Explanation Given: The given boundary value problem is k ∂ 2 u ∂ x 2 − h u = ∂ u ∂ t , 0 < x < π , t > 0 and the given boundary conditions are u ( 0 , t ) = 0 , u ( π , t ) = u 0 and u ( x , 0 ) = 0 . Calculation: The given boundary value problem is, k ∂ 2 u ∂ x 2 − h u = ∂ 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 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 ) a n d ∂ u ∂ t = ∂ v ∂ t in equation (1), k ∂ 2 v ∂ x 2 + k ψ ″ ( x ) − h v − h ψ = ∂ v ∂ t . (3) Second derivative of ψ is zero in the equation (3) by homogeneous system, k ψ ″ ( x ) − h ψ = 0 The solution of the above equation is expressed as follows. ψ ( x ) = c 1 cosh h k x + c 2 sinh h k x . (4) At given boundary conditions u ( 0 , t ) = 0 and u ( π , t ) = u 0 , ψ ( 0 ) = 0 and ψ ( π ) = 0 . Substitute ψ ( 0 ) = 0 and ψ ( π ) = 0 in equation (4), c 1 = 0 and c 2 = u 0 sinh h k π Substitute c 1 = 0 and c 2 = u 0 sinh h k π in equation (4), ψ ( x ) = u 0 sinh h k x sinh h k π . (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 ′ + h T k T = − λ The equation of the variable X is, X ″ + λ X = 0 . (7) The equation of the variable T is, T ′ + ( h + λ k ) 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

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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