International Journal of Analysis and Applications
ISSN 2291-8639
Volume 14, Number 1 (2017), 42-51
http://www.etamaths.com

OSCILLATION CRITERIA FOR SECOND-ORDER NONLINEAR FUNCTIONAL

DYNAMIC EQUATIONS WITH DAMPING ON TIME SCALES

EMİNE TUǦLA AND FATMA SERAP TOPAL∗

Abstract. In this paper, we study oscillatory behavior of second-order dynamic equations with

damping under some assumptions on time scales. New theorems extend and improve the results in
the literature. Illustrative examples are given.

1. Introduction

During the past decades, the questions regarding the study of oscillatory properties of differential
equations with damping or distributed deviating arguments have become an important area of research
due to the fact that such equations arise in many real life problems.

In 1988, Hilger introduced the theory of time scales in his Ph.D. thesis [1] in order to unify continuous
and discrete analysis; see also [4]. Preliminaries about time scale calculus can be found in [2, 3] and
omitted here.

There has been much research achievement about the oscillation of dynamic equations on time scales
in the last few years; see the papers [5-8, 10,11, 13-16, 18-20] and the references therein.

In [9], Chen et al. investigated the oscillation of a second-order nonlinear dynamic equation with
positive and negative coefficients of the form

(r(t)x∆(t))4 + p(t)f(x(ξ(t))) −q(t)f(x(δ(t))) = 0.
In [17], S. enel concerned with the oscillatory behavior of all solutions of nonlinear second order

damped dynamic equation
(r(t)Ψ(x∆(t)))∆ + p(t)Ψ(x∆(t)) + q(t)f(xσ(t)) = 0.
In [12], Erbe et al. studied the oscillatory behavior of the solutions of the second order nonlinear

functional dynamic equation

(a(t)(x∆(t))γ)∆ +

n∑
i=0

pi(t)Φαi(x(gi(t))) = 0,

on an arbitrary time scale T.
In this study, we are concerned with the oscillation of solutions of second order dynamic equations

with damping terms of the following form
(r(t)g(x(t),x∆(t)))∆ +p(t)g(x(t),x∆(t)) +q1(t)f1(x(τ1(t))) +q2(t)f2(x(τ2(t))) = 0 (1.1)

on a time scale T such that inf T = t0 and sup T = ∞.
This paper is organized as follows. In this section we give some assumptions and lemmas that we

need through our work. In Section 2, we establish some new sufficient conditions for oscillation of
(1.1). Finally, in Section 3, we present some examples to illustrate our results.

Now, we mention some definitions and lemmas from calculus on time scales which can be found in
[2-3].

Lemma 1.1. Assume that g : T → R is strictly increasing and that T̃ := g(T) = {g(t) : t ∈ T} is a
time scale. If f : T → R is an rd-continuous functions, g is differentiable with rd-continuous derivative,

Received 10th January, 2017; accepted 17th March, 2017; published 2nd May, 2017.
2010 Mathematics Subject Classification. 4K11; 39A10; 39A99.

Key words and phrases. nonlinear dynamic equations; damping equations; oscillation solutions; time scales.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

42



OSCILLATION FOR NONLINEAR FUNCTIONAL DYNAMIC EQUATIONS 43

and a,b ∈ T, then ∫ b
a

f(t)g4(t)4t =
∫ g(b)
g(a)

(f ◦g−1)(s)4̃s,

where g−1 is the inverse function of g and 4̃ denotes the derivative on T̃.

Lemma 1.2. Every rd-continuous function has an antiderivative. In particular if t0 ∈ T, then F
defined by

F(t) :=

∫ t
t0

f(τ)4τ for t ∈ T

is an antiderivative of f.

Lemma 1.3. Assume that f : T → R is strictly increasing and that T̃ := f(T) = {f(t) : t ∈ T} is a
time scale. Let g : T̃ → R. If f4(t) and g4̃(f(t)) exist for t ∈ Tκ, then

(g ◦f)4 = (g4̃ ◦f)f4,
where 4̃ denotes the derivative on T̃.

