Explanation Given: The boundary conditions are as follows: a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 0 < X < L , t > 0 u ( 0 , t ) = 0 u ( x , 0 ) = 0 E ∂ u ∂ x | x = L = F 0 and ∂ u ∂ t | x = L = 0 Calculation: The boundary conditions are given as follows: a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 0 < x < L , t > 0 u ( 0 , t ) = 0 u ( x , 0 ) = 0 E ∂ u ∂ x | x = L = F 0 and ∂ u ∂ t | x = L = 0 The differential equation is given as follows: a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 … … ( 1 ) Substitute u ( x , t ) = v ( x , t ) + ψ ( x ) equation ( 1 ) . a 2 ∂ 2 ( v ( x , t ) + ψ ( x ) ) ∂ x 2 = ∂ 2 ( v ( x , t ) + ψ ( x ) ) ∂ t 2 a 2 ( ∂ 2 v ∂ x 2 + ψ ″ ( x ) ) = ∂ 2 v ∂ t 2 The above equation is homogenous only if the value of ψ ″ ( x ) is 0 . The general solution of the equation ψ ″ ( x ) = 0 is as follows: ψ ( x ) = c 1 x + c 2 … … ( 2 ) Substitute 0 for x in u ( x , t ) = v ( x , t ) + ψ ( x ) . u ( 0 , t ) = v ( 0 , t ) + ψ ( 0 ) ψ ( 0 ) = 0 ( u ( 0 , t ) = 0 ) Substitute 0 for x in equation ( 2 ) . ψ ( 0 ) = ( c 1 ⋅ 0 ) + c 2 c 2 = 0 Substitute 0 for c 2 in equation ( 1 ) . ψ ( x ) = c 1 x … … ( 3 ) Differentiate the equation u ( x , t ) = v ( x , t ) + ψ ( x ) with respect to x . u ′ ( x , t ) = v ′ ( x , t ) + ψ ′ ( x ) Substitute the value of u ′ ( x , t ) in E ∂ u ∂ x | x = L = F 0 . E ( v ′ ( x , t ) + ψ ′ ( x ) ) | x = L = F 0 E ( v ′ ( L , t ) + ψ ′ ( L ) ) = F 0 … … ( 4 ) For the above equation to be homogenous the value v ′ ( L , t ) is 0 . Substitute v ′ ( L , t ) = 0 in equation ( 4 ) . E ψ ′ ( L ) = F 0 ψ ′ ( L ) = F 0 E Differentiate equation ( 3 ) with respect to x . ψ ′ ( x ) = c 1 Substitute L for x in the above equation. ψ ′ ( L ) = c 1 c 1 = F 0 E Substitute c 1 = F 0 E in equation ( 3 ) . ψ ( x ) = x F 0 E Differentiate the equation u ( x , t ) = v ( x , t ) + ψ ( x ) with respect to t . u ′ t ( x , t ) = v ′ t ( x , t ) Substitute L for x in the above equation. u ′ t ( L , t ) = v ′ t ( L , t ) v ′ t ( L , t ) = 0 The boundary for v ( x , t ) are as follows: a 2 ∂ 2 v ∂ x 2 = ∂ 2 v ∂ t 2 0 < x < L , t > 0 v ( 0 , t ) = 0 v ( x , 0 ) = − x F 0 E ∂ v ∂ x | x = L = 0 and ∂ v ∂ t | x = L = 0 Separate the variables with the help of separation constant and take the separation constant λ as follows: X ″ X = T ″ a 2 T = − λ The separated equations are written as follows: X ″ + λ X = 0 … … ( 5 ) T ″ + a 2 λ T = 0 … … ( 6 ) The product solution is as follows: v ( x , t ) = X ( x ) T ( t ) … … ( 7 ) Substitute 0 for x in equation ( 7 ) . v ( 0 , t ) = X ( 0 ) T ( t ) X ( 0 ) T ( t ) = 0 ( v ( 0 , t ) = 0 ) X ( 0 ) = 0 Differentiate equation ( 7 ) with respect to x . v ′ x ( x , t ) = X ′ ( x ) T ( t ) Substitute L for x in the above equation. v ′ x ( L , t ) = X ′ ( L ) T ( t ) X ′ ( L ) T ( t ) = 0 ( ∂ v ∂ x | x = L = 0 ) X ′ ( L ) = 0 There are three cases for λ . First case is λ = 0 . Substitute 0 for λ in equation ( 5 ) . X ″ = 0 The general solution of the above equation is as follows: X ( x ) = c 3 + c 4 x ( 8 ) Substitute 0 for x in equation ( 8 ) . X ( 0 ) = c 3 + c 4 ⋅ 0 c 3 = 0 ( X ( 0 ) = 0 ) Substitute 0 for c 3 in equation ( 8 ) . X ( x ) = c 4 x Differentiate the above equation with respect to x . X ′ ( x ) = c 4 Substitute L for x in the above equation. X ′ ( L ) = c 4 c 4 = 0 ( X ′ ( L ) = 0 ) Since, the value of c 3 and c 4 is 0 so, for λ = 0 the non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute − α 2 for λ in equation ( 5 ) . X ″ − α 2 X = 0 The general solution of the above equation is as follows: X ( x ) = c 5 cosh ( α x ) + c 6 sinh ( α x ) … … ( 9 ) Substitute 0 for x in equation ( 9 ) . X ( 0 ) = c 5 cosh ( 0 ) + c 6 sinh ( 0 ) c 5 = 0 ( X ( 0 ) = 0 ) Substitute 0 for c 5 in equation ( 9 ) . X ( x ) = c 6 sinh ( α x ) Differentiate the above equation with respect to x . X ′ ( x ) = a c 6 cosh ( α x ) Substitute L for x in the above equation. X ′ ( L ) = α c 6 cosh ( α L ) α c 6 cosh ( α L ) = 0 ( X ′ ( L ) = 0 ) c 6 = 0 Since, the value of c 5 and c 6 is 0 so, for λ = − α 2 the non-trivial solution does not exist. Third case is λ = α 2 ( λ > 0 ) . Substitute α 2 for λ in equation ( 5 ) . X ″ + α 2 X = 0 The general solution of the above equation is as follows: X ( x ) = c 7 cos ( α x ) + c 8 sin ( α x ) … … ( 10 ) Substitute 0 for x in equation ( 10 )

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 12.7, Problem 6E

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The longitudinal displacement u(x,t) of a vibrating elastic bar such that a2∂2u∂x2=∂2u∂t2, u(0,t)=0,u(x,0)=0,E∂u∂x|x=L=F0 and ∂u∂t|x=L=0.

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Chapter 12.7, Problem 5E

Chapter 12.7, Problem 7E

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

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The longitudinal displacement u ( x , t ) of a vibrating elastic bar such that a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , u ( 0 , t ) = 0 , u ( x , 0 ) = 0 , E ∂ u ∂ x | x = L = F 0 and ∂ u ∂ t | x = L = 0 .

Solution Summary: The author explains the longitudinal displacement of a vibrating elastic bar, which is u(x,t).

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The longitudinal displacement u(x,t) of a vibrating elastic bar such that a2∂2u∂x2=∂2u∂t2, u(0,t)=0,u(x,0)=0,E∂u∂x|x=L=F0 and ∂u∂t|x=L=0.

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Chapter 12 Solutions