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A oscillation theorem for a class of even order neutral differential equations
J. Math. Anal. Appl. 273 (2002) 172–189 www.academicpress.com
A oscillation theorem for a class of even order neutral differential equations Satoshi Tanaka Department of Electrical Engineering, Hachinohe National College of Technology, Hachinohe 039-1192, Japan Received 8 May 2001 Submitted by R.P. Agarwal
Abstract The even order neutral differential equation dn x(t) + h(t)x(t − τ ) + f t, x g(t) = 0 (1) dt n is considered under the following conditions: n 2 is even; τ > 0; h ∈ C(R); g ∈ C[t0 , ∞), limt →∞ g(t) = ∞; f ∈ C([t0 , ∞) × R), uf (t, u) 0 for (t, u) ∈ [t0 , ∞) × R, and f (t, u) is nondecreasing in u ∈ R for each fixed t t0 . It is shown that (1) is oscillatory if and only if the certain non-neutral differential equation is oscillatory, for the case where 0 µ h(t) λ < 1 or 1 < λ h(t) µ. 2002 Elsevier Science (USA). All rights reserved.
1. Introduction and main results We shall be concerned with the oscillatory behavior of solutions of the even order neutral differential equation dn x(t) + h(t)x(t − τ ) + f t, x g(t) = 0. (1) dt n Throughout this paper, the following conditions are assumed to hold: n 2 is even; τ > 0; h ∈ C(R); g ∈ C[t0 , ∞), limt →∞ g(t) = ∞; f ∈ C([t0 , ∞) × R), E-mail address: [email protected]. 0022-247X/02/$ – see front matter 2002 Elsevier Science (USA). All rights reserved. PII: S 0 0 2 2 - 2 4 7 X ( 0 2 ) 0 0 2 3 5 - 4
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uf (t, u) 0 for (t, u) ∈ [t0 , ∞) × R, and f (t, u) is nondecreasing in u ∈ R for each fixed t t0 . By a solution of (1), we mean a function x(t) that is continuous and satisfies (1) on [tx , ∞) for some tx t0 . A solution is said to be oscillatory if it has arbitrarily large zeros; otherwise it is said to be nonoscillatory. Eq. (1) is said to be oscillatory if every solution of (1) is oscillatory. Neutral differential equations find numerous applications in natural science and technology. For instance, they are frequently used for the study of distributed networks containing lossless transmission lines. See Hale [10]. Oscillation properties of even order neutral differential equations have been investigated by many authors. We refer the reader to [1–9,11,12,18–24]. For odd order neutral differential equations, it has been shown by Zhang and Yang [24] that the equation γ −1 dN x(t − σ ) = 0 x(t) − x(t − τ ) + p(t)x(t − σ ) N dt is oscillatory if and only the ordinary differential equation γ −1 x (N+1) (t) + τ −1 p(t)x(t) x(t) = 0 is oscillatory, where N 1 is odd, γ > 0, σ ∈ R, p ∈ C[t0 , ∞), p(t) 0 for t t0 . (See also Tang and Shen [23].) For even order neutral differential equations, recently, the following result has been established in [22]. Theorem 1.0. Let c > 0. Then the even order neutral differential equation dn x(t) + cx(t − τ ) + f t, x g(t) = 0 n dt is oscillatory if and only if the even order non-neutral differential equation 1 f t, x g(t) = 0 x (n) (t) + 1+c is oscillatory.
(2)
The purpose of this paper is to generalize Theorem 1.0 with c = 1 for (1) with the following cases (H1) and (H2): (H1) 0 µ h(t) λ < 1 for t ∈ R; (H2) 1 < λ h(t) µ for t ∈ R. Here, µ and λ are constants. It is convenience only that the parts of µ and λ in (H1) and (H2) are opposite each other. Throughout this paper we use the notation: Hi (t) = h(t)h(t − τ ) · · · h t − (i − 1)τ , i = 1, 2, . . . . H0 (t) = 1;
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We define the function S(t) on R by ∞ i=0 (−1)i Hi (t) if (H1) holds, S(t) = ∞ (−1)i+1 if (H2) holds, i=1 Hi (t +iτ )
for t ∈ R.
It is easy to see that S(t) is converges uniformly on R, and hence S(t) is continuous on R. In Section 2 we will show that 1−µ 1−λ S(t) , t ∈ R. (3) 0< 1 − µ2 1 − λ2 We note that if µ = λ = c = 1, then 1−λ 1 1−µ 1 , and S(t) = . = = 2 2 1+c 1+c 1−µ 1−λ Main result of this paper is the following theorem. Theorem 1.1. Suppose that (H1) or (H2) holds. Then (1) is oscillatory if and only if x (n) (t) + f t, S g(t) x g(t) = 0 (4) is oscillatory. Theorem 1.1 means that (1) has a nonoscillatory solution if and only if (4) has a nonoscillatory solution. Theorem 1.1 is a generalization of Theorem 1.0 with c = 1. Indeed, for the case where h(t) ≡ c > 0 and c = 1, we see that S(t) = (1 + c)−1 and that (2) is oscillatory if and only if y (n) (t) + f t, (1 + c)−1 y g(t) = 0 is oscillatory. (Put x(t) = (1 + c)y(t).) Now we assume that h(t + τ ) = h(t),
h(t) = 1 and h(t) 0
for t ∈ R.
Then it is easy to verify that (H1) or (H2) holds, and S(t) = sequently, from Theorem 1.1, we have the following result.
[1 + h(t)]−1 .
(5) Con-
Corollary 1.1. Suppose that (5) holds. Then (1) is oscillatory if and only if
x(g(t)) x (n) (t) + f t, =0 1 + h(g(t)) is oscillatory. The oscillatory behavior of solutions of non-neutral differential equations of the form (6) x (n) (t) + f t, x g(t) = 0
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has been intensively studied in the last three decades. We refer the reader to [3, 9,13–17,19] and the references cited therein. Combining Theorem 1.1 with the known oscillation results for non-neutral differential equations of the form (6), we can derive various oscillation results for neutral differential equations of the form (1). In Section 2 we obtain oscillation criteria for the linear neutral differential equation dn x(t) + h(t)x(t − τ ) + p(t)x(t − σ ) = 0, (7) dt n and for the nonlinear neutral differential equation γ −1 dn x(t) + h(t)x(t − τ ) + p(t)x(t − σ ) x(t − σ ) = 0, (8) n dt where γ > 0, γ = 1 and the following conditions are assumed to hold: σ ∈ R;
p ∈ C[t0 , ∞),
p(t) > 0
for t t0 .
(9)
It is possible to obtain oscillation results for more general equations such as (1). However, for simplicity, we have restricted our attention to equations (7) and (8) in Section 2. In Section 3 we prove that the function u(t) behaves like the function S(t)[u(t) + h(t)u(t − τ )] as t → ∞, under some conditions, which plays a crucial part in the proof of Theorem 1.1. We show the “if” part and the “only if” part of Theorem 1.1 in Sections 4 and 5, respectively.
2. Oscillation criteria In this section we establish oscillation criteria for neutral differential equations of the form (1). First let us show that S(t) satisfies (3). Lemma 2.1. If (H1) or (H2) holds, then S(t) satisfies (3). Proof. Assume that (H1) holds. Let t ∈ R. Then S(t) =
∞
H2j (t) 1 − h(t − 2j τ ) .
j =0
We see that S(t)
∞
λ2j (1 − µ) =
1−µ , 1 − λ2
µ2j (1 − λ) =
1−λ . 1 − µ2
j =0
and S(t)
∞ j =0
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In the same way, the conclusion follows for the case (H2), by using S(t) =
∞ j =1
1 h(t + 2j τ ) − 1 . H2j (t + 2j τ )
✷
We need the following result which was obtained by Kusano and M. Naito [16]. Lemma 2.2. If the differential inequality x (n) (t) + f t, x g(t) 0 has an eventually positive solution, then the differential equation x (n) (t) + f t, x g(t) = 0 has an eventually positive solution. From Theorem 1.1, Lemmas 2.1 and 2.2, we have the following result. Corollary 2.1. Suppose that (H1) or (H2) holds. If 1−λ x (n) (t) + f t, x g(t) = 0 2 1−µ is oscillatory, then (1) is oscillatory. If 1−µ f t, x g(t) = 0 x (n) (t) + 2 1−λ has a nonoscillatory solution, then (1) has a nonoscillatory solution.
