**1. Introduction**

During the past few decades, an active worldwide research on the oscillation and nonoscillation for dynamic equations on time scales has been carried out by many mathematicians. Some interesting monographs [1–5] contain many important works in this area. In particular, many researchers have studied oscillation of second order dynamic equations. For some recent results on the topic, we refer the reader to the works [6–20] and the references cited therein.

Consider the second order dynamic equation on time scales

$$g\left(p(t)x^{\Lambda}(t)\right)^{\Lambda} + g\left(t, x(\eta(t))\right) = 0, \quad t \in \mathbb{T}\_0 \subseteq \mathbb{T},\tag{1}$$

where *p* ∈ *Crd*(<sup>T</sup>0, <sup>R</sup>+), *η* ∈ *Crd*(<sup>T</sup>0,<sup>T</sup>), *g* : T0 × R → R is continuous and sgn *g*(*<sup>t</sup>*, *x*) = sgn *x* for *t* ∈ T0, lim*<sup>t</sup>*→∞ *η*(*t*) = ∞.

Oscillation of the Equation (1) has been studied by Doslý and Hilger [ ˘ 6], Grace, Agarwal, Bohner and O'Regan [7], Zhou, Ahmad and Alsaedi [20]. A non-oscillatory of Equation (1) is also considered by Graef and Hill [21], Erbe, Baoguo and Peterson [22]. For more details, we refer the reader to see the references cited therein. However, to the authors' knowledge, there are no papers dealing with the analysis of structure of non-oscillatory solutions and sufficient and necessary conditions for existence of all kinds of non-oscillatory solutions for dynamic equations on time scales.

Our aim is to give a classification of non-oscillatory solutions to second order superlinear and sublinear dynamic equations on time scales, which is presented in Section 2. Then, we obtain the sufficient and necessary conditions for existence of some kinds of non-oscillatory solutions in Section 3.

#### **2. Classification of Non-Oscillatory Solutions**

Let T be a time scale (i.e., a closed subset of the real numbers R) with supT = ∞. We assume throughout that T has the topology that it inherits from the standard topology on the real numbers R. For *t* ∈ T, we define the forward jump operator *σ* : T → T by *σ*(*t*) := inf{*s* ∈ T : *s* > *t*}. Denote by *Crd*(<sup>T</sup>, R) the space consisting of all functions which are right-dense points in T and its left-sided limits exist (finite) at left-dense points in T. Furthermore, let us put [*<sup>t</sup>*0, ∞) := T0 = {*t* ∈ T : *t*0 ≤ *t* < <sup>∞</sup>}.

*Mathematics* **2019**, *7*, 680

**Definition 1.** *If*

$$\frac{f(t,x)}{x} \ge \frac{f(t,y)}{y} \text{ for } x \ge y > 0 \text{ or } x \le y < 0, \ t \in \mathbb{T}\_{0,\delta}$$

*then f is said to be superlinear. If*

$$\frac{f(t,x)}{x} \le \frac{f(t,y)}{y} \text{ for } x \ge y > 0 \text{ or } x \le y < 0, \ t \in \mathbb{T}\_{0,\delta}$$

*then f is said to be sublinear.*

> *Next, for convenience, we set*

$$P(t) = \int\_{t\_0}^{t} \frac{1}{p(s)} \Delta s, \ \not P(t) = \int\_{t}^{\infty} \frac{1}{p(s)} \Delta s.$$

**Lemma 1.** *Assume that* ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* = ∞ *and x*(*t*) *is an eventually positive solution of Equation (1). Then, there exist c*1 > 0, *c*2 > 0 *and t*1 ∈ T0 *such that*

$$x^{\Lambda}(t) > 0, \ c\_1 \le x(t) \le c\_2 P(t), \ t \ge t\_1.$$

**Proof.** Choose *t* ≥ *t*0 sufficiently large such that *x*(*η*(*t*)) > 0. Suppose that there exists *t*1 > *t*0 such that *<sup>x</sup>*<sup>Δ</sup>(*<sup>t</sup>*1) ≤ 0. Integrating Equation (1) from *t*1 to *t*, we ge<sup>t</sup>

$$p(t)\mathbf{x}^{\Delta}(t) - p(t\_1)\mathbf{x}^{\Delta}(t\_1) + \int\_{t\_1}^{t} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} = 0. \tag{2}$$

Dividing Equation (2) by *p*(*t*), and then integrating from *<sup>t</sup>*2(<sup>&</sup>gt; *<sup>t</sup>*1) to *t*, we have

$$\mathbf{x}(t) - \mathbf{x}(t\_2) - p(t\_1)\mathbf{x}^{\Lambda}(t\_1) \int\_{t\_2}^{t} \frac{1}{p(\mathbf{s})} \Delta \mathbf{s} + \int\_{t\_2}^{t} \left[ \frac{1}{p(\mathbf{s})} \int\_{t\_1}^{\mathbf{s}} \mathbf{g}(\theta, \mathbf{x}(\eta(\theta))) \Delta \theta \right] \Delta \mathbf{s} = \mathbf{0}. \tag{3}$$

Noting that sgn *g*(*<sup>θ</sup>*, *x*) = sgn *x* and *<sup>x</sup>*<sup>Δ</sup>(*<sup>t</sup>*1) ≤ 0, after the transposition of terms, letting *t* → <sup>∞</sup>, we ge<sup>t</sup> *x*(*t*) → <sup>−</sup>∞, which is a contradiction. Hence, *x*<sup>Δ</sup>(*t*) > 0. Therefore, there exists *c*1 > 0 such that *x*(*t*) ≥ *c*1. By Equation (3), there exists *c*2 > 0 such that *x*(*t*) ≤ *<sup>c</sup>*2*<sup>P</sup>*(*t*). The proof is complete.

**Lemma 2.** *Assume that* ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* < ∞ *and x*(*t*) *is an eventually positive solution of Equation (1). Then, there exist c*1 > 0, *c*2 > 0 *and t*1 ∈ T0 *such that*

$$x(t) \ge -p(t)x^{\Delta}(t)\mathcal{P}(t),\ c\_1\mathcal{P}(t) \le x(t) \le c\_{2\prime} \ t \ge t\_1.s$$

**Proof.** Let *t* ≥ *t*0 be sufficiently large so that *x*(*η*(*t*)) > 0. Then, it follows from Equation (1) and sgn *g*(*<sup>t</sup>*, *x*) = sgn *x* that (*p*(*t*)*x*<sup>Δ</sup>(*t*))<sup>Δ</sup> < 0, for *t* ≥ *t*1. Hence,

$$p(s)\mathbf{x}^{\Lambda}(s) \le p(t)\mathbf{x}^{\Lambda}(t) \quad s > t \ge t\_1. \tag{4}$$

Dividing Equation (4) by *p*(*s*), and then integrating from *t* to *<sup>t</sup>*2(<sup>&</sup>gt; *<sup>t</sup>*1), we have

$$
\varkappa(t\_2) - \varkappa(t) \le p(t)\varkappa^\Delta(t) \int\_t^{t\_2} \frac{1}{p(s)} \Delta s.
$$

We now show that lim*<sup>t</sup>*→∞ *x*(*t*) < ∞. If not, let lim*<sup>t</sup>*→∞ *x*(*t*) = ∞. Integrating Equation (1) from *t*0 to *t*, we ge<sup>t</sup>

$$p(t)\mathbf{x}^{\Lambda}(t) - p(t\_0)\mathbf{x}^{\Lambda}(t\_0) + \int\_{t\_0}^{t} \mathbf{g}(s, \mathbf{x}(\eta(s))) \Delta s = 0,$$

Dividing the above equation by *p*(*t*), and then integrating from *t*0 to *t* yields

$$\mathbf{x}(t) = \mathbf{x}(t\_0) + \hat{P}(t\_0)p(t\_0)\mathbf{x}^\Lambda(t\_0) - \hat{P}(t)p(t\_0)\mathbf{x}^\Lambda(t\_0) - \int\_{t\_0}^t \left[\frac{1}{p(s)} \int\_{t\_0}^s \mathbf{g}(\theta, \mathbf{x}(\eta(\theta))) \Delta \theta\right] \Delta s.$$

Hence, we obtain that *P* <sup>ˆ</sup>(*t*) → ∞ as *t* → <sup>∞</sup>, which is a contradiction. Consequently, we ge<sup>t</sup>

$$\mathbf{x}(t) \ge -p(t)\mathbf{x}^{\Delta}(t) \int\_{t}^{\infty} \frac{1}{p(\mathbf{s})} \Delta \mathbf{s}.$$

Thus, the first part of the lemma holds. On the other hand, dividing Equation (4) by *p*(*s*), and then integrating from *t*1 to *t*, we have

$$\mathbf{x}(t) \le \mathbf{x}(t\_1) + p(t\_1)\mathbf{x}^{\Delta}(t\_1) \int\_{t\_1}^{t} \frac{1}{p(\mathbf{s})} \Delta \mathbf{s} \le \mathbf{x}(t\_1) + |p(t\_1)\mathbf{x}^{\Delta}(t\_1)| \mathcal{P}(t\_1) \stackrel{\Delta}{=} c\_2.$$

Since *p*(*t*)*x*<sup>Δ</sup>(*t*) is decreasing, we ge<sup>t</sup>

$$x(t) + p(t\_1)x^{\Lambda}(t\_1) \int\_t^{\infty} \frac{1}{p(s)} \Delta s \ge x(t) + p(t)x^{\Lambda}(t) \int\_t^{\infty} \frac{1}{p(s)} \Delta s \ge 0.$$

If *<sup>x</sup>*<sup>Δ</sup>(*<sup>t</sup>*1) < 0, then *x*(*t*) ≥ |*p*(*<sup>t</sup>*1)*x*<sup>Δ</sup>(*<sup>t</sup>*1)|*P*<sup>ˆ</sup>(*t*) = *<sup>c</sup>*1*P*<sup>ˆ</sup>(*t*). If *<sup>x</sup>*<sup>Δ</sup>(*<sup>t</sup>*1) ≥ 0, then we can assume that *x*<sup>Δ</sup>(*t*) ≥ 0, for *t* ≥ *t*1. Otherwise, by choosing *t*2 = *t*1, we repeat the above process. Thus, *x*(*t*) is nondecreasing for *t* ≥ *t*1. Therefore,

$$\mathbf{x}(t) \ge \mathbf{x}(t\_1) = \frac{\mathbf{x}(t\_1)}{\mathcal{P}(t\_1)} \mathcal{P}(t\_1) \ge \frac{\mathbf{x}(t\_1)}{\mathcal{P}(t\_1)} \mathcal{P}(t) \triangleq c\_1 \mathcal{P}(t).$$

The proof is complete.

**Remark 1.** *If x*(*t*) *is an eventually negative solution of Equation (1), then there are analogous conclusions to Lemma 1 and Lemma 2, in which we just need to change the sign of constants c*1*, c*2 *into negative values and inverse the sign of inequalities.*

**Theorem 1.** *Let S denote the set of all non-oscillatory solutions of Equation (1). Assume that* ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* = ∞*. Then, any non-oscillatory solutions of Equation (1) must belong to one of the following classes:*

$$\begin{aligned} A^0\_\mathcal{c} &= \{ \mathbf{x}(t) \in \mathbb{S} : \lim\_{t \to \infty} \mathbf{x}(t) = \mathbf{c} \neq \mathbf{0}, \lim\_{t \to \infty} p(t) \mathbf{x}^\Delta(t) = \mathbf{0} \}, \\ A^\mathcal{c}\_\infty &= \{ \mathbf{x}(t) \in \mathbb{S} : \lim\_{t \to \infty} \mathbf{x}(t) = \infty, \lim\_{t \to \infty} p(t) \mathbf{x}^\Delta(t) = \mathbf{c} \neq \mathbf{0} \}, \\ A^0\_\infty &= \{ \mathbf{x}(t) \in \mathbb{S} : \lim\_{t \to \infty} \mathbf{x}(t) = \infty, \lim\_{t \to \infty} p(t) \mathbf{x}^\Delta(t) = \mathbf{0} \}. \end{aligned}$$

**Proof.** Without loss of generality, let *x*(*t*) be an eventually positive solution of Equation (1). By Lemma 1, it is easy to see that either lim*<sup>t</sup>*→∞ *x*(*t*) = *c* > 0, or lim*<sup>t</sup>*→∞ *<sup>x</sup>*(*t*)=+<sup>∞</sup>.

