>0$.

For the nonlinear third-order periodic boundary value problems, we recall the following results. In [1], by using Schauder fixed point theorem, together with perturbation technique, Kong and his collaborators established the existence of positive periodic solutions for (1.1)–(1.2) with $k_1=k_2=0$, $k_3\in (0,\frac{1}{\sqrt{3}}))$ and $\lambda =1$, where a(t)f(u) is a nonnegative function defined on $[0,\omega ]\times (0,\infty )$ and is nonincreasing in $u>0$ for $t\in [0,\omega ]$ and

$$\begin{aligned} \lim \limits _{u\rightarrow 0^+}f(u)=+\infty ,~~\lim \limits _{u\rightarrow +\infty }f(u)=0. \end{aligned}$$

Typical example of nonlinearity in [1] is of the form $f(u)=\frac{1}{u^{\nu }}$. Afterwards, Sun [2] and Chu [3] used different methods about fixed point theorems to generalize the results of [1] under the a(t)f(u) satisfies suitable conditions.

In [4], Feng proved the existence of positive periodic solutions of (1.1)–(1.2) with $k_1>0$, $k_2>0$, $k_3=0$ and $\lambda =1$, by using the well-known Guo-Krasnoselskii’s fixed point theorem where a(t)f(u) is nonnegative. Recently, Ren and her collaborators [5, 6] made an exhaustive study of Green’s function of the third-order linear differential equation $u'''(t)+\breve{a} u''(t)+\breve{b} u'(t)+\breve{c}u(t)=\breve{h}(t)$. Also, by applying the properties of the Green’s function, the authors showed that the existence of positive periodic solutions corresponding to third-order singular differential equation

$$\begin{aligned} u'''(t)+\breve{a} u''(t)+\breve{b} u'(t)+\breve{c}u(t)=\breve{f}(t,u)+\breve{e}(t),~~t\in (0,\omega ), \end{aligned}$$

where $\breve{f}\in \text{ Car }~(\mathbb {R}\times (0,+\infty ),(0,+\infty ))$ is nonnegative and has a singularity at $u=0$, $\breve{e}\in L^{1}(\mathbb {R})$ is an $\omega$-periodic function.

Notice that all the aforementioned results in these works use a key condition that the nonlinear term is nonnegative, that is, a is nonnegative. The terminology “indefinite” was probably introduced by Hess and Kato [7] under the framework of linear eigenvalue problems. In a recent series of papers, it is worth mentioning that the qualitative study of the solutions to the indefinite equations has been treated by many authors in both ODEs and PDEs. For example, by using Schauder fixed point theorem, Hai [8] showed the existence of positive solutions of the semilinear elliptic boundary value problem with $\hat{a}$ changing-sign

$$\begin{aligned} \left\{ \begin{aligned}&-\Delta y=\hat{\lambda } \hat{a}(x)\hat{f}(y)~~~\text{ in }~~~\Omega ,\\&y=0~~~~~~~~~~~~~~~~~~~\text{ on }~~~\partial \Omega . \end{aligned} \right. \end{aligned}$$

And the study the existence of positive periodic solutions for singular differential equations with an indefinite weight began from 2010, Bravo and Torres [9] discussed a special second-order differential equation (Emden–Flower equation) with $\hat{b}$ changing-sign

$$\begin{aligned} x''(t)=\frac{\hat{b}(t)}{x^{3}(t)},~~~t\in (0,\omega ). \end{aligned}$$

Recently, the study of existence of positive periodic solutions for second-order singular differential equations has attracted by many researchers’ attention, see [10,11,12,13].

Motivated by the above works and focus on the indefinite weight, our main results are the following.

Theorem 1.1

Assume that $(A_1)-(A_2)$ are satisfied.

$(A_1)$ If f(0) is bounded, then there exists two constants $\epsilon$ and $\delta \in (0,1)$ such that $f(x)\ge \delta f(0)$ for $x\in (0,\epsilon )$.

$(A_2)$ $a\in C([0,\omega ],\mathbb {R})$, $a\not \equiv 0$, and there exists a number $k>1$ such that

$$\begin{aligned} \int ^\omega _0G(t,s)a^{+}(s)ds\ge k\int ^\omega _0G(t,s)a^{-}(s)ds,~~~t\in [0,\omega ]. \end{aligned}$$

(1.3)

Then there exists a positive constant $\lambda ^{*}>0$ such that (1.1)–(1.2) has at least one positive periodic solution for $\lambda \in (0,\lambda ^{*})$.

Theorem 1.2

If f(0) is unbounded, that is, f has a singularity at $u=0$ in the form of $f(u):=\frac{1}{u^{\nu }}$ where $\nu \in (0,\infty )$ is a constant. Then suppose that $1-\sigma ^*\omega >0$ and

$$\begin{aligned} \Vert a^{+}\Vert _{\infty }>\frac{\Vert a^{-}\Vert _{\infty }}{\sigma ^{\nu +1}\omega \sigma ^*} \end{aligned}$$

are satisfied. Then there exists a positive constant $\lambda _0$ such that (1.1)–(1.2) admits at least one positive periodic solution for $\lambda \in (0,\lambda _0)$.

Remark 1.3

In the case that f(0) is bounded, condition $(A_2)$ covers the case that a(t) is positive for all $t\in [0,\omega ]$. In this paper, we show the global structure of positive periodic solutions for (1.1)–(1.2) with the positive weight by using bifurcation theory and we mention that our result is new in the case the weight function is positive.

Remark 1.4

In the case that f(0) is unbounded, that is, this problem can be regarded as a class of indefinite singular problems.

