Explanation Given: The given function is u = f ( x − c t ) + g ( x + c t ) Formula Used: If u = f ( z ) and z = v ( x , y ) then by chain rule ∂ u ∂ v = ∂ u ∂ z ⋅ ∂ z ∂ x + ∂ u ∂ z ⋅ ∂ z ∂ y Calculation: Consider the given equation. u = f ( x − c t ) + g ( x + c t ) …… (1) Let us take y = x − c t …… (2) And z = x + c t …… (3) So, equation (1) becomes u = f ( y ) + g ( z ) ……(4) Differentiate equation (4) partially with respect to ‘ x’ . ∂ u ∂ x = f ′ ( y ) ∂ y ∂ x + g ′ ( z ) ∂ z ∂ x = f ′ ( y ) ∂ ∂ x ( x − c t ) + g ′ ( z ) ∂ ∂ x ( x + c t ) [ from (2) and (3) ] = f ′ ( y ) ( 1 ) + g ′ ( z ) ( 1 ) = f ′ ( y ) + g ′ ( z ) Differentiating again with respect to ‘ x’ we get ∂ 2 u ∂ x 2 = f ′ ′ ( y ) ∂ y ∂ x + g ′ ′ ( z ) ∂ z ∂ x ∂ 2 u ∂ x 2 = f ′ ′ ( y ) ∂ ∂ x ( x − c t ) + g ′ ′ ( z ) ∂ ∂ x ( x + c t ) [ from (2) and (3) ] ∂ 2 u ∂ x 2 = f ′ ′ ( y ) ( 1 ) + g ′ ′ ( z ) ( 1 ) ∂ 2 u ∂ x 2 = f ′ ′ ( y ) + g ′ ′ ( z ) …… (5) Differentiate equation (4) partially with respect to ‘ t’

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 4.7, Problem 23P

To determine

To show: That ∂2u∂x2=1c2∂2u∂t2 for the given function.

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Chapter 4.7, Problem 22P

Chapter 4.7, Problem 24P

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Chapter 4 Solutions

Mathematical Methods in the Physical Sciences

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Knowledge Booster

That ∂ 2 u ∂ x 2 = 1 c 2 ∂ 2 u ∂ t 2 for the given function.

Solution Summary: The author explains that the given function is u=f(x-ct) + z=v

To show: That ∂2u∂x2=1c2∂2u∂t2 for the given function.

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Chapter 4 Solutions