(i) If lim*<sup>t</sup>*→∞ *x*(*t*) = *c* > 0, then *x*(*t*) and *p*(*t*)*x*<sup>Δ</sup>(*t*) are eventually positive. From Equation (1), since (*p*(*t*)*x*<sup>Δ</sup>(*t*))<sup>Δ</sup> ≤ 0, that is, *p*(*t*)*x*<sup>Δ</sup>(*t*) is nonincreasing, so lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) exists. Now, we will show that lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = 0. On the contrary, suppose that lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = *c* > 0. Furthermore, *tt*0*x*<sup>Δ</sup>(*s*)<sup>Δ</sup>*s* ≥ *tt*0*c p*(*s*)Δ*s*. Thus, lim*<sup>t</sup>*→∞ *x*(*t*) = <sup>∞</sup>, which leads to a contradiction.

(ii) If lim*<sup>t</sup>*→∞ *<sup>x</sup>*(*t*)=+<sup>∞</sup>, then, in view of the fact that lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) exists, it follows by L'Hôpital's rule that

$$\lim\_{t \to \infty} \frac{x(t)}{P(t)} = \lim\_{t \to \infty} p(t)x^{\Lambda}(t).$$

On the other hand, by Lemma 1, we ge<sup>t</sup>

$$0 \le \frac{\mathfrak{x}(t)}{P(t)} \le c\_2.$$

Therefore, either lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = *c* = 0 or lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = 0.

**Theorem 2.** *Let S denote the set of all non-oscillatory solutions of Equation (1). Assume that* ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* < ∞*. Then, any non-oscillatory solutions of Equation (1) must belong to one of the following classes:*

$$\begin{aligned} A\_{\mathfrak{c}} &= \{ \mathfrak{x}(t) \in \mathcal{S} : \lim\_{t \to \infty} \mathfrak{x}(t) = \mathfrak{c} \neq \mathbf{0} \}, \\ A\_{\mathfrak{d}}^{\mathfrak{c}} &= \{ \mathfrak{x}(t) \in \mathcal{S} : \lim\_{t \to \infty} \mathfrak{x}(t) = \mathfrak{0}, \lim\_{t \to \infty} p(t) \mathfrak{x}^{\mathfrak{A}}(t) = \mathfrak{c} \neq \mathbf{0} \}, \\ A\_{\mathfrak{d}}^{\infty} &= \{ \mathfrak{x}(t) \in \mathcal{S} : \lim\_{t \to \infty} \mathfrak{x}(t) = \mathfrak{0}, \lim\_{t \to \infty} p(t) \mathfrak{x}^{\mathfrak{A}}(t) = \infty \}. \end{aligned}$$

**Proof.** Without loss of generality, let *x*(*t*) be an eventually positive solution of Equation (1). By Equation (1), for sufficiently large *t*, we have that (*p*(*t*)*x*<sup>Δ</sup>(*t*))<sup>Δ</sup> < 0. Then, *p*(*t*)*x*<sup>Δ</sup>(*t*) and *x*<sup>Δ</sup>(*t*) are monotone and have eventually the same sign (either positive or negative). Firstly, we show that lim*<sup>t</sup>*→∞ *x*(*t*) = ∞ does not hold. Indeed, if lim*<sup>t</sup>*→∞ *x*(*t*) = <sup>∞</sup>, then we ge<sup>t</sup> by integrating Equation (1) from *t*0 to *t* that

$$p(t)\mathbf{x}^{\Lambda}(t) - p(t\_0)\mathbf{x}^{\Lambda}(t\_0) + \int\_{t\_0}^{t} \mathbf{g}(s, \mathbf{x}(\eta(s))) \Delta s = 0.$$

Dividing the above equation by *p*(*t*), and then integrating from *t*0 to *t*, we obtain

$$\begin{split} \mathbf{x}(t) &= \quad \mathbf{x}(t\_0) + p(t\_0)\mathbf{x}^\mathbf{A}(t\_0) \int\_{t\_0}^t \frac{1}{p(\theta)} \Delta \theta \\ &- \int\_{t\_0}^t \frac{1}{p(\theta)} \int\_{t\_0}^\theta \mathbf{g}(s, \mathbf{x}(\eta(s))) \Delta s \Delta \theta \\ &= \quad \mathbf{x}(t\_0) + p(t\_0)\mathbf{x}^\mathbf{A}(t\_0) \left[ \int\_{t\_0}^\infty \frac{1}{p(\theta)} \Delta \theta - \int\_t^\infty \frac{1}{p(\theta)} \Delta \theta \right] \\ &- \int\_{t\_0}^t \frac{1}{p(\theta)} \int\_{t\_0}^\theta \mathbf{g}(s, \mathbf{x}(\eta(s))) \Delta s \Delta \theta. \end{split}$$

Therefore, lim*<sup>t</sup>*→∞ ∞*t* 1 *p*(*θ*)Δ*<sup>θ</sup>* = <sup>−</sup>∞, which is a contradiction. Hence, either lim*<sup>t</sup>*→∞ *x*(*t*) = *c* = 0 or lim*<sup>t</sup>*→∞ *x*(*t*) = 0. Since *p*(*t*)*x*<sup>Δ</sup>(*t*) has a deterministic sign and it is monotone, this means that either lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) exists or lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = ∞. If lim*<sup>t</sup>*→∞ *x*(*t*) = 0 and lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) exists, then, by L'Hôpital's rule,

$$\lim\_{t \to \infty} \frac{\mathbf{x}(t)}{\mathcal{P}(t)} = -\lim\_{t \to \infty} p(t)\mathbf{x}^{\mathbf{A}}(t).$$

By Lemma 2, either lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = *c* = 0, or lim*<sup>t</sup>*→∞ *p*(*t*)*x*<sup>Δ</sup>(*t*) = ∞.

#### **3. Existence of Non-Oscillatory Solutions**

In this section, we establish sufficient and necessary conditions for existence of some kinds of non-oscillatory solutions for Equation (1).

**Theorem 3.** *Assume that* (i) ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* = ∞*;* (ii) *g*(*<sup>t</sup>*, *x*) *is superlinear or sublinear.* *Then, Equation (1) has a non-oscillatory solution x*(*t*) ∈ *A*0*c if and only if*

$$\int\_{t\_0}^{\infty} P(\sigma(s)) |g(s,k)| \Delta s < \infty, \text{ for some } k \neq 0. \tag{5}$$

**Proof.** *Necessity*. Without loss of generality, let *x*(*t*) ∈ *A*0*c* be eventually positive. By Lemma 1, there exist *c*1 > 0, *c*2 > 0 and *t*1 ∈ T0 such that

$$x^{\Lambda}(t) > 0, \ c\_1 \le x(\eta(t)) \le c\_{2\prime} \ t \ge t\_1.$$

Multiplying Equation (1) by *<sup>P</sup>*(*σ*(*t*)), and then integrating from *t*1 to *t*, we have

$$P(t)p(t)\mathbf{x}^{\Lambda}(t) - P(t\_1)p(t\_1)\mathbf{x}^{\Lambda}(t\_1) - \mathbf{x}(t) + \mathbf{x}(t\_1) + \int\_{t\_1}^{t} P(\sigma(\mathbf{s})) \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} = 0.1$$

Since sgn *g*(*<sup>s</sup>*, *x*) = sgn *x*, it follows that the first term of above identity is finite as *t* tends to infinity. Therefore,

$$\int\_{t\_1}^{\infty} P(\sigma(\mathbf{s})) \mathbf{g}(\mathbf{s}, \mathfrak{x}(\eta(\mathbf{s}))) \Delta s < \infty.$$

If *g* is superlinear, then

$$\mathcal{g}(t, c\_1) \le \frac{c\_1 \mathcal{g}(t, \mathfrak{x}(\eta(s)))}{\mathfrak{x}(\eta(s))} \le \mathcal{g}(t, \mathfrak{x}(\eta(s))).$$

which implies that

$$\int\_{t\_1}^{\infty} P(\sigma(s)) g(s, c\_1) \Delta s < \infty.$$

Similarly, if *g* is sublinear, then

$$\int\_{t\_1}^{\infty} P(\sigma(s)) g(s, c\_2) \Delta s < \infty.$$

*Sufficiency*. Without loss of generality, we let *k* > 0. If *g* is superlinear, then let *c* = *k*/2; if *g* is sublinear, then let *c* = *k*.

Choose *T* > *t*0 so large that

$$\int\_{T}^{\infty} P(\sigma(s)) |\mathcal{g}(s, c)| \Delta s < \frac{\mathcal{L}}{2}.$$

Let

$$X = \left\{ \mathbf{x} \mid \mathbf{x} \in \mathbb{C}\_{rd}(\mathbb{T}\_{0}, \mathbb{R}), \sup\_{t \in \mathbb{T}\_{0}} |\mathbf{x}(t)| < \infty \right\}.$$

Endowed on *X* with the norm *x* = sup*<sup>t</sup>*∈T0 |*x*(*t*)|, *X* is a Banach space. We define the set

$$\Omega = \{ \mathbf{x} = \mathbf{x}(t) : \mathbf{x} \in X, \ c \le \mathbf{x}(t) \le 2c, \ t \in \mathbb{T}\_0 \}.$$

Clearly, Ω is a bounded, closed and convex subset of *X*. Define the map S on Ω as follows:

$$(\mathcal{S}x)(t) = \begin{cases} \displaystyle \mathbf{c} + \int\_{T}^{t} \mathbf{P}(\sigma(\mathbf{s})) \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} + \mathbf{P}(t) \int\_{t}^{\infty} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s}, & t \ge T, \\\ (\mathcal{S}x)(T), & t \le t \le T. \end{cases}$$

Step I. S maps Ω into Ω. Obviously, letting *x* = *x*(*t*) ∈ Ω, we have *c* ≤ *x*(*t*) ≤ 2*c* for *t* ≥ *T*. Then,

$$\mathcal{L} \le (\mathcal{S}\mathbf{x})(t) \le \mathfrak{c} + \int\_{T}^{\infty} P(\sigma(\mathbf{s})) \underline{\chi}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} \\ < \mathfrak{c} + 2 \int\_{T}^{\infty} P(\sigma(\mathbf{s})) \underline{\chi}(\mathbf{s}, \mathbf{c}) \Delta \mathbf{s} \le 2 \mathfrak{c}.$$

Hence, *c* ≤ (S*x*)(*t*) ≤ 2*c*, for *t* ∈ T0. Therefore, SΩ ⊆ Ω.

Step II. S is completely continuous.

