Fifth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 10, 2003, pp. 79–88.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
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OSCILLATION OF THIRD ORDER FUNCTIONAL
DIFFERENTIAL EQUATIONS WITH DELAY
TUNCAY CANDAN & RAJBIR S. DAHIYA
Abstract. We consider third order functional differential equations with discrete and continuous delay. We then develop several theorems related to the
oscillatory behavior of these differential equations.
1. Introduction
Our goal in this paper is to study functional differential equations of the form
(b(t)(a(t)x0 (t))0 )0 +
m
X
qi (t)f (x(σi (t))) = h(t),
(1.1)
i=1
where a, b, h ∈ C([t0 , ∞), R), a(t), b(t) > 0, f : R → R continuous, σi (t) → ∞, as
t → ∞, i = 1, 2, . . . , m, and
Z d
0
0 0
(b(t)(a(t)x (t)) ) +
q(t, ξ)f (x(σ(t, ξ)))dξ = 0,
(1.2)
c
where a, b ∈ C([t0 , ∞), R), f ∈ C(R, R). The oscillations of solutions of third order
equations were studied by Rao and Dahiya [8], Tantawy [9], Waltman [10] and
Zafer and Dahiya [11]. The results in this paper for equation (1.1) are more general
comparing to Zafer and Dahiya [11]. The results for equation (1.2) are essentially
new.
As is customary, a solution of equations (1.1) and (1.2) is called oscillatory if it
has arbitrarily large zeros, otherwise it is called nonoscillatory. The solution of equations (1.1) and (1.2) is called almost oscillatory if it is oscillatory or lim x(i) (t) = 0,
t→∞
i = 0, 1, 2.
2. Main Results
Oscillatory behavior of third order differential equations with discrete
delay. Assume that xf (x) > 0, x 6= 0, qi (t) ≥ 0 is not identically zero in any half
1991 Mathematics Subject Classification. 34K11.
Key words and phrases. Oscillation theory.
c
2003
Southwest Texas State University.
Published February 28, 2003.
79
80
T. CANDAN & R. S. DAHIYA
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line of the form (τ, ∞) for some τ ≥ 0, i = 1, 2, . . . , m and σi (t) < t, σi0 (t) > 0,
i=1,2,. . . ,m, b0 (t) ≥ 0, and
Z ∞
Z ∞
dt
dt
= ∞,
= ∞.
(2.1)
b(t)
a(t)
Theorem 2.1. Let f (x) = x and h(t) = 0. Suppose that there exist a differentiable
function p ∈ C([t0 , ∞), R), p(t) > 0 such that
Z ∞h
i
b(t)(p0 (t))2
q(t)p(t) − Pm (σ (t)−T )
dt = ∞,
(2.2)
i
0
i=1 a(σi (t)) σi (t)4p(t)
where q(t) = min{q1 (t), q2 (t), . . . , qm (t)}, for every T ≥ 0, and that
Z r
iX
Z t hZ r
m
1
1
du
dv
qi (r)dr > 1,
σ(t)
σ(t) a(u)
u b(v)
i=1
(2.3)
where σ(t) = max{σ1 (t), σ2 (t), . . . , σm (t)}. Then the equation (1.1) is oscillatory.
Proof. Let x(t) be a non-oscillatory solution of (1.1). Assume x(t) is eventually
positive. Since σi (t) → ∞ as t → ∞ for i = 1, 2, . . . , m, there exist a t1 ≥ t0 such
that x(t) > 0 and x(σi (t)) > 0 for t ≥ t1 . From (1.1), we have
(b(t)(a(t)x0 (t))0 )0 = −
m
X
qi (t)x(σi (t)).
(2.4)
i=1
Since qi (t) is not negative and x(σi (t)) > 0 is positive for t ≥ t1 , the right-hand
side becomes non-positive. Therefore, we have
(b(t)(a(t)x0 (t))0 )0 ≤ 0
for t ≥ t1 . Thus, x(t), x0 (t), (a(t)x0 (t))0 are monotone and eventually one-signed.
Now we want to show that there is a t2 ≥ t1 such that for t ≥ t2
(a(t)x0 (t))0 > 0.
0
(2.5)
0
Suppose this is not true, then (a(t)x (t)) ≤ 0. Since qi (t), i = 1, 2, . . . , m are
not identically zero and b(t) > 0, it is clear that there is t3 ≥ t2 such that
b(t3 )(a(t3 )x0 (t3 ))0 < 0. Then, for t > t3 we have
b(t)(a(t)x0 (t))0 ≤ b(t3 )(a(t3 )x0 (t3 ))0 < 0.
