Explanation Given: The boundary conditions are as follows: ∂ 2 u ∂ x 2 − u = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 1 , t ) = 0 ∂ u ∂ x | x = 0 = 0 u ( x , 0 ) = 1 − x 2 Calculation: The boundary conditions are given as follows: ∂ 2 u ∂ x 2 − u = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 1 , t ) = 0 ∂ u ∂ x | x = 0 = 0 u ( x , 0 ) = 1 − x 2 Consider the product solution as follows: u ( x , t ) = X ( x ) T ( t ) … … ( 1 ) Substitute u = X T in ∂ 2 u ∂ x 2 − u = ∂ u ∂ t . ∂ 2 ( X T ) ∂ x 2 − ( X T ) = ∂ ( X T ) ∂ t X ″ T − X T = X T ′ X ″ T = X T ′ + X T X ″ T = X ( T ′ + T ) Divide the above equation by X T . X ″ T X T = X ( T ′ + T ) X T X ″ X = T ′ + T T Separate the variables with the help of separation constant and take the separation constant λ as follows: X ″ X = T ′ + T T = − λ The separated equations are written as follows: X ″ + λ X = 0 … … ( 2 ) T ′ + ( 1 + λ ) T = 0 … … ( 3 ) Substitute 1 for x in equation ( 1 ) . u ( 1 , t ) = X ( 1 ) T ( t ) X ( 1 ) T ( t ) = 0 ( u ( 1 , t ) = 0 ) X ( 1 ) = 0 Differentiate equation ( 1 ) with respect to x . u ′ x ( x , t ) = X ′ ( x ) T ( t ) Substitute 0 for x in the above equation. u ′ x ( 0 , t ) = X ′ ( 0 ) T ( t ) X ′ ( 0 ) T ( t ) = 0 ( u ′ x ( 0 , t ) = 0 ) X ′ ( 0 ) = 0 There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 2 ) . X ″ = 0 The general solution of the equation is as follows: X ( x ) = c 1 + c 2 x … … ( 4 ) Substitute 1 for x in equation ( 4 ) . X ( 1 ) = c 1 + c 2 c 1 + c 2 = 0 ( X ( 1 ) = 0 ) c 1 = − c 2 Substitute c 1 = − c 2 in equation ( 4 ) . X ( x ) = − c 2 + c 2 x = c 2 ( x − 1 ) Differentiate the above equation with respect to x . X ′ ( x ) = c 2 Substitute 0 for x in the above equation. X ′ ( 0 ) = c 2 c 2 = 0 ( X ′ ( 0 ) = 0 ) Since, the value of c 1 and c 2 is 0 so, for λ = 0 the non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute λ = − α 2 in equation ( 2 ) . X ″ − α 2 X = 0 The general solution of the equation is as follows: X ( x ) = c 3 cos h ( α x ) + c 4 sinh ( α x ) … … ( 5 ) Substitute 1 for x in equation ( 5 ) . X ( 1 ) = c 3 cos h ( α ) + c 4 sinh ( α ) c 3 cos h ( α ) + c 4 sinh ( α ) = 0 ( X ( 1 ) = 0 ) c 3 = − c 4 sinh ( α ) cos h ( α ) Substitute the value of c 3 in equation ( 5 ) . X ( x ) = − ( c 4 sinh ( α ) cos h ( α ) ) cos h ( α x ) + c 4 sinh ( α x ) = c 4 ( sinh ( α x ) − sinh ( α ) cos h ( α x ) cos h ( α ) ) Differentiate the above equation with respect to x . X ′ ( x ) = c 4 ( α cosh ( α x ) − α sinh ( α ) sinh ( α x ) cos h ( α ) ) Substitute 1 for x in the above equation. X ′ ( 1 ) = c 4 ( α cosh ( α ) − α sinh ( α ) sinh ( α ) cos h ( α ) ) α c 4 ( ( cosh ( α ) ) 2 − ( sinh ( α ) ) 2 cos h ( α ) ) = 0 ( X ′ ( 1 ) = 0 ) α c 4 ( 1 cos h ( α ) ) = 0 c 4 = 0 Substitute the value of c 4 in c 3 = − c 4 sinh ( α ) cos h ( α ) . c 3 = − ( 0 ) ⋅ sinh ( α ) cos h ( α ) c 3 = 0 Since, the value of c 3 and c 4 is 0 so for λ = − α 2 the non-trivial solution does not exist. The third case is λ = α 2 ( λ > 0 ) . Substitute λ = α 2 in equation ( 2 ) . X ″ + α 2 X = 0 The general solution of the equation is as follows: X ( x ) = c 5 cos ( α x ) + c 6 sin ( α x ) … … ( 6 ) Substitute 1 for x in equation ( 6 ) . X ( 1 ) = c 5 cos ( α ) + c 6 sin ( α ) c 5 cos ( α ) + c 6 sin ( α ) = 0 ( X ( 1 ) = 0 ) c 5 = − c 6 sin ( α ) cos ( α ) Substitute the value of c 5 in equation ( 6 ) . X ( x ) = − c 6 sin ( α ) cos ( α ) cos ( α x ) + c 6 sin ( α x ) = c 6 ( sin ( α x ) − c 6 sin ( α ) cos ( α ) cos ( α x ) ) Differentiate the above equation with respect to 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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Chapter 12.7, Problem 8E

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The solution u(x,t) of the differential equation ∂2u∂x2−u=∂u∂t such that u(1,t)=0, ∂u∂x|x=0=0 and u(x,0)=1−x2.

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Chapter 12.7, Problem 7E

Chapter 12.7, Problem 9E

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

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The solution u ( x , t ) of the differential equation ∂ 2 u ∂ x 2 − u = ∂ u ∂ t such that u ( 1 , t ) = 0 , ∂ u ∂ x | x = 0 = 0 and u ( x , 0 ) = 1 − x 2 .

Solution Summary: The author calculates the boundary conditions of the differential equation partial.

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The solution u(x,t) of the differential equation ∂2u∂x2−u=∂u∂t such that u(1,t)=0, ∂u∂x|x=0=0 and u(x,0)=1−x2.

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