© E.R. Korfanty et al., Published by EDP Sciences, 2019

1 Introduction

In this paper, we study existence of non-trivial solutions, or positive solutions of the following second-order periodic Boundary Value Problem (BVP):

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

$$x\left(0\right)=x\left(T\right), x\mathrm{\prime }\left(0\right)=x\mathrm{\prime }\left(T\right),$$(1.2)where $g:\mathbb{R}\to \mathbb{R}$ and $f:[0,T]\to \mathbb{R}$ are continuous functions.

Equation (1.1) can be considered a special case of the Liénard-type equations, of the general form [1, 2],

$$x\mathrm{″}+f\left(x\right)x\mathrm{\prime }=g(x,t).$$(1.3)

This class of differential equations describes physical dynamical systems of one time-dependant variable x, subject to a damping force f, and a restoring force g. Thus, (1.1) represents a class of conservative systems, where there is no damping force. It considers the special case where the restoring force can be expressed by a t-dependent linear part, and an x-dependent non-linear part.

Since the early 1920’s, when Liénard first studied equations of the form x″ + f(x)x′ + x = 0 with periodic boundary conditions [3], many generalizations fitting the form (1.3) have been investigated (see, e.g., [1, 4–8] and the references therein). This is due to the wide application of these models to oscillatory systems arising in Physics and Engineering. In some cases, only positive solutions are relevant, and so effort has gone into investigating existence of positive periodic solutions to equations like (1.3); for example, see [9–11].

Equation (1.1) can also be considered a generalization of the Duffing-type equations, of the form:

$$x\mathrm{″}+g\left(x\right)=p\left(t\right),$$(1.4)where $g:\mathbb{R}\to \mathbb{R}$ is continuous, and $p:\mathbb{R}\to \mathbb{R}$ is both continuous and T-periodic [12]. Other generalizations of the Duffing-type equations have been considered, such as those of the form:

$$x\mathrm{″}+bx\mathrm{\prime }+g(x,t)=0,$$(1.5) or

$$x\mathrm{″}+\mathrm{bx}\mathrm{\prime }+g(x,t)=p\left(t\right).$$(1.6)

A direct motivation for considering BVP (1.1)–(1.2) is the Mathieu-Duffing type equations, of the form:

$$x\mathrm{\text{'}\text{'}}\pm b(1+\cos t)x=\pm x^{\beta },$$(1.7)taken with 2π-periodic boundary conditions, and where b, β > 0 [13]. Since 1831, when Faraday was investigating wave motion in fluids [14], there have been several models of oscillatory systems developed which are of the form (1.7). For example, see [15–17]. Obviously, the study of periodic solutions for (1.7) is a concrete case of system (1.1)–(1.2).

Since the linear operator Lx = x″ is non-invertible under the periodic boundary condition (1.2), BVP (1.1)–(1.2) represents the case of resonance [18]. Therefore, the common method of converting a BVP to a fixed point problem using L ⁻¹ and then applying fixed point theorems or Leray-Schauder degree theory would not work. To consider the existence of solutions for (1.1)–(1.2), we use the coincidence degree theory developed in [19]. The proofs take inspiration from the ideas in [2, 18, 20, 21]. Due to characteristics of equation (1.1), the requirements of the obtained results on existence of at least one solution for (1.1)–(1.2) are weaker than previous study on solvability of the general Liénard-type equations (1.3). In particular, the conditions contained on functions f and g in (1.1) are only centred on a linear growth for g.

In general, the study of positive solutions for BVPs requires different techniques and stronger conditions than solution existence. We prove both existence of a positive solution and existence of at least one solution using the same theorem from the coincidence degree theory for semilinear operator equations. This approach has the advantage of direct comparison between sufficient conditions for a positive solution and a non-trivial solution of BVP (1.1)–(1.2).

In Section 2, we introduce preliminaries, operators, and space setting for coincidence degree theory [19] that will be used in the sequel. Section 3 proves existence of a non-trivial periodic solution for BVP (1.1)–(1.2). Solvability of system (1.1)–(1.2) with a positive solution is studied in Section 4. Section 5 provides visual interpretation for conditions required in the obtained results. Finally, examples and numerical solutions are given in Section 6 to show applications of the theorems.

2 Preliminaries

The following theorem on solvability of the semilinear equation Lx + Nx = 0, where L is a linear operator and N is nonlinear, is a main result from the coincidence degree theory [19].

Theorem 2.1. Suppose that X and Z are normed vector spaces over R, and that $L:D\left(L\right)\subset X\to Z$ is a (linear) Fredholm operator of index zero. Let Ω ⊂ X be open and bounded, and suppose that $N:\bar{\mathrm{\Omega }}\subset X\to Z$ is an L-compact mapping. Then, if the following conditions hold, the equation Lx + Nx = 0 has at least one solution in $D\left(L\right)\cap \bar{\mathrm{\Omega }}$ :

• (H₁) Lx + λNx ≠ 0 for all x ∈ ∂Ω ∩ (D(L)\ker(L)), λ ∈ (0, 1);

• (H₂) Nx ∉ im(L) for each x ∈ ∂Ω ∩ ker(L);

• (H₃) $D_0(\mathrm{QN}{|}_{\ker \left(L\right)},\mathrm{\Omega }\cap \ker (L\left)\right)\ne 0$ , for some continuous projection operator Q: Z → Z, such that ker(Q) = im(L).

