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Week 15 - Solution

# Week 15 - Solution - Tutorial Note Week 15 Solution Wave...

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Tutorial Note - Week 15 - Solution Wave Equation 1. I. Let the solution be u ( x, t ) = X ( x ) T ( t ), then u tt = 2 ∂t 2 [ X ( x ) T ( t )] = X ( x ) T ′′ ( t ) and u xx = 2 ∂x 2 [ X ( x ) T ( t )] = X ′′ ( x ) T ( T ) . The PDE becomes u tt = c 2 u xx = X ( x ) T ′′ ( t ) = c 2 X ′′ ( x ) T ( t ) = X ′′ ( x ) X ( x ) = T ′′ ( t ) c 2 T ( t ) = λ , where λ is a constant. Form the boundary conditions, u (0 , t ) = X (0) T ( t ) = 0 = X (0) = 0 u ( L, t ) = X ( L ) T ( t ) = 0 = X ( L ) = 0 since we do not want T ( t ) = 0. Then we have two diferential equations: b X ′′ ( x ) + λX ( x ) = 0, X (0) = X ( L ) = 0 ··· SLP T ′′ ( t ) + λc 2 T ( t ) = 0 II. From the SLP analysis that we have done be±ore, X n ( x ) = β n sin p L x P and λ n = p L P 2 , where n = 1, 2, . . . III. Solving ±or the T ′′ ( T ) + λc 2 T ( t ) = 0 with λ = α 2 > 0, λ n = ( L ) 2 and T ( t ) = e mt , T ′′ ( t ) + λc 2 T ( t ) = 0 = m 2 e mt + λc 2 e mt = 0 = m 2 + α 2 c 2 = 0 = m = ± αci . Hence, T ( t ) = a n cos p nπc L t P + b n sin p nπc L t P , where a n and b n are constants. IV. From Step II and III, u ( x, t ) = s n =1 X n ( x ) T n ( t ) = s n =1 β n sin p L x PB a n cos p nπc L t P + b n sin p nπc L t = s n =1 sin p L x A n cos p nπc L t P + B n sin p nπc L t , where A n = a n β n and B n = b n β n . 1

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V. Using the initial conditions u ( x, 0) = f ( x ) and u t ( x, 0) = g ( x ), u ( x, 0) = s n =1 A n sin p L x P = f ( x ) ; u t ( x, t ) = ∂t b s n =1 sin p L x PB A n cos p nπc L t P + B n sin p nπc L t ² = s n =1 nπc L sin p L x A n sin p nπc L t P + B n cos p nπc L t ; u t ( x, 0) = s n =1 nπc L B n sin p L x P = g ( x ) .
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