The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 3. APPROXIMATE SOLUTIONS 67
Uniform Continuity
A function y = f(t) : D ↦→ R is continuous if small changes in t make only small
changes in y,
(∀t 1 , t 2 ∈ D)(∀ɛ > 0)(∃δ > 0) (|t 1 − t 2 | < δ =⇒ |f(t 1 ) − f(t 2 )| < ɛ) (3.211)
A function y = f(t) : D ↦→ R is uniformly continuous if small changes in t make
only small changes in y, and furthermore, the size of the changes in y do not depend
on the size of the changes in t.
(∀ɛ > 0)(∃δ > 0) (∀t 1 , t 2 ∈ D)(|t 1 − t 2 | < δ =⇒ |f(t 1 ) − f(t 2 )| < ɛ) (3.212)
Uniform Continuity implies continuity but not vice-versa.
On compact (closed and bounded) domains, continuity implies uniform continuity.
Lipshitz Continuity implies uniform continuity.
The following result tells us that solutions that start “together” stay together.
Theorem 3.12. Let f(t, y) ∈ C(D), f ∈ L(y, K)(D), and let y be a solution to
y ′ = f(t, y) (3.213)
y(t 0 ) = y 0 (3.214)
on a rectangle R = [t 0 − h, t 0 + h] × [ỹ 0 − l, ỹ 0 + l]. Then y(t, y 0 ) is continuous in
both t and y 0 on R.
Proof. Suppose we have two solutions y(t, y 01 ), y(t, y 02 ) on R. Then by the Fundamental
Inequality (equation 3.75)
|y(t 0 , y 01 ) − y(t, y 02 )| ≤ |y 01 − y 02 |e Kh for |t − t 0 | ≤ h (3.215)
The y(t 0 , y 0 ) is continuous in y 0 uniformly in t.
Theorem 3.13. Under the same conditions as Theorem 3.12, and, in addition, if
∂f/∂y exists and is continuous in both t and y on D, then
exists and is continuous in both t and y.
∂y(t, y 0 )
∂y 0
Proof. Pick a ỹ 0 , and write ỹ(t) = y(t, ỹ 0 ) and y(t) = y(t, y 0 ). Define
p(t, y 0 ) =
y(t) − ỹ(t)
y 0 − ỹ 0
, for y 0 ≠ ỹ 0 (3.216)
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge