M257-316Notes_Lecture25

# M257-316Notes_Lecture25 - Chapter 21 Lecture 25 Solution by...

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

Chapter 21 Lecture 25 Solution by separation of variables Example 21.1 u tt = c 2 u xx 0 <x<L, t> 0 (21.1) BC: u (0 ,t )=0 ,u ( L, t (21.2) IC: u ( x, 0) = f ( x ) t ( x, 0) = g ( x ) (21.3) For a guitar string c = s T 0 ρ 0 . Separate Variables u ( x, t )= X ( x ) T ( t ) ¨ T ( t ) c 2 T ( t ) = X 0 ( x ) X ( x ) = λ 2 (21.4) ¨ T ( t )+ λ 2 c 2 T ( t T ( t c 1 cos( λct c 2 sin( λct ) (21.5) X 0 + λ 2 X =0 X (0) = 0 = X ( L ) ¾ X ( x A cos( λx B sin λx X (0) = A X ( L B sin λL ¾ λ n = L n =1 , 2 ,... X n = sin ³ nπx L ´ . 125

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

View Full Document
Lecture 25 Solution by separation of variables Therefore u ( x, t )= X n =1 A n cos µ nπct L sin ³ nπx L ´ + B n sin µ nπct L sin ³ nπx L ´ (21.6) u ( x, 0) = X n =1 A n sin ³ nπx L ´ = f ( x ) A n = 2 L L R 0 f ( x ) sin ( nπx L ) (21.7) u t ( x, t X n =1 A n ³ nπc L ´ sin µ nπct L sin ³ nπx L ´ + B n ³ nπc L ´ cos µ nπct L sin ³ nπx L ´ (21.8) u t ( x, 0) = X n =1 B n ³ nπc L ´ sin ³ nπx L ´ = g ( x ) B n ( nπc L ) = 2 L L R 0 g ( x ) sin ( nπx L ) dx . (21.9) Therefore u ( x, t X n =1 ½ A n cos µ nπct L + B n sin µ nπct L ¶¾ sin ³ nπx L ´ . (21.10) 21.1 Notes 1. Period and Frequency: cos ³ nπc L ( t + T ) ´
This is the end of the preview. Sign up to access the rest of the 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 / 4

M257-316Notes_Lecture25 - Chapter 21 Lecture 25 Solution by...

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

View Full Document
Ask a homework question - tutors are online