Explanation Given: The boundary value problem is ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 , the boundary conditions are u ( 0 , t ) = 0 , u ( π , t ) = 0 , u ( x , 0 ) = { x , 0 < x < π 2 π − x , π 2 ≤ x < π and u t ( x , 0 ) = 0 . Calculation: The given boundary value problem is as follows: ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u Consider the equation as follows: u ( x , t ) = X ( x ) T ( t ) … … ( 1 ) For the boundary condition u ( 0 , t ) = 0 . u ( 0 , t ) = 0 X ( 0 ) T ( t ) = 0 X ( 0 ) = 0 For the boundary condition u ( π , t ) = 0 . u ( π , t ) = 0 X ( π ) T ( t ) = 0 X ( π ) = 0 Separate the variables of equation ( 1 ) and consider separation constant as − λ . The equation of the variable X is as follows: X ″ + λ X = 0 … … ( 2 ) The equation of the variable T is as follows: T ″ + ( 1 + λ ) T = 0 … … ( 3 ) There are three cases for the value of λ . First case is when λ = 0 . Substitute 0 for λ in equation ( 2 ) . X ″ = 0 The general solution for the above differential equation is as follows: X ( x ) = c 1 + c 2 x … … ( 4 ) Use the boundary condition X ( 0 ) = 0 . X ( 0 ) = c 1 c 1 = 0 Substitute 0 for c 1 in equation ( 4 ) . X ( x ) = x c 2 Use the boundary condition X ( π ) = 0 . X ( π ) = π c 2 π c 2 = 0 c 2 = 0 Therefore, for the case λ = 0 the non-trivial solution does not exist. Second case is when λ = − α 2 for which λ < 0 . Substitute − α 2 for λ in equation ( 2 ) . X ″ − α 2 X = 0 The general solution for the above differential equation is as follows: X ( x ) = c 3 cos h ( α x ) + c 4 sin h ( α x ) … … ( 5 ) Use the boundary condition X ( 0 ) = 0 . X ( 0 ) = c 3 cos h ( 0 ) + c 4 sin h ( 0 ) c 3 = 0 Substitute 0 for c 1 in equation ( 5 ) . X ( x ) = c 4 sin h ( α x ) Use the boundary condition X ( π ) = 0 . X ( π ) = c 4 sin h ( α π ) c 4 sin h ( α π ) = 0 c 4 = 0 Therefore, for the case λ = − α 2 the non-trivial solution does not exist. Third case is λ = α 2 for which λ > 0 . Substitute λ = α 2 in equation ( 3 ) . X ″ + α 2 X = 0 The general solution of the above equation is, X ( x ) = c 5 cos α x + c 6 sin α x … … ( 6 ) Use the boundary condition X ( 0 ) = 0

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 16E

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To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Chapter 12.4, Problem 15E

Chapter 12.4, Problem 17E

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Chapter 12 Solutions

Differential Equations with Boundary-Value Problems

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The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .

Solution Summary: The author explains the solution of the boundary value problem, which is partial, 2u,. The boundary conditions are: u(x,t)=

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To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

Want to see the full answer?

Chapter 12 Solutions

The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 &lt; x &lt; π , t &gt; 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .

Solution Summary: The author explains the solution of the boundary value problem, which is partial, 2u,. The boundary conditions are: u(x,t)=

Videos

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

Want to see the full answer?

Chapter 12 Solutions

The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .