Distributions of zeros of solutions to first order delay dynamic equations

Abstract

This paper is concerned with the distributions of zeros of solutions to first order delay dynamic equations on time scales. The results are obtained using iterative sequences.

1 Introduction

The oscillation and distributions of zeros of solutions of first order delay differential and difference equations are studied widely in the literature; see [1–18] and the references therein. However, there are only a few papers considering the distribution of zeros of solutions of first order delay and advanced dynamic equations on time scales (see [8, 19]). In [12], Zhou considered the first order delay differential equation

$$ x^{\prime} ( t ) +p ( t ) x ( t-\tau ) =0,\quad\text{for }t\in [ t_{0},\infty ) , $$

(1.1)

where

$$ \int_{t-\tau}^{t}p ( s ) \,ds\geq\rho>\frac{1}{e}, \quad\text{and}\quad \rho < 1, $$

(1.2)

and established lower and upper bounds for the quotient $x ( t-\tau ) /x ( t )$. In particular the author proved that $x ( t-\tau ) /x ( t ) \geq f_{n} ( \rho )$ and $x ( t-\tau ) /x ( t ) < g_{m} ( \rho )$, where the sequences $f_{n} ( \rho )$ and $g_{n} ( \rho )$ are defined by

$$ \textstyle\begin{cases} f_{1} ( \rho ) =e^{\rho},\qquad f_{n+1} ( \rho ) =e^{\rho f_{n} ( \rho ) },& n=1,2,\ldots, \\ g_{1} ( \rho ) =\frac{2 ( 1-\rho ) }{\rho ^{2}},\qquad g_{m+1} ( \rho ) =\frac{2 ( 1-\rho ) g_{m}^{2} ( \rho ) }{g_{m}^{2} ( \rho ) \rho^{2}+2},& m=1,2,\ldots, \end{cases} $$

and using these sequences the author studied the distribution of zeros of solutions of (1.1). In [13], Zhang and Zhou considered the first order delay differential equation

$$ x^{\prime} ( t ) +p ( t ) x \bigl( \tau ( t ) \bigr) =0,\quad\text{for }t\in [ t_{0},\infty ) , $$

(1.3)

and studied the distribution of zeros of solutions using the two sequences $f_{n} ( \rho )$ and $g_{m} ( \rho )$ where

$$ \textstyle\begin{cases} f_{0} ( \rho ) =1,\qquad f_{n+1} ( \rho ) =e^{\rho f_{n} ( \rho ) },& n=0,1,2,\ldots, \\ g_{1} ( \rho ) =\frac{2 ( 1-\rho ) }{\rho ^{2}},\qquad g_{m+1} ( \rho ) =\frac{2 ( 1-\rho ) g_{m}^{2} ( \rho ) }{g_{m}^{2} ( \rho ) \rho^{2}+2},& m=1,2,\ldots,\end{cases} $$

and

$$ \int_{\tau ( t ) }^{t}p ( s ) \,ds\geq\rho>0, \quad\text{and}\quad 0< \rho< 1. $$

(1.4)

Zhang and Lian in [19] initiated the study of the distribution of zeros of dynamic equations on time scales and in particular, they considered the first order delay dynamic equation

$$ x^{\Delta} ( t ) +p ( t ) x \bigl( \tau ( t ) \bigr) =0,\quad\text{for }t\in [ t_{0},\infty ) _{\mathbb{T}}, $$

(1.5)

on a time scale $\mathbb{T}$, where $p\in\mathbf{C}_{\mathrm{rd}} ( \mathbb{T},\mathbb{R}^{+} )$ is a non-negative rd-continuous function, $\tau\in\mathbf{C}_{\mathrm{rd}} ( \mathbb{T},\mathbb{T} )$ is strictly increasing, $\tau ( t ) < t$ for $t\in\mathbb{T}$ and $\lim_{t\rightarrow\infty}\tau ( t ) =\infty$. In [19] the authors established lower and the upper bounds for the quotient $x ( \tau ( t ) ) /x ( t )$ using the sequences $f_{n}$ and $g_{m}$ where

$$ f_{0} ( \rho ) =1,\qquad f_{n} ( \rho ) =e^{ ( 1-\rho ) f_{n-1} ( \rho ) },\quad n=1,2,\ldots, $$

(1.6)

and

$$ \textstyle\begin{cases} g_{1} ( \rho ) =\frac{2\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) },\text{ } \\ g_{m} ( \rho ) =\frac{2\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) +\frac{2}{g_{m-1}^{2} ( \rho ) }},\quad m=2,3,\ldots,\end{cases} $$

(1.7)

and where $M<(1-\rho)/2$ and $0\leq\rho<1$ satisfies the condition

$$ \sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho, $$

where $E= \{ \lambda:\lambda>0, 1-\lambda p ( t ) \mu ( t ) >0 \}$; $\zeta_{\mu ( s ) }$ and $\mu ( s )$ will be defined later.

Motivated by these papers, we study the distribution of zeros of oscillatory solutions of the delay dynamic equation (1.5) on a time scale $\mathbb{T}$ by considering new sequences $f_{n}$ and $g_{m}$. In the next section, we present some basic ideas on time scales. In Section 3, we establish lower and upper bounds for $x ( \tau ( t ) ) /x ( t )$ and in Section 4, we study the distribution of zeros of solutions of (1.5).

2 Some preliminaries and lemmas

In this section, we present some preliminaries; see [20, 21]. A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers $\mathbb{R}$. The forward and backward jump operators are defined by

$$ \sigma(t):=\inf \{ s\in\mathbb{T}: s>t \} ,\quad\text{and}\quad \rho(t):=\sup \{ s\in\mathbb{T}: s< t \}, $$

with $\inf \emptyset=\sup \mathbb{T}$ and $\sup \emptyset =\inf \mathbb{T}$. The graininess function μ on a time scale $\mathbb{T}$ is defined by $\mu ( t ) :=\sigma(t)-t$. For a function $f:\mathbb{T\rightarrow R}$ the (delta) derivative is defined by

$$ f^{\Delta} ( t ) =\frac{f ( \sigma(t) ) -f ( t ) }{\sigma(t)-t}, $$

if f is continuous at t and t is right-scattered. If t is not right-scattered then the derivative is defined by

$$ f^{\Delta} ( t ) =\lim_{s\rightarrow t}\frac{f ( t ) -f ( s ) }{t-s}, $$

provided this limit exists. A function f is said to be Δ-differentiable if its Δ-derivative exists. A useful formula is $f^{\sigma}=f ( \sigma ( t ) ) =f ( t ) +\mu ( t ) f^{\Delta} ( t )$. We will make use of the following product and quotient rules for the derivative of the product fg and the quotient $f/g$ (where $gg^{\sigma}\neq0$, and here $g^{\sigma }=g\circ\sigma$) of two Δ-differentiable functions f and g:

$$ ( fg ) ^{\Delta}=f^{\Delta}g+f^{\sigma}g^{\Delta }=fg^{\Delta }+f^{\Delta}g^{\sigma}, \qquad \biggl( \frac{f}{g} \biggr) ^{\Delta }=\frac{f^{\Delta}g-fg^{\Delta}}{gg^{\sigma}}. $$

