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In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)+ 9ny+1 = 0, n> no 20, (1.1) where zn = yn + Pnyo(n), a is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pn} is a nonnegative real sequence with 0 < Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) >n for all n > no; +0o and Ln=no Va =+00, (H4) En=no un b

In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)+ 9ny+1 = 0, n> no 20, (1.1) where zn = yn + Pnyo(n), a is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pn} is a nonnegative real sequence with 0 < Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) >n for all n > no; +0o and Ln=no Va =+00, (H4) En=no un b

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)) +9ny%+1 =0, n> no > 0, (1.1) where zn = yn + PnYo(n), a is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pn} is a nonnegative real sequence with 0 < Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) >n for all n > no; (H4) En=no dn = +0o and En=no Va '00+ = D/1
Transcribed Image Text:In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)) +9ny%+1 =0, n> no > 0, (1.1) where zn = yn + PnYo(n), a is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pn} is a nonnegative real sequence with 0 < Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) >n for all n > no; (H4) En=no dn = +0o and En=no Va '00+ = D/1
Lemma 2. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E S2 for n > N > no and assume that 1 E-Eq,E+1B5+1= 0. (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; *Azn (ii) { } is decreasing for all n> N; Zn (iii) {} is increasing for all n >N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} € S2 for all n > N. Since a,A(b,(Azn)") is decreasing, we have n-1 ba(Azn)“ > E aşA(b,(Azs)“) > AnanA(bn(Azn)"), n>N. s=N as From the last inequality, we obtain ´bn(Azn)ª AnA(bn(Azn)") – a(Azn)“ 1 An A„An+1 for all n>N > no. Thus, ba(Azn)"} is decreasing for all n > N, so (ii) holds and An п-1 1/a1/a Azs bn 1/a Azn Zn 2 E Cn, n N. 1/a An (2.2) 1/a AS s=N Hence, 1/a (2) CrAzn - Zna <0, Zn C„Cn+1 which implies that {} is decreasing for all n> N, so (i) holds. 1/a Since bi"Azn is positive and strictly increasing for any n >N, it is easy to see that for all n > N1 >N, n-1 1 Zn < ZN, + BCAZ. Σ 1/a s=N bs 1/a N-1 1 = ZNI - bi"Azn E s=N b!/a 1/a 1/a 1 Azn (2.3) s=N b/a 1/a We claim that b, as n→ o From the definition of Zn and using the fact that { } is decreasing, we have Azn → 00 as n→ 0, If this is not the case, then b,Az, → 2d < ∞ Co(n) Yn 2 Zn 1- Pn Cn = Enzn. Summing equation (1.1) from n to o and using the last inequality, we obtain 1 A(bn(Azn)ª) >-Ë An s=n /a Now b"Azn → 2d as n → o implies b"Azn > d for n large enough, which in turn /a implies zn > dBn. Combining the last two inequalities and summing once more, we obtain 1 (2d)ª > dª E £ n=N, an Ps+1, S=n which contradicts (2.1). Thus, b Hence, in view of (2.3) and (H4), we obtain Azn → 00 as n → 00 as we claimed. Zn S b/a From the last inequality, we see that "AznBn, n2 N. B„Azn Zn >0 Bn B„Bn+1
Transcribed Image Text:Lemma 2. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E S2 for n > N > no and assume that 1 E-Eq,E+1B5+1= 0. (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; *Azn (ii) { } is decreasing for all n> N; Zn (iii) {} is increasing for all n >N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} € S2 for all n > N. Since a,A(b,(Azn)") is decreasing, we have n-1 ba(Azn)“ > E aşA(b,(Azs)“) > AnanA(bn(Azn)"), n>N. s=N as From the last inequality, we obtain ´bn(Azn)ª AnA(bn(Azn)") – a(Azn)“ 1 An A„An+1 for all n>N > no. Thus, ba(Azn)"} is decreasing for all n > N, so (ii) holds and An п-1 1/a1/a Azs bn 1/a Azn Zn 2 E Cn, n N. 1/a An (2.2) 1/a AS s=N Hence, 1/a (2) CrAzn - Zna <0, Zn C„Cn+1 which implies that {} is decreasing for all n> N, so (i) holds. 1/a Since bi"Azn is positive and strictly increasing for any n >N, it is easy to see that for all n > N1 >N, n-1 1 Zn < ZN, + BCAZ. Σ 1/a s=N bs 1/a N-1 1 = ZNI - bi"Azn E s=N b!/a 1/a 1/a 1 Azn (2.3) s=N b/a 1/a We claim that b, as n→ o From the definition of Zn and using the fact that { } is decreasing, we have Azn → 00 as n→ 0, If this is not the case, then b,Az, → 2d < ∞ Co(n) Yn 2 Zn 1- Pn Cn = Enzn. Summing equation (1.1) from n to o and using the last inequality, we obtain 1 A(bn(Azn)ª) >-Ë An s=n /a Now b"Azn → 2d as n → o implies b"Azn > d for n large enough, which in turn /a implies zn > dBn. Combining the last two inequalities and summing once more, we obtain 1 (2d)ª > dª E £ n=N, an Ps+1, S=n which contradicts (2.1). Thus, b Hence, in view of (2.3) and (H4), we obtain Azn → 00 as n → 00 as we claimed. Zn S b/a From the last inequality, we see that "AznBn, n2 N. B„Azn Zn >0 Bn B„Bn+1
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