Explanation Given: The boundary conditions are as follows: u ( 0 , t ) = 1 lim x → ∞ u ( x , t ) = 0 u ( x , 0 ) = e − x ∂ u ∂ t | t = 0 =0 or u t ( x , 0 ) = 0 Calculation: Take Laplace transform on both sides of equation ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 as follows: L { ∂ 2 u ∂ x 2 } = L { ∂ 2 u ∂ t 2 } d 2 U d x 2 = s 2 U ( x , s ) − s u ( x , 0 ) − u t ( x , 0 ) ...... ( 1 ) Take the Laplace transform on both sides of the boundary condition u ( 0 , t ) = 1 as follows: L { u ( 0 , t ) } = L { 1 } U ( 0 , s ) = 1 s ...... ( 2 ) Take the Laplace transform on both sides of the boundary condition lim x → ∞ u ( x , t ) = 0 as follows: L { lim x → ∞ u ( x , t ) } = L { 0 } lim x → ∞ U ( x , s ) = 0 ...... ( 3 ) Now substitute u ( x , 0 ) = e − x and u t ( x , 0 ) = 0 in equation (1). d 2 U d x 2 = s 2 U ( x , s ) − s x e − x d 2 U d x 2 − s 2 U ( x , s ) = − s e − x Solve the differential equation by the method undetermined coefficient. The auxiliary equation for the above differential equation as follows: r 2 − s 2 = 0 Solve above auxiliary equation as follows: r 2 − s 2 = 0 r 2 = s 2 r = ± s ( Square root ) Therefore, the complementary function of the for differential equation d 2 U d x 2 − s 2 U ( x , s ) = − s e − x is shown below. C .F = c 1 e − s x + c 2 e s x The particular integral of the differential equation d 2 U d x 2 − s 2 U ( x , s ) = − s e − x is obtained as follows: ( D 2 − s 2 ) U ( x , s ) = − s e − x P .I = − s e − x D 2 − s 2 ( P

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 15.2, Problem 10E

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The solution of boundary value problem subjected to wave equation ∂2u∂x2=∂2u∂t2.

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The solution of boundary value problem subjected to wave equation ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 .

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The solution of boundary value problem subjected to wave equation ∂2u∂x2=∂2u∂t2.

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