Electronic Journal of Differential Equations, Vol. 2006(2006), No. 18, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
OSCILLATION FOR HIGHER ORDER NONLINEAR ORDINARY
DIFFERENTIAL EQUATIONS WITH IMPULSES
CHAOLONG ZHANG, WEIZHEN FENG
Abstract. In this paper, we study the oscillation of solutions to higher order
nonlinear ordinary differential equations with impulses. Several criteria for
the oscillations of solutions are given. We find some suitable impulse functions
such that all solutions are oscillatory under the impulse control.
1. Introduction
There are many publication on the oscillation of solutions to classical second
order nonlinear ordinary differential equations; see for example [1, 2, 3, 8, 9, 10,
13, 14, 15, 16]. There are also some publications on the oscillation of second order
ODEs with impulses [4, 7, 12], and some on higher order [5, 6]. In this paper, we
study higher order nonlinear ODEs with impulses. Under conditions (A) (B) (C)
stated below, we can always find some suitable impulse functions such that all the
solutions of the equation become oscillatory under the impulse control. We believe
that this oscillation result, under the impulse control, is significant both for the
theory and the applications.
2. Main results
We consider the system
x(2n) (t) + f (t, x(t)) = 0,
(i)
x
(t+
k)
(i)
= gk(i) (x (tk )),
t ≥ t0 , t 6= tk ,
i = 0, 1, . . . , 2n − 1, k = 1, 2 . . . ,
(2.1)
(i)
x(i) (t+
0 ) = x0 ,
where
x(i−1) (tk + h) − x(i−1) (tk )
,
h
h→0
x(i−1) (tk + h) − x(i−1) (t+
k)
x(i) (t+
)
=
lim
,
k
h
h→0+
x(i) (tk ) = lim−
2000 Mathematics Subject Classification. 34A37, 34K06, 34K11, 34K25.
Key words and phrases. Higher order; impulses; oscillation; ODE.
c
2006
Texas State University - San Marcos.
Submitted October 19, 2005. Published February 2, 2006.
Supported by grant 011471 from the Natural Science Foundation of Guangdong,
grant 0120 from the Natural Science Foundation of Guangdong Higher Education,
and grant Z03052 from the Natural Science Foundation of Guangdong Education Bureau.
1
2
C. ZHANG, W. FENG
EJDE-2006/18
0 < t0 < t1 < t2 < · · · < tk < . . . , k = 1, 2, . . . , limk→∞ tk = +∞, x(0) (t) = x(t),
and n is a natural number. In this article, we assume that the following conditions:
(A) f (t, x) is continuous on [t0 , +∞) × (−∞, +∞); xf (t, x) > 0 for x 6= 0;
f (t,x)
ϕ(x) ≥ p(t) for x 6= 0, where p(t) is positive and continuous on [t0 , +∞);
xϕ(x) > 0 for x 6= 0; ϕ0 (x) ≥ 0.
(B) gk(i) (x) is continuous on (−∞, +∞), and there exist positive numbers
(i) (i)
ak , bk such that
(i)
ak ≤
gk(i) (x)
(i)
≤ bk , i = 0, 1, . . . , 2n − 1.
x
(C)
(i)
(t1 − t0 ) +
(i) (i)
a1
(t − t1 ) +
(i−1) 2
b1
(i) (i)
(i−1) (i−1)
b2
b1
(i−1) (i−1)
b2
b1
(t3 − t2 )
(i)
a1 a2 . . . am
+ ··· +
a1 a2
(i−1)
. . . bm
(2.2)
(tm+1 − tm ) + · · · = +∞,
Definition 2.1. A function x : [t0 , t0 + α) → R, t0 > 0, α > 0 is said to be a
solution of (2.1), if
(i)
(i) x(i) (t+
0 ) = x0 , i = 0, 1, . . . 2n − 1
(ii) for t ∈ [t0 , t0 + α) and t 6= tk , x(t) satisfies x(2n) (t) + f (t, x(t)) = 0
(i)
(iii) x(i) (t) is left continuous on tk ∈ [t0 , t0 + α), and x(i) (t+
k ) = gk(i) x (tk ),
i = 0, 1, . . . 2n − 1.
Definition 2.2. A solution of (2.1) is said to be non-oscillatory if it is eventually
positive or eventually negative. Otherwise,this solution is said to be oscillatory.
Since (2.1) can be transformed into a first-order impulsive differential system,
theorems on the existence of solutions, the uniqueness of solutions and the existence
of global solutions can be seen in [11]. In the following, we always assume the
solutions of (2.1) exists on [t0 , +∞).
Lemma 2.3. Let x(t) be a solution of (2.1), and conditions (A), (B), (C) be
satisfied. Suppose that there exists an i ∈ {1, 2, . . . , 2n − 1} and some T ≥ t0 , such
that x(i) (t) > 0 (< 0), x(i+1) (t) ≥ 0 (≤ 0) for t ≥ T . Then there exists some
T1 ≥ T , such that x(i−1) (t) > 0 (< 0), for t ≥ T1 .
Proof. Without loss of generality, let T = t0 , x(i) (t) > 0, x(i+1) (t) ≥ 0 for t ≥ T .
Assume that for any tk > T , x(i−1) (tk ) < 0. By x(i+1) (t) ≥ 0, x(i) (t) > 0,
t ∈ (tk , tk+1 ], we have that x(i) (t) is monotonically nondecreasing on (tk , tk+1 ].
For t ∈ (t1 , t2 ], we have
x(i) (t) ≥ x(i) (t+
1)
Integrating the above inequality, we have
(i) +
x(i−1) (t2 ) ≥ x(i−1) (t+
1 ) + x (t1 )(t2 − t1 )
(2.3)
(i) +
x(i−1) (t3 ) ≥ x(i−1) (t+
2 ) + x (t2 )(t3 − t2 )
(2.4)
Similarly,
(i)
(i)
From x (t2 ) ≥ x
(i−1)
x
(t+
1)
and (2.3), (2.4), we have
(i−1)
(t3 ) ≥ x
(i) +
(t+
2 ) + x (t2 )(t3 − t2 )
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OSCILLATION FOR HIGHER ORDER ODES
(i−1) (i−1)
≥ b2
x
(i)
(t2 ) + a2 x(i) (t2 )(t3 − t2 )
(i−1)
(i) +
(i)
[x(i−1) (t+
1 ) + x (t1 )(t2 − t1 )] + a2 x (t2 )(t3 − t2 )
(i−1)
(i) +
[x(i−1) (t+
1 ) + x (t1 )(t2 − t1 ) +
≥ b2
3
(i)
(i)
≥ b2
a2
(i−1)
b2
x(i) (t+
1 )(t3 − t2 )]
Applying induction, we have that for any natural number m,
(i−1)
(i−1) (i−1) (i−1) +
x(i−1) (tm ) ≥ bm−1 . . . b3
b2
x
(t1 ) + x(i) (t+
1 )[(t2 − t1 )
(i) (i)
(i)
+
(t − t2 ) + · · · +
(i−1) 3
b2
(i)
ak
(i)
a2 a3 . . . am−1
a2
(i−1) (i−1)
b3
b2
(t − tm−1 )]
(i−1) m
(2.5)
. . . bm−1
(i−1)
bk
By condition (C) and
> 0,
> 0, for all sufficiently large m, we have
x(i−1) (tm ) > 0. Which is contrary to the assumption. Hence, there exists some j
such that tj > T and x(i−1) (tj ) ≥ 0. Then
(i−1) (i−1)
x(i−1) (t+
j ) ≥ aj
(i)
(i−1)
Note that x (t) > 0 yields x
For t ∈ (tj , tj+1 ], we have
x
(tj ) ≥ 0.
(t) being monotonically increasing on (tj , tj+1 ].
x(i−1) (t) > x(i−1) (t+
j ) ≥ 0.
Especially,
x(i−1) (tj+1 ) > x(i−1) (t+
j ) > 0.
Similarly, for t ∈ (tj+1 , tj+2 ], we have
(i−1)
(i−1)
x(i−1) (t) > x(i−1) (t+
(tj+1 ) > 0.
j+1 ) ≥ aj+1 x
By induction,for t ∈ (tj+m−1 , tj+m ], we have x(i−1) (t) > 0. So for t ≥ tj+1 , we have
x(i−1) (t) > 0.
Summing up the above discussion, there exists some T1 ≥ T such that x(i−1) (t) > 0,
t ≥ T1 . The proof of the other case in this theorem is similar; so we omit it. The
proof of Lemma 2.3 is complete.
Lemma 2.4. Let x(t) be a solution of (2.1) and conditions (A), (B), (C) be satisfied. Suppose that there exist an i ∈ {1, 2, . . . , 2n} and some T ≥ t0 such that
x(t) > 0, x(i) (t) ≤ 0, for t ≥ T , and x(i) (t) is not always equal to 0 in [t, +∞).
Then x(i−1) (t) > 0 for all sufficiently large t.
Proof. Without loss of generality, let T = t0 . We claim that x(i−1) (tk ) > 0 for any
tk ≥ T . If it is not true, then there exists some tj ≥ T , such that x(i−1) (tj ) ≤ 0.
Since x(i) (t) ≤ 0, x(i−1) (t) is monotonically non-increasing in (tk , tk+1 ] for k ≥ j.
Also because x(i) (t) is not always equal to 0 in [t, +∞), there exists some tl ≥ tj
such that x(i) (t) is not always equal to 0 in (tl , tl+1 ]. Without loss of generality, we
can assume l = j, that is, x(i) (t) is not always equal to 0 in (tj , tj+1 ]. So we have
(i−1) (i−1)
x(i−1) (tj+1 ) < x(i−1) (t+
j ) ≤ aj
x
(tj ) ≤ 0.
For t ∈ (tj+1 , tj+2 ], we have
(i−1)
(i−1)
x(i−1) (tj+2 ) < x(i−1) (t+
(tj+1 ) < 0.
j+1 ) ≤ aj+1 x
4
C. ZHANG, W. FENG
EJDE-2006/18
By induction, for t ∈ (tj+m , tj+m+1 ], we have x(i−1) (t) < 0. So we have x(i−1) (t) <
0, x(i) (t) ≤ 0, t ∈ (tj+1 , +∞). By Lemma 2.3,for all sufficiently large t,we have
x(i−2) (t) < 0. Similarly, we can conclude, using Lemma 2.3 repeatedly, that for all
sufficiently large t, we have x(t) < 0. This is a contradiction to x(t) > 0 (t ≥ T ).
Hence, we have x(i−1) (tk ) > 0 for any tk ≥ T . So we have x(i−1) (t) > 0 for all
sufficiently large t. The proof of Lemma 2.4 is complete.
Lemma 2.5. Let x(t) be a solution of (2.1) and conditions (A), (B), (C) be satisfied. Suppose T ≥ t0 , x(t) > 0 for t ≥ T . Then there exist some T 0 ≥ T and
l ∈ {1, 3, . . . , 2n − 1} such that for t ≥ T 0 ,
x(i) (t) > 0,
i−1 (i)
(−1)
i = 0, 1, . . . , l;
x (t) > 0,
i = l + 1, . . . , 2n − 1;
(2.6)
x(2n) (t) ≤ 0.
Proof. Let T = t0 . Since x(t) > 0(t ≥ t0 ), by (2.1) and that p(t) is nonnegative
and is not always equal to 0 in any (t, +∞), we have
x(2n) (t) = −f (t, x(t)) ≤ −p(t)ϕ(x(t)) ≤ 0
and x(2n) (t) is not always equal to 0 in (t, +∞). By Lemma 2.4, we have x(2n−1) (t) >
0. Without loss of generality, let x(2n−1) (t) > 0 for t ≥ t0 . So x(2n−2) (t) > 0 is
monotonically nondecreasing on (tk , tk+1 ]. If for any tk , x(2n−2) (tk ) < 0, then
x(2n−2) (t) < 0(t ≥ t0 ). If there exists some tj such that x(2n−2) (tj ) ≥ 0, by that
(2n−2)
x(2n−2) (t) is monotonically increasing and ak
> 0,we get x(2n−2) (t) > 0 for
t > tj . So there exists some T1 ≥ T , such that one of the following statements hold
x(2n−1) (t) > 0,
(2n−1)
x
x(2n−2) (t) > 0,
(2n−2)
(t) > 0,
x
(t) < 0,
for t ≥ T1
(2.7)
for t ≥ T1
(2.8)
When (2.7) holds, Lemma 2.3 yields that x(2n−3) (t) > 0 for all sufficiently large t.
Using Lemma 2.3 repeatedly, for all sufficiently large t,we can conclude that
x(2n−2) (t) > 0, . . . , x0 (t) > 0,
x(2n−1) (t) > 0,
(2n−3)
When (2.8) holds, by Lemma 2.4, we have x
t. Hence,there exists some T2 ≥ T1 such that
x(t) > 0.
(t) > 0, for all sufficiently large
x(2n−3) (t) > 0,
x(2n−4) (t) > 0,
for t ≥ T2
(2.9)
x(2n−3) (t) > 0,
x(2n−4) (t) < 0,
for t ≥ T2
(2.10)
Repeating the discussion above, we can get, eventually, that there exist some T 0 ≥ T
and l ∈ {1, 3, . . . , 2n − 1}, such that for t ≥ T 0 ,
x(i) (t) > 0,
i−1 (i)
(−1)
x (t) > 0,
i = 0, 1, . . . , l;
i = l + 1, l + 2, . . . , 2n − 1;
x(2n) (t) ≤ 0.
