Download Full Issue in PDF - Academy Publisher

JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1533
t
∗
limsup ∫ [ β ( s, T ) g( s)( q( s) − p( s))
t→∞
T
Δ γ + 1
α()(( s g ())) s
+
− ] Δ s =∞.
γ+
1 γ
( γ + 1) g ( s)
(29)
Then every solution of (5) is either oscillatory on [ t , ∞)
0 T
or tends to zero.
proof Suppose that x is an eventually positive solution
of (5), say xt () > 0, x( τ ( t)) > 0, x( δ ( t)) > 0for all t ≥ t for
1
some t ≥ t . We consider only this case, because the
1 0
proof for the case that x is eventually negative is similar.
In the view of (5), by (H2) - (H5), and there exists
t ≥ t such that for all t ≥ t , we have
2 1
2
Δ
( ( z ) γ
γ
α )
Δ
≤−( q( t) − p( t)) x ( δ( t)) < 0 , (30)
γ
then α( z
Δ ) is strictly decreasing on [ t 2
, ∞ ). Hence z (t)
Δ
and z () t are of constant sign eventually. We claim that
x()
t is bounded. If not, there exists { t k
} ⊆ [ t 2
, ∞ ), such
that lim t =∞ ,lim x( t ) =∞,
and
k→∞
k
k→∞
k
x( t ) = max{ x( s): t ≤ s≤
t }.
k
Since lim τ ( t ) =∞, we can choose a large k such that
k→∞
0
k
τ ( t ) > t , and by (H2), we obtain that
k
x( τ( t )) = max{ x( s) : t ≤ s≤τ( t )}
Therefore, for all large k,
k
0
0
≤ max{ x( s) : t ≤ s ≤ t } = x( t ).
z( τ ( t )) ≥ x( t ) −c x( τ ( t )) ≥ (1 − c ) x( t )
k k 0 k
0 k
,
and lim zt ( ) =∞. From (H1) and (30), as in the proof of
k→∞
k
Theorem 1 [4], there exists t 3
≥ t 2
such that for all t ≥ t ,
3
we have
In view of (5), (30) and (31), we get
0
k
Δ
zt () > 0, z () t > 0. (31)
Δ
( ( z ) γ
γ
α )
Δ
≤−( q( t) − p( t)) z ( δ( t)) < 0. (32)
Now by using the same proof of Theorem 1, we get a
contradiction with (29). Thus x (t) is bounded and hence z
(t) is bounded.
Also, by using (H1) and the same proof of Theorem 1
Δ
in [4], there exist t 4 ≥ t 3 such that z () t > 0on [t 4 , ∞).
There are two cases.
Δ
Case 1 zt () > 0 and z () t > 0. As in the proof of
Theorem 1, we get a contradiction with (29).
Δ
Case 2 zt () < 0and z ( t) > 0 . We claim lim xt ( ) = 0 .
k
k
k
t→∞
Assume not, then there exists { t } ⊆[ t , ∞)
such that
k 5
lim t =∞,
lim xt ( ) = : b> 0 and x( t ) = max{ x( s):
t ≤ s
k
k
k
0
k→∞
t k
k→∞
≤ }. But, by x( τ ( t )) ≤ x( t ), we get
k
k
0 > zt ( ) ≥ xt ( )(1 −c) →b(1 − c) > 0, as k→∞.
k
k
0 0
Which is a contradiction. This completes the proof.
Corollary 4 Assume that (H1) - (H5) hold, Furthermore,
suppose that for all sufficiently large T ∗ , and for δ ( T ) >
T ∗
,
we have
t
∗
α()
s
limsup ∫ ( sβ
( s, T )( q( s) − p( s)) − ) Δ s =∞.
t→∞
T
γ+
1 γ
( γ + 1) s
Then every solution of (5) is either oscillatory on [ t , ∞)
0 T
or tends to zero.
Corollary 5 Assume that (H1) - (H5) hold, Furthermore,
suppose that for all sufficiently large T ∗ , and for δ ( T ) >
T ∗
,
we have
t
∗
limsup ∫ β (, sT)(() qs − ps ()) Δ s=∞.
t→∞
T
Then every solution of (5) is either oscillatory on [ t , ∞)
0 T
or tends to zero.
We next study a Philos-type oscillation criteria for (5).
Theorem 4 Assume that (H1) - (H5) hold. Let g (t) be
as defined in Theorem 1, and H, h∈C rd
( D, )
such that
H ∈R . Furthermore, suppose that there exists a
positive rd-continuous function ϕ () t such that (24), (25)
hold, and for all sufficiently largeT ∗ , we have
1 t
∗
limsup ∫ { β ( s, T ) g( s)( q( s) − p( s)) H( t, s)
t→∞
t0
Htt (, )
0
α()( s h (,)) t s
− } Δ s =∞.
γ + 1
−
γ+
1 γ
( γ + 1) g ( s)
(33)
Then every solution of (5) is either oscillatory on [ t , ∞)
0 T
or tends to zero.
Proof Suppose that (5) has a nonoscillatory solution x
(t), without loss of generality, say xt () > 0, x( τ ( t)) > 0,
x( δ ( t)) > 0, for all t ≥ t , for some t
1
1
≥ t 0
. By (H2) - (H5),
we obtain that (30) holds for all t ≥ t , and zt () and
1
Δ
z () t are of constant sign eventually. Similar to the proof
of Theorem 3, we claim that x()
t is bounded. If not, there
exists{ t } ⊆[ t , ∞ ), for all large k, there exists t
k 1
2
≥ t 1
, such
that (31) and (32) hold for t ≥ t . Again we define wt () as
2
in the proof of Theorem 1, then there exists t 3
≥ t 2
,
sufficiently large such that for t
∗
≥ t and for t ≥ t ∗
, we
3
find
β (, tt) gt ()( qt () − pt ())
3
Δ g () t γ g()
t
≤− w () t + w () t −
( w ()). t
g t t g t
Δ
σ σ λ
σ 1 γ σ λ
() α ()( ())
And similar to the proof of the theorem 3, we obtain
1 t
∫ [ β ( s, t ) g( s)( q( s) − p( s)) H( t, s)
t
Htt (, )
0 3
0
γ + 1
( h ( t, s)) α( s) −
γ+
1 γ ] s (
∗
ϕ t ) w (
∗
t )
− Δ ≤ +
( γ + 1) g ( s)
(34)
© 2013 ACADEMY PUBLISHER