(10)
(11)
Proof. Assume that there exists a nonoscillatory solution of (1). Then Theorem 1.1 implies that (4) has a nonoscillatory solution x(t). Without loss of generality, we may assume that x(t) > 0 for all large t, since the case x(t) < 0 can be treated similarly. Put y(t) = (1 − λ)(1 − µ2 )−1 x(t). From Lemma 2.1 we see that 1 − λ (n) 1−λ x (t) = f t, S g(t) x g(t) 1 − µ2 1 − µ2
1−λ 1−λ 1−λ f t, x g(t) = f t, y g(t) 1 − µ2 1 − µ2 1 − µ2
−y (n) (t) = −
for all large t. From Lemma 2.2 it follows that (10) has a nonoscillatory solution. Consequently, if (10) is oscillatory, then (1) is oscillatory. Let y(t) be an eventually positive solution of (11). Then Lemma 2.1 implies that x(t) = (1 − λ2 )(1 − µ)−1 y(t) is an eventually positive solution of x (n) (t) + f t, S g(t) x g(t) 0, and hence (1) has a nonoscillatory solution, by Lemma 2.2 and Theorem 1.1. This completes the proof. ✷
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Now let us derive oscillation criteria for (7) and (8). The following oscillation result was established by Kitamura [15, Corollaries 5.1 and 3.1]. Lemma 2.3. Assume that (9) holds. If
∞ t n−1−ε p(t) dt = ∞
for some ε > 0,
(12)
then the equation x (n) (t) + p(t)x(t − σ ) = 0
(13)
is oscillatory. If
∞ t n−1 p(t) dt < ∞,
(14)
then (13) has a nonoscillatory solution. Lemma 2.4. Let γ > 0 and γ = 1. Assume that (9) holds. Then the equation γ −1 x (n) (t) + p(t)x(t − σ ) x(t − σ ) = 0 is oscillatory if and only if
∞ t min{γ ,1}(n−1)p(t) dt = ∞.
(15)
Combining Corollary 2.1 with Lemmas 2.3 and 2.4, we have the following oscillation criteria for (7) and (8). Corollary 2.2. If (12) holds, then (7) is oscillatory. If (14) holds, then (7) has a nonoscillatory solution. Corollary 2.3. Eq. (8) is oscillatory if and only if (15) holds. Remark 2.1. Corollary 2.2 with (H1) have been already established by Jaroš and Kusano [11, Theorems 3.1 and 4.1]. Corollary 2.2 with (H2) extends the results in [5, Theorem 1] and [8, Theorem 7]. Remark 2.2. Corollary 2.3 with (H1) has been obtained by Y. Naito [19] in the case where h(t) is locally Lipschitz continuous.
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3. Relation between u(t) and u(t) + h(t)u(t − τ ) In this section we study the relation between functions u(t) and u(t) + h(t) × u(t − τ ). We use the notation: (∆u)(t) = u(t) + h(t)u(t − τ ). We require the following results which play crucial parts in the proof of Theorem 1.1. Lemma 3.1. Suppose that (H1) and the following condition (16) hold: 1 u ∈ C[T − τ, ∞), ∆u ∈ C [T , ∞), (∆u)(t) 0, (∆u) (t) 0 for t T , and limt →∞ (∆u) (t)/t l = 0 for some l ∈ {0, 1, 2, . . .}.
(16)
Then u(t) = S(t)(∆u)(t) + o t l
(t → ∞).
(17)
Lemma 3.2. Suppose that (H2) and (16) hold. Assume, moreover, that limt →∞ λ−t /τ u(t) = 0. Then (17) holds. Proof of Lemma 3.1. From u(t) = (∆u)(t) − h(t)u(t − τ ),
t T,
it follows that u(t) = (∆u)(t) − h(t) (∆u)(t − τ ) − h(t − τ )u(t − 2τ ) = H0 (t)(∆u)(t) − H1 (t)(∆u)(t − τ ) + H2 (t)u(t − 2τ ) for t T + τ , and u(t) =
m
(−1)i Hi (t)(∆u)(t − iτ )
i=0
+ (−1)m+1 Hm+1 (t)u t − (m + 1)τ
for t T + mτ , m = 0, 1, 2, . . . .
(18)
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Let t ∈ [T + mτ, T + (m + 1)τ ] (m = 0, 1, 2, . . .). Then u s − (m + 1)τ u t − (m + 1)τ max s∈[T +mτ,T +(m+1)τ ] = max u(s) ≡ M. s∈[T −τ,T ]
179
(19)
Observe that m (−1)i Hi (t)(∆u)(t − iτ ) i=0
t m i = (−1) Hi (t) (∆u)(t) − (∆u) (s) ds i=0
t −iτ
∞
= S(t)(∆u)(t) −
−
(−1)i Hi (t) (∆u)(t)
i=m+1
t
m
(∆u) (s) ds,
(−1)i Hi (t)
i=1
and
(20)
t −iτ
∞ ∞ λm+1 i . (−1) Hi (t) λi = 1−λ i=m+1
(21)
i=m+1
Since t ∈ [T + mτ, T + (m + 1)τ ], we see that (t − T )τ −1 − 1 m (t − T )τ −1 , so that λm+1 λ(t −T )τ
−1 −1+1
= λ(t −T )/τ .
(22)
Combining (18) with (19)–(22), we have u(t) − S(t)(∆u)(t)
t m λ(t −T )/τ i (∆u)(t) + λ (∆u) (s) ds + λ(t −T )/τ M. 1−λ i=1
t −iτ
An easy computation shows that m i=1
t
(∆u) (s) ds =
i
λ
m i=1
t −iτ
=
i
λ
i
t −(j
−1)τ
(∆u) (s) ds
j =1 t −j τ
m m j =1 i=j
t −(j
−1)τ
(∆u) (s) ds
i
λ
t −j τ
(23)
180
S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189 m λj − λm+1 = 1−λ
t −(j
−1)τ
(∆u) (s) ds
j =1
1 j λ 1−λ m
j =1
t −j τ
t −(j
−1)τ
(∆u) (s) ds.
(24)
t −j τ
If s ∈ [t − j τ, t − (j − 1)τ ], then (t − s)τ −1 j (t − s)τ −1 + 1. This implies that m
t −(j
−1)τ
λ
j =1
(∆u) (s) ds
j
m
t −(j
−1)τ
λ(t −s)/τ (∆u) (s) ds
j =1 t −j τ
t −j τ
t =λ
t /τ
λ−s/τ (∆u) (s) ds
t −mτ
t λ
t /τ
λ−s/τ (∆u) (s) ds.
(25)
T
In view of (23)–(25), we obtain (t −T )/τ λt /τ u(t) − S(t)(∆u)(t) λ (∆u)(t) + 1−λ 1−λ (t −T )/τ
+λ
M,
t
λ−s/τ (∆u) (s) ds
T
t T.
From limt →∞ (∆u) (t)/t l = 0, we conclude that limt →∞ λ(t −T )/τ (∆u)(t) = 0, and t −s/τ d t −s/τ (∆u) (s) ds (∆u) (s) ds dt T λ T λ = lim lim d −t /τ t l ] t →∞ t →∞ λ−t /τ t l dt [λ (∆u) (t) = 0, t →∞ [log λ−1/τ + lt −1 ]t l
= lim
so that (17) holds. The proof is complete. ✷ Proof of Lemma 3.2. Let t T be fixed. Since u(t) =
(∆u)(t + τ ) u(t + τ ) − , h(t + τ ) h(t + τ )
S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
we see that u(t) = = =
=
181
1 (∆u)(t + 2τ ) u(t + 2τ ) (∆u)(t + τ ) − − h(t + τ ) h(t + τ ) h(t + 2τ ) h(t + 2τ ) u(t + 2τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) − + H1 (t + τ ) H2 (t + 2τ ) H2 (t + 2τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) − H1 (t + τ ) H (t + 2τ ) 2 (∆u)(t + 3τ ) u(t + 3τ ) 1 − + H2 (t + 2τ ) h(t + 3τ ) h(t + 3τ ) u(t + 3τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) (∆u)(t + 3τ ) − + − , H1 (t + τ ) H2 (t + 2τ ) H3 (t + 3τ ) H3 (t + 3τ )
so that u(t) =
m (−1)m (−1)i+1 (∆u)(t + iτ ) + u(t + mτ ) Hi (t + iτ ) Hm (t + mτ )
(26)
i=1
for m = 1, 2, . . . . From limt →∞ λ−t /τ u(t) = 0 it follows that (−1)m u(t + mτ ) lim λ−m u(t + mτ ) lim m→∞ Hm (t + mτ ) m→∞ = λt /τ lim λ−(t +mτ )/τ u(t + mτ ) m→∞ = λt /τ lim λ−s/τ u(s) = 0. s→∞
(∆u) (t)/t l
Since limt →∞ = 0, we can take a constant K > 0 such that |(∆u)(t)| Kt l+1 for t T . Then (−1)i+1 −i l+1 H (t + iτ ) (∆u)(t + iτ ) λ K(t + iτ ) , i = 1, 2, . . . . i Therefore, letting m → ∞ in (26), we obtain u(t) =
∞ (−1)i+1 (∆u)(t + iτ ). Hi (t + iτ ) i=1
The equality t +iτ
(∆u) (s) ds,
(∆u)(t + iτ ) = (∆u)(t) + t
shows that t +iτ ∞ (−1)i+1 u(t) − S(t)(∆u)(t) = (∆u) (s) ds. Hi (t + iτ ) i=1
t
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S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
Observe that t +iτ
∞ u(t) − S(t)(∆u)(t) λ−i
(∆u) (s) ds
i=1
=
∞
t −i
λ
∞ ∞
=
j =1
t +j τ
−i
λ
j =1 i=j ∞
(∆u) (s) ds
j =1 t +(j −1)τ
i=1
=
t +j τ
i
(∆u) (s) ds
t +(j −1)τ
λ−j 1 − λ−1
t +j τ
(∆u) (s) ds.
