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Existence for continuous nonoscillatory solutions of second-order nonlinear difference equations with continuous variable
Mathematical and Computer Modelling 46 (2007) 670–679 www.elsevier.com/locate/mcm
Existence for continuous nonoscillatory solutions of second-order nonlinear difference equations with continuous variableI Jiqin Deng School of Mathematics and Computational Science, Xiangtan University, Hunan 411105, PR China Received 14 November 2006; accepted 29 November 2006
Abstract In this paper, by using fixed point theory, under quite general condition on the nonlinear term, we obtain a existence result of continuous nonoscillatory solutions of second-order nonlinear difference equations with continuous variable. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Nonoscillatory solution; Nonlinear term; Fixed point theory; Continuous variable
1. Introduction In this paper, we study the existence of continuous nonoscillatory solutions on t of the second-order difference equation with continuous variable 42τ x(t) + f (t, x(t − σ )) = 0,
t ≥ t0
(1)
where τ, σ > 0, 4τ x(t) = x(t + τ ) − x(t), 42τ x(t) = 4τ (4τ x(t)) and f ∈ C([t0 , ∞) × R, R). A continuous solution of (1) on t is called nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is called oscillatory. For the discussion of oscillation of (1), the reader is referred to the papers [1–6]. To my knowledge, so far, very few existence results for continuous nonoscillatory solutions of (1) on t have been obtained since in this case the known methods in discussing the existence of solutions of differential equations or difference equations are not useful. Therefore, it is a significant work to investigate the existence of continuous nonoscillatory solutions of (1) on t under quite general conditions on the nonlinear term, f . This is also our aim in this paper. In Section 2, by using the Schauder–Tikhonov point theorem, we obtain a sufficient condition for the existence R t R fixed ∞ of positive global solutions of x(t) = κ + 1t t1 s g(h − ς (h), (h − σ )x(h − σ )) dh ds and its corollary. In Section 3, we apply the results of the second section to second-order nonlinear difference equation with continuous variable (1) and prove that there exists a continuous nonoscillatory solution to (1) on t under quite general conditions on the nonlinear term, f . This is our main result in this paper. I Supported by Research Fund of Hunan Provincial Education Department (No. 05C101).
E-mail address: [email protected]. c 2006 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2006.11.028
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
2. Lemmas In this section, we consider the existence of positive global solutions of the following equation Z Z 1 t ∞ x(t) = κ + g(h − ς (h), (h − σ )x(h − σ )) dh ds, t ≥ t1 t t1 s
(2)
where κ is a positive constant, g ∈ C([t0 , ∞) × R, R), ς (t) ∈ C([t1 , ∞), [−ς0 , ς0 ]) where ς0 is a positive constant and t1 is a constant with t1 > 1 + σ + τ + ς0 . Lemma 2.1. If there exist κ > 0 and 0 < α < 1 such that for each x(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ t ≥ t1 − σ , Z ∞ κ(1 − α) g(h − ς (h), (h − σ )x(h − σ )) dh ≤ for t ≥ t1 , tα t
κ tα
for
(3)
then Eq. (2) has a positive global solution xκ (t) with xκ (t) > 0 for t ≥ t1 and lim xκ (t) = κ.
(4)
t→∞
Proof of Lemma 2.1. Let X be the linear space of all continuous functions x ∈ C([t0 − σ, ∞)) such that sup |x(t)| < +∞
t≥t0 −σ
with norm kxk = sup |x(t)|. t≥t0 −σ
It follows that X is Banach space. Define a nonempty, bounded, convex and closed subset B of X and the operator T as follows: o n κ B = x ∈ X : |x(t) − κ| ≤ α for t ≥ t1 − σ. t and T x(t) = T x(t1 ), t0 − σ ≤ t < t1 , Z Z 1 t ∞ T x(t) = κ + g(h − ς (h), (h − σ )x(h − σ )) dh ds, t t1 s
t ≥ t1 .
Then, from (3), we have
Z tZ ∞
1
|T x(t) − κ| = g(h − ς (h), (h − σ )x(h − σ )) dh ds
t t1 s Z t 1 κ(1 − α) ≤ ds t t1 sα κ ≤ α , t ≥ t1 t which yields T u(∞) = κ.
(5)
(6)
(7)
Now, we divide the proof into several steps. First, from (5) and (6), it is obvious that T : B → B. Next, for any > 0, from (6), it is easy to see that there exists t2 ≥ t1 such that for any x(t), y(t) ∈ B,
(8)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
Z tZ ∞ 1 [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds t t1 s Z tZ ∞ Z tZ ∞ 1 1 ≤ g(h − ς (h), (h − σ )x(h − σ )) dh ds + g(h − ς (h), (h − σ )y(h − σ )) dh ds t t t1
s
t1
s
2κ ≤ , t ≥ t2 , tα i.e., for any x(t), y(t) ∈ B,
(9)
≤
|T x(t) − T y(t)| ≤ ,
t ≥ t2 .
(10)
On the other hand, from g ∈ C([t0 , ∞)×R, R), it is easy to see that there exists δ > 0 such that for any x(t), y(t) ∈ B with kx − yk ≤ δ, |g(t − ς (t), (t − σ )x(t − σ )) − g(t − ς (t), (t − σ )y(t − σ ))| ≤ , t1 ≤ t ≤ t2 t2 which, together with (3) and (9), yields Z tZ ∞ 1 |T x(t) − T y(t)| ≤ [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds t t s Z 1t Z ∞ 1 [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds ≤ t t1 t2 Z Z 1 t t2 |g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))| dh ds + t t1 s ≤ 2, t2 > t ≥ t1 which, together with (10), yields that for any x(t), y(t) ∈ B with kx − yk ≤ δ, |T x(t) − T y(t)| ≤ 2,
t ≥ t1 ,
i.e., T is continuous on B.
(11)
Next, let {xn }n≥1 ⊂ B. Then, from (3), we have |T xn (t3 ) − T xn (t4 )| Z t Z ∞ Z Z 3 1 1 t4 ∞ = g(h − ς (h), (h − σ )xn (h − σ )) dh ds − g(h − ς (h), (h − σ )xn (h − σ )) dh ds t3 t1 s t4 t1 s Z Z t4 − t3 t3 ∞ ≤ g(h − ς (h), (h − σ )xn (h − σ )) dh dm t3 t4 t s Z t 1Z ∞ 4 1 g(h − ς (h), (h − σ )xn (h − σ )) dh dm + t4 t3 s ≤ 2κ(t4 − t3 ), n ≥ 1, t4 ≥ t3 ≥ t1 which yields that {T xn }n≥1
is equicontinuous in X.
