Lecture Notes in Differential Equations - Bruce E. Shapiro

103
But on the square −1 ≤ t ≤ 1, −1 ≤ y ≤ 1,
|t||p + q| ≤ 1 × 2 = 2 (12.17)
So we need to find a K ≥ 2. We can pick any such K, e.g., K = 2. Then
every stop follows as a consequence reading from the bottom (12.16) to the
top (12.13). Hence f is Lipshitz.
Theorem 12.3. Boundedness =⇒ Lipshitz. If f(t, y) is continuously
differentiable and there exists some positive number K such that
∂f
∣ ∂y ∣ < K (12.18)
for all y, y ∈ R (for some rectangle R), then f is Lipshitz in y on R with
Lipshitz constant K.
Proof. By the mean value theorem, for any p, q, there is some number c
between p and q such that
∂ f(t, p) − f(t, q)
f(t, c) =
∂y p − q
(12.19)
By the assumption (12.18),
f(t, p) − f(t, q)
∣ p − q ∣ < K (12.20)
hence
|f(t, p) − f(t, q)| < K|p − q| (12.21)
for all p, q, which is the definition of Lipshitz. Hence f is Lipshitz.
Example 12.2. Show that f(t, y) = ty 2 is Lipshitz in −1 < t < 1, −1 <
y < 1.
Since
∣ ∣∣∣ ∂f
∂y ∣ = |2ty| ≤ 2 × 1 ≤ 1 = 2 (12.22)
on the square, the function is Lipshitz with K = 2.
Lemma 12.4. If f is Lipshitz in y with Lipshitz constant K, then each of
the φ i (t) are defined on R and satisfy
|φ k (t) − y 0 | ≤ M|t − t 0 | (12.23)