We recall the definition of a L-compact mapping in Theorem 2.1.

Definition 2.2. Suppose that L: X → Z is a Fredholm operator of index zero, where $X=\ker \left(L\right)\oplus X_1$ and $Z=\mathrm{im}\left(L\right)\oplus Z_1$ are normed vector spaces. Denote the associated projection operators by P: X → ker(L) and Q: Z → Z ₁ .

Let L _P be the operator L restricted to the domain X ₁ . Then, L _P is invertible, and we can define $K_{\mathrm{PQ}}=L_P^{-1}(I-Q)$ . We say that an operator $N:\bar{\mathrm{\Omega }}\subset X\to Z$ is L-compact if the following conditions hold:

1. $\mathrm{QN}:\bar{\mathrm{\Omega }}\to Z_1$ is continuous, and $\mathrm{QN}\left(\bar{\mathrm{\Omega }}\right)$ is bounded;

2. $K_{\mathrm{PQ}}N:\bar{\mathrm{\Omega }}\to X$ is compact.

Throughout the rest of the paper, the following definitions of the relevant spaces and operators will be used.

Fix T > 0, and let $X=\{x\in C^1([0,T],\mathbb{R}):x(0)=x(T),x\mathrm{\prime }(0)=x\mathrm{\prime }(T\left)\right\}$. Equipped with the standard C ¹-norm, $\Vert x\Vert =\max \{{\Vert x\Vert }_{\mathrm{\infty }}+{\Vert x\text{'}\Vert }_{\mathrm{\infty }}\}$, X is a Banach space. Also, let $Z=L^1\left(\right[0,T],\mathbb{R})$, with the standard L ¹-norm.

Let L be the linear operator on X defined by

$$\begin{array}{cc}\left(\mathrm{Lx}\right)\left(t\right)=x\mathrm{″}\left(t\right)& \mathrm{a.e}.\hspace{0.5em}t\in [0,T]\end{array}$$on the domain $D\left(L\right)=\{x\in X:x^{\mathrm{\prime }}\mathrm{is} \mathrm{absolutely} \mathrm{continuous} \mathrm{on}\hspace{0.5em}[0,\mathrm{T}\left]\right\}$. Then, L maps D(L) into Z.

It is straightforward to show that $\ker \left(L\right)\cong \mathbb{R}$, and that the image of L is:

$$\mathrm{im}\left(L\right)=\left\{h\in Z:{\int }_0^T\mathrm{}h\left(t\right) \mathrm{d}t=0\right\},$$(2.1)which is obviously closed in Z.

Furthermore, for every $h\in \mathrm{im}\left(L\right)$, the unique $x\in D\left(L\right)$ such that h = Lx is given by:

$$\begin{array}{cc}x\left(t\right)={\int }_0^t\mathrm{}y\left(s\right)\mathrm{d}s-\frac{t}{T}{\int }_0^T\mathrm{}y\left(s\right)\mathrm{d}s& \mathrm{a.e}.\hspace{0.5em}t\in [0,T]\end{array},$$(2.2) where,

$$\begin{array}{cc}y\left(t\right)={\int }_0^t\mathrm{}h\left(s\right) \mathrm{d}s& \mathrm{a.e}.\hspace{0.5em}t\in [0,T]\end{array}.$$

This will give an explicit form for the operator $L_P^{-1}$ from Definition 2.2. Let P: X → ker(L) be defined by P(x) = x(0). This projection gives $X=\ker \left(L\right)\oplus X_1$, where $X_1=\{x\in X:x(0)=0\}$. It is clear that any x of the form (2.2) satisfies x(0) = 0. Therefore, for each $h\in \mathrm{im}\left(L\right)$ ,

$$\begin{array}{cc}\left(L_P^{-1}h\right)\left(t\right)={\int }_0^t\mathrm{}{y}_h\left(s\right) \mathrm{d}s-\frac{t}{T}{\int }_0^T\mathrm{}{y}_h\left(s\right) \mathrm{d}s& \mathrm{a.e}.\hspace{0.5em}t\in [0,T]\end{array},$$(2.3)where

$$\begin{array}{cc}{y}_h\left(t\right)={\int }_0^t\mathrm{}h\left(s\right) \mathrm{d}s& \mathrm{a.e}.\hspace{0.5em}t\in [0,T]\end{array}.$$

Next, we define the following projection operator Q: Z → im(L),

$$Q\left(h\right)=\frac{1}{T}{\int }_0^T\mathrm{}h\left(t\right) \mathrm{d}t.$$(2.4)Then, $Z=\mathrm{im}\left(L\right)\oplus Z_1$, where $Z_1=\mathrm{coker}\left(L\right)\cong \mathbb{R}$ consists of the constant functions. Therefore, L is a Fredholm operator of index zero.

We consider equation (1.1) as the operator equation Lx + Nx = 0, where N: X → Z is given by:

$$\left(\mathrm{Nx}\right)\left(t\right)=f\left(t\right)x\left(t\right)-g\left(x\left(t\right)\right),$$(2.5)for $f\in C\left(\right[0,T],\mathbb{R})$ and $g\in C(\mathbb{R},\mathbb{R})$. We assume that f, g ≠ 0.

Lemma 2.3. When restricted to a bounded domain $\mathrm{\Omega }\subset X$ , $N:\bar{\mathrm{\Omega }}\to Z$ is L-compact.

Proof. Let M _Ω be such that ||x|| ≤ M _Ω for all $x\in \bar{\mathrm{\Omega }}$. We show that the conditions in Definition 2.2 are satisfied. First, we consider $\mathrm{QN}:\bar{\mathrm{\Omega }}\to \mathbb{R}$,

$$\mathrm{QN}\left(x\right)=\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right)x\left(t\right)-g\left(x\left(t\right)\right) \mathrm{d}t.$$(2.6)

Consider $\left|\mathrm{QN}\right(x)-\mathrm{QN}(x_0\left)\right|$ for $x,x_0\in X$,

$$\begin{array}{ll}\left|\mathrm{QN}\right(x)-\mathrm{QN}(x_0\left)\right|& \le \frac{1}{T}{\int }_0^T\mathrm{}\left|f\right(t\left)\right|\cdot \left|x\right(t)-x_0(t\left)\right| \mathrm{d}t+\frac{1}{T}{\int }_0^T\mathrm{}\left|g\right(x\left(t\right))-g(x_0\left(t\right)\left)\right| \mathrm{d}t\\ & \le \left|\right|x-x_0\left|{|}_{\mathrm{\infty }}\right|\left|f\right|{|}_{\mathrm{\infty }}+\frac{1}{T}{\int }_0^T\mathrm{}\left|g\right(x\left(t\right))-g(x_0\left(t\right)\left)\right| \mathrm{d}t.\\ & \end{array}$$

Let ϵ > 0 and $\delta =\min \{{\delta }_0,\frac{ϵ}{2\left|\right|f|{|}_{\mathrm{\infty }}}\}$, where δ ₀ is such that $|x-x_0|<{\delta }_0$ implies $\left|g\right(x)-g(x_0\left)\right|<\frac{ϵ}{2}$. Then, $\left|\right|x-x_0\left|\right|<\delta$ implies $\left|\mathrm{QN}\right(x)-\mathrm{QN}(x_0\left)\right|<\epsilon$. Therefore, QN is continuous on X.

Next, to see that $\mathrm{QN}\left(\bar{\mathrm{\Omega }}\right)$ is bounded, for $x\in \bar{\mathrm{\Omega }}$,

$$\left|\right|\mathrm{QNx}\left|\right|\le T\left|\right|f\left|{|}_{\mathrm{\infty }}\right|\left|x\right|{|}_{\mathrm{\infty }}+{\int }_0^T\mathrm{}\left|g\right(x\left(t\right)\left)\right| \mathrm{d}t.$$Using the bound on $\bar{\mathrm{\Omega }}$, we have,

$$\begin{array}{ll}\left|\right|\mathrm{QNx}\left|\right|& \le T\left|\right|f|{|}_{\mathrm{\infty }}{M}_{\mathrm{\Omega }}+{\int }_0^T\mathrm{}|g\left(x\right(t\left)\right)| \mathrm{d}t\\ & =T\left|\right|f|{|}_{\mathrm{\infty }}{M}_{\mathrm{\Omega }}+T\max \{\left|g\right(u\left)\right|:u\in [-M_{\mathrm{\Omega }},M_{\mathrm{\Omega }}]\}.\end{array}$$Because g is continuous, this bounds $N\left(\bar{\mathrm{\Omega }}\right)$.

Lastly, we can use the Ascoli-Arzelá Theorem to show that $K_{\mathrm{PQ}}N:\bar{\mathrm{\Omega }}\to X$ is compact. The verification is omitted.

3 Existence of T-periodic solutions

In this section, we establish criteria on solvability of BVP (1.1) and (1.2).

Theorem 3.1. Suppose the following conditions hold for f and g:

• (A₁) There exists positive constants k ₁ and k ₂ such that $\left|g\right(x\left)\right|\le k_1\left|x\right|+k_2$ for each $x\in \mathbb{R}$ ;

• (A₂) There are constants R ₁ , R ₂ > 0 such that ${\int }_0^{T}g\left(x\right(t\left)\right) \mathrm{d}t={\int }_0^{T}f\left(t\right)x\left(t\right) \mathrm{d}t\exists \tau \in [0,T]$ such that $R_1\le \left|x\right(\tau \left)\right|\le R_2$ ;

• (A₃) $k_1<\frac{\sqrt{12}}{T^2}-\left|\right|f|{|}_{\infty }$ $\left(\mathrm{and} \mathrm{so} \left|\right|f|{|}_{\infty }<\frac{\sqrt{12}}{T^2}\right)$ ;

• (A₄) $\underset{x\to \pm \infty }{\lim }\frac{g\left(x\right)}{x}<\frac{1}{T}{\int }_0^{T}f\left(t\right) \mathrm{d}t$ .

