Explanation Given: The given partial differential equation is ∂ 2 u ∂ x 2 + x 2 = ∂ 2 u ∂ t 2 , t > 0 and the given boundary conditions are u ( 0 , t ) = 0 , u ( 1 , t ) = 0 , u ( x , 0 ) = f ( x ) and ∂ u ∂ t | t = 0 = 0 . Calculation: The given partial differential equation is, ∂ 2 u ∂ x 2 + x 2 = ∂ 2 u ∂ t 2 , t > 0 . (1) Consider the equation, u ( x , t ) = v ( x , t ) + ψ ( x , t ) . (2) Partially differentiate both sides of the equation (2) with respect to x as follows. ∂ u ∂ x = ∂ v ∂ x + ψ ′ ( x ) Partially differentiate both sides of the above equation with respect to x as follows. ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) Substitute ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) and ∂ 2 u ∂ t 2 = ∂ 2 v ∂ t 2 in equation (1), ∂ 2 v ∂ x 2 + ψ ″ ( x ) + x 2 = ∂ 2 v ∂ t 2 Second derivative of ψ is zero in the above equation by homogeneous system, ψ ″ ( x ) + x 2 = 0 Integrate the above equation with respect to x as follows. ψ ′ ( x ) = − x 3 3 + c 1 Integrate the above equation with respect to x as follows. ψ ( x , t ) = − x 4 12 + c 1 x + c 2 . (3) At boundary conditions u ( 0 , t ) = 0 , u ( 1 , t ) = 0 and ∂ u ∂ t | t = 0 = 0 , ψ ( 0 ) = 0 and ψ ( 1 ) = 0 c 1 = − 11 12 and c 2 = 1 Substitute c 1 = − 11 12 and c 2 = 1 in equation (3), ψ ( x , t ) = − x 4 12 − 11 12 x + 1 Substitute ψ ( x , t ) = − x 4 12 − 11 12 x + 1 in equation (2), u ( x , t ) = − x 4 12 − 11 12 x + 1 + v ( x , t ) . (4) When λ = α 2 , α > 0 . Consider the equation, v ( x , t ) = X ( x ) T ( t ) . (5) Separate the variables of equation (5), X ″ X = − λ and T ′ k T = − λ The equation of the variable X is, X ″ + λ X = 0 . (6) The equation of the variable T is, T ′ + λ k T = 0

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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The displacement u ( x , t ) under given boundary conditions.

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