Structure of Non-Oscillatory Solutions for Second Order Dynamic Equations on Time Scales

Abstract

In this paper, we make a detailed analysis of the structure of non-oscillatory solutions for second order superlinear and sublinear dynamic equations on time scales. The sufficient and necessary conditions for existence of non-oscillatory solutions are presented.

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 [1,2,3,4,5] 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 [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] and the references cited therein.

Consider the second order dynamic equation on time scales

$${\left[p\left(t\right)x^{\Delta }\left(t\right)\right]}^{\Delta }+g(t,x\left(\eta \left(t\right)\right))=0,t\in {\mathbb{T}}_0\subseteq \mathbb{T},$$

(1)

where $p\in C_{rd}({\mathbb{T}}_0,{\mathbb{R}}^{+}),\eta \in C_{rd}({\mathbb{T}}_0,\mathbb{T})$, $g:{\mathbb{T}}_0\times \mathbb{R}\to \mathbb{R}$ is continuous and $\mathrm{sgn}g(t,x)=\mathrm{sgn}x$ for $t\in {\mathbb{T}}_0$, ${lim}_{t\to \infty }\eta \left(t\right)=\infty$.

Oscillation of the Equation (1) has been studied by Dosšlý 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 $\mathbb{T}$ be a time scale (i.e., a closed subset of the real numbers $\mathbb{R}$) with $sup\mathbb{T}=\infty$. We assume throughout that $\mathbb{T}$ has the topology that it inherits from the standard topology on the real numbers $\mathbb{R}$. For $t\in \mathbb{T}$, we define the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by $\sigma \left(t\right):=inf\{s\in \mathbb{T}:s>t\}$. Denote by $C_{rd}(\mathbb{T},\mathbb{R})$ the space consisting of all functions which are right-dense points in $\mathbb{T}$ and its left-sided limits exist (finite) at left-dense points in $\mathbb{T}$. Furthermore, let us put $[t_0,\infty ):={\mathbb{T}}_0=\{t\in \mathbb{T}:t_0\le t<\infty \}$.

Definition 1.

If

$$\frac{f(t,x)}{x}\ge \frac{f(t,y)}{y}\mathrm{for}x\ge y>0\mathrm{or}x\le y<0,t\in {\mathbb{T}}_0,$$

then f is said to be superlinear. If

$$\frac{f(t,x)}{x}\le \frac{f(t,y)}{y}\mathrm{for}x\ge y>0\mathrm{or}x\le y<0,t\in {\mathbb{T}}_0,$$

then f is said to be sublinear.

Next, for convenience, we set

$$P\left(t\right)={\int }_{t_0}^t\frac{1}{p\left(s\right)}\Delta s,\widehat{P}\left(t\right)={\int }_t^{\infty }\frac{1}{p\left(s\right)}\Delta s.$$

Lemma 1.

Assume that ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s=\infty$ and $x\left(t\right)$ is an eventually positive solution of Equation (1). Then, there exist $c_1>0,c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that

$$x^{\Delta }\left(t\right)>0,c_1\le x\left(t\right)\le c_{2}P\left(t\right),t\ge t_1.$$

Proof. 

Choose $t\ge t_0$ sufficiently large such that $x\left(\eta \right(t\left)\right)>0$. Suppose that there exists $t_1>t_0$ such that $x^{\Delta }\left(t_1\right)\le 0$. Integrating Equation (1) from $t_1$ to t, we get

$$p\left(t\right)x^{\Delta }\left(t\right)-p\left(t_1\right)x^{\Delta }\left(t_1\right)+{\int }_{t_1}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s=0.$$

(2)

Dividing Equation (2) by $p\left(t\right)$, and then integrating from $t_2(>t_1)$ to t, we have

$$x\left(t\right)-x\left(t_2\right)-p\left(t_1\right)x^{\Delta }\left(t_1\right){\int }_{t_2}^t\frac{1}{p\left(s\right)}\Delta s+{\int }_{t_2}^t[\frac{1}{p\left(s\right)}{\int }_{t_1}^{s}g(\theta ,x\left(\eta \left(\theta \right)\right))\Delta \theta ]\Delta s=0.$$

(3)

Noting that $\mathrm{sgn}g(\theta ,x)=\mathrm{sgn}x$ and $x^{\Delta }\left(t_1\right)\le 0$, after the transposition of terms, letting $t\to \infty$, we get $x\left(t\right)\to -\infty$, which is a contradiction. Hence, $x^{\Delta }\left(t\right)>0$. Therefore, there exists $c_1>0$ such that $x\left(t\right)\ge c_1$. By Equation (3), there exists $c_2>0$ such that $x\left(t\right)\le c_{2}P\left(t\right)$. The proof is complete. □

Lemma 2.