We first claim that S is continuous. Let *xn* ∈ Ω and *xn* − *x* → 0 as *n* → ∞. Since Ω is a closed set, *x* ∈ Ω. For *t* ≥ *T*, we ge<sup>t</sup>

$$|(\mathcal{S}x\_n)(t) - (\mathcal{S}x)(t)| \le \int\_T^{\infty} P(\sigma(s)) |\mathcal{g}(s, x\_n(s)) - \mathcal{g}(s, x(s))| \, |\Delta s|$$

Since |*g*(*<sup>s</sup>*, *xn*(*s*)) − *g*(*<sup>s</sup>*, *x*(*s*))| → 0 as *n* → <sup>∞</sup>, so, by using the Lebesgue dominated convergence theorem, we conclude that lim*n*→∞ S*xn* − S*x* = 0, which implies that S is continuous in Ω.

Next, we show that SΩ is relatively compact. It suffices to prove that the family of functions {S*x* : *x* ∈ Ω} is bounded and uniformly Cauchy, and {S*x* : *x* ∈ Ω} is equi-continuous on [*<sup>t</sup>*0, *<sup>T</sup>*1] for any *T*1 ∈ [*<sup>t</sup>*0, <sup>∞</sup>). The boundedness is obvious. By Equation (5), for any *ε* > 0, let *T*∗ ≥ *T* be so large that 

$$\int\_{T^\*}^{\infty} P(\sigma(s)) g(s, c) \,\Delta s < \frac{\varepsilon}{6}.$$

Then, for *x* ∈ Ω, *t*2 > *t*1 ≥ *T*<sup>∗</sup>, we have

$$\begin{split} & \left| (\mathcal{S}\mathbf{x})(t\_{2}) - (\mathcal{S}\mathbf{x})(t\_{1}) \right| \\ & \leq \left| \int\_{t\_{1}}^{t\_{2}} P(\sigma(s)) g(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ & + \left| P(t\_{2}) \int\_{t\_{2}}^{\infty} g(s, \mathbf{x}(\eta(s))) \Delta s \right| + \left| P(t\_{1}) \int\_{t\_{1}}^{\infty} g(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ & \leq \left| \int\_{t\_{1}}^{t\_{2}} P(\sigma(s)) g(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ & + \left| \int\_{t\_{2}}^{\infty} P(\sigma(s)) g(s, \mathbf{x}(\eta(s))) \Delta s \right| + \left| \int\_{t\_{1}}^{\infty} P(\sigma(s)) g(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ & \leq 3 \left| \int\_{T^{\*}}^{\infty} P(\sigma(s)) g(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ & \leq 6 \left| \int\_{T^{\*}}^{\infty} P(\sigma(s)) g(s, \mathbf{c}) \Delta s \right| < \varepsilon. \end{split}$$

Hence, {S*x* : *x* ∈ Ω} is uniformly Cauchy. Furthermore, for any *T*1 ∈ [*<sup>t</sup>*0, ∞) and *x* ∈ Ω with *T* ≤ *t*1 < *t*2 ≤ *T*1, we ge<sup>t</sup>

$$\begin{aligned} &| (\mathcal{S}x)(t\_2) - (\mathcal{S}x)(t\_1) | \\ &\le \left| \int\_{t\_1}^{t\_2} P(\sigma(s)) \mathcal{g}(s, \mathbf{x}(\eta(s))) \Delta s \right. \\ &\left. + [P(t\_2) - P(t\_1)] \int\_{t\_1}^{\infty} \mathcal{g}(s, \mathbf{x}(\eta(s))) \Delta s - P(t\_2) \int\_{t\_1}^{t\_2} \mathcal{g}(s, \mathbf{x}(\eta(s))) \Delta s \right| \\ &\le \left| \frac{c |P(t\_2) - P(t\_1)|}{P(T)} + P(T\_1) \int\_{t\_1}^{t\_2} \mathcal{g}(s, c) \Delta s . \end{aligned} \right| $$

Hence, there exists a *δ* > 0 such that

$$|(\mathcal{S}x)(t\_2) - (\mathcal{S}x)(t\_1)| < \varepsilon, \text{ when } 0 < t\_2 - t\_1 < \delta.$$

From the definition of operator S, clearly, we have

$$|(\mathcal{S}\mathfrak{x})(t\_2) - (\mathcal{S}\mathfrak{x})(t\_1)| = 0 < \varepsilon, \text{ when } t\_0 \le t\_1 < t\_2 \le T.$$

Thus, it follows that {S*x* : *x* ∈ Ω} is equi-continuous on [*<sup>t</sup>*0, *<sup>T</sup>*1]. Hence, S is completely continuous. By Schauder's fixed point theorem, we deduce that there exists a *x*0 ∈ Ω such that S*<sup>x</sup>*0 = *x*0, which is a non-oscillatory solution of Equation (1) with *x*0 ∈ *A*0*c* . The proof is completed.

#### **Theorem 4.** *Assume that*

(i) ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* = ∞*;*

(ii) *g*(*<sup>t</sup>*, *x*) *is superlinear or sublinear.*

> *Then, Equation (1) has a non-oscillatory solution x*(*t*) ∈ *Ac*∞ *if and only if*

$$\int\_{t\_0}^{\infty} |g(s, kP(\eta(s)))| \, \text{As} < \infty, \text{ for some } k \neq 0. \tag{6}$$

**Proof.** *Necessity*. Let *x*(*t*) ∈ *Ac*∞ be eventually positive. By Lemma 1 and *p*(*t*)*x*<sup>Δ</sup>(*t*) → *c* as *t* → <sup>∞</sup>, there exist *c*1 > 0, *c*2 > 0 and *t*1 ∈ T0 such that

$$\mathfrak{x}^{\Delta}(t) > 0,\ c\_1 P(\eta(t)) \le \mathfrak{x}(\eta(t)) \le c\_2 P(\eta(t)),\ t \ge t\_1.$$

Integrating Equation (1) from *t*1 to *t*, we have

$$p(t)\mathbf{x}^{\Delta}(t) - p(t\_1)\mathbf{x}^{\Delta}(t\_1) + \int\_{t\_1}^{t} \mathbf{g}(s, \mathbf{x}(\eta(s))) \Delta s = 0,$$

which implies that

$$\int\_{t\_1}^{\infty} g(s, x(\eta(s))) \Delta s < \infty.$$

If *g* is superlinear, then

$$\mathcal{g}(\mathbf{s}, \mathbf{c}\_1 P(\eta(\mathbf{s}))) \le \frac{\mathbf{c}\_1 P(\eta(\mathbf{s}))}{\mathbf{x}(\eta(\mathbf{s}))} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \le \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))),$$

which implies that

$$\int\_{t\_1}^{\infty} \lg(s\_\prime c\_1 P(\eta(s))) \,\Delta s < \infty.$$

Similarly, if *g* is sublinear, then

$$\int\_{t\_1}^{\infty} \lg(s\_\prime c\_2 P(\eta(s))) \Delta s < \infty.$$

*Sufficiency*. Without loss of generality, let *k* > 0. If *g* is superlinear, then let *c* = *k*/2; if *g* is sublinear, then let *c* = *k*.

Choose *T* > *t*0 so large that

$$\int\_{T}^{\infty} |g(s, cP(\eta(s)))| \text{As} < \frac{c}{2}.$$

Let

$$X = \left\{ \mathbf{x} \,|\, \mathbf{x} \in \mathbb{C}\_{rd}(\mathbb{T}\_{0}, \mathbb{R}), \sup\_{t \in \mathbb{T}\_{0}} \frac{|\mathbf{x}(t)|}{P(t)} < \infty \right\}. \,\, \mathbf{x}$$

Endowed on *X* with the norm *x* = sup*<sup>t</sup>*∈T0 |*x*(*t*)| *P*(*t*) , *X* is a Banach space. Introduce a set

$$\Omega = \{ \mathbf{x} = \mathbf{x}(t) : \mathbf{x} \in X, \ cP(t) \le \mathbf{x}(t) \le 2cP(t), \ t \in \mathbb{T}\_0 \}$$

.

Clearly, Ω is a bounded, closed and convex subset of *X*. Define a map S on Ω by *Mathematics* **2019**, *7*, 680

$$\mathcal{S}(\mathcal{S}\mathbf{x})(t) = \begin{cases} cP(t) + \int\_{T}^{t} P(\sigma(\mathbf{s})) \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} + P(t) \int\_{t}^{\infty} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s}, & t \ge T, \\\ (\mathcal{S}\mathbf{x})(T), & t \le T. \end{cases}$$

Step I. S maps Ω into Ω. Let *x* = *x*(*t*) ∈ Ω. Then, *cP*(*t*) ≤ *x*(*t*) ≤ 2*cP*(*t*) for *t* ≥ *T*, and

$$cP(t) \le (\mathcal{S}x)(t) \le cP(t) + P(t) \int\_T^{\infty} \mathcal{g}(s, x(\eta(s))) \Delta s \\
< P(t) \left[ c + 2 \int\_T^{\infty} \mathcal{g}(s, c) \Delta s \right] \\
\le 2cP(t).$$

Hence, SΩ ⊆ Ω.

Step II. S is completely continuous.

We first prove that S is continuous. Let *xn* ∈ Ω and *xn* − *x* → 0 as *n* → ∞. Since Ω is a closed set, *x* ∈ Ω. For *t* ≥ *T*, we ge<sup>t</sup>

$$|(\mathcal{S}\mathbf{x}\_n)(t) - (\mathcal{S}\mathbf{x})(t)| \le P(t) \int\_T^\infty |\mathbf{g}(\mathbf{s}, \mathbf{x}\_n(\mathbf{s})) - \mathbf{g}(\mathbf{s}, \mathbf{x}(\mathbf{s}))| \Delta \mathbf{s}.$$

Since |*g*(*<sup>s</sup>*, *xn*(*s*)) − *g*(*<sup>s</sup>*, *x*(*s*))| → 0 as *n* → <sup>∞</sup>, lim*n*→∞ S*xn* − S*x* = 0, which implies that S is continuous in Ω.

Next, we show SΩ is relatively compact. By Equation (6), for any *ε* > 0, let *t*∗ ≥ *T* be sufficiently large such that

$$\int\_{\mathfrak{r}^\*}^{\infty} |\lg(\mathfrak{s}\_\prime c P(\eta(\mathfrak{s})))| \Delta \mathfrak{s} < \frac{\varepsilon}{8}.$$

Since lim*<sup>t</sup>*→∞ *P*(*t*) = <sup>∞</sup>, there exists a *T*∗ ≥ *t*∗ such that

$$\frac{1}{P(t)} \left| \int\_{T}^{t^\*} P(\sigma(s)) \mathcal{g}(s, \mathfrak{x}(\eta(s)) \Delta s) \right| < \frac{\varepsilon}{8'} \quad \text{for } t \ge T^\*.$$

Hence, for *x* ∈ Ω, *t*2 > *t*1 ≥ *T*<sup>∗</sup>, we have

$$\begin{split} & \quad \left| (P^{-1} \mathcal{S} \boldsymbol{x})(t\_{2}) - (P^{-1} \mathcal{S} \boldsymbol{x})(t\_{1}) \right| \\ & \leq \quad \frac{1}{P(t\_{2})} \left| \int\_{T}^{t^{\*}} P(\sigma(s)) g(s, \boldsymbol{x}(\eta(s))) \Delta s \right| + \frac{1}{P(t\_{1})} \left| \int\_{T}^{t^{\*}} P(\sigma(s)) g(s, \boldsymbol{x}(\eta(s))) \Delta s \right| \\ & \quad + \left| \frac{1}{P(t\_{2})} \int\_{t^{\*}}^{t\_{2}} P(\sigma(s)) g(s, \boldsymbol{x}(\eta(s))) \Delta s \right| + \left| \frac{1}{P(t\_{1})} \int\_{t^{\*}}^{t\_{1}} P(\sigma(s)) g(s, \boldsymbol{x}(\eta(s))) \Delta s \right| \\ & \quad + \left| \int\_{t\_{1}}^{t\_{2}} g(s, \boldsymbol{x}(\eta(s))) \Delta s \right| \\ & \leq \quad \frac{\varepsilon}{4} + 3 \int\_{t^{\*}}^{\infty} |g(s, \boldsymbol{x}(\eta(s)))| \Delta s \\ & \leq \quad \frac{\varepsilon}{4} + 6 \int\_{t^{\*}}^{\infty} |g(s, \boldsymbol{c} P(\eta(s)))| \Delta s < \varepsilon. \end{split}$$