Definition 1.1. A function p : T → R is said to be regressive provided 1 + µ(t)p(t) 6= 0 for all t ∈ Tκ,
where µ(t) = σ(t) − t. The set of all regressive rd-continuous functions p : T → R is denoted by R.

Let p ∈R for all t ∈ T. The exponential function on T is defined by

ep(t,s) = exp

(∫ t
s

ζµ(r)(p(r))∆r

)
where ζµ(s) is the cylinder transformation given by

ζµ(r)(p(r)) :=

{
1

µ(r)
Log(1 + µ(r)p(r)), if µ(r) > 0;

p(r), if µ(r) = 0.

The exponential function y(t) = ep(t,s) is the solution to the initial value problem y
∆ = p(t)y,

y(s) = 1. Other properties of the exponential function are given in the following lemma [3, Theorem
1.39].

Lemma 1.4. Let p,q ∈R. Then
i. e0(t,s) ≡ 1 and ep(t,t) ≡ 1;
ii. ep(σ(t),s) = (1 + µ(t)p(t))ep(t,s);

iii. 1
ep(t,s)

= e	(t,s) where, 	p(t) = −
p(t)

1+µ(t)p(t)
;

iv. ep(t,s) =
1

ep(s,t)
= e	p(s,t);

v. ep(t,s)ep(s,r) = ep(t,r);
vi. ep(t,s)eq(t,s) = ep⊕q(t,s);

vii.
ep(t,s)
eq(t,s)

= ep	q(t,s);

viii.
(

1
ep(.,s)

)∆
= − p(t)

eσp (.,s)
.

Throughout this paper we assume that the followings:
(C1) t0 ∈ T and [t0,∞)T = {t ∈ T : t ≥ t0},
(C2) r ∈ Crd([t0,∞)T, (0,∞)) and

∫∞
t0

1
r(t)

∆t = ∞,
(C3) p,q1,q2 ∈ Crd([t0,∞)T, [0,∞))
(C4) τ1,τ2 ∈ Crd(T,T), lim

t→∞
τ1(t) = lim

t→∞
τ2(t) = ∞, τ2 has inverse function τ−12 ∈ Crd(T,T), v :=

τ−12 ◦τ1 ∈ Crd(T,T), τ
∆
1 ,v

∆ ∈ Crd([t0,∞)T, (0,∞)), τ1(t),v(t) ≤ t for t ∈ [t0,∞)T, τ1([t0,∞)T) =
[τ1(t0),∞)T, v([t0,∞)T) = [v(t0),∞)T, where τ1([t0,∞)T) = {τ1(t) : t ∈ [t0,∞)T} and v([t0,∞)T) =
{v(t) : t ∈ [t0,∞)T},
(C5) f1,f2 ∈ C(R,R), there exist positive constants L1,L2,M such that

f1(u)
u
≥ L1, 0 <

f2(u)
u
≤

L2 and | f2(u) |≤ M for u 6= 0 and
q1(t)

e−p
r

(σ(t),t0)
L1 −q2(v(t))L2v∆(t) > 0 for t ∈ [t0,∞)T,

(C6) g ∈ C(R × R,R), there exist positive constants L3 such that
g(u,v)
v
≤ L3 and vg(u,v) >



44 TUǦLA AND TOPAL

0 for v 6= 0,

(C7)

∫ ∞
t

[
1

r(s)

∫ s
v(s)

q2(u)∆u

]
∆s < ∞ for every sufficiently large t ∈ T,

(C8) −
p(t)
r(t)

is positively regressive, which means 1 −µ(t)p(t)
r(t)

> 0.

The following lemma has an important role to prove our main results.

Lemma 1.5. Assume that (C1)−(C7) hold. Furthermore, suppose that x is a positive solution of (1.1)

on [t0,∞)T, then
(
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

)∆
< 0 and x∆(t) > 0 on [t0,∞)T.

Proof. Easily we get(
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

)∆
=

(r(t)g(x(t),x∆(t)))∆e−p
r

(t,t0) − (e−pr (t,t0))
∆r(t)g(x(t),x∆(t))

e−p
r

(t,t0)e−pr (σ(t),t0)

=
(r(t)g(x(t),x∆(t)))∆ + p(t)g(x(t),x∆(t))

e−p
r

(σ(t),t0)

=
−q1(t)f1(x(τ1(t))) −q2(t)f2(x(τ2(t)))

e−p
r

(σ(t),t0)

< 0.

This implies that
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

is decreasing.

We claim that x∆(t) > 0 on [t0,∞)T.

If not, then there is t ≥ t1 such that
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

≤
r(t1)g(x(t1),x

∆(t1))

e−p
r

(t1,t0)

:= a < 0. From (C6),

we get x∆(t) ≤ a
L3

e−p
r

(t,t0)

r(t)
. Integrating from t1 to t and using decreasing of e−p

r
(.,t0), we have

x(t) −x(t1) ≤
ae−p

r
(t1,t0)

L3

∫ t
t1

1

r(s)
∆s.

So x(t) ≤−∞. This implies that x(t) is eventually negative which is a contradiction.
Hence, x∆(t) > 0 on [t0,∞)T. �

2. Main Results

In this section, we’ll obtain some new oscillation criteria of second-order dynamic equation (1.1)
with damping by using the generalized Riccati transformation and the inequality technique.

Theorem 2.1. Assume that (C1) − (C8) hold. Furthermore, suppose that there exists a positive
function α ∈ Crd([t0,∞)T,R) such that for every sufficiently large T,

lim sup
t→∞

∫ t
T

([
q1(s)L1

e−p
r

(σ(s),t0)

−q2(v(s))L2v∆(s)
]
α(s) −

(α∆+ (−s))2(r(τ1(s)))L3
4α(s)τ∆1 (s)

)
∆s = ∞, (2.1)

where α∆+ (s) = max{α∆(s), 0}. Then every solution of (1.1) is oscillatory.

Proof. Assume that x is a nonoscillatory solution of (1.1). Without loss of generality, we may assume
x is an eventually positive solution of (1.1). That is, there exists t1 ∈ T for t ≥ t1 and x(t) > 0. We
defined the function z by

z(t) =

∫ t
t1

g(x(s),x∆(s))

e−p
r

(s,t0)

∆s +

∫ ∞
t

1

r(s)

∫ s
v(s)

q2(u)f2(x(τ2(u)))∆u∆s

≥
∫ t
t1

g(x(s),x∆(s))

e−p
r

(s,t0)

∆s

≥ 0.



OSCILLATION FOR NONLINEAR FUNCTIONAL DYNAMIC EQUATIONS 45

Thus, we get

z∆(t) =

(∫ t
t1

g(x(s),x∆(s))

e−p
r

(s,t0)

∆s

)∆
+

(∫ ∞
t

1

r(s)

∫ s
v(s)

q2(u)f2(x(τ2(u)))∆u∆s

)∆
=

g(x(t),x∆(t))

e−p
r

(t,t0)

−
1

r(t)

∫ t
v(t)

q2(u)f2(x(τ2(u)))∆u

and

r(t)z∆(t) =
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

−
∫ t
v(t)

q2(u)f2(x(τ2(u)))∆u

=
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

−
∫ t
v(t1)

q2(u)f2(x(τ2(u)))∆u +

∫ v(t)
v(t1)

q2(u)f2(x(τ2(u)))∆u.

Making substitution s = v(u), we have∫ t
t1

q2(v(u))f2(x(τ1(u)))v
∆(u)∆u =

∫ v(t)
v(t1)

q2(s)f2(x(τ1(v
−1(s))))∆̃s

=

∫ v(t)
v(t1)

q2(s)f2(x(τ2(s)))∆̃s for t ∈ [t1,∞)T. (2.2)

According to condition v([t0,∞)T) = [v(t0),∞)T in (C4), we get that the derivative ∆ on T is equal
to the derivative 4̃ on T̃ := v([t0,∞)T) in (2.2). Hence, we conclude∫ t

t1

q2(v(u))f2(x(τ1(u)))v
∆(u)∆u =

∫ v(t)
v(t1)

q2(s)f2(x(τ2(s)))∆s for t ∈ [t1,∞)T.

