P384_Assignment_8_SOLUTIONS - 1 Physics 384 2016 Assignment 8 SOLUTIONS Problem I Consider the inhomogeneous ODE d2 y(x k 2 y(x = f(x dx2 where the

P384_Assignment_8_