Then, equation (1.1) has at least one solution. Furthermore, if g satisfies g(0) ≠ 0, then the solutions are non-trivial.

The proof of Theorem 3.1 follows directly Lemmas 3.2–3.5.

Lemma 3.2. Let $h\in L^1[0,T]$ be such that ${\int }_0^{T}h\left(t\right) \mathrm{d}t=0$ . Then, if x(t) is T-periodic, such that −x″ = h(t) a.e. $t\in [0,T]$ , then $\left|\right|x|{|}_{\infty }\le |x\left(\tau \right)|+\frac{T}{\sqrt{12}}|\left|h\right|{|}_1$ for all $\tau \in [0,T]$ . Furthermore, $\left|\right|\mathrm{x\text{'}}|{|}_{\infty }\le |\left|h\right|{|}_1$ .

Proof. Suppose that x is a continuous and T-periodic function. Let y be the function such that $x\left(t\right)=\overline{x}+y\left(t\right)$ for all $t\in [0,T]$, where

$$\overline{x}=\frac{1}{T}{\int }_0^T\mathrm{}x\left(t\right) \mathrm{d}t.$$

Then, y is also continuous, and T-periodic. Consider the following integral,

$${\int }_0^T-\mathrm{x″}\left(t\right)x\left(t\right) \mathrm{d}t=-x\mathrm{\prime }\left(t\right)x\left(t\right){|}_0^T+{\int }_0^T\mathrm{}\left[x\mathrm{\prime }\right(t){]}^2 \mathrm{d}t=|\left|y\mathrm{\prime }\right|{|}_2^2.$$

Because −x″(t) = h(t) a.e. $t\in [0,T]$, we have

$$\begin{array}{ll}\left|\right|y\mathrm{\prime }|{|}_2^2& ={\int }_0^T\mathrm{}h\left(t\right)x\left(t\right) \mathrm{d}t\\ & ={\int }_0^T\mathrm{}h\left(t\right)[\overline{x}+y(t\left)\right] \mathrm{d}t\\ & =\overline{x}{\int }_0^T\mathrm{}h\left(t\right) \mathrm{d}t+{\int }_0^T\mathrm{}h\left(t\right)y\left(t\right) \mathrm{d}t\\ & ={\int }_0^T\mathrm{}h\left(t\right)y\left(t\right) \mathrm{d}t, \mathrm{because} \bar{h}=0\\ & \le \left|\right|h\left|{|}_1\right|\left|y\right|{|}_{\mathrm{\infty }}.\end{array}$$

Next, applying the Sobolev inequality [2],

$$\left|\right|y|{|}_{\mathrm{\infty }}^2\le \frac{T}{12}|\left|y\right|{|}_2^2,$$along with $\left|\right|y\mathrm{\prime }|{|}_2^2\le |\left|h\right|{|}_1\left|\right|y|{|}_{\mathrm{\infty }}$, one can deduce the following:

$$\left|\right|y|{|}_{\mathrm{\infty }}\le \frac{T}{12}|\left|h\right|{|}_1, \hspace{1em}\left|\right|y\mathrm{\prime }|{|}_2\le \sqrt{\frac{T}{12}}|\left|h\right|{|}_1.$$

Now, for any fixed $\tau \in [0,T]$,

$$x\left(t\right)=x\left(\tau \right)+{\int }_t^{\tau }\mathrm{}x\mathrm{\prime }\left(s\right) \mathrm{d}s.$$

The Cauchy inequality gives that,

$${\int }_{\tau }^t\mathrm{}x\mathrm{\prime }\left(s\right) \mathrm{d}s\le \sqrt{T}\sqrt{{\int }_0^T\mathrm{}\left|x\mathrm{\prime }\right(s){|}^2 \mathrm{d}s}=\sqrt{T}\left|\right|y\mathrm{\prime }|{|}_2.$$We have

$$\left|x\right(t\left)\right|\le \left|x\right(\tau \left)\right|+\sqrt{T}\left|\right|y\mathrm{\prime }|{|}_2\le |x\left(\tau \right)|+\frac{T}{\sqrt{12}}|\left|h\right|{|}_1,$$and therefore, $\left|\right|x|{|}_{\mathrm{\infty }}\le |x\left(\tau \right)|+\frac{T}{\sqrt{12}}|\left|h\right|{|}_1$ for all $\tau \in [0,T]$.

To find the upper bound on $\left|\right|x\mathrm{\prime }\left|\right|$, we consider,

$$\left|\right|x\mathrm{″}|{|}_1=||-h|{|}_1=\left|\right|h|{|}_1.$$

Since x is T-periodic, there exists $t^{\mathrm{*}}\in [0,T]$ such that x′(t*) = 0, and so,

$$\left|x\mathrm{\prime }\right(t\left)\right|=\left|{\int }_{t^{\mathrm{*}}}^t\mathrm{}x\mathrm{″}\left(s\right)\mathrm{d}s\right|\le \left|\right|x\mathrm{″}|{|}_1=|\left|h\right|{|}_1.$$

We obtain that $\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}\le |\left|h\right|{|}_1$.

