Explanation Given: The given boundary conditions are, w ( 0 , y ) = 0 , w ( a , y ) = 0 , ∂ 2 w ∂ x 2 | x = 0 = 0 and ∂ 2 w ∂ x 2 | x = a = 0 , 0 < y < b and w ( x , 0 ) = 0 , w ( x , b ) = 0 , ∂ 2 w ∂ y 2 | y = 0 = 0 and ∂ 2 w ∂ y 2 | y = b = 0 , 0 < x < a . Calculation: The given linear fourth order differential equation is, ∂ 4 w ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 w ∂ y 4 = q 0 D sin π x a sin π y b . (1) The given particular solution is, w ( x , y ) = C sin π x a sin π y b . (2) Partially differentiate both sides of the equation (2) with respect to x , ∂ w ∂ x = C π a cos π x a sin π y b Again, partially differentiate both sides of the above equation with respect to x , ∂ 2 w ∂ x 2 = − C ( π a ) 2 sin π x a sin π y b . (3) Differentiate both sides of the above equation partially with respect to x , ∂ 3 w ∂ x 3 = − C ( π a ) 3 cos π x a sin π y b Further differentiate both sides of the above equation partially with respect to x , ∂ 4 w ∂ x 4 = C ( π a ) 4 sin π x a sin π y b . (4) Now, partially differentiate both sides of the equation (3) with respect to y , ∂ 3 w ∂ x 2 ∂ y = − C ( π a ) 2 π b sin π x a cos π y b Again, partially differentiate both sides of the above equation with respect to y , ∂ 4 w ∂ x 2 ∂ y 2 = C ( π a ) 2 ( π b ) 2 sin π x a sin π y b Multiply by 2 in the above equation, 2 ∂ 4 w ∂ x 2 ∂ y 2 = 2 C ( π a ) 2 ( π b ) 2 sin π x a sin π y b . (5) Now, partially differentiate both sides of the equation (2) with respect to y , ∂ w ∂ y = C π b sin π x a cos π y b Further differentiate both sides of the above equation partially with respect to y , ∂ 2 w ∂ y 2 = − C ( π b ) 2 sin π x a sin π y b Partially differentiate both sides of the above equation with respect to y , ∂ 3 w ∂ y 3 = − C ( π b ) 3 sin π x a cos π y b On differentiating both sides of the above equation partially with respect to y gives, ∂ 4 w ∂ y 4 = C ( π b ) 4 sin π x a sin π y b . (6) Add the equations (4), (5) and (6), ∂ 4 w ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 w ∂ y 4 = C [ ( π a ) 4 + 2 ( π a ) 2 ( π b ) 2 + ( π b ) 4 ] sin π x a sin π y b Compare the above equation with the equation (1), C [ ( π a ) 4 + 2 ( π a ) 2 ( π b ) 2 + ( π b ) 4 ] sin π x a sin π y b = q 0 D sin π x a sin π y b Equate the coefficients of the above equation, q 0 D = C [ ( π a ) 4 + 2 ( π a ) 2 ( π b ) 2 + ( π b ) 4 ] C = q 0 D π 4 ( 1 a 2 + 1 b 2 ) 2 = q 0 a 4 b 4 π 4 D ( a 2 + b 2 ) 2 Substitute the value of C in equation (2), w ( x , y ) = q 0 a 4 b 4 π 4 D ( a 2 + b 2 ) 2 sin π x a sin π y b

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 13, Problem 20CR

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To prove: The given particular solution satisfies the given boundary conditions.

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Chapter 13, Problem 19CR

Chapter 14.1, Problem 1E

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Chapter 13 Solutions

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Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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