The convolution theorem is very useful for example to solve certain types of

# The convolution theorem is very useful for example to

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The convolution theorem is very useful, for example to solve certain types of integral equa- tions. f ( x ) = g ( x ) + λ ˆ -∞ d x 0 h ( x - x 0 ) f ( x 0 ) (26a) F ( k ) = G ( k ) + 2 πλH ( k ) F ( k ) . (26b) Assuming that g and h are known, and we can calculate G and H . F ( k ) = G ( k ) 1 - 2 πλH ( k ) (27a) f ( x ) = 1 2 π ˆ -∞ d k G ( k ) 1 - 2 πλH ( k ) e ikx . (27b) 7 Wave equation on infinite string 2 u ( x, t ) ∂x 2 - 1 c 2 2 u ( x, t ) ∂t 2 = 0 . (28) Inspired by the complex exponential form of the Fourier series, we can combine our previously found trigonometric solutions into complex exponential solutions u ( x, t ) = e ik ( x ± ct ) . An infinite string only has initial conditions, no boundary conditions, and thus no string lengthscale to select certain normal modes, so any k is valid: u ( x, t ) = 1 2 π ˆ -∞ d k [ A ( k ) e ik ( x - ct ) + B ( k ) e ik ( x + ct ) ] . (29) We scaled A ( k ) , B ( k ) are arbitrary (so far), so we can pull out a factor of 1 / 2 π , no problem. So the initial conditions are φ ( x ) = u ( x, 0) (30a) = 1 2 π ˆ -∞ d k [ A ( k ) + B ( k )] e ikx (30b) ψ ( x ) = ∂u ∂t t =0 (30c) = - ic 2 π ˆ -∞ d k k [ A ( k ) - B ( k )] e ikx . (30d) We can invert the inverse Fourier transforms above to find relations between the Fourier coef- 6 ficients and the initial conditions. A ( k ) + B ( k ) = 1 2 π ˆ -∞ d x φ ( x ) e - ikx (31a) = Φ( k ) (31b) A ( k ) - B ( k ) = i 2 πck ˆ -∞ d x ψ ( x ) e - ikx (31c) = i ck Ψ( k ) (31d) A ( k ) = 1 2 Φ( k ) + i ck Ψ( k ) (31e) B ( k ) = 1 2 Φ( k ) - i ck Ψ( k ) . (31f) A ( k ) , B ( k ) determine everything. 8 Summary FT of Lorentzian: F 1 x 2 + a 2 = r π 2 e -| k | a a . (32) Convolution theorem: F ˆ -∞ d y f ( x - y ) g ( y ) = 2 πF ( k ) G ( k ) . (33) FT solution of infinite string: coefficients are functions of FTs of ICs. #### You've reached the end of your free preview.

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