Lemma 3.3. $U_1=\{x\in D(L)\backslash \ker (L):\mathrm{Lx}+\lambda \mathrm{Nx}=0\hspace{0.5em}\mathrm{for} \mathrm{some}\hspace{0.5em}\lambda \in [\mathrm{0,1}\left]\right\}$ is bounded.

Proof. Let $x\in U_1$ and $h\left(t\right)=\lambda f\left(t\right)x\left(t\right)-\lambda g\left(x\right(t\left)\right)$. Then, h(t) = −x″, and so ${\int }_0^T\mathrm{}h\left(t\right) \mathrm{d}t=0$. We can apply Lemma 3.2.

First, by continuity, and assumption (A ₁),

$$\begin{array}{cc}\left|h\right(t\left)\right|& \le \lambda \left|f\right(t\left)\right|\cdot \left|x\right(t\left)\right|+\lambda k_1\left|x\right(t\left)\right|+\lambda k_2\\ & \le \lambda \left(\right|f\left(t\right)|+k_1)\left|\right|x|{|}_{\mathrm{\infty }}+\lambda k_2.\end{array}$$

Integrating both sides yields,

$$\begin{array}{cc}\left|\right|h|{|}_1& \le \lambda \left({\int }_0^T\mathrm{}\left|f\right(t\left)\right| \mathrm{d}t+k_{1}T\right){\Vert x\Vert }_{\infty }+\lambda k_{2}T\\ & \le \lambda T\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1){\Vert x\Vert }_{\infty }+\lambda k_{2}T.\end{array}$$

Thus,

$$\begin{array}{ll}\left|\right|x|{|}_{\mathrm{\infty }}& \le \left|x\right(\tau \left)\right|+\frac{T}{\sqrt{12}}\left|\right|h|{|}_1\\ & \le \left|x\left(\tau \right)\right|+\frac{T}{\sqrt{12}}\left[\lambda T({\left|\left|f\right|\right|}_{\mathrm{\infty }}+k_1){\Vert x\Vert }_{\infty }+\lambda k_{2}T\right]\\ & \le \left|x\left(\tau \right)\right|+\frac{T^2}{\sqrt{12}}\left[({\left|\left|f\right|\right|}_{\mathrm{\infty }}+k_1){\Vert x\Vert }_{\infty }+k_2\right].\\ & \end{array}$$

The above holds for any $\tau \in [0,T]$. Because ${\int }_0^T\mathrm{}h\left(t\right) \mathrm{d}t=0$, we have that ${\int }_0^T\mathrm{}g\left(x\right(t\left)\right) \mathrm{d}t={\int }_0^T\mathrm{}f\left(t\right)x\left(t\right) \mathrm{d}t$. By assumption (A ₂), there exists ${\tau }^{\mathrm{*}}\in [0,T]$ for which we know that $R_1\le \left|x\right({\tau }^{\mathrm{*}}\left)\right|\le R_2$. Thus, for τ = τ*, we have,

$$\left|\right|x|{|}_{\mathrm{\infty }}\le R_2+\frac{T^2}{\sqrt{12}}[\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1\left)\right|\left|x\right|{|}_{\mathrm{\infty }}+k_2],$$and therefore,

$$\left(1-\frac{T^2}{\sqrt{12}}\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)\right)\left|\right|x|{|}_{\mathrm{\infty }}\le R_2+\frac{k_2{T}^2}{\sqrt{12}}.$$(3.1)

Using assumption (A ₃), we obtain:

$$\left|\right|x|{|}_{\mathrm{\infty }}\le {\left(1-\frac{T^2}{\sqrt{12}}\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)\right)}^{-1}\left(R_2+\frac{k_2{T}^2}{\sqrt{12}}\right),$$(3.2)which shows that $\left|\right|x|{|}_{\mathrm{\infty }}$ is bounded.

Next, from Lemma 3.2, we have $\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}\le |\left|h\right|{|}_1$, and so,

$$\begin{array}{ll}\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}& \le \left(\lambda T\left|\right|f|{|}_{\mathrm{\infty }}+k_{1}T\right)\left|\right|x|{|}_{\mathrm{\infty }}+\lambda k_{2}T\\ & \le T\left(\left|\right|f|{|}_{\mathrm{\infty }}+k_1\right){\left(1-\frac{T^2}{\sqrt{12}}\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)\right)}^{-1}\left(R_2+\frac{k_2{T}^2}{\sqrt{12}}\right)+k_{2}T.\end{array}$$

Therefore, $\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}$ is also bounded. The proof is complete.

Lemma 3.4. $U_2=\{x\in \ker (L):\mathrm{Nx}\in \mathrm{im}(L\left)\right\}$ is bounded.