2 Preliminary Results

Lemma 2.1

([15, Lemma 2.4]) Let $\tilde{E}$ be a Banach space, and let $\{C_{n}\}$ be a family of closed connected subsets of $\tilde{E}$. Assume that

• (i) there exists $z_{n}\in C_{n}$, $n=1,2,\ldots$, and $z_{*}\in \tilde{E}$ such that $z_{n}\rightarrow z_{*}$;

• (ii) $\lim \nolimits _{n\rightarrow \infty }r_{n}=\infty$, where ${r_{n}=\sup \{\Vert x\Vert _{\infty }|x\in C_{n}\}}$;

• (iii) for every $R>0$, $(\bigcup ^{\infty }_{n=1}C_{n})\cap B_{R}$ is a relatively compact set of $\tilde{E}$, where

  $$\begin{aligned} B_{R}=\{x\in \tilde{E}|~\Vert x\Vert _{\infty }\le R\}. \end{aligned}$$

  Then $\mathcal {D}:=\limsup \nolimits _{n\rightarrow +\infty } C_{n}$ contains an unbounded component $\mathcal {C}$ with $z_{*}\in \mathcal {C}$.

Lemma 2.2

Assume that $(A_1)$ is satisfied and $\delta \in (0,1)$. Then there exists a number $\bar{\lambda }>0$, such that (1.1)–(1.2) has a positive periodic solution $u_{\lambda }$ for all $\lambda \in (0,\bar{\lambda })$ with

$$\begin{aligned} u_{\lambda }\ge \lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds. \end{aligned}$$

To use bifurcation theorem to prove Lemma 2.2, we extend f to

$$\begin{aligned} \tilde{f}(s)=\left\{ \begin{array}{c} f(s), s\le 0\\ f(0), s<0 \end{array} \right. \end{aligned}$$

Let $X=C[0,\omega ]$ be the Banach space with its usual normal $\Vert u\Vert _{\infty }:=\max \limits _{u\in [0,\omega ]}|u(t)|$. Let us consider

$$\begin{aligned} \begin{aligned}&u'''(t)+k_1u''(t)+k_2u'(t)+k_3u(t)=\lambda a^{+}(t)\tilde{f}(u(t)),~~~t\in (0,\omega ),\\&u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega ). \end{aligned} \end{aligned}$$

(2.1)

Define a linear operator L: $E\rightarrow X$

$$\begin{aligned} Lu:=u'''(t)+k_1u''(t)+k_2u'(t)+k_3u(t),~u\in E \end{aligned}$$

with the Domain

$$\begin{aligned} E:=\{u\in C^3[0,1]:~u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega )\}. \end{aligned}$$

Next, we show that the conditions of $k_1,~k_2,~k_3$ of inverse positive operator L.

Lemma 2.3

If (2.1) has nontrivial solutions, then $k_1,k_3\in (0,\infty )$ and $k_2\in (0,(\frac{\pi }{\omega })^{2})$.

Proof

Let y be a nontrivial solution of (2.1) and

$$\begin{aligned} y'''(t)+k_1y''(t)+k_2y'(t)+k_3y(t)=(y'+k_1y)(y''+k_2y)=0 \end{aligned}$$

(2.2)

for some $k_1,~k_2\in \mathbb {R}$ and $k_3=k_1k_2$.

If $y''+k_2y=0$, $y(0)=y(\omega ),~y'(0)-y'(\omega )=1$, then we have $k_2\in (0,(\frac{\pi }{\omega })^2)$ and consequently

$$\begin{aligned} y(t)=\frac{\cos (\frac{\sqrt{k_2}}{2}\omega )}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}}{2}\omega )} \cos (\sqrt{k_2}t)+\frac{1}{2\sqrt{k_2}}\sin (\sqrt{k_2}t). \end{aligned}$$

In this case we obtain that $k_1,k_3\in \mathbb {R}$ and $k_2\in (0,\left( \frac{\pi }{\omega }\right) ^2)$.

If $y''+k_2y\ne 0$, then

$$\begin{aligned} y''+k_2y\in \text{ ker }(y'+k_1y). \end{aligned}$$

This implies that $y'+k_1y=0$, $y(0)-y(\omega )=1$, i.e., $k_1\in (0,\infty )$ and

$$\begin{aligned} y''+k_2y=\frac{e^{-k_1t}}{1-e^{-k_1\omega }},~~y(0)=y(\omega ),~~y'(0)=y'(\omega ). \end{aligned}$$

A simple computation yields

$$\begin{aligned} y(t)= & {} \int ^t_0 \frac{\cos (\sqrt{k_2}(t-s-\frac{\omega }{2}))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})} \frac{e^{-k_1t}}{1-e^{-k_1\omega }}ds\\{} & {} +\int ^\omega _t\frac{\cos (\sqrt{k_2}(t-s+\frac{\omega }{2}))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})} \frac{e^{-k_1t}}{1-e^{-k_1\omega }}ds, \end{aligned}$$

where $k_2\in (0,\left( \frac{\pi }{\omega }\right) ^2)$ from [14, Lemma 3.2]. Since $k_3=k_1k_2$, then $k_3\in (0,\infty )$. $\square$

It is worth mentioning that if L is inverse positive, then L is invertible on E and the Green’s function G(t, s) related to the problem

$$\begin{aligned} Ly=h(t),~~(t,s)\in [0,\omega ]\times [0,\omega ] \end{aligned}$$

is positive, here $h\in X$ and $h\not \equiv 0$.

