Explanation Given: The given boundary value problem is ∂ 2 u ∂ x 2 + sin x cos t = ∂ 2 u ∂ t 2 , 0 < x < π , t > 0 and the given boundary conditions are u ( 0 , t ) = 0 , u ( π , t ) = 0 , u ( x , 0 ) = 0 and ∂ u ∂ t | t = 0 = 0 . Calculation: The given boundary value problem is, ∂ 2 u ∂ x 2 + sin x cos t = ∂ 2 u ∂ t 2 . (1) When λ = α 2 , α > 0 . Consider the equation, u ( x , t ) = X ( x ) . (2) Separate the variable of equation (2), X ″ X = − λ The equation of the variable X is, X ″ + λ X = 0 Substitute λ = α 2 in the above equation, 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 . (3) At boundary conditions u ( 0 , t ) = 0 , u ( π , t ) = 0 , u ( x , 0 ) = 0 and ∂ u ∂ t | t = 0 = 0 , c 1 = 0 For non-trivial solution c 2 ≠ 0 , α = n Substitute c 1 = 0 and α = n in the equation (3), X ( x ) = sin n x For fixed t the variable u can be written as, u ( x , t ) = ∑ n = 1 ∞ u n ( t ) sin n x . (4) Partially differentiate both sides of the equation (4) with respect to x , ∂ u ∂ x = ∑ n = 1 ∞ u n ( t ) ( n ) cos n x Partially differentiate both sides of the above equation with respect to x , ∂ 2 u ∂ x 2 = ∑ n = 1 ∞ u n ( t ) ( − n 2 ) sin n x . (5) Partially differentiate both sides of the equation (4) with respect to t , ∂ u ∂ t = ∑ n = 1 ∞ u ′ n ( t ) sin n x Partially differentiate both sides of the above equation with respect to t , ∂ 2 u ∂ t 2 = ∑ n = 1 ∞ u ″ n ( t ) sin n x

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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