Proof. Let $x\in \ker \left(L\right)$, x is a constant function, so let x(t) = c. If x is such that $\mathrm{Nx}\in \mathrm{im}\left(L\right)$, we have,

$${\int }_0^T\mathrm{}\left(\mathrm{Nx}\right)\left(t\right)={\int }_0^T\mathrm{}\mathrm{cf}\left(t\right)-g\left(c\right) \mathrm{d}t=0,$$and

$$c{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t=g\left(c\right)T.$$

This implies that,

$$U_2=\left\{c\in \mathbb{R} :g\left(c\right)=\frac{c}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t\right\}.$$

Assume that U ₂ is unbounded. Then there exists a sequence $\left\{c_n\right\}\subset \mathbb{R}$ such that for each n, $g\left(c_n\right)=\frac{c_n}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t$, and $\left|c_n\right|\to \mathrm{\infty }$. This produces,

$$\frac{g\left(c_n\right)}{c_n}=\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t=\mathrm{constant}.$$

Therefore, $\underset{n\to \mathrm{\infty }}{\lim }\frac{g\left(c_n\right)}{c_n}=\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t$, which is a contradiction to assumption (A ₄).

Lemma 3.5. $U_3=\{x\in \ker (L):H(\lambda ,x)=\lambda \mathrm{Jx}+(1-\lambda )\mathrm{QNx}=0, \lambda \in [0,1\left]\right\}$ is bounded.

Proof. Again, consider $x\in \ker \left(L\right)$, $x=c\in \mathbb{R}$. In this case,

$$\mathrm{QN}\left(x\right)=\frac{c}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t-g\left(c\right).$$

Therefore, $H(\lambda ,x)=\lambda c+(1-\lambda )\left(\frac{c}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t-g\left(c\right)\right)$, which is a constant function for each λ. First, consider the special cases λ = 0 and λ = 1.

When λ = 0, H(λ,x) = 0 clearly implies that $g\left(c\right)=\frac{c}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t$. This is exactly the case of Lemma 3.4, and so the values of c contributed by λ = 0 are bounded.

When λ = 1, H(λ,x) = x, so H(λ,x) = 0 implies that y = 0; so λ = 1 contributes only c = 0, and will not affect the boundedness of U ₃.

Lastly, for $\lambda \in \left(\mathrm{0,1}\right)$, we show that the following set is bounded,

$$\mathcal{C}≔\left\{c\in \mathbb{R}:g\left(c\right)=\left(\frac{\lambda }{1-\lambda }+\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t\right)c, \lambda \in \left(\mathrm{0,1}\right)\right\}.$$

Otherwise, there exists a sequence $\left\{c_n\right\}$ such that $\left|c_n\right|\to \mathrm{\infty }$ with corresponding $\left\{{\lambda }_n\right\}\in \left(\mathrm{0,1}\right)$, such that:

$$g\left(c_n\right)=\left(\frac{{\lambda }_n}{1-{\lambda }_n}+\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t\right)c_n,$$or

$$\frac{g\left(c_n\right)}{c_n}-\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t=\frac{{\lambda }_n}{1-{\lambda }_n}.$$

Since ${\lambda }_n\in \left(\mathrm{0,1}\right)$, there exists a subsequence {${\lambda }_{n_k}\}\subset \{{\lambda }_n\}$ such that ${\lim }_{k\to \mathrm{\infty }}{\lambda }_{n_k}=r<1$. Therefore, ${\lim }_{k\to \mathrm{\infty }}\frac{{\lambda }_{n_k}}{1-{\lambda }_{n_k}}\ge 0$. From assumption (A ₄), we see that ${\lim }_{\left|x\right|\to \mathrm{\infty }}\frac{g\left(x\right)}{x}<\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right)\mathrm{d}t$, which implies that,

$$\underset{k\to \mathrm{\infty }}{\lim }\left(\frac{g\left(c_{n_k}\right)}{c_{n_k}}-\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t\right)<0.$$

This contradiction shows that $\mathcal{C}$ is bounded. As result, U ₃ is also bounded.

Proof of Theorem 3.1: Let Ω be an open set such that $\{U_1\cup U_2\cup U_3\}\subset \mathrm{\Omega }$. Theorem 2.1 ensures the existence of at least one T-periodic solution to equation (1.1) in $D\left(L\right)\cap \bar{\mathrm{\Omega }}$. Since g(0) ≠ 0, the solution is non-trivial.

4 Existence of positive T-periodic solutions

Existence of positive solutions is significant in particular applications. In this section, we establish conditions on existence of positive periodic solutions for equation (1.1).

Theorem 4.1. Suppose the following conditions hold for f and g:

• (B ₁) There exist positive constants k ₁ and k ₂ such that $0\le \left|g\right(x\left)\right|\le k_{1}x+k_2$ for x > 0;

• (B ₂ ) There are constants R ₁ , R ₂ > 0 such that,

$${\int }_0^T\mathrm{}g\left(x\right(t\left)\right) \mathrm{d}t={\int }_0^T\mathrm{}f\left(t\right)x\left(t\right) \mathrm{d}t\Rightarrow \exists \tau \in [0,T], R_1\le x\left(\tau \right)\le R_2;$$

• (B ₃) $k_1<\frac{\sqrt{12}}{T^2}-\left|\right|f|{|}_{\mathrm{\infty }}$;

• (B ₄) ${\int }_0^{R_1}\mathrm{}g\left(u\right) \mathrm{d}u>R_4\left(\left|\right|f|{|}_{\mathrm{\infty }}{R}_{3}T+\frac{R_4}{2}\right)$, where R ₃ and R ₄ are the following constants:

$$\begin{array}{ll}{R}_3& =\frac{R_2+\frac{k_2{T}^2}{\sqrt{12}}}{1-\frac{T^2}{\sqrt{12}}\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)},\\ R_4& =\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)R_{3}T+k_{2}T;\end{array}$$

• This requires g to be non-negative on (0, R ₁), but in addition, we will require g to be non-negative on (0, R ₂).

