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

CHAPTER 3. APPROXIMATE SOLUTIONS 65
where h = min(a, b/M). We know by the Weierstrass approximation theorem that
such a sequence exists.
The set of functions {P n (t, y)} is uniformly bounded. Denote by M the common
bound of both f and {P n (t, y)}. Since P n are polynomials, they satisfy the Lipshitz
condition on any bounded domain such as D, hence by the fundamental existence
theorem already proven, there exists a sequence of functions y n (t), t 0 − h ≤ t ≤
t 0 + h, such that
where
y ′ n(t) = P n (t, y n (t)) (3.193)
y n (t 0 ) = y 0 (3.194)
|y n (t) − y n (t 0 )| ≤ Mh, t 0 − h ≤ t ≤ t 0 + h (3.195)
i.e., the set of functions y n (t) are uniformly bounded on [t 0 − h, t 0 + h].
By the mean value theorem, for any t 1 , t 2 , there exists some c between t 1 and t 2
such that
y n (t 2 ) − y n (t 1 ) = y ′ n(c)(t 2 − t 1 ) (3.196)
But by definition, the y ′ n = P n , hence
|y n (t 2 ) − y n (t 1 )| = |P n (c)||t 2 − t 1 | ≤ M|t 2 − t 1 | (3.197)
Let ɛ > 0 and choose δ = ɛ/M. Then if |t 2 − t 1 | ≤ δ,
Therefore S = {y n } is equicontinuous.
|y n (t 2 ) − y n (t 1 )| ≤ ɛ (3.198)
Since S is uniformly bounded and equicontinuous on [t 0 − h, t 0 + h], it has a
uniformly convergent sequence. Call this sequence φ n (t) → φ. By definition, φ n ∈ S
hence there exists some P k(n) such that φ ′ n(t) = P k(n) (t, φ n (t)) and therefore
∣ φ n(t) − y 0 −
∫ t
φ n (t) − y 0 =
∫ t
t 0
P k(n) (s, φ n (s))ds (3.199)
∫ t
f(s, φ n (s))ds
∣ ≤ ∣ P k(n) (t, φ n (t)) − f(t, φ n (t)) ∣ ds
t 0 t 0
(3.200)
≤ ɛ n h (3.201)
Taking the limit at n → ∞ gives
∣ φ(t) − y 0 −
f(s, φ(s))ds
∣ = 0 (3.202)
t 0
∫ t
Hence φ is a solution of the IVP, proving existence.
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge