Explanation Given: The boundary value problem is a ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 and the boundary conditions are u ( 0 , t ) = 0 , u ( L , t ) = 0 , ∂ u ∂ t | t = 0 = 0 and u ( x , 0 ) = f ( x ) . Calculation: The boundary value problem is a ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 . The expression for superposition principle is given below. u ( x , t ) = u 1 ( x , t ) + u 2 ( x , t ) The first problem is a ∂ 2 u 1 ∂ x 2 = ∂ 2 u 1 ∂ t 2 . The given boundary conditions are u 1 ( 0 , t ) = 0 , u 1 ( L , t ) = 0 , ∂ u 1 ∂ t | t = 0 = 0 and u 1 ( x , 0 ) = f ( x ) . Consider the equation u 1 ( x , t ) = X ( x ) T ( t ) . Separate the variables of above equation and take separation constant − λ . X ″ X = − λ and T ″ a T = − λ The equation of the variable X is X ″ + λ X = 0 . The equation of the variable T is T ″ + k λ T = 0 . If λ = α 2 > 0 , substitute α 2 for λ in equation X ″ + λ X = 0 . 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 . At boundary condition u 1 ( 0 , t ) = 0 , u 1 ( L , t ) = 0 , ∂ u 1 ∂ t | t = 0 = 0 and u 1 ( x , 0 ) = f ( x ) , c 1 = 0 . For non trivial solution c 2 ≠ 0 , α = n π L , substitute c 1 = 0 and α = n π L in equation X ( x ) = c 1 cos α x + c 2 sin α x . X ( x ) = c 2 sin n π L x Substitute α 2 for λ in equation T ″ + k λ T = 0 . T ″ + a α 2 T = 0 The general solution of the above equation is, T ( t ) = c 3 cos α a t + c 4 sin α a t ...... ( 1 ) At the given boundary conditions, c 4 = 0 . Substitute 0 for c 4 and n π L for α in the equation (1). T ( t ) = c 3 cos n π a L t Substitute c 2 sin n π L x for X ( x ) and c 3 cos n π a L t for T ( t ) in equation u 1 ( x , t ) = X ( x ) T ( t ) . u 1 ( x , t ) = c 2 c 3 cos n π a L t sin n π L x Rewrite the term c 2 c 3 = A n in the above equation and apply superposition principle as follows: u 1 ( x , t ) = ∑ n = 1 ∞ ( A n cos n π a L t ) sin n π L x At the given boundary conditions the value of A n is A n = 2 L ∫ 0 L f ( x ) sin n π L x d x . Substitute 2 L ∫ 0 L f ( x ) sin n π L x d x for A n in equation u 1 ( x , t ) = ∑ n = 1 ∞ ( A n cos n π a L t ) sin n π L x . u 1 ( x , t ) = ∑ n = 1 ∞ ( 2 L ∫ 0 L f ( x ) sin n π L x d x ) cos n π a L t sin n π L x The given first problem is a ∂ 2 u 2 ∂ x 2 = ∂ 2 u 2 ∂ t 2

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 12.4, Problem 14E

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The derivation of u(x,t)=∑n=1∞[(2L∫0Lf(x)sinnπLxdx)cosnπaLt+(2nπa∫0Lg(x)sinnπLxdx)sinnπaLt]sinnπLx using superposition principle.

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Chapter 12.4, Problem 13E

Chapter 12.4, Problem 15E

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Chapter 12 Solutions

Differential Equations with Boundary-Value Problems

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The derivation of u ( x , t ) = ∑ n = 1 ∞ [ ( 2 L ∫ 0 L f ( x ) sin n π L x d x ) cos n π a L t + ( 2 n π a ∫ 0 L g ( x ) sin n π L x d x ) sin n π a L t ] sin n π L x using superposition principle.

Solution Summary: The author calculates the boundary value problem using superposition principle.

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The derivation of u(x,t)=∑n=1∞[(2L∫0Lf(x)sinnπLxdx)cosnπaLt+(2nπa∫0Lg(x)sinnπLxdx)sinnπaLt]sinnπLx using superposition principle.

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Chapter 12 Solutions