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bAzn is decreasing for all n> N; 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 2 0, (1.1) where zn = yn + PnYo(n), ¤ 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 n for all n> no; ( H4) Σno a = +00 and Ln=no = +00. 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 9sEs+1Bs+1 =. (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; 1/a (ii) { 1/a An Zn (iii) {} is increasing for all n 2 N. Bn Proof. Let {Yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E S2 for all n> N. Since a,A(b,(Azn)ª) is decreasing, we have n-1 ba(Azn)ª > 4,A(b.(Az.)ª) > AņanA(bn(Azn)ª), n2N. s=N as From the last inequality, we obtain bn(Azn)ª A„A(b,(Azn)ª) – bn(Azn)ª an <0 An AnAn+1

bAzn is decreasing for all n> N; 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 2 0, (1.1) where zn = yn + PnYo(n), ¤ 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 n for all n> no; ( H4) Σno a = +00 and Ln=no = +00. 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 9sEs+1Bs+1 =. (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; 1/a (ii) { 1/a An Zn (iii) {} is increasing for all n 2 N. Bn Proof. Let {Yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E S2 for all n> N. Since a,A(b,(Azn)ª) is decreasing, we have n-1 ba(Azn)ª > 4,A(b.(Az.)ª) > AņanA(bn(Azn)ª), n2N. s=N as From the last inequality, we obtain bn(Azn)ª A„A(b,(Azn)ª) – bn(Azn)ª an <0 An AnAn+1

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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/d Azn is decreasing for all n> N; 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 <p< 1; (H3) {o(n)} is a sequence of integers such that o(n) 2n for all n2 no; (H4) Σ. = +00 and E=no Va = +00, %3D 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 = 0, (2.1) n=N An S=n Then: (i) {} is decreasing for all n > N; ug (ii) { 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)" > 4;A(b,(Azs)ª) > AnanA(bn(Azn)“), n2N. as s=N From the last inequality, we obtain A„A(b,(Azn)") – bn(Azn)ª1 a(ba(Ača)“) = AnA(b,(Az,)ª) – b.(Az.)ª! An AnAn+1
Transcribed Image Text:/d Azn is decreasing for all n> N; 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 <p< 1; (H3) {o(n)} is a sequence of integers such that o(n) 2n for all n2 no; (H4) Σ. = +00 and E=no Va = +00, %3D 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 = 0, (2.1) n=N An S=n Then: (i) {} is decreasing for all n > N; ug (ii) { 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)" > 4;A(b,(Azs)ª) > AnanA(bn(Azn)“), n2N. as s=N From the last inequality, we obtain A„A(b,(Azn)") – bn(Azn)ª1 a(ba(Ača)“) = AnA(b,(Az,)ª) – b.(Az.)ª! An AnAn+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 b,(Azn)“ > E aşA(b,(Azs)“) > AnanA(bn(Azn)"), n>N. s=N as From the last inequality, we obtain AnA(bn(Azn)") – a(Azn)“ 1 bn(Azn)ª An A„An+1 ba(Azn)"} for all n>N > no. Thus, is decreasing for all n > N, so (ii) holds and An n-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 CrAzn - Zna <0, Zn C„Cn+1 which implies that {} is decreasing for all n > N, so (i) holds. 1/a Since b,"Az, is positive and strictly increasing for any n >N, it is easy to see that for all n> N1 >N, n-1 1/a Azn Zn < ZN, +bn 1 Σ 1/a s=N bs N-1 1 = ZN - bi"Azn £ s=N b!/a 1/a 1/a 1 + bn"Azn (2.3) s=N b/a 1/a 1/a We claim that b Azn → oo as n → 0, If this is not the case, then bn"Azn → 2d <00 as n→ o From the definition of Zn and using the fact that { } is decreasing, we have 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 b,(Azn)“ > E aşA(b,(Azs)“) > AnanA(bn(Azn)"), n>N. s=N as From the last inequality, we obtain AnA(bn(Azn)") – a(Azn)“ 1 bn(Azn)ª An A„An+1 ba(Azn)"} for all n>N > no. Thus, is decreasing for all n > N, so (ii) holds and An n-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 CrAzn - Zna <0, Zn C„Cn+1 which implies that {} is decreasing for all n > N, so (i) holds. 1/a Since b,"Az, is positive and strictly increasing for any n >N, it is easy to see that for all n> N1 >N, n-1 1/a Azn Zn < ZN, +bn 1 Σ 1/a s=N bs N-1 1 = ZN - bi"Azn £ s=N b!/a 1/a 1/a 1 + bn"Azn (2.3) s=N b/a 1/a 1/a We claim that b Azn → oo as n → 0, If this is not the case, then bn"Azn → 2d <00 as n→ o From the definition of Zn and using the fact that { } is decreasing, we have 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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