Chem Differential Eq HW Solutions Fall 2011 2

# Chem Differential Eq HW Solutions Fall 2011 2 - 2 Chapter 1...

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Unformatted text preview: 2 Chapter 1 A Preview of Applications and Techniques Exercises 1.2 1. We have ∂ ∂t ∂u ∂t =- ∂ ∂t ∂v ∂x and ∂ ∂x ∂v ∂t =- ∂ ∂x ∂u ∂x . So ∂ 2 u ∂t 2 =- ∂ 2 v ∂t∂x and ∂ 2 v ∂x∂t =- ∂ 2 u ∂x 2 . Assuming that ∂ 2 v ∂t∂x = ∂ 2 v ∂x∂t , it follows that ∂ 2 u ∂t 2 = ∂ 2 u ∂x 2 , which is the one dimensional wave equation with c = 1. A similar argument shows that v is a solution of the one dimensional wave equation. 5. (a) We have u ( x, t ) = F ( x + ct ) + G ( x- ct ). To determine F and G , we use the initial data: u ( x, 0) = 1 1 + x 2 ⇒ F ( x ) + G ( x ) = 1 1 + x 2 ; (1) ∂u ∂t ( x, 0) = 0 ⇒ cF ( x )- cG ( x ) = 0 ⇒ F ( x ) = G ( x ) ⇒ F ( x ) = G ( x ) + C, (2) where C is an arbitrary constant. Plugging this into (1), we find 2 G ( x ) + C = 1 1 + x 2 ⇒ G ( x ) = 1 2 1 1 + x 2- C ; and from (2) F ( x ) = 1 2 1 1 + x 2 + C ....
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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