t +(j −1)τ
Since (s − t)τ −1 j (s − t)τ −1 + 1 for s ∈ [t + (j − 1)τ, t + j τ ], we have ∞
−j
t +j τ
(∆u) (s) ds
λ
j =1
t +(j −1)τ
t +j τ
∞
λ−(s−t )/τ (∆u) (s) ds
j =1 t +(j −1)τ
∞ =λ
t /τ
λ−s/τ (∆u) (s) ds.
t
We note here that λ−t /τ (∆u) (t) is integrable on [T , ∞), because of limt →∞ (∆u) (t)/t l = 0. Using the same argument as in the proof of Lemma 3.1, we conclude that
∞ t /τ λ λ−s/τ (∆u) (s) ds = o t l (t → ∞). t
This completes the proof. ✷
4. Proof of the “if” part of Theorem 1.1 In this section we prove the “if” part of Theorem 1.1, namely we show that if (1) has a nonoscillatory solution, then (4) has a nonoscillatory solution. Let x(t) be a nonoscillatory solution of (1). Then we easily see that (∆x)(t) = 0 and (∆x)(t)(∆x)(n)(t) 0 for all large t t0 . For such functions as (∆x)(t), we have the following well-known lemma of Kiguradze [13]. (See also [14].)
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Lemma 4.1. Let w ∈ C n [T , ∞) satisfy w(t) = 0 and w(t)w(n) (t) 0 for t T . Then there exist an integer k ∈ {1, 3, . . . , n − 1} and a number T1 T such that w(t)w(i) (t) > 0, 0 i k − 1, (27) (−1)i−k w(t)w(i) (t) 0, k i n, for t T1 . In particular, w (t) 0 for t T1 . A function w(t) satisfying (27) for all large t is called a function of Kiguradze degree k. Let w(t) be a function of Kiguradze degree k ∈ {1, 3, . . . , n − 1} satisfying w(t) > 0 for all large t. It can be shown (cf. [11–13,19]) that lim w(i) (t) = 0,
t →∞
i = k + 1, k + 2, . . . , n − 1,
and that one of the following three cases holds: lim w(k) (t) = const > 0
and
t →∞
lim w(k) (t) = 0
and
lim w(k) (t) = 0
and
t →∞ t →∞
lim w
t →∞
lim w(k−1) (t) = ∞;
t →∞ (k−1)
(t) = ∞;
lim w(k−1) (t) = const > 0.
t →∞
(28a) (28b) (28c)
If (28a) holds, then we put l = k, and if (28b) or (28c) holds, then we put l = k − 1. Then it is easy to verify that l ∈ {0, 1, 2, . . ., n − 1}, w (t) w(t) = 0 and lim l = L (L > 0 or L = ∞). (29) l t →∞ t t →∞ t From Lemmas 2.1, 3.1 and 3.2, and the above observation, we obtain the next result. lim
Lemma 4.2. Suppose that (H1) or (H2) holds. Let u ∈ C[T − τ, ∞) satisfy ∆u ∈ C n [T , ∞) and (∆u)(t) > 0 for t T . Assume, moreover, that limt →∞ λ−t /τ u(t) = 0 if (H2) holds. If (∆u)(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n − 1}, then there exist a constant α > 0 and an integer l ∈ {0, 1, 2, . . ., n − 1} such that (30) u(t) S(t) (∆u)(t) − αt l > 0 for all large t T . Proof. Recalling (29), we have (∆u) (t) (∆u)(t) = 0 and lim = L (L > 0 or L = ∞) t →∞ t →∞ tl tl for some l ∈ {0, 1, 2, . . . , n − 1}. Choose a constant α > 0 so small that α < L. By Lemmas 2.1, 3.1 and 3.2, we conclude that (30) holds. ✷ lim
Now let us prove the “if” part of Theorem 1.1.
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S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
Proof of the “if” part of Theorem 1.1. Let x(t) be a nonoscillatory solution of (1). Without loss of generality, we may assume that x(t) > 0 for all large t. Then (∆x)(t) > 0 and (∆x)(n) (t) 0 for all large t. By virtue of Lemma 4.1, we see that (∆x)(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n−1}. From Lemma 4.2, there are a constant α > 0 and an integer l ∈ {0, 1, 2, . . . , n − 1} such that x(t) S(t) (∆x)(t) − αt l > 0 for all large t. Set w(t) = (∆x)(t) − αt l . Then x(t) S(t)w(t) > 0 for all large t. We see that −w(n) (t) = −(∆x)(n)(t) = f t, x g(t) f t, S g(t) w g(t) for all large t. Lemma 2.2 implies that (4) has a nonoscillatory solution. The proof is complete. ✷
5. Proof of the “only if” part of Theorem 1.1 In this section we show the “only if” part of Theorem 1.1. To this end, we require the following mapping Φ, which is an “inverse” of the operator ∆: i ∞ i=0 (−1) Hi (t)v(t − iτ ) if (H1) holds, (Φv)(t) = ∞ (−1)i+1 v(t + iτ ) if (H2) holds, i=1 Hi (t +iτ ) for t ∈ R, where v ∈ v ∈ C(R): |v(t)| M max t k , 1 , t ∈ R ≡ V ,
(31)
M > 0 and k ∈ {1, 2, . . .}. The properties of the mapping Φ are as follows. Lemma 5.1. The mapping Φ is well-defined on V and has the following properties (i)–(iv): (i) (ii) (iii) (iv)
Φ maps V into C(R); Φ is continuous on V in the C(R)-topology; Φ satisfies (∆(Φv))(t) = v(t) for t ∈ R and v ∈ V ; if (H2) holds, then limt →∞ λ−t /τ (Φv)(t) = 0 for v ∈ V .
Here and hereafter, C(R) is regarded as the Fréchet space of all continuous functions on R with the topology of uniform convergence on every compact subinterval of R.
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Proof of Lemma 5.1. First we assume that (H1) holds. Let v ∈ V . We see that (−1)i Hi (t)v(t − iτ ) λi M max (t − iτ )k , 1 λi M max t k , 1 , i = 0, 1, 2, . . ., (32) ∞ for t ∈ R, so that the series i=0 (−1)i Hi (t)v(t − iτ ) converges uniformly on every compact subinterval of R. This implies that Φ is well-defined and Φ satisfies (i). It can be show that Φ satisfies (iii). In fact, h(t)(Φv)(t − τ ) − v(t) =
∞ (−1)i h(t)Hi (t − τ )v t − (i + 1)τ − v(t) i=0 ∞
=−
(−1)i+1 Hi+1 (t)v t − (i + 1)τ − v(t)
i=0
=−
∞
(−1)i Hi (t)v(t − iτ ) − v(t)
i=1
= −(Φv)(t),
t ∈ R, v ∈ V .
