M257-316Notes_Lecture24

M257-316Notes_Lecture24 - x − ct u x ct(20.5 Let c = 1 t...

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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 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
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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
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20.3. DOMAIN OF DEPENDENCE Example 20.1 Special Case: u ( x, 0) = 1 | x | < 1 0 | x | > 1 (20.4) u ( x, t ) =
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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...
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