Let $f:\mathbb{R\rightarrow R}$ be continuously differentiable and suppose $g:\mathbb{T\rightarrow R}$ is delta differentiable. Then $f\circ g:\mathbb{T\rightarrow R}$ is delta differentiable and the chain rule,

$$ ( f\circ g ) ^{\Delta} ( t ) = \biggl\{ \int_{0}^{1}f^{\prime} \bigl( g ( t ) +h\mu ( t ) g^{\Delta} ( t ) \bigr) \,dh \biggr\} g^{\Delta} ( t ) , $$

(2.1)

holds. A special case of (2.1) is

$$ \bigl[ x^{\lambda} ( t ) \bigr] ^{\Delta}=\lambda \int _{0}^{1} \bigl[ hx^{\sigma}+ ( 1-h ) x \bigr] ^{\lambda -1}x^{\Delta } ( t ) \,dh. $$

For $s, t\in\mathbb{T}$, a function $F:\mathbb{T\rightarrow R}$ is called an antiderivative of $f :\mathbb{T\rightarrow R}$ provided $F^{\Delta }=f(t)$ holds for all $t\in\mathbb{T}$. In this case we define the integral of f by $\int_{s}^{t}f(\tau)\Delta\tau=F(t)-F(s)$. For $a, b\in \mathbb{T}$, and a Δ-differentiable function f, the Cauchy integral of $f^{\Delta}$ is defined by $\int_{a}^{b}f^{\Delta}(\tau )\Delta\tau=f(b)-f(a)$. The integration by parts formula reads

$$ \int_{a}^{b}f ( t ) g^{\Delta} ( t ) \Delta t= \bigl[ f ( t ) g ( t ) \bigr] _{a}^{b}- \int _{a}^{b}f^{\Delta } ( t ) g^{\sigma} ( t ) \Delta t, $$

and infinite integrals are defined as

$$ \int_{a}^{\infty}f ( t ) \Delta t=\lim _{b\rightarrow \infty } \int_{a}^{b}f ( t ) \Delta t. $$

A function p: $\mathbb{T\rightarrow R}$ is called regressive if $1+\mu ( t ) p ( t ) \neq0$ for $t\in\mathbb{T}$. A function $p:\mathbb{T\rightarrow R}$ is called positively regressive (we write $p\in\mathcal{R}^{+}$) if it is rd-continuous function and satisfies $1+\mu ( t ) p ( t ) >0$ for all $t\in \mathbb{T}$. Hilger in [20] showed that for $p ( t )$ rd-continuous and regressive, the solution of the initial value problem

$$ y^{\Delta} ( t ) =p ( t ) y ( t ) ,\quad y ( t_{0} ) =1, $$

is given by the generalized exponential function $e_{p} ( t,t_{0} )$, which is defined by

$$ e_{p} ( t,t_{0} ) =\exp \biggl\{ \int_{t_{0}}^{t}\zeta _{\mu ( s ) } \bigl( p ( s ) \bigr) \Delta s \biggr\} , $$

where $t_{0}$, $t\in\mathbb{T}$, and the cylinder transformation $\zeta _{h} ( z )$ is defined by

$$ \zeta_{h} ( z ) = \textstyle\begin{cases} \frac{\log ( 1+hz ) }{h}, & \text{if }h\neq0, \\ z, &\text{if }h=0,\end{cases} $$

where $z\in\mathbb{R}$ and $h\in\mathbb{R}^{+}$.

The next lemma can be found in [11].

Lemma 2.1

Assume $t_{0}$, $t\in\mathbb{T}$.

(i) For a non-negative φ with $-\varphi\in\mathcal{R}^{+}$, we have the following inequality:

$$ 1- \int_{t_{0}}^{t}\varphi ( s ) \Delta s\leq e_{-\varphi }(t,t_{0})\leq\exp \biggl\{ - \int_{t_{0}}^{t}\varphi(s)\Delta s \biggr\} . $$

(2.2)

(ii) If φ is rd-continuous and non-negative, then

$$ 1+ \int_{t_{0}}^{t}\varphi ( s ) \Delta s\leq e_{\varphi }(t,t_{0})\leq\exp \biggl\{ \int_{t_{0}}^{t}\varphi(s)\Delta s \biggr\} . $$

(2.3)

Lemma 2.2

Assume that $\mathbb{T}$ is a time scale with $t_{0}\in \mathbb{T}$. If $f ( t ) >0$ on $[ t_{0},\infty ) _{\mathbb{T}}$, then

$$ \bigl[ \ln f ( t ) \bigr] ^{\Delta}\leq\frac{f^{\Delta } ( t ) }{f ( t ) },\quad\textit{for }t\in [ t_{0},\infty ) _{\mathbb{T}}. $$

Proof

Fix t. We consider two cases: (i) $f^{\Delta} ( t ) \leq0$ and (ii) $f^{\Delta} ( t ) \geq0$.

In the first case, we see that

$$ h\mu ( t ) f^{\Delta} ( t ) +f ( t ) \leq f ( t ). $$

Now recall $f ( \sigma ( t ) ) =f ( t ) +\mu ( t ) f^{\Delta} ( t )$ so $h\mu ( t ) f^{\Delta} ( t ) +f ( t )=h f ( \sigma ( t ) )+(1-h) f ( t )>0$ and as a result

$$\frac{1}{h\mu ( t ) f^{\Delta} ( t ) +f ( t )} \geq \frac{1}{f ( t )}. $$

Apply the chain rule (2.1), and we get (note $f^{\Delta } ( t ) \leq0$)

$$ \bigl[ \ln f ( t ) \bigr] ^{\Delta}= \biggl\{ \int _{0}^{1}\frac{1}{h\mu ( t ) f^{\Delta} ( t ) +f ( t ) }\,dh \biggr\} f^{\Delta} ( t ) \leq\frac{f^{\Delta} ( t ) }{f ( t ) }. $$

In the second case, we see that

$$ h\mu ( t ) f^{\Delta} ( t ) +f ( t ) \geq f ( t ) . $$

Applying the chain rule (2.1), we get

$$ \bigl[ \ln f ( t ) \bigr] ^{\Delta}= \biggl\{ \int _{0}^{1}\frac{1}{h\mu ( t ) f^{\Delta} ( t ) +f ( t ) }\,dh \biggr\} f^{\Delta} ( t ) \leq\frac{f^{\Delta} ( t ) }{f ( t ) }. $$

Thus, we deduce in both cases that

$$ \bigl[ \ln f ( t ) \bigr] ^{\Delta}\leq\frac{f^{\Delta } ( t ) }{f ( t ) }. $$

The proof is complete. □

Lemma 2.3

Assume that $\mathbb{T}$ is a time scale with $t_{0}\in \mathbb{T}$. If $f(t)>0$ and $f^{\Delta} ( t ) \geq0$ for $t\in [ t_{0},\infty ) _{\mathbb{T}}$, then for $\alpha>0$

$$ \bigl[ e^{-\alpha f ( t ) } \bigr] ^{\Delta}\geq-\alpha e^{-\alpha f ( t ) }f^{\Delta} ( t ) . $$

Proof

Since $f ( t ) >0$ and $f^{\Delta} ( t ) \geq0$ for $t\in [ t_{0},\infty ) _{\mathbb{T}}$, we have for $h\in ( 0,1 )$

$$ f ( t ) \leq h\mu ( t ) f^{\Delta} ( t ) +f ( t ) . $$

(2.4)

