Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
362 LESSON 33. NUMERICAL METHODS We then observe that since y n ≈ y(t n ) and y n+1 ≈ y(t n+1 ), then (33.12) follows immediately from (33.16). If the scalar initial value problem of equation (33.4) is replaced by a systems of equations y ′ = f(t, y), y(t 0 ) = y 0 (33.17) then the Forward Euler’s Method has the obvious generalization y n+1 = yn + hf(t n , y n ) (33.18) Example 33.1. Solve y ′ = y, y(0) = 1 on the interval [0, 1] using h = 0.25. The exact solution is y = e x . We compute the values using Euler’s method. For any given time point t k , the value y k depends purely on the values of t k1 and y k1 . This is often a source of confusion for students: although the formula y k+1 = y k + hf(t k , y k ) only depends on t k and not on t k+1 it gives the value of y k+1 . We are given the following information: ⎫ (t 0 , y 0 ) = (0, 1) ⎪⎬ f(t, y) = y ⎪⎭ h = 0.25 We first compute the solution at t = t 1 . (33.19) y 1 = y 0 + hf(t 0 , y 0 ) = 1 + (0.25)(1) = 1.25 (33.20) t 1 = t 0 + h = 0 + 0.25 = 0.25 (33.21) (t 1 , y 1 ) = (0.25, 1.25) (33.22) Then we compute the solutions at t = t 1 , t 2 , . . . until t k+1 = 1. y 2 = y 1 + hf(t 1 , y 1 ) (33.23) = 1.25 + (0.25)(1.25) = 1.5625 (33.24) t 2 = t 1 + h = 0.25 + 0.25 = 0.5 (33.25) (t 2 , y 2 ) = (0.5, 1.5625) (33.26) y 3 = y 2 + hf(t 2 , y 2 ) (33.27) = 1.5625 + (0.25)(1.5625) = 1.953125 (33.28) t 3 = t 2 + h = 0.5 + 0.25 = 0.75 (33.29) (t 3 , y 3 ) = (0.75, 1.953125) (33.30)
363 y 4 = y 3 + hf(t 3 , y 3 ) (33.31) = 1.953125 + (0.25)(1.953125) = 2.44140625 (33.32) t 4 = t 3 + 0.25 = 1.0 (33.33) (t 4 , y 4 ) = (1.0, 2.44140625) (33.34) Since t 4 = 1 we are done. The solutions are tabulated in table ?? for this and other step sizes. t h = 1/2 h = 1/4 h = 1/8 h = 1/16 exact solution 0.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.0625 1.0625 1.0645 0.1250 1.1250 1.1289 1.1331 0.1875 1.1995 1.2062 0.2500 1.2500 1.2656 1.2744 1.2840 0.3125 1.3541 1.3668 0.3750 1.4238 1.4387 1.4550 0.4375 1.5286 1.5488 0.5000 1.5000 1.5625 1.6018 1.6242 1.6487 0.5625 1.7257 1.7551 0.6250 1.8020 1.8335 1.8682 0.6875 1.9481 1.9887 0.7500 1.9531 2.0273 2.0699 2.1170 0.8125 2.1993 2.2535 0.8750 2.2807 2.3367 2.3989 0.9375 2.4828 2.5536 1.0000 2.2500 2.4414 2.5658 2.6379 2.7183
- 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: 361 Figure 33.1: Illustration of Eu
- 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
363<br />
y 4 = y 3 + hf(t 3 , y 3 ) (33.31)<br />
= 1.953125 + (0.25)(1.953125) = 2.44140625 (33.32)<br />
t 4 = t 3 + 0.25 = 1.0 (33.33)<br />
(t 4 , y 4 ) = (1.0, 2.44140625) (33.34)<br />
S<strong>in</strong>ce t 4 = 1 we are done. The solutions are tabulated <strong>in</strong> table ?? for this<br />
and other step sizes.<br />
t h = 1/2 h = 1/4 h = 1/8 h = 1/16 exact solution<br />
0.0000 1.0000 1.0000 1.0000 1.0000 1.0000<br />
0.0625 1.0625 1.0645<br />
0.1250 1.1250 1.1289 1.1331<br />
0.1875 1.1995 1.2062<br />
0.2500 1.2500 1.2656 1.2744 1.2840<br />
0.3125 1.3541 1.3668<br />
0.3750 1.4238 1.4387 1.4550<br />
0.4375 1.5286 1.5488<br />
0.5000 1.5000 1.5625 1.6018 1.6242 1.6487<br />
0.5625 1.7257 1.7551<br />
0.6250 1.8020 1.8335 1.8682<br />
0.6875 1.9481 1.9887<br />
0.7500 1.9531 2.0273 2.0699 2.1170<br />
0.8125 2.1993 2.2535<br />
0.8750 2.2807 2.3367 2.3989<br />
0.9375 2.4828 2.5536<br />
1.0000 2.2500 2.4414 2.5658 2.6379 2.7183
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