Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro

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102 LESSON 12. EXISTENCE OF SOLUTIONS* 2. f(t, y) is bounded by some number M on a rectangle R, i.e, for all t, y in R. 3. f(t, y) is differentiable on R. |f(t, y)| < M (12.8) 4. ∂f/∂y is also bounded by M on R, i.e., ∂f(t, y) ∣ ∂y ∣ < M (12.9) 5. Define the Picard iterates as φ 0 (t) = y 0 and for all k = 1, 2, . . . . ∫ t φ k (t) = y 0 + f(s, φ k−1 (s))ds t 0 (12.10) Definition 12.2. Lipshitz Condition. A function f(t, y) is said to satisfy a Lipshitz Condition on y in the rectangle R or be Lipshitz in y on R if there exists some number K > 0, that we will call the Lipshitz Constant, such that |f(t, y 1 ) − f(t, y 2 )| ≤ K|y 1 − y 2 | (12.11) for all t, y 1 , y 2 in some rectangle R. Example 12.1. Show that f(t, y) = ty 2 is Lipshitz on the square −1 < t < 1, −1 < y < 1. We need to find a K such that for f(t, y) = ty 2 , i.e., we need to show But so we need to find a K such that |f(t, p) − f(t, q)| ≤ K|p − q| (12.12) |tp 2 − tq 2 | ≤ K|p − q| (12.13) |tp 2 − tq 2 | = |t(p − q)(p + q)| (12.14) |t(p − q)(p + q)| ≤ K|p − q| (12.15) |t||p + q| ≤ K (12.16)

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)

102 LESSON 12. EXISTENCE OF SOLUTIONS*<br />

2. f(t, y) is bounded by some number M on a rectangle R, i.e,<br />

for all t, y <strong>in</strong> R.<br />

3. f(t, y) is differentiable on R.<br />

|f(t, y)| < M (12.8)<br />

4. ∂f/∂y is also bounded by M on R, i.e.,<br />

∂f(t, y)<br />

∣ ∂y ∣ < M (12.9)<br />

5. Def<strong>in</strong>e the Picard iterates as φ 0 (t) = y 0 and<br />

for all k = 1, 2, . . . .<br />

∫ t<br />

φ k (t) = y 0 + f(s, φ k−1 (s))ds<br />

t 0<br />

(12.10)<br />

Def<strong>in</strong>ition 12.2. Lipshitz Condition. A function f(t, y) is said to satisfy<br />

a Lipshitz Condition on y <strong>in</strong> the rectangle R or be Lipshitz <strong>in</strong> y on R if<br />

there exists some number K > 0, that we will call the Lipshitz Constant,<br />

such that<br />

|f(t, y 1 ) − f(t, y 2 )| ≤ K|y 1 − y 2 | (12.11)<br />

for all t, y 1 , y 2 <strong>in</strong> some rectangle R.<br />

Example 12.1. Show that f(t, y) = ty 2 is Lipshitz on the square −1 <<br />

t < 1, −1 < y < 1.<br />

We need to f<strong>in</strong>d a K such that<br />

for f(t, y) = ty 2 , i.e., we need to show<br />

But<br />

so we need to f<strong>in</strong>d a K such that<br />

|f(t, p) − f(t, q)| ≤ K|p − q| (12.12)<br />

|tp 2 − tq 2 | ≤ K|p − q| (12.13)<br />

|tp 2 − tq 2 | = |t(p − q)(p + q)| (12.14)<br />

|t(p − q)(p + q)| ≤ K|p − q| (12.15)<br />

|t||p + q| ≤ K (12.16)

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