Applying the chain rule (2.1) and using (2.4), we see that

$$ \bigl[ e^{-\alpha f ( t ) } \bigr] ^{\Delta}= \biggl\{ -\alpha \int_{0}^{1}e^{-\alpha ( f ( t ) +h\mu ( t ) f^{\Delta} ( t ) ) }\,dh \biggr\} f^{\Delta} ( t ) \geq-\alpha e^{-\alpha f ( t ) }f^{\Delta} ( t ) . $$

The proof is complete. □

3 Lower and upper bounds for $x ( \tau (t ) ) /x ( t )$

In this section, we establish lower and upper bounds for $x ( \tau ( t ) ) /x ( t )$ where $x(t)$ is a solution of equation (1.5). We use the notation $\tau ^{0} ( t ) =t$ and inductively define the iterates of $\tau ^{-i} ( t )$ by

$$ \tau^{-i} ( t ) =\bigl(\tau^{-1}\circ\tau^{- ( i-1 ) }\bigr) ( t ) ,\quad\text{for }i=1, 2,\ldots , $$

where $\tau^{-1} ( t )$ is the inverse function of $\tau ( t )$. From the definition it is clear that

$$ \tau ( t ) < t< \tau^{-1} ( t ) < \cdots< \tau ^{- ( n-1 ) } ( t ) < \tau^{- n } ( t ) < \cdots . $$

To find the lower bound for $x ( \tau ( t ) ) /x ( t )$ we define for $0<\rho<1$ a sequence $f_{n} ( \rho )$ by

$$ \textstyle\begin{cases} f_{0} ( \rho ) =1,\qquad f_{1} ( \rho ) =1/\rho , \\ f_{n+2} ( \rho ) =\frac{f_{n} ( \rho ) }{f_{n} ( \rho ) +1-e^{ ( 1-\rho ) f_{n} ( \rho ) }}, &n=0,1,2,3,\ldots. \end{cases} $$

(3.1)

We note some properties of $f_{n} ( \rho )$ for the reader’s interest (see [9] or use an elementary argument using $\frac{x}{x+1-e^{(1-\rho)x} }$). For $0<1-\rho\leq1/e$, we have

$$ 1\leq f_{n} ( \rho ) \leq f_{n+2} ( \rho ) \leq e,\quad n=0,1,2,\ldots, $$

so there exists a function $f ( \rho )$ such

$$ \lim_{n\rightarrow\infty}f_{n} ( \rho ) =f ( \rho ) , \quad 1\leq f ( \rho ) \leq e, $$

where $f ( \rho )$ satisfies

$$ f ( \rho ) =e^{ ( 1-\rho ) f ( \rho ) }. $$

(3.2)

If $( 1-\rho ) >1/e$, then either $f_{n} ( \rho )$ is nondecreasing and $\lim_{n\rightarrow\infty}f_{n} ( \rho ) =+\infty$ or $f_{n} ( \rho )$ is negative or $f_{n} ( \rho )$ is ∞ after a finite numbers of terms.

Theorem 3.1

Assume that $\mathbb{T}$ is a time scale and $t^{\prime}$, $t_{0}$, $t_{1} \in\mathbb{T}$, $t_{0}\geq t^{\prime}$, $t_{1}\geq\tau^{-3} ( t_{0} )$, $x(t)$ is a solution of (1.5) on $[t^{\prime},\infty)_{\mathbb{T}}$, $x(t)$ is positive on $[ t_{0},t_{1} ] _{\mathbb{T}}$ and there exists $\rho\in(0,1)$ with $\infty>f_{n} ( \rho )> 0$ for $n\in\{2,3,\ldots\}$ and

$$ \sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho\quad\textit{for } t \in \bigl[ \tau^{-3} ( t_{0} ) ,t_{1} \bigr] _{\mathbb{T}}; $$

(3.3)

here $E= \{ \lambda:\lambda >0, 1-\lambda p ( t ) \mu ( t ) >0\textit{ for }t\in [ \tau^{-2} ( t_{0} ) ,t_{1} ] _{\mathbb{T}} \}$. Then for $n\geq0$ when $\tau^{- ( 2+n )} ( t_{0} ) \leq t_{1}$ we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }\geq f_{n} ( \rho ) ,\quad\textit{for }t\in \bigl[ \tau^{- ( 2+n ) } ( t_{0} ) ,t_{1} \bigr]_{\mathbb{T}} , $$

where $f_{n} ( \rho )$ is defined in (3.1).

Proof

From (1.5), we see that

$$ x^{\Delta} ( t ) =-p ( t ) x \bigl( \tau ( t ) \bigr) \leq0,\quad\text{for }t \in \bigl[ \tau^{-1} ( t_{0} ) ,t_{1} \bigr] _{\mathbb{T}}, $$

(3.4)

so since $x ( t )$ is nonincreasing on $[ \tau ^{-1} ( t_{0} ) , t_{1} ] _{\mathbb{T}}$ we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }\geq1,\quad\text{for }t\in \bigl[ \tau^{-2} ( t_{0} ) ,t_{1} \bigr] _{\mathbb{T}}. $$

(3.5)

Note (3.5) and the fact that x is positive on $[ t_{0},t_{1} ] _{\mathbb{T}}$, so for $t\in [ \tau ^{-2} ( t_{0} ) , t_{1} ] _{\mathbb{T}}$ we have (note $x ( \sigma ( t ) )>0$ since $\sigma(t) \geq t \geq\tau^{-2} ( t_{0} ) >t_{0}$)

$$\begin{aligned} 0 =&-\mu ( t ) \bigl[ x^{\Delta} ( t ) +p ( t ) x \bigl( \tau ( t ) \bigr) \bigr] \\ =&x ( t ) -x \bigl( \sigma ( t ) \bigr) -\mu ( t ) p ( t ) x \bigl( \tau ( t ) \bigr) \\ < &x ( t ) -\mu ( t ) p ( t ) x ( t ) \\ =& \bigl[ 1-\mu ( t ) p ( t ) \bigr] x ( t ) . \end{aligned}$$

Hence $1-\mu ( t ) p ( t ) >0$ for $t\in [ \tau^{-2} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$ so $-p\in\mathcal{R}^{+}$ on the interval $[ \tau ^{-2} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$. Using Lemma 2.1 (with the time scale $[ \tau^{-2} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$) and (3.3), we have for $t\in [ \tau^{-3} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$ (note $\tau(t) \in [ \tau^{-2} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$)

$$\begin{aligned} \int_{\tau ( t ) }^{t}p ( s ) \Delta s \geq &1-\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -p ( s ) \bigr) \Delta s \biggr\} \\ \geq&1-\sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int _{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \\ \geq&1-\rho. \end{aligned}$$

(3.6)

Integrating (1.5) from $\tau ( t )$ to t, we get

$$ x \bigl( \tau ( t ) \bigr) =x ( t ) + \int _{\tau ( t ) }^{t}p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s, $$

(3.7)

and hence, for $t\in [ \tau^{-3} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$, we get

$$x \bigl( \tau ( t ) \bigr) =x ( t ) + \int _{\tau ( t ) }^{t}p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s \geq x ( t ) +x \bigl( \tau ( t ) \bigr) \int_{\tau ( t ) }^{t}p ( s ) \Delta s\geq x ( t ) +x \bigl( \tau ( t ) \bigr) ( 1-\rho ) , $$

so

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }\geq\frac{1}{\rho}=f_{1} ( \rho ) >0,\quad\text{for }t\in \bigl[ \tau ^{-3} ( t_{0} ) ,t_{1} \bigr] . $$