Dividing (2.6) by b(t) and then integrating between t3 and t, we obtain
Z t
1
ds.
a(t)x0 (t) − a(t3 )x0 (t3 ) < b(t3 )(a(t3 )x(t3 )0 )0
b(s)
t3
(2.6)
(2.7)
Letting t → ∞ in (2.7), and because of (2.1) we see that a(t)x0 (t) → −∞ as t → ∞.
Thus there is a t4 ≥ t3 such that a(t4 )x0 (t4 ) < 0. Using (a(t)x0 (t))0 ≤ 0, we have
for t ≥ t4
a(t)x0 (t) ≤ a(t4 )x0 (t4 ).
(2.8)
If we divide (2.8) by a(t) and integrate from t4 to t with t → ∞, the right-hand
side becomes negative. Thus, we have x(t) → −∞. But this is a contradiction x(t)
being eventually positive and therefore it proves that (2.5) holds. Now we consider
two cases.
Suppose x0 (t) is eventually positive, say x0 (t) > 0 for t ≥ t2 . Define the function
z(t) by
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OSCILLATION OF THIRD ORDER FUNCTIONAL D. E.
81
b(t)(a(t)x0 (t))0
p(t).
z(t) = Pm
i=1 x(σi (t))
It is obvious that z(t) > 0 for t ≥ t2 and z 0 (t) is
Pm
Pm 0
x (σ (t))σi0 (t)
p0 (t)
0
i=1 qi (t)x(σi (t))
P
Pm i
z (t) = −
z(t) − i=1
p(t) +
z(t).
m
p(t)
i=1 x(σi (t))
i=1 x(σi (t))
Then
Pm 0
x (σ (t))σi0 (t)
p0 (t)
Pm i
z(t),
z (t) ≤ −q(t)p(t) +
z(t) − i=1
p(t)
i=1 x(σi (t))
where q(t) = min{q1 (t), q2 (t), . . . , qm (t)}. On the other hand, using
0
0 0
(b(t)(a(t)x (t)) ) ≤ 0, b0 (t) ≥ 0 and (2.5), we can find that
0
(a(t)x0 (t))00 ≤ 0.
(2.9)
(2.10)
Using (2.10) and the equality
Z
a(t)x0 (t) = a(T )x0 (T ) +
t
(a(s)x0 (s))0 ds,
(2.11)
T
we have
a(t)x0 (t) ≥ (t − T )(a(t)x0 (t))0
0
(2.12)
0
for T ≥ t2 . Since (a(t)x (t)) is non-increasing, we obtain
a(σi (t))x0 (σi (t)) ≥ (σi (t) − T )(a(t)x0 (t))0
f or
i = 1, 2, . . . , m.
(2.13)
Multiplying both sides of (2.13) by
σi0 (t)
a(σi (t))
and taking the summation from 1 to m, we have
m
X
σi0 (t)x0 (σi (t)) ≥
i=1
m
X
(σi (t) − T )
a(σi (t))
i=1
σi0 (t)(a(t)x0 (t))0 .
(2.14)
Then, using (2.14) in (2.9), it follows that
p0 (t)
z (t) ≤ −q(t)p(t) +
z(t) −
p(t)
Pm
0
i=1
(σi (t)−T ) 0
a(σi (t)) σi (t) 2
z (t),
b(t)p(t)
and completing the square will leads to
z 0 (t) ≤ −q(t)p(t) + Pm
i=1
b(t)(p0 )2 (t)
(σi (t)−T ) 0
a(σi (t)) σi (t)4p(t)
.
(2.15)
Integrating (2.15) between T and t and letting t → ∞, we see that lim z(t) = −∞.
t→∞
This contradicts z(t) being eventually positive.
If x0 (t) is eventually negative. We integrate (1.1) from t to ∞ and since
b(t)(a(t)x0 (t))0 > 0,
we have
0
0
Z
−b(t)(a(t)x (t)) +
t
m
∞X
i=1
qi (r)x(σi (r))dr ≤ 0.
(2.16)
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Now integrating (2.16) from t to ∞ after dividing by b(t) and using a(t)x0 (t) < 0,
will lead to
Z ∞Z r
m
X
1
a(t)x0 (t) +
du
qi (r)x(σi (r))dr ≤ 0.
(2.17)
t
t b(u)
i=1
Dividing (2.17) by a(t) and integrating again from t to ∞ gives
Z ∞hZ r
m
Z r 1
i X
1
du
dv
qi (r)x(σi (r))dr ≤ x(t).
u b(v)
t
t a(u)
i=1
(2.18)
Replacing t by σ(t) in (2.18), where σ(t) = max{σ1 (t), σ2 (t), . . . , σm (t)}, will give
Z t hZ r
m
Z r 1
i X
1
du
dv
qi (r)x(σi (r))dr ≤ x(σ(t)).
(2.19)
u b(v)
σ(t)
σ(t) a(u)
i=1
Using the fact that σi (t) < t and x(t) is decreasing in (2.19), we obtain
Z t hZ r
m
Z r 1
i X
1
du
dv
qi (r)dr ≤ 1.
σ(t)
σ(t) a(u)
u b(v)
i=1
This is a contradiction to (2.3). Therefore, the proof is complete
Example 2.2. Consider the following functional differential equation
(e−t x0 )00 +
2
X
(2i − 1)e−t x(t − (i + 1)π) = 0.
i=1
−t
Now a(t) = e , b(t) = 1, q1 (t) = e−t , q2 (t) = 3e−t σ1 (t) = t − 2π, σ2 (t) = t − 3π,
p(t) = et .
We can easily see that the conditions of Theorem 2.1 are satisfied. It is easy to
verify that x(t) = cost is a solution of this problem.
Theorem 2.3. Let f 0 (x) ≥ λ for some λ > 0, and h(t) = 0. Suppose that there
exist a differentiable function p ∈ C([t0 , ∞), R), p(t) > 0 such that
Z ∞h
i
b(t)(p0 (t))2
dt = ∞,
(2.20)
q(t)p(t) − Pm (σ (t)−T )
i
0
i=1 a(σi (t)) σi (t)4λp(t)
where q(t) = min{q1 (t), q2 (t), . . . , qm (t)}, for every T ≥ 0, and that
Z t hZ r
m
Z r 1
i X
1
lim sup
du
dv
qi (r)dr = ∞,
t→∞
σ(t)
σ(t) a(u)
u b(v)
i=1
(2.21)
where σ(t) = max{σ1 (t), σ2 (t), . . . , σm (t)}. Then the equation (1.1) is oscillatory.
Proof. The beginning part of the proof is similar to the proof of Theorem 2.1 until
we reach at two possible cases. Suppose x0 (t) is eventually positive. Then, we can
define
b(t)(a(t)x0 (t))0
z(t) = Pm
p(t) > 0.
i=1 f (x(σi (t)))
It is obvious that z(t) > 0 for t ≥ t2 and z 0 (t) is
Pm
0
i=1 qi (t)f (x(σi (t)))
z (t) = − P
p(t)
m
i=1 f (x(σi (t)))
Pm 0
f (x(σi (t)))x0 (σi (t))σi0 (t)
p0 (t)
+
z(t) − i=1 Pm
z(t).
p(t)
i=1 f (x(σi (t)))
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OSCILLATION OF THIRD ORDER FUNCTIONAL D. E.
83
Then
z 0 (t) ≤ −q(t)p(t) +
p0 (t)
z(t) −
p(t)
Pm
i=1
f 0 (x(σi (t))x0 (σi (t))σi0 (t)
Pm
z(t),
i=1 f (x(σi (t)))
(2.22)
where q(t) = min{q1 (t), q2 (t), . . . , qm (t)}. On the other hand, since
0
0 0
(b(t)(a(t)x (t)) ) ≤ 0, (2.5) holds and b0 (t) ≥ 0, we can obtain
(a(t)x0 (t))00 ≤ 0.
(2.23)
Using (2.23) and the equality
0
Z
0
t
a(t)x (t) = a(T )x (T ) +
(a(s)x0 (s))0 ds
(2.24)
T
will lead to
a(t)x0 (t) ≥ (t − T )(a(t)x0 (t))0 .
Now using non-increasing nature of (a(t)x0 (t))0 , we obtain
a(σi (t))x0 (σi (t)) ≥ (σi (t) − T )(a(t)x0 (t))0
f or
(2.25)
i = 1, 2, . . . , m.
