How to deduce this equation from Equation 1.1 Explain to me the method. Show me the steps of determine yellow and inf is here Theorem 7 For any values of the quotient E-1 , If A < 1, then thepositive equilibrium point ỹ of Eq.(1.1) is a global attractor and the followingconditions holda1 b2 2 a2B1, a1ß3 > azß1, a184 2 a4B1, a1ß5 > a5,B1, a2ß3 > a3B2, a2ß4 > a4B2,a2B5 > a5,B2, a3ß4 > a4B3, a3ß5 > a5ß3, a4ß5 > a5,B4 and ag > (a1 + a2 + a3 + a4).(5.25)proof: Let {ym}=-5 be a positive solution of Eq.(1.1). and let H :(0, 0)6 –→ (0, 0) be a continuous function which is defined byH(uo, ..., U5Auo +E1(B;u;)By differentiating the function H(uo,..., u5) with respect to u; (i = 0,..., 5),we obtainА,Huo(5.26)(@1b2 – a2B1) u2 + (a1B3 – a3B1) uz + (a1B4 – a4B1) u4 + (a1B5 – a5,B1) ugHu2(5.27)(a132 – a261) u1 + (@2B3 – a3B2) uz + (a2B4 – 04B2) u4 + (a2B5 – a5B2) uzHuz(5.28)16 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 7 For any values of the…

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How to deduce this equation from Equation 1.1 Explain to me the method. Show me the steps of determine yellow and inf is here

Theorem 7 For any values of the quotient E-1 , If A < 1, then the
positive equilibrium point ỹ of Eq.(1.1) is a global attractor and the following
conditions hold
a1 b2 2 a2B1, a1ß3 > azß1, a184 2 a4B1, a1ß5 > a5,B1, a2ß3 > a3B2, a2ß4 > a4B2,
a2B5 > a5,B2, a3ß4 > a4B3, a3ß5 > a5ß3, a4ß5 > a5,B4 and ag > (a1 + a2 + a3 + a4).
(5.25)
proof: Let {ym}=-5 be a positive solution of Eq.(1.1). and let H :
(0, 0)6 –
→ (0, 0) be a continuous function which is defined by
H(uo, ..., U5
Auo +
E1(B;u;)
By differentiating the function H(uo,..., u5) with respect to u; (i = 0,..., 5),
we obtain
А,
Huo
(5.26)
(@1b2 – a2B1) u2 + (a1B3 – a3B1) uz + (a1B4 – a4B1) u4 + (a1B5 – a5,B1) ug
Hu
2
(5.27)
(a132 – a261) u1 + (@2B3 – a3B2) uz + (a2B4 – 04B2) u4 + (a2B5 – a5B2) uz
Huz
(5.28)
16

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

Explanation:

In the theorem it is clearly mentioned that they have used positive equilibrium point of $\tilde{y}$ to prove the conditions.

And the positive equilibrium point of $\tilde{y}$ is:

$\tilde{y} = \frac{\sum_{i =}^{5} α_{i}}{(1 - A) \sum_{i = 1}^{5} β_{i}}$

Steps:

1) We have taken the continuous function H and by differentiating this function w.r.t. u_i(i=0,1,2,3,4,5).

2)While differentiating w.r.t u₀ we got A as mentioned is (5.26).

3) While differentiating w.r.t u₁ we got an equation as mentioned in (5.27) now see this equation, the terms in bracket. As it is positive solution of (1.1) The Terms in the bracket should also be positive for that the positive terms has to be greater than equals to negative terms that follows -

$α_{1} β_{2} \geq α_{2} β_{1}$ ; $α_{1} β_{3} \geq α_{3} β_{1}$ ; $α_{1} β_{4} \geq α_{4} β_{1}$ ; $α_{1} β_{5} \geq α_{5} β_{1}$

4) While differentiating w.r.t u₂we got an equation (5.28) apply the same process here also of terms in the bracket to be positive we got

$α_{2} β_{3} \geq α_{3} β_{2}$ ; $α_{2} β_{4} \geq α_{4} β_{2}$ ; $α_{2} β_{5} \geq α_{5} β_{2}$

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