When $\tau^{-4} ( t_{0} ) \leq t_{1}$, note, for $t\in [ \tau^{-4} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$ and $\tau(t) \leq s\leq t$, that

$$ \int_{\tau ( s ) }^{\tau ( t ) }\frac {x^{\Delta } ( \xi ) }{x ( \xi ) }\Delta\xi+ \int_{\tau ( s ) }^{\tau ( t ) }p ( \xi ) \frac{x ( \tau ( \xi ) ) }{x ( \xi ) }\Delta \xi=0, $$

(3.8)

so from Lemma 2.2 we have

$$ \int_{\tau ( s ) }^{\tau ( t ) } \bigl[ \ln x ( \xi ) \bigr] ^{\Delta}\Delta\xi+ \int_{\tau ( s ) }^{\tau ( t ) }p ( \xi ) \frac {x ( \tau ( \xi ) ) }{x ( \xi ) }\Delta\xi \leq0, $$

(3.9)

which implies that

$$ \frac{x ( \tau ( s ) ) }{x ( \tau ( t ) ) }\geq\exp \biggl\{ \int_{\tau ( s ) }^{\tau ( t ) }p ( \xi ) \frac{x ( \tau ( \xi ) ) }{x ( \xi ) }\Delta\xi \biggr\} , $$

and so using (3.5), we have (note $\xi\in [ \tau ^{-2} ( t_{0} ) ,t_{1} ] _{ \mathbb{T}}$ since $\tau(t) \leq s \leq t$ and $t\in [ \tau^{-4} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$)

$$ \frac{x ( \tau ( s ) ) }{x ( \tau ( t ) ) }\geq\exp \biggl\{ f_{0} ( \rho ) \int_{\tau ( s ) }^{\tau ( t ) }p ( \xi ) \Delta\xi \biggr\} ; $$

(3.10)

we write $f_{0} ( \rho )$ (which is of course 1 here) to indicate the general procedure. Now applying Lemma 2.3 and using (3.6), (3.7) and (3.10), we get (here $t\in [ \tau^{-4} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$)

$$\begin{aligned} x \bigl( \tau ( t ) \bigr) =&x ( t ) + \int _{\tau ( t ) }^{t}p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s \\ \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) \int_{\tau ( t ) }^{t}p ( s ) \exp \biggl\{ f_{0} ( \rho ) \int_{\tau ( s ) }^{\tau ( t ) }p ( \xi ) \Delta\xi \biggr\} \Delta s \\ =&x ( t ) +x \bigl( \tau ( t ) \bigr) \int _{\tau ( t ) }^{t}p ( s ) \exp \biggl\{ f_{0} ( \rho ) \biggl( \int_{\tau ( s ) }^{s}p ( \xi ) \Delta\xi- \int_{\tau ( t ) }^{s}p ( \xi ) \Delta\xi \biggr) \biggr\} \Delta s \\ \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) e^{ ( 1-\rho ) f_{0} ( \rho ) } \int_{\tau ( t ) }^{t}p ( s ) \exp \biggl\{ -f_{0} ( \rho ) \int_{\tau ( t ) }^{s}p ( \xi ) \Delta\xi \biggr\} \Delta s \\ \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) e^{ ( 1-\rho ) f_{0} ( \rho ) } \int_{\tau ( t ) }^{t}\frac{- [ \exp \{ -f_{0} ( \rho ) \int_{\tau ( t ) }^{s}p ( \xi ) \Delta\xi \} ] ^{\Delta}}{f_{0} ( \rho ) }\Delta s \\ =&x ( t ) +x \bigl( \tau ( t ) \bigr) e^{ ( 1-\rho ) f_{0} ( \rho ) } \biggl[ \frac{1-\exp \{ -f_{0} ( \rho ) \int_{\tau ( t ) }^{t}p ( \xi ) \Delta\xi \} }{f_{0} ( \rho ) } \biggr] \Delta s \\ \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) \biggl( \frac{e^{ ( 1-\rho ) f_{0} ( \rho ) }-1}{f_{0} ( \rho ) } \biggr) . \end{aligned}$$

Thus, for $t\in [ \tau^{-4} ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$, we get

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }\geq\frac{f_{0} ( \rho ) }{f_{0} ( \rho ) +1-e^{ ( 1-\rho ) f_{0} ( \rho ) }}=f_{2} ( \rho ) >0. $$

Repeating the above procedure, when $\tau^{- ( 2+n ) } ( t_{0} ) \leq t_{1}$ we get for $t\in [ \tau ^{- ( 2+n ) } ( t_{0} ) ,t_{1} ] _{\mathbb{T}}$

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }\geq\frac{f_{n-2} ( \rho ) }{f_{n-2} ( \rho ) +1-e^{ ( 1-\rho ) f_{n-2} ( \rho ) }}=f_{n} ( \rho ) >0. $$

The proof is complete. □

Remark 3.1

From the proof of Theorem 3.1 notice in the statement of Theorem 3.1 we could replace $\infty>f_{n} ( \rho )> 0$ for $n\in\{2,3, \ldots\}$ with $\infty>f_{n} ( \rho )> 0$ for $n\in\{2,3,\ldots, N-2\}$ if $\tau^{- ( 2+N ) } ( t_{0} )< t_{1} < \tau^{- ( 3+N ) } ( t_{0} )$ or $\infty>f_{n} ( \rho )> 0$ for $n\in \{2,3,\ldots, N-3\}$ if $\tau^{- ( 2+N ) } ( t_{0} )=t_{1} < \tau^{- ( 3+N ) } ( t_{0} )$.

To establish the upper bound for $x ( \tau ( t ) ) /x ( t )$, we define a sequence $g_{m} ( \rho )$ by

$$ \textstyle\begin{cases} g_{1} ( \rho ) :=\frac{2\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }, \\ g_{m+1} ( \rho ) :=\frac{2 ( \rho-\frac {1}{g_{m} ( \rho ) } ) }{ [ ( 1-\rho ) ^{2}-2M ( 1-\rho ) ] },\end{cases} $$

(3.11)

where $0\leq\rho<1$, $m=1,2,3,\ldots$ , and $0\leq M<(1-\rho)/2$.

We note some properties of $g_{m} ( \rho )$ for the reader’s interest. Note $g_{m+1} ( \rho ) < g_{m} ( \rho )$, for $m=1,2,3,\ldots$ , and trivially

$$ g_{1} ( \rho ) >\frac{\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }. $$

More generally when $0<1-\rho\leq1/e$ using an induction argument (i.e. assuming $g_{m} ( \rho )>\frac{\rho }{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }$) then

$$\begin{aligned} g_{m+1} ( \rho ) =&\frac{2 ( \rho g_{m} ( \rho ) -1 ) }{g_{m} ( \rho ) [ ( 1-\rho ) ^{2}-2M ( 1-\rho ) ] } \\ >&\frac{2\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }-\frac{2}{\rho}>\frac{\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }; \end{aligned}$$

thus $g_{k} ( \rho )>\frac{\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }$ where $k=1,3,\ldots$ . Then there exists a function $g ( \rho )$ with

$$ \lim_{m\rightarrow\infty}g_{m} ( \rho ) =g ( \rho ) =\frac{2}{\rho-\sqrt{2 ( 2M-1 ) -4 ( M-1 ) \rho -\rho^{2}}}. $$

for $0<1-\rho\leq1/e$ (note $2 ( 2M-1 ) -4 ( M-1 ) \rho-\rho^{2}>0$ if $0<1-\rho\leq1/e$).

