Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
226 LESSON 26. GENERAL EXISTENCE THEORY* to a system. We create a system by defining the variables Then the differential equation becomes which we can rewrite as We then define functions f, and g, so that our system can be written as x 1 = y x 2 = y ′ (26.62) x ′ 2 + 4t 3 x 2 + x 3 1 = sin t (26.63) x ′ 2 = sin t − x 3 1 − 4t 3 x 2 (26.64) f(x 1 , x 2 ) = sin t − x 3 1 − 4t 3 x 2 g(x 1 , x 2 ) = x 2 (26.65) x ′ 1 = f(x 1 , x 2 ) x ′ 2 = g(x 1 , x 2 ) (26.66) with initial condition x 1 (0) = 1 x 2 (0) = 1 (26.67) It is common to define a vector x = (x 1 , x 2 ) and a vector function F(x) = (f(x 1 , x 2 ), g(x 1 , x 2 )) (26.68) Then we have a vector initial value problem } x ′ (t) = F(x) = (sin t − x 3 1 − 4t 3 x 2 , x 2 ) x(0) = (1, 1) (26.69) Since the set of all differentiable functions on R 2 is a vector space, our theorem on vector spaces applies. Even though we proved theorem (26.50) for first order equations every step in the proof still works when y and f become vectors. On any closed rectangle surrounding the initial condition F and ∂F/∂x i is bounded, continuous, and differentiable. So there is a unique solution to this initial value problem.
Lesson 27 Higher Order Linear Equations Constant Coefficients and the Linear Operator Generalizing the previous sections we can write the general nth-order linear equation with constant coefficients as a n y (n) + a n−1 y (n−1) + · · · + a 1 y ′ + a 0 y = f(t) (27.1) The corresponding characteristic polynomial is P n (r) = a n r n + a n−1 r n−1 + · · · + a 0 (27.2) = a n (r − r 1 )(r − r 2 ) · · · (r − r n ) (27.3) = 0 (27.4) and the corresponding n-th order linear operator is L n = a n D n + a n−1 D n−1 + · · · + a 0 (27.5) = a n (D − r 1 )(D − r 2 ) · · · (D − r n ) (27.6) where a 0 , ..., a n ∈ R are constants and r 1 , r 2 , ..., r n ∈ C are the roots of P n (r) = 0 (some or all of which may be repeated). The corresponding 227
- Page 183 and 184: 175 Example 20.3. Show that y = sin
- Page 185 and 186: 177 and therefore the system of equ
- Page 187 and 188: Lesson 21 Reduction of Order The me
- Page 189 and 190: 181 The method of reduction of orde
- Page 191 and 192: 183 Plugging these into Bessel’s
- Page 193 and 194: 185 Example 21.5. Find a second sol
- Page 195 and 196: Lesson 22 Non-homogeneous Equations
- Page 197 and 198: 189 where r 1 and r 2 are the roots
- Page 199 and 200: 191 This is a first order linear eq
- Page 201 and 202: 193 Theorem 22.5. Properties of the
- Page 203 and 204: 195 where (∫ ν(t) = exp ) −r 2
- Page 205 and 206: 197 The characteristic equation is
- Page 207 and 208: Lesson 23 Method of Annihilators In
- Page 209 and 210: 201 Theorem 23.5. (D 2 − 2aD + (a
- Page 211 and 212: 203 The method of annihilators is r
- Page 213 and 214: Lesson 24 Variation of Parameters T
- Page 215 and 216: 207 Substituting into equation (24.
- Page 217 and 218: 209 Example 24.3. Solve the initial
- Page 219 and 220: Lesson 25 Harmonic Oscillations If
- Page 221 and 222: 213 It is standard to define a new
- Page 223 and 224: 215 As with the unforced case, we c
- Page 225 and 226: Lesson 26 General Existence Theory*
- Page 227 and 228: 219 In the case just proven, there
- Page 229 and 230: 221 Theorem 26.5. Under the same co
- Page 231 and 232: 223 Since K n /(1 − K) → 0 as n
- Page 233: 225 for any φ ∈ V. Let g, h be f
- Page 237 and 238: 229 L n+1 (e rt y) = e rt a n (D +
- Page 239 and 240: 231 Example 27.2. Find the general
- Page 241 and 242: 233 Differentiating, u ′ (t) = d
- Page 243 and 244: 235 Integrating, − 2K |t − t 0
- Page 245 and 246: 237 a closed form expression for a
- Page 247 and 248: 239 Example 27.6. Find the general
- Page 249 and 250: 241 The characteristic equation is
- Page 251 and 252: 243 The Wronskian In this section w
- Page 253 and 254: 245 Certainly every φ(t) given by
- Page 255 and 256: 247 the differential equation. Over
- Page 257 and 258: 249 By the lemma, to obtain the der
- Page 259 and 260: 251 Example 27.14. Find the general
- Page 261 and 262: 253 So that f(t) = a n (t)y[ (n) +
- Page 263 and 264: Lesson 28 Series Solutions In many
- Page 265 and 266: 257 Changing the index of the secon
- Page 267 and 268: 259 Since the first two terms (corr
- Page 269 and 270: 261 Hence ∞∑ ∞∑ ∞∑ 0 =
- Page 271 and 272: 263 has an analytic solution at t =
- Page 273 and 274: 265 By the triangle inequality, |(k
- Page 275 and 276: 267 Table 28.1: Table of Special Fu
- Page 277 and 278: 269 Thus y = a 0 ( 1 + 1 6 t3 + 1 +
- Page 279 and 280: 271 into (28.114) and collect terms
- Page 281 and 282: 273 Summary of Power series method.
- Page 283 and 284: Lesson 29 Regular Singularities The
Lesson 27<br />
Higher Order L<strong>in</strong>ear<br />
<strong>Equations</strong><br />
Constant Coefficients and the L<strong>in</strong>ear Operator<br />
Generaliz<strong>in</strong>g the previous sections we can write the general nth-order l<strong>in</strong>ear<br />
equation with constant coefficients as<br />
a n y (n) + a n−1 y (n−1) + · · · + a 1 y ′ + a 0 y = f(t) (27.1)<br />
The correspond<strong>in</strong>g characteristic polynomial is<br />
P n (r) = a n r n + a n−1 r n−1 + · · · + a 0 (27.2)<br />
= a n (r − r 1 )(r − r 2 ) · · · (r − r n ) (27.3)<br />
= 0 (27.4)<br />
and the correspond<strong>in</strong>g n-th order l<strong>in</strong>ear operator is<br />
L n = a n D n + a n−1 D n−1 + · · · + a 0 (27.5)<br />
= a n (D − r 1 )(D − r 2 ) · · · (D − r n ) (27.6)<br />
where a 0 , ..., a n ∈ R are constants and r 1 , r 2 , ..., r n ∈ C are the roots of<br />
P n (r) = 0 (some or all of which may be repeated). The correspond<strong>in</strong>g<br />
227
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