This work aims to investigate the equilibria, local stability, global attractivity and the exact solutions of the following difference equations Bun-1un-5 Un+1 = aun-1+ n = 0,1, .., (1) Yun-3 - dun-5 Bun-1un-5 Un+1 = aUn-1 n = 0,1, .., (2) yun-3 + dun-5' where the coefficients a, B, y, and d are positive real numbers and the initial con- ditions ui for all i = present the numerical solutions via some 2D graphs. -5, -4, ..., 0, are arbitrary non-zero real numbers. We also

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.7: Problem Solving: Consecutive Integers
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This work aims to investigate the equilibria, local stability, global attractivity
and the exact solutions of the following difference equations
Bun-1un-5
dun-5
Un+1 = aun-1+
п 3D0, 1,...,
(1)
Yun-3
Bun-1un-5
Yun-3 + dun-5
Un+1 = Qun-1
n = 0,1, ..,
(2)
where the coefficients a, B, y, and & are positive real numbers and the initial con-
ditions ui for all i = -5, -4, .., 0, are arbitrary non-zero real numbers. We also
present the numerical solutions via some 2D graphs.
2. ON THE EQUATION Un+1 = QUn-1+
Bun-1un-5
yun-3-dun-5
This section is devoted to study the qualitative behaviors of Eq. (1). The
equilibrium point of Eq. (1) is given by
11. EXACT SOLUTION OF EQ. (2) WHEN a = B =y= 6 = 1
This section shows the exact solutions of the following equation:
Un-1Un-5
Un+1 = Un-1
n = 0, 1,...,
(18)
Un-3 + Un-5
where the initial conditions are selected to be positive real numbers.
Theorem 9 Let {un}-5 be a solution to Eq. (18) and suppose that u-5
а, и-4 %3D 6, и-з 3D с, и-2 %3D d, и-1 %3D е, ио 3D f. Then, for n %3D 0, 1,2, ..., the
solutions of Eq. (18) are given by the following formulas:
%3D
e2n 2n-1
II (ie +c) (ic+a)'
f2n d2n-1
IIT (if+ d)(id + b)'
c2n e2n
U8n-5
2n-1
i=1
U8n-4 =
-2n-1
U8n-3 =
2n-
IT (ic + a) I (ie + c)
d2n f2n
U8n-2 =
II (id + b) IIT (if + d)'
e2n+1 2n
i=1
U8n-1 =
II, (ie + c)(ic + a)'
U8n =
IT (if + d)(id + b)'
c2n+1e2n+1
U8n+1 =
72n+1
2n
IT (ic + a) I, (ie + c)
d2n+1 f2n+1
IT (id + b) II(if + d)"
=D1
U8n+2 =
n+1
Proof.
It can be easily seen that the solutions are true for n = 0. We suppose
that n >0 and assume that our assumption holds for n- 1. That is,
e2n-2,2n-3
2n-3
U8n-13 =
II (ie + c) (ic + a)
f2n-2 d2n-3
2n-3
II" (if + d)(id + b)
c2n-22n-2
U8n-12 =
U8n-11 =
2n-2
II (ic + a) I (ie + c)
d2n-2 f2n-2
II (id + b) IT (if + d)
e2n-12n-2
U8n-10 =
2n-3
li=1
2n-2
U8n-9 =
T2n-2
I (ie + c)(ic + a)
f2n-12n-2
IT(if + d)(id+b)
U8n-8
li=1
can-1e2n-1
U8n-7 =
-2n-1
2n-2
:1
II (ic+ a) II (ie+c)
d2n-1 f2n-1
U8n-6 =
2n-1
II (id + b) II (if + d)
Transcribed Image Text:This work aims to investigate the equilibria, local stability, global attractivity and the exact solutions of the following difference equations Bun-1un-5 dun-5 Un+1 = aun-1+ п 3D0, 1,..., (1) Yun-3 Bun-1un-5 Yun-3 + dun-5 Un+1 = Qun-1 n = 0,1, .., (2) where the coefficients a, B, y, and & are positive real numbers and the initial con- ditions ui for all i = -5, -4, .., 0, are arbitrary non-zero real numbers. We also present the numerical solutions via some 2D graphs. 2. ON THE EQUATION Un+1 = QUn-1+ Bun-1un-5 yun-3-dun-5 This section is devoted to study the qualitative behaviors of Eq. (1). The equilibrium point of Eq. (1) is given by 11. EXACT SOLUTION OF EQ. (2) WHEN a = B =y= 6 = 1 This section shows the exact solutions of the following equation: Un-1Un-5 Un+1 = Un-1 n = 0, 1,..., (18) Un-3 + Un-5 where the initial conditions are selected to be positive real numbers. Theorem 9 Let {un}-5 be a solution to Eq. (18) and suppose that u-5 а, и-4 %3D 6, и-з 3D с, и-2 %3D d, и-1 %3D е, ио 3D f. Then, for n %3D 0, 1,2, ..., the solutions of Eq. (18) are given by the following formulas: %3D e2n 2n-1 II (ie +c) (ic+a)' f2n d2n-1 IIT (if+ d)(id + b)' c2n e2n U8n-5 2n-1 i=1 U8n-4 = -2n-1 U8n-3 = 2n- IT (ic + a) I (ie + c) d2n f2n U8n-2 = II (id + b) IIT (if + d)' e2n+1 2n i=1 U8n-1 = II, (ie + c)(ic + a)' U8n = IT (if + d)(id + b)' c2n+1e2n+1 U8n+1 = 72n+1 2n IT (ic + a) I, (ie + c) d2n+1 f2n+1 IT (id + b) II(if + d)" =D1 U8n+2 = n+1 Proof. It can be easily seen that the solutions are true for n = 0. We suppose that n >0 and assume that our assumption holds for n- 1. That is, e2n-2,2n-3 2n-3 U8n-13 = II (ie + c) (ic + a) f2n-2 d2n-3 2n-3 II" (if + d)(id + b) c2n-22n-2 U8n-12 = U8n-11 = 2n-2 II (ic + a) I (ie + c) d2n-2 f2n-2 II (id + b) IT (if + d) e2n-12n-2 U8n-10 = 2n-3 li=1 2n-2 U8n-9 = T2n-2 I (ie + c)(ic + a) f2n-12n-2 IT(if + d)(id+b) U8n-8 li=1 can-1e2n-1 U8n-7 = -2n-1 2n-2 :1 II (ic+ a) II (ie+c) d2n-1 f2n-1 U8n-6 = 2n-1 II (id + b) II (if + d)
In addition, Eq. (18) leads to
U8n-5U8n-9
U8n-3 = U8n-5
И8п-7 + u8n-9
e2n 2n –1
e2n-1,2n-2
e2n c2n-1
(ie+c)(ic+a) [I",?(ie+c)(ic+a)
e2n-1c2n-2
IT",?(ie+c)(ic+a)
c2n-2e2n
c2n-1e2n-1
II?",(ic+a) [I?",²(ie+c)
2n-1
П (е + c) (iс + а)
i=D1
1.
e2n c2n-1
II ie + e)(ic + a) II, "(ie + c)(ic +a) II (ic + a) (TT Geta) + eIlie-
2n-1
1
li=1
II (ic+a)
i=1
cII"?(ic+a)
e2n 2n-1
c2n-22n
IT (ie + c)(ic +a) II'(ie +c)(ic + a) ( en-ie+a +)
2n-1
li=1
2n-1
c2n-1e2n ((2n – 1)c+a)
ПТ (iе + с) (іс +а) ПТ (ie + с) (iс + а) (2m)с + а)
(2n — 1)с + а
(2n)с + а
e2n c2n-1
2n-1
li=1
2п-1
li=1
e2n c2n-1
(1
II", (ie + c)(ic +a)
2n-1
li=1
e2n 2n
c2n e2n
2n-1
Li=1
2n-1
-2n
П (iе + с) (ic+ta)((2n)с + а)
П (с + а) П-T (ie + c)
i=1
i=1
Similarly, one can prove other solutions.
Transcribed Image Text:In addition, Eq. (18) leads to U8n-5U8n-9 U8n-3 = U8n-5 И8п-7 + u8n-9 e2n 2n –1 e2n-1,2n-2 e2n c2n-1 (ie+c)(ic+a) [I",?(ie+c)(ic+a) e2n-1c2n-2 IT",?(ie+c)(ic+a) c2n-2e2n c2n-1e2n-1 II?",(ic+a) [I?",²(ie+c) 2n-1 П (е + c) (iс + а) i=D1 1. e2n c2n-1 II ie + e)(ic + a) II, "(ie + c)(ic +a) II (ic + a) (TT Geta) + eIlie- 2n-1 1 li=1 II (ic+a) i=1 cII"?(ic+a) e2n 2n-1 c2n-22n IT (ie + c)(ic +a) II'(ie +c)(ic + a) ( en-ie+a +) 2n-1 li=1 2n-1 c2n-1e2n ((2n – 1)c+a) ПТ (iе + с) (іс +а) ПТ (ie + с) (iс + а) (2m)с + а) (2n — 1)с + а (2n)с + а e2n c2n-1 2n-1 li=1 2п-1 li=1 e2n c2n-1 (1 II", (ie + c)(ic +a) 2n-1 li=1 e2n 2n c2n e2n 2n-1 Li=1 2n-1 -2n П (iе + с) (ic+ta)((2n)с + а) П (с + а) П-T (ie + c) i=1 i=1 Similarly, one can prove other solutions.
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