(2.26)
Multiplying both sides of (2.26) by
σi0 (t)
a(σi (t))
and taking the summation from 1 to m, we have
m
m
X
X
0
(σi (t) − T ) 0
0
0
σi (t)(a(t)x0 (t)) .
σi (t)x (σi (t)) ≥
a(σ
(t))
i
i=1
i=1
(2.27)
Then, using (2.27) in (2.22), it follows that
p0 (t)
z (t) ≤ −q(t)p(t) +
z(t) − λ
p(t)
0
Pm
i=1
(σi (t)−T ) 0
a(σi (t)) σi (t) 2
z (t),
b(t)p(t)
and then completing the square leads to
b(t)(p0 (t))2
z 0 (t) ≤ −q(t)p(t) +
λ
Pm
i=1
(σi (t)−T ) 0
a(σi (t)) σi (t)4p(t)
.
(2.28)
Integrating (2.28) between T to t and letting t → ∞, we see that lim z(t) = −∞.
t→∞
This contradicts z(t) being eventually positive.
If x0 (t) is eventually negative and proceeding as in the proof of Theorem 2.1 we
will end up with
Z ∞ hZ r
m
Z r 1
i X
1
du
dv
qi (r)f (x(σi (r)))dr ≤ x(σ(t)),
σ(t)
σ(t) a(u)
u b(v)
i=1
where σ(t) = max{σ1 (t), σ1 (t), . . . , σn (t)}. Thus we have
Z t hZ r
m
Z r 1
i X
1
du
dv
qi (r)f (x(σi (r)))dr ≤ x(σ(t)).
σ(t)
σ(t) a(u)
u b(v)
i=1
(2.29)
Using the fact that σi (t) < t, f (x) is increasing and x(t) is decreasing in (2.29), we
obtain
Z t hZ r
m
Z r 1
i X
1
x(σ(t))
du
dv
qi (r)dr ≤
.
a(u)
b(v)
f
(x(σ(t)))
σ(t)
σ(t)
u
i=1
84
T. CANDAN & R. S. DAHIYA
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Since x(t) is decreasing and positive, it is approaching a finite non-negative number
as t → ∞. In view of (2.21) and the last equation, it is not possible that lim x(t) >
t→∞
0. Suppose lim x(t) = 0, then
t→∞
lim
t→∞
x(σ(t))
1
1
1
= lim
= 0
≤ .
f (x(σ(t))) t→∞ f 0 (x(σ(t)))
f (0)
λ
This is a contradiction to (2.21) . Therefore, the proof is complete.
Theorem 2.4. Suppose that f 0 (x) ≥ λ for some λ > 0 and
Z t hZ r
m
Z r 1
i X
1
lim sup
du
dv
qi (r)dr = ∞.
t→∞
σ(t)
σ(t) a(u)
u b(v)
i=1
(2.30)
In addition, suppose that there exist a continuously differentiable function p ∈
C([t0 , ∞), R), p(t) > 0 and an oscillatory function ψ(t) such that
Z ∞h
i
b(t)(p0 (t))2
dt = ∞
(2.31)
q(t)p(t) − Pm (σ (t)−T )
i
0
i=1 a(σi (t)) σi (t)4λdp(t)
for some d ∈ (0, 1) and for every T ≥ 0, and
(b(t)(a(t)ψ 0 (t))0 )0 = h(t),
lim ψ (i) (t) = 0,
t→∞
i = 0, 1, 2.
(2.32)
Then the equation (1.1) is almost oscillatory.
Proof. Let x(t) be a non-oscillatory solution of (1.1). Without loss of generality
we may assume that x(t) is eventually positive. Consider
y(t) = x(t) − ψ(t).
(2.33)
Obviously y(t) is eventually positive, otherwise, x(t) < ψ(t) and it is a contradiction
with oscillatory behavior of ψ(t). We know that,
(b(t)(a(t)y 0 (t))0 )0 ≤ 0.
(2.34)
Proceeding as in the proof of Theorem 2.1, there is a t1 ≥ 0 such that for t ≥ t1
(a(t)y 0 (t))0 > 0 and
(a(t)y 0 (t))00 ≤ 0.
Consider again two cases. Suppose that y 0 (t) is eventually positive, then y(t) is
increasing and eventually positive. On the other hand, since ψ(t) → 0 as t → ∞
and y(t) = x(t) − ψ(t), there exists a t2 ≥ t1 such that
x(σi (t)) ≥ dy(σi (t)) f or
t ≥ t2
and
d ∈ (0, 1),
i = 1, 2, . . . , m.