Thus, for t ∈ [t1,∞)T, we get

r(t)z∆(t) =
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

−
∫ t
v(t1)

q2(u)f2(x(τ2(u)))∆u +

∫ t
t1

q2(v(u))f2(x(τ1(u)))v
∆(u)∆u

(r(t)(z∆(t))∆ =

(
r(t)g(x(t),x∆(t))

e−p
r

(t,t0)

)∆
−q2(t)f2(x(τ2(t))) + q2(v(t))f2(x(τ1(t)))v∆(t)

=
(r(t)g(x(t),x∆(t)))∆e−p

r
(t,t0) − (e−pr (t,t0))

∆r(t)g(x(t),x∆(t))

e−p
r

(t,t0)e−pr (σ(t),t0)
−q2(t)f2(x(τ2(t)))

+q2(v(t))f2(x(τ1(t)))v
∆(t)

=
(r(t)g(x(t),x∆(t)))∆ + p(t)g(x(t),x∆(t))

e−p
r

(σ(t),t0)

−q2(t)f2(x(τ2(t)))

+q2(v(t))f2(x(τ1(t)))v
∆(t)

=
−q1(t)f1(x(τ1(t))) −q2(t)f2(x(τ2(t)))

e−p
r

(σ(t),t0)

−q2(t)f2(x(τ2(t)))

+q2(v(t))f2(x(τ1(t)))v
∆(t)

≤
−q1(t)L1x(τ1(t))
e−p

r
(σ(t),t0)

+ q2(v(t))L2x(τ1(t))v
∆(t)

= −
[

q1(t)L1
e−p

r
(σ(t),t0)

−q2(v(t))L2v∆(t)
]
x(τ1(t))

= −Q(t)x(τ1(t)) < 0,

where Q(t) =
q1(t)L1

e−p
r

(σ(t),t0)

−q2(v(t))L2v∆(t).



46 TUǦLA AND TOPAL

Thus, there exists t2 ∈ [t1,∞)T such that r(t)z∆(t) strictly decreasing on [t2,∞)T and either even-
tually positive or eventually negative. Since r(t) > 0 for t ∈ [t0,∞)T, z∆(t) is also either eventually
positive or eventually negative.

We claim

z∆(t) > 0 for t ∈ [t2,∞)T. (2.3)
Assume that (2.3) does not hold, then there exists tξ ∈ [t2,∞)T such that z∆(tξ) < 0. Since

r(t)z∆(t) is strictly decreasing on [t2,∞)T, it is clear that r(t)z∆(t) ≤ r(tξ)z∆(tξ) = −c < 0 for
t ∈ [tξ,∞)T. Thus, we obtain z∆(t) ≤ −c 1r(t) for t ∈ [tξ,∞)T. By integrating both sides of the last
inequality from tξ to t, we get

z(t) −z(tξ) ≤−c
∫ t
tξ

1

r(s)
∆(s) for t ∈ [tξ,∞)T.

Noticing (C2), we have lim
t→∞

z(t) = −∞. This contradicts z(t) ≥ 0. Therefore, (2.3) holds. Thus, we
have z∆(t) > 0 on [t2,∞)T.

Define the function w by generalized Riccati substitution

w(t) := α(t)
r(t)z∆(t)

x(τ1(t))
.

There exist t3 ∈ [t2,∞)T such that w(t) > 0 for t ∈ [t3,∞)T.
Easily, we get

w∆ = (rz∆)∆
α

x◦ τ1
+ (rz∆)σ

(
α

x◦ τ1

)∆
= (rz∆)∆

α

x◦ τ1
+ (rz∆)σ

(
α∆

(x◦ τ1)σ
−

(x◦ τ1)∆α
x◦ τ1(x◦ τ1)σ

)
≤ (rz∆)∆

α

x◦ τ1
+ α∆+

(rz∆)σ

(x◦ τ1)σ
−α

(rz∆)σ

(x◦ τ1)σ
(x◦ τ1)∆

x◦ τ1

= (rz∆)∆
α

x◦ τ1
+ α∆+

wσ

ασ
−α

wσ

ασ
(x◦ τ1)∆

x◦ τ1
,

where α∆+ (s) = max{α∆(s), 0}.
Thus, we have

w∆ ≤−Qα + α∆+
wσ

ασ
−α

wσ

ασ
(x◦ τ1)∆

x◦ τ1
.

From the chain rule, we know that

(x◦ τ1)∆ = (x∆̃ ◦ τ1)τ∆1 .
According to condition τ1([t0,∞)T) = [τ1(t0),∞)T in (C4), we get that the derivative ∆ on T is

equal to the derivative 4̃ on T̃ := τ1([t0,∞)T). So, we have

w∆ ≤−Qα + α∆+
wσ

ασ
−α

wσ

ασ
(x∆ ◦ τ1)τ∆1

x◦ τ1
.

Also

z∆(t) = g(x(t),x∆(t)) −
1

r(t)

∫ t
v(t)

q2(u)f2(x(τ2(u)))∆u

≤ g(x(t),x∆(t))
≤ L3x∆(t),

implies that

−x∆(t) ≤−
z∆(t)

L3
and so

w∆ ≤−Qα + α∆+
wσ

ασ
−α

wσ

ασ
(z∆ ◦ τ1)
x◦ τ1

τ∆1
L3

.



OSCILLATION FOR NONLINEAR FUNCTIONAL DYNAMIC EQUATIONS 47

Since τ1(t) ≤ t ≤ σ(t) and r(t)z∆(t) is strictly decreasing on [t2,∞)T, we get

(r ◦ τ1)(z∆ ◦ τ1) ≥ (rz∆)σ

and

(z∆ ◦ τ1) ≥
(rz∆)σ

(r ◦ τ1)
.

Thus, we get

w∆ ≤ −Qα + α∆+
wσ

ασ
−

α

L3

wσ

ασ
(rz∆)σ

(r ◦ τ1)
τ∆1
x◦ τ1

= −Qα + α∆+
wσ

ασ
−

α

L3

(
wσ

ασ

)2
(x◦ τ1)σ

x◦ τ1
τ∆1
r ◦ τ1

. (2.4)

From (C4) we see that τ1(t) is strictly increasing on [t0,∞)T. Since t ≤ σ(t), we have τ1(t) ≤ τσ1 (t).
Since x∆(t) > 0, we get x ◦ τ1(t) ≤ x ◦ τσ1 (t). Hence, from (2.4) there exist a sufficiently large
t4 ∈ [t3,∞)T such that

w∆ ≤ −Qα + α∆+
wσ

ασ
−

α

L3

(
wσ

ασ

)2
τ∆1
r ◦ τ1

= −Qα−
[
α∆+
2

√
(r ◦ τ1)L3
ατ∆1

−
wσ

ασ

√
ατ∆1

(r ◦ τ1)L3

]2
+

(α∆+ )
2(r ◦ τ1)L3
4ατ∆1

≤ −Qα +
(α∆+ )

2(r ◦ τ1)L3
4ατ∆1

. (2.5)

Integrating both sides of the last inequality from t4 to t, we get

w(t) −w(t4) ≤−
∫ t
t4

[
(Q(s)α(s)) −

(α∆+ (s))
2(r(τ1(s)))L3

4α(s)τ∆1 (s)

]
∆s

Since w(t) > 0 for t ∈ [t3,∞)T we have∫ t
t4

[
(Q(s)α(s)) −

(α∆+ (s))
2(r(τ1(s)))L3

4α(s)τ∆1 (s)

]
∆s ≤ w(t4) −w(t) ≤ w(t4)

and

lim sup
t→∞

∫ t
t4

[
(Q(s)α(s)) −

(α∆+ (s))
2(r(τ1(s)))L3

4α(s)τ∆1 (s)

]
∆s ≤ w(t4) < ∞,

which is a contradiction to (2.1). The proof is completed. �

Theorem 2.2. Assume that (C1) − (C8) hold. Let H be an rd-continuous function defined as follows:

H : DT ≡{(t,s) : t ≥ s ≥ t0, t,s ∈ [t0,∞)T}→ R,

such that

H(t,t) = 0, for t ≥ t0,
H(t,s) > 0, for t > s ≥ t0,

and H has a nonpositive rd-continuous delta partial derivative H∆s with respect to the second variable
and

lim sup
t→∞

1

H(t,T)

∫ t
T

H(t,s)

(
Q(s)α(s) −

(α∆+ (−s))2(r(τ1(s)))L3
4α(s)τ∆1 (s)

)
∆s = ∞ (2.6)

for every sufficiently large T , where Q(t) =
q1(t)L1

e−p
r

(σ(t),t0)
−q2(v(t))L2v∆(t) and α∆+ (s) = max{α∆(s), 0},

then all solutions of (1.1) are oscillatory.



48 TUǦLA AND TOPAL

Proof. Assume that x is a nonoscillatory solution of (1.1). Without loss of generality, we may assume
x is an eventually positive solution of (1.1). Proceeding as in the proof of the Theorem 2.1, we get

w∆(t) ≤−Q(t)α(t) +
(α∆+ )

2(t)(r ◦ τ1(t))L3
4α(t)τ∆1 (t)

.

Multiplying by H(t,s) and then integrating from t4 to t, we obtain∫ t
t4

H(t,s)w∆(s)4s ≤
∫ t
t4

H(t,s)

(
−Q(s)α(s) +

(α∆+ )
2(s)(r ◦ τ1(s))L3
4α(s)τ∆1 (s)

)
∆s.

Since ∫ t
t4

H(t,s)w∆(s)∆s = H(t,s)w(s) |s=ts=t4 −
∫ t
t4

H∆s (t,s)w
σ(s)∆s,

we get

−H(t,t4)w(t4) ≤
∫ t
t4

H(t,s)

(
−Q(s)α(s) +

(α∆+ )
2(s)(r ◦ τ1(s))L3
4α(s)τ∆1 (s)

)
∆s.

Thus, we have∫ t
t4

H(t,s)

(
Q(s)α(s) −

(α∆+ )
2(s)(r ◦ τ1(s))L3
4α(s)τ∆1 (s)

)
∆s ≤ H(t,t4)w(t4)

and so

1

H(t,t4)

∫ t
t4

H(t,s)

(
Q(s)α(s) −

(α∆+ )
2(s)(r ◦ τ1(s))L3
4α(s)τ∆1 (s)

)
∆s ≤ w(t4) < ∞,

which contradicts with (2.6). This completes the proof. �

Theorem 2.3. Assume that (C1)−(C8) hold. Let H be an rd-continuous function defined as follows:
H : DT ≡{(t,s) : t ≥ s ≥ t0, t,s ∈ [t0,∞)T}→ R,

such that

H(t,t) = 0, for t ≥ t0,
H(t,s) > 0, for t > s ≥ t0,

and H has an rd-continuous ∆−partial derivative H∆s on DT with respect to the second variable.
Let h : DT → R be an rd-continuous function satisfying

H∆s (t,s) + H(t,s)
α∆+ (s)

ασ(s)
=
h(t,s)

ασ(s)

√
H(t,s), (t,s) ∈ DT

and

lim sup
t→∞

1

H(t,T)

∫ t
T

(
H(t,s)Q(s)α(s) −

[h(t,s)]2(r ◦ τ1)(s)L3
4α(s)τ∆1 (s)

)
∆s = ∞ (2.7)

for every sufficiently large T , where Q(t) =
q1(t)L1

e−p
r

(σ(t),t0)
−q2(v(t))L2v∆(t) and α∆+ (s) = max{α∆(s), 0},

then all the solutions of (1.1) are oscillatory.

