Explanation Given: Partial differential equation ∂ 2 w ∂ x 2 − ∂ 2 w ∂ y 2 = 1 Make the substitutions as below x = u + v , y = u − v The equation can be re-written as below ∂ 2 w ∂ x 2 − ∂ 2 w ∂ y 2 = 1.......... ( 1 ) ∂ w ∂ x = ∂ w ∂ u ∂ u ∂ x + ∂ w ∂ v ∂ v ∂ x Differentiating x = u + v , y = u − v u = ( x + y ) 2 v = ( y − x ) 2 ∂ u ∂ x = 1 2 , ∂ v ∂ x = 1 2 ∂ w ∂ x = ∂ w ∂ u ( 1 2 ) + ∂ w ∂ v ( 1 2 ) Therefore ∂ w ∂ x = ∂ w ∂ u ( 1 2 ) + ∂ w ∂ v ( 1 2 ) ∂ w ∂ x = { ∂ ∂ u ( 1 2 ) + ∂ ∂ v ( 1 2 ) } w Similarly ∂ w ∂ y = ∂ w ∂ u ( − 1 2 ) + ∂ w ∂ v ( 1 2 ) ∂ w ∂ y = ∂ w ∂ u ( − 1 2 ) + ∂ w ∂ v ( 1 2 ) ∂ w ∂ y = 1 2 ( − ∂ ∂ u + ∂ ∂ v ) w ∂ 2 w ∂ x 2 = ∂ ∂ x

Mathematical Methods in the Physical Sciences

3rd Edition

ISBN: 9780471198260

Author: Mary L. Boas

Publisher: Wiley, John & Sons, Incorporated

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Question

Chapter 4.11, Problem 3P

To determine

To prove: ∂2w∂u∂v=1 for the given conditions.

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Chapter 4.11, Problem 2P

Chapter 4.11, Problem 4P

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Chapter 4 Solutions

Mathematical Methods in the Physical Sciences

Ch. 4.1 - If u=x2/x2+y2, find u/x,u/y.Ch. 4.1 - Prob. 2P Ch. 4.1 - If z=lnu2+v2+w2, find z/u,z/v,z/w.Ch. 4.1 - For w=x3y32xy+6, find 2w/ax2 and 2w/ay2 at the...Ch. 4.1 - For w=8x4+y42xy2, find 2w/x2 and 2w/y2 at the...Ch. 4.1 - For u=excosy, (a) verify that 2u/xy=2u/yx; (b)...Ch. 4.1 - Prob. 7P Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...

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To prove : ∂ 2 w ∂ u ∂ v = 1 for the given conditions.

Solution Summary: The author explains that the equation can be re-written as below.

To prove: ∂2w∂u∂v=1 for the given conditions.

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Chapter 4 Solutions