To do that, let us define a linear operator

$$\begin{aligned} L_1:=y''(t)+k_2y(t),~D(L_1):=\{y\in C^2[0,\omega ]:y(0)=y(\omega ),~y'(0)=y'(\omega )\}. \end{aligned}$$

Then the Green function of $L_1y=\hat{h}_1(t)$ is

$$\begin{aligned} \begin{aligned} G_1(t,s)= \left\{ \begin{array}{cc} \frac{\cos (\sqrt{k_2}(t-s-\frac{\omega }{2}))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})}, &{} 0\le s\le t\le \omega , \\ \frac{\cos (\sqrt{k_2}(t-s+\frac{\omega }{2}))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})}, &{} 0\le t\le s\le \omega . \end{array} \right. \end{aligned} \end{aligned}$$

Define a linear operator

$$\begin{aligned} L_2:=y'+k_1y,~D(L_2):=\{y\in C^1[0,\omega ]:y(0)=y(\omega )\}. \end{aligned}$$

Then the Green’s function of $L_2y=\hat{h}_2(t)$ is

$$\begin{aligned} \begin{aligned} G_2(t,s)= \left\{ \begin{array}{cc} \frac{e^{k_{1}(\omega +s-t)}}{e^{k_1\omega }-1}, &{} 0\le s\le t\le \omega ,\\ \frac{e^{k_1(s-t)}}{e^{k_1\omega }-1}, &{} 0\le t\le s\le \omega . \end{array} \right. \end{aligned} \end{aligned}$$

Obviously,

$$\begin{aligned} Ly=L_2(L_1y), \end{aligned}$$

then Green’s function of $Ly=h(t)$ is

$$\begin{aligned} G(t,s):=\int ^\omega _0G_2(t,x)G_1(x,s)dt,~~~~(t,s)\in [0,\omega ]\times [0,\omega ], \end{aligned}$$

i.e.,

$$\begin{aligned} \begin{aligned} G(t,s)= \left\{ \begin{array}{ll} \dfrac{1}{k^2_1+k_2}\left[ \frac{k_2\cos (\sqrt{k_2}(\frac{\omega }{2}-s+t)) +\sqrt{k_2}\sin (\sqrt{k_2}(\frac{\omega }{2}-s+t))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})}+\frac{e^{k_1(s-t)}}{e^{k_1\omega }-1}\right] , &{} 0\le t\le s\le \omega ,\\ \dfrac{1}{k^2_1+k_2}\left[ \frac{k_2\cos (\sqrt{k_2}(\frac{\omega }{2}+s-t)) -\sqrt{k_2}\sin (\sqrt{k_2}(\frac{\omega }{2}+s-t))}{2\sqrt{k_2}\sin (\frac{\sqrt{k_2}\omega }{2})}+\frac{e^{k_1(\omega -t+s)}}{e^{k_1\omega }-1}\right] , &{} 0\le s\le t\le \omega . \end{array} \right. \end{aligned}\nonumber \\ \end{aligned}$$

(2.3)

Lemma 2.4

Let $k_2\in (0,(\frac{\pi }{\omega })^2)$ and $k_1,k_3\in (0,\infty )$. Then

$$\begin{aligned} G(t,s)>0,~~~~~~\text{ for }~~~(t,s)\in [0,\omega ]\times [0,\omega ]. \end{aligned}$$

Proof

It is an immediate consequence of the fact that for $j=1,2$,

$$\begin{aligned} G_{j}(t,s)>0,~~~~~~\text{ for }~~~(t,s)\in [0,\omega ]\times [0,\omega ]. \end{aligned}$$

$\square$

Lemma 2.5

(Maximum principle) Let $k_2\in (0,(\frac{\pi }{\omega })^2)$, $k_1,k_3\in (0,\infty )$ and $L(\tilde{y}(t))\ge 0$ for $\tilde{y}\in E$, then $\tilde{y}(t)\ge 0$ on $[0,\omega ]$.

Proof

Let $\tilde{y}(t)\in E$ satisfies $L(\tilde{y}(t))\ge 0$ on $[0,\omega ]$. The $\tilde{y}$ is a solution of BVP $L(\tilde{y}(t))=\tilde{h}(t)$ and $\tilde{h}(t)$ is a nonnegative continuous function. Thus the function $\tilde{y}$ is a solution of an integral equation

$$\begin{aligned} \tilde{y}=\int ^\omega _0G(t,s)\tilde{h}(s)ds\ge 0, \end{aligned}$$

(2.4)

since $G(t,s)>0$ on $[0,\omega ]\times [0,\omega ]$. $\square$

So (2.1) is equivalent to the integral equation

$$\begin{aligned} u(t)=\lambda \int ^\omega _0G(t,s)a^{+}(s)\tilde{f}(u(s))ds,~~~t\in (0,\omega ). \end{aligned}$$

Let P be the cone in X defined by

$$\begin{aligned} P:=\{u\in X|u(t)\ge 0,~~u(t)\ge \sigma \Vert u\Vert _{\infty },~~t\in [0,\omega ]\}, \end{aligned}$$

where $\sigma :=\frac{\sigma _*}{\sigma ^*}$, $\sigma ^*=\max \nolimits _{0\le t,s\le \omega }G(t,s)$, $\sigma _*=\min \nolimits _{0\le t,s\le \omega }G(t,s)$ and $A:P\rightarrow P$ be the map defined by

$$\begin{aligned} Au(t):=\lambda \int ^\omega _0G(t,s)a^{+}(s)\tilde{f}(u(s))ds,~~~t\in [0,\omega ]. \end{aligned}$$

(2.5)

Then from condition $(A_1)$ and the positivity of G(t, s), $A(P)\subset P$ and $A:P\rightarrow P$ is compact and continuous. So the existence of positive solutions for (2.1) is equivalent to the existence of a fixed point of A in P.

Next, we consider the principal eigenvalue of the linear problem

$$\begin{aligned} \begin{aligned}&u'''(t)+k_1u''(t)+k_2u'(t)+k_3u(t)=\lambda a^{+}(t)u(t),~~t\in (0,\omega ),\\&u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega ). \end{aligned} \end{aligned}$$

(2.6)

Lemma 2.6

Suppose that $a\in C([0,\omega ], \mathbb {R})$ with $a(t)\not \equiv 0$ on any subinterval of $[0,\omega ]$, and $a^{+}(t)>0$ for all $t\in [0,\omega ]$. Then (2.6) has a principal eigenvalue $\lambda _{1}$, which is positive and simple, and the corresponding eigenfunction $\varphi _1(t)$ is positive on $[0,\omega ]$.

Proof

From the definition of cone P, we know that P is normal and has nonempty interior. Obviously, $X=P-P$. Since

$$\begin{aligned} G(t,s)>0,~~~~(t,s)\in [0,\omega ]\times [0,\omega ], \end{aligned}$$

it follows that the linear operator

$$\begin{aligned} \bar{L}u(t):=\int ^\omega _0G(t,s)a^{+}(s)u(s)ds \end{aligned}$$

(2.7)

is a strong positive operator, that is, $\bar{L}u(t)\in \text{ int }~P$. By Krein–Rutman Theorem [16], Theorem 19.3(a)], the spectral radius $r(\bar{L})$ is positive, and there exists a function $\varphi _1\in E$ such that $\varphi _1>0$ on $[0,\omega ]$ and $\bar{L}\varphi _1=r(\bar{L})\varphi _1$. Thus, $\lambda _1=(r(\bar{L}))^{-1}>0$. Let $\bar{L}^*$ is conjugate operator of $\bar{L}$, and $\bar{L}^*\psi _1=\lambda _1\psi _1$, where $\psi _1\in E$ such that $\psi _1>0$ on $[0,\omega ]$ corresponding to $\lambda _1$. Since

$$\begin{aligned} \langle \bar{L}\varphi _1,\psi _1\rangle =\lambda _1\langle \varphi _1,\psi _1\rangle =\langle \varphi _1,\bar{L}^*\psi _1\rangle , \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ denotes the standard $L^2(0,\omega )$ inner product, then the eigenvalue $\lambda _1$ has multiplicity 1. Thus, $\lambda _1=(r(\bar{L}))^{-1}$ is the principal eigenvalue of (2.6). $\square$

Lemma 2.7

Assume that $(A_1)$ holds. If $u\in \partial \Omega _r$, $r>0$ and $\Omega _r:=\{t\in [0,\omega ]|~\Vert u\Vert _{\infty }<r\}$, then

$$\begin{aligned} \Vert Lu\Vert _{\infty }\le \lambda \widehat{M_r}\int ^\omega _0G(t,s)a^{+}(s)ds, \end{aligned}$$

where $\widehat{M_r}=1+\max _{0\le s\le r}\{\tilde{f}(s)\}$.

Proof

Since $\tilde{f}(u(t))\le \widehat{M_r}$ for $t\in [0,\omega ]$, it follows that

$$\begin{aligned} \begin{aligned} \Vert Lu\Vert _{\infty }&\le \lambda \int ^\omega _0G(t,s)a^{+}(s)\tilde{f}(u(s))ds\le \lambda \widehat{M_r}\int ^\omega _0G(t,s)a^{+}(s)ds. \end{aligned} \end{aligned}$$

$\square$

Lemma 2.8

Assume that $\{(\mu _k,y_{k})\}\subset (0,\infty )\times P$ is a sequence of positive solutions of (2.1). Suppose that $\mu _k<C_0$ for some constant $C_0>0$ and

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\Vert y''_k\Vert _{\infty }=\infty . \end{aligned}$$

(2.8)

Then

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\Vert y_k\Vert _{\infty }=\infty . \end{aligned}$$

(2.9)

Proof

Suppose by the way of contradiction that $\{\Vert y_{k}\Vert _{\infty }\}$ is bounded. It follows that

$$\begin{aligned} \Vert \mu _ka^{+}\tilde{f}(y_k)\Vert _{\infty }\le \beta , \end{aligned}$$

where $\beta$ is independent of k and is a constant. Thus,

$$\begin{aligned} y'''_k(t)+k_1y''_k(t)+k_2y'_k(t)+k_3y_k(t)=\mu _k a^{+}(t)\tilde{f}(y_k(t)) \end{aligned}$$

that $\{y'''_k\}$ is uniformly bounded in C[0, 1], and subsequently $\{y''_k\}$ is uniformly bounded in C[0, 1], which contradicts (2.8). $\square$

Proof of Lemma 2.2

Let $\Sigma$ be the closure of the set of positive periodic solutions of (2.1) in E. To prove Lemma 2.2, we use bifurcation approach to treat the case $\tilde{f}_{0}:=\lim \nolimits _{u\rightarrow 0^+}\frac{\tilde{f}(u)}{u}=\infty$. In this approach, it is crucial to construct a sequence of function $\{\tilde{f}^{[n]}\}$ such that $\{\tilde{f}^{[n]}\}$ is asymptotic linear at 0 and satisfies

$$\begin{aligned} \{\tilde{f}^{[n]}\}\rightarrow \tilde{f},~~~~(\tilde{f}^{[n]})_{0}\rightarrow \infty . \end{aligned}$$

From corresponding auxiliary equations, we obtain a sequence of unbounded components $\{C^{[n]}\}$ via nonlinear Krein–Rutman bifurcation theorem, see Dancer [17] and Zeidler [18]. So we can find unbounded components $\mathcal {C}$ which satisfies

$$\begin{aligned} \mathcal {C}\subset \limsup \limits _{n\rightarrow \infty } C^{[n]} \end{aligned}$$

and joins (0, 0) with $(0,\infty )$.