{vj }∞ j =1
be a sequence in V converging Now we prove that Φ satisfies (ii). Let to v ∈ V uniformly on every compact subinterval of R. It suffices to verify that Φvj converges to Φv uniformly on every compact subinterval of R. Take an arbitrary compact subinterval I of R. Let ε > 0. There is an integer q 1 such that ∞ ε (33) λi M max t k , 1 < , t ∈ I. 3 i=q+1
There exists an integer j0 1 such that q ε λi vj (t − iτ ) − v(t − iτ ) < , 3
t ∈ I, j j0 .
i=0
It follows from (32)–(34) that q (Φvj )(t) − (Φv)(t) λi vj (t − iτ ) − v(t − iτ )
∞ i (−1) Hi (t)vj (t − iτ ) + i=q+1 ∞ (−1)i Hi (t)v(t − iτ ) + i=q+1 ε ε + 2 · = ε, t ∈ I, j j0 , 3 3 which implies that Φvj converges Φv uniformly on I . i=0
(34)
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Now we suppose that (H2) holds. Let v ∈ V . Then (−1)i+1 −i k H (t + iτ ) v(t + iτ ) λ M max (t + iτ ) , 1 i λ−i M max 2k−1 t k + i k τ k , 1
(35)
for t ∈ R and i ∈ {1, 2, . . .}. Consequently, by the same arguments as in the case (H1), we can show that Φ is well-defined and satisfies (i)–(iii). It is easy to see that Φ satisfies (iv). Indeed, from (35), we have (Φv)(t) 2k−1 M S1 t k + S2 for t > 1, ∞ −i k −i and S = τ k where S1 = ∞ 2 i=1 λ i=1 λ i . The proof is complete. ✷ Now we prove the “only if” part of Theorem 1.1. Proof of the “only if” part of Theorem 1.1. We show that if (4) has a nonoscillatory solution, then (1) has a nonoscillatory solution. Let w(t) be a nonoscillatory solution of (4). We may assume that w(t) is eventually positive. Lemma 4.1 implies that w(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n − 1}. Hence, (27) holds and one of the cases (28a)–(28c) is satisfied. We can take a sufficiently large number T > 1 such that w(i) (t) > 0 (i = 0, 1, 2, . . ., k − 1), w(g(t)) > 0 for t T . Integrating (4), we have
t w(t) − P (t) = T
(t − s)k−1 (k − 1)!
∞ s
(r − s)n−k−1 F (r) dr ds 0 (n − k − 1)!
(36)
for t T , where F (t) = f (t, S(g(t))w(g(t))), (t − T )i (t − T )k (k) w (∞) + w(i) (T ), k! i! k−1
P (t) =
i=0
= limt →∞ ∈ [0, ∞). Note that P (T ) = w(T ) and P (t) and w(T ) > 0 for t T . Consider the set Y of functions y ∈ C(R) which satisfies w(k) (∞)
y(t) = 0
w(k) (t)
for t T
and 0 y(t) w(t) − P (t)
for t T .
Then Y is closed and convex. Set (1/2)P (t), t T , η(t) = (1/2)P (T ), t < T . By w(k) (∞) ∈ [0, ∞), there is a constant M P (T ) such that w(t) Mt k and P (t) Mt k for t T . Define the set V by (31). We easily see that Y ⊂ V and η∈V.
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For each y ∈ Y , we define the mapping F : Y −→ C(R) as follows: t (t −s)k−1 ∞ (r−s)n−k−1 T (k−1)! s (n−k−1)! (F y)(t) = × f r, (Φy) g(r) + (Φη) g(r) dr ds, t T , 0, t T , where
F (t), u S g(t) w g(t) , f (t, u) = f (t, u), 0 u S g(t) w g(t) , 0, u 0,
for t T and u ∈ R. In view of (36) and the fact that 0 f (t, u) F (t),
(t, u) ∈ [T , ∞) × R,
we see that F is well-defined on Y and maps Y into itself. Since Φ is continuous on Y , by the Lebesgue dominated convergence theorem, we can show that F is continuous on Y as a routine computation. Now we claim that F (Y ) is relatively compact. We note that F (Y ) is uniformly bounded on every compact subinterval of R, because of F (Y ) ⊂ Y . By the Ascoli–Arzelà theorem, it suffices to verify that F (Y ) is equicontinuous on every compact subinterval of R. Let I be an arbitrary compact subinterval of [T , ∞). If k = 1, then
∞
0 (F y) (t) t
(s − t)n−2 F (s) ds, (n − 2)!
t T , y ∈ Y.
If k 3, then
t
0 (F y) (t) T
(t − s)k−2 (k − 2)!
∞ s
(r − s)n−k−1 F (r) dr ds (n − k − 1)!
for t T and y ∈ Y . Thus we see that {(F y) (t): y ∈ Y } is uniformly bounded on I . The mean value theorem implies that F (Y ) is equicontinuous on I . Since |(F y)(t1 ) − (F y)(t2 )| = 0 for t1 , t2 ∈ (−∞, T ], we conclude that F (Y ) is equicontinuous on every compact subinterval of R. By applying the Schauder–Tychonoff fixed point theorem to the operator F , there exists a y˜ ∈ Y such that y˜ = F y. ˜ Put x(t) = (Φ y)(t) ˜ + (Φη)(t). Then we obtain P (t) P (t) = (F y)(t) ˜ + > 0, t T , (37) 2 2 by (iii) of Lemma 5.1, and hence (∆x)(t) is a function of Kiguradze degree k. (∆x)(t) = y(t) ˜ +
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We will show that 0 < x(t) S(t)w(t)
for all large t.
(38)
Then the proof of Theorem 1.1 will be complete, since (37) and (38) imply that dn x(t) + h(t)x(t − τ ) = (F y) ˜ (n) (t) = −f¯ t, x g(t) n dt = −f t, x g(t) for all large t, which means that x(t) is a nonoscillatory solution of (1). If w(k) (∞) > 0, then we put l = k, and if w(k) (∞) = 0, then we put l = k − 1. By (37), we find that limt →∞ (∆x)(k) (t) = (1/2)w(k) (∞), so that limt →∞ (∆x) (t)/t l = 0. From Lemma 4.2 it follows that x(t) > 0 for all large t T . In view of Lemmas 2.1, 3.1 and 3.2, and the fact that limt →∞ P (t)/t l = const > 0, we have P (t) x(t) (∆x)(t) + S(t) 2 for all large t. Hence, by (37), we obtain P (t) P (t) P (t) P (t) x(t) y(t) ˜ + + w(t) − P (t) + + = w(t) S(t) 2 2 2 2 for all large t. This completes the proof. ✷
References [1] Y. Chen, Existence of nonoscillatory solutions of nth order neutral delay differential equations, Funkcial. Ekvac. 35 (1992) 557–570. [2] Q. Chuanxi, G. Ladas, Oscillations of higher order neutral differential equations with variable coefficients, Math. Nachr. 150 (1991) 15–24. [3] L.H. Erbe, Q. Kong, B.G. Zhang, Oscillation Theory for Functional Differential Equations, Dekker, New York, 1995. [4] J.R. Graef, M.K. Grammatikopoulos, P.W. Spikes, Asymptotic properties of solutions of nonlinear neutral delay differential equations of the second order, Rad. Mat. 4 (1988) 133–149. [5] J.R. Graef, P.W. Spikes, On the oscillation of an nth-order nonlinear neutral delay differential equation, J. Comput. Appl. Math. 41 (1992) 35–40. [6] M.K. Grammatikopoulos, G. Ladas, A. Meimaridou, Oscillations of second order neutral delay differential equations, Rad. Mat. 1 (1985) 267–274. [7] M.K. Grammatikopoulos, G. Ladas, A. Meimaridou, Oscillation and asymptotic behavior of second order neutral differential equations, Ann. Mat. Pura Appl. 148 (1987) 29–40. [8] M.K. Grammatikopoulos, G. Ladas, A. Meimaridou, Oscillation and asymptotic behavior of higher order neutral equations with variable coefficients, Chinese Ann. Math. Ser. B 9 (1988) 322–338. [9] I. Györi, G. Ladas, Oscillation Theory of Delay Differential Equations, Clarendon, Oxford, 1991. [10] J.K. Hale, Theory of Functional Differential Equations, Springer-Verlag, New York, 1977. [11] J. Jaroš, T. Kusano, Oscillation theory of higher order linear functional differential equations of neutral type, Hiroshima Math. J. 18 (1988) 509–531.
