Explanation Given: The given boundary conditions are u ( 0 , 1 ) = 100 and u ( π , y ) = e − y . Calculation: Consider the equation u ( x , y ) as follows: u ( x , y ) = u 1 ( x , y ) + u 2 ( x , y ) + u 3 ( x , y ) ...... ( 1 ) The boundary value problem to be solved is as follows: ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 ...... ( 2 ) Take Fourier sine transform with respect to x on both sides of equation (2). F { ∂ 2 u 1 ∂ x 2 } + F { ∂ 2 u 1 ∂ y 2 } = 0 d 2 U 1 d x 2 − α 2 U 1 ( α , y ) = 0 The particular solution of the equation d 2 U 1 d x 2 − α 2 U 1 ( α , y ) = 0 is obtained follows: U 1 ( α , y ) = c 1 e − α y + c 2 e α y Using variable separation method the solution of the above equation is written as below: u 1 ( x , y ) = ∑ n = 1 ∞ A n e − n y sin n x ...... ( 3 ) At the given boundary conditions u ( 0 , 1 ) = 100 and u ( π , y ) = e − y , the value of A n is obtained as follows: A n = 2 π ∫ 0 π f ( x ) sin n x d x Substitute the value of A n in equation (3). u 1 ( x , y ) = ∑ n = 1 ∞ ( 2 π ∫ 0 π f ( x ) sin n x d x ) e − n y sin n x ...... ( 4 ) Further, take Fourier sine transform with respect to y on both sides of equation (2). F { ∂ 2 u 2 ∂ x 2 } + F { ∂ 2 u 2 ∂ y 2 } = 0 d 2 U 2 d x 2 − α 2 U 2 ( x , α ) = 0 The particular solution of the equation d 2 U 2 d x 2 − α 2 U 2 ( x , α ) = 0 is obtained follows: U 2 ( x , α ) = c 1 cosh α x + c 2 sinh α x At boundary condition u 2 ( 0 , 1 ) = 100 the expression of U 2 ( x , α ) is obtained as follows: U 2 ( x , α ) = 200 π ( 1 − cos α ) sinh α ( π − x ) α sinh α π Take inverse Fourier transform of the above equation and apply Fourier sine transform as follows: F − 1 { U 2 ( x , α ) } = F − 1 { 200 π ( 1 − cos α ) sinh α ( π − x ) α sinh α π } u 2 ( x , y ) = 200 π ∫ 0 ∞ ( 1 − cos α ) sinh α ( π − x ) α sinh α π sin α y d α ...... ( 5 ) For second boundary condition take Fourier sine transform with respect to y of both sides of equation (2). F { ∂ 2 u 3 ∂ x 2 } + F { ∂ 2 u 3 ∂ y 2 } = 0 d 2 U 3 d x 2 − α 2 U 3 ( x , α ) = 0 The particular solution of the above equation is obtained as follows: U 3 ( x , α ) = c 1 cosh α x + c 2 sinh α x

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 15.4, Problem 18E

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The steady state temperature u(x,y) for the plate under given boundary conditions.

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Chapter 15.4, Problem 17E

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Chapter 15 Solutions

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The steady state temperature u ( x , y ) for the plate under given boundary conditions.

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The steady state temperature u(x,y) for the plate under given boundary conditions.

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