To apply the bifurcation theorem, we extend $\tilde{f}$ to a function $\{g^{[n]}\}:\mathbb {R}\rightarrow \mathbb {R}$ by

$$\begin{aligned} g^{[n]}(u)=\left\{ \begin{array}{c} nu,~~~u\in [0,\frac{1}{n}],\\ \tilde{f}(u),~~~u\in (-\infty ,0)\cup (\frac{1}{n},\infty ), \end{array} \right. \end{aligned}$$

then

$$\begin{aligned} ug^{[n]}(u)>0,~~~\forall ~|u|\in (0,\infty ), ~~\text{ and }~~(g^{[n]})_{0}=\lim \limits _{u\rightarrow 0}\frac{g^{[n]}(u)}{u}=n. \end{aligned}$$

Thus,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }(g^{[n]})_{0}=\infty . \end{aligned}$$

Let us consider the auxiliary family of the equations

$$\begin{aligned} \begin{aligned}&u'''+k_1u''+k_2u'+k_3u=\lambda a^{+}(t)g^{[n]}(u),~~~t\in (0,\omega ),\\&u(0)=u(\omega ),~~u'(0)=u'(\omega ),~~u''(0)=u''(\omega ). \end{aligned} \end{aligned}$$

(2.10)

Let $\zeta \in C(\mathbb {R})$ be such that

$$\begin{aligned} g^{[n]}(u)=(g^{[n]})_0u+\zeta (u)=nu+\zeta ^{[n]}(u), \end{aligned}$$

(2.11)

then

$$\begin{aligned} \lim \limits _{|u|\rightarrow 0}\frac{\zeta ^{[n]}(u)}{|u|}=0{.} \end{aligned}$$

Let us consider

$$\begin{aligned} Lu=\lambda a^{+}(t)(g^{[n]})_0u+\lambda a^{+}(t)\zeta ^{[n]}(u) \end{aligned}$$

(2.12)

as a bifurcation problem from the trivial solution $u\equiv 0$.

(2.12) can be converted to the equivalent the equation

$$\begin{aligned} \begin{aligned} u(t)&=\lambda \int ^\omega _0G(t,s)[\lambda a^{+}(s)(g^{[n]})_0u(s)+\lambda a^{+}(s)\zeta ^{[n]}(u(s))]\\&:=\lambda L^{-1}[a^{+}(\cdot )(g^{[n]})_0u(\cdot )](t)+\lambda L^{-1}[a^{+}(\cdot )\zeta ^{[n]}(u(\cdot ))](t). \end{aligned} \end{aligned}$$

(2.13)

Further we note that $\Vert L^{-1}[a^{+}(\cdot )\zeta ^{[n]}(u(\cdot ))]\Vert _{\infty }=o(\Vert u\Vert _{\infty })$ for u near 0 in E.

From the fact $(g^{[n]})_0>0$, the results of nonlinear Krein–Rutman theorem for (2.12) can be stated as follows: there exists a continuum $C^{[n]}$ of positive solutions of (2.12) joining $(\frac{\lambda _1}{(g^{[n]})_0})$ to infinity in P. Moreover, $C^{[n]}\backslash \{(\frac{\lambda _1}{(g^{[n]})_0},0)\}\subset \text{ int }~P$ and $(\frac{\lambda _1}{(g^{[n]})_0},0)$ is the only positive bifurcation point of (2.12) lying on trivial solutions line $u\equiv 0$.

Next, we show that $\{C^{[n]}\}$ satisfies all of the conditions of Lemma 2.1. Since

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{\lambda _1}{(g^{[n]})_0} =\lim \limits _{n\rightarrow \infty }\frac{\lambda _1}{n}=0, \end{aligned}$$

(2.14)

condition (i) in Lemma 2.1 is satisfied with $z^*=(0,0)$. Obviously

$$\begin{aligned} r_{n}=\sup \{|\lambda |+\Vert u\Vert _{\infty }|(\lambda ,u)\in C^{[n]}\}=\infty , \end{aligned}$$

so (ii) holds. From the Arzela–Ascoli Theorem and the definition of $g^{[n]}$, we can get that (iii) is satisfied. Therefore, the superior limit of $\{C^{[n]}\}$, that is, $\mathcal {D}$ contains an unbounded connected component $\mathcal {C}$ with $(0,0)\in \mathcal {C}$.

Let $\{\mu _{k},y_{k}\}\subset \mathcal {C}$ be such that $|\mu _{k}|+\Vert y_{k}\Vert _{\infty }\rightarrow \infty$ as $k\rightarrow \infty$. Then

$$\begin{aligned} \begin{array}{c} y'''_k(t)+k_1y''_k(t)+k_2y'_k(t)+k_3y_k(t)=\mu _ka^{+}(t)g^{[n]}(y_k(t)),~~t\in (0,\omega ),\\ y_k(0)=y_k(\omega ),~~y'_k(0)=y'_k(\omega ),~~y''_k(0)=y''_k(\omega ). \end{array} \end{aligned}$$

We divide the proof into two steps.

Step 1. We show that $\sup \{\Vert u\Vert _{\infty }|(\lambda ,u)\in C^{[n]}\}=\infty$.

Suppose by the way of contradiction that $\sup \{\Vert u\Vert _{\infty }|(\lambda ,u)\in C^{[n]}\}=:\beta _1<\infty$. Since $\{\mu _{k},y_{k}\}\subset \mathcal {C}$ is such that $|\mu _{k}|+\Vert y_{k}\Vert _{\infty }\rightarrow \infty$, then we have

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\mu _k=\infty . \end{aligned}$$

(2.15)

Note that

$$\begin{aligned} \frac{g^{[n]}(y_k(t))}{y_{k}(t)}>\inf \left\{ \frac{g^{[n]}(s)}{s}|0<s<\beta _1\right\} . \end{aligned}$$

Since $a^{+}(t)>0$ for $t\in [a,b]\subset [0,\omega ]$. So, there exists a constant $\beta _2>0$, such that

$$\begin{aligned} a^{+}(t)\frac{g^{[n]}(y_k(t))}{y_{k}(t)}>\beta _2>0,~~~t\in [a,b]. \end{aligned}$$

(2.16)

Combining (2.15) and (2.16) with the relation

$$\begin{aligned} y'''_k(t)+k_1y''_k(t)+k_2y'_k(t)+k_3y_k(t)=\mu _ka^{+}(t)\frac{g^{[n]}(y_k(t))}{y_{k}},~~~t\in [0,\omega ]. \end{aligned}$$

(2.17)

Thus, from (2.17) and the remarks in the final paragraph on P. 56 of [19], we deduce that $y_k$ must change its sign on [a, b] if k is large enough. This is a contradiction. Hence $\{\Vert y_{k}\Vert _{\infty }\}$ is unbounded.

