Unformatted text preview: ∂f 1 ∂u ∂u ∂t =c f f 1 , ∂f 2 ∂t = ∂f 2 ∂v ∂v ∂t =c f f 2 , ∂ 2 f 1 ∂x 2 = n 2 x f 00 1 , ∂ 2 f 2 ∂x 2 = n 2 x f 00 2 , ∂ 2 f 1 ∂y 2 = n 2 y f 00 1 , ∂ 2 f 2 ∂y 2 = n 2 y f 00 2 , ∂ 2 f 1 ∂z 2 = n 2 z f 00 1 , ∂ 2 f 2 ∂z 2 = n 2 z f 00 2 , ∂ 2 f 1 ∂t 2 = c 2 f f 00 1 , ∂ 2 f 2 ∂t 2 = c 2 f f 00 2 . Therefore, we ﬁnd ∇ 2 p = ∂ 2 p ∂x 2 + ∂ 2 p ∂y 2 + ∂ 2 p ∂z 2 = ( n 2 x + n 2 y + n 2 z ) ( f 00 1 + f 00 2 ) = f 00 1 + f 00 2 , where we have made used of the fact that n is a unit vector  n  = 1. On the other hand, we also ﬁnd 1 c 2 ∂ 2 p ∂t 2 = 1 c 2 f ( c 2 f f 00 1 + c 2 f f 00 2 ) = f 00 1 + f 00 2 . Thus, equation (1) is indeed a solution of the acoustic wave equation. 1...
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 Fall '04
 POLLACK
 Physics, Magnetism, Partial differential equation, wave equation, plane wave

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