The proof of Lemma 2.5 is complete.
We remark that if x(t) is an eventually negative solution of (2.1), then there are
conclusions similar to Lemma 2.4 and Lemma 2.5.
EJDE-2006/18
OSCILLATION FOR HIGHER ORDER ODES
5
(0)
Theorem 2.6. If conditions (A),(B),(C) hold, ak ≥ 1 and
Z
t1
p(t)dt +
Z
1
(2n−1)
(2n−1) (2n−1)
b2
t2
(2.11)
tm+1
Z
b1
t3
p(t)dt + . . .
(2n−1) (2n−1)
b2
b1
t1
1
+
Z
1
p(t)dt +
b1
t0
t2
p(t)dt + · · · = +∞
(2n−1)
. . . bm
tm
then every solution of (2.1) is oscillatory.
Proof. Let x(t) be a non-oscillatory solution of (2.1). Without loss of generality,
let x(t) > 0(t ≥ t0 ), By Lemma 2.5 and (2.1), there exists T 0 ≥ t0 such that, for
t ≥ T 0 , we have
x(2n) (t) ≤ 0,
x0 (t) > 0,
x(2n−1) (t) > 0,
x(t) > 0.
So x(2n−1) (t) is monotonically non-increasing on (tk , tk+1 ] and x(t) is monotonically
increasing on (tk , tk+1 ]. Let
u(t) =
x(2n−1) (t)
.
ϕ(x(t))
0
Then u(t+
k ) ≥ 0 (k = 1, 2, . . . ), u(t) ≥ 0 (t ≥ t0 ). Since ϕ (x) ≥ 0, for t 6= tk ,
u0 (t) = −
f (t, x(t)) h x(2n−1) (t)x0 (t) i 0
−
ϕ (x(t)) ≤ −p(t)
ϕ(x(t))
ϕ2 (x(t))
(2n−1)
u(t+
k)=
(2.12)
(2n−1) (2n−1)
b
x(2n−1) (tk )
x(2n−1) (t+
b
k)
≤ k
≤ k
+
(0)
ϕ(x(tk ))
ϕ(ak x(tk ))
x
(tk )
(2n−1)
≤ bk
u(tk )
ϕ(x(tk ))
(2.13)
Integrating (2.12) from t0 to t1 we have
u(t1 ) ≤ u(t+
0)−
Z
t1
p(t)dt ,
(2.14)
t0
u(t+
1)
≤
(2n−1)
u(t1 )
b1
≤
(2n−1)
b1
[u(t+
0)
Z
t1
−
p(t)dt] .
(2.15)
t0
Similar to the above inequality, we have
(2n−1)
u(t2 )
(2n−1)
[u(t+
1)−
u(t+
2 ) ≤ b2
≤ b2
Z
t2
p(t)dt]
t1
≤
(2n−1) (2n−1)
u(t+
b2
[b1
0)
−
(2n−1)
b1
Z
t1
p(t)dt −
t0
(2n−1) (2n−1)
b2
[u(t+
0)
≤ b1
Z
p(t)dt −
t0
p(t)dt]
t1
t1
−
t2
Z
1
(2n−1)
b1
Z
t2
p(t)dt]
t1
(2.16)
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C. ZHANG, W. FENG
EJDE-2006/18
By induction, for any natural number m, we have
Z t1
(2n−1) (2n−1)
+
(2n−1)
u(t+
)
≤
b
b
.
.
.
b
[u(t
p(t)dt −
)
−
m
m
1
2
0
1
(2n−1)
b1
t0
− ··· −
Z
1
(2n−1) (2n−1)
b1
b2
(2n−1)
. . . bm−2
(2n−1) (2n−1)
b1
b2
(2n−1) (2n−1)
. . . bm−2 bm−1
t2
p(t)dt
t1
tm−1
p(t)dt
(2.17)
tm−2
Z
1
−
Z
tm
p(t)dt]
tm−1
By (2.11) and (2.17), for all sufficiently large m, u(t+
m ) < 0. This contradicts
u(t+
m ) ≥ 0. So every solution of (2.1) is oscillatory. The proof of Theorem 2.6 is
complete.