Hence, {S*x* : *x* ∈ Ω} is uniformly Cauchy. Furthermore, for any *T*1 ∈ [*<sup>t</sup>*0, ∞) and *x* ∈ Ω, if *T* ≤ *t*1 < *t*2 ≤ *T*1, then

$$\begin{aligned} &\left| (P^{-1}\mathcal{S}\boldsymbol{x})(t\_2) - (P^{-1}\mathcal{S}\boldsymbol{x})(t\_1) \right| \\ &= \left| \left[ \frac{1}{P(t\_2)} - \frac{1}{P(t\_1)} \right] \int\_{T}^{t\_1} P(\sigma(s)) \boldsymbol{g}(s, \boldsymbol{x}(\boldsymbol{\eta}(s))) \Delta s \\ &+ \frac{1}{P(t\_2)} \int\_{t\_1}^{t\_2} P(\sigma(s)) \boldsymbol{g}(s, \boldsymbol{x}(\boldsymbol{\eta}(s))) \Delta s + \int\_{t\_1}^{t\_2} \boldsymbol{g}(s, \boldsymbol{x}(\boldsymbol{\eta}(s))) \Delta s \right]. \end{aligned}$$

Hence, there exists a *δ* > 0 such that

$$|(P^{-1}\mathcal{S}\mathfrak{x})(t\_2) - (P^{-1}\mathcal{S}\mathfrak{x})(t\_1)| < \varepsilon,\text{ if } 0 < t\_2 - t\_1 < \delta.$$

From the definition of operator S, it is clear that

$$|(P^{-1}\mathcal{S}x)(t\_2) - (P^{-1}\mathcal{S}x)(t\_1)| = 0 < \varepsilon, \text{ if } t\_0 \le t\_1 < t\_2 \le T.$$

From the foregoing arguments, we deduce that {S*x* : *x* ∈ Ω} is equi-continuous on [*<sup>t</sup>*0, *<sup>T</sup>*1]. Hence, S is completely continuous. Consequently, by Schauder's fixed point theorem, there exists a *x*0 ∈ Ω such that S*<sup>x</sup>*0 = *x*0, which is a non-oscillatory solution of Equation (1) with *x*0 ∈ *Ac*∞. The proof is completed.

**Theorem 5.** *Assume that*

(i) ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* < ∞*;* (ii) *g*(*<sup>t</sup>*, *x*) *is superlinear or sublinear. Then,Equation(1)hasa non-oscillatorysolutionx*(*t*) ∈

$$\int\_{t\_0}^{\infty} \hat{P}(s) |g(s,k)| \Delta s < \infty, \text{ for some } k \neq 0.$$

*Ac if and only if*

**Proof.** *Necessity*. Let *x*(*t*) ∈ *Ac* be eventually positive. Firstly, we show that ∞*t*0 *<sup>P</sup>*<sup>ˆ</sup>(*s*)|*g*(*<sup>s</sup>*, *x*(*η*(*t*)))|<sup>Δ</sup>*s* < ∞. If not, multiplying Equation (1) by *P* <sup>ˆ</sup>(*t*), and then integrating from *t*0 to *t*, we have

$$
\hat{P}(t)p(t)\mathbf{x}^{\Delta}(t) - \hat{P}(t\_0)p(t\_0)\mathbf{x}^{\Delta}(t\_0) - \mathbf{x}(t) + \mathbf{x}(t\_0) + \int\_{t\_0}^{t} \hat{P}(s)\mathbf{g}(s, \mathbf{x}(\eta(s)))\Delta s = 0.
$$

Then, lim*<sup>t</sup>*→∞ *P* <sup>ˆ</sup>(*t*)*p*(*t*)*x*<sup>Δ</sup>(*t*) = <sup>−</sup>∞. Therefore, there exist *t*1 ≥ *t*0, *M* > 0 such that *P* <sup>ˆ</sup>(*t*)*p*(*t*)*x*<sup>Δ</sup>(*t*) ≤ <sup>−</sup>*M*, for *t* ≥ *t*1. Hence,

$$
\pi(t) - \pi(t\_1) \le M \ln \left( \frac{\vec{P}(t)}{\vec{P}(t\_1)} \right).
$$

Since lim*<sup>t</sup>*→∞ *P* <sup>ˆ</sup>(*t*) = 0, lim*<sup>t</sup>*→∞ ln *P*<sup>ˆ</sup>(*t*) *<sup>P</sup>*<sup>ˆ</sup>(*<sup>t</sup>*1) = <sup>−</sup>∞. This is a contradiction.

By Lemma 2, there exist *c*1 > 0, *c*2 > 0 and *t*1 ∈ T0 such that *c*1 ≤ *x*(*η*(*t*)) ≤ *c*2, *t* ≥ *t*1. Therefore, if *g* is superlinear, then 

$$\int\_{t\_1}^{\infty} \mathcal{P}(t) \mathcal{g}(s, c\_1) \Delta s < \infty.$$

If *g* is sublinear, then

$$\int\_{t\_1}^{\infty} \hat{P}(t) g(s, c\_2) \Delta s < \infty.$$

*Sufficiency*. Without loss of generality, we let *k* > 0. If *g* is superlinear, then let *c* = *k*/2; if *g* is sublinear, then let *c* = *k*.

