Explanation Given: The given boundary value problem is ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 and the given boundary condition are u ( 0 , t ) = u 0 , ∂ u ∂ x | x = 1 = − u ( 1 , t ) + u 1 and u ( x , 0 ) = u 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 ) (2) Partially differentiate both sides of the equation (2) with respect to x , ∂ u ∂ x = ∂ v ∂ x + ψ ′ ( x ) Partially differentiate both sides of the above equation with respect to x , ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) Substitute ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) a n d ∂ u ∂ t = ∂ v ∂ t in equation (1), ∂ 2 v ∂ x 2 + ψ ″ ( x ) = ∂ v ∂ t (3) Second derivative of ψ is zero in the equation (3) by homogeneous system, ψ ″ ( x ) = 0 Integrate the above equation with respect to x , ψ ′ ( x ) = c 1 Integrate the above equation with respect to x , ψ ( x ) = c 1 x + c 2 (4) At boundary conditions u ( 0 , t ) = u 0 , ∂ u ∂ x | x = 1 = − u ( 1 , t ) + u 1 and u ( x , 0 ) = u 0 , ψ ( 0 ) = 0 and ψ ′ ( 1 ) + ψ ( 1 ) = u 1 Substitute ψ ( 0 ) = 0 and ψ ′ ( 1 ) + ψ ( 1 ) = u 1 in equation (4), c 1 = 1 2 ( u 1 − u 0 ) and c 2 = u 0 Substitute c 1 = 1 2 ( u 1 − u 0 ) and c 2 = u 0 in equation (4), ψ ( x ) = 1 2 ( u 1 − u 0 ) x + u 0 (5) When λ = α 2 , α > 0 Consider the equation, v ( x , t ) = X ( x ) T ( t ) (6) Separate the variables of equation (6), X ″ X = − λ and T ′ T = − λ The equation of the variable X is, X ″ + λ X = 0 (7) The equation of the variable T is, T ′ + λ T = 0 (8) 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 ) = u 0 , ∂ u ∂ x | x = 1 = − u ( 1 , t

Bundle: Differential Equations with Boundary-Value Problems, 9th + WebAssign Printed Access Card for Zill's Differential Equations with Boundary-Value Problems, 9th Edition, Single-Term

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ISBN: 9781337604918

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 12, Problem 15RE

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