(i)
(0)
(0)
Theorem 2.7. If conditions (A), (B), (C) hold, bk ≤ 1, ak ≥ 1, bk ≥ 1
R +∞ 2n−1
(i = 1, 2, . . . , 2n − 1, k = 1, 2, . . . ) and
t
p(t)dt = +∞, then every bounded
solution of (2.1) is oscillatory.
Proof. Let x(t) be a non-oscillatory solution of (2.1). Without loss of generality,
let x(t) > 0 for t ≥ t0 . By Lemma 2.5, we can divided (2.6) into two cases:
Case (i): If l = 1, then x(t) > 0, x0 (t) > 0, x00 (t) < 0, x000 (t) > 0, x(4) (t) < 0, . . . ,
x(2n−1) (t) > 0, x(2n) (t) ≤ 0.
Case (ii): If l ≥ 3, then x(t) > 0, x0 (t) > 0, x00 (t) > 0, x000 (t) > 0, . . . , x(l) (t) > 0,
x(l+1) (t) < 0, . . . , x(2n−1) (t) > 0, x(2n) (t) ≤ 0.
Both cases tells us that x0 (t) > 0, t ∈ (tk , tk+1 ], k = 1, 2, . . . . So x(t) is monoton(0)
ically increasing on (tk , tk+1 ]. Since ak ≥ 1, x(t) is monotonically increasing on
[t0 , +∞), that is, x(t) ≥ x(t0 ) for t ≥ t0 . By (2.1), we have
x(2n) (s) = −f (s, x(s)) ≤ −p(s)ϕ(x(t0 )) = −cp(s),
s ∈ (tk , tk+1 ]
(2.18)
where c = ϕ(x(t0 )) > 0. Multiplying (2.18) by s2n−1 and then integrating it from
tk to t, we have
Z t
Z t
s2n−1 x(2n) (s)ds < −c
s2n−1 p(s)ds, t ∈ (tk , tk+1 ] .
(2.19)
tk
tk
We will consider the following two cases:
(a) if the case (i) holds, then for t ∈ (tk , tk+1 ] we have,
Z t
s2n−1 x(2n) (s)ds
tk
Z
t
=
s2n−1 dx(2n−1) (s)
tk
= t2n−1 x(2n−1) (t) − tk2n−1 x(2n−1) (t+
k ) − (2n − 1)
Z
t
s2n−2 x(2n−1) (s)ds
tk
= ...
=
2n−1
X
i=0
(−1)i+1
2n−1
X
(2n − 1)! i (i) +
(2n − 1)! i (i)
t x (t) +
(−1)i
tk x (tk ).
i!
i!
i=0
EJDE-2006/18
OSCILLATION FOR HIGHER ORDER ODES
7
Especially, for any natural number k,
Z tk+1
s2n−1 x(2n) (s)ds
tk
=
2n−1
X
2n−1
X
(2n − 1)! i
(2n − 1)! i (i) +
tk+1 x(i) (tk+1 ) +
tk x (tk ).
(−1)i
i!
i!
i=0
(−1)i+1
i=0
No matter if i is odd or even, for i = 1, 2, . . . 2n − 1,
(i)
(i)
i
(i)
(−1)i (x(i) (t+
k ) − x (tk )) ≥ (−1) (bk − 1)x (tk ) ≥ 0.
For any natural number m and t ∈ (tm , tm+1 ], we have
Z t
s2n−1 x(2n) (s)ds
t1
Z
t2
s2n−1 x(2n) (s)ds +
=
Z
t1
t3
s2n−1 x(2n) (s)ds
t2
Z
tm
2n−1 (2n)
+ ··· +
s
x
2n−1
X
(−1)i+1
i=0
+
m 2n−1
X
X
tm
2n−1
X
(2n − 1)! i (i) +
(2n − 1)! i (i)
t x (t) +
(−1)i
t1 x (t1 )
i!
i!
i=0
(−1)i
k=2 i=0
(2n − 1)! i (i) +
tk (x (tk ) − x(i) (tk ))
i!
≥ −(2n − 1)!x(t) +
2n−1
X
(−1)i
i=0
+
m 2n−1
X
X
s2n−1 x(2n) (s)ds
(s)ds +
tm−1
=
t
Z
(−1)i
k=2 i=0
(2n − 1)! i (i) +
t1 x (t1 )
i!
(2n − 1)! i (i)
tk (bk − 1)x(i) (tk )
i!
≥ −(2n − 1)!x(t) +
2n−1
X
(−1)i
i=0
(2n − 1)! i (i) +
t1 x (t1 ).
i!