Step 2. We show that $\sup \{\lambda |(\lambda ,u)\in C^{[n]}\}=\infty$.

Now, taking $\{\mu _{k},y_{k}\}\subset \mathcal {C}$ be such that

$$\begin{aligned} \Vert y_{k}\Vert _{\infty }\rightarrow \infty ,~~\text{ as }~~k\rightarrow \infty . \end{aligned}$$

(2.18)

We show that $\lim \nolimits _{k\rightarrow \infty }\mu _{k}=0.$ Suppose on the contrary that, choosing a subsequence and relabelling if necessary, $\mu _k\ge \mu _1$ for some constant $\mu _1>0$. Then from (2.18) and $(A_2)$, we have

$$\begin{aligned} a^{+}(t)>0,~~~~\lim \limits _{k\rightarrow \infty }\mu _k\frac{g^{[n]}(y_k(t))}{y_k(t)}=\infty , ~~\text{ for } \text{ all }~~t\in [a_1,b_1]. \end{aligned}$$

Thus, from (2.17) and from the remarks in the final paragraph on P. 56 of [19], we deduce that $y_k$ must change its sign on [a, b] if k is large enough. This is a contradiction. Therefore, $\lim \nolimits _{k\rightarrow \infty }\mu _k=0$.

We rewrite (2.10) to

$$\begin{aligned} u_{\lambda }=\lambda \int ^\omega _0G(t,s)a^{+}(s)g^{[n]}(u(s))ds=:L_{\lambda }u. \end{aligned}$$

By Lemma 2.7, for every $r>0$ and $u\in \partial \Omega _r$,

$$\begin{aligned} \Vert Lu\Vert _{\infty }\le \lambda \widehat{M}_r\int ^\omega _0G(t,s)a^{+}(s)ds, \end{aligned}$$

where $\widehat{M_r}=1+\max _{0\le s\le r}\{g^{[n]}(u(s))\}$.

Let there exists a constant $\bar{\lambda }$ be such that

$$\begin{aligned} \bar{\lambda }\widehat{M}_r\int ^\omega _0G(t,s)a^{+}(s)ds=r. \end{aligned}$$

Then for $\lambda \in (0,\bar{\lambda })$ and $u\in \partial \Omega _r$, $\Vert Lu\Vert _{\infty }<\Vert u\Vert _{\infty }$. This mean that

$$\begin{aligned} \Sigma \cap \{(\lambda ,u)\times P|0<\lambda <\bar{\lambda }, u\in P:\Vert u\Vert _{\infty }=r\}=\emptyset , \end{aligned}$$

i.e., $\mathcal {C}$ is also an unbounded component joining (0, 0) with $(0,\infty )$ in $[0,\infty )\times X$. Thus, for $\lambda \in (0,\bar{\lambda })$, (2.1) has at least one positive solution. Moreover, from $(A_1)$ we have $u_{\lambda }\ge \lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds$. $\square$

3 Proof of Theorem 1.1

From $(A_2)$, there exist two positive constants $\alpha ,\gamma \in (0,1)$ such that

$$\begin{aligned} \int ^\omega _0G(t,s)a^{-}(s)ds|f(x)|\le \gamma f(0) \int ^\omega _0G(t,s)a^{+}(s)ds \end{aligned}$$

(3.1)

for $x\in [0,\alpha ]$, $t\in [0,1]$. Fixed $\delta \in (\gamma ,1)$ and let $\lambda ^*>0$ be such that

$$\begin{aligned} \Vert \tilde{u}_{\lambda }\Vert _{\infty } +\lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds\le \alpha \end{aligned}$$

(3.2)

for $\lambda <\lambda ^*$, where $\tilde{u}_{\lambda }$ is given by Lemma 2.2, from $(A_1)$,

$$\begin{aligned} \Vert \tilde{u}_{\lambda }\Vert _{\infty }\ge \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds, \end{aligned}$$

and

$$\begin{aligned} |f(x)-f(y)|\le f(0)\left( \frac{\delta -\gamma }{2}\right) \end{aligned}$$

(3.3)

for $x,y\in [-\alpha ,\alpha ]$ with $|x-y|\le \lambda ^*\delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds$.

