Explanation Given: The given boundary value problem is ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 and the given boundary conditions are u ( 0 , t ) = 1 − e − t , u ( 1 , t ) = 1 − e − t and u ( x , 0 ) = 0 . Calculation: The given boundary value problem is, ∂ 2 u ∂ x 2 = ∂ u ∂ t . (1) Consider the equation, u ( x , t ) = v ( x , t ) + ψ ( x , t ) . (2) By using the formula, ψ ( x , t ) = u ( 0 , t ) + x L [ u ( 1 , t ) − u ( 0 , t ) ] In the given boundary value problem L = 1 At given boundary conditions u ( 0 , t ) = 1 − e − t , u ( 1 , t ) = 1 − e − t and u ( x , 0 ) = 0 , ψ ( x , t ) = 1 − e − t . (3) Substitute ψ ( x , t ) = 1 − e − t in equation (2), u ( x , t ) = v ( x , t ) + 1 − e − t . (4) Partially differentiate both sides of the equation (4) with respect to t , ∂ u ∂ t = ∂ v ∂ t + e − t Substitute ∂ u ∂ t = ∂ v ∂ t + e − t and ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 in equation (1), ∂ 2 v ∂ x 2 − e − t = ∂ v ∂ t . (5) When λ = α 2 , α > 0 Consider the equation, v ( x , t ) = X ( x ) . (6) Separate the variable of equation (6), X ″ X = − λ The equation of the variable X is, X ″ + λ X = 0 . (7) Substitute λ = α 2 in equation (7), X ″ + α 2 X = 0 The general solution of the above equation is, X ( x ) = c 1 cos α x + c 2 sin α x (9) At boundary conditions u ( 0 , t ) = 1 − e − t , u ( 1 , t ) = 1 − e − t and u ( x , 0 ) = 0 , c 1 = 0 For non-trivial solution c 2 ≠ 0 , α = n π Substitute c 1 = 0 and α = n π in the equation (9), X ( x ) = c 2 sin n π x . (10) The equation with G ( x , t ) is expressed as, G ( x , t ) = − e − t For fixed t the variables v and G can be written as, v ( x , t ) = ∑ n = 1 ∞ v n ( t ) sin n π x . (11) And the second equation is, G ( x , t ) = ∑ n = 1 ∞ G n ( t ) sin n π x . (12) Consider t as a parameter then the coefficient G n is, G n = 2 ∫ 0 1 e − t sin n π x d x = 2 e − t ∫ 0 1 sin n π x d x = 2 e − t ( − 1 ) n − 1 n π Substitute G ( x , t ) = − e − t and G n = 2 e − t ( − 1 ) n − 1 n π in equation (12), − e − t = 2 e − t ∑ n = 1 ∞ ( − 1 ) n − 1 n π sin n π x

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 13.6, Problem 16E

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