• (B ₅) $\frac{g\left(x\right)}{x}>\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t$ for $x\in (0,R_1)$, and $\frac{g\left(x\right)}{x}<\frac{1}{T}{\int }_0^T\mathrm{}f\left(t\right) \mathrm{d}t$ for $x>R_2$.

Then, equation (1.1) has at least one non-trivial, positive and T-periodic solution.

Remark 4.2.Condition (B ₃) ensures that R ₃ is well-defined and positive.

We will show the following properties of equation (1.1), then use them to construct Ω.

Suppose that x is a positive and T-periodic solution to Lx + λNx = 0. The following lemmas show that $\left|\right|x|{|}_{\mathrm{\infty }}\le R_3$, $\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}\le \lambda R_4$, and that there exists an R ₅ > 0 such that x(t) ≥ R ₅ for all $t\in [0,T]$.

Remark 4.3.If x is a positive and T-periodic solution for Lx + λNx = 0, then there exists $\tau \in [0,T]$ such that R ₁ ≤ x(τ) ≤ R ₂ , where R ₁ and R ₂ are as in assumption (B ₂). This can be shown by direct integration of:

$$x\mathrm{\text{'}\text{'}}\left(t\right)+\lambda f\left(t\right)x\left(t\right)-\lambda g\left(x\right)=0,$$which gives

$${\int }_0^T\mathrm{}f\left(t\right)x\left(t\right) \mathrm{d}t={\int }_0^T\mathrm{}g\left(x\right(t\left)\right) \mathrm{d}t.$$

Lemma 4.4. There are constants R ₃ , R ₄ > 0 such that whenever x is a positive, T-periodic solution to Lx + λNx = 0, we have $\left|\right|x|{|}_{\infty }\le R_3$ , and $\left|\right|\mathrm{x\text{'}}|{|}_{\infty }\le \lambda R_4$ .

Proof. Let x be such a solution, and define h(t) = λf(t)x(t) − λg(x(t)). Following the proof of Lemma 3.3,

$$\begin{array}{l}\left|h\right(t\left)\right|\le \lambda T\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1\left)\right|\left|x\right|{|}_{\mathrm{\infty }}+\lambda k_{2}T,\end{array}$$and

$$\begin{array}{cc}\left|\right|x|{|}_{\mathrm{\infty }}& \le x\left(\tau \right)+\frac{T^2}{\sqrt{12}}\left[\right(\left|\right|f|{|}_{\mathrm{\infty }}+k_1)\left|\right|x|{|}_{\mathrm{\infty }}+k_2]\\ & \le R_2+\frac{T^2}{\sqrt{12}}\left[\left(\left|f\right|{|}_{\mathrm{\infty }}+k_1\right)\left|x\right|{|}_{\mathrm{\infty }}+k_2\right].\end{array}$$

Therefore, we arrive at

$$\left(1-\frac{T^2}{\sqrt{12}}\left(\right|\left|f\right|{|}_{\mathrm{\infty }}+k_1)\right)\left|\right|x|{|}_{\mathrm{\infty }}\le R_2+\frac{k_2{T}^2}{\sqrt{12}}.$$

By assumption (B ₃),

$$\left|\right|x|{|}_{\mathrm{\infty }}\le {\left(1-\frac{a{T}^2}{\sqrt{12}}({\left|\right|f\left|\right|}_{\mathrm{\infty }}+k_1)\right)}^{-1}\left(R_2+\frac{k_2{T}^2}{\sqrt{12}}\right)=R_3.$$

Following the proof of Lemma 3.3, we have,

$$\left|\right|x\mathrm{\prime }|{|}_{\mathrm{\infty }}\le \lambda T\left(\left|\right|f|{|}_{\mathrm{\infty }}+k_1\right)R_3+\lambda k_{2}T=\lambda R_4.$$□

Lemma 4.5. There is a constant R ₅ > 0 such that for any positive and T-periodic solutions of Lx + λNx = 0, we have that x(t) ≥ R ₅ for all $t\in [0,T]$ .