Next, by (6) and (7), we have Z tZ ∞ 1 κ |T xn (t) − T xn (∞)| = g(h − ς (h), (h − σ )x(h − σ )) dh ds ≤ α , t t1 s t which yields that for any > 0, there exists t ≥ t1 such that |T xn (t) − T xn (∞)| < ,
n ≥ 1, t ≥ t ,
(12)
n ≥ 1, t ≥ t1
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
i.e., {T xn }n≥1
is equiconvergent in X
which, together with (12), the assumption that {xn }n≥1 ∈ B and the Arzela-Ascoli theorem (See [7]) yields that {T xn }n≥1
is relatively compact.
(13)
Finally, from Schauder–Tikhonov fixed point theorem, (8), (11) and (13), it is easy to see that T has a fixed point xκ in B, i.e., Z Z 1 t ∞ xκ (t) = κ + g(h − ς (h), (h − σ )xκ (h − σ )) dh ds, t ≥ t1 . t t1 s Clearly, xκ (t) for t ≥ t1 is a positive global solution of (2) which satisfies xκ (t) > 0 for t ≥ t1 and (4). The proof is complete. Now we consider the equations 1 t
Z tZ
1 x(t) = κ + t
Z tZ
x(t) = κ +
t1
l(h − ς (h), (h − σ )x(h − σ )) dh ds −
2κ[t 1−α − t11−α ] , 3t
t ≥ t1
(14)
l(h − ς (h), (h − σ )x(h − σ )) dh ds +
2κ[t 1−α − t11−α ] , 3t
t ≥ t1
(15)
∞ s
and
t1
∞ s
where t1 , κ and ς are as in (2) and Lemma 2.1 and l ∈ C([t0 , ∞) × R, R). For them, we have the following result. Corollary 2.2. If there exist κ > 0 and 0 < α < 1 such that for each x(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ tκα for t ≥ t1 − σ , Z ∞ κ(1 − α) l(h − ς (h), (h − σ )x(h − σ )) dh ≤ for t ≥ t1 (16) 3t α t then Eqs. (14) and (15) has a positive global solution x1 (t) and x2 (t) with xi (t) > 0 (i = 1, 2) for t ≥ t1 , respectively and satisfying x1 (t) ≤ x2 (t),
(17)
t ≥ t1
and lim x1 (t) = lim x2 (t) = κ.
t→∞
(18)
t→∞
Proof of Corollary 2.2. In Lemma 2.1, take g(t − ς (t), (t − σ )x(t − σ )) = l(t − ς (t), (t − σ )x(t − σ )) −
2ακ(1 − α) 3t 1+α
and 2ακ(1 − α) , 3t 1+α respectively. Then, from (16) and Lemma 2.1, it is easy to see that Eqs. (14) and (15) has a positive global solution x1 (t) and x2 (t) with xi (t) > 0 (i = 1, 2) for t > t1 , respectively and satisfying g(t − ς (t), (t − σ )x(t − σ )) = l(t − ς (t), (t − σ )x(t − σ )) +
κ− and
κ[t 1−α − t11−α ] κ[t 1−α − t11−α ] ≤ x1 (t) ≤ κ − , t 3t
t ≥ t1
(19)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
κ[t 1−α − t11−α ] κ[t 1−α − t11−α ] ≥ x2 (t) ≥ κ + , t ≥ t1 t 3t which, together with (19), yields that (17) and (18) hold. The proof is complete. κ+
3. Main result and example In this section, we study the existence of continuous nonoscillatory solutions of the second-order difference equation with continuous variable (1). Our main result is as follows. Theorem 3.1. If there exist κ > 0, 0 < α < 1, ς (t) ∈ C([t1 , ∞), [−ς0 , ς0 ]) where ς0 is a positive constant, β ∈ R and t1 ≥ 1 + σ + τ + ς0 such that for any x(t), y(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ tκα and |y(t) − κ| ≤ tκα for t ≥ t1 − σ , Z ∞ κ(1 − α) for t ≥ t1 , (20) β f (h − ς (h), (h − σ )x(h − σ )) dh ≤ 3t α t 1 for some n ∈ N with (n+1)(t+τ ≥ τ2t(1+α) 2+α for t ≥ t1 , )1+α Z Z t+τ Z ∞ t+2τ Z ∞ β f (h − ς (h), (h − σ )x(h − σ )) dh ds − β f (h − ς (h), (h − σ )x(h − σ )) dh ds t+τ s t s καn(1 − α)τ 2 + f (t, (t − σ )x(t − σ )) ≤ , t ≥ t1 (21) 3(n + 1)t 1+α
and | f (t, (t − σ )[(1 − ξ2 )x(t − σ ) + ξ2 y(t − σ )]) − f (t, (t − σ )[(1 − ξ1 )x(t − σ ) + ξ1 y(t − σ )])| καn(ξ2 − ξ1 )(1 − α)τ 2 , t ≥ t1 3(n + 1)t 1+α where 1 ≥ ξ2 ≥ ξ1 ≥ 0, then Eq. (1) has a continuous nonoscillatory solution x(t) with
(22)
≤
x(t) = κ. (23) t Proof of Theorem 3.1. From (20), Lemma 2.1 and Corollary 2.2, it is easy to see that there exists x1 (t) ∈ B where B is as in the proof of Lemma 2.1 such that lim
t→∞
x1 (t) = κ,
t0 − σ ≤ t < t1 and Z Z 2κ[t 1−α − t11−α ] 1 t ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds − , x1 (t) = κ + t t1 s 3t
t ≥ t1
i.e., t+τ
Z t
s+τ
Z s
hx1 (h) dh ds =
t+τ
Z t
Z
s t+τ
Z
s+τ
κh dh ds
s+τ
Z
Z
h
∞
Z
β f (y − ς (y), (y − σ )x1 (y − σ )) dy dz dh ds
+ t
s t+τ
Z
Z
− t
t1
z s+τ 2κ[h 1−α
− t11−α ] dh ds 3
s
= κτ 2 t + κτ 3 Z t+τ Z s+τ Z + t
Z
s t+τ
t
t1 s+τ
Z
− s
h
∞
Z
β f (y − ς (y), (y − σ )x1 (y − σ )) dy dz dh ds z
2κτ 2 t11−α 2κh 1−α dh ds + , 3 3
t ≥ t1
J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
675
which, together with (21), yields Z t+2τ Z ∞ 42τ (t x1 (t)) = β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds t+τ
s
t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
− t
s
4κ(t + τ )1−α 2κt 1−α 2κ(t + 2τ )1−α + − − 3 3 3 Z t+2τ Z ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds = t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds # " 1−α 2κ 2κ 1−α τ τ 1−α 1−α − (t + τ ) −1 + 1+ t 1+ −1 3 t +τ 3 t Z t+2τ Z ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds = −
t
s
t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
− t
s
i ∞ X (1 − α)(−α) · · · (−α − i + 2) 2κ τ 1−α − (t + τ ) 3 i! t +τ i=1 ∞ X (1 − α)(−α) · · · (−α − i + 2) τ i 2κ + t 1−α 3 i! t i=1 Z t+2τ Z ∞ > β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds 1 1 κα(1 − α)τ 2 1 2κ(1 − α)τ 1 − − − + 3 tα (t + τ )α 3 t 1+α (t + τ )1+α Z t+2τ Z ∞ > β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds −
t
s
t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds +
− t t+2τ
Z
s ∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
≥ t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds +
− t
2κα(1 − α)τ 2 κα(1 − α 2 )τ 3 − 1+α 3(t + τ ) 3t 2+α
s
≥ − f (t, (t − σ )x1 (t − σ )) +
καn(1 − α)τ 2 , 3(n + 1)(t + τ )1+α
2καn(1 − α)τ 2 3(n + 1)(t + τ )1+α
t ≥ t1 ,
(24)
i.e., 42τ x11 (t) + f (t, x11 (t − σ )) > where x11 (t) = t x1 (t).
