# 10. - Using the Fourier Transform to Solve PDEs In these...

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Using the Fourier Transform to Solve PDEs In these notes we are going to solve the wave and telegraph equations on the full real line by Fourier transforming in the spatial variable. We start with The Wave Equation If u ( x, t ) is the displacement from equilibrium of a string at position x and time t and if the string is undergoing small amplitude transverse vibrations, then we have seen that 2 u ∂ t 2 ( x, t ) = c 2 2 u ∂x 2 ( x, t ) (1) for a constant c . We are now going to solve this equation by multiplying both sides by e - ikx and integrating with respect to x . That is, we shall Fourier transform with respect to the spatial variable x . Denote the Fourier transform with respect to x , for each fixed t , of u ( x, t ) by ˆ u ( k, t ) = Z -∞ u ( x, t ) e - ikx dx We have already seen (in property (D) in the notes “Fourier Transforms”) that the Fourier transform of the derivative f 0 ( x ) is Z -∞ f 0 ( x ) e - ikx dx = ik Z -∞ f ( x ) e - ikx dx = ik ˆ f ( k ) (2) (by integration by parts with u = e - ikx , dv = f 0 ( x ) dx , du = - ike - ikx dx , v ( x ) = f ( x ) and assuming that f ( x ) 0 as x → ±∞ ). Applying this with f ( x ) = ∂x u ( x, t ) and a second time with f ( x ) = u ( x, t ), gives that the Fourier transform of 2 u ∂x 2 ( x, t ) is - k 2 ˆ u ( k, t ). Computation of the Fourier transform of 2 u ∂ t 2 ( x, t ) is even easier. For the first t –derivative, Z -∞ u t ( x, t ) e - ikx dx = Z -∞ lim h 0 u ( x,t + h ) - u ( x,t ) h e - ikx dx = lim h 0 1 h Z -∞ u ( x, t + h ) e - ikx dx - Z -∞ u ( x, t ) e - ikx dx = lim h 0 1 h ˆ u ( k, t + h ) - ˆ u ( k, t ) = ∂t ˆ u ( k, t ) (3) To get two t -derivatives, we just apply this twice (with u replaced by u t the first time) Z -∞ 2 u ∂ t 2 ( x, t ) e - ikx dx = ∂ t Z -∞ ∂ t u ( x, t ) e - ikx dx = 2 ∂ t 2 ˆ u ( k, t ) So applying the Fourier transform to both sides of (1) gives 2 ∂ t 2 ˆ u ( k, t ) = - c 2 k 2 ˆ u ( k, t ) (4) This has not yet led to the solution for u ( x, t ) or ˆ u ( k, t ), but it has led to a considerable simplification. We now have, for each fixed k , a constant coefficient, homogeneous, second order ordinary differential equation for ˆ u ( k, t ). February 21, 2007 Using the Fourier Transform to Solve PDEs 1

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To emphasise that each k may now be treated independently, fix any k and write ˆ u ( k, t ) = U ( t ). The differential equation (4) now is U 00 ( t ) + c 2 k 2 U ( t ) = 0. From earlier courses, we know that this equation can
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