How to deduce this equation from Equation1 Explain to me the method. Show me the steps of determine green and the inf is here

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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How to deduce this equation from Equation1 Explain to me the method. Show me the steps of determine green and the inf is here

Theorem 4.Assume that e+ d, A+B+C+D+1 and
е
|A (е — d)| +|В(е - d) — е [(А+ B+С+D) —1]| +
|C(e – d) +e [(A+B+C+D) 1]|+|D(e – d)|< |e – d,
(11)
then the equilibrium point (7) of Eq.(1) is locally
asymptotically stable.
Proof.From
(9)
and
(11)
we
deduce
that
po|+\P1|+|p2|+|P3| < 1, and hence the proof follows
with the aid of Theorem 2.O
Transcribed Image Text:Theorem 4.Assume that e+ d, A+B+C+D+1 and е |A (е — d)| +|В(е - d) — е [(А+ B+С+D) —1]| + |C(e – d) +e [(A+B+C+D) 1]|+|D(e – d)|< |e – d, (11) then the equilibrium point (7) of Eq.(1) is locally asymptotically stable. Proof.From (9) and (11) we deduce that po|+\P1|+|p2|+|P3| < 1, and hence the proof follows with the aid of Theorem 2.O
difference equation
bxp-k
[dxp-k– eXp-i]
Xp+1 = Axn+ Bxp-k+ CXp-1+ DXp-o+
n = 0,1,2,..
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X_0...,X_],..., X_k, ..., X_1, Xo are arbitrary positive real
numbers such that k <1< o. Note that the special cases
of Eq.(1) have been studied in [1] when B= C= D=0,
and k = 0,1= 1, b is replaced by – b and in [27] when
B= C= D= 0, and k= 0, b is replaced by – b and in
[33] when B = C = D = 0, 1 = 0 and in [32] when
A=C=D=0, 1=0, b is replaced by – b.
(1)
%3D
where d + e. If [(A+B+C+D)– 1] (e – d) > 0, then
the only positive equilibrium point x of Eq.(1) is given by
X=
[(A+B+C+D)– 1] (e – d) *
(7)
Theorem 2./[18]. Assume that po, Pi, P2 and p3 E R. Then
|pol+]p1|+|P2|+|p3| < 1,
(5)
is a sufficient condition for the asymptotic stability of
Eq. (2).
provided du̟ # eu2. Consequently, we get
ƏF(x,x,X,X)
= A= Po,
One
ƏF(x,X,x,X)
e [(A+B+C+D) –1]
(e - d)
= B-
= P1,
Ine
(9)
= C+ e(A+B+C+D) –1]
(e - d)
ƏF(x,x,X,X)
P2,
aF(x,X,X,X)
= D= P3,
Transcribed Image Text:difference equation bxp-k [dxp-k– eXp-i] Xp+1 = Axn+ Bxp-k+ CXp-1+ DXp-o+ n = 0,1,2,.. where the coefficients A, B, C, D, b, d, e E (0,0), while k, 1 and o are positive integers. The initial conditions X_0...,X_],..., X_k, ..., X_1, Xo are arbitrary positive real numbers such that k <1< o. Note that the special cases of Eq.(1) have been studied in [1] when B= C= D=0, and k = 0,1= 1, b is replaced by – b and in [27] when B= C= D= 0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1 = 0 and in [32] when A=C=D=0, 1=0, b is replaced by – b. (1) %3D where d + e. If [(A+B+C+D)– 1] (e – d) > 0, then the only positive equilibrium point x of Eq.(1) is given by X= [(A+B+C+D)– 1] (e – d) * (7) Theorem 2./[18]. Assume that po, Pi, P2 and p3 E R. Then |pol+]p1|+|P2|+|p3| < 1, (5) is a sufficient condition for the asymptotic stability of Eq. (2). provided du̟ # eu2. Consequently, we get ƏF(x,x,X,X) = A= Po, One ƏF(x,X,x,X) e [(A+B+C+D) –1] (e - d) = B- = P1, Ine (9) = C+ e(A+B+C+D) –1] (e - d) ƏF(x,x,X,X) P2, aF(x,X,X,X) = D= P3,
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