Lecture Notes in Differential Equations - Bruce E. Shapiro

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44 LESSON 6. EXPONENTIAL RELAXATION appropriately first, so that neither t nor y appear in either integrand) 1 ∫ y y 0 ∫ du t u − C = t 0 ds τ (6.3) ln |y − C| − ln |y 0 − C| = t τ − t 0 (6.4) τ ln y − C ∣y 0 − C ∣ = 1 τ (t − t 0) (6.5) Exponentiating both sides of the equation y − C ∣y 0 − C ∣ = e(t−t0)/τ (6.6) hence |y − C| = |y 0 − C|e (t−t0)/τ (6.7) The absolute value poses a bit of a problem of interpretation. However, the only way that the fraction F = y − C (6.8) y 0 − C can change signs is if it passes through zero. This can only happen if 0 = e (t−t0)/τ (6.9) which has no solution. So whatever the sign of F , it does not change. At t = t 0 we have y − C = y 0 − C (6.10) hence and so by continuity with the initial condition ∣ ∣∣∣ y − C y 0 − C ∣ = 1 (6.11) y − C y 0 − C = 1 (6.12) y = C + (y 0 − C)e (t−t0)/τ (6.13) It is convenient to consider the two cases τ > 0 and τ < 0 separately. 1 We must do this because we are using both t and y as endpoints of our integration, and hence must change the name of the symbols we use in the equation, e.g, first we turn (6.2) into du/(u − C) = ds/τ.
45 Exponential Runaway First we consider τ > 0. For convenience we will choose t 0 = 0. Then y = C + (y 0 − C)e t/τ (6.14) The solution will either increase exponentially to ∞ or decrease to −∞ depending on the sign of y 0 − C. Example 6.1. Compound interest. If you deposit an amount y 0 in the bank and it accrues interest at a rate r, where r is measured in fraction per year (i.e., r = 0.03 means 3% per annum, and it has units of 1/year), and t is measured in years, then the rate at which the current amount on deposit increases is given by dy = ry (6.15) dt Then we have C = 0 and τ = 1/r, so y = y 0 e rt (6.16) So the value increases without bound over time (we don’t have a case where y 0 < because that would mean you owe money). Suppose we also have a fixed salary S per year, and deposit that entirely into our account. Then instead of (6.15), we have dy dt = ry + S = r(y + S/r) (6.17) In this case we see that C = −S/r, the negative ratio of the fixed and interest-based additions to the account. The solution is then y = − S ( r + y 0 + S ) e rt (6.18) r We still increase exponentially without bounds. Now consider what happens if instead of depositing a salary we instead withdraw money at a fixed rate of W per year. Since W causes the total amount to decrease, the differential equation becomes dy dt = ry − W = r(y − W/r) (6.19) Now C = W/r. If W/r < y 0 then the rate of change will be positive initially and hence positive for all time, and we have y = W ( r + y 0 − W ) e rt (6.20) r
Page 1 and 2: Lecture Notes in Differential Equat
Page 3 and 4: Contents Front Cover . . . . . . .
Page 5 and 6: CONTENTS v Dedicated to the hundred
Page 7 and 8: Preface These lecture notes on diff
Page 9 and 10: Lesson 1 Basic Concepts A different
Page 11 and 12: 3 Definition 1.2 (Solution, ODE). A
Page 13 and 14: 5 1.22 is restricted to being a pos
Page 15 and 16: 7 Figure 1.1 illustrates what this
Page 17 and 18: 9 We will study linear equations in
Page 19 and 20: Lesson 2 A Geometric View One way t
Page 21 and 22: 13 We can extend this geometric int
Page 23 and 24: 15 that since the slope of the solu
Page 25 and 26: Lesson 3 Separable Equations An ODE
Page 27 and 28: 19 Since it is not possible to solv
Page 29 and 30: 21 where M(t) = −a(t) and N(y) =
Page 31 and 32: 23 Example 3.10. Find a general sol
Page 33 and 34: Lesson 4 Linear Equations Recall th
Page 35 and 36: 27 So far any function µ will work
Page 37 and 38: 29 Example 4.1. Solve the different
Page 39 and 40: 31 Since p(t) = 1 (the coefficient
Page 41 and 42: 33 Multiplying equation 4.68 by µ
Page 43 and 44: 35 Substituting y = t = 0 in this i
Page 45 and 46: 37 ∫ t t 0 Evaluating the integra
Page 47 and 48: Lesson 5 Bernoulli Equations The Be
Page 49 and 50: 41 This is a Bernoulli equation wit
Page 51: Lesson 6 Exponential Relaxation One
Page 55 and 56: 47 Figure 6.2: Illustration of the
Page 57 and 58: 49 This is identical to with Theref
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 and 110:
101 • Interchangeability of Limit
Page 111 and 112:
103 But on the square −1 ≤ t
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 =
Page 161 and 162:
Lesson 18 Complex Roots We know for
Page 163 and 164:
155 Theorem 18.2. Euler’s Formula
Page 165 and 166:
157 For k = 0, 1, 2, . . . , n −
Page 167 and 168:
159 and its roots are given by The
Page 169 and 170:
161 The motivation for equation 18.
Page 171 and 172:
Lesson 19 Method of Undetermined Co
Page 173 and 174:
165 3. If f(t) = e rt and r is a ro
Page 175 and 176:
167 Example 19.4. Solve ⎫ y ′
Page 177 and 178:
169 Adding the two equations gives
Page 179 and 180:
Lesson 20 The Wronskian We have see
Page 181 and 182:
173 Definition 20.1. The Wronskian
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 and 234:
225 for any φ ∈ V. Let g, h be f
Page 235 and 236:
Lesson 27 Higher Order Linear Equat
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:
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:
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
Page 285 and 286:
277 ∑ ∞ ∞∑ ∞∑ 0 = t 2 a
Page 287 and 288:
279 Case 2: Two equal real roots. S
Page 289 and 290:
281 Example 29.6. Solve t 2 y ′
Page 291 and 292:
Lesson 30 The Method of Frobenius I
Page 293 and 294:
285 This is a homogeneous linear eq
Page 295 and 296:
287 Example 30.4. Find a Frobenius
Page 297 and 298:
289 Thus a Frobenius solution is y
Page 299 and 300:
291 Example 30.6. Find the form of
Page 301 and 302:
293 term by term to (30.97). Starti
Page 303 and 304:
295 Let j = n − k. Then |n − 1
Page 305 and 306:
297 is a solution of (t − t 0 ) 2
Page 307 and 308:
299 Evaluation of the integral depe
Page 309 and 310:
301 Example 30.8. In example 30.4 w
Page 311 and 312:
Lesson 31 Linear Systems The genera
Page 313 and 314:
305 is where λ 2 − T λ + ∆ =
Page 315 and 316:
307 (b) If λ 1 ≠ λ 2 ∈ R, i.e
Page 317 and 318:
309 We will verify (31.54) by induc
Page 319 and 320:
311 The Jordan Form Let A be a squa
Page 321 and 322:
313 y = 1 [( ) ( ) ] ( ) 4 1 e 2t 1
Page 323 and 324:
315 Theorem 31.8. (Abel’s Formula
Page 325 and 326:
317 By a similar argument, the seco
Page 327 and 328:
319 we can replace (31.118) with a
Page 329 and 330:
321 Corollary 31.12. The generalize
Page 331 and 332:
323 where (A − λI)w 2 = w 1 , i.
Page 333 and 334:
325 so that λ = 3, −5. The eigen
Page 335 and 336:
327 Non-constant Coefficients We ca
Page 337 and 338:
329 we find that ∫ M(t)g(t)dt = (
Page 339 and 340:
Lesson 32 The Laplace Transform Bas
Page 341 and 342:
333 Figure 32.1: A piecewise contin
Page 343 and 344:
335 Example 32.4. From integral A.1
Page 345 and 346:
337 apply this result iteratively.
Page 347 and 348:
L [ t x−1] [ ] 1 d = L x dt tx =
Page 349 and 350:
341 Equating numerators and expandi
Page 351 and 352:
343 Derivatives of the Laplace Tran
Page 353 and 354:
345 can be written as as illustrate
Page 355 and 356:
347 Translations in the Laplace Var
Page 357 and 358:
349 Summary of Translation Formulas
Page 359 and 360:
351 The inverse transform is [ ] f(
Page 361 and 362:
353 Example 32.18. Find the Laplace
Page 363 and 364:
355 Similarly, we can express a uni
Page 365 and 366:
357 Figure 32.7: Solution of exampl
Page 367 and 368:
Lesson 33 Numerical Methods Euler
Page 369 and 370:
361 Figure 33.1: Illustration of Eu
Page 371 and 372:
363 y 4 = y 3 + hf(t 3 , y 3 ) (33.
Page 373 and 374:
365 Figure 33.3: Illustration of th
Page 375 and 376:
367 result with a smaller step size
Page 377 and 378:
369 Expanding the final term in a T
Page 379 and 380:
371 k 2 = y 0 + h 2 f(t 0, k 1 ) (3
Page 381 and 382:
Lesson 34 Critical Points of Autono
Page 383 and 384:
375 Since both f and g are differen
Page 385 and 386:
377 Using the cos π/4 = √ 2/2 an
Page 387 and 388:
379 values, of the matrix. We find
Page 389 and 390:
381 Distinct Real Nonzero Eigenvalu
Page 391 and 392:
383 eigendirection {λ 1 , v 1 }dom
Page 393 and 394:
385 Figure 34.5: Phase portraits ty
Page 395 and 396:
387 Complex Conjugate Pair with non
Page 397 and 398:
389 The angular change is described
Page 399 and 400:
391 Figure 34.8: Topological instab
Page 401 and 402:
393 Figure 34.10: phase portraits f
Page 403 and 404:
Appendix A Table of Integrals Basic
Page 405 and 406:
397 ∫ x √ x − adx = 2 3 a(x
Page 407 and 408:
399 ∫ x √ ax2 + bx + c dx = 1 a
Page 409 and 410:
401 ∫ ∫ ∫ ∫ e ax2 dx = −
Page 411 and 412:
403 ∫ tan 3 axdx = 1 a ln cos ax
Page 413 and 414:
405 Products of Trigonometric Funct
Page 415 and 416:
Appendix B Table of Laplace Transfo
Page 417 and 418:
409 e at cosh kt t sin kt t cos kt
Page 419 and 420:
Appendix C Summary of Methods First
Page 421 and 422:
413 The resulting equation is linea
Page 423 and 424:
415 for y once z is known. Method o
Page 425 and 426:
Bibliography [1] Bear, H.S. Differe
Page 427 and 428:
BIBLIOGRAPHY 419
Page 429 and 430:
BIBLIOGRAPHY 421
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