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 1, 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) + H, (23) + H, (24) + e, (25) a2 - 4 (a1 + a2) (B1 + B2) B2 = +E. (26) By some calculations from equations (23)-(26), we get 2 µe (e – 1) e2 + € – 1 µ² (e? – (e + 1)) (² +e – 1) (e? + € – 1)2 a1 + a2 = and a a2 = B1 + B2 = 1 and B1 B2 = 0, hence, we see that (a1 + a2)² – 4 a1 a2 # 0 and (B1 + B2)² – 4 B1 B2 # 0, which implies that a1 7 a2 and B1 7 B2. Thus, µ (e² – (e + 1)) e2 + € – 1 µ (e² – e + 1) B1 = 0 and B2 = 1. (27) %3D e2 + € – 1
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 1, 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) + H, (23) + H, (24) + e, (25) a2 - 4 (a1 + a2) (B1 + B2) B2 = +E. (26) By some calculations from equations (23)-(26), we get 2 µe (e – 1) e2 + € – 1 µ² (e? – (e + 1)) (² +e – 1) (e? + € – 1)2 a1 + a2 = and a a2 = B1 + B2 = 1 and B1 B2 = 0, hence, we see that (a1 + a2)² – 4 a1 a2 # 0 and (B1 + B2)² – 4 B1 B2 # 0, which implies that a1 7 a2 and B1 7 B2. Thus, µ (e² – (e + 1)) e2 + € – 1 µ (e² – e + 1) B1 = 0 and B2 = 1. (27) %3D e2 + € – 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the steps of determine yellow and all information is here
Transcribed Image Text: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, B1), (a2, B2), ...
where a1 + a2 and B1 + B2. Then, from (11) and (12), we obtain
(a1 + a2) (B1 + B2)
+ H,
B2 – €
(23)
(a1 + a2) (B1 + B2)
B1 – €
(a1 + a2) (B1 + B2)
X2 =
(24)
+ €,
(25)
X2 - l
(a1 + a2) (B1 + B2)
+ €.
(26)
By some calculations from equations (23)-(26), we get
2 με (ε-1)
€2 + € – 1
µ? (e? – (e + 1)) (e² +€ – 1)
(e2 + € – 1)2
ai + a2 =
and α1 α2
B1 + B2 = 1
and B1 B2 = 0,
hence, we see that (a1 + a2)² – 4 a1 a2 # 0 and (B1 + B2)² – 4 B1 B2 # 0, which implies that
ai 7 az and B1 # B2. Thus,
u (e? – (e + 1))
e? + € – 1
µ (e? – e + 1)
e2 + € – 1
, a2 =
B1 = 0 and B2 = 1.
(27)
Transcribed Image Text:In this paper, we solve and study the properties of the following system
Wn-p 2 Zn-h
p=0
Wn-p
h=0
E zn-h
h=1
p=1
Wn+1
+u and zn+1 =
+e,
(4)
Zn - €
Wn - 4
where u and e are arbitrary positive real numbers with initial conditions w; and z; for i =
-2, –1,0.
Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then
6+ V – 4 ko k1
A
- V2 – 4 ko ki
1- ki + ko
+
w2n-2
+B
Ty – Oy -
- V2 - 4 ko ki
-4 ko ki
+ V2 – 4 ko ki
V2 - 4 ko ki
A
ko
+ B
ko
w2n-1
- 1
2
2
•((금-1) (는)
1– ko + k1
+ V2 - 4 vo V1
- V2 - 4 VI
1- v + vo
22n-2
+D
%3D
1
+ Vu2 – 4 o VI
gb + V2 – 4 vo vỊ
1
1.
+D
- V2 – 4 v vỊ
V2 – 4 vo vi
22n-1
- v + vo
1- 0 - VI
where A, B, C and D are constants defined as
wo -H
2-1 +z-2
ko
ki =
0 = ko + ki + ko ki,
w-1 + w-2
20 - e
20 - e
w-1 +w-2
, = vo + vi + vo V1,
2-1 +2-2
1-Om
Vo2 – 4 ko k1
-)") (v - 4ko ki – 4) + 2 (wo - (G ) -).
2 62 - 8 ko ki
V2- 4 ko ki
262 - 8 ko ki
1- ko + ki
- ko – k1
(1 – kj + ko
1- k1 - ko
B =
u- w-2
+2
wn-
- vi +u
C =
-2
+ 2
22 - 8 0 vI
- o - VI
) [(는
1- v1 + o
- v0 - v1
1- v + o
2) (V2 – 4vo vI +
+2
2 2
-- 0 - v1
since ko + ki #1 and vo + vı #1 for n e N.
Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow
form
2 zn-h
Wn-p
p=1
Wn+1
h=1
Zn+1
and
(5)
1
E Wn-p
1
> Zn-h
h=0
wn -u
p-0
4
Then, we assume that
Wn+1 - l
Wn - H
kn+1 =
1
kn =
2
Wn-p
p=0
Wn-p
p=1
(6)
Zn+1 -€
Zn - €
Vn+1 =
Vn
2
E Zn-h
> Zn-h
h=1
h=0
Substituting (6) in (5), we have
Zn-h
h=1
1
== kn-1,
Vn
kn+1 =
(7)
E Wn-p
p=1
Vn+1 =
= Vn-1.
kn
%3D
Wn -
Hence, we see that
1
ki = -,
Vo
wo - H
20 - €
ko
w-1+ w_2
Vo =
Z-1+ z-2
ko
and
at n = 1, k2 = ko and
V2 = Vo,
at n = 2, k3 = k1 and
V3 = V1,
at n = 3, k4 = ko and
V4 = Vo,
at n = 4, ks = k and
V5 = V1,
(8)
at n = 2 n, k2n = ko and
V2n = Vo,
at n = 2n + 1, k2n+1 = k, and
V2n+1 = V1.
Now, from the relations in (6) we get
Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2).
(9)
Using (8) in (9), we have
W2n = H+ ko (w2n-1 + w2n-2),
W2n+1 = µ+ k1 (w2n + w2n-1),
(10)
Z2n = €+ vo (z2n-1 + 2n-2),
22n+1 = €+ V1 (22n + 22n-1).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 5 steps with 5 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
