Explanation Given: The given heat equation is k ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) = ∂ u ∂ t and the given boundary conditions are ∂ u ∂ x | x = 0 = 0 , ∂ u ∂ x | x = 1 = 0 , ∂ u ∂ y | y = 0 = 0 , ∂ u ∂ y | y = 1 = 0 and u ( x , y , 0 ) = x y . Calculation: The given heat equation is, k ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) = ∂ u ∂ t . (1) Consider the equation, u ( x , y , t ) = X ( x ) Y ( y ) T ( t ) . (2) Separate the variables of equation (2), X ″ X = − α 2 and − Y ″ Y + T ′ k T = − α 2 The equation of the variable X is, X ″ + α 2 X = 0 . (3) The equation of the variable Y and T is, Y ″ Y = T ′ k T + α 2 . (4) Separate the variables of equation (4), Y ″ Y = − β 2 and T ′ k T + α 2 = − β 2 The equation of the variable Y is, Y ″ + β 2 Y = 0 . (5) The equation of the variable T is, T ′ + k ( α 2 + β 2 ) T = 0 . (6) The general solution of the equation (3) is, X ( x ) = c 1 cos α x + c 2 sin α x . (7) The general solution of the equation (5) is, Y ( y ) = c 3 cos β y + c 4 sin β y . (8) The general solution of the equation (6) is, T ( t ) = c 5 e − k ( α 2 + β 2 ) t . (9) At boundary condition ∂ u ∂ x | x = 0 = 0 , X ′ ( x ) = − α c 1 sin α x + α c 2 cos α x c 2 = 0 At boundary condition ∂ u ∂ y | y = 0 = 0 , Y ′ ( y ) = − β c 3 sin β y + β c 4 cos β y c 4 = 0 At boundary conditions ∂ u ∂ x | x = 1 = 0 and ∂ u ∂ y | y = 1 = 0 , α = m π and β = n π Substitute c 2 = 0 and α = m π in equation (7), X ( x ) = c 1 cos m π x . (10) Substitute c 4 = 0 and β = n π in equation (8), Y ( y ) = c 3 cos n π y . (11) Substitute α = m π and β = n π in equation (9) T ( t ) = c 5 e − k π 2 ( m 2 + n 2 ) t

WEB ASSIGN FOR ZILL'S DIFFERENTIAL EQUAT

9th Edition

ISBN: 9780357539545

Author: ZILL

Publisher: CENGAGE L

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Chapter 12.8, Problem 2E

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The heat equation under given boundary conditions.

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