Assume that ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s<\infty$ and $x\left(t\right)$ is an eventually positive solution of Equation (1). Then, there exist $c_1>0,c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that

$$x\left(t\right)\ge -p\left(t\right)x^{\Delta }\left(t\right)\widehat{P}\left(t\right),c_1\widehat{P}\left(t\right)\le x\left(t\right)\le c_2,t\ge t_1.$$

Proof. 

Let $t\ge t_0$ be sufficiently large so that $x\left(\eta \right(t\left)\right)>0$. Then, it follows from Equation (1) and $\mathrm{sgn}g(t,x)=\mathrm{sgn}x$ that ${\left(p\left(t\right)x^{\Delta }\left(t\right)\right)}^{\Delta }<0$, for $t\ge t_1$. Hence,

$$p\left(s\right)x^{\Delta }\left(s\right)\le p\left(t\right)x^{\Delta }\left(t\right)s>t\ge t_1.$$

(4)

Dividing Equation (4) by $p\left(s\right)$, and then integrating from t to $t_2(>t_1)$, we have

$$x\left(t_2\right)-x\left(t\right)\le p\left(t\right)x^{\Delta }\left(t\right){\int }_t^{t_2}\frac{1}{p\left(s\right)}\Delta s.$$

We now show that ${lim}_{t\to \infty }x\left(t\right)<\infty$. If not, let ${lim}_{t\to \infty }x\left(t\right)=\infty$. Integrating Equation (1) from $t_0$ to t, we get

$$p\left(t\right)x^{\Delta }\left(t\right)-p\left(t_0\right)x^{\Delta }\left(t_0\right)+{\int }_{t_0}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s=0,$$

Dividing the above equation by $p\left(t\right)$, and then integrating from $t_0$ to t yields

$$x\left(t\right)=x\left(t_0\right)+\widehat{P}\left(t_0\right)p\left(t_0\right)x^{\Delta }\left(t_0\right)-\widehat{P}\left(t\right)p\left(t_0\right)x^{\Delta }\left(t_0\right)-{\int }_{t_0}^t\left[\frac{1}{p\left(s\right)}{\int }_{t_0}^{s}g(\theta ,x\left(\eta \left(\theta \right)\right))\Delta \theta \right]\Delta s.$$

Hence, we obtain that $\widehat{P}\left(t\right)\to \infty$ as $t\to \infty$, which is a contradiction. Consequently, we get

$$x\left(t\right)\ge -p\left(t\right)x^{\Delta }\left(t\right){\int }_t^{\infty }\frac{1}{p\left(s\right)}\Delta s.$$

Thus, the first part of the lemma holds. On the other hand, dividing Equation (4) by $p\left(s\right)$, and then integrating from $t_1$ to t, we have

$$x\left(t\right)\le x\left(t_1\right)+p\left(t_1\right)x^{\Delta }\left(t_1\right){\int }_{t_1}^t\frac{1}{p\left(s\right)}\Delta s\le x\left(t_1\right)+\left|p\left(t_1\right)x^{\Delta }\left(t_1\right)\right|\widehat{P}\left(t_1\right)\triangleq c_2.$$

Since $p\left(t\right)x^{\Delta }\left(t\right)$ is decreasing, we get

$$x\left(t\right)+p\left(t_1\right)x^{\Delta }\left(t_1\right){\int }_t^{\infty }\frac{1}{p\left(s\right)}\Delta s\ge x\left(t\right)+p\left(t\right)x^{\Delta }\left(t\right){\int }_t^{\infty }\frac{1}{p\left(s\right)}\Delta s\ge 0.$$

If $x^{\Delta }\left(t_1\right)<0$, then $x\left(t\right)\ge |p\left(t_1\right)x^{\Delta }\left(t_1\right)|\widehat{P}\left(t\right)=c_1\widehat{P}\left(t\right)$. If $x^{\Delta }\left(t_1\right)\ge 0$, then we can assume that $x^{\Delta }\left(t\right)\ge 0$, for $t\ge t_1$. Otherwise, by choosing $t_2=t_1$, we repeat the above process. Thus, $x\left(t\right)$ is nondecreasing for $t\ge t_1$. Therefore,

$$x\left(t\right)\ge x\left(t_1\right)=\frac{x\left(t_1\right)}{\widehat{P}\left(t_1\right)}\widehat{P}\left(t_1\right)\ge \frac{x\left(t_1\right)}{\widehat{P}\left(t_1\right)}\widehat{P}\left(t\right)\triangleq c_1\widehat{P}\left(t\right).$$

The proof is complete. □

Remark 1.

If $x\left(t\right)$ 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 ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s=\infty$. Then, any non-oscillatory solutions of Equation (1) must belong to one of the following classes:

$$\begin{array}{c}{\displaystyle A_c^0=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=c\ne 0,\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right)=0\},}\hfill \\ {\displaystyle A_{\infty }^c=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=\infty ,\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right)=c\ne 0\},}\hfill \\ {\displaystyle A_{\infty }^0=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=\infty ,\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right)=0\}.}\hfill \end{array}$$

Proof. 

Without loss of generality, let $x\left(t\right)$ be an eventually positive solution of Equation (1). By Lemma 1, it is easy to see that either ${lim}_{t\to \infty }x\left(t\right)=c>0$, or ${lim}_{t\to \infty }x\left(t\right)=+\infty$.

(i) If ${lim}_{t\to \infty }x\left(t\right)=c>0$, then $x\left(t\right)$ and $p\left(t\right)x^{\Delta }\left(t\right)$ are eventually positive. From Equation (1), since ${\left(p\left(t\right)x^{\Delta }\left(t\right)\right)}^{\Delta }\le 0$, that is, $p\left(t\right)x^{\Delta }\left(t\right)$ is nonincreasing, so ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)$ exists. Now, we will show that ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=0$. On the contrary, suppose that ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=c^{\prime }>0$. Furthermore, ${\int }_{t_0}^t{x}^{\Delta }\left(s\right)\Delta s\ge {\int }_{t_0}^t\frac{c^{\prime }}{p\left(s\right)}\Delta s$. Thus, ${lim}_{t\to \infty }x\left(t\right)=\infty$, which leads to a contradiction.

(ii) If ${lim}_{t\to \infty }x\left(t\right)=+\infty$, then, in view of the fact that ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)$ exists, it follows by L’Hôpital’s rule that

$$\underset{t\to \infty }{lim}\frac{x\left(t\right)}{P\left(t\right)}=\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right).$$

On the other hand, by Lemma 1, we get

$$0\le \frac{x\left(t\right)}{P\left(t\right)}\le c_2.$$

Therefore, either ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=c\ne 0$ or ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=0$. □

Theorem 2.

Let S denote the set of all non-oscillatory solutions of Equation (1). Assume that ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s<\infty$. Then, any non-oscillatory solutions of Equation (1) must belong to one of the following classes:

$$\begin{array}{c}{\displaystyle A_c=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=c\ne 0\},}\hfill \\ {\displaystyle A_0^c=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=0,\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right)=c\ne 0\},}\hfill \\ {\displaystyle A_0^{\infty }=\{x\left(t\right)\in S:\underset{t\to \infty }{lim}x\left(t\right)=0,\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right)=\infty \}.}\hfill \end{array}$$

Proof. 

Without loss of generality, let $x\left(t\right)$ be an eventually positive solution of Equation (1). By Equation (1), for sufficiently large t, we have that ${\left(p\left(t\right)x^{\Delta }\left(t\right)\right)}^{\Delta }<0$. Then, $p\left(t\right)x^{\Delta }\left(t\right)$ and $x^{\Delta }\left(t\right)$ are monotone and have eventually the same sign (either positive or negative). Firstly, we show that ${lim}_{t\to \infty }x\left(t\right)=\infty$ does not hold. Indeed, if ${lim}_{t\to \infty }x\left(t\right)=\infty$, then we get by integrating Equation (1) from $t_0$ to t that

$$p\left(t\right)x^{\Delta }\left(t\right)-p\left(t_0\right)x^{\Delta }\left(t_0\right)+{\int }_{t_0}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s=0.$$

Dividing the above equation by $p\left(t\right)$, and then integrating from $t_0$ to t, we obtain

$$\begin{array}{ccc}\hfill x\left(t\right)& =& x\left(t_0\right)+p\left(t_0\right)x^{\Delta }\left(t_0\right){\int }_{t_0}^t\frac{1}{p\left(\theta \right)}\Delta \theta \hfill \\ & & -{\int }_{t_0}^t\frac{1}{p\left(\theta \right)}{\int }_{t_0}^{\theta }g(s,x\left(\eta \left(s\right)\right))\Delta s\Delta \theta \hfill \\ & =& x\left(t_0\right)+p\left(t_0\right)x^{\Delta }\left(t_0\right)\left[{\int }_{t_0}^{\infty }\frac{1}{p\left(\theta \right)}\Delta \theta -{\int }_t^{\infty }\frac{1}{p\left(\theta \right)}\Delta \theta \right]\hfill \\ & & -{\int }_{t_0}^t\frac{1}{p\left(\theta \right)}{\int }_{t_0}^{\theta }g(s,x\left(\eta \left(s\right)\right))\Delta s\Delta \theta .\hfill \end{array}$$