Since f is an increasing function, we obtain
f (x(σi (t))) ≥ f (dy(σi (t))) f or
t ≥ t2 ,
i = 1, 2, . . . , m.
Define z(t) by
b(t)(a(t)y 0 (t))0
p(t),
z(t) = Pm
i=1 f (dy(σi (t)))
then obviously z(t) > 0 for t ≥ t2 and z 0 (t) is
Pm
p0 (t)
0
i=1 qi (t)f (x(σi (t)))
z (t) = − P
p(t) +
z(t)
m
p(t)
i=1 f (dy(σi (t)))
Pm 0
f (dy(σi (t)))y 0 (σi (t))σi0 (t)
−d i=1 Pm
z(t).
i=1 f (dy(σi (t)))
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OSCILLATION OF THIRD ORDER FUNCTIONAL D. E.
85
Then, using f 0 (x) ≥ λ > 0, we obtain
z 0 (t) ≤ −q(t)p(t) +
Pm 0
y (σi (t))σi0 (t)
p0 (t)
z(t) − dλ Pi=1
z(t),
m
p(t)
i=1 f (dy(σi (t)))
(2.35)
where q(t) = min{q1 (t), q2 (t), . . . , qm (t)}. We can now show that
m
X
σi0 (t)y 0 (σi (t)) ≥
i=1
m
X
(σi (t) − T )
a(σi (t))
i=1
σi0 (t)(a(t)y 0 (t))0
(2.36)
as in proof of Theorem 2.1. Using (2.35) and (2.36), we have
Pm (σi (t)−T ) 0
p0 (t)
i=1 a(σi (t)) σi (t) 2
0
z(t) − dλ
z (t).
z (t) ≤ −q(t)p(t) +
p(t)
b(t)p(t)
Completing the square in the above equation leads to
z 0 (t) ≤ −q(t)p(t) + Pm
i=1
b(t)(p0 )2 (t)
(σi (t)−T ) 0
a(σi (t)) σi (t)4λdp(t)
.
(2.37)
Integrating (2.37) from T to t and letting t → ∞, we see that lim z(t) = −∞.
t→∞
This contradicts z(t) being eventually positive.
Now suppose y 0 (t) is eventually negative. Since y is eventually positive and
decreasing, lim y(t) = c, where c is a nonnegative number. Therefore, lim x(t) =
t→∞
t→∞
c. Integrating (1.1) three times as we did in the proof of Theorem 2.1, we will end
up with
Z ∞ hZ r
m
Z r 1
i X
1
du
dv
qi (r)f (x(σi (r)))dr ≤ y(t),
u b(v)
σ(t)
σ(t) a(u)
i=1
where σ(t) = max{σ1 (t), σ1 (t), . . . , σn (t)}. Thus we have
Z t hZ r
m
Z r 1
i X
1
du
dv
qi (r)f (x(σi (r)))dr ≤ y(t).
σ(t)
σ(t) a(u)
u b(v)
i=1
(2.38)
Hence, we conclude that lim inf x(t) = 0. But x(t) is monotone, so we have
t→∞
lim x(t) = 0. Thus c = 0 and by (2.32) and (2.33) lim x(i) (t) = 0, i = 0, 1, 2,
t→∞
t→∞
which means that x(t) is almost oscillatory. This completes the proof.
Oscillatory behavior of third order differential equations with continuous
deviating arguments. Suppose that the following conditions hold unless stated
otherwise
R ∞ dt
R ∞ dt
(a) a(t) > 0, b(t) > 0, b0 (t) ≥ 0,
a(t) = ∞,
b(t) = ∞,
(b) q(t, ξ) ∈ C([t0 , ∞) × [c, d], R), q(t, ξ) > 0,
(c) f (x)
x ≥ > 0, for x 6= 0, is a constant,
(d) σ(t, ξ) ∈ C([t0 , ∞) × [c, d], R), σ(t, ξ) < t, ξ ∈ [c, d], σ(t, ξ) is nondecreasing
with respect to t and ξ and
lim min σ(t, ξ) = ∞.
t→∞ ξ∈[c,d]
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T. CANDAN & R. S. DAHIYA
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Theorem 2.5. If
Z
∞
d
Z
q(s, ξ)dξds = ∞
t1
and
Z
t
g(t)
hZ
r
g(t)
(2.39)
c
1
du
a(u)
Z
r
u
i
1
dv
b(v)
d
Z
q(r, ξ)dξdr > 1,
(2.40)
c
where g(t) = σ(t, d). Then the equation (1.2) is oscillatory.
