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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 (2.1) "s+1Ds+1 = . n=N Un s=n Then: Zn (i) {} is decreasing for all n> N; bn (ii) "} is decreasing for all n> N; 1/a Zn (iii) -} is increasing for all n> N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z,} E S2 for all n > N. Since a„A(b,(Azn)ª) is decreasing, we have n-1 b,(Azn)ª > £ a,A(b,(Azs)“) > ApdnA(b,(Azn)“), n>N. s=N as From the last inequality, we obtain a 1 (b,(Azn)ª` A,A(b,(Atn)ª) – ,(Azn)ª Un An A„An+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 (2.1) "s+1Ds+1 = . n=N Un s=n Then: Zn (i) {} is decreasing for all n> N; bn (ii) "} is decreasing for all n> N; 1/a Zn (iii) -} is increasing for all n> N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z,} E S2 for all n > N. Since a„A(b,(Azn)ª) is decreasing, we have n-1 b,(Azn)ª > £ a,A(b,(Azs)“) > ApdnA(b,(Azn)“), n>N. s=N as From the last inequality, we obtain a 1 (b,(Azn)ª` A,A(b,(Atn)ª) – ,(Azn)ª Un An A„An+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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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 = o. (2.1) s+1 n=N An s=n Then: Zn -} is decreasing for all n> N; 1/a (ii) 'Azn “} 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 {z,} E S2 for all n > N. Since a„A(bn(Azn)¤) is decreasing, we have n-1 a,A(b,(Azs)“) b,(Azn)ª > E > A,a,A(b,(Azn)“), n>N. as s=N From the last inequality, we obtain b(Azn) \ A A„A(b,(Azn)ª) – b,(Azn)ª1 an <0 An A„An+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 = o. (2.1) s+1 n=N An s=n Then: Zn -} is decreasing for all n> N; 1/a (ii) 'Azn “} 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 {z,} E S2 for all n > N. Since a„A(bn(Azn)¤) is decreasing, we have n-1 a,A(b,(Azs)“) b,(Azn)ª > E > A,a,A(b,(Azn)“), n>N. as s=N From the last inequality, we obtain b(Azn) \ A A„A(b,(Azn)ª) – b,(Azn)ª1 an <0 An A„An+1
/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
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