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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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Problem 5 (Bonus: Banach fixed point theorem). Let f be a continuous
function such that
\f (x) – f(y)| < a\x – y|
(о.1)
for a constant a satisfying 0 < a < 1. The purpose of this exercise is to prove
that there is an x such that f(x)
point theorem.
[Comment. While this theorem does not seem directly related to differen-
= x. This is known as the Banach fixed
tial equations, the method of its proof is used in the proof of the existence
and uniqueness theorem for differential equations. You can read more about
this in Boyce-DiPrima section 2.8 and Braun section 1.10.]
Write a complete proof of the theorem by filling the gaps in these steps:
1. Choose any xo and define a sequence xn inductively by
Xn = f(xn-1).
That is: x1 = f(x0), x2 = f (x1), and so on.
2. Using (o.1), show that for every n
|Xn – Xp-1|< a"-1|x1 – xol-
3. Write x, as
Xn =
(Xn – Xn-1) + (Xn–1 – Xn–2) +.. +(x1 – xo) + xo.
4. By comparing the above sum to the convergent series Eo a" conclude
that the sequence xn converges as n → ∞,
5. Let x = lim,00 Xn. Justify that we can pass to the limit n → 0 in
Xn = f(xn-1)
and conclude that x satisfies f(x)
6. Using (0.1), show that if y is any number satisfying f(y) = y, then
y = x, that is: there is only one solution to the equation f (x)
Transcribed Image Text:Problem 5 (Bonus: Banach fixed point theorem). Let f be a continuous function such that \f (x) – f(y)| < a\x – y| (о.1) for a constant a satisfying 0 < a < 1. The purpose of this exercise is to prove that there is an x such that f(x) point theorem. [Comment. While this theorem does not seem directly related to differen- = x. This is known as the Banach fixed tial equations, the method of its proof is used in the proof of the existence and uniqueness theorem for differential equations. You can read more about this in Boyce-DiPrima section 2.8 and Braun section 1.10.] Write a complete proof of the theorem by filling the gaps in these steps: 1. Choose any xo and define a sequence xn inductively by Xn = f(xn-1). That is: x1 = f(x0), x2 = f (x1), and so on. 2. Using (o.1), show that for every n |Xn – Xp-1|< a"-1|x1 – xol- 3. Write x, as Xn = (Xn – Xn-1) + (Xn–1 – Xn–2) +.. +(x1 – xo) + xo. 4. By comparing the above sum to the convergent series Eo a" conclude that the sequence xn converges as n → ∞, 5. Let x = lim,00 Xn. Justify that we can pass to the limit n → 0 in Xn = f(xn-1) and conclude that x satisfies f(x) 6. Using (0.1), show that if y is any number satisfying f(y) = y, then y = x, that is: there is only one solution to the equation f (x)
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