Combining the inequality above and (2.19), we have
−(2n − 1)!x(t) +
2n−1
X
i (2n
(−1)
i=0
− 1)! i (i) +
t1 x (t1 ) ≤ −c
i!
Z
t
s2n−1 p(s)ds.
t1
So x(t) → +∞, as t → +∞. This contradicts that x(t) is bounded.
(b) If the case (ii) holds, then x(t) is non-negative and strictly increasing on t ∈
[t1 , +∞). Hence, for any natural number m, we have
Z t
+
x(t) = x(tm ) +
x0 (s)ds, t ∈ (tm , tm+1 ],
tm
x(tm ) = x(t+
m−1 ) +
tm
Z
x0 (s)ds,
tm−1
...
x(t2 ) = x(t+
1)+
Z
t2
t1
x0 (s)ds
8
C. ZHANG, W. FENG
EJDE-2006/18
and
x(t) =
m
X
+
(x(t+
k ) − x(tk )) + x(t1 ) +
m−1
X Z tk+1
k=2
k=1
x0 (s)ds +
Z
tk
t
x0 (s)ds
(2.20)
tm
Since x00 (t) > 0, t ∈ (tk , tk+1 ], k ≥ 1, we can get
(1)
0
x0 (t) > x0 (t+
1 ) ≥ a1 x (t1 ),
0
0
x (t) > x
(t+
2)
≥
(1)
a2 x0 (t2 )
>
t ∈ (t1 , t2 ]
(1) (1)
a2 a1 x0 (t1 ),
t ∈ (t2 , t3 ] .
Applying induction, for any natural number k,
(1) (1)
(1)
0
x0 (t) > x0 (t+
k ) ≥ ak ak−1 . . . a1 x (t1 ),
t ∈ (tk , tk+1 ] .
(0)
Combining (2.20) and ak ≥ 1, we have
x(t) > x0 (t1 )
m−1
X
(1) (1)
(1)
ak ak−1 . . . a1 (tk+1 − tk ),
t ∈ (tm , tm+1 ]
k=1
(0)
From the condition (C) and bk ≥ 1, we have
+∞
X
(1) (1)
(1)
ak ak−1 . . . a1 (tk+1 − tk ) = +∞
k=1
Then x(t) → +∞ (t → +∞), which contradicts that x(t) is bounded. Therefore,
every solution of (2.1) is oscillatory. The proof of Theorem 2.7 is complete.
Qm (0)
Theorem 2.8. If conditions (A), (B), (C) hold, k=1 ak > b > 0 (m = 1, 2, . . . ),
(2n−1)
bk
≤ 1, and for any δ > 0,
Z
+∞
inf
f (t, x) dt = +∞
(2.21)
δ≤|x|<+∞
then every solution of (2.1) is oscillatory.
Proof. Let x(t) be a non-oscillatory solution of (2.1). Without loss of generality,
let x(t) > 0, t ≥ t0 . By Lemma 2.5, x0 (t) ≥ 0, t ≥ t0 . So x(t) is monotonically
nondecreasing on (t0 , +∞).
(0)
(0)
+
+
x(t1 ) ≥ x(t+
0 ), x(t2 ) ≥ x(t1 ) ≥ a1 x(t1 ) ≥ a1 x(t0 ),
(0)
(0) (0)
+
x(t3 ) ≥ x(t+
2 ) ≥ a2 x(t2 ) ≥ a2 a1 x(t0 )
By induction, we have
(0) (0)
+
+
(0)
(0)
x(tm+1 ) ≥ x(t+
m ) ≥ am x(tm ) ≥ · · · ≥ a1 a2 . . . am x(t0 ) > bx(t0 ).
We can assume that x(t) ≥ bx(t+
0 ), t ∈ (t0 , +∞). By (2.21), as t → +∞, we have
Z t
Z t
f (s, x(s))ds ≥
inf
f (s, x)ds → +∞ ;
+
t0
that is,
Rt
t0
t0 bx(t0 )≤|x|<+∞
f (s, x(s))ds → +∞. Integrating (2.1) from t0 to t1 , we have
Z t1
x(2n−1) (t1 ) +
f (s, x(s))ds = x(2n−1) (t+
0)
t0
EJDE-2006/18
OSCILLATION FOR HIGHER ORDER ODES
9
Similar to the above formula, for any natural number integrating (2.1) from tk−1
to tk , we have
Z tk
(2n−1)
x
(tk ) +
f (s, x(s))ds = x(2n−1) (t+
k−1 )
tk−1
So, we have
Z
x(2n−1) (t1 ) +
t1
f (s, x(s))ds = x(2n−1) (t+
0 ),
t0
Z t2
x(2n−1) (t2 ) +
f (s, x(s))ds = x(2n−1) (t+
1 ),
t1
...
