M257-316Notes_Lecture24

# M257-316Notes_Lecture24 - x − ct ) + u ( x + ct )] (20.5)...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 20 Lecture 24 - Space-Time Interpretation of D’Alembert’s Solution u ( x, t )= 1 2 [ u 0 ( x ct )+ u 0 ( x + ct )] + 1 2 c x + ct Z x ct v 0 ( s ) ds (20.1) 20.1 Characteristics In the x t plane the lines x ct = x 0 and x + ct = x 0 (20.2) are called characteristics. x ct = x 0 t = 1 c x 1 c x 0 x + ct = x 0 t = 1 c x + 1 c x 0 (20.3) 121

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

View Full Document
Lecture 24 - Space-Time Interpretation of D’Alembert’s Solution x + ct = x 0 thus t = 1 c x 0 + 1 c x 0 1 c x ct = x 0 thus t = 1 c x 1 c x 0 1 c 20.2 Region of Inﬂuence The lines x + ct = x 0 and x ct = x 0 bound the region of inﬂuence. 20.3 Domain of Dependence The lines x = x 0 ct 0 and x = x 0 + ct 0 that pass through the point ( x 0 ,t 0 ) bound the domain of dependence. 122
20.3. DOMAIN OF DEPENDENCE Example 20.1 Special Case: u ( x, 0) = ½ 1 | x | < 1 0 | x | > 1 (20.4) u ( x, t )= 1 2 [ u 0 (
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x − ct ) + u ( x + ct )] (20.5) Let c = 1. t = 1 2 : x r − 1 2 = 1 ⇒ x r = 3 2 x R + 1 2 = 1 x R = 1 2 x ` − 1 2 = − 1 ⇒ x ` = − 1 2 x L + 1 2 = − 1 x L = − 3 2 (20.6) t = 1: x r − 1 = 1 ⇒ x r = 2 x R + 1 = 1 ⇒ x R = 1 x ` − 1 = − 1 ⇒ x ` = 0 x ` + 1 = − 1 ⇒ x L = − 2 (20.7) t = 2: x r − 2 = 1 ⇒ x r = 3 x R + 2 = 1 ⇒ x R = − 1 x ` − 2 = − 1 ⇒ x ` = 1 x L + 2 = − 1 ⇒ x L = − 3 (20.8) 123...
View Full Document

## This note was uploaded on 05/04/2011 for the course MATH 25 taught by Professor Lo during the Spring '11 term at BC.

### Page1 / 3

M257-316Notes_Lecture24 - x − ct ) + u ( x + ct )] (20.5)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online