Proof. Suppose that x(t) is non-oscillatory solution of (1.2). Without loss of
generality we may assume that x(t) is eventually positive. (If x(t) is eventually
negative solution, it can be proved by the same arguments). From (1.2), we have
Z d
0
0 0
(b(t)(a(t)x (t)) ) = −
q(t, ξ)f (x(σ(t, ξ)))dξ.
(2.41)
c
Proceeding as in the proof of Theorem 2.1, we have
(b(t)(a(t)x0 (t))0 )0 ≤ 0,
(a(t)x0 (t))0 > 0 and (a(t)x0 (t))00 ≤ 0
for large enough t. Thus, x(t), x0 (t) and (a(t)x0 (t))0 are monotone and eventually
one-signed. From condition (c),
f (x(σ(t, ξ))) ≥ x(σ(t, ξ)) > 0.
Therefore,
0
0 0
Z
0 ≥ (b(t)(a(t)x (t)) ) + d
q(t, ξ)x(σ(t, ξ))dξ.
(2.42)
c
Now consider again two cases.
Suppose that x0 (t) is eventually positive, say x0 (t) > 0 for t > t2 . Now we can
choose a constant k > 0 such that x(k) > 0. By (d), there exist a sufficiently large
T such that σ(t, ξ) > k for t > T , ξ ∈ [c, d]. Therefore,
x(σ(t, ξ)) ≥ x(k).
Thus,
(b(t)(a(t)x0 (t))0 )0 + x(k)
Z
d
q(t, ξ)dξ ≤ 0.
(2.43)
c
Integrating this last equation from t1 to t, we get
b(t)(a(t)x0 (t))0 ≤ b(t1 )(a(t1 )x0 (t1 ))0 − x(k)
Z tZ
d
q(s, ξ)dξds.
t1
(2.44)
c
Taking the limit of both sides as t → ∞ and using (2.39), the last inequality
above leads to a contradiction to (a(t)x0 (t))0 > 0. Now suppose x0 (t) is eventually
negative. Proceeding as in the proof of Theorem 2.1 and integrating equation (1.2)
three times, we get
Z ∞hZ r
Z r 1
i Z d
1
du
dv
q(r, ξ)f (x(σ(r, ξ)))dξdr ≤ x(t)
(2.45)
u b(v)
c
t
t a(u)
Using (c) in (2.45), we obtain
Z ∞hZ r
Z r 1
i Z d
1
du
dv q(r, ξ)x(σ(r, ξ))dξdr ≤ x(t).
t
t a(u)
u b(v)
c
(2.46)
EJDE/CONF/10
OSCILLATION OF THIRD ORDER FUNCTIONAL D. E.
Replace t by g(t) in (2.46), where g(t) = σ(t, d), then we have
Z t hZ r
Z r 1
i Z d
1
du
dv
q(r, ξ)x(σ(r, ξ))dξdr ≤ x(g(t)).
g(t)
g(t) a(u)
u b(v)
c
87
(2.47)
Since x(t) is decreasing and positive,
Z t hZ r
Z r 1
i Z d
1
du
dv
q(r, ξ)dξdr ≤ 1.
g(t)
g(t) a(u)
u b(v)
c
This is a contradiction to (2.40). Therefore, the proof is complete.
Example 2.6. Consider the following functional differential equation
Z 1/2π −1/ξ
2e
1
x(t − )dξ = 0
x000 +
2
ξ
ξ
2/7π
−1/ξ
so that a(t) = 1, b(t) = 1, f (x) = x, q(t, ξ) = 2e ξ2 , σ(t, ξ) = t − 1ξ . We can
easily see that the conditions of Theorem 2.5 are satisfied. It is easy to verify that
x(t) = e−t sin t is a solution of this problem.
Theorem 2.7. Suppose (2.40) holds. In addition to that suppose there exist p ∈
C([t0 , ∞), R), p(t) > 0 such that
Z ∞h
a(σ(t, c))b(t)(p0 (t))2 i
Γ(t)p(t) −
dt = ∞,
(2.48)
(σ(t, c) − T )σ 0 (t, c)4p(t)
Rd
where Γ(t) = c q(t, ξ)dξ. Then the equation (1.2) is oscillatory.