καn(1 − α)τ 2 > 0, 3(n + 1)(t + τ )1+α
t ≥ t1
(25)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
Similarly, there exists x2 (t) ∈ B such that 42τ x22 (t) + f (t, x22 (t − σ )) < −
καn(1 − α)τ 2 < 0, 3(n + 1)(t + τ )1+α
(26)
t ≥ t1
where x22 (t) = t x2 (t). Now, take s(t) = max{ξ : 1 ≥ ξ ≥ 0, 42τ [(1 − ξ )x11 (t) + ξ x22 (t)] + f (t, (1 − ξ )x11 (t − σ ) + ξ x22 (t − σ )) = 0.},
t ≥ t1 .
(27)
It follows that 42τ [(1 − s(t))x11 (t) + s(t)x22 (t)] + f (t, (1 − s(t))x11 (t − σ ) + s(t)x22 (t − σ )) = 0,
t ≥ t1
(28)
i.e., x(t) = (1 − s(t))x11 (t) + s(t)x22 (t) is a nonoscillatory solution of (1) and satisfies (23). We claim that s(t) ∈ C([t1 , ∞), [0, 1]), i.e., x(t) = (1 − s(t))x11 (t) + s(t)x22 (t) is a continuous nonoscillatory solution of (1). If there exist δ > 0, T ≥ t1 and {tm }∞ m=1 with tm ≥ t1 for m ≥ 1 and limm→∞ tm = T such that |s(tm ) − s(T )| ≥ δ,
m ≥ 1,
then, there exists {s(tm k )} ∈ {s(tm )} such that one of the following two inequalities holds: lim s(tm k ) = ξ0 ≥ s(T ) + δ
(29)
lim s(tm k ) = ξ0 ≤ s(T ) − δ.
(30)
k→∞
or k→∞
If (29) holds, then, 0 = lim {42τ [(1 − s(tm k ))x11 (tm k ) + s(tm k )x22 (tm k )] k→∞
+ f (tm k , (1 − s(tm k ))x11 (tm k − σ ) + s(tm k )x22 (tm k − σ ))} = 42τ [(1 − ξ0 )x11 (T ) + ξ0 x22 (T )] + f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) which, together with (27), yields a contradiction. If (30) holds, then, 0 = lim {42τ [(1 − s(tm k ))x11 (tm k ) + s(tm k )x22 (tm k )] k→∞
+ f (tm k , (1 − s(tm k ))x11 (tm k − σ ) + s(tm k )x22 (tm k − σ ))} = 42τ [(1 − ξ0 )x11 (T ) + ξ0 x22 (T )] + f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) which, together with (24)–(26), yields 42τ [(1 − s(T ))x11 (T ) + s(T )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) = 42τ [(ξ0 − s(T ))x11 (T ) + (s(T ) − ξ0 )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) − f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) καn(s(T ) − ξ0 )(1 − α)τ 2 ≤ (s(T ) − ξ0 ) 42τ [x22 (T ) − x11 (T )] + 3(n + 1)t 1+α ≤ (s(T ) − ξ0 )[(42τ x22 (T ) + f (T, x22 (T − σ ))) − (42τ x11 (T ) + f (T, x11 (T − σ )))] + <0
2καn(s(T ) − ξ0 )(1 − α)τ 2 3(n + 1)t 1+α
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
which yields a contradiction, since 42τ [(1 − s(T ))x11 (T ) + s(T )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) = 0. Therefore, s(t) ∈ C([t1 , ∞), [0, 1]). The proof is complete.
Example 3.2. Consider the following equation 421/2 x(t) +
sin t = 0, 300 + x(t − 3)
t ≥ 0.
(31)
Take τ=
1 , 2
σ = 3,
ς (t) =
1 − π, 2
β=
1 4 sin2
1 4
,
Then, for any x(t), y(t) ∈ C([0, ∞)) with |x(t) − 100| ≤ for π2 ≥ t ≥ 0, we have
t0 = 0, 100 √ t
t1 = 102 ,
and |y(t) − 100| ≤
κ = 100, 100 √ t
α=
1 , 2
n = 1.
