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# hw5_soln - Problems from Strauss 2.1 3 The disturbance will...

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Problems from Strauss, 2.1: # 3: The disturbance will reach the flea after time t = ρ T ( l 4 - a ). # 5: As in the hint we have: u ( x, t ) = 1 2 c { length of ( x - ct, x + ct ) ( - a, a ) } When t = a 2 c : u ( x, t ) = 1 2 c { length of ( x - a 2 , x + a 2 ) ( - a, a ) } = 1 2 c × a | x | ≤ a 2 3 a 2 - | x | a 2 ≤ | x | ≤ 3 a 2 0 | x | ≥ 3 a 2 When t = a c : u ( x, t ) = 1 2 c { length of ( x - a, x + a ) ( - a, a ) } = 1 2 c × 2 a - | x | | x | ≤ 2 a 0 | x | ≥ 2 a When t = 3 a 2 c : u ( x, t ) = 1 2 c { length of ( x - 3 a 2 , x + 3 a 2 ) ( - a, a ) } = 1 2 c × 2 a | x | ≤ a 2 5 a 2 - | x | a 2 ≤ | x | ≤ 5 a 2 0 | x | ≥ 5 a 2 When t = 2 a c : u ( x, t ) = 1 2 c { length of ( x - 2 a, x + 2 a ) ( - a, a ) } = 1 2 c × 2 a | x | ≤ a 3 a - | x | a ≤ | x | ≤ 3 a 0 | x | ≥ 3 a Similar formulas can be developed for t = 5 a 2 c . # 9 Following the hint in the book should lead to search for solutions of the form: u ( x, t ) = f ( x + 4 t ) + g ( x - t ) Match this up with the initial data to get a formula that solves the PDE. Problems for Haberman, 4.4: # 3b: Separating variables u ( x, t ) = h ( t ) φ ( x ) leads you to solve φ ( x ) = - λφ ( x ) φ (0) = φ ( L ) = 0 h ( t ) + β ρ 0 h ( t ) + λ T 0 ρ 0 h ( t ) = 0 The solutions to φ ( x ) take the form λ = ( L ) 2 , φ ( x ) = C n sin( nπx L ). To solve the ODE in time, plug in λ = ( L ) 2 and consider the characteristic equation r 2 + β ρ 0 r + L 2 T 0 ρ 0 = 0 1

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2 By assumption β
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