By Jerrold E. Marsden

Easy complicated research skillfully combines a transparent exposition of middle concept with a wealthy number of applications. Designed for undergraduates in arithmetic, the actual sciences, and engineering who've accomplished years of calculus and are taking advanced research for the 1st time..

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**Example text**

54) for n, m ≥ Nε , which shows that f˜n is Cauchy with respect to the maximum norm. More precisely, pick δ, T > 0 such that V = [t0 , t0 + T ] × Bδ (x0 ) ⊂ U and let M = max |f (t, x)|. 55) (t,x)∈V δ }, and Then xh (t) ∈ Bδ (x0 ) for t ∈ [t0 , t0 + T0 ], where T0 = min{T, M |xh (t) − xh (s)| ≤ M |t − s|. 56) Hence the family x1/n (t) is equicontinuous and there is a uniformly convergent subsequence φn (t) → φ(t). 12). 13) holds. Using that f is uniformly continuous on V , we can find δ(h) → 0 as h → 0 such that |f (s, y) − f (t, x)| ≤ δ(h) |y − x| ≤ M h, |s − t| ≤ h.

Initial value problems Now we come to the proof of our anticipated result. 12. Suppose U = R × Rn and for every T > 0 there are constants M (T ), L(T ) such that |f (t, x)| ≤ M (T ) + L(T )|x|, (t, x) ∈ [−T, T ] × Rn . 12) are defined for all t ∈ R. Proof. Using the above estimate for f we have (t0 = 0 without loss of generality) t |φ(t)| ≤ |x0 | + (M + L|φ(s)|)ds, t ∈ [0, T ] ∩ I. 5) M LT (e − 1). 49) L Thus φ lies in a compact ball and the result follows by the previous lemma. 9). 11. 12 is false (in general) if the estimate is replaced by |f (t, x)| ≤ M (T ) + L(T )|x|α with α > 1.

Since fn (x1 ) is bounded, we can choose a subsequence (1) (1) fn (x) such that fn (x1 ) converges (Bolzano-Weierstraß). Similarly we can (2) (1) extract a subsequence fn (x) from fn (x) which converges at x2 (and hence (1) also at x1 since it is a subsequence of fn (x)). By induction we get a se(j) (n) quence fn (x) converging at x1 , . . , xj . ). We will show that it converges uniformly for all x: Fix ε > 0 and chose δ such that |fn (x) − fn (y)| ≤ 3ε for |x − y| < δ. The balls Bδ (xj ) cover our interval and by compactness even finitely many, say 1 ≤ j ≤ p suffice.