Show me the steps of determine blue and the inf is here Theorem 4 Every solution of Eq.(1.1) is bounded if A< 1.proof Let {ym} 5 be a solution of Eq.(1.1). It follows from Eq.(1.1)m=-5thata1Ym-1 +a2Ym-2+ a3ym-3 + a4Ym-4 + a5Ym-5Ут+1Aym +B1ym–1+ B2Ym-2 + B3Ym–3 + BaYm-4+ B5Ym–5a1Ym-1Аут +B1ym-1 + B2Ym-2 + B3ym-3 + B4Ym-4 + B5Ym-5a2Ym-2B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5azYm-3В1Ут-1 + В2Ут-2 + Взут-3 + ВАУт-4 + В53т-5a4Ym-4Віут-1 + В2Ут-2 + Взут-3 + ВАУт-4 + Взут-5a5Ym-5B1ym–1 + B2Ym-2 + B3ym-3 + B4Ym-4 + B5Ym-5Thena1Ym–1a2Ym-2a3Ym-3+B3ym-3a4Ym-4a5Ym-5+B5Ym-5Ym+1 < Aym +Biym-1B2Ym-2B4Ym-49.aiAym +B1a2a3for all m >1.B2B3B4By using a comparison, we can write the right hand side as followsa2Aym +B1a3+B3Ym+1 =B2B4thenYm = a"yo + constant,and this equation is locally asymptotically stable because A < 1, and con-verges to the equilibrium pointa1b2B3B4B5 + a2ß1B3B4B5 + a3ß132B4B5 + a4B1B2B3ß5 + a5ß1B2B3ß4B1B2B3B4B5 (1 – A)Therefore,a1 B2B3B4B5 + a2ß1B3B4B5 + a3ß1B2B4B5 + ¤4B1B2B3B5 + a5ß1B2ß3ß4lim sup ymB182B3B4B5 (1 – A)|Thus, the solution of Eq.(1.1) is bounded and the proof is complete.+ The main focus of this article is to discuss some qualitative behavior ofthe solutions of the nonlinear difference equationa1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5Ут+1 — Аутtт 3 0, 1, 2, ...,B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5(1.1)where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi-tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note thatthe special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 == a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special caseB4when a4 =B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in thespecial case when az = B5 = 0.

Answered: Theorem 4 Every solution of Eq.(1.1) is…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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^{Show me the steps of determine blue and the inf is here}

Theorem 4 Every solution of Eq.(1.1) is bounded if A< 1.
proof Let {ym} 5 be a solution of Eq.(1.1). It follows from Eq.(1.1)
m=-5
that
a1Ym-1 +a2Ym-2+ a3ym-3 + a4Ym-4 + a5Ym-5
Ут+1
Aym +
B1ym–1+ B2Ym-2 + B3Ym–3 + BaYm-4+ B5Ym–5
a1Ym-1
Аут +
B1ym-1 + B2Ym-2 + B3ym-3 + B4Ym-4 + B5Ym-5
a2Ym-2
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5
azYm-3
В1Ут-1 + В2Ут-2 + Взут-3 + ВАУт-4 + В53т-5
a4Ym-4
Віут-1 + В2Ут-2 + Взут-3 + ВАУт-4 + Взут-5
a5Ym-5
B1ym–1 + B2Ym-2 + B3ym-3 + B4Ym-4 + B5Ym-5
Then
a1Ym–1
a2Ym-2
a3Ym-3
+
B3ym-3
a4Ym-4
a5Ym-5
+
B5Ym-5
Ym+1 < Aym +
Biym-1
B2Ym-2
B4Ym-4
9.
ai
Aym +
B1
a2
a3
for all m >1.
B2
B3
B4
By using a comparison, we can write the right hand side as follows
a2
Aym +
B1
a3
+
B3
Ym+1 =
B2
B4
then
Ym = a"yo + constant,
and this equation is locally asymptotically stable because A < 1, and con-
verges to the equilibrium point
a1b2B3B4B5 + a2ß1B3B4B5 + a3ß132B4B5 + a4B1B2B3ß5 + a5ß1B2B3ß4
B1B2B3B4B5 (1 – A)
Therefore,
a1 B2B3B4B5 + a2ß1B3B4B5 + a3ß1B2B4B5 + ¤4B1B2B3B5 + a5ß1B2ß3ß4
lim sup ym
B182B3B4B5 (1 – A)
|
Thus, the solution of Eq.(1.1) is bounded and the proof is complete.
+

The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi-
tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.

Expert Solution

Step 1

The proof to the theorem 4 is to show that every solution of equation (1.1) is bounded if A<1

Here, the equation (1.1) is the following.

Let ${\{y_{m}\}}_{m = - 5}^{\infty}$ be a solution of Eq. (1.1).

It follows from Eq. (1.1) that

$y_{m + 1} = A y_{m} + \frac{α_{1} y_{m - 1} + α_{2} y_{m - 2} + α_{3} y_{m - 3} + α_{4} y_{m - 4} + α_{5} y_{m - 5}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} = A y_{m} + \frac{α_{1} y_{m - 1}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} + \frac{α_{2} y_{m - 2}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} + \frac{α_{3} y_{m - 3}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} + \frac{α_{4} y_{m - 4}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} + \frac{α_{5} y_{m - 5}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}}$

We are given that the coefficients $A, α_{i}, β_{i} \in (0, \infty), i = 1, . . ., 5,$ and $y_{- 5}, y_{- 4}, y_{- 3}, y_{- 2}, y_{- 1}, y_{0}$ are arbitrary positive real numbers.

So,

$\frac{1}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}} \leq \frac{1}{β_{i} y_{m - i}}, i = 1, 2, 3, 4, 5 .$

Thus,

$y_{m + 1} \leq A y_{m} + \frac{α_{1} y_{m - 1}}{β_{1} y_{m - 1}} + \frac{α_{2} y_{m - 2}}{β_{2} y_{m - 2}} + \frac{α_{3} y_{m - 3}}{β_{3} y_{m - 3}} + \frac{α_{4} y_{m - 4}}{β_{4} y_{m - 4}} + \frac{α_{5} y_{m - 5}}{β_{5} y_{m - 5}} = A y_{m} + \frac{α_{1}}{β_{1}} + \frac{α_{2}}{β_{2}} + \frac{α_{3}}{β_{3}} + \frac{α_{4}}{β_{4}} + \frac{α_{5}}{β_{5}}$

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