Proof. Assume that x is a nonoscillatory solution of (1.1). Without loss of generality, we may assume
x is an eventually positive solution of (1.1). Proceeding as in the proof of the Theorem 2.1, we have
(2.5). Multiplying (2.5) by H(t,s) and then integrating from t4 to t, we obtain∫ t

t4

H(t,s)Q(s)α(s)∆s ≤ −
∫ t
t4

H(t,s)w∆(s)∆s +

∫ t
t4

H(t,s)
α∆+ (s)

ασ(s)
wσ(s)∆s

−
∫ t
t4

H(t,s)
α(s)τ∆1 (s)

(ασ(s))2(r ◦ τ1)(s)L3
[wσ(s)]2∆s.

Thus, using ∫ t
t4

H(t,s)w∆(s)∆s = [H(t,s)w(s)]s=ts=t4 −
∫ t
t4

H∆s (t,s)w
σ(s)∆s,



OSCILLATION FOR NONLINEAR FUNCTIONAL DYNAMIC EQUATIONS 49

we have ∫ t
t4

H(t,s)Q(s)α(s)∆s ≤ H(t,t4)w(t4) +
∫ t
t4

([
H∆s (t,s) + H(t,s)

α∆+ (s)

ασ(s)

]
wσ(s)

−H(t,s)
α(s)τ∆1 (s)

(ασ(s))2(r ◦ τ1)(s)L3
[wσ(s)]2

)
∆s

≤ H(t,t4)w(t4) +
∫ t
t4

(
h(t,s)

ασ(s)

√
H(t,s)wσ(s)

−H(t,s)
α(s)τ∆1 (s)

(ασ(s))2(r ◦ τ1)(s)L3
[wσ(s)]2

)
∆s

= H(t,t4)w(t4) +

∫ t
t4

(
[h(t,s)]2(r ◦ τ1)(s)L3

4α(s)τ∆1 (s)

−
[
h(t,s)

2ασ(s)

√
(ασ(s))2(r ◦ τ1)(s)L3

α(s)τ∆1 (s)
−

√
H(t,s)

α(s)τ∆1 (s)

(ασ(s))2(r ◦ τ1)(s)L3
wσ(s)

]2
∆s

≤ H(t,t4)w(t4) +
∫ t
t4

[h(t,s)]2(r ◦ τ1)(s)L3
4α(s)τ∆1 (s)

∆s.

So, we get

1

H(t,t4)

∫ t
t4

(
H(t,s)Q(s)α(s) −

[h(t,s)]2(r ◦ τ1)(s)L3
4α(s)τ∆1 (s)

)
∆s ≤ w(t4) < ∞,

which is a contradiction to (2.7). The proof is completed. �

3. Examples

Example 3.1. Let T = R. Consider the equation

(
1

t

x′(t)

2 + sin2(x(t))

)′
+ t

1
3

x′(t)

2 + sin2(x(t))
+

1

t2
x(t

1
5 − 3)(x(t

1
5 − 3)2 + 4) +

10

t21
x(t2 − 3)

x2(t2 − 3) + 1
= 0, (3.1)

for t ≥ t0 := 4
Here r(t) = 1

t
,p(t) = t

1
3 ,q1(t) =

1
t2
,q2(t) =

10
t21
,τ1(t) = t

1
5 − 3,τ2(t) = t2 − 3, g(x(t),x′(t)) =

x′(t)
2+sin2(x(t))

, f1(u) = u(u
2 + 4) and f2(u) =

u
u2+1

,
f1(u)
u

= u2 + 4 ≥ 4 := L1 and
f2(u)
u

= 1
u2+1

≤ 1 :=
L2 for u 6= 0, | f2(u) |≤ 12 := M,

g(x(t),x′(t))
x′(t)

≤ 1
2

:= L3.

Thus, we obtain
∫∞
t0

1
r(t)

dt =
∫∞
t0
tdt = ∞,v(t) = t

1
10 < t for t ∈ [4,∞) and 1 − µ(t)p(t)

r(t)
= 1 >

0 for t ∈ [4,∞).
Also, we get

q1(t)L1
e−p
r

(σ(t),t0)

−q2(v(t))L2v′(t) =
4e(t−4)

4
3

t2
−

1

t3
=

4te(t−4)
4
3 − 1

t3
> 0 for t ∈ [4,∞)

and∫ ∞
t

[
1

r(s)

∫ s
v(s)

q2(u)du

]
ds =

∫ ∞
t

[
s

∫ s
s

1
10

10

u21
du

]
ds =

∫ ∞
t

s17 − 1
2s19

ds < ∞ for t ∈ [4,∞).

