M257-316Notes_Lecture23

M257-316Notes_Lecture23 - Chapter 19 Lecture 23 - 1D Wave...

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Chapter 19 Lecture 23 - 1D Wave Equation 2 u ∂t 2 = c 2 2 u ∂x 2 (19.1) 2 u 2 expect 2 initial conditions u ( x, 0) = f ( x ) ∂u ( x, 0) = g ( x ) 2 u 2 expect 2 boundary conditions u (0 ,t )=0 u ( L, t . (19.2) 19.1 Guitar String Note: µ + c ¶µ c u ( x, t . (19.3) 117
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Lecture 23 - 1D Wave Equation Let w = µ ∂t c ∂x u then ∂w + c =0 . 1st order wave eq. (19.4) Thus + c right moving wave (19.5) c left moving wave . (19.6) Claim: u 1 ( x, t )= G ( x + ct ) is a solution to (19.1) u t = cG 0 u tt = c 2 G 0 (19.7) u x = G 0 u xx = G 0 . (19.8) Therefore u tt c 2 u xx = c 2 G 0 c 2 G 0 . (19.9) Similarly u 2 ( x, t F ( x ct ) is also a solution to (19.1). Is the sum of two solutions also a solution? Claim u ( x, t α 1 u 1 ( x, t )+ α 2 u 2 ( x, t ) is a solution of (19.1) if u 1 and u 2 are solutions. 2 2 ( α 1 u 1 + α 2 u 2 α 1 2 u 1 2 + α 2 2 u 2 2 = α 1 c 2 2 u 1 2 + α 2 c 2 2 u 2 2 since u 1 and u 2 solve (19.1) (19.11) Thus 2 2 ( α 1 u 1 + α 2 u 2 c 2 2 2 ( α 1 u 1 + α 2 u 2 ) .
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M257-316Notes_Lecture23 - Chapter 19 Lecture 23 - 1D Wave...

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