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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Chapter 4.11, Problem 3P

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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

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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