Chem Differential Eq HW Solutions Fall 2011 2

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 tx and 2 v xt =- 2 u x 2 . Assuming that 2 v tx = 2 v xt , 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 ....
View Full Document

Ask a homework question - tutors are online