for t ≥ t1 − 3, from sin t ≥
2t π
100θ2 (t) 100θ1 (t) and y(t) = 100 + √ , t ≥ t1 − 3 √ t t where maxt1 ≤t<∞ {|θ1 (t)|, |θ2 (t)|} ≤ 1, Z ∞ Z ∞ sin(π + h − 12 ) = β f (h − ς (h), (h − σ )x(h − σ )) dh dh 4 sin2 41 (300 + (h − 3)x(h − 3)) t t Z ∞ sin h − 12 dh = 1 (h−3) t 400 sin2 14 3 + (h − 3) 1 + θ√ h−3 Z ∞ sin h − 12 dh = t 400 sin2 14 1 + θ1 (h−3)(h−3) √ h h h−3 Z ∞ sin h − 1 i ! ∞ X 2 θ (h − 3)(h − 3) 1 1+ = (−1)i dh √ 2 1 h h−3 t 400h sin 4 i=1 Z ∞ sin h − 1 Z ∞ cos h − 21 cos h − 12 2 1 ≤ + − dh + dh 2 1 2 1 2 1 2 2 3/2 400h sin 4 400h sin 4 400h sin 4 360h sin2 14 t t Z ∞ cos t − 1 Z ∞ cos h − 1 2 2 1 + dh + dh ≤ 2 1 2 1 2 3/2 360h sin2 14 t 400t sin 4 t 400h sin 4 x(t) = 100 +
κ(1 − α) for t ≥ t1 , 3t α t+2τ Z ∞ β f (h − ς (h), (h − σ )x(h − σ )) dh ds
≤
Z t+τ
s
Z
t+τ
∞
Z
− t
s
Z t+2τ Z ∞ = s t+τ
β f (h − ς (h), (h − σ )x(h − σ )) dh ds + f (t, (t − σ )x(t − σ )) i ! ∞ sin π + h − 21 X i θ1 (h − 3)(h − 3) 1+ (−1) dh ds √ h h−3 400h sin2 14 i=1
(32)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
i ! θ (h − 3)(h − 3) 1 i − 1+ dh ds + f (t, (t − σ )x(t − σ )) (−1) √ 2 1 h h−3 400h sin 4 t s i=1 Z t+1 Z ∞ sin h − 1 cos h − 21 sin h − 12 cos h − 12 sin h − 12 2 dh ds + − − + = − 400h sin2 41 200h 2 sin2 41 200h 3 sin2 14 200h 2 sin2 41 200h 3 sin2 14 t+ 12 s 1 1 1 1 Z t+ 1 Z ∞ sin h − 1 cos h − sin h − cos h − sin h − 2 2 2 2 2 2 dh ds + + − − + 2 1 2 1 2 1 2 1 2 1 2 3 2 3 400h sin 4 200h sin 4 200h sin 4 200h sin 4 200h sin 4 t s Z t+1 Z ∞ sin h − 1 ∞ 2 X sin t θ1 (h − 3)(h − 3) i i + + dh ds (−1) √ 2 1 h h−3 t+ 12 s 400h sin 4 i=1 √ 300 + 100(t − 3) 1 + θ(t−3) Z
t+τ
Z
sin(π + h − 21 )
∞
∞ X
t
θ (h − 3)(h − 3) i 1 i − (−1) ( ) dh ds √ 2 1 h h−3 400h sin 4 i=1 t s Z t+1 Z t+1 cos h − 1 sin h − 12 2 dh ds ≤ − 200h 2 sin2 41 200h 3 sin2 41 t+ 12 s Z t+ 1 Z t+1 cos h − 1 sin h − 12 sin t 2 2 sin t dh ds + − − − 2 1 2 1 2 3 200h sin 4 200h sin 4 t s 100t 300 + 100(t − 3) 1 + Z t+1 Z t+1 sin h − 1 ∞ i 2 X 1 i θ1 (h − 3)(h − 3) + dh ds + (−1) √ h h−3 100t 2 sin2 41 t+ 21 s 400h sin2 14 i=1 i Z t+ 1 Z t+1 sin h − 1 X ∞ 2 2 i θ1 (h − 3)(h − 3) dh ds − (−1) √ 2 1 h h−3 400h sin 4 i=1 t s ! Z t+ 1 Z t+1 2 1 1 1 ≤ dh ds + + 3/2 2 1 2 1 2 3 90t sin sin 100h 100h t s 4 4 Z
+
≤
t+ 12
Z
∞
sin h −
1 100t 2 sin2 41
t+ 21
Z +
καn(1 − α)τ 2 , 3(n + 1)t 1+α
1 2
t+1
Z
t
∞ X
s
1 180h 3/2 sin2 14
θ√ 1 (t−3) t−3
dh ds
(33)
t ≥ t1
and | f (t, (t − σ )[(1 − ξ2 )x(t − σ ) + ξ2 y(t − σ )]) − f (t, (t − σ )[(1 − ξ1 )x(t − σ ) + ξ1 y(t − σ )])| sin t sin t = − 300 + (t − 3)[(1 − ξ2 )x(t − 3) + ξ2 y(t − 3)] 300 + (t − 3)[(1 − ξ1 )x(t − 3) + ξ1 y(t − 3)] 1 1 h i− h i = θ2 (t−3) θ2 (t−3) 300 + 100(t − 3) 1 + (1−ξ√2 )θ1 (t−3) + ξ2√ 300 + 100(t − 3) 1 + (1−ξ√1 )θ1 (t−3) + ξ1√ t−3
t−3
t−3
t−3
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
≤
<
(1−ξ2 )θ√ 1 (t−3)(t−3) t t−3 καn(ξ2 − ξ1 )(1 − α)τ 2 , t 3(n + 1)t 1+α
100t 2 1 +
√ 2(ξ2 − ξ1 ) t − 3 1 (t−3)(t−3) √ + ξ2 θ2 (t−3)(t−3) 1 + (1−ξ1 )θ√ + t t−3
t t−3
ξ1 θ2 (t−3)(t−3) √ t t−3
≥ t1
which, together with (32), (33), yields that (20) and (21) and (22) hold. Hence, from Theorem 3.1, it is easy to see that (31) has a nonoscillatory solution x(t) satisfying (23). References [1] [2] [3] [4] [5]
R.P. Agarwal, Y. Zhou, Oscillation of partial difference equations with continuous variables, Math. Comput. Modelling 31 (2–3) (2000) 17–29. R.P. Agarwal, Difference Equations and Inequalities: Second Edition, Revised and Expended, Marcel Dekker, New York, 2000, p. 980. Y. Zhang, J. Yan, Oscillation criteria for difference equations with continuous argument, Acta. Math. Sinica 38 (1995) 406–411. B.G. Zhang, J. Yan, S.K. Chou, Oscillation of difference equations with continuous variable, Comput. Math. Appl. 36 (1998) 11–18. Z.G. Zhang, P. Bi, J.F. Chen, Oscillation of second-order nonlinear difference equations with continuous variable, J. Math. Anal. Appl. 255 (2001) 349–357. [6] J. Deng, A note of oscillation of second-order nonlinear difference equations with continuous variable, J. Math. Anal. Appl. 280 (2003) 188–194. [7] J.B. Conway, A Course in Functional Analysis, Springer-Verlag, New York, 1990.
