# hw1ans - MATH 438 1 Homework 1(due at 11:00 am on Problem 1...

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Unformatted text preview: MATH 438 1 Homework 1 (due at 11:00 am on January 29, 2008) Problem 1. Based on the understanding of the D’Alembert solution and the method of characteristics, solve the following problem for semi-infinite string: u tt- a 2 u xx = 0 , x > , t > ,a ∈ R , u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , x > , u (0 ,t ) = 0 , t > . Solution . Let u ( x,t ) = f ( x- at ) + g ( x + at ). Recall that in D’Alembert’s solution: g ( ξ ) = 1 2 φ ( ξ ) + 1 2 a Z ξ ψ ( s )d s + C, f ( ξ ) = 1 2 φ ( ξ )- 1 2 a Z ξ ψ ( s )d s- C, so that the D’Alembert solution is u ( x,t ) = φ ( x- at ) + φ ( x + at ) 2 + 1 2 a Z x + at x- at ψ ( s )d s. In our case we have a restriction at x = 0, from which we find that u (0 ,t ) dictates: f ( z ) + g (- z ) = 0 , z =- at < , since t > 0. So, for x- at < 0: f ( x- at ) =- g ( at- x ) =- φ ( at- x ) 2- 1 2 a Z at- x ψ ( s )d s- C. Hence, the solution is given by x > at : D Alembert s solution , < x < at : u ( x,t ) =- φ ( at- x ) + φ ( x + at ) 2 + 1 2 a Z x + at at- x ψ ( s )d s. Problem 2. Solve u tt- a 2 u xx = 0 , x > , t > ,a ∈ R , u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , x > , u (0 ,t ) = μ ( t ) , t > . MATH 438 2 Solution . It is enough to consider the case φ ( x ) = 0 and ψ ( x ) = 0 (Why?). In this case the only “disturbance” is coming from the boundary condition at x = 0, thus once can seek for a solution in the form...
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hw1ans - MATH 438 1 Homework 1(due at 11:00 am on Problem 1...

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