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1530 JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013
The derivative and the forward jump operator are
related by the useful formula
σ
f f μ f
Δ
σ
= + , where f : = f σ.
We will also make use of the following product and
quotient rules for the derivative of the product f g and the
quotient f g( gg σ ≠ 0) of two differentiable functions f
and g :
Δ
⎛ f ⎞ f g−
f g
( f g) Δ = f Δ g+ f σ
g
Δ
, and ⎜ ⎟ =
⎝ g ⎠ gg σ
Δ
Δ
. (6)
By using the product rule, the derivative of f () t =
( t − α) m
for m ∈ andα ∈ T can be calculated as
m−1
Δ
v
m−v−1
() = ( σ() −α)( −α) .
v=
0
f t ∑ t t
(7)
For a, b∈ T and a differentiable function f , the
Cauchy integral of f Δ is defined by
b Δ
∫a f () t Δ t = f( b) − f( a)
.
The integration by parts formula follows from (6) and
reads
b Δ
b b σ Δ
∫ f () tgt () Δ t= ftgt () ()| −∫ f tg()
tΔt.
a
To prove our main results, we will use the formula
1 1
( x γ Δ
( t)) [ hx σ (1 h) x] γ − Δ
= γ + − dhx ( t)
0
a
a
∫ . (8)
which is a simple consequence of Keller′s chain rule [2].
Also, we need the following lemma [5].
Lemma 1 Assume A and B are nonnegtive constants, λ
> 1, then
λ
−
− ≤( − 1) .
1
AB λ A λ λ B
λ
The reader is referred to [2] for more detailed and
extensive developments in calculus on time scales.
III. MAIN RESULTS
First, we state the oscillation criteria for (4).
Set
yt () = xt () + ctx () ( τ ()). t
(9)
Theorem 1 Assume that (H1) - (H5) hold, Furthermore,
suppose that there exists a positive Δ−differentiable
function g()
t such that for all sufficiently large T ∗
, and
∗
for all δ (T) > T , we have
t
∗
limsup ∫ ( β (, sT ) gsQs () () −
t→∞
T
α( s)(( g ( s)) )
( γ + 1) g ( s)
Δ γ + 1
+
γ+
1 γ
) Δ s =∞.
Then every solution of (4) is oscillatory on [ t , ∞)
0 T
.
(10)
proof Suppose (4) has a nonoscillatory solution x (t).
without loss of generality, there exists some t 1
≥ t 0
,
sufficiently large such that xt () > 0, x( τ ( t)) > 0, x( δ ( t))
> 0 for all t ≥ t . Hence In the view of (9), by (H1) we
1
get yt () > 0. from (4) and by (H2) - (H5), we have that
Δ
( y )) γ
γ
α
Δ
≤ −( q() t − p()) t x ( δ()) t < 0,
and using the same proof of Theorem 1 [4], there exists
t ≥ t such that for all t ≥ t , we have
2 1
2
Δ
⎧yt () > 0, y () t > 0,
(11)
⎨
Δ
( ( y ) γ Δ
γ
⎩ α ) ≤ −( q( t) − p( t))(1 − c( δ( t))) y ( δ( t)) < 0.
By the definition of Qt (), we get
Δ
( ( y ) γ
γ
α )
Δ
≤ − Q( t) y ( δ( t)) < 0. (12)
Make the generalized Riccati substitution
Δ γ
α()( t y ()) t
wt () = gt ()
. (13)
γ
y () t
By the product and quotient rules, we have for all t ≥ t2
Δ γ Δ
Δ gt ()( α()( t y())) t ⎛ gt () ⎞
Δ γ
w () t = ( ()( t y ())) t
γ
+⎜ γ ⎟ α
y () t ⎝ y () t ⎠
Δ γ Δ
gt ()( α()( t y()))
t
= +
γ
y () t
Δ
γ Δ
g () t g()( t y ()) t
Δ γ σ
( −
)( α( t)( y ( t)) ) .
γσ γ γσ
y () t y () t y () t
From (12) - (14), we obtain
Δ
Δ
⎛ y( δ ( t)) ⎞ g ( t)
σ
w () t ≤− g() t Q() t ⎜ ⎟ + w () t
σ
⎝ yt () ⎠ g () t
σ γ Δ
gtw () ()( t y()) t
−
σ
γ .
g () t y () t
γ
Δ
σ
(14)
(15)
First consider the case when δ () t ≥ t . For all large t,
Δ
from y () t > 0, we have
which implies that
y( δ ( t))
≥ 1 ,
yt ()
Δ
σ γ Δ
Δ
g () t σ g() t w ()( t y ()) t
w () t ≤− g() t Q() t + w () t − . (16)
σ σ γ
g () t g () t y () t
Next consider the case when δ () t ≤ t, for all large t. By
using α( y
Δ )
γ
is strictly decreasing on [ t 2
, ∞ ) , we can
choose t ≥ t such that δ () t ≥ t , for t ≥ t . Then we
3 2
2
3
obtain
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