Existence for continuous nonoscillatory solutions of second-order nonlinear difference equations with continuous variableI Jiqin Deng School of Mathematics and Computational Science, Xiangtan University, Hunan 411105, PR China Received 14 November 2006; accepted 29 November 2006
Abstract In this paper, by using fixed point theory, under quite general condition on the nonlinear term, we obtain a existence result of continuous nonoscillatory solutions of second-order nonlinear difference equations with continuous variable. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Nonoscillatory solution; Nonlinear term; Fixed point theory; Continuous variable
1. Introduction In this paper, we study the existence of continuous nonoscillatory solutions on t of the second-order difference equation with continuous variable 42τ x(t) + f (t, x(t − σ )) = 0,
t ≥ t0
(1)
where τ, σ > 0, 4τ x(t) = x(t + τ ) − x(t), 42τ x(t) = 4τ (4τ x(t)) and f ∈ C([t0 , ∞) × R, R). A continuous solution of (1) on t is called nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is called oscillatory. For the discussion of oscillation of (1), the reader is referred to the papers [1–6]. To my knowledge, so far, very few existence results for continuous nonoscillatory solutions of (1) on t have been obtained since in this case the known methods in discussing the existence of solutions of differential equations or difference equations are not useful. Therefore, it is a significant work to investigate the existence of continuous nonoscillatory solutions of (1) on t under quite general conditions on the nonlinear term, f . This is also our aim in this paper. In Section 2, by using the Schauder–Tikhonov point theorem, we obtain a sufficient condition for the existence R t R fixed ∞ of positive global solutions of x(t) = κ + 1t t1 s g(h − ς (h), (h − σ )x(h − σ )) dh ds and its corollary. In Section 3, we apply the results of the second section to second-order nonlinear difference equation with continuous variable (1) and prove that there exists a continuous nonoscillatory solution to (1) on t under quite general conditions on the nonlinear term, f . This is our main result in this paper. I Supported by Research Fund of Hunan Provincial Education Department (No. 05C101).
E-mail address: [email protected]. c 2006 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2006.11.028
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
2. Lemmas In this section, we consider the existence of positive global solutions of the following equation Z Z 1 t ∞ x(t) = κ + g(h − ς (h), (h − σ )x(h − σ )) dh ds, t ≥ t1 t t1 s
(2)
where κ is a positive constant, g ∈ C([t0 , ∞) × R, R), ς (t) ∈ C([t1 , ∞), [−ς0 , ς0 ]) where ς0 is a positive constant and t1 is a constant with t1 > 1 + σ + τ + ς0 . Lemma 2.1. If there exist κ > 0 and 0 < α < 1 such that for each x(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ t ≥ t1 − σ , Z ∞ κ(1 − α) g(h − ς (h), (h − σ )x(h − σ )) dh ≤ for t ≥ t1 , tα t
κ tα
for
(3)
then Eq. (2) has a positive global solution xκ (t) with xκ (t) > 0 for t ≥ t1 and lim xκ (t) = κ.
(4)
t→∞
Proof of Lemma 2.1. Let X be the linear space of all continuous functions x ∈ C([t0 − σ, ∞)) such that sup |x(t)| < +∞
t≥t0 −σ
with norm kxk = sup |x(t)|. t≥t0 −σ
It follows that X is Banach space. Define a nonempty, bounded, convex and closed subset B of X and the operator T as follows: o n κ B = x ∈ X : |x(t) − κ| ≤ α for t ≥ t1 − σ. t and T x(t) = T x(t1 ), t0 − σ ≤ t < t1 , Z Z 1 t ∞ T x(t) = κ + g(h − ς (h), (h − σ )x(h − σ )) dh ds, t t1 s
t ≥ t1 .
Then, from (3), we have
Z tZ ∞
1
|T x(t) − κ| = g(h − ς (h), (h − σ )x(h − σ )) dh ds
t t1 s Z t 1 κ(1 − α) ≤ ds t t1 sα κ ≤ α , t ≥ t1 t which yields T u(∞) = κ.
(5)
(6)
(7)
Now, we divide the proof into several steps. First, from (5) and (6), it is obvious that T : B → B. Next, for any > 0, from (6), it is easy to see that there exists t2 ≥ t1 such that for any x(t), y(t) ∈ B,
(8)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
Z tZ ∞ 1 [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds t t1 s Z tZ ∞ Z tZ ∞ 1 1 ≤ g(h − ς (h), (h − σ )x(h − σ )) dh ds + g(h − ς (h), (h − σ )y(h − σ )) dh ds t t t1
s
t1
s
2κ ≤ , t ≥ t2 , tα i.e., for any x(t), y(t) ∈ B,
(9)
≤
|T x(t) − T y(t)| ≤ ,
t ≥ t2 .
(10)
On the other hand, from g ∈ C([t0 , ∞)×R, R), it is easy to see that there exists δ > 0 such that for any x(t), y(t) ∈ B with kx − yk ≤ δ, |g(t − ς (t), (t − σ )x(t − σ )) − g(t − ς (t), (t − σ )y(t − σ ))| ≤ , t1 ≤ t ≤ t2 t2 which, together with (3) and (9), yields Z tZ ∞ 1 |T x(t) − T y(t)| ≤ [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds t t s Z 1t Z ∞ 1 [g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))] dh ds ≤ t t1 t2 Z Z 1 t t2 |g(h − ς (h), (h − σ )x(h − σ )) − g(h − ς (h), (h − σ )y(h − σ ))| dh ds + t t1 s ≤ 2, t2 > t ≥ t1 which, together with (10), yields that for any x(t), y(t) ∈ B with kx − yk ≤ δ, |T x(t) − T y(t)| ≤ 2,
t ≥ t1 ,
i.e., T is continuous on B.
(11)
Next, let {xn }n≥1 ⊂ B. Then, from (3), we have |T xn (t3 ) − T xn (t4 )| Z t Z ∞ Z Z 3 1 1 t4 ∞ = g(h − ς (h), (h − σ )xn (h − σ )) dh ds − g(h − ς (h), (h − σ )xn (h − σ )) dh ds t3 t1 s t4 t1 s Z Z t4 − t3 t3 ∞ ≤ g(h − ς (h), (h − σ )xn (h − σ )) dh dm t3 t4 t s Z t 1Z ∞ 4 1 g(h − ς (h), (h − σ )xn (h − σ )) dh dm + t4 t3 s ≤ 2κ(t4 − t3 ), n ≥ 1, t4 ≥ t3 ≥ t1 which yields that {T xn }n≥1
is equicontinuous in X.
Next, by (6) and (7), we have Z tZ ∞ 1 κ |T xn (t) − T xn (∞)| = g(h − ς (h), (h − σ )x(h − σ )) dh ds ≤ α , t t1 s t which yields that for any > 0, there exists t ≥ t1 such that |T xn (t) − T xn (∞)| < ,
n ≥ 1, t ≥ t ,
(12)
n ≥ 1, t ≥ t1
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
i.e., {T xn }n≥1
is equiconvergent in X
which, together with (12), the assumption that {xn }n≥1 ∈ B and the Arzela-Ascoli theorem (See [7]) yields that {T xn }n≥1
is relatively compact.
(13)
Finally, from Schauder–Tikhonov fixed point theorem, (8), (11) and (13), it is easy to see that T has a fixed point xκ in B, i.e., Z Z 1 t ∞ xκ (t) = κ + g(h − ς (h), (h − σ )xκ (h − σ )) dh ds, t ≥ t1 . t t1 s Clearly, xκ (t) for t ≥ t1 is a positive global solution of (2) which satisfies xκ (t) > 0 for t ≥ t1 and (4). The proof is complete. Now we consider the equations 1 t
Z tZ
1 x(t) = κ + t
Z tZ
x(t) = κ +
t1
l(h − ς (h), (h − σ )x(h − σ )) dh ds −
2κ[t 1−α − t11−α ] , 3t
t ≥ t1
(14)
l(h − ς (h), (h − σ )x(h − σ )) dh ds +
2κ[t 1−α − t11−α ] , 3t
t ≥ t1
(15)
∞ s
and
t1
∞ s
where t1 , κ and ς are as in (2) and Lemma 2.1 and l ∈ C([t0 , ∞) × R, R). For them, we have the following result. Corollary 2.2. If there exist κ > 0 and 0 < α < 1 such that for each x(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ tκα for t ≥ t1 − σ , Z ∞ κ(1 − α) l(h − ς (h), (h − σ )x(h − σ )) dh ≤ for t ≥ t1 (16) 3t α t then Eqs. (14) and (15) has a positive global solution x1 (t) and x2 (t) with xi (t) > 0 (i = 1, 2) for t ≥ t1 , respectively and satisfying x1 (t) ≤ x2 (t),
(17)
t ≥ t1
and lim x1 (t) = lim x2 (t) = κ.
t→∞
(18)
t→∞
Proof of Corollary 2.2. In Lemma 2.1, take g(t − ς (t), (t − σ )x(t − σ )) = l(t − ς (t), (t − σ )x(t − σ )) −
2ακ(1 − α) 3t 1+α
and 2ακ(1 − α) , 3t 1+α respectively. Then, from (16) and Lemma 2.1, it is easy to see that Eqs. (14) and (15) has a positive global solution x1 (t) and x2 (t) with xi (t) > 0 (i = 1, 2) for t > t1 , respectively and satisfying g(t − ς (t), (t − σ )x(t − σ )) = l(t − ς (t), (t − σ )x(t − σ )) +
κ− and
κ[t 1−α − t11−α ] κ[t 1−α − t11−α ] ≤ x1 (t) ≤ κ − , t 3t
t ≥ t1
(19)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
κ[t 1−α − t11−α ] κ[t 1−α − t11−α ] ≥ x2 (t) ≥ κ + , t ≥ t1 t 3t which, together with (19), yields that (17) and (18) hold. The proof is complete. κ+
3. Main result and example In this section, we study the existence of continuous nonoscillatory solutions of the second-order difference equation with continuous variable (1). Our main result is as follows. Theorem 3.1. If there exist κ > 0, 0 < α < 1, ς (t) ∈ C([t1 , ∞), [−ς0 , ς0 ]) where ς0 is a positive constant, β ∈ R and t1 ≥ 1 + σ + τ + ς0 such that for any x(t), y(t) ∈ C([t0 − σ, ∞)) with |x(t) − κ| ≤ tκα and |y(t) − κ| ≤ tκα for t ≥ t1 − σ , Z ∞ κ(1 − α) for t ≥ t1 , (20) β f (h − ς (h), (h − σ )x(h − σ )) dh ≤ 3t α t 1 for some n ∈ N with (n+1)(t+τ ≥ τ2t(1+α) 2+α for t ≥ t1 , )1+α Z Z t+τ Z ∞ t+2τ Z ∞ β f (h − ς (h), (h − σ )x(h − σ )) dh ds − β f (h − ς (h), (h − σ )x(h − σ )) dh ds t+τ s t s καn(1 − α)τ 2 + f (t, (t − σ )x(t − σ )) ≤ , t ≥ t1 (21) 3(n + 1)t 1+α
and | f (t, (t − σ )[(1 − ξ2 )x(t − σ ) + ξ2 y(t − σ )]) − f (t, (t − σ )[(1 − ξ1 )x(t − σ ) + ξ1 y(t − σ )])| καn(ξ2 − ξ1 )(1 − α)τ 2 , t ≥ t1 3(n + 1)t 1+α where 1 ≥ ξ2 ≥ ξ1 ≥ 0, then Eq. (1) has a continuous nonoscillatory solution x(t) with
(22)
≤
x(t) = κ. (23) t Proof of Theorem 3.1. From (20), Lemma 2.1 and Corollary 2.2, it is easy to see that there exists x1 (t) ∈ B where B is as in the proof of Lemma 2.1 such that lim
t→∞
x1 (t) = κ,
t0 − σ ≤ t < t1 and Z Z 2κ[t 1−α − t11−α ] 1 t ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds − , x1 (t) = κ + t t1 s 3t
t ≥ t1
i.e., t+τ
Z t
s+τ
Z s
hx1 (h) dh ds =
t+τ
Z t
Z
s t+τ
Z
s+τ
κh dh ds
s+τ
Z
Z
h
∞
Z
β f (y − ς (y), (y − σ )x1 (y − σ )) dy dz dh ds
+ t
s t+τ
Z
Z
− t
t1
z s+τ 2κ[h 1−α
− t11−α ] dh ds 3
s
= κτ 2 t + κτ 3 Z t+τ Z s+τ Z + t
Z
s t+τ
t
t1 s+τ
Z
− s
h
∞
Z
β f (y − ς (y), (y − σ )x1 (y − σ )) dy dz dh ds z
2κτ 2 t11−α 2κh 1−α dh ds + , 3 3
t ≥ t1
J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
675
which, together with (21), yields Z t+2τ Z ∞ 42τ (t x1 (t)) = β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds t+τ
s
t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
− t
s
4κ(t + τ )1−α 2κt 1−α 2κ(t + 2τ )1−α + − − 3 3 3 Z t+2τ Z ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds = t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds # " 1−α 2κ 2κ 1−α τ τ 1−α 1−α − (t + τ ) −1 + 1+ t 1+ −1 3 t +τ 3 t Z t+2τ Z ∞ β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds = −
t
s
t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
− t
s
i ∞ X (1 − α)(−α) · · · (−α − i + 2) 2κ τ 1−α − (t + τ ) 3 i! t +τ i=1 ∞ X (1 − α)(−α) · · · (−α − i + 2) τ i 2κ + t 1−α 3 i! t i=1 Z t+2τ Z ∞ > β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds 1 1 κα(1 − α)τ 2 1 2κ(1 − α)τ 1 − − − + 3 tα (t + τ )α 3 t 1+α (t + τ )1+α Z t+2τ Z ∞ > β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds −
t
s
t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds +
− t t+2τ
Z
s ∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds
≥ t+τ
s t+τ
Z
∞
Z
β f (h − ς (h), (h − σ )x1 (h − σ )) dh ds +
− t
2κα(1 − α)τ 2 κα(1 − α 2 )τ 3 − 1+α 3(t + τ ) 3t 2+α
s
≥ − f (t, (t − σ )x1 (t − σ )) +
καn(1 − α)τ 2 , 3(n + 1)(t + τ )1+α
2καn(1 − α)τ 2 3(n + 1)(t + τ )1+α
t ≥ t1 ,
(24)
i.e., 42τ x11 (t) + f (t, x11 (t − σ )) > where x11 (t) = t x1 (t).