Next, note that $\lambda <\lambda ^*$. We find a solution $u_{\lambda }$ of (1.1)–(1.2) has the form $\tilde{u}_{\lambda }+v_{\lambda }$. Thus $v_{\lambda }$ solves

$$\begin{aligned} v'''_{\lambda }+k_1v''_{\lambda }+k_2v'_{\lambda }+k_3v_{\lambda }= & {} \lambda a^{+}(t)(f(\tilde{u}_{\lambda }+v_{\lambda })-f(\tilde{u}_{\lambda }))\\{} & {} -\lambda a^{-}(t)f(\tilde{u}_{\lambda }+v_{\lambda }),~~~~t\in (0,\omega ),\\ v_{\lambda }(0)= & {} v_{\lambda }(\omega ),~v'_{\lambda }(0)=v'_{\lambda }(\omega ),~v''_{\lambda }(0)=v''_{\lambda }(\omega ). \end{aligned}$$

For all of $w\in C([0,\omega ])$, we suppose that $v=Aw$ is the solution of

$$\begin{aligned} v'''_{\lambda }+k_1v''_{\lambda }+k_2v'_{\lambda }+k_3v_{\lambda }= & {} \lambda a^{+}(t)(f(\tilde{u}_{\lambda }+w)-f(\tilde{u}_{\lambda }))\\{} & {} -\lambda a^{-}(t)f(\tilde{u}_{\lambda }+w),~~~~t\in (0,\omega ),\\ v_{\lambda }(0)= & {} v_{\lambda }(\omega ),~v'_{\lambda }(0)=v'_{\lambda }(\omega ),~v''_{\lambda }(0)=v''_{\lambda }(\omega ). \end{aligned}$$

Then A is completely continuous. Let $v\in C([0,\omega ])$ and $\theta \in (0,1)$ be such that $v=\theta Av$. Then

$$\begin{aligned} v'''_{\lambda }+k_1v''_{\lambda }+k_2v'_{\lambda }+k_3v_{\lambda }= & {} \lambda \theta a^{+}(t)(f(\tilde{u}_{\lambda }+w)-f(\tilde{u}_{\lambda }))-\lambda \theta a^{-}(t)f(\tilde{u}_{\lambda }+w),\\{} & {} ~~~~t\in (0,\omega ). \end{aligned}$$

Next, We show that $\Vert v\Vert _{\infty }\ne \lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds$. Suppose not. Then, from (3.1) and (3.2), it follows that

$$\begin{aligned} \Vert \tilde{u}_{\lambda }+v\Vert _{\infty }\le \Vert \tilde{u}_{\lambda }\Vert _{\infty }+\Vert v\Vert _{\infty }\le \alpha \end{aligned}$$

and

$$\begin{aligned} |f(\tilde{u}_{\lambda }+v)-f(\tilde{u}_{\lambda })|\le f(0)\left( \frac{\delta -\gamma }{2}\right) , \end{aligned}$$

which together with (3.1) implies that

$$\begin{aligned} \Vert v\Vert _{\infty }\le \lambda \frac{\delta +\gamma }{2}f(0)\int ^\omega _0G(t,s)a^{+}(s)ds<\lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds, \end{aligned}$$

(3.4)

which is a contradiction. From Leray–Schauder fixed point theorem, A has a fixed point

$$\begin{aligned} v_{\lambda }\le \lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds. \end{aligned}$$

Hence $v_{\lambda }$ satisfies (3.4) and

$$\begin{aligned} \begin{aligned} u_{\lambda }&\ge \tilde{u}_{\lambda }(t)-v_{\lambda }(t)\\&\ge \lambda \delta f(0)\int ^\omega _0G(t,s)a^{+}(s)ds-\lambda \frac{\delta +\gamma }{2}f(0)\int ^\omega _0G(t,s)a^{+}(s)ds\\&=\lambda \frac{\delta -\gamma }{2}f(0)\int ^\omega _0G(t,s)a^{+}(s)ds\\&>0, \end{aligned} \end{aligned}$$

i.e., $u_{\lambda }$ is a positive solution of (1.1)–(1.2). This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

In this section, we discuss the case that f(0) is unbounded, that is, f has a singularity at $u=0$ and can be defined as $f=\frac{a(t)}{u^{\nu }}$ where $\nu \in (0,\infty )$ is a constant. Then, since a is sign-changing, the maximum principle does not hold. Hence, we need to choose appropriate domain so that $\lambda \frac{a(t)}{u^{\mu }}+u$ becomes positive. (1.1)-(1.2) can rewrite

$$\begin{aligned}{} & {} u'''(t)+k'_1u''(t)+k'_2u'(t)+k'_3u(t)=\lambda \frac{a(t)}{u^{\nu }}+u(t),~~~t\in (0,\omega ), \nonumber \\{} & {} u(0)=u(\omega ),~u'(0)=u'(\omega ),~u''(0)=u''(\omega ), \end{aligned}$$

(4.1)

where $k'_1,k'_2,k'_3$ are constants and satisfy Lemma 2.3. Let

$$\begin{aligned} (\Phi u)(t):=\lambda \int ^\omega _0G(t,s)\frac{a(s)}{u^{\nu }(s)}ds+\int ^\omega _0G(t,s)u(s)ds. \end{aligned}$$

Now, we define two open sets

$$\begin{aligned} \Omega _1:=\{u\in C_{\omega }:\Vert u\Vert _{\infty }<r_1\}~~\text{ and }~~\Omega _2:=\{u\in C_{\omega }:\Vert u\Vert _{\infty }<r_2\}, \end{aligned}$$

where $r_1$ and $r_2$ are two constants and

$$\begin{aligned} r_2>r_1=(\lambda \sigma ^*\omega \Vert a^{+}\Vert _{\infty })^{\frac{1}{\nu +1}}> \frac{1}{\sigma }\left( \lambda \Vert a^{-}\Vert _{\infty }\right) ^{\frac{1}{\nu +1}}. \end{aligned}$$

First, we show that $\Phi (P\cap (\bar{\Omega }_2\backslash \Omega _1))\subset P$. In fact, for any $u\in P\cap (\bar{\Omega }_2\backslash \Omega _1)$,

$$\begin{aligned} \sigma r_1\le u(t)\le r_2,~~t\in \mathbb {R}. \end{aligned}$$

Since $r_1>\frac{1}{\sigma }(\lambda \Vert a^{-}\Vert _{\infty })^{\frac{1}{\nu +1}}$, it follows that

$$\begin{aligned} \begin{aligned} \lambda \frac{a(t)}{u^{\nu }}+u(t)=&\lambda \frac{a^{+}(t)}{u^{\nu }}-\lambda \frac{a^{-}(t)}{u^{\nu }}+u(t)\\>&-\lambda \frac{a^{-}(t)}{u^{\nu }}+\sigma r_1\\>&-\lambda \frac{\Vert a^{-}\Vert }{(\sigma r_1)^{\nu }}+\sigma r_1>0. \end{aligned} \end{aligned}$$

(4.2)

By (4.2), then

$$\begin{aligned} \begin{aligned} \min \limits _{t\in \mathbb {R}}(\Phi u)(t)&=\min \limits _{t\in \mathbb {R}}\lambda \int ^{\omega }_0G(t,s)\frac{a(s)}{u^{\nu }}ds +\min \limits _{t\in \mathbb {R}}\int ^{\omega }_0G(t,s)u(s)ds\\&\ge \lambda \sigma _*\int ^{\omega }_0\frac{a(s)}{u^{\nu }}ds +\sigma _*\int ^{\omega }_0u(s)ds\\&=\lambda \sigma \sigma ^*\int ^{\omega }_0\frac{a(s)}{u^{\nu }}ds +\sigma \sigma ^*\int ^{\omega }_0u(s)ds\\&\ge \lambda \sigma \max \limits _{t\in \mathbb {R}}\int ^{\omega }_0G(t,s)\frac{a(s)}{u^{\nu }}ds +\sigma \max \limits _{t\in \mathbb {R}}\int ^{\omega }_0G(t,s)u(s)ds\\&= \sigma \Vert \Phi u\Vert _{\infty }, \end{aligned} \end{aligned}$$

which implies $\Phi (P\cap (\bar{\Omega }_2\backslash \Omega _1))\subset P$. By using Arzela–Ascoli theorem, it is easy to show that $\Phi$ is a completely continuous operator.

Next, we show that

$$\begin{aligned} \Vert \Phi u\Vert _{\infty }\le \Vert u\Vert _{\infty },~~~\text{ for }~~u\in P\cap \partial \Omega _2. \end{aligned}$$

In fact, for all $u\in P\cap \partial \Omega _2$, we obtain that $\Vert u\Vert _{\infty }=r_2$ and

$$\begin{aligned} \sigma r_2\le u(t)\le r_2,~~\text{ for } \text{ all }~~t\in \mathbb {R}. \end{aligned}$$

From (4.2), then

$$\begin{aligned} \begin{aligned} (\Phi u)(t)&=\lambda \int ^{\omega }_0G(t,s)\frac{a(s)}{u^{\nu }}ds+\int ^{\omega }_0G(t,s)u(s)ds\\&=\lambda \int ^{\omega }_0G(t,s)\left( \frac{a^{+}(s)}{u^{\nu }}-\frac{a^{-}(s)}{u^{\nu }}\right) ds+\int ^{\omega }_0G(t,s)u(s)ds\\&\le \lambda \frac{\sigma ^*\omega \Vert a^{+}\Vert _{\infty }}{(\sigma r_2)^{\nu }}+\sigma ^*r_2\omega . \end{aligned} \end{aligned}$$

(4.3)

Then we can choose that $r_2$ larger enough such that

$$\begin{aligned} \frac{\lambda \omega \sigma ^*\Vert a^{+}\Vert _{\infty }}{(\sigma r_2)^{\nu }}+\sigma ^*r_2\omega \le r_2. \end{aligned}$$

Therefore, $\Vert \Phi u\Vert _{\infty }\le \Vert u\Vert _{\infty }$.

Finally, we show that

$$\begin{aligned} \Vert \Phi u\Vert _{\infty }\ge \Vert u\Vert _{\infty },~~\text{ for }~u\in P\cap \partial \Omega _1. \end{aligned}$$

(4.4)

In fact, for all $u\in P\cap \partial \Omega _1$, then $\Vert u\Vert _{\infty }=r_1$ and

$$\begin{aligned} \sigma r_1\le u(t)\le r_1,~~~t\in \mathbb {R}. \end{aligned}$$

It follows from (4.2) that

$$\begin{aligned} \begin{aligned} (\Phi u)(t)&=\lambda \int ^{\omega }_0G(t,s)\left( \frac{a(s)}{u^{\nu }}\right) ds+\int ^{\omega }_0G(t,s)u(s)ds\\&=\lambda \int ^{\omega }_0G(t,s)\left( \frac{a^{+}(s)}{u^{\nu }}-\frac{a^{-}(s)}{u^{\nu }}\right) ds +\int ^{\omega }_0G(t,s)u(s)ds\\&\ge \lambda \omega \frac{\sigma _*\Vert a^{+}\Vert _{\infty }}{r_1^{\nu }}=r_1, \end{aligned} \end{aligned}$$

(4.5)

since $r_1=(\lambda \omega \sigma _*\Vert a^{+}\Vert _{\infty })^{\frac{1}{\nu +1}}$. Hence, $\Vert \Phi u\Vert _{\infty }\ge \Vert u\Vert _{\infty }$ holds.

In conclusion, (1.1)–(1.2) has an $\omega$-periodic solution u satisfying $u\in [\sigma r_1,r_2]$. Furthermore,

$$\begin{aligned} \begin{aligned} 0<\lambda \le \frac{r^{1+\nu }_2\sigma ^{\nu }(1-\sigma ^*\omega )}{\omega \sigma ^*\Vert a^{+}\Vert _{\infty }},~~~~~ 0<\lambda \le \frac{(r_1\sigma )^{1+\nu }}{\Vert a^{-}\Vert _{\infty }}, \end{aligned} \end{aligned}$$

so there exists a constant $\lambda _0$ such that (1.1)–(1.2) has at least one positive periodic solution for all $\lambda \in (0,\lambda _0)$.