show me steps of the determine green and all information is here explain to me step by step i need every details Theorem 3.1. The system (4) has a prime period-two solutions if and only if? +€-1+ 0,(22)7is true.Proof. First, assume that the system (4) has a distinct prime period-two solution in the follow-ing form..., (a1, B1), (a2, B2), (a1, B1), (a2, B2), ...where a1 + a2 and B1 + B2. Then, from (11) and (12), we obtain(a1 + a2) (B1 + B2)B2 – €(a1 + a2) (B1 + B2)B1 – €(a1 + a2) (B1 + B2)(23)a2 =(24)+ €,(25)a2 - H(a1 + a2) (B1 + B2)B2+ E.(26)By some calculations from equations (23)-(26), we get2με (ε-1)e2 + € – 1µ? (e? - (e+1)) (e² +e – 1)(e? + € – 1)2ai + a2 =and a1 a2 =B1 + B2 = 1and B1 B2 = 0,hence, we see that (a1 + a2)² – 4a1 a2 # 0 and (B1 + B2)2 – 4 B1 B2 # 0, which implies thata1 7 a2 and B1 # B2. Thus,µ (e² – (e + 1))e2 + € – 1µ (e? – e+ 1)e2 + € – 1, &2 =B1 = 0 and B2 = 1.(27)Clearly a; exist if and only if e²+e-1 7 0 and they are positive if e > 1 for i = 1, 2. Secondly,suppose that e²+ € – 1 7 0 is satisfied. Then, the system (4) has a prime period two solution(az, 61) = (4H(2² – (e + 1))e2 + € – 1H (e? – e + 1) ,1)and (a2, B2)e2 + € – 1Now, we show that (w1, z1) = (a1, B1). From (4), we have(wo + w-1) (z-1+2-2)Wi20 – €(a1 + a2) (B1 + B2)B2 - €(2εμ(ε -+ µH,1+ l,e2 + € – 12εμ(ε-1)+μ (1-) (2+-1)(1 – e) (e² + e – 1)8 In this paper, we solve and study the properties of the following systemWn-pZn-hWn-pZn-hp=0h=1p=1h=0Wn+1+andZn+1+ €,(4)%3DZn - €Wn - Lwhere u ande are arbitrary positive real numbers with initial conditions w; and z; for i =-2, –1,0.Let w2n-1 = Un, w2n-2 = Vn, 22n-1=Хn and z2m-2—Yn, implies thatVn+1 – ko Un – ko Vn = µ,(11)Un+1 – ki Vn+1 – kį Un = H,andYn+1 - Vo Xn – v Yn = e,(12)Xn+1 – Vị Yn+1 - Vị Xn = €.It is observe that, the equations (11) and (12) are system of linear non-homogeneous differenceequations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is(E - ko) Vn – ko Un(E – ki ) Un – ki E V,(13)l,(14)where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko,respectively,(E – ki) (E – ko) Vn – (E – ki) ko Un(E – k1) µ,(15)ko (E – ki) Un – ko ki E V,ko H.(16)Thus, the combination of (15) and (16) is(E – (ko + ki + ko k1) E + ko ki) Va = (1– kı + ko) µ,(E² – Ø E + ko ki) V = (1– kị + ko) H,(17)where ø = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it iseasy to obtain the solution,0 + Vo? – 4 ko kı(1– k + ko- k14 ko kiVn = A+ B(18)2- ko -since A and B are constants. By Substituting (18) in (13), we give1= Ako$² – 4 ko k1- 1$+ V62 - 4 ko kiUn22 – 4 ko ki16² – 4 ko ki(19)+ B2• ((-) ( .- ki + ko- ko – kikoko6

Given: From the equation (23) to (26) we get, α1+α2=2μεε-1ε2+ε-1 and α1α2=μ2ε2-ε+1ε2+ε-1ε2+ε-12…

Theorem 3.1. The system (4) has a prime period-two solutions if and only if ? +€ -1 + 0, (22) 7 is true. Proof. First, assume that the system (4) has a distinct prime period-two solution in the follow- ing form ., (a1, B1), (a2, B2), (a1, 31), (a2, B2), ... where a + a2 and B1 + B2. Then, from (11) and (12), we obtain (@1 + a2) (B1 + B2) B2 – € (a1 + a2) (B1 + B2) B1 - € (@1 + a2) (B1 + B2) + Hy (23) a2 = + Hy (24) + €, (25) a2 – µ (a1 + a2) (B1 + B2) B2 = + €. (26) By some calculations from equations (23)-(26), we get 2 µ e (e – 1) e2 + € – 1 µ² (e? – (e + 1)) (e² +e – 1) (e² + € – 1)² - a1 + a2 = and a1 ¤2 = B1 + B2 = 1 and B1 B2 = 0, hence, we see that (a1 + a2)² – 4 a1 a2 # 0 and (ß1 + B2)² – 4 B1 32 # 0, which implies that a1 # a2 and ß1 # B2. Thus, µ (e? – (e + 1)) H (e² – e + 1) e2 + € – 1 , d2 = B1 = 0 and 6B2 = 1. (27) e2 + € – 1 Clearly a; exist if and only if e2 +e-1#0 and they are positive if e > 1 for i = 1, 2. Secondly, suppose that e² + € – 1 7 0 is satisfied. Then, the system (4) has a prime period two solution (e1, 5) = (" - (-+ 1),) and (a2, 52) = ("-+D). e2 + € – e2 + € – 1 Now, we show that (w1, z1) (a1, B1). From (4), we have (wo+ w_1) (z–1 + z-2) Wi %3D 20 – € (a1 + a2) (B1 + B2) + H, B2 - € (2εμ(ε-1) 1- € 1 + µ, e? + € – 1 2 e µ(e – 1) + µ (1 – e) (e² + e – 1) (1 – e) (e² + e – 1) 8

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