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Theorem 1. If condition (2.1) holds and Ca n+1 2 q,E+B1 } > 1, Ba 00 1 lim sup (2.5) An+1 s+' s+1Ps+1 s=N n+1 s=n+1 then S2 = Ø. Proof. Let {yn} be a positive solution of equation (1.1), with the corresponding sequence {zn} € S2 for all n > N. Using (2.4) in equation (1.1), we obtain CA(a,A(b, (Azm)*))+q„E <0, n2N. (2.6) n+1 E „a (2.7) - as s=N From (2.7), it follows that ba(Azn)°> と t=Ds n-1 1 n-1 1 と-とqE14+と-24E++1 | as as s=N t=S s=N t=n n-1 E 9,EAs+11+An E s+1

Theorem 1. If condition (2.1) holds and Ca n+1 2 q,E+B1 } > 1, Ba 00 1 lim sup (2.5) An+1 s+' s+1Ps+1 s=N n+1 s=n+1 then S2 = Ø. Proof. Let {yn} be a positive solution of equation (1.1), with the corresponding sequence {zn} € S2 for all n > N. Using (2.4) in equation (1.1), we obtain CA(a,A(b, (Azm)*))+q„E <0, n2N. (2.6) n+1 E „a (2.7) - as s=N From (2.7), it follows that ba(Azn)°> と t=Ds n-1 1 n-1 1 と-とqE14+と-24E++1 | as as s=N t=S s=N t=n n-1 E 9,EAs+11+An E s+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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Theorem 1. If condition (2.1) holds and Ca 00 1 n+1 Σ > 1, lim sup s+1Cs+jAs+1+ Ca Ba 'n+1 s=n+1 s+1Ps+1 (2.5) An+1 s=N n00 then S2 = Ø. Proof. Let {yn} be a positive solution of equation (1.1), with the corresponding sequence {zn} € S2 for all n > N. Using (2.4) in equation (1.1), we obtain CA(a,A(b,(Azn)“))+q„E+1+ <0, n>N. (2.6) Summing inequality (2.6) from n to o, we have n-1 00 1 A(b, (Azn)“) > E ; (2.7) s=N Us t=s From (2.7), it follows that n-1 1 00 b„(Azn)“ > E÷ (Eq.E as s=N t3Ds n-1 n-1 n-1 1 1 Fa „a s=N Us t=s s=N as t=n n-1 E 9,EAs+12+1+An L 9sEs+15+1• s=N s=n Using (2.2), in the last inequality, we obtain An+1%+1 n >2 9,As+1Es+1<s+1 z" +An+1 2 (2.8) Ps+1<s+1• Ca ´n+1 s=N s=n+1
Transcribed Image Text:Theorem 1. If condition (2.1) holds and Ca 00 1 n+1 Σ > 1, lim sup s+1Cs+jAs+1+ Ca Ba 'n+1 s=n+1 s+1Ps+1 (2.5) An+1 s=N n00 then S2 = Ø. Proof. Let {yn} be a positive solution of equation (1.1), with the corresponding sequence {zn} € S2 for all n > N. Using (2.4) in equation (1.1), we obtain CA(a,A(b,(Azn)“))+q„E+1+ <0, n>N. (2.6) Summing inequality (2.6) from n to o, we have n-1 00 1 A(b, (Azn)“) > E ; (2.7) s=N Us t=s From (2.7), it follows that n-1 1 00 b„(Azn)“ > E÷ (Eq.E as s=N t3Ds n-1 n-1 n-1 1 1 Fa „a s=N Us t=s s=N as t=n n-1 E 9,EAs+12+1+An L 9sEs+15+1• s=N s=n Using (2.2), in the last inequality, we obtain An+1%+1 n >2 9,As+1Es+1<s+1 z" +An+1 2 (2.8) Ps+1<s+1• Ca ´n+1 s=N s=n+1
In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Az,)") +9ny+1 =0, n2 no 2 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: (Hi) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pm} is a nonnegative real sequence with 0< Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) > n for alln> no: (H4) Eno = +0o and E-no Va = +o0, %3D %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 9sEs+1Bs+1 = 00, Ea pa (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; Cn i/a va"} is decreasing for all n> N; (ii) Zn (iii) } is increasing for all n2 N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E Sz for all n> N. Since anA(b, (Azn)) is decreasing, we have n-1 ba(Azn)ª > 4,A(b:(Az.)ª) > AnanA(bn(Azn)“), n2N. as s=N From the last inequality, we obtain A„A(b,(Azn)a) – bn(Azn)ª+ %3D An A,An+1 Lemma 3. Assume that (2.1) holds and let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} € S2 for all n> N. Then Enzn S yn S Zn for n2 N. (2.4) Proof. From the definition of Zn, we have zn Yn for all n> N. Furthermore, Yn 2 Zn - PnZa(n) 2 Enzn where we have used the fact that {} is decreasing for all n>N. This completes the proof of the lemma. Next, we obtain sufficient condition for all nonoscillatory solutions of (1.1) to be of the Kneser type. for all n >N> no. Thus, { } is decreasing for all n> N, so (ii) holds and An 1/a AZzn 1/a 1/a Ca, n2N. (2.2) s=N Hence,
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,(Az,)") +9ny+1 =0, n2 no 2 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: (Hi) {an}, {bn}, and {qn} are positive real sequences for all n> no; (H2) {Pm} is a nonnegative real sequence with 0< Pn Sp< 1; (H3) {o(n)} is a sequence of integers such that o(n) > n for alln> no: (H4) Eno = +0o and E-no Va = +o0, %3D %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 9sEs+1Bs+1 = 00, Ea pa (2.1) n=N Un s=n Then: (i) {} is decreasing for all n> N; Cn i/a va"} is decreasing for all n> N; (ii) Zn (iii) } is increasing for all n2 N. Bn Proof. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} E Sz for all n> N. Since anA(b, (Azn)) is decreasing, we have n-1 ba(Azn)ª > 4,A(b:(Az.)ª) > AnanA(bn(Azn)“), n2N. as s=N From the last inequality, we obtain A„A(b,(Azn)a) – bn(Azn)ª+ %3D An A,An+1 Lemma 3. Assume that (2.1) holds and let {yn} be a positive solution of equation (1.1) with the corresponding sequence {zn} € S2 for all n> N. Then Enzn S yn S Zn for n2 N. (2.4) Proof. From the definition of Zn, we have zn Yn for all n> N. Furthermore, Yn 2 Zn - PnZa(n) 2 Enzn where we have used the fact that {} is decreasing for all n>N. This completes the proof of the lemma. Next, we obtain sufficient condition for all nonoscillatory solutions of (1.1) to be of the Kneser type. for all n >N> no. Thus, { } is decreasing for all n> N, so (ii) holds and An 1/a AZzn 1/a 1/a Ca, n2N. (2.2) s=N Hence,
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