Example 1.2. Let f(t, y) = ty². Then since f(t, y2) — f(t, y₁)| t|y2 + Y1||Y2 y₁ is not bounded by any constant times |y

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Example 1 2 Let F T Y Ty Then Since F T Y2 F T Y T Y2 Y1 Y2 Y Is Not Bounded By Any Constant Times Y 1 (186 KiB) Viewed 32 times

Example 1.2. Let f(t, y) = ty². Then since f(t, y2) — f(t, y₁)| t|y2 + Y1||Y2 y₁ is not bounded by any constant times |y2 − y₁|, f is not Lipschitz with respect to y on the domain Rx R. However f is Lipschitz on any rectangle R = [a, b] × [c, d] since we have t|y₁ + y2 ≤ 2 max{|a|, |b|} · max{|c|, |d|} on R. =
Problem 69.4 Show that f(x, y) = xy² (a) satisfies a Lipschitz condition on the rectangle a ≤ x ≤ b and c≤ y ≤d. (b) does not satisfy a Lipschitz condition on any strip a ≤ x ≤ b and -∞0 ≤ y ≤ ∞. (a) Note that |x| ≤ max(|a|, |b|) = A and |y| ≤ max(|c|, |d|) = C. f(x, y₁) — ƒ (x, y2) 91 – 92 xy³ – xy² 91 – 2 = |x(y₁ + Y₂)| ≤ |x|(|y1| + |y2|) < 2AC is a bound. =
EXAMPLE 5.8.1 The function f(t,y) = ty² satisfies a Lipschitz condition for 0 ≤ t ≤ b < ∞, \y - yol ≤ c <∞0, since Ity, |ty₁² - ty₂²| \t\ \y₁ + y₂| \Y₁ − Y₂] ≤ b(2c + 2[yol)|y₁ - y₂| because |Y₁ + Y₂| = |V₁ — Yo + Y₂ − yo + 2yol ≤ |y₁ − yol + |32 − Yol + 2|yo| \y1 ≤ 2c + 2|yo| Therefore, we can take K - b(2c + 2|yol), and we have obtained the result.