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[12] J. Jaroš, T. Kusano, Asymptotic behavior of nonoscillatory solutions of nonlinear functional differential equations of neutral type, Funkcial. Ekvac. 32 (1989) 251–263. [13] I.T. Kiguradze, On the oscillatory character of solutions of the equation d m u/dt m + a(t) × |u|n sign u = 0, Mat. Sb. 65 (1964) 172–187, in Russian. [14] I.T. Kiguradze, T.A. Chanturia, Asymptotic Properties of Solutions of Nonautonomous Ordinary Differential Equations, Kluwer, Dordrecht, 1993. [15] Y. Kitamura, Oscillation of functional differential equations with general deviating arguments, Hiroshima Math. J. 15 (1985) 445–491. [16] T. Kusano, M. Naito, Comparison theorems for functional-differential equations with deviating arguments, J. Math. Soc. Japan 33 (1981) 509–532. [17] M. Naito, On strong oscillation of retarded differential equations, Hiroshima Math. J. 11 (1981) 553–560. [18] M. Naito, An asymptotic theorem for a class of nonlinear neutral differential equations, Czechoslovak Math. J. 48 (1998) 419–432. [19] Y. Naito, Nonoscillatory solutions of neutral differential equations, Hiroshima Math. J. 20 (1990) 231–258. [20] S. Tanaka, Existence of positive solutions of higher order nonlinear neutral differential equations, Rocky Mountain J. Math. 30 (2000) 1139–1149. [21] S. Tanaka, Oscillation of solutions of even order neutral differential equations, Dynam. Systems Appl. 9 (2000) 353–360. [22] S. Tanaka, A necessary and sufficient condition for the oscillation in a class of even order neutral differential equations, Electron. J. Qual. Theory Differential Equations 4 (2000) 1–27. [23] X.H. Tang, J.H. Shen, Oscillation and existence of positive solutions in a class of higher order neutral equations, J. Math. Anal. Appl. 213 (1997) 662–680. [24] B.G. Zhang, B. Yang, New approach of studying the oscillation of neutral differential equations, Funkcial. Ekvac. 41 (1998) 79–89.
A oscillation theorem for a class of even order neutral differential equations Satoshi Tanaka Department of Electrical Engineering, Hachinohe National College of Technology, Hachinohe 039-1192, Japan Received 8 May 2001 Submitted by R.P. Agarwal
Abstract The even order neutral differential equation dn x(t) + h(t)x(t − τ ) + f t, x g(t) = 0 (1) dt n is considered under the following conditions: n 2 is even; τ > 0; h ∈ C(R); g ∈ C[t0 , ∞), limt →∞ g(t) = ∞; f ∈ C([t0 , ∞) × R), uf (t, u) 0 for (t, u) ∈ [t0 , ∞) × R, and f (t, u) is nondecreasing in u ∈ R for each fixed t t0 . It is shown that (1) is oscillatory if and only if the certain non-neutral differential equation is oscillatory, for the case where 0 µ h(t) λ < 1 or 1 < λ h(t) µ. 2002 Elsevier Science (USA). All rights reserved.
1. Introduction and main results We shall be concerned with the oscillatory behavior of solutions of the even order neutral differential equation dn x(t) + h(t)x(t − τ ) + f t, x g(t) = 0. (1) dt n Throughout this paper, the following conditions are assumed to hold: n 2 is even; τ > 0; h ∈ C(R); g ∈ C[t0 , ∞), limt →∞ g(t) = ∞; f ∈ C([t0 , ∞) × R), E-mail address: [email protected]. 0022-247X/02/$ – see front matter 2002 Elsevier Science (USA). All rights reserved. PII: S 0 0 2 2 - 2 4 7 X ( 0 2 ) 0 0 2 3 5 - 4
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173
uf (t, u) 0 for (t, u) ∈ [t0 , ∞) × R, and f (t, u) is nondecreasing in u ∈ R for each fixed t t0 . By a solution of (1), we mean a function x(t) that is continuous and satisfies (1) on [tx , ∞) for some tx t0 . A solution is said to be oscillatory if it has arbitrarily large zeros; otherwise it is said to be nonoscillatory. Eq. (1) is said to be oscillatory if every solution of (1) is oscillatory. Neutral differential equations find numerous applications in natural science and technology. For instance, they are frequently used for the study of distributed networks containing lossless transmission lines. See Hale [10]. Oscillation properties of even order neutral differential equations have been investigated by many authors. We refer the reader to [1–9,11,12,18–24]. For odd order neutral differential equations, it has been shown by Zhang and Yang [24] that the equation γ −1 dN x(t − σ ) = 0 x(t) − x(t − τ ) + p(t)x(t − σ ) N dt is oscillatory if and only the ordinary differential equation γ −1 x (N+1) (t) + τ −1 p(t)x(t) x(t) = 0 is oscillatory, where N 1 is odd, γ > 0, σ ∈ R, p ∈ C[t0 , ∞), p(t) 0 for t t0 . (See also Tang and Shen [23].) For even order neutral differential equations, recently, the following result has been established in [22]. Theorem 1.0. Let c > 0. Then the even order neutral differential equation dn x(t) + cx(t − τ ) + f t, x g(t) = 0 n dt is oscillatory if and only if the even order non-neutral differential equation 1 f t, x g(t) = 0 x (n) (t) + 1+c is oscillatory.
(2)
The purpose of this paper is to generalize Theorem 1.0 with c = 1 for (1) with the following cases (H1) and (H2): (H1) 0 µ h(t) λ < 1 for t ∈ R; (H2) 1 < λ h(t) µ for t ∈ R. Here, µ and λ are constants. It is convenience only that the parts of µ and λ in (H1) and (H2) are opposite each other. Throughout this paper we use the notation: Hi (t) = h(t)h(t − τ ) · · · h t − (i − 1)τ , i = 1, 2, . . . . H0 (t) = 1;
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We define the function S(t) on R by ∞ i=0 (−1)i Hi (t) if (H1) holds, S(t) = ∞ (−1)i+1 if (H2) holds, i=1 Hi (t +iτ )
for t ∈ R.
It is easy to see that S(t) is converges uniformly on R, and hence S(t) is continuous on R. In Section 2 we will show that 1−µ 1−λ S(t) , t ∈ R. (3) 0< 1 − µ2 1 − λ2 We note that if µ = λ = c = 1, then 1−λ 1 1−µ 1 , and S(t) = . = = 2 2 1+c 1+c 1−µ 1−λ Main result of this paper is the following theorem. Theorem 1.1. Suppose that (H1) or (H2) holds. Then (1) is oscillatory if and only if x (n) (t) + f t, S g(t) x g(t) = 0 (4) is oscillatory. Theorem 1.1 means that (1) has a nonoscillatory solution if and only if (4) has a nonoscillatory solution. Theorem 1.1 is a generalization of Theorem 1.0 with c = 1. Indeed, for the case where h(t) ≡ c > 0 and c = 1, we see that S(t) = (1 + c)−1 and that (2) is oscillatory if and only if y (n) (t) + f t, (1 + c)−1 y g(t) = 0 is oscillatory. (Put x(t) = (1 + c)y(t).) Now we assume that h(t + τ ) = h(t),
h(t) = 1 and h(t) 0
for t ∈ R.
Then it is easy to verify that (H1) or (H2) holds, and S(t) = sequently, from Theorem 1.1, we have the following result.
[1 + h(t)]−1 .
(5) Con-
Corollary 1.1. Suppose that (5) holds. Then (1) is oscillatory if and only if
x(g(t)) x (n) (t) + f t, =0 1 + h(g(t)) is oscillatory. The oscillatory behavior of solutions of non-neutral differential equations of the form (6) x (n) (t) + f t, x g(t) = 0
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175
has been intensively studied in the last three decades. We refer the reader to [3, 9,13–17,19] and the references cited therein. Combining Theorem 1.1 with the known oscillation results for non-neutral differential equations of the form (6), we can derive various oscillation results for neutral differential equations of the form (1). In Section 2 we obtain oscillation criteria for the linear neutral differential equation dn x(t) + h(t)x(t − τ ) + p(t)x(t − σ ) = 0, (7) dt n and for the nonlinear neutral differential equation γ −1 dn x(t) + h(t)x(t − τ ) + p(t)x(t − σ ) x(t − σ ) = 0, (8) n dt where γ > 0, γ = 1 and the following conditions are assumed to hold: σ ∈ R;
p ∈ C[t0 , ∞),
p(t) > 0
for t t0 .
(9)
It is possible to obtain oscillation results for more general equations such as (1). However, for simplicity, we have restricted our attention to equations (7) and (8) in Section 2. In Section 3 we prove that the function u(t) behaves like the function S(t)[u(t) + h(t)u(t − τ )] as t → ∞, under some conditions, which plays a crucial part in the proof of Theorem 1.1. We show the “if” part and the “only if” part of Theorem 1.1 in Sections 4 and 5, respectively.
2. Oscillation criteria In this section we establish oscillation criteria for neutral differential equations of the form (1). First let us show that S(t) satisfies (3). Lemma 2.1. If (H1) or (H2) holds, then S(t) satisfies (3). Proof. Assume that (H1) holds. Let t ∈ R. Then S(t) =
∞
H2j (t) 1 − h(t − 2j τ ) .
j =0
We see that S(t)
∞
λ2j (1 − µ) =
1−µ , 1 − λ2
µ2j (1 − λ) =
1−λ . 1 − µ2
j =0
and S(t)
∞ j =0
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In the same way, the conclusion follows for the case (H2), by using S(t) =
∞ j =1
1 h(t + 2j τ ) − 1 . H2j (t + 2j τ )
✷
We need the following result which was obtained by Kusano and M. Naito [16]. Lemma 2.2. If the differential inequality x (n) (t) + f t, x g(t) 0 has an eventually positive solution, then the differential equation x (n) (t) + f t, x g(t) = 0 has an eventually positive solution. From Theorem 1.1, Lemmas 2.1 and 2.2, we have the following result. Corollary 2.1. Suppose that (H1) or (H2) holds. If 1−λ x (n) (t) + f t, x g(t) = 0 2 1−µ is oscillatory, then (1) is oscillatory. If 1−µ f t, x g(t) = 0 x (n) (t) + 2 1−λ has a nonoscillatory solution, then (1) has a nonoscillatory solution.
(10)
(11)
Proof. Assume that there exists a nonoscillatory solution of (1). Then Theorem 1.1 implies that (4) has a nonoscillatory solution x(t). Without loss of generality, we may assume that x(t) > 0 for all large t, since the case x(t) < 0 can be treated similarly. Put y(t) = (1 − λ)(1 − µ2 )−1 x(t). From Lemma 2.1 we see that 1 − λ (n) 1−λ x (t) = f t, S g(t) x g(t) 1 − µ2 1 − µ2
1−λ 1−λ 1−λ f t, x g(t) = f t, y g(t) 1 − µ2 1 − µ2 1 − µ2
−y (n) (t) = −
for all large t. From Lemma 2.2 it follows that (10) has a nonoscillatory solution. Consequently, if (10) is oscillatory, then (1) is oscillatory. Let y(t) be an eventually positive solution of (11). Then Lemma 2.1 implies that x(t) = (1 − λ2 )(1 − µ)−1 y(t) is an eventually positive solution of x (n) (t) + f t, S g(t) x g(t) 0, and hence (1) has a nonoscillatory solution, by Lemma 2.2 and Theorem 1.1. This completes the proof. ✷
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Now let us derive oscillation criteria for (7) and (8). The following oscillation result was established by Kitamura [15, Corollaries 5.1 and 3.1]. Lemma 2.3. Assume that (9) holds. If
∞ t n−1−ε p(t) dt = ∞
for some ε > 0,
(12)
then the equation x (n) (t) + p(t)x(t − σ ) = 0
(13)
is oscillatory. If
∞ t n−1 p(t) dt < ∞,
(14)
then (13) has a nonoscillatory solution. Lemma 2.4. Let γ > 0 and γ = 1. Assume that (9) holds. Then the equation γ −1 x (n) (t) + p(t)x(t − σ ) x(t − σ ) = 0 is oscillatory if and only if
∞ t min{γ ,1}(n−1)p(t) dt = ∞.
(15)
Combining Corollary 2.1 with Lemmas 2.3 and 2.4, we have the following oscillation criteria for (7) and (8). Corollary 2.2. If (12) holds, then (7) is oscillatory. If (14) holds, then (7) has a nonoscillatory solution. Corollary 2.3. Eq. (8) is oscillatory if and only if (15) holds. Remark 2.1. Corollary 2.2 with (H1) have been already established by Jaroš and Kusano [11, Theorems 3.1 and 4.1]. Corollary 2.2 with (H2) extends the results in [5, Theorem 1] and [8, Theorem 7]. Remark 2.2. Corollary 2.3 with (H1) has been obtained by Y. Naito [19] in the case where h(t) is locally Lipschitz continuous.
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3. Relation between u(t) and u(t) + h(t)u(t − τ ) In this section we study the relation between functions u(t) and u(t) + h(t) × u(t − τ ). We use the notation: (∆u)(t) = u(t) + h(t)u(t − τ ). We require the following results which play crucial parts in the proof of Theorem 1.1. Lemma 3.1. Suppose that (H1) and the following condition (16) hold: 1 u ∈ C[T − τ, ∞), ∆u ∈ C [T , ∞), (∆u)(t) 0, (∆u) (t) 0 for t T , and limt →∞ (∆u) (t)/t l = 0 for some l ∈ {0, 1, 2, . . .}.
(16)
Then u(t) = S(t)(∆u)(t) + o t l
(t → ∞).
(17)
Lemma 3.2. Suppose that (H2) and (16) hold. Assume, moreover, that limt →∞ λ−t /τ u(t) = 0. Then (17) holds. Proof of Lemma 3.1. From u(t) = (∆u)(t) − h(t)u(t − τ ),
t T,
it follows that u(t) = (∆u)(t) − h(t) (∆u)(t − τ ) − h(t − τ )u(t − 2τ ) = H0 (t)(∆u)(t) − H1 (t)(∆u)(t − τ ) + H2 (t)u(t − 2τ ) for t T + τ , and u(t) =
m
(−1)i Hi (t)(∆u)(t − iτ )
i=0
+ (−1)m+1 Hm+1 (t)u t − (m + 1)τ
for t T + mτ , m = 0, 1, 2, . . . .
(18)
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Let t ∈ [T + mτ, T + (m + 1)τ ] (m = 0, 1, 2, . . .). Then u s − (m + 1)τ u t − (m + 1)τ max s∈[T +mτ,T +(m+1)τ ] = max u(s) ≡ M. s∈[T −τ,T ]
179
(19)
Observe that m (−1)i Hi (t)(∆u)(t − iτ ) i=0
t m i = (−1) Hi (t) (∆u)(t) − (∆u) (s) ds i=0
t −iτ
∞
= S(t)(∆u)(t) −
−
(−1)i Hi (t) (∆u)(t)
i=m+1
t
m
(∆u) (s) ds,
(−1)i Hi (t)
i=1
and
(20)
t −iτ
∞ ∞ λm+1 i . (−1) Hi (t) λi = 1−λ i=m+1
(21)
i=m+1
Since t ∈ [T + mτ, T + (m + 1)τ ], we see that (t − T )τ −1 − 1 m (t − T )τ −1 , so that λm+1 λ(t −T )τ
−1 −1+1
= λ(t −T )/τ .
(22)
Combining (18) with (19)–(22), we have u(t) − S(t)(∆u)(t)
t m λ(t −T )/τ i (∆u)(t) + λ (∆u) (s) ds + λ(t −T )/τ M. 1−λ i=1
t −iτ
An easy computation shows that m i=1
t
(∆u) (s) ds =
i
λ
m i=1
t −iτ
=
i
λ
i
t −(j
−1)τ
(∆u) (s) ds
j =1 t −j τ
m m j =1 i=j
t −(j
−1)τ
(∆u) (s) ds
i
λ
t −j τ
(23)
180
S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189 m λj − λm+1 = 1−λ
t −(j
−1)τ
(∆u) (s) ds
j =1
1 j λ 1−λ m
j =1
t −j τ
t −(j
−1)τ
(∆u) (s) ds.
(24)
t −j τ
If s ∈ [t − j τ, t − (j − 1)τ ], then (t − s)τ −1 j (t − s)τ −1 + 1. This implies that m
t −(j
−1)τ
λ
j =1
(∆u) (s) ds
j
m
t −(j
−1)τ
λ(t −s)/τ (∆u) (s) ds
j =1 t −j τ
t −j τ
t =λ
t /τ
λ−s/τ (∆u) (s) ds
t −mτ
t λ
t /τ
λ−s/τ (∆u) (s) ds.
(25)
T
In view of (23)–(25), we obtain (t −T )/τ λt /τ u(t) − S(t)(∆u)(t) λ (∆u)(t) + 1−λ 1−λ (t −T )/τ
+λ
M,
t
λ−s/τ (∆u) (s) ds
T
t T.
From limt →∞ (∆u) (t)/t l = 0, we conclude that limt →∞ λ(t −T )/τ (∆u)(t) = 0, and t −s/τ d t −s/τ (∆u) (s) ds (∆u) (s) ds dt T λ T λ = lim lim d −t /τ t l ] t →∞ t →∞ λ−t /τ t l dt [λ (∆u) (t) = 0, t →∞ [log λ−1/τ + lt −1 ]t l
= lim
so that (17) holds. The proof is complete. ✷ Proof of Lemma 3.2. Let t T be fixed. Since u(t) =
(∆u)(t + τ ) u(t + τ ) − , h(t + τ ) h(t + τ )
S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
we see that u(t) = = =
=
181
1 (∆u)(t + 2τ ) u(t + 2τ ) (∆u)(t + τ ) − − h(t + τ ) h(t + τ ) h(t + 2τ ) h(t + 2τ ) u(t + 2τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) − + H1 (t + τ ) H2 (t + 2τ ) H2 (t + 2τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) − H1 (t + τ ) H (t + 2τ ) 2 (∆u)(t + 3τ ) u(t + 3τ ) 1 − + H2 (t + 2τ ) h(t + 3τ ) h(t + 3τ ) u(t + 3τ ) (∆u)(t + τ ) (∆u)(t + 2τ ) (∆u)(t + 3τ ) − + − , H1 (t + τ ) H2 (t + 2τ ) H3 (t + 3τ ) H3 (t + 3τ )
so that u(t) =
m (−1)m (−1)i+1 (∆u)(t + iτ ) + u(t + mτ ) Hi (t + iτ ) Hm (t + mτ )
(26)
i=1
for m = 1, 2, . . . . From limt →∞ λ−t /τ u(t) = 0 it follows that (−1)m u(t + mτ ) lim λ−m u(t + mτ ) lim m→∞ Hm (t + mτ ) m→∞ = λt /τ lim λ−(t +mτ )/τ u(t + mτ ) m→∞ = λt /τ lim λ−s/τ u(s) = 0. s→∞
(∆u) (t)/t l
Since limt →∞ = 0, we can take a constant K > 0 such that |(∆u)(t)| Kt l+1 for t T . Then (−1)i+1 −i l+1 H (t + iτ ) (∆u)(t + iτ ) λ K(t + iτ ) , i = 1, 2, . . . . i Therefore, letting m → ∞ in (26), we obtain u(t) =
∞ (−1)i+1 (∆u)(t + iτ ). Hi (t + iτ ) i=1
The equality t +iτ
(∆u) (s) ds,
(∆u)(t + iτ ) = (∆u)(t) + t
shows that t +iτ ∞ (−1)i+1 u(t) − S(t)(∆u)(t) = (∆u) (s) ds. Hi (t + iτ ) i=1
t
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Observe that t +iτ
∞ u(t) − S(t)(∆u)(t) λ−i
(∆u) (s) ds
i=1
=
∞
t −i
λ
∞ ∞
=
j =1
t +j τ
−i
λ
j =1 i=j ∞
(∆u) (s) ds
j =1 t +(j −1)τ
i=1
=
t +j τ
i
(∆u) (s) ds
t +(j −1)τ
λ−j 1 − λ−1
t +j τ
(∆u) (s) ds.
t +(j −1)τ
Since (s − t)τ −1 j (s − t)τ −1 + 1 for s ∈ [t + (j − 1)τ, t + j τ ], we have ∞
−j
t +j τ
(∆u) (s) ds
λ
j =1
t +(j −1)τ
t +j τ
∞
λ−(s−t )/τ (∆u) (s) ds
j =1 t +(j −1)τ
∞ =λ
t /τ
λ−s/τ (∆u) (s) ds.
t
We note here that λ−t /τ (∆u) (t) is integrable on [T , ∞), because of limt →∞ (∆u) (t)/t l = 0. Using the same argument as in the proof of Lemma 3.1, we conclude that
∞ t /τ λ λ−s/τ (∆u) (s) ds = o t l (t → ∞). t
This completes the proof. ✷
4. Proof of the “if” part of Theorem 1.1 In this section we prove the “if” part of Theorem 1.1, namely we show that if (1) has a nonoscillatory solution, then (4) has a nonoscillatory solution. Let x(t) be a nonoscillatory solution of (1). Then we easily see that (∆x)(t) = 0 and (∆x)(t)(∆x)(n)(t) 0 for all large t t0 . For such functions as (∆x)(t), we have the following well-known lemma of Kiguradze [13]. (See also [14].)
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183
Lemma 4.1. Let w ∈ C n [T , ∞) satisfy w(t) = 0 and w(t)w(n) (t) 0 for t T . Then there exist an integer k ∈ {1, 3, . . . , n − 1} and a number T1 T such that w(t)w(i) (t) > 0, 0 i k − 1, (27) (−1)i−k w(t)w(i) (t) 0, k i n, for t T1 . In particular, w (t) 0 for t T1 . A function w(t) satisfying (27) for all large t is called a function of Kiguradze degree k. Let w(t) be a function of Kiguradze degree k ∈ {1, 3, . . . , n − 1} satisfying w(t) > 0 for all large t. It can be shown (cf. [11–13,19]) that lim w(i) (t) = 0,
t →∞
i = k + 1, k + 2, . . . , n − 1,
and that one of the following three cases holds: lim w(k) (t) = const > 0
and
t →∞
lim w(k) (t) = 0
and
lim w(k) (t) = 0
and
t →∞ t →∞
lim w
t →∞
lim w(k−1) (t) = ∞;
t →∞ (k−1)
(t) = ∞;
lim w(k−1) (t) = const > 0.
t →∞
(28a) (28b) (28c)
If (28a) holds, then we put l = k, and if (28b) or (28c) holds, then we put l = k − 1. Then it is easy to verify that l ∈ {0, 1, 2, . . ., n − 1}, w (t) w(t) = 0 and lim l = L (L > 0 or L = ∞). (29) l t →∞ t t →∞ t From Lemmas 2.1, 3.1 and 3.2, and the above observation, we obtain the next result. lim
Lemma 4.2. Suppose that (H1) or (H2) holds. Let u ∈ C[T − τ, ∞) satisfy ∆u ∈ C n [T , ∞) and (∆u)(t) > 0 for t T . Assume, moreover, that limt →∞ λ−t /τ u(t) = 0 if (H2) holds. If (∆u)(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n − 1}, then there exist a constant α > 0 and an integer l ∈ {0, 1, 2, . . ., n − 1} such that (30) u(t) S(t) (∆u)(t) − αt l > 0 for all large t T . Proof. Recalling (29), we have (∆u) (t) (∆u)(t) = 0 and lim = L (L > 0 or L = ∞) t →∞ t →∞ tl tl for some l ∈ {0, 1, 2, . . . , n − 1}. Choose a constant α > 0 so small that α < L. By Lemmas 2.1, 3.1 and 3.2, we conclude that (30) holds. ✷ lim
Now let us prove the “if” part of Theorem 1.1.
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S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
Proof of the “if” part of Theorem 1.1. Let x(t) be a nonoscillatory solution of (1). Without loss of generality, we may assume that x(t) > 0 for all large t. Then (∆x)(t) > 0 and (∆x)(n) (t) 0 for all large t. By virtue of Lemma 4.1, we see that (∆x)(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n−1}. From Lemma 4.2, there are a constant α > 0 and an integer l ∈ {0, 1, 2, . . . , n − 1} such that x(t) S(t) (∆x)(t) − αt l > 0 for all large t. Set w(t) = (∆x)(t) − αt l . Then x(t) S(t)w(t) > 0 for all large t. We see that −w(n) (t) = −(∆x)(n)(t) = f t, x g(t) f t, S g(t) w g(t) for all large t. Lemma 2.2 implies that (4) has a nonoscillatory solution. The proof is complete. ✷
5. Proof of the “only if” part of Theorem 1.1 In this section we show the “only if” part of Theorem 1.1. To this end, we require the following mapping Φ, which is an “inverse” of the operator ∆: i ∞ i=0 (−1) Hi (t)v(t − iτ ) if (H1) holds, (Φv)(t) = ∞ (−1)i+1 v(t + iτ ) if (H2) holds, i=1 Hi (t +iτ ) for t ∈ R, where v ∈ v ∈ C(R): |v(t)| M max t k , 1 , t ∈ R ≡ V ,
(31)
M > 0 and k ∈ {1, 2, . . .}. The properties of the mapping Φ are as follows. Lemma 5.1. The mapping Φ is well-defined on V and has the following properties (i)–(iv): (i) (ii) (iii) (iv)
Φ maps V into C(R); Φ is continuous on V in the C(R)-topology; Φ satisfies (∆(Φv))(t) = v(t) for t ∈ R and v ∈ V ; if (H2) holds, then limt →∞ λ−t /τ (Φv)(t) = 0 for v ∈ V .
Here and hereafter, C(R) is regarded as the Fréchet space of all continuous functions on R with the topology of uniform convergence on every compact subinterval of R.
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185
Proof of Lemma 5.1. First we assume that (H1) holds. Let v ∈ V . We see that (−1)i Hi (t)v(t − iτ ) λi M max (t − iτ )k , 1 λi M max t k , 1 , i = 0, 1, 2, . . ., (32) ∞ for t ∈ R, so that the series i=0 (−1)i Hi (t)v(t − iτ ) converges uniformly on every compact subinterval of R. This implies that Φ is well-defined and Φ satisfies (i). It can be show that Φ satisfies (iii). In fact, h(t)(Φv)(t − τ ) − v(t) =
∞ (−1)i h(t)Hi (t − τ )v t − (i + 1)τ − v(t) i=0 ∞
=−
(−1)i+1 Hi+1 (t)v t − (i + 1)τ − v(t)
i=0
=−
∞
(−1)i Hi (t)v(t − iτ ) − v(t)
i=1
= −(Φv)(t),
t ∈ R, v ∈ V .
{vj }∞ j =1
be a sequence in V converging Now we prove that Φ satisfies (ii). Let to v ∈ V uniformly on every compact subinterval of R. It suffices to verify that Φvj converges to Φv uniformly on every compact subinterval of R. Take an arbitrary compact subinterval I of R. Let ε > 0. There is an integer q 1 such that ∞ ε (33) λi M max t k , 1 < , t ∈ I. 3 i=q+1
There exists an integer j0 1 such that q ε λi vj (t − iτ ) − v(t − iτ ) < , 3
t ∈ I, j j0 .
i=0
It follows from (32)–(34) that q (Φvj )(t) − (Φv)(t) λi vj (t − iτ ) − v(t − iτ )
∞ i (−1) Hi (t)vj (t − iτ ) + i=q+1 ∞ (−1)i Hi (t)v(t − iτ ) + i=q+1 ε ε + 2 · = ε, t ∈ I, j j0 , 3 3 which implies that Φvj converges Φv uniformly on I . i=0
(34)
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Now we suppose that (H2) holds. Let v ∈ V . Then (−1)i+1 −i k H (t + iτ ) v(t + iτ ) λ M max (t + iτ ) , 1 i λ−i M max 2k−1 t k + i k τ k , 1
(35)
for t ∈ R and i ∈ {1, 2, . . .}. Consequently, by the same arguments as in the case (H1), we can show that Φ is well-defined and satisfies (i)–(iii). It is easy to see that Φ satisfies (iv). Indeed, from (35), we have (Φv)(t) 2k−1 M S1 t k + S2 for t > 1, ∞ −i k −i and S = τ k where S1 = ∞ 2 i=1 λ i=1 λ i . The proof is complete. ✷ Now we prove the “only if” part of Theorem 1.1. Proof of the “only if” part of Theorem 1.1. We show that if (4) has a nonoscillatory solution, then (1) has a nonoscillatory solution. Let w(t) be a nonoscillatory solution of (4). We may assume that w(t) is eventually positive. Lemma 4.1 implies that w(t) is a function of Kiguradze degree k for some k ∈ {1, 3, . . . , n − 1}. Hence, (27) holds and one of the cases (28a)–(28c) is satisfied. We can take a sufficiently large number T > 1 such that w(i) (t) > 0 (i = 0, 1, 2, . . ., k − 1), w(g(t)) > 0 for t T . Integrating (4), we have
t w(t) − P (t) = T
(t − s)k−1 (k − 1)!
∞ s
(r − s)n−k−1 F (r) dr ds 0 (n − k − 1)!
(36)
for t T , where F (t) = f (t, S(g(t))w(g(t))), (t − T )i (t − T )k (k) w (∞) + w(i) (T ), k! i! k−1
P (t) =
i=0
= limt →∞ ∈ [0, ∞). Note that P (T ) = w(T ) and P (t) and w(T ) > 0 for t T . Consider the set Y of functions y ∈ C(R) which satisfies w(k) (∞)
y(t) = 0
w(k) (t)
for t T
and 0 y(t) w(t) − P (t)
for t T .
Then Y is closed and convex. Set (1/2)P (t), t T , η(t) = (1/2)P (T ), t < T . By w(k) (∞) ∈ [0, ∞), there is a constant M P (T ) such that w(t) Mt k and P (t) Mt k for t T . Define the set V by (31). We easily see that Y ⊂ V and η∈V.
S. Tanaka / J. Math. Anal. Appl. 273 (2002) 172–189
187
For each y ∈ Y , we define the mapping F : Y −→ C(R) as follows: t (t −s)k−1 ∞ (r−s)n−k−1 T (k−1)! s (n−k−1)! (F y)(t) = × f r, (Φy) g(r) + (Φη) g(r) dr ds, t T , 0, t T , where
F (t), u S g(t) w g(t) , f (t, u) = f (t, u), 0 u S g(t) w g(t) , 0, u 0,
for t T and u ∈ R. In view of (36) and the fact that 0 f (t, u) F (t),
(t, u) ∈ [T , ∞) × R,
we see that F is well-defined on Y and maps Y into itself. Since Φ is continuous on Y , by the Lebesgue dominated convergence theorem, we can show that F is continuous on Y as a routine computation. Now we claim that F (Y ) is relatively compact. We note that F (Y ) is uniformly bounded on every compact subinterval of R, because of F (Y ) ⊂ Y . By the Ascoli–Arzelà theorem, it suffices to verify that F (Y ) is equicontinuous on every compact subinterval of R. Let I be an arbitrary compact subinterval of [T , ∞). If k = 1, then
∞
0 (F y) (t) t
(s − t)n−2 F (s) ds, (n − 2)!
t T , y ∈ Y.
If k 3, then
t
0 (F y) (t) T
(t − s)k−2 (k − 2)!
∞ s
(r − s)n−k−1 F (r) dr ds (n − k − 1)!
for t T and y ∈ Y . Thus we see that {(F y) (t): y ∈ Y } is uniformly bounded on I . The mean value theorem implies that F (Y ) is equicontinuous on I . Since |(F y)(t1 ) − (F y)(t2 )| = 0 for t1 , t2 ∈ (−∞, T ], we conclude that F (Y ) is equicontinuous on every compact subinterval of R. By applying the Schauder–Tychonoff fixed point theorem to the operator F , there exists a y˜ ∈ Y such that y˜ = F y. ˜ Put x(t) = (Φ y)(t) ˜ + (Φη)(t). Then we obtain P (t) P (t) = (F y)(t) ˜ + > 0, t T , (37) 2 2 by (iii) of Lemma 5.1, and hence (∆x)(t) is a function of Kiguradze degree k. (∆x)(t) = y(t) ˜ +
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We will show that 0 < x(t) S(t)w(t)
for all large t.
(38)
Then the proof of Theorem 1.1 will be complete, since (37) and (38) imply that dn x(t) + h(t)x(t − τ ) = (F y) ˜ (n) (t) = −f¯ t, x g(t) n dt = −f t, x g(t) for all large t, which means that x(t) is a nonoscillatory solution of (1). If w(k) (∞) > 0, then we put l = k, and if w(k) (∞) = 0, then we put l = k − 1. By (37), we find that limt →∞ (∆x)(k) (t) = (1/2)w(k) (∞), so that limt →∞ (∆x) (t)/t l = 0. From Lemma 4.2 it follows that x(t) > 0 for all large t T . In view of Lemmas 2.1, 3.1 and 3.2, and the fact that limt →∞ P (t)/t l = const > 0, we have P (t) x(t) (∆x)(t) + S(t) 2 for all large t. Hence, by (37), we obtain P (t) P (t) P (t) P (t) x(t) y(t) ˜ + + w(t) − P (t) + + = w(t) S(t) 2 2 2 2 for all large t. This completes the proof. ✷
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