Proof. Suppose that x(t) is non-oscillatory solution of (1.2). We can assume that
x(t) is eventually positive. The case of x(t) is eventually negative can be proved by
the same arguments. Proceeding as in the proof of Theorem 2.1, we have
(b(t)(a(t)x0 (t))0 )0 ≤ 0,
(a(t)x0 (t))0 > 0 and (a(t)x0 (t))00 ≤ 0.
Thus, x(t), x0 (t) and (a(t)x0 (t))0 are monotone and eventually one-signed. From
condition (c),
f (x(σ(t, ξ))) ≥ x(σ(t, ξ)) > 0.
Z d
0
0 0
(b(t)(a(t)x (t)) ) + q(t, ξ)x(σ(t, ξ))dξ ≤ 0.
(2.49)
c
If x0 (t) is eventually positive, then we can define
z(t) =
b(t)(a(t)x0 (t))0
p(t).
x(σ(t, c))
It is obvious that z(t) > 0 for t ≥ t2 and z 0 (t) is
z 0 (t) =
(b(t)(a(t)x0 (t))0 )0
p0 (t)
x0 (σ(t, c))σ 0 (t, c)
p(t) +
z(t) −
z(t).
x(σ(t, c))
p(t)
x(σ(t, c))
From proof of Theorem 2.1, we have
a(t)x0 (t) ≥ (t − T )(a(t)x0 (t))0 .
Since (a(t)x(t))0 is non-increasing, we have
a(σ(t, c))x0 (σ(t, c)) ≥ (σ(t, c) − T )(a(t)x0 (t))0 ,
(2.50)
88
T. CANDAN & R. S. DAHIYA
EJDE/CONF/10
then
(σ(t, c) − T )(a(t)x0 (t))0
.
a(σ(t, c))
Plug (2.51) in (2.50), then we obtain
x0 (σ(t, c)) ≥
(2.51)
(b(t)(a(t)x0 (t))0 )0
p0 (t)
(σ(t, c) − T )σ 0 (t, c) 2
p(t) +
z(t) −
z (t).
x(σ(t, c))
p(t)
p(t)b(t)a(σ(t, c)))
Completing the square leads to
z 0 (t) =
z 0 (t) ≤ −Γ(t)p(t) +
b(t)a(σ(t, c))(p0 (t))2
.
(σ(t, c) − T )σ 0 (t, c))4p(t)
(2.52)
Integrating (2.52) from T to t and letting t → ∞, we see that lim z(t) = −∞.
t→∞
This contradicts z(t) being eventually positive.
If x0 (t) is eventually negative, the proof is exactly the same as in the second part
of the proof of previous Theorem.
References
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[3] T. Candan and R. S. Dahiya, Oscillatory behavior of n−th order nonlinear neutral differential
equations with forcing terms, Math. J. Toyama Univ. 24, (2001), 9-19.
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[5] I. Györi and G. Ladas, Oscillation Theory of Delay Differential Equations with Applications,
Clarendon Press, Oxford (1991).
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Mat. Sb. 65 (1965), 172-187.
[7] G. S. Ladde, V. Lakshmikantham and B. G. Zhang, Oscillation Theory of Differential Equations with Deviating Arguments, Marcel Dekker, Inc. (1987).
[8] V. Sree Hari Rao and R. S. Dahiya, Properties of solutions of a class of third-order linear
differential equations, Period. Math. Hungarica 20(3) (1989), 177-184.
[9] A. S. Tantawy, “Oscillation of Third and Higher Order Functional Differential Equations,
Ph.D Thesis, ISU (1989).
[10] P. Waltman, Oscillation criteria for third-order nonlinear differential equations, Pasific J.
Math. 18 (1966), 385-389.
[11] A. Zafer and R. S. Dahiya, Oscillatory and Asymptotic Behavior of Solutions of Third Order
Delay Differential Equations With Forcing Terms, Differential Equations and Dynamical
Systems. 1(2), (1993) 123-136.
Tuncay Candan
Department of Mathematics, Iowa State University
Ames, IA 50011, USA
E-mail address: tuncayca@iastate.edu
Rajbir S. Dahiya
Department of Mathematics, Iowa State University
Ames, IA 50011, USA
E-mail address: dahiya@math.iastate.edu