Proof. Let x be such a solution. Then:

$$x\mathrm{″}\left(t\right)+\lambda f\left(t\right)x\left(t\right)=\lambda g\left(x\left(t\right)\right).$$

Multiplying by x′, we have,

$$x\mathrm{\text{'}\text{'}}\left(t\right)x\mathrm{\prime }\left(t\right)+\lambda f\left(t\right)x\left(t\right)x\mathrm{\prime }\left(t\right)=\lambda g\left(x\right(t\left)\right)x\mathrm{\prime }\left(t\right).$$

By assumption (B ₂), we can choose $\tau \in [0,T]$ such that R ₁ ≤ x(τ) ≤ R ₂. Integrating from τ to t, for arbitrary $t\in [0,T]$,

$$\lambda {\int }_{x\left(\tau \right)}^{x\left(t\right)}\mathrm{}g\left(u\right) \mathrm{d}u={\int }_{x\mathrm{\prime }\left(\tau \right)}^{x\mathrm{\prime }\left(t\right)}\mathrm{}u \mathrm{d}u+\lambda {\int }_{\tau }^t\mathrm{}f\left(s\right)x\left(s\right)x\mathrm{\prime }\left(s\right) \mathrm{d}s,$$and so

$$\begin{array}{cc}\frac{1}{2}\left(x\mathrm{\prime }\right(t){)}^2-\frac{1}{2}(x\mathrm{\prime }\left(\tau \right){)}^2-\lambda {\int }_{x\left(\tau \right)}^{x\left(t\right)}\mathrm{}g\left(u\right) \mathrm{d}u& =-\lambda {\int }_{\tau }^t\mathrm{}f\left(s\right)x\left(s\right)x\mathrm{\prime }\left(s\right) \mathrm{d}t\\ & \le \lambda \left|\right|f\left|{|}_{\mathrm{\infty }}{R}_3{R}_4\right(t-\tau )\\ & \le \lambda \left|\right|f|{|}_{\mathrm{\infty }}{R}_3{R}_{4}T.\end{array}$$

This tells us that,

$$\begin{array}{cc}\lambda {\int }_{x\left(t\right)}^{x\left(\tau \right)}\mathrm{}g\left(u\right) \mathrm{d}u& \le \lambda \left|\right|f|{|}_{\mathrm{\infty }}{R}_3{R}_{4}T+\frac{1}{2}(x\mathrm{\prime }\left(\tau \right){)}^2-\frac{1}{2}\left(x\mathrm{\prime }\right(t){)}^2\\ & \le \lambda \left|\right|f|{|}_{\mathrm{\infty }}{R}_3{R}_{4}T+\frac{1}{2}(x\mathrm{\prime }\left(\tau \right){)}^2\\ & \le \lambda \left|\right|f|{|}_{\mathrm{\infty }}{R}_3{R}_{4}T+\frac{{\lambda }^2{R}_4^2}{2}.\end{array}$$

Dividing by λ yields,

$$\begin{array}{ll}{\int }_{x\left(t\right)}^{x\left(\tau \right)}\mathrm{}g\left(u\right) \mathrm{d}u\le \left|\right|f|{|}_{\mathrm{\infty }}{R}_3{R}_{4}T+\frac{\lambda R_4^2}{2}\le R_4\left(\left|\right|f|{|}_{\mathrm{\infty }}{R}_{3}T+\frac{R_4}{2}\right).& \end{array}$$

Now, because g is continuous and non-negative on (0, R ₁), (B ₄) guarantees that there exits a constant R ₅ > 0 such that,

$${\int }_{x_0}^{R_1}\mathrm{}g\left(x\right)\mathrm{d}x>R_4\left(\left|\right|f|{|}_{\mathrm{\infty }}{R}_{3}T+\frac{R_4}{2}\right),$$for all $x_0\in (0, R_5)$.

We know,

$${\int }_{x\left(t\right)}^{x\left(\tau \right)}\mathrm{}g\left(u\right)\mathrm{d}u={\int }_{x\left(t\right)}^{R_1}\mathrm{}g\left(u\right)\mathrm{d}u+{\int }_{R_1}^{x\left(\tau \right)}\mathrm{}g\left(u\right)\mathrm{d}u,$$and $R_1\le x\left(\tau \right)\le R_2$, so ${\int }_{R_1}^{x\left(\tau \right)}\mathrm{}g\left(u\right) \mathrm{d}u\ge 0$. Thus,

$${\int }_{x\left(t\right)}^{x\left(\tau \right)}\mathrm{}g\left(u\right)\mathrm{d}u\ge {\int }_{x\left(t\right)}^{R_1}\mathrm{}g\left(u\right)\mathrm{d}u.$$

Assume, by way of contradiction, that x(t) < R ₅. If this is the case, then $x\left(t\right)\in (0,R_5)$ and,

$${\int }_{x\left(t\right)}^{R_1}\mathrm{}g\left(x\right)\mathrm{d}x>R_4\left(\left|\right|f|{|}_{\mathrm{\infty }}{R}_{3}T+\frac{R_4}{2}\right),$$but

$${\int }_{x\left(t\right)}^{R_1}\mathrm{}g\left(u\right)\mathrm{d}u\le {\int }_{x\left(t\right)}^{x\left(\tau \right)}\mathrm{}g\left(u\right)\mathrm{d}u\le R_4\left(\left|\right|f|{|}_{\mathrm{\infty }}{R}_{3}T+\frac{R_4}{2}\right),$$a contradiction. This means that x(t) ≥ R ₅ > 0.