καn(1 − α)τ 2 > 0, 3(n + 1)(t + τ )1+α
t ≥ t1
(25)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
Similarly, there exists x2 (t) ∈ B such that 42τ x22 (t) + f (t, x22 (t − σ )) < −
καn(1 − α)τ 2 < 0, 3(n + 1)(t + τ )1+α
(26)
t ≥ t1
where x22 (t) = t x2 (t). Now, take s(t) = max{ξ : 1 ≥ ξ ≥ 0, 42τ [(1 − ξ )x11 (t) + ξ x22 (t)] + f (t, (1 − ξ )x11 (t − σ ) + ξ x22 (t − σ )) = 0.},
t ≥ t1 .
(27)
It follows that 42τ [(1 − s(t))x11 (t) + s(t)x22 (t)] + f (t, (1 − s(t))x11 (t − σ ) + s(t)x22 (t − σ )) = 0,
t ≥ t1
(28)
i.e., x(t) = (1 − s(t))x11 (t) + s(t)x22 (t) is a nonoscillatory solution of (1) and satisfies (23). We claim that s(t) ∈ C([t1 , ∞), [0, 1]), i.e., x(t) = (1 − s(t))x11 (t) + s(t)x22 (t) is a continuous nonoscillatory solution of (1). If there exist δ > 0, T ≥ t1 and {tm }∞ m=1 with tm ≥ t1 for m ≥ 1 and limm→∞ tm = T such that |s(tm ) − s(T )| ≥ δ,
m ≥ 1,
then, there exists {s(tm k )} ∈ {s(tm )} such that one of the following two inequalities holds: lim s(tm k ) = ξ0 ≥ s(T ) + δ
(29)
lim s(tm k ) = ξ0 ≤ s(T ) − δ.
(30)
k→∞
or k→∞
If (29) holds, then, 0 = lim {42τ [(1 − s(tm k ))x11 (tm k ) + s(tm k )x22 (tm k )] k→∞
+ f (tm k , (1 − s(tm k ))x11 (tm k − σ ) + s(tm k )x22 (tm k − σ ))} = 42τ [(1 − ξ0 )x11 (T ) + ξ0 x22 (T )] + f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) which, together with (27), yields a contradiction. If (30) holds, then, 0 = lim {42τ [(1 − s(tm k ))x11 (tm k ) + s(tm k )x22 (tm k )] k→∞
+ f (tm k , (1 − s(tm k ))x11 (tm k − σ ) + s(tm k )x22 (tm k − σ ))} = 42τ [(1 − ξ0 )x11 (T ) + ξ0 x22 (T )] + f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) which, together with (24)–(26), yields 42τ [(1 − s(T ))x11 (T ) + s(T )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) = 42τ [(ξ0 − s(T ))x11 (T ) + (s(T ) − ξ0 )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) − f (T, (1 − ξ0 )x11 (T − σ ) + ξ0 x22 (T − σ )) καn(s(T ) − ξ0 )(1 − α)τ 2 ≤ (s(T ) − ξ0 ) 42τ [x22 (T ) − x11 (T )] + 3(n + 1)t 1+α ≤ (s(T ) − ξ0 )[(42τ x22 (T ) + f (T, x22 (T − σ ))) − (42τ x11 (T ) + f (T, x11 (T − σ )))] + <0
2καn(s(T ) − ξ0 )(1 − α)τ 2 3(n + 1)t 1+α
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
which yields a contradiction, since 42τ [(1 − s(T ))x11 (T ) + s(T )x22 (T )] + f (T, (1 − s(T ))x11 (T − σ ) + s(T )x22 (T − σ )) = 0. Therefore, s(t) ∈ C([t1 , ∞), [0, 1]). The proof is complete.
Example 3.2. Consider the following equation 421/2 x(t) +
sin t = 0, 300 + x(t − 3)
t ≥ 0.
(31)
Take τ=
1 , 2
σ = 3,
ς (t) =
1 − π, 2
β=
1 4 sin2
1 4
,
Then, for any x(t), y(t) ∈ C([0, ∞)) with |x(t) − 100| ≤ for π2 ≥ t ≥ 0, we have
t0 = 0, 100 √ t
t1 = 102 ,
and |y(t) − 100| ≤
κ = 100, 100 √ t
α=
1 , 2
n = 1.
for t ≥ t1 − 3, from sin t ≥
2t π
100θ2 (t) 100θ1 (t) and y(t) = 100 + √ , t ≥ t1 − 3 √ t t where maxt1 ≤t<∞ {|θ1 (t)|, |θ2 (t)|} ≤ 1, Z ∞ Z ∞ sin(π + h − 12 ) = β f (h − ς (h), (h − σ )x(h − σ )) dh dh 4 sin2 41 (300 + (h − 3)x(h − 3)) t t Z ∞ sin h − 12 dh = 1 (h−3) t 400 sin2 14 3 + (h − 3) 1 + θ√ h−3 Z ∞ sin h − 12 dh = t 400 sin2 14 1 + θ1 (h−3)(h−3) √ h h h−3 Z ∞ sin h − 1 i ! ∞ X 2 θ (h − 3)(h − 3) 1 1+ = (−1)i dh √ 2 1 h h−3 t 400h sin 4 i=1 Z ∞ sin h − 1 Z ∞ cos h − 21 cos h − 12 2 1 ≤ + − dh + dh 2 1 2 1 2 1 2 2 3/2 400h sin 4 400h sin 4 400h sin 4 360h sin2 14 t t Z ∞ cos t − 1 Z ∞ cos h − 1 2 2 1 + dh + dh ≤ 2 1 2 1 2 3/2 360h sin2 14 t 400t sin 4 t 400h sin 4 x(t) = 100 +
κ(1 − α) for t ≥ t1 , 3t α t+2τ Z ∞ β f (h − ς (h), (h − σ )x(h − σ )) dh ds
≤
Z t+τ
s
Z
t+τ
∞
Z
− t
s
Z t+2τ Z ∞ = s t+τ
β f (h − ς (h), (h − σ )x(h − σ )) dh ds + f (t, (t − σ )x(t − σ )) i ! ∞ sin π + h − 21 X i θ1 (h − 3)(h − 3) 1+ (−1) dh ds √ h h−3 400h sin2 14 i=1
(32)
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J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
i ! θ (h − 3)(h − 3) 1 i − 1+ dh ds + f (t, (t − σ )x(t − σ )) (−1) √ 2 1 h h−3 400h sin 4 t s i=1 Z t+1 Z ∞ sin h − 1 cos h − 21 sin h − 12 cos h − 12 sin h − 12 2 dh ds + − − + = − 400h sin2 41 200h 2 sin2 41 200h 3 sin2 14 200h 2 sin2 41 200h 3 sin2 14 t+ 12 s 1 1 1 1 Z t+ 1 Z ∞ sin h − 1 cos h − sin h − cos h − sin h − 2 2 2 2 2 2 dh ds + + − − + 2 1 2 1 2 1 2 1 2 1 2 3 2 3 400h sin 4 200h sin 4 200h sin 4 200h sin 4 200h sin 4 t s Z t+1 Z ∞ sin h − 1 ∞ 2 X sin t θ1 (h − 3)(h − 3) i i + + dh ds (−1) √ 2 1 h h−3 t+ 12 s 400h sin 4 i=1 √ 300 + 100(t − 3) 1 + θ(t−3) Z
t+τ
Z
sin(π + h − 21 )
∞
∞ X
t
θ (h − 3)(h − 3) i 1 i − (−1) ( ) dh ds √ 2 1 h h−3 400h sin 4 i=1 t s Z t+1 Z t+1 cos h − 1 sin h − 12 2 dh ds ≤ − 200h 2 sin2 41 200h 3 sin2 41 t+ 12 s Z t+ 1 Z t+1 cos h − 1 sin h − 12 sin t 2 2 sin t dh ds + − − − 2 1 2 1 2 3 200h sin 4 200h sin 4 t s 100t 300 + 100(t − 3) 1 + Z t+1 Z t+1 sin h − 1 ∞ i 2 X 1 i θ1 (h − 3)(h − 3) + dh ds + (−1) √ h h−3 100t 2 sin2 41 t+ 21 s 400h sin2 14 i=1 i Z t+ 1 Z t+1 sin h − 1 X ∞ 2 2 i θ1 (h − 3)(h − 3) dh ds − (−1) √ 2 1 h h−3 400h sin 4 i=1 t s ! Z t+ 1 Z t+1 2 1 1 1 ≤ dh ds + + 3/2 2 1 2 1 2 3 90t sin sin 100h 100h t s 4 4 Z
+
≤
t+ 12
Z
∞
sin h −
1 100t 2 sin2 41
t+ 21
Z +
καn(1 − α)τ 2 , 3(n + 1)t 1+α
1 2
t+1
Z
t
∞ X
s
1 180h 3/2 sin2 14
θ√ 1 (t−3) t−3
dh ds
(33)
t ≥ t1
and | f (t, (t − σ )[(1 − ξ2 )x(t − σ ) + ξ2 y(t − σ )]) − f (t, (t − σ )[(1 − ξ1 )x(t − σ ) + ξ1 y(t − σ )])| sin t sin t = − 300 + (t − 3)[(1 − ξ2 )x(t − 3) + ξ2 y(t − 3)] 300 + (t − 3)[(1 − ξ1 )x(t − 3) + ξ1 y(t − 3)] 1 1 h i− h i = θ2 (t−3) θ2 (t−3) 300 + 100(t − 3) 1 + (1−ξ√2 )θ1 (t−3) + ξ2√ 300 + 100(t − 3) 1 + (1−ξ√1 )θ1 (t−3) + ξ1√ t−3
t−3
t−3
t−3
679
J. Deng / Mathematical and Computer Modelling 46 (2007) 670–679
≤
<
(1−ξ2 )θ√ 1 (t−3)(t−3) t t−3 καn(ξ2 − ξ1 )(1 − α)τ 2 , t 3(n + 1)t 1+α
100t 2 1 +
√ 2(ξ2 − ξ1 ) t − 3 1 (t−3)(t−3) √ + ξ2 θ2 (t−3)(t−3) 1 + (1−ξ1 )θ√ + t t−3
t t−3
ξ1 θ2 (t−3)(t−3) √ t t−3
≥ t1
which, together with (32), (33), yields that (20) and (21) and (22) hold. Hence, from Theorem 3.1, it is easy to see that (31) has a nonoscillatory solution x(t) satisfying (23). References [1] [2] [3] [4] [5]
R.P. Agarwal, Y. Zhou, Oscillation of partial difference equations with continuous variables, Math. Comput. Modelling 31 (2–3) (2000) 17–29. R.P. Agarwal, Difference Equations and Inequalities: Second Edition, Revised and Expended, Marcel Dekker, New York, 2000, p. 980. Y. Zhang, J. Yan, Oscillation criteria for difference equations with continuous argument, Acta. Math. Sinica 38 (1995) 406–411. B.G. Zhang, J. Yan, S.K. Chou, Oscillation of difference equations with continuous variable, Comput. Math. Appl. 36 (1998) 11–18. Z.G. Zhang, P. Bi, J.F. Chen, Oscillation of second-order nonlinear difference equations with continuous variable, J. Math. Anal. Appl. 255 (2001) 349–357. [6] J. Deng, A note of oscillation of second-order nonlinear difference equations with continuous variable, J. Math. Anal. Appl. 280 (2003) 188–194. [7] J.B. Conway, A Course in Functional Analysis, Springer-Verlag, New York, 1990.