Explanation Given: The given partial differential equation is k ∂ 2 u ∂ x 2 + r = ∂ u ∂ t , 0 < x < 1 , t > 0 and the given boundary conditions are u ( 0 , t ) = u 0 , u ( 1 , t ) = u 0 and u ( x , 0 ) = 0 . Calculation: The given partial differential equation is, k ∂ 2 u ∂ x 2 + r = ∂ 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), k ∂ 2 v ∂ x 2 + k ψ ″ ( x ) + r = ∂ v ∂ t . (3) Second derivative of ψ is zero in the equation (3) by homogeneous system, k ψ ″ ( x ) + r = 0 So, the equation is, ψ ″ ( x ) = − r k Integrate the above equation, ψ ′ ( x ) = − r k x + c 1 Integrate the above equation, ψ ( x ) = − r 2 k x 2 + c 1 x + c 2 . (4) At given boundary conditions u ( 0 , t ) = u 0 and u ( 1 , t ) = u 0 , ψ ( 0 ) = u 0 and ψ ( 1 ) = u 0 Substitute ψ ( 0 ) = u 0 and ψ ( 1 ) = u 0 in equation (4), c 1 = r 2 k and c 2 = u 0 Substitute c 1 = r 2 k and c 2 = u 0 in equation (4), ψ ( x ) = u 0 + r 2 k ( x − x 2 ) . (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 ′ k T = − λ The equation of the variable X is, X ″ + λ X = 0 . (7) The equation of the variable T is, T ′ + 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 . (9) At boundary conditions u ( 0 , t ) = u 0 , u ( 1 , t ) = u 0 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

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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