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 ∂t 2 expect 2 initial conditions u ( x, 0) = f ( x ) ∂u ∂t ( x, 0) = g ( x ) 2 u ∂x 2 expect 2 boundary conditions u (0 , t ) = 0 u ( L, t ) = 0 . (19.2) 19.1 Guitar String Note: ∂t + c ∂x ∂t c ∂x u ( x, t ) = 0 . (19.3) 117
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Lecture 23 - 1D Wave Equation Let w = ∂t c ∂x u then ∂w ∂t + c ∂w ∂x = 0 . 1st order wave eq. (19.4) Thus ∂t + c ∂x right moving wave (19.5) ∂t c ∂x left moving wave . (19.6) Claim: u 1 ( x, t ) = G ( x + ct ) is a solution to (19.1) u t = cG u tt = c 2 G (19.7) u x = G u xx = G . (19.8) Therefore u tt c 2 u xx = c 2 G 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 ∂t 2 ( α 1 u 1 + α 2 u 2 ) = α 1 2 u 1 ∂t 2 + α 2 2 u 2 ∂t 2 = α 1 c 2 2 u 1 ∂x 2 + α 2 c 2 2 u 2 ∂x 2 since u 1 and u 2 solve (19.1) (19.11) Thus 2 ∂t 2 ( α 1 u 1 + α 2 u 2 ) = c 2
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