Theorem 3.2

Assume that $\mathbb{T}$ is a time scale and $t^{\prime }, t_{0} \in\mathbb{T}$, $t_{0}\geq t^{\prime}$, $x(t)$ is a solution of (1.5) on $[t^{\prime},\infty)_{\mathbb{T}}$, there exists a positive integer $N \geq4$ such that $x ( t )$ is positive on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$ and there exists $\rho\in(0,1)$ with $g_{m} ( \rho )> 0$ for $m\in\{2,3,\ldots, N-3\}$ and

$$ \sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho\quad\textit{for } t \in \bigl[ \tau^{-3} ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

(3.12)

where $E= \{ \lambda:\lambda >0,1-\lambda p ( t ) \mu ( t ) >0\textit{ for }t\in [ \tau^{-2} ( t_{0} ) , \tau^{-N} ( t_{0} ) ] _{\mathbb{T}} \}$ and

$$ M=\sup_{s\in [ \tau^{-3} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}}p ( s ) \mu ( s ) < \frac{1-\rho}{2}. $$

Then for $m\in\{1,\ldots,N-3\}$ we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }< g_{m} ( \rho ) ,\quad\textit{for }t\in \bigl[ \tau ^{-3} ( t_{0} ),\tau^{- ( N-m ) } ( t_{0} ) \bigr] _{\mathbb{T}}, $$

where $g_{m} ( \rho )$ is defined in (3.11).

Proof

From (1.5), we see that

$$ x^{\Delta} ( t ) \leq0,\quad\text{for }t\in \bigl[ \tau ^{-1} ( t_{0} ),\tau^{-N} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

(3.13)

and as in Theorem 3.1 notice $1-\mu ( t ) p ( t ) >0$ for $t\in [ \tau^{-2} ( t_{0} ) , \tau ^{-N} ( t_{0} ) ] _{\mathbb{T}}$ so $-p\in \mathcal{R}^{+}$ on the interval $[ \tau^{-2} ( t_{0} ) , \tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$. From Lemma 2.1 (with the time scale $[ \tau^{-2} ( t_{0} ) , \tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$) and (3.12), we have for $t\in [ \tau^{-3} ( t_{0} ) , \tau^{-(N-1)} ( t_{0} ) ]_{\mathbb{T}}$ (note $\tau^{-1}(t) \leq\tau^{-N} ( t_{0} )$)

$$ \int_{\tau ( t ) }^{t}p ( s ) \Delta s\geq 1-\rho\quad\text{and}\quad \int_{t}^{\tau^{-1} ( t ) }p ( s ) \Delta s\geq1-\rho. $$

(3.14)

Let $t\in [ \tau^{-3} ( t_{0} ) , \tau ^{-(N-1)} ( t_{0} ) ] _{\mathbb{T}}$ and consider

$$ G ( r ) := \int_{t}^{r}p ( s ) \Delta s-1+\rho ,\quad\text{for }r \in \bigl[ t,\tau^{-1} ( t ) \bigr] _{\mathbb{T}}. $$

Note $G: [ t,\tau^{-1} ( t ) ] \rightarrow \mathbb{R}$ is nondecreasing, $G ( t ) =-1+\rho<0$, and

$$ G \bigl( \tau^{-1} ( t ) \bigr) = \int_{t}^{\tau ^{-1} ( t ) }p ( s ) \Delta s-1+\rho\geq1-\rho-1+ \rho=0. $$

If $G ( \tau^{-1} ( t ) ) =0$, then

$$ \int_{t}^{\tau^{-1} ( t ) }p ( s ) \Delta s=G \bigl( \tau^{-1} ( t ) \bigr) +1-\rho=1-\rho, $$

whereas if $G ( \tau^{-1} ( t ) ) >0$ then $G ( t ) <0<G ( \tau^{-1} ( t ) )$.

In either case (from the intermediate value theorem [20]) there exists $t^{\ast}\in [ t,\tau^{-1} ( t ) ] _{ \mathbb{T}}$ with $\sigma ( t^{\ast} ) \in [ t,\tau ^{-1} ( t ) ] _{ \mathbb{T}}$ such that $G(t^{\ast }) G(\sigma ( t^{\ast} )) \leq0$ and so

$$ \int_{t}^{t^{\ast}}p ( s ) \Delta s\leq1-\rho\quad\text{and}\quad \int_{t}^{\sigma ( t^{\ast} ) }p ( s ) \Delta s\geq1-\rho. $$

(3.15)

Integrating both sides of (1.5) from t to $\sigma ( t^{\ast} )$, for $t\in [ \tau^{-3} ( t_{0} ),\tau^{- ( N-1 ) } ( t_{0} ) ] _{\mathbb{T}}$, we have

$$ x ( t ) =x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + \int_{t}^{\sigma ( t^{\ast} ) }p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s. $$

(3.16)

Fix $t\in [ \tau^{-3} ( t_{0} ) , \tau ^{-(N-1)} ( t_{0} ) ] _{\mathbb{T}}$. Let $s\in \mathbb{T}$ be such that $t\leq s\leq\sigma ( t^{\ast} ) \leq\tau^{-1} ( t )$ (here $t^{\ast}$ is as described above, and note $\tau ( t ) \leq\tau ( s ) \leq t$) and integrating (1.5) from $\tau ( s )$ to t yields

$$ x \bigl( \tau ( s ) \bigr) =x ( t ) + \int _{\tau ( s ) }^{t}p ( u ) x \bigl( \tau ( u ) \bigr) \Delta u, $$

and this together with x being nonincreasing on $[ \tau ^{-1} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$ and (3.14) will give

$$\begin{aligned} x \bigl( \tau ( s ) \bigr) \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) \int_{\tau ( s ) }^{t}p ( u ) \Delta u \\ =&x ( t ) +x \bigl( \tau ( t ) \bigr) \biggl\{ \int_{\tau ( s ) }^{s}p ( u ) \Delta u- \int_{t}^{s}p ( u ) \Delta u \biggr\} \\ \geq&x ( t ) +x \bigl( \tau ( t ) \bigr) \biggl\{ 1-\rho- \int_{t}^{s}p ( u ) \Delta u \biggr\} , \end{aligned}$$

(3.17)

so from (3.15), (3.16) and (3.17), we obtain

$$\begin{aligned} x ( t ) =&x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + \int_{t}^{\sigma ( t^{\ast} ) }p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s \\ \geq&x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + \int _{t}^{\sigma ( t^{\ast} ) }p ( s ) \biggl\{ x ( t ) +x \bigl( \tau ( t ) \bigr) \biggl\{ 1-\rho- \int_{t}^{s}p ( u ) \Delta u \biggr\} \biggr\} \Delta s \\ \geq&x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + ( 1-\rho ) x ( t ) + ( 1-\rho ) ^{2}x \bigl( \tau ( t ) \bigr) \\ &{}-x \bigl( \tau ( t ) \bigr) \biggl\{ \int_{t}^{t^{\ast }}p ( s ) \biggl\{ \int_{t}^{s}p ( u ) \Delta u \biggr\} \Delta s \\ &{} + \int_{t^{\ast}}^{\sigma ( t^{\ast} ) }p ( s ) \biggl\{ \int_{t}^{s}p ( u ) \Delta u \biggr\} \Delta s \biggr\} . \end{aligned}$$

(3.18)

Let $F ( s )=\int_{t}^{s}p ( u ) \Delta u$, and note

$$\begin{aligned} \bigl[ F^{2} ( s ) \bigr] ^{\Delta} =&2 \int _{0}^{1} \bigl[ hF^{\sigma} ( s ) + ( 1-h ) F ( s ) \bigr] F^{\Delta} ( s ) \,dh \\ =&2 \int_{0}^{1} \bigl[ hF^{\sigma} ( s ) + ( 1-h ) F ( s ) \bigr] p ( s ) \,dh \\ \geq&2F ( s ) p ( s ) . \end{aligned}$$

Hence,

$$\begin{aligned} \int_{t}^{t^{\ast}}p ( s ) \biggl\{ \int_{t}^{s}p ( u ) \Delta u \biggr\} \Delta s =& \int_{t}^{t^{\ast}}p ( s ) F ( s ) \Delta s\leq \frac{1}{2}F^{2} \bigl( t^{\ast} \bigr) \\ =&\frac{1}{2} \biggl( \int_{t}^{t^{\ast}}p ( u ) \Delta u \biggr) ^{2} \leq\frac{ ( 1-\rho ) ^{2}}{2}, \end{aligned}$$

(3.19)

and so we obtain

$$\begin{aligned} &{\int_{t}^{t^{\ast}}p ( s ) \Delta s \int_{t}^{s}p ( u ) \Delta u+ \int_{t^{\ast}}^{\sigma ( t^{\ast} ) }p ( s ) \Delta s \int_{t}^{s}p ( u ) \Delta u} \\ &{\quad \leq\frac{ ( 1-\rho ) ^{2}}{2}+\mu \bigl( t^{\ast} \bigr) p \bigl( t^{\ast} \bigr) \int _{t}^{t^{\ast }}p ( u ) \Delta u} \\ &{\quad \leq\frac{ ( 1-\rho ) ^{2}}{2}+ ( 1-\rho ) M.} \end{aligned}$$

(3.20)

Note $\sigma ( t^{\ast} ) \in [ t,\tau^{-1} ( t ) ] _{\mathbb{T}}$, $t\in [ \tau^{-3} ( t_{0} ),\tau^{- ( N-1 ) } ( t_{0} ) ] _{\mathbb{T}}$, and x is positive on $[ t_{0},\tau ^{-N} ( t_{0} ) ] _{\mathbb{T}}$ (so $x ( \sigma ( t^{\ast} ) ) >0$). Thus from (3.18) and (3.20), we obtain

$$\begin{aligned} x ( t ) \geq&x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + ( 1-\rho ) x ( t ) \\ &{}+ ( 1-\rho ) ^{2}x \bigl( \tau ( t ) \bigr) - \biggl[ \frac{ ( 1-\rho ) ^{2}}{2}+ ( 1-\rho ) M \biggr] x \bigl( \tau ( t ) \bigr) \\ =&x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) + ( 1-\rho ) x ( t ) \\ &{} + \biggl[ \frac{ ( 1-\rho ) ^{2}}{2}- ( 1-\rho ) M \biggr] x \bigl( \tau ( t ) \bigr) , \end{aligned}$$

(3.21)

and so we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }< \frac{2\rho}{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }=g_{1} ( \rho ) ,\quad\text{for }t \in \bigl[ \tau ^{-3} ( t_{0} ),\tau^{- ( N-1 ) } ( t_{0} ) \bigr] _{\mathbb{T}}. $$

(3.22)

Fix $t\in [ \tau^{-3} ( t_{0} ),\tau^{- ( N-2 ) } ( t_{0} ) ] _{\mathbb{T}}$ and with $t^{\ast}$ as described above we have $t\leq\sigma ( t^{\ast } ) \leq\tau^{-1} ( t ) \leq\tau^{- ( N-1 ) } ( t_{0} )$, so from (3.22) we have

$$ x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) >\frac {1}{g_{1} ( \rho ) }x \bigl( \tau \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) \bigr), $$

and since x is nonincreasing on $[ \tau^{-1} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$ and $\tau ( \sigma ( t^{\ast} ) ) \leq t\leq \tau^{- ( N-1 ) } ( t_{0} )$ we have

$$ x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) >\frac {1}{g_{1} ( \rho ) }x \bigl( \tau \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) \bigr) \geq \frac{1}{g_{1} ( \rho ) }x ( t ) . $$

(3.23)

Substituting (3.23) into (3.21), we obtain for $t\in [ \tau^{-3} ( t_{0} ),\tau^{- ( N-2 ) } ( t_{0} ) ] _{\mathbb{T}}$ that

$$ x ( t ) >\frac{1}{g_{1} ( \rho ) }x ( t ) + ( 1-\rho ) x ( t ) + \biggl[ \frac{ ( 1-\rho ) ^{2}}{2}- ( 1-\rho ) M \biggr] x \bigl( \tau ( t ) \bigr) , $$

and so we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }< \frac{2 ( \rho-\frac{1}{g_{1} ( \rho ) } ) }{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }:=g_{2} ( \rho ) . $$

Repeating the above procedure, we obtain for $t\in [ \tau ^{-3} ( t_{0} ),\tau ^{- ( N-m ) } ( t_{0} ) ] _{\mathbb{T}}$

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }< \frac{2 ( \rho-\frac{1}{g_{m-1} ( \rho ) } ) }{ ( 1-\rho ) ^{2}-2M ( 1-\rho ) }:=g_{m} ( \rho ) . $$

The proof is complete. □

4 Distributions of zeros of solutions

In this section, we study the distribution of zeros of solutions of (1.5) using the lower and upper bounds for $x ( \tau ( t ) ) /x ( t )$ in Section 3.

Theorem 4.1

Assume that $\mathbb{T}$ is a time scale and $t^{\prime }, t_{0}\in\mathbb{T}$, $t_{0}\geq t^{\prime}$, $x(t)$ is a solution of (1.5) on $[t^{\prime},\infty)_{\mathbb{T}}$, and there exist $\rho\in(0,1)$ and $n_{0}, m_{0} \in \{1,2,\ldots\}$ with $f_{n_{0}} (\rho ) \geq g_{m_{0}} ( \rho )$, and with

$$N=2+\min_{n\geq1, m\geq1} \bigl\{ n+m: f_{n} (\rho ) \geq g_{m} ( \rho ) \bigr\} =2+n^{\star }+m^{\star} $$

assume $\infty>f_{k} ( \rho )> 0$, $g_{k} ( \rho )> 0$ for $n\in\{2,3,\ldots, N-3\}$ and

$$\sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho\quad\textit{for } t \in \bigl[ \tau^{-3} ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

where $E= \{ \lambda:\lambda >0, 1-\lambda p ( t ) \mu ( t ) >0\textit{ for }t\in [ \tau^{-2} ( t_{0} ) , \tau ^{-N} ( t_{0} ) ] _{\mathbb{T}} \}$ and

$$ M=\sup_{s\in [ \tau^{-3} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}}p ( s ) \mu ( s ) < \frac{1-\rho}{2}. $$

Then every solution of (1.5) cannot be totally positive or totally negative on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$.