Therefore, ${lim}_{t\to \infty }{\int }_t^{\infty }\frac{1}{p\left(\theta \right)}\Delta \theta =-\infty$, which is a contradiction. Hence, either ${lim}_{t\to \infty }x\left(t\right)=c\ne 0$ or ${lim}_{t\to \infty }x\left(t\right)=0$. Since $p\left(t\right)x^{\Delta }\left(t\right)$ has a deterministic sign and it is monotone, this means that either ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)$ exists or ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=\infty$. If ${lim}_{t\to \infty }x\left(t\right)=0$ and ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)$ exists, then, by L’Hôpital’s rule,

$$\underset{t\to \infty }{lim}\frac{x\left(t\right)}{\widehat{P}\left(t\right)}=-\underset{t\to \infty }{lim}p\left(t\right)x^{\Delta }\left(t\right).$$

By Lemma 2, either ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=c\ne 0$, or ${lim}_{t\to \infty }p\left(t\right)x^{\Delta }\left(t\right)=\infty$. □

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) ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s=\infty$;

(ii) $g(t,x)$ is superlinear or sublinear.

Then, Equation (1) has a non-oscillatory solution $x\left(t\right)\in A_c^0$ if and only if

$${\int }_{t_0}^{\infty }P\left(\sigma \left(s\right)\right)\left|g(s,k)\right|\Delta s<\infty ,forsomek\ne 0.$$

(5)

Proof. 

Necessity. Without loss of generality, let $x\left(t\right)\in A_c^0$ be eventually positive. By Lemma 1, there exist $c_1>0,c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that

$$x^{\Delta }\left(t\right)>0,c_1\le x\left(\eta \left(t\right)\right)\le c_2,t\ge t_1.$$

Multiplying Equation (1) by $P\left(\sigma \right(t\left)\right)$, and then integrating from $t_1$ to t, we have

$$P\left(t\right)p\left(t\right)x^{\Delta }\left(t\right)-P\left(t_1\right)p\left(t_1\right)x^{\Delta }\left(t_1\right)-x\left(t\right)+x\left(t_1\right)+{\int }_{t_1}^{t}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s=0.$$

Since $\mathrm{sgn}g(s,x)=\mathrm{sgn}x$, it follows that the first term of above identity is finite as t tends to infinity. Therefore,

$${\int }_{t_1}^{\infty }P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s<\infty .$$

If g is superlinear, then

$$g(t,c_1)\le \frac{c_{1}g(t,x\left(\eta \left(s\right)\right))}{x\left(\eta \right(s\left)\right)}\le g(t,x\left(\eta \left(s\right)\right)),$$

which implies that

$${\int }_{t_1}^{\infty }P\left(\sigma \left(s\right)\right)g(s,c_1)\Delta s<\infty .$$

Similarly, if g is sublinear, then

$${\int }_{t_1}^{\infty }P\left(\sigma \left(s\right)\right)g(s,c_2)\Delta s<\infty .$$

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

$${\int }_T^{\infty }P\left(\sigma \left(s\right)\right)\left|g(s,c)\right|\Delta s<\frac{c}{2}.$$

Let

$$X=\left\{x|x\in C_{rd}({\mathbb{T}}_0,\mathbb{R}),\underset{t\in {\mathbb{T}}_0}{sup}\left|x\left(t\right)\right|<\infty \right\}.$$

Endowed on X with the norm $\parallel x\parallel ={sup}_{t\in {\mathbb{T}}_0}\left|x\left(t\right)\right|$, X is a Banach space. We define the set

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

Clearly, $\Omega$ is a bounded, closed and convex subset of X. Define the map $\mathcal{S}$ on $\Omega$ as follows:

$$\left(\mathcal{S}x\right)\left(t\right)=\left\{\begin{array}{cc}{\displaystyle c+{\int }_T^{t}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s+P\left(t\right){\int }_t^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s,}\hfill & t\ge T,\hfill \\ \left(\mathcal{S}x\right)\left(T\right),\hfill & t_0\le t\le T.\hfill \end{array}\right.$$

Step I. $\mathcal{S}$ maps $\Omega$ into $\Omega$. Obviously, letting $x=x\left(t\right)\in \Omega$, we have $c\le x\left(t\right)\le 2c$ for $t\ge T$. Then,

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

Hence, $c\le \left(\mathcal{S}x\right)\left(t\right)\le 2c$, for $t\in {\mathbb{T}}_0$. Therefore, $\mathcal{S}\Omega \subseteq \Omega$.

Step II. $\mathcal{S}$ is completely continuous.

We first claim that $\mathcal{S}$ is continuous. Let $x_n\in \Omega$ and $\parallel x_n-x\parallel \to 0$ as $n\to \infty$. Since $\Omega$ is a closed set, $x\in \Omega$. For $t\ge T$, we get

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

Since $|g(s,x_n\left(s\right))-g(s,x\left(s\right))|\to 0$ as $n\to \infty$, so, by using the Lebesgue dominated convergence theorem, we conclude that ${lim}_{n\to \infty }\parallel \mathcal{S}{x}_n-\mathcal{S}x\parallel =0$, which implies that $\mathcal{S}$ is continuous in $\Omega$.

