Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
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)
- Page 59 and 60: 51 this becomes a first-order ODE i
- Page 61 and 62: Lesson 7 Autonomous Differential Eq
- Page 63 and 64: 55 Figure 7.1: A plot of the right-
- Page 65 and 66: 57 Figure 7.2: Solutions of the log
- Page 67 and 68: 59 Figure 7.4: Solutions of the thr
- Page 69 and 70: Lesson 8 Homogeneous Equations Defi
- Page 71 and 72: 63 where z = y/t, the differential
- Page 73 and 74: Lesson 9 Exact Equations We can re-
- Page 75 and 76: 67 Now compare equation (9.2) with
- Page 77 and 78: 69 Hence dg dy = 0 =⇒ g = C′ (9
- Page 79 and 80: 71 From the first of equations (9.5
- Page 81 and 82: 73 Differentiating equations (9.81)
- Page 83 and 84: 75 This has the form Mdt + Ndy = 0
- Page 85 and 86: Lesson 10 Integrating Factors Defin
- Page 87 and 88: 79 Differentiating with respect to
- Page 89 and 90: 81 Proof. In each of the five cases
- Page 91 and 92: 83 as required by equation (10.31).
- Page 93 and 94: 85 Since M y ≠ N t , equation (10
- Page 95 and 96: 87 the revised equation (10.100) is
- Page 97 and 98: 89 Substituting (10.129) into (10.1
- Page 99 and 100: Lesson 11 Method of Successive Appr
- Page 101 and 102: 93 because the integral is zero (th
- Page 103 and 104: 95 Example 11.1. Construct the Pica
- Page 105 and 106: 97 We can then plug this expression
- Page 107 and 108: Lesson 12 Existence of Solutions* I
- Page 109: 101 • Interchangeability of Limit
- Page 113 and 114: 105 Thus lim φ n = φ 0 + lim n→
- Page 115 and 116: 107 because the right hand side doe
- Page 117 and 118: Lesson 13 Uniqueness of Solutions*
- Page 119 and 120: 111 The proof of theorem (13.1) is
- Page 121 and 122: 113 But δ(t) is an absolute value,
- Page 123 and 124: 115 Substituting (13.66) into (13.6
- Page 125 and 126: Lesson 14 Review of Linear Algebra
- Page 127 and 128: 119 Definition 14.10. An m × n (or
- Page 129 and 130: 121 Definition 14.19. Matrix Multip
- Page 131 and 132: 123 In practical terms, computation
- Page 133 and 134: 125 Simplifying 4x − 2 + 3z = 0 (
- Page 135 and 136: Lesson 15 Linear Operators and Vect
- Page 137 and 138: 129 Example 15.3. By a similar argu
- Page 139 and 140: 131 Therefore ‖y + z‖ 2 ≤ ‖
- Page 141 and 142: 133 Definition 15.5. Two vectors y,
- Page 143 and 144: Lesson 16 Linear Equations With Con
- Page 145 and 146: 137 Hence both r = 1 and r = 3. Thi
- Page 147 and 148: 139 The second order linear initial
- Page 149 and 150: 141 The general solution to is give
- Page 151 and 152: Lesson 17 Some Special Substitution
- Page 153 and 154: 145 Therefore since z = y ′ , Int
- Page 155 and 156: 147 Example 17.5. Solve yy ′′ +
- Page 157 and 158: 149 where I is the identity matrix.
- Page 159 and 160: 151 can be rewritten by solving a =
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)