chapter10 - “book” — 2005/4/7 — 22:31 — page S-84...

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Unformatted text preview: “book” — 2005/4/7 — 22:31 — page S-84 — #105 Answers to Selected Exercises for Chapter 10 Section 10.1 (page 815) 1. (a) With t f = 0 . 1 and ∆ t = 10 − 5 ∆ x Maximum Error Error Ratio rms Error Error Ratio 1/5 0.0037159688 0.0027628082 1/10 0.0007940482 4.6798 0.0005903717 4.6798 1/20 0.0001971695 4.0272 0.0001394199 4.2345 1/40 0.0000452154 4.3607 0.0000319721 4.3607 (b) With t f = 0 . 1 and ∆ x = 10 − 3 ∆ t Maximum Error Error Ratio rms Error Error Ratio 1/50 0.0112003009 0.0079198087 1/100 0.0052808980 2.1209 0.0037341588 2.1209 1/200 0.0025299535 2.0873 0.0017889473 2.0873 1/400 0.0012326399 2.0525 0.0008716080 2.0525 1/800 0.0006076075 2.0287 0.0004296434 2.0287 3. (c) Time discretization: with t f = 1 and ∆ x = π/ 1000 ∆ t Maximum Error Error Ratio rms Error Error Ratio 1/5 0.0009542298 0.0006533598 1/10 0.0002453850 3.8887 0.0001629630 4.0093 1/20 0.0000617417 3.9744 0.0000406191 4.0120 1/40 0.0000153004 4.0353 0.0000100349 4.0478 1/80 0.0000036557 4.1854 0.0000023887 4.2009 Spatial discretization: with t f = 1 and ∆ t = 10 − 3 ∆ x Maximum Error Error Ratio rms Error Error Ratio π/ 5 0.0088353502 0.0064419887 π/ 10 0.0022991404 3.8429 0.0016060232 4.0111 π/ 20 0.0005729541 4.0128 0.0004012384 4.0027 π/ 40 0.0001431207 4.0033 0.0001002815 4.0011 π/ 80 0.0000357555 4.0028 0.0000250572 4.0021 5. The solutions computed with 800 times steps and with 777 time steps are shown in Figure S.5(a). With 800 time steps, λ = 1 / 2, but with 777 time steps, λ ≈ . 515. 7. The solutions shown in Figure S.5(b) were computed using the Crank-Nicolson scheme with ∆ t = 10 − 2 . S-84 “book” — 2005/4/7 — 22:31 — page S-85 — #106 Selected answers for Section 10.1 S-85 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 800 times steps 777 time steps 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 t=1 t=2 t=3 t=4 Figure S.5 (a, left) Figure for solution to Exercise 5 (Section 10.1). (b, right) Figure for solution to Exercise 7 (Section 10.1). 9. The solutions shown in Figure S.6 were computed using the Crank-Nicolson scheme with ∆ T = 2 × 10 − 3 and ∆ Z = 1 / 20.-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 T=0.05 T=0.1 T=0.25 T=1 Figure S.6 Figure for solution to Exercise 9 (Section 10.1). 11. (c) With t f = 0 . 1 ∆ x ∆ t Maximum Error Error Ratio rms Error Error Ratio 1/5 1/150 0.0001238856 0.0000921084 1/10 1/600 0.0000068990 17.9569 0.0000051294 17.9569 1/20 1/2400 0.0000004413 15.6335 0.0000003120 16.4380 1/40 1/9600 0.0000000274 16.1067 0.0000000194 16.1067 1/80 1/38400 0.0000000017 16.0260 0.0000000012 16.0267 “book” — 2005/4/7 — 22:31 — page S-86 — #107 S-86 Answers to Selected Exercises for Chapter 10 Section 10.2 (page 826) 1. (a) between 205 and 210 time steps, t ≈ 1 . 11 (b) between 300 and 305 time steps, t ≈ 1 . 58 (c) between 555 and 560 time steps, t ≈ 2 . 86 3. Note that r = [1 −...
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chapter10 - “book” — 2005/4/7 — 22:31 — page S-84...

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