x(2n−1) (tm ) +
Z
tm
f (s, x(s))ds = x(2n−1) (t+
m−1 ),
tm−1
t
x(2n−1) (t) +
Z
f (s, x(s))ds = x(2n−1) (t+
m) .
tm
For t ∈ (tm , tm+1 ], we have
Z t
m
m
X
X
(2n−1)
(2n−1)
x
(t) +
x
(ti ) +
f (s, x(s))ds =
x(2n−1) (t+
i ).
t0
i=1
i=0
Then
x(2n−1) (t) +
m
X
x(2n−1) (ti ) − x(2n−1) (t+
i ) +
Z
t
f (s, x(s))ds = x(2n−1) (t+
0 ).
t0
i=1
(2n−1)
Lemma 2.5 shows that x
(t) > 0 for sufficiently large t. Hence,
m
Z t
X
(2n−1) (2n−1)
x(2n−1) (t) ≤ −
(1−bk
)x
(ti ) −
f (s, x(s))ds+x(2n−1) (t+
0 ) (2.22)
t0
i=1
Rt
By condition
≤ 1 and (2.22), we have x(2n−1) (t) ≤ − t0 f (s, x(s))ds +
(2n−1)
x(2n−1) (t+
(t) < 0. This
0 ) → −∞ as t → +∞. So, for all sufficiently large t, x
(2n−1)
contradicts that x
(t) > 0. So every solution of (2.1) is oscillatory. The proof
of Theorem 2.8 is complete.
(2n−1)
bk
(0)
(2n−1)
Corollary 2.9. Assume the conditions (A), (B), (C) hold, and ak ≥ 1, bk
R +∞
1. If
p(t)dt = +∞, then every solution of (2.1) is oscillatory.
(2n−1)
≤ 1, we have
Z t2
Z t3
1
1
p(t)dt + (2n−1) (2n−1)
p(t)dt + . . .
p(t)dt + (2n−1)
t1
t2
t0
b1
b1
b2
Z tm+1
1
+ (2n−1) (2n−1)
p(t)dt
(2n−1)
tm
. . . bm
b1
b2
Z t1
Z t2
Z t3
Z tm+1
≥
p(t)dt +
p(t)dt +
p(t)dt + · · · +
p(t)dt
Proof. By bk
Z t1
t0
tm+1
t1
Z
=
p(t)dt
t0
t2
tm
≤
10
C. ZHANG, W. FENG
EJDE-2006/18
Rt
and t0m+1 p(t)dt → +∞ as m → +∞. Then (2.11) holds. By Theorem 2.6, every
solution of (2.1) is oscillatory.
Corollary 2.10. Assume conditions (A), (B), (C) hold, and that there exists a
R +∞ α
(0)
1
α
positive number α > 0, such that ak ≥ 1, (2n−1)
≥ ( tk+1
t p(t)dt =
tk ) . If
bk
+∞, then every solution of (2.1) is oscillatory.
α
≥ ( tk+1
tk ) , we have
1
Proof. By
(2n−1)
bk
t1
Z
p(t)dt +
+
(2n−1) (2n−1)
b2
Z t2
1
(2n−1)
b1
≥
=
(2n−1)
. . . bm
t1
(2n−1) (2n−1)
b2
b1
t1
p(t)dt +
Z
1
p(t)dt +
Z
1
+
≥
t2
1
b1
≥
(2n−1)
b1
t0
Z
1
t3
p(t)dt + . . .
t2
tm+1
p(t)dt
tm
Z
1
(2n−1) (2n−1)
b1
b2
Z tm+1
(2n−1) (2n−1)
(2n−1)
tm
b1
b2
. . . bm
Z t2
Z t3
1
[
tα
tα
2 p(t)dt +
3 p(t)dt
tα
t1
t2
1
Z t2
Z t3
1
α
[
t
p(t)dt
+
tα p(t)dt
tα
t1
t2
1
Z tm+1
1
tα p(t)dt
tα
1 t1
t3
p(t)dt + . . .
t2
p(t)dt
Z
tm+1
+ ··· +
tm
Z tm+1
+ ··· +
tα
m+1 p(t)dt]
tα p(t)dt]
tm
Rt
and t1m+1 p(t)dt → +∞ as m → +∞. Then (2.11) holds. By Theorem 2.6, we
every solution of (2.1) is oscillatory.
3. Examples
subsection*Example 3.1 Consider the equation
x(2n) (t) +
x(k + ) =
(0)
(0)
1 3
x = 0,
4t
t≥
1
, t 6= k, k = 1, 2, . . .
2
k+1
x(k), x(i) (k + ) = x(i) (k), i = 1, . . . , 2n − 1,
k
1
1
(i)
x( ) = x0 , x(i) ( ) = x0 ,
2
2
(i)
(i)
(3.1)
1
where ak = bk = k+1
> 1, ak = bk = 1, i = 1, 2, . . . , 2n − 1, p(t) = 4t
,
k
1 3
1
3
ϕ(x) = x , f (t, x) = 4t x , tk = k, t0 = 2 . It is obvious that the conditions (A) and
(i)
(i−1)
(B) are satisfied. For condition (C),we have: For i > 1, ak = bk
= 1,
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) + · · · + (tm+1 − tm ) + . . .
1
= + 1 + · · · + 1 + · · · = +∞.
2
EJDE-2006/18
OSCILLATION FOR HIGHER ORDER ODES
(1)
(0)
For i = 1, ak = 1, bk =
11
k+1
k ,
1
1
1
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) + · · · +
(tm+1 − tm ) + . . .
2
3
m+1
1 1 1
1
= + + + ··· +
+ · · · = +∞.
2 2 3
m+1
(2n−1)
Therefore, condition (C) holds. Since bk
= 1, we have
Z t1
Z t2
Z t3
1
1
p(t)dt + (2n−1)
p(t)dt + (2n−1) (2n−1)
p(t)dt + . . .
t0
t1
t2
b1
b1
b2
Z tm+1
1
+ (2n−1) (2n−1)
p(t)dt
(2n−1)
tm
b1
b2
. . . bm
Z tm+1
Z t3
Z t2
Z t1
p(t)dt
p(t)dt + · · · +
p(t)dt +
p(t)dt +
=
=
t2
t1
t0
tm+1
Z
Z
tm+1
p(t)dt =
t0
t0
tm
1
dt
4t
1
1
t
= ln t|t0m+1 = (ln tm+1 − ln t0 )
4
4
Since ln tm+1 → +∞ as m → +∞, we get that the condition of Theorem 2.6 hold.
So every solution of (3.1) is oscillatory.
Example 3.2. Consider the sub-linear system
1 1
1
x(2n) (t) + 2 x 3 = 0, t ≥ , t 6= k, k = 1, 2, . . . ,
t
2
k
x(k + ) = x(k), x(i) (k + ) =
x(i) (k), i = 1, . . . , 2n − 1 ,
k+1
1
1
(i)
x( ) = x0 , x(i) ( ) = x0 ,
2
2
(0)
(0)
(i)
(3.2)
(i)
k
where ak = bk = 1, ak = bk = k+1
, i = 1, 2, . . . , 2n − 1, p(t) = t12 , tk = k,
1
1
ϕ(x) = x 3 , f (t, x(t)) = t12 x 3 (t), t0 = 21 . It is obvious that the condition (A) and
(i)
(i−1)
k
(B) hold. For condition (C), we have: For i > 1 and ak = bk
= k+1
,
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) + · · · + (tm+1 − tm ) + · · · =
(1)
For i = 1 and ak =
(0)
k
k+1 , bk
1
+ 1 + · · · + 1 + · · · = +∞.
2
= 1,
1
1
1
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) + · · · +
(tm+1 − tm ) + . . .
2
3
m+1
1 1 1
1
= + + + ··· +
+ · · · = +∞.
2 2 3
m+1
So, condition (C) holds. Let α = 1. Then
Z
Z +∞
Z +∞
1
k+1
tk+1
k + 1 +∞
1
1
=
≥
=
tp(t)dt =
t 2 dt =
dt = +∞ .
(2n−1)
k
t
k
t
t
k
b
k
Therefore, the conditions of Corollary 2.10 are satisfied. Then every solution of
(3.2) is oscillatory.
12
C. ZHANG, W. FENG
EJDE-2006/18
Acknowledgements. The authors are grateful to the anonymous referee for his
(her) suggestions and comments on the original manuscript.
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Chaolong Zhang
Department of Computation Science,, Zhongkai University of Agriculture and Technology, Guangzhou, 510225, China
E-mail address: zhcl88@126.com
Weizhen Feng
School of Mathematical Sciences, South China Normal University, Guangzhou 510631,
China
E-mail address: wsy@scnu.edu.cn