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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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(a1 + a2) (B1 + B2) (23) B2 – € (a1 + a2) (B1 + B2) + µ, (24) B1 – € (a1 + a2) (B1 + B2) + €, (25) B1 a2 – l (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 + € – 1) (e2 + € – 1)2 ai + a2 = and aj a2 = B1 + B2 = 1 and B1 B2 = 0, hence, we see that (a1 + a2)² – 4a1 a2 + 0 and (B1 + B2)² – 4 B1 B2 7 0, which implies that a1 7 a2 and B1 # B2. Thus, µ (e? – (e + 1)) µ (e? – e + 1) B1 0 and B2 = 1. (27) e2 + € – 1 e2 + € – 1 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 e? + € – 1+ 0 is satisfied. Then, the system (4) has a prime period two solution 4 (e² – (e + 1)) e? + € – 1 and (a2, 32) = ("( –-e+1) ;) (a1, B1) €2 + € – 1 Now, we show that (w1, z1) = (a1, B1). From (4), we have (wo + w_1) (z–1 + 2-2) W1 20 - € (a1 + a2) (B1 + B2) B2 - € (2εμ(ε-1) 1- € e2 + €– 1 2 e µ(e – 1) + µ (1 – €) (e² + € – 1) (1 – e) (e? + € – 1) 8 µ(1 – e)(-2 € + e? +e – 1) (1 – €) (e² + e – 1) µ(e? – e – 1) (e? + € – 1) ' and (w-1 + w-2) (2o + z–1) 21 + €, wo – H (a1 + a2) (B1 + B2) + €, a2 - l (2 ε με- + 1 a2 - µ e2 + € – 1 e? + € –1 (2 εμ(ε -1) + €, -2μ (ε = 1) , -e +€ = B1. €2 + € – 1 Similarly, we can prove (w2, z2) are equal to (a2, B2). Thus, we conclude that the system (4) has a prime period two solution if and only if e² + € – 1 7 0. || ||
Transcribed Image Text:(a1 + a2) (B1 + B2) (23) B2 – € (a1 + a2) (B1 + B2) + µ, (24) B1 – € (a1 + a2) (B1 + B2) + €, (25) B1 a2 – l (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 + € – 1) (e2 + € – 1)2 ai + a2 = and aj a2 = B1 + B2 = 1 and B1 B2 = 0, hence, we see that (a1 + a2)² – 4a1 a2 + 0 and (B1 + B2)² – 4 B1 B2 7 0, which implies that a1 7 a2 and B1 # B2. Thus, µ (e? – (e + 1)) µ (e? – e + 1) B1 0 and B2 = 1. (27) e2 + € – 1 e2 + € – 1 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 e? + € – 1+ 0 is satisfied. Then, the system (4) has a prime period two solution 4 (e² – (e + 1)) e? + € – 1 and (a2, 32) = ("( –-e+1) ;) (a1, B1) €2 + € – 1 Now, we show that (w1, z1) = (a1, B1). From (4), we have (wo + w_1) (z–1 + 2-2) W1 20 - € (a1 + a2) (B1 + B2) B2 - € (2εμ(ε-1) 1- € e2 + €– 1 2 e µ(e – 1) + µ (1 – €) (e² + € – 1) (1 – e) (e? + € – 1) 8 µ(1 – e)(-2 € + e? +e – 1) (1 – €) (e² + e – 1) µ(e? – e – 1) (e? + € – 1) ' and (w-1 + w-2) (2o + z–1) 21 + €, wo – H (a1 + a2) (B1 + B2) + €, a2 - l (2 ε με- + 1 a2 - µ e2 + € – 1 e? + € –1 (2 εμ(ε -1) + €, -2μ (ε = 1) , -e +€ = B1. €2 + € – 1 Similarly, we can prove (w2, z2) are equal to (a2, B2). Thus, we conclude that the system (4) has a prime period two solution if and only if e² + € – 1 7 0. || ||
E Wn-p Wn-p Zn-h Zn-h h=1 p=1 +u and zn+1 = p=0 h=0 Wn+1 + €, (4) Zn - € Wn - l where u and e 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 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = 4, (11) Un+1 – ki Vn+1 – kị Un = µ, and Yn+1 - vo Xn - vo Yn = €, Xn+1 – Vi Yn+1 – vi Xn (12) = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko U, (E – ki) Un – kį E V, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – kı) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko Un (E – ki) µ, (15) ko (E – k1) Un – ko ki E Va ko H. (16) Thus, the combination of (15) and (16) is (E² – (ko + k1 + ko ki) E + ko ki) Vn = (1 – kị + ko) µ, (E² – ¢ E + ko ki) Vn = (1 – k1 + ko) µ, (17) where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, $ + Vo2 – 4 ko k1 $ - Vo2 – 4 ko ki + B (1-k+ ko + (18) Vn = A 2 1- ko – - k1, since A and B are constants. By Substituting (18) in (13), we give 2 – 4 ko ki 0 + Vo? – 4 ko ki 1 Un = A ko 2 -4 ko k1 $ - Vø2 – 4 ko k1 1 (19) + B 2 (-) ()) - - ki + ko - ko – k1 + ko ko 6
Transcribed Image Text:E Wn-p Wn-p Zn-h Zn-h h=1 p=1 +u and zn+1 = p=0 h=0 Wn+1 + €, (4) Zn - € Wn - l where u and e 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 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = 4, (11) Un+1 – ki Vn+1 – kị Un = µ, and Yn+1 - vo Xn - vo Yn = €, Xn+1 – Vi Yn+1 – vi Xn (12) = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko U, (E – ki) Un – kį E V, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – kı) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko Un (E – ki) µ, (15) ko (E – k1) Un – ko ki E Va ko H. (16) Thus, the combination of (15) and (16) is (E² – (ko + k1 + ko ki) E + ko ki) Vn = (1 – kị + ko) µ, (E² – ¢ E + ko ki) Vn = (1 – k1 + ko) µ, (17) where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, $ + Vo2 – 4 ko k1 $ - Vo2 – 4 ko ki + B (1-k+ ko + (18) Vn = A 2 1- ko – - k1, since A and B are constants. By Substituting (18) in (13), we give 2 – 4 ko ki 0 + Vo? – 4 ko ki 1 Un = A ko 2 -4 ko k1 $ - Vø2 – 4 ko k1 1 (19) + B 2 (-) ()) - - ki + ko - ko – k1 + ko ko 6
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