Explanation Given: The boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , u ( 0 , t ) = 0 , ∂ u ∂ x | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h > 0 , t > 0 and u ( x , 0 ) = f ( x ) , 0 < x < 1 . Calculation: Consider a function u as a function of ( x , t ) as follows: u ( x , t ) = v ( x , t ) + ψ ( x ) Substitute v ( x , t ) + ψ ( x ) for u ( x , t ) into partial differential equation as follows: k ∂ 2 v ∂ x 2 + k ψ ″ = ∂ v ∂ t The above equation will be homogeneous only if ψ ″ = 0 . The general solution of the above equation is ψ ( x ) is c 1 x + c 2 . The given boundary condition is u ( 0 , t ) = 0 implies that the value of ψ ( 0 ) is zero. Substitute 0 for ψ ( x ) and 0 for x in equation ψ ( x ) = c 1 x + c 2 . 0 = c 1 ( 0 ) + c 2 0 = c 2 Substitute 0 for c 2 in equation ψ ( x ) = c 1 x + c 2 . ψ ( x ) = c 1 x + 0 = c 1 x Apply the given boundary condition ∂ u ∂ x | x = 1 = − h ( u ( 1 , t ) − u 0 ) in u ( x , t ) = v ( x , t ) + ψ ( x ) . ( ∂ v ∂ x + ψ ′ ) | x = 1 = − h [ v ( 1 , t ) + ψ ( 1 ) − u 0 ] The above expression will be homogeneous if ψ ′ ( 1 ) = − h ψ ( 1 ) + h u 0 . Since the value of ψ ( 1 ) = ψ ′ ( 1 ) is c 1 , substitute c 1 for ψ ′ ( 1 ) in equation ψ ′ ( 1 ) = − h ψ ( 1 ) + h u 0 . c 1 = − h c 1 + h u 0 c 1 + h c 1 = h u 0 c 1 = h u 0 h + 1 Substitute h u 0 h + 1 for c 1 in equation ψ ( x ) = c 1 x . ψ ( x ) = h u 0 h + 1 x The new boundary- value problem is k ∂ 2 v ∂ x 2 = ∂ v ∂ t , 0 < x < 1 , t > 0 , v ( 0 , t ) = 0 , ∂ v ∂ x | x = 1 = − h v ( 1 , t ) , h > 0 , t > 0 and v ( x , 0 ) = f ( x ) − h u 0 h + 1 x x , 0 < x < 1 . The equation is given below: k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 ...... ( 1 ) Consider u ( x , t ) is X ( x ) T ( t ) . Substitute X ( x ) T ( t ) for u ( x , t ) in equation (1). k ∂ 2 ∂ x 2 ( X ( x ) T ( t ) ) = ∂ ∂ t ( X ( x ) T ( t ) ) k X ″ ( x ) T ( t ) = X ( x ) T ′ ( t ) k X ″ ( x ) X ( x ) = T ′ ( t ) T ( t ) Consider k X ″ ( x ) X ( x ) = T ′ ( t ) T ( t ) as − λ . Therefore, the equation is reduced as follows: k X ″ ( x ) X ( x ) = − λ k X ″ ( x ) X ( x ) + λ = 0 k X ″ ( x ) + λ X ( x ) X ( x ) = 0 k X ″ ( x ) + λ X ( x ) = 0 The characteristic equation of above expression is given below. k m 2 + λ = 0 m 2 = − λ k m = ± i λ k m = ± i α , where α = λ k Therefore, the solution is X ( x ) = c 1 cos α x + c 2 sin α x . Consider the equation T ′ ( t ) T ( t ) as − λ . Therefore, the equation is reduced as follows: T ′ ( t ) T ( t ) = − λ T ′ ( t ) T ( t ) + λ = 0 T ′ ( t ) + λ T ( t ) T ( t ) = 0 T ′ ( t ) + λ T ( t ) = 0 The characteristic equation of above expression is given below. m + λ = 0 m = − λ Therefore, the solution is given below: T ( t ) = c 3 e − λ t T ( t ) = c 3 e − k α 2 t ( k α 2 = λ ) Now, differentiate the expression X ( x ) = c 1 cos α x + c 2 sin α x as follows: X ′ ( x ) = − c 1 α sin α x + c 2 α cos α x Substitute 0 for x in X ( x ) = c 1 cos α x + c 2 sin α x

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 12.7, Problem 2E

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The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂x|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Chapter 12.7, Problem 1E

Chapter 12.7, Problem 3E

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Chapter 12 Solutions

Differential Equations with Boundary-Value Problems

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The steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 &lt; x &lt; 1 , t &gt; 0 , u ( 0 , t ) = 0 , ∂ u ∂ x | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h &gt; 0 , t &gt; 0 and u ( x , 0 ) = f ( x ) , 0 &lt; x &lt; 1 .

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The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂x|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Chapter 12 Solutions

The steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , u ( 0 , t ) = 0 , ∂ u ∂ x | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h > 0 , t > 0 and u ( x , 0 ) = f ( x ) , 0 < x < 1 .