Let *T* > *t*0 be large so that

$$\int\_{T}^{\infty} \hat{P}(\eta(s)) |g(s, c)| \Delta s < \frac{c}{2}.$$

Let

$$X = \left\{ x \mid x \in \mathbb{C}\_{rd}(\mathbb{T}\_{0}, \mathbb{R}), \sup\_{t \in \mathbb{T}\_{0}} |x(t)| < \infty \right\}. $$

Endowed on *X* with the norm *x* = sup*<sup>t</sup>*∈T0 |*x*(*t*)|, *X* is a Banach space. Define the set

$$\Omega = \{ \mathbf{x} = \mathbf{x}(t) : \mathbf{x} \in X, \ c \le \mathbf{x}(t) \le 2c, \ t \in \mathbb{T}\_0 \}\dots$$

Define a map S on Ω as follows:

$$\mathbf{f}(\mathcal{S}\mathbf{x})(t) = \begin{cases} \mathbf{c} + \hat{\mathbf{P}}(t) \int\_{T}^{t} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} + \int\_{t}^{\infty} \hat{\mathbf{P}}(\mathbf{s}) \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s}, & t \ge T, \\\ (\mathcal{S}\mathbf{x})(T), & t \le t \le T. \end{cases}$$

Similarly to the previous process, we can show that S has a fixed point *x*0, which is a non-oscillatory solution of Equation (1) with *x*0 ∈ *Ac*. The proof is complete.

#### **Theorem 6.** *Assume that*

(i) ∞*t*0 1*p*(*s*)Δ*<sup>s</sup>* < ∞*;* (ii) *g*(*<sup>t</sup>*, *x*) *is superlinear or sublinear.*

*Then,Equation(1)hasanon-oscillatorysolutionx*(*t*) ∈ *Ac*0*ifandonly*

$$\int\_{t\_0}^{\infty} |g(s, k\varPhi(\eta(s)))| \, |\Delta s| < \infty, \text{ for some } k \neq 0.$$

 *if*

**Proof.** *Necessity*. Let *x*(*t*) ∈ *Ac*0 be eventually positive. Then, lim*<sup>t</sup>*→∞ *x*(*t*) = 0, lim*<sup>t</sup>*→∞ *x*(*t*) *P*<sup>ˆ</sup>(*t*) = *c* = 0. Assume that *c* > 0. Then, there exist *c*1 > 0, *c*2 > 0 and *t*1 ∈ T0 such that *<sup>c</sup>*1*P* <sup>ˆ</sup>(*η*(*t*)) ≤ *x*(*η*(*t*)) ≤ *<sup>c</sup>*2*P* <sup>ˆ</sup>(*η*(*t*)), *t* ≥ *t*1. By Lemma 2, we have that *x*(*t*) ≥ <sup>−</sup>*p*(*t*)*x*<sup>Δ</sup>(*t*)*P*<sup>ˆ</sup>(*t*), *t* ≥ *t*1. Thus, −*p*(*t*)*x*<sup>Δ</sup>(*t*) ≤ *c*2, *t* ≥ *t*1. On the other hand,

$$\int\_{t\_1}^{t} \mathbf{g}(\mathbf{s}, \mathbf{x}(\eta(\mathbf{s}))) \Delta \mathbf{s} = p(t\_1) \mathbf{x}^{\Delta}(t\_1) - p(t) \mathbf{x}^{\Delta}(t) \le |p(t\_1) \mathbf{x}^{\Delta}(t\_1)| + c\_{2\Delta}$$

therefore

$$\int\_{t\_1}^{t} \lg(s\_\prime \ge (\eta(s))) \,\mathrm{\Delta s} < \infty.$$

If *g* is superlinear, then

$$\int\_{t\_1}^{\infty} \mathbf{g}\left(\mathbf{s}, c\_1 \mathcal{P}(\eta(\mathbf{s}))\right) \Delta \mathbf{s} < \infty.$$

If *g* is sublinear, then

$$\int\_{t\_1}^{\infty} g(s, c\_2 \hat{P}(\eta(s))) \Delta s < \infty.$$

*Sufficiency*. Without loss of generality, we let *k* > 0. If *g* is superlinear, then let *c* = *k*/2; if *g* is sublinear, then let *c* = *k*.

Let *T* > *t*0 be large so that

$$\int\_{T}^{\infty} \mathcal{P}(\eta(s)) |g(s, c)| \Delta s < \frac{c}{2}.$$

Let

$$X = \left\{ \mathbf{x} \,|\, \mathbf{x} \in \mathbb{C}\_{rd}(\mathbb{T}\_0, \mathbb{R}), \sup\_{t \in \mathbb{T}\_0} \frac{|\mathbf{x}(t)|}{\mathcal{P}(t)} < \infty \right\}.$$

Endowed on *X* with the norm *x* = sup*<sup>t</sup>*∈T0 |*x*(*t*)| *P*<sup>ˆ</sup>(*t*) , *X* is a Banach space. Define the set

$$\Omega = \left\{ \mathbf{x} = \mathbf{x}(t) : \mathbf{x} \in X, \ c\hat{P}(t) \le \mathbf{x}(t) \le 2c\hat{P}(t), \ t \in \mathbb{T}\_0 \right\}.$$

Define a map S on Ω as follows

$$\mathcal{S}(\mathcal{S}x)(t) = \begin{cases} c\mathcal{P}(t) + \mathcal{P}(t) \int\_{T}^{t} \mathcal{g}(s, x(\eta(s))) \Delta s + \int\_{t}^{\infty} \mathcal{P}(s) \mathcal{g}(s, x(\eta(s))) \Delta s, & t \ge T\_{\prime} \\\ (\mathcal{S}x)(T), & t\_{0} \le t \le T. \end{cases}$$

Using the earlier arguments, one can show that Equation (1) has a non-oscillatory solution in *Ac*0. The proof is complete.
