{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 −...
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online