Hence, we have∫ ∞
T

([
q1(s)L1

e−p
r

(σ(s),t0)

−q2(v(s))L2v′(s)
]
α(s) −

(α′+(s))
2(r(τ1(s)))L3

4α(s)τ∆1 (s)

)
ds

=

∫ ∞
T

(
4se(s−4)

4
3 − 1

s3
s3 −

3s2 1
s

1
5 −3

1
2

4s3 1
5s

4
5

)
ds



50 TUǦLA AND TOPAL

=

∫ ∞
T

(
4se(s−4)

4
3 − 1 −

15

8(s
2
5 − 3s

1
5 )

)
ds

=

∫ ∞
T

(
32(s

7
5 − 3s

6
5 )e(s−4)

4
3 − 8(s

2
5 − 3s

1
5 ) − 15

8(s
2
5 − 3s

1
5 )

)
ds = ∞.

Therefore, according to Theorem 2.1, every solution of (3.1) is oscillatory on [4,∞).

Example 3.2. Let T = 2N0 . Consider the equation(
tx∆(t)

)∆
+

1

t2
x∆(t) +

t + 1

2
x(
t

2
)(x2(

t

2
) + 2) +

1

2t2
x(2t)

2 + x2(2t)
= 0, (3.2)

for t ∈ [t0,∞)T, t ≥ t0 := 2
Here, r(t) = t, p(t) = 1

t2
, q1(t) =

t+1
2
, q2(t) =

1
2t2
, τ1(t) =

t
2
< t, τ2(t) = 2t, g(x(t),x

∆(t)) =

x∆(t), f1(u) = u(u
2 + 2) and f2(u) =

u
u2+2

,
f1(u)
u

= u2 + 2 ≥ 2 := L1 and
f2(u)
u

= 1
u2+2

≤ 1
2

:=

L2 for u 6= 0, | f2(u) |≤ 12 := M,
g(x(t),x∆(t))

x∆(t)
= 1, L3 = 1.

Therefore, we obtain
∫∞
t0

1
r(t)

∆t =
∫∞

2
1
t
∆t = ∞, v(t) = t

4
< t,v∆(t) = 1

4
for t ∈ [2,∞)T and

1 −µ(t)p(t)
r(t)

= 1 − t 1
t3

= 1 − 1
t2
> 0 for t ∈ [2,∞)T.

Also, we get

q1(t)L1
e−p
r

(σ(t),t0)

−q2(v(t))L2v∆(t) > q1(t)L1 −q2(v(t))L2v∆(t) =
t3 + t2 − 1

t2
> 0 for t ∈ [2,∞)T

and∫ ∞
t

[
1

r(s)

∫ s
v(s)

q2(u)∆u

]
∆s =

∫ ∞
t

[
1

s

∫ s
s
4

1

2u2
∆u

]
∆s =

∫ ∞
t

[
1

s

(
−1
u

)∣∣∣∣s
s
4

∆u

]
∆s =

∫ ∞
t

3

s2
∆s < ∞.

Hence we have∫ ∞
T

([
q1(s)L1

e−p
r

(σ(s),t0)

−q2(v(s))L2v∆(s)
]
α(s) −

(α∆+ (s))
2(r(τ1(s)))L3

4α(s)τ∆1 (s)

)
∆s

=

∫ ∞
T

([
s3 + s2 − 1

s2

]
s−

s
2

4s1
2

)
∆s

=

∫ ∞
T

(
s3 + s2 − 1

s
−

1

4

)
∆s = ∞.

Thus, according to Theorem 2.1, every solution of (3.2) is oscillatory on [2,∞)T.

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Department of Mathematics, Ege University,35100 Bornova, Izmir-Turkey

∗Corresponding author: f.serap.topal@ege.edu.tr


	1. Introduction
	2. Main Results
	3. Examples
	References