Proof

Note

$$ f_{n^{\star}} (\rho ) \geq g_{m^{\star}} ( \rho ). $$

(4.1)

Without loss of generality assume x is positive on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$. From Theorem 3.1 we have

$$\frac{x ( \tau ( t ) ) }{x ( t ) }\geq f_{n^{\star}} ( \rho ) ,\quad\text{for }t\in \bigl[ \tau ^{- ( 2+n^{\star} ) } ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr]_{\mathbb{T}} $$

and from Theorem 3.2 we have (note $m^{\star}=N-(2+n^{\star}) \leq N-3$)

$$\frac{x ( \tau ( t ) ) }{x ( t ) }< g_{m^{\star}} ( \rho ) ,\quad\text{for }t\in \bigl[ \tau^{-3} ( t_{0} ),\tau^{- ( N-m^{\star} ) } ( t_{0} ) \bigr] _{\mathbb{T}}. $$

Note since $N=2+n^{\star}+m^{\star}$ we have (take $t=\tau^{- ( N-m^{\star} ) } ( t_{0} )$)

$$f_{n^{\star}} ( \rho ) \leq\frac{x ( \tau^{- ( 1+n^{\star} ) } ( t_{0} ) ) }{x ( \tau ^{- ( 2+n^{\star} ) } ( t_{0} ) ) }< g_{m^{\star}} ( \rho ), $$

which contradicts (4.1). The proof is complete. □

Theorem 4.2

Assume that $\mathbb{T}$ is a time scale and $t^{\prime }, t_{0}\in\mathbb{T}$, $t_{0}\geq t^{\prime}$, $x(t)$ is a solution of (1.5) on $[t^{\prime},\infty)_{\mathbb{T}}$, and there exist $\rho\in(0,1)$ and a positive integer $N\geq4$ and $m_{0} \in\{1,2,\ldots,N-3\}$ with

$$\int_{\tau ( t_{m_{0}} ) }^{t_{m_{0}}}p ( s ) \Delta s>1-\frac{1}{g_{m_{0}} ( \rho ) } \quad\textit{where } t_{m_{0}}=\tau^{- ( N-m_{0} ) } ( t_{0} ), $$

and with

$$m^{\star}=\min_{m \in\{1,\ldots,N-3\}} \biggl\{ m: \int_{\tau ( t_{m} ) }^{t_{m}}p ( s ) \Delta s>1-\frac{1}{g_{m} ( \rho ) } \biggr\} \quad \textit{where } t_{m}=\tau^{- ( N-m ) } ( t_{0} ) $$

assume $\infty>f_{k} ( \rho )> 0$, $g_{k} ( \rho )> 0$ for $n\in\{2,3,\ldots, N-3\}$ and

$$\sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho\quad\textit{for } t \in \bigl[ \tau^{-3} ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

where $E= \{ \lambda:\lambda >0, 1-\lambda p ( t ) \mu ( t ) >0\textit{ for }t\in [ \tau^{-2} ( t_{0} ) , \tau ^{-N} ( t_{0} ) ] _{\mathbb{T}} \}$ and

$$ M=\sup_{s\in [ \tau^{-3} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}}p ( s ) \mu ( s ) < \frac{1-\rho}{2}. $$

Then every solution of (1.5) cannot be totally positive or totally negative on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$.

Proof

Note

$$ \int_{\tau ( t_{m^{\star}} ) }^{t_{m^{\star}}}p ( s ) \Delta s>1-\frac{1}{g_{m^{\star}} ( \rho ) }\quad\hbox{where } t_{m^{\star}}=\tau^{- ( N-m^{\star} ) } ( t_{0} ). $$

(4.2)

Without loss of generality assume x is positive on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$. From Theorem 3.2, we have

$$\frac{x ( \tau ( t ) ) }{x ( t ) }< g_{m^{\star}} ( \rho ) ,\quad\text{for }t\in \bigl[ \tau^{-3} ( t_{0} ),\tau^{- ( N-m^{\star} ) } ( t_{0} ) \bigr] _{\mathbb{T}}, $$

so in particular

$$ \frac{x ( \tau ( t_{m^{\star}} ) ) }{x ( t_{m^{\star}} ) }< g_{m^{\star}} ( \rho ). $$

(4.3)

Integrating (1.5) from $\tau ( t_{m^{\star}} )$ to $t_{m^{\star}}$, we obtain

$$ x \bigl( \tau ( t_{m^{\star}} ) \bigr) -x ( t_{m^{\star}} ) = \int_{\tau ( t_{m^{\star}} ) }^{t_{m^{\star}}}p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s\geq x \bigl( \tau ( t_{m^{\star}} ) \bigr) \int_{\tau ( t_{m^{\star}} ) }^{t_{m^{\star}}}p ( s ) \Delta s, $$

and this together with (4.3) gives

$$ \int_{\tau ( t_{m^{\star}} ) }^{t_{m^{\star}}}p ( s ) \Delta s\leq1-\frac{x ( t_{m^{\star}} ) }{x ( \tau ( t_{m^{\star}} ) ) }\leq1-\frac{1}{g_{m^{\star}} ( \rho ) }, $$

which contradicts (4.2). The proof is complete. □

Theorem 4.3

Assume that $\mathbb{T}$ is a time scale and $t^{\prime }, t_{0}\in\mathbb{T}$, $t_{0}\geq t^{\prime}$, $x(t)$ is a solution of (1.5) on $[t^{\prime},\infty)_{\mathbb{T}}$, and there exist $\rho\in(0,1)$, a constant L and $n_{0}, m_{0} \in\{1,2,\ldots\}$ with

$$\frac{1+\ln f_{n_{0}-1} ( \rho ) }{f_{n_{0}-1} ( \rho ) }-\frac {1}{g_{m_{0}} ( \rho ) } < L $$

and with

$$N=2 + \min_{n\geq1, m\geq1} \biggl\{ n+m: L > \biggl( \frac{1+\ln f_{n-1} ( \rho ) }{f_{n-1} ( \rho ) }- \frac{1}{ g_{m} ( \rho ) } \biggr) \biggr\} =2+n^{\star}+m^{\star}, $$

assume $\infty>f_{k} ( \rho )> 0$, $g_{k} ( \rho )> 0$ for $n\in\{2,3,\ldots, N-3\}$ and

$$\sup_{\lambda\in E} \biggl\{ \lambda\exp \biggl\{ \int_{\tau ( t ) }^{t}\zeta_{\mu ( s ) } \bigl( -\lambda p ( s ) \bigr) \Delta s \biggr\} \biggr\} \leq\rho\quad\textit{for } t \in \bigl[ \tau^{-3} ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

where $E= \{ \lambda:\lambda >0, 1-\lambda p ( t ) \mu ( t ) >0\textit{ for } t\in [ \tau^{-2} ( t_{0} ) , \tau ^{-N} ( t_{0} ) ] _{\mathbb{T}} \}$ and

$$ M=\sup_{s\in [ \tau^{-3} ( t_{0} ),\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}}p ( s ) \mu ( s ) < \frac{1-\rho}{2}. $$

Suppose $f_{n^{\star}-1} ( \rho ) \geq1$, $f_{n^{\ast }} ( \rho )> f_{n^{\ast}-1} ( \rho )$ and for $t^{\ast}\in [ \tau ( t_{1} ) ,t_{1} ] _{\mathbb{T}}$ (here $t_{1}=\tau^{- ( N-m^{\star} ) } ( t_{0} )$) that

$$ \int_{\tau ( t_{1} ) }^{t^{\ast}}p ( s ) \Delta s+ \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) \Delta s \geq L. $$

(4.4)

Then every solution of (1.5) cannot be totally positive or totally negative on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$.

Proof

Note

$$ L> \biggl( \frac{1+\ln f_{n^{\ast}-1} ( \rho ) }{f_{n^{\ast }-1} ( \rho ) }-\frac{1}{g_{m^{\ast}} ( \rho ) } \biggr). $$

(4.5)

Without loss of generality assume x is positive on $[ t_{0},\tau^{-N} ( t_{0} ) ] _{\mathbb{T}}$. From Theorem 3.1, we have

$$\begin{aligned} &{\frac{x ( \tau ( t ) ) }{x ( t ) }\geq f_{n^{\ast}} ( \rho ) ,\quad t\in \bigl[ \tau ^{- ( 2+n^{\star} ) } ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr]_{\mathbb{T}},} \end{aligned}$$

(4.6)

$$\begin{aligned} &{\frac{x ( \tau ( t ) ) }{x ( t ) }\geq f_{n^{\ast}-1} ( \rho ) ,\quad t\in \bigl[ \tau ^{- ( 1+n^{\star} ) } ( t_{0} ) , \tau ^{-N} ( t_{0} ) \bigr]_{\mathbb{T}},} \end{aligned}$$

(4.7)

and from Theorem 3.2, we have

$$ \frac{x ( \tau ( t ) ) }{x ( t ) }< g_{m^{\ast}} ( \rho ) ,\quad t\in \bigl[ \tau^{-3} (t_{0} ),\tau^{-(N-m^{\ast})} ( t_{0} ) \bigr] _{\mathbb{T}}, $$

so in particular (with $t_{1}=\tau^{- ( N-m^{\star} ) } ( t_{0} )=\tau^{- ( 2+n^{\star} ) } ( t_{0} )$) we have

$$ \frac{x ( \tau ( t_{1} ) ) }{x ( t_{1} ) }< g_{m^{\ast}} ( \rho ) . $$

(4.8)

From (4.6) and $f_{n^{\ast}} ( \rho )> f_{n^{\ast }-1} ( \rho )$ we have

$$ \frac{x ( \tau ( t_{1} ) ) }{x ( t_{1} ) }>f_{n^{\ast}-1} ( \rho ) . $$

Now since x is nonincreasing on $[ \tau^{-1}( t_{0}),\tau ^{-N } ( t_{0} ) ] _{\mathbb{T}}$ and $f_{n^{\star}-1} ( \rho ) \geq1$ (and trivially note $\frac{x ( \tau ( t_{1} ) ) }{ x (\tau ( t_{1} ) ) }=1$) there exists a $t^{\ast}\in [ \tau ( t_{1} ) ,t_{1} ] _{\mathbb{T}}$ with

$$ \frac{x ( \tau ( t_{1} ) ) }{x ( t^{\ast } ) }\leq f_{n^{\ast}-1} ( \rho )\quad\text{and}\quad \frac{x ( \tau ( t_{1} ) ) }{x ( \sigma ( t^{\ast} ) ) }\geq f_{n^{\ast}-1} ( \rho ) . $$

(4.9)

Integrating (1.5) from $\sigma ( t^{\ast} )$ to $t_{1}$, we obtain

$$ x \bigl( \sigma \bigl( t^{\ast} \bigr) \bigr) -x ( t_{1} ) = \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) x \bigl( \tau ( s ) \bigr) \Delta s\geq x \bigl( \tau ( t_{1} ) \bigr) \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) \Delta s, $$

which implies

$$ \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) \Delta s\leq \biggl( \frac{x ( \sigma ( t^{\ast} ) ) }{x ( \tau ( t_{1} ) ) }-\frac{x ( t_{1} ) }{x ( \tau ( t_{1} ) ) } \biggr) . $$

(4.10)

From (4.8), (4.9) and (4.10), we obtain

$$ \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) \Delta s\leq \biggl( \frac{1}{f_{n^{\ast}-1} ( \rho ) }-\frac {1}{g_{m^{\ast }} ( \rho ) } \biggr) . $$

(4.11)

Divide (1.5) by x and integrate from $\tau ( t_{1} )$ to $t^{\ast}$, and we get

$$ \int_{\tau ( t_{1} ) }^{t^{\ast}}\frac{x^{\Delta} ( s ) }{x ( s ) }\Delta s=- \int_{\tau ( t_{1} ) }^{t^{\ast }}p ( s ) \frac{x ( \tau ( s ) ) }{x ( s ) }\Delta s \leq-f_{n^{\ast}-1} ( \rho ) \int _{\tau ( t_{1} ) }^{t^{\ast}}p ( s ) \Delta s, $$

which implies

$$ \int_{\tau ( t_{1} ) }^{t^{\ast}}p ( s ) \Delta s\leq-\frac{1}{f_{n^{\ast}-1} ( \rho ) } \int_{\tau ( t_{1} ) }^{t^{\ast}}\frac{x^{\Delta} ( s ) }{x ( s ) }\Delta s. $$

(4.12)

From (4.9), (4.12) and Lemma 2.2, we obtain

$$\begin{aligned} \int_{\tau ( t_{1} ) }^{t^{\ast}}p ( s ) \Delta s \leq &- \frac{1}{f_{n^{\ast}-1} ( \rho ) } \int_{\tau ( t_{1} ) }^{t^{\ast}} \bigl[ \ln x ( s ) \bigr] ^{\Delta }\Delta s=\frac{1}{f_{n^{\ast}-1} ( \rho ) }\ln \biggl( \frac{x ( \tau ( t_{1} ) ) }{x ( t^{\ast} ) } \biggr) \\ \leq&\frac{\ln f_{n^{\ast}-1} ( \rho ) }{f_{n^{\ast }-1} ( \rho ) }, \end{aligned}$$

(4.13)

and from (4.5), (4.11) and (4.13) we have

$$ \int_{\tau ( t_{1} ) }^{t^{\ast}}p ( s ) \Delta s+ \int_{\sigma ( t^{\ast} ) }^{t_{1}}p ( s ) \Delta s\leq \biggl( \frac{1+\ln f_{n^{\ast}-1} ( \rho ) }{f_{n^{\ast }-1} ( \rho ) }-\frac{1}{g_{m^{\ast}} ( \rho ) } \biggr) < L, $$

which contradicts (4.4). The proof is complete. □

Remark 4.1

When $\mathbb{T=R}$ equation (1.5) is the delay differential equation

$$ x^{\prime} ( t ) +p ( t ) x \bigl( \tau ( t ) \bigr) =0, \quad t\in \mathbb{R}. $$

Theorem 3.1 and Theorem 3.2 are related to the results in [9], Lemma 2.1 and Lemma 2.2, and Theorem 4.3 is motivated from results in [13], Theorem 3.