Next, we show that $\mathcal{S}\Omega$ is relatively compact. It suffices to prove that the family of functions $\{\mathcal{S}x:x\in \Omega \}$ is bounded and uniformly Cauchy, and $\{\mathcal{S}x:x\in \Omega \}$ is equi-continuous on $[t_0,T_1]$ for any $T_1\in [t_0,\infty )$. The boundedness is obvious. By Equation (5), for any $\epsilon >0$, let $T^{*}\ge T$ be so large that

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

Then, for $x\in \Omega ,t_2>t_1\ge T^{*}$, we have

$$\begin{array}{cc}& |\left(\mathcal{S}x\right)\left(t_2\right)-\left(\mathcal{S}x\right)\left(t_1\right)|\hfill \\ \le & |{\int }_{t_1}^{t_2}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ & +|P\left(t_2\right){\int }_{t_2}^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s|+|P\left(t_1\right){\int }_{t_1}^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ \le & |{\int }_{t_1}^{t_2}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ & +|{\int }_{t_2}^{\infty }P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|+|{\int }_{t_1}^{\infty }P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ \le & 3|{\int }_{T^{*}}^{\infty }P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ \le & 6|{\int }_{T^{*}}^{\infty }P\left(\sigma \left(s\right)\right)g(s,c)\Delta s|<\epsilon .\hfill \end{array}$$

Hence, $\{\mathcal{S}x:x\in \Omega \}$ is uniformly Cauchy.

Furthermore, for any $T_1\in [t_0,\infty )$ and $x\in \Omega$ with $T\le t_1<t_2\le T_1$, we get

$$\begin{array}{cc}& |\left(\mathcal{S}x\right)\left(t_2\right)-\left(\mathcal{S}x\right)\left(t_1\right)|\hfill \\ \le & |{\int }_{t_1}^{t_2}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s\hfill \\ & +[P\left(t_2\right)-P\left(t_1\right)]{\int }_{t_1}^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s-P\left(t_2\right){\int }_{t_1}^{t_2}g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ \le & \frac{c|P\left(t_2\right)-P\left(t_1\right)|}{P\left(T\right)}+P\left(T_1\right){\int }_{t_1}^{t_2}g(s,c)\Delta s.\hfill \end{array}$$

Hence, there exists a $\delta >0$ such that

$$|\left(\mathcal{S}x\right)\left(t_2\right)-\left(\mathcal{S}x\right)\left(t_1\right)|<\epsilon ,\mathrm{when}0<t_2-t_1<\delta .$$

From the definition of operator $\mathcal{S}$, clearly, we have

$$|\left(\mathcal{S}x\right)\left(t_2\right)-\left(\mathcal{S}x\right)\left(t_1\right)|=0<\epsilon ,\mathrm{when}{t}_0\le t_1<t_2\le T.$$

Thus, it follows that $\{\mathcal{S}x:x\in \Omega \}$ is equi-continuous on $[t_0,T_1]$. Hence, $\mathcal{S}$ is completely continuous. By Schauder’s fixed point theorem, we deduce that there exists a $x_0\in \Omega$ such that $\mathcal{S}{x}_0=x_0$, which is a non-oscillatory solution of Equation (1) with $x_0\in A_c^0$. The proof is completed.

Theorem 4.

Assume that

(i) ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s=\infty$;

(ii) $g(t,x)$ is superlinear or sublinear.

Then, Equation (1) has a non-oscillatory solution $x\left(t\right)\in A_{\infty }^c$ if and only if

$${\int }_{t_0}^{\infty }\left|g\right(s,kP\left(\eta \left(s\right)\right)|\Delta s<\infty ,forsomek\ne 0.$$

(6)

Proof. 

Necessity. Let $x\left(t\right)\in A_{\infty }^c$ be eventually positive. By Lemma 1 and $p\left(t\right)x^{\Delta }\left(t\right)\to c$ as $t\to \infty$, there exist $c_1>0,c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that

$$x^{\Delta }\left(t\right)>0,c_{1}P\left(\eta \left(t\right)\right)\le x\left(\eta \left(t\right)\right)\le c_{2}P\left(\eta \left(t\right)\right),t\ge t_1.$$

Integrating Equation (1) from $t_1$ to t, we have

$$p\left(t\right)x^{\Delta }\left(t\right)-p\left(t_1\right)x^{\Delta }\left(t_1\right)+{\int }_{t_1}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s=0,$$

which implies that

$${\int }_{t_1}^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s<\infty .$$

