Prove the most general version of the inequality: for u(t), k(t), C(t) ≥ 0, C(t) smooth, u(t) ≤ C(t)+ › + √² ["C'(s) e' k(r) dr ds k k(s)u(s) ds ⇒ u(t) ≤ C'(0)elő k(s) ds +
Prove the most general version of the inequality: for u(t), k(t), C(t) ≥ 0, C(t) smooth, u(t) ≤ C(t)+ › + √² ["C'(s) e' k(r) dr ds k k(s)u(s) ds ⇒ u(t) ≤ C'(0)elő k(s) ds +
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:1. In class we proved Gronwall's inequality, Ch.8 §4 Lemma, and §7 problem
8. Namely, for u(t) ≥ 0, and C, k ≥ 0
u(t) ≤ C + k k s t u
and for u(t), k(t) ≥ 0, and C ≥ 0
u(t) ≤ C(t)+
u(s) ds ⇒ u(t) ≤ Cekt,
u(t) ≤ C +
+ ſª k(s)u(s) ds ⇒ u(t) ≤ Celő k(s) ds
Prove the most general version of the inequality: for u(t), k(t), C(t) ≥ 0,
C(t) smooth,
So k(s)u(s) ds ⇒ u(t) ≤ C(0)elő k(s) ds +
*+ *
C'(s)ek(r) dr ds
![Theorem Let WCE be open and suppose f: WE has Lipschitz constant K.
Let y(t), z(t) be solutions to
(1)
z' = f(x)
on the closed interval [to, t₁]. Then, for all t = [to, tr]:
|y(t) z(t)| ≤y(to) - z(to) exp(K(t-to)).
The proof depends on a useful inequality (Gronwall's) which we prove first.
Lemma Let u: [0, a]→R be continuous and nonnegative. Suppose C ≥0, K 20
are such that
u(t) ≤ C + [* Ku(s) de
for all t€ [0, a]. Then
for all t € [0, a].
170
then
Proof. First, suppose C > 0, let
hence
By differentiation of U we find
Hence
so
u(t) ≤ Ceki
and so
U(t) = C +
c + √²
Since
we have
- [* Ku(s) ds > 0;
u(t) ≤U(t).
U'(t) = Ku(t);
U' (t)
U (1)
d
dt
=
Ku(t)
U (1)
log U(t) log U (0) + Kt
by integration. Since U (0) C, we have by exponentiation
U(t) ≤ Cekt,
We turn to the proof of the theorem.
Define
SK.
(log U(t)) ≤ K
u(t) ≤ Cekt.
If C0, then apply the above argument for a sequence of positive c; that tend
to 0 asio. This proves the lemma.
8. FUNDAMENTAL THEORY
v(t)= y(t) z(t).
y(t) — 2(t) = y(t) — 2(to) + √ [f(s, v(s)) — ƒ(s, z(s))] ds,
-
v(t) ≤ v(to) + Kv(s) ds.
Now apply the lemma to the function u(t)= v(to + 1) to get
v(t) ≤ v(to) exp (K (t-to)),
which is just the conclusion of the theorem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2eee5fef-24c0-42d2-a99f-ec7ca4df2e97%2F29267a1a-1ec8-4897-99cb-8e868599df0b%2Fjz0yp9d_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem Let WCE be open and suppose f: WE has Lipschitz constant K.
Let y(t), z(t) be solutions to
(1)
z' = f(x)
on the closed interval [to, t₁]. Then, for all t = [to, tr]:
|y(t) z(t)| ≤y(to) - z(to) exp(K(t-to)).
The proof depends on a useful inequality (Gronwall's) which we prove first.
Lemma Let u: [0, a]→R be continuous and nonnegative. Suppose C ≥0, K 20
are such that
u(t) ≤ C + [* Ku(s) de
for all t€ [0, a]. Then
for all t € [0, a].
170
then
Proof. First, suppose C > 0, let
hence
By differentiation of U we find
Hence
so
u(t) ≤ Ceki
and so
U(t) = C +
c + √²
Since
we have
- [* Ku(s) ds > 0;
u(t) ≤U(t).
U'(t) = Ku(t);
U' (t)
U (1)
d
dt
=
Ku(t)
U (1)
log U(t) log U (0) + Kt
by integration. Since U (0) C, we have by exponentiation
U(t) ≤ Cekt,
We turn to the proof of the theorem.
Define
SK.
(log U(t)) ≤ K
u(t) ≤ Cekt.
If C0, then apply the above argument for a sequence of positive c; that tend
to 0 asio. This proves the lemma.
8. FUNDAMENTAL THEORY
v(t)= y(t) z(t).
y(t) — 2(t) = y(t) — 2(to) + √ [f(s, v(s)) — ƒ(s, z(s))] ds,
-
v(t) ≤ v(to) + Kv(s) ds.
Now apply the lemma to the function u(t)= v(to + 1) to get
v(t) ≤ v(to) exp (K (t-to)),
which is just the conclusion of the theorem.
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