If g is superlinear, then

$$g(s,c_{1}P\left(\eta \left(s\right)\right))\le \frac{c_{1}P\left(\eta \left(s\right)\right)}{x\left(\eta \right(s\left)\right)}g(s,x\left(\eta \left(s\right)\right))\le g(s,x\left(\eta \left(s\right)\right)),$$

which implies that

$${\int }_{t_1}^{\infty }g(s,c_{1}P\left(\eta \left(s\right)\right))\Delta s<\infty .$$

Similarly, if g is sublinear, then

$${\int }_{t_1}^{\infty }g(s,c_{2}P\left(\eta \left(s\right)\right))\Delta s<\infty .$$

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

$${\int }_T^{\infty }\left|g\right(s,cP\left(\eta \left(s\right)\right)|\Delta s<\frac{c}{2}.$$

Let

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

Endowed on X with the norm $\parallel x\parallel ={sup}_{t\in {\mathbb{T}}_0}\frac{\left|x\right(t\left)\right|}{P\left(t\right)}$, X is a Banach space. Introduce a set

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

Clearly, $\Omega$ is a bounded, closed and convex subset of X. Define a map $\mathcal{S}$ on $\Omega$ by

$$\left(\mathcal{S}x\right)\left(t\right)=\left\{\begin{array}{cc}{\displaystyle cP\left(t\right)+{\int }_T^{t}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s+P\left(t\right){\int }_t^{\infty }g(s,x\left(\eta \left(s\right)\right))\Delta s,}\hfill & t\ge T,\hfill \\ \left(\mathcal{S}x\right)\left(T\right),\hfill & t_0\le t\le T.\hfill \end{array}\right.$$

Step I. $\mathcal{S}$ maps $\Omega$ into $\Omega$. Let $x=x\left(t\right)\in \Omega$. Then, $cP\left(t\right)\le x\left(t\right)\le 2cP\left(t\right)$ for $t\ge T$, and

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

Hence, $\mathcal{S}\Omega \subseteq \Omega$.

Step II. $\mathcal{S}$ is completely continuous.

We first prove that $\mathcal{S}$ is continuous. Let $x_n\in \Omega$ and $\parallel x_n-x\parallel \to 0$ as $n\to \infty$. Since $\Omega$ is a closed set, $x\in \Omega$. For $t\ge T$, we get

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

Since $|g(s,x_n\left(s\right))-g(s,x\left(s\right))|\to 0$ as $n\to \infty$, ${lim}_{n\to \infty }\parallel \mathcal{S}{x}_n-\mathcal{S}x\parallel =0$, which implies that $\mathcal{S}$ is continuous in $\Omega$.

Next, we show $\mathcal{S}\Omega$ is relatively compact. By Equation (6), for any $\epsilon >0$, let $t^{*}\ge T$ be sufficiently large such that

$${\int }_{t^{*}}^{\infty }\left|g\right(s,cP\left(\eta \left(s\right)\right)|\Delta s<\frac{\epsilon }{8}.$$

Since ${lim}_{t\to \infty }P\left(t\right)=\infty$, there exists a $T^{*}\ge t^{*}$ such that

$$\frac{1}{P\left(t\right)}|{\int }_T^{t^{*}}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right)\Delta s|<\frac{\epsilon }{8},\mathrm{for}t\ge T^{*}.$$

Hence, for $x\in \Omega ,t_2>t_1\ge T^{*}$, we have

$$\begin{array}{cc}& |\left(P^{-1}\mathcal{S}x\right)\left(t_2\right)-\left(P^{-1}\mathcal{S}x\right)\left(t_1\right)|\hfill \\ \le & \frac{1}{P\left(t_2\right)}|{\int }_T^{t^{*}}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|+\frac{1}{P\left(t_1\right)}|{\int }_T^{t^{*}}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ & +|\frac{1}{P\left(t_2\right)}{\int }_{t^{*}}^{t_2}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|+|\frac{1}{P\left(t_1\right)}{\int }_{t^{*}}^{t_1}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ & +|{\int }_{t_1}^{t_2}g(s,x\left(\eta \left(s\right)\right))\Delta s|\hfill \\ \le & \frac{\epsilon }{4}+3{\int }_{t^{*}}^{\infty }\left|g(s,x\left(\eta \left(s\right)\right))\right|\Delta s\hfill \\ \le & \frac{\epsilon }{4}+6{\int }_{t^{*}}^{\infty }\left|g(s,cP\left(\eta \left(s\right)\right))\right|\Delta s<\epsilon .\hfill \end{array}$$

Hence, $\{\mathcal{S}x:x\in \Omega \}$ is uniformly Cauchy. Furthermore, for any $T_1\in [t_0,\infty )$ and $x\in \Omega$, if $T\le t_1<t_2\le T_1$, then

$$\begin{array}{cc}& |\left(P^{-1}\mathcal{S}x\right)\left(t_2\right)-\left(P^{-1}\mathcal{S}x\right)\left(t_1\right)|\hfill \\ =& |\left[\frac{1}{P\left(t_2\right)}-\frac{1}{P\left(t_1\right)}\right]{\int }_T^{t_1}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s\hfill \\ & +\frac{1}{P\left(t_2\right)}{\int }_{t_1}^{t_2}P\left(\sigma \left(s\right)\right)g(s,x\left(\eta \left(s\right)\right))\Delta s+{\int }_{t_1}^{t_2}g(s,x\left(\eta \left(s\right)\right))\Delta s|.\hfill \end{array}$$

Hence, there exists a $\delta >0$ such that

$$|\left(P^{-1}\mathcal{S}x\right)\left(t_2\right)-\left(P^{-1}\mathcal{S}x\right)\left(t_1\right)|<\epsilon ,\mathrm{if}0<t_2-t_1<\delta .$$

From the definition of operator $\mathcal{S}$, it is clear that

$$|\left(P^{-1}\mathcal{S}x\right)\left(t_2\right)-\left(P^{-1}\mathcal{S}x\right)\left(t_1\right)|=0<\epsilon ,\mathrm{if}{t}_0\le t_1<t_2\le T.$$

From the foregoing arguments, we deduce that $\{\mathcal{S}x:x\in \Omega \}$ is equi-continuous on $[t_0,T_1]$. Hence, $\mathcal{S}$ is completely continuous. Consequently, by Schauder’s fixed point theorem, there exists a $x_0\in \Omega$ such that $\mathcal{S}{x}_0=x_0$, which is a non-oscillatory solution of Equation (1) with $x_0\in A_{\infty }^c$. The proof is completed. □

Theorem 5.

Assume that

(i) ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s<\infty$;

(ii) $g(t,x)$ is superlinear or sublinear.

Then, Equation (1) has a non-oscillatory solution $x\left(t\right)\in A_c$ if and only if

$${\int }_{t_0}^{\infty }\widehat{P}\left(s\right)\left|g(s,k)\right|\Delta s<\infty ,forsomek\ne 0.$$

Proof. 

Necessity. Let $x\left(t\right)\in A_c$ be eventually positive. Firstly, we show that ${\int }_{t_0}^{\infty }\widehat{P}\left(s\right)\left|g(s,x\left(\eta \left(t\right)\right))\right|\Delta s<\infty$. If not, multiplying Equation (1) by $\widehat{P}\left(t\right)$, and then integrating from $t_0$ to t, we have

$$\widehat{P}\left(t\right)p\left(t\right)x^{\Delta }\left(t\right)-\widehat{P}\left(t_0\right)p\left(t_0\right)x^{\Delta }\left(t_0\right)-x\left(t\right)+x\left(t_0\right)+{\int }_{t_0}^t\widehat{P}\left(s\right)g(s,x\left(\eta \left(s\right)\right))\Delta s=0.$$

Then, ${lim}_{t\to \infty }\widehat{P}\left(t\right)p\left(t\right)x^{\Delta }\left(t\right)=-\infty$. Therefore, there exist $t_1\ge t_0$, $M>0$ such that $\widehat{P}\left(t\right)p\left(t\right)x^{\Delta }\left(t\right)\le -M$, for $t\ge t_1$. Hence,

$$x\left(t\right)-x\left(t_1\right)\le Mln\left(\frac{\widehat{P}\left(t\right)}{\widehat{P}\left(t_1\right)}\right).$$

Since ${lim}_{t\to \infty }\widehat{P}\left(t\right)=0$, ${lim}_{t\to \infty }ln\left(\frac{\widehat{P}\left(t\right)}{\widehat{P}\left(t_1\right)}\right)=-\infty$. This is a contradiction.

By Lemma 2, there exist $c_1>0,c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that $c_1\le x\left(\eta \left(t\right)\right)\le c_2,t\ge t_1$. Therefore, if g is superlinear, then

$${\int }_{t_1}^{\infty }\widehat{P}\left(t\right)g(s,c_1)\Delta s<\infty .$$

If g is sublinear, then

$${\int }_{t_1}^{\infty }\widehat{P}\left(t\right)g(s,c_2)\Delta s<\infty .$$

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

$${\int }_T^{\infty }\widehat{P}\left(\eta \left(s\right)\right)\left|g(s,c)\right|\Delta s<\frac{c}{2}.$$

Let

$$X=\left\{x|x\in C_{rd}({\mathbb{T}}_0,\mathbb{R}),\underset{t\in {\mathbb{T}}_0}{sup}\left|x\left(t\right)\right|<\infty \right\}.$$

Endowed on X with the norm $\parallel x\parallel ={sup}_{t\in {\mathbb{T}}_0}\left|x\left(t\right)\right|$, X is a Banach space. Define the set

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

Define a map $\mathcal{S}$ on $\Omega$ as follows:

$$\left(\mathcal{S}x\right)\left(t\right)=\left\{\begin{array}{cc}{\displaystyle c+\widehat{P}\left(t\right){\int }_T^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s+{\int }_t^{\infty }\widehat{P}\left(s\right)g(s,x\left(\eta \left(s\right)\right))\Delta s,}\hfill & t\ge T,\hfill \\ \left(\mathcal{S}x\right)\left(T\right),\hfill & t_0\le t\le T.\hfill \end{array}\right.$$

Similarly to the previous process, we can show that $\mathcal{S}$ has a fixed point $x_0$, which is a non-oscillatory solution of Equation (1) with $x_0\in A_c$. The proof is complete. □

Theorem 6.

Assume that

(i) ${\int }_{t_0}^{\infty }\frac{1}{p\left(s\right)}\Delta s<\infty$;

(ii) $g(t,x)$ is superlinear or sublinear.

Then, Equation (1) has a non-oscillatory solution $x\left(t\right)\in A_0^c$ if and only if

$${\int }_{t_0}^{\infty }\left|g\right(s,k\widehat{P}\left(\eta \left(s\right)\right)|\Delta s<\infty ,forsomek\ne 0.$$

Proof. 

Necessity. Let $x\left(t\right)\in A_0^c$ be eventually positive. Then, ${lim}_{t\to \infty }x\left(t\right)=0$, ${lim}_{t\to \infty }\frac{x\left(t\right)}{\widehat{P}\left(t\right)}=c\ne 0$. Assume that $c>0$. Then, there exist $c_1>0$, $c_2>0$ and $t_1\in {\mathbb{T}}_0$ such that $c_1\widehat{P}\left(\eta \left(t\right)\right)\le x\left(\eta \left(t\right)\right)\le c_2\widehat{P}\left(\eta \left(t\right)\right),t\ge t_1$. By Lemma 2, we have that $x\left(t\right)\ge -p\left(t\right)x^{\Delta }\left(t\right)\widehat{P}\left(t\right),t\ge t_1$. Thus, $-p\left(t\right)x^{\Delta }\left(t\right)\le c_2,t\ge t_1$. On the other hand,

$${\int }_{t_1}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s=p\left(t_1\right)x^{\Delta }\left(t_1\right)-p\left(t\right)x^{\Delta }\left(t\right)\le \left|p\left(t_1\right)x^{\Delta }\left(t_1\right)\right|+c_2,$$

therefore

$${\int }_{t_1}^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s<\infty .$$

If g is superlinear, then

$${\int }_{t_1}^{\infty }g(s,c_1\widehat{P}\left(\eta \left(s\right)\right))\Delta s<\infty .$$

If g is sublinear, then

$${\int }_{t_1}^{\infty }g(s,c_2\widehat{P}\left(\eta \left(s\right)\right))\Delta s<\infty .$$

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

$${\int }_T^{\infty }\widehat{P}\left(\eta \left(s\right)\right)\left|g(s,c)\right|\Delta s<\frac{c}{2}.$$

Let

$$X=\left\{x|x\in C_{rd}({\mathbb{T}}_0,\mathbb{R}),\underset{t\in {\mathbb{T}}_0}{sup}\frac{\left|x\right(t\left)\right|}{\widehat{P}\left(t\right)}<\infty \right\}.$$

Endowed on X with the norm $\parallel x\parallel ={sup}_{t\in {\mathbb{T}}_0}\frac{\left|x\right(t\left)\right|}{\widehat{P}\left(t\right)}$, X is a Banach space. Define the set

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

Define a map $\mathcal{S}$ on $\Omega$ as follows

$$\left(\mathcal{S}x\right)\left(t\right)=\left\{\begin{array}{cc}{\displaystyle c\widehat{P}\left(t\right)+\widehat{P}\left(t\right){\int }_T^{t}g(s,x\left(\eta \left(s\right)\right))\Delta s+{\int }_t^{\infty }\widehat{P}\left(s\right)g(s,x\left(\eta \left(s\right)\right))\Delta s,}\hfill & t\ge T,\hfill \\ \left(\mathcal{S}x\right)\left(T\right),\hfill & t_0\le t\le T.\hfill \end{array}\right.$$

Using the earlier arguments, one can show that Equation (1) has a non-oscillatory solution in $A_0^c$. The proof is complete. □

4. Conclusions

In the current paper, the structure of non-oscillatory solutions for a class of second order dynamic equations on time scales is considered. Under the differentiable assumptions, we first establish two classifications of non-oscillatory solutions. Forthermore, by using the assumptions of superlinear and sublinear function, we obtain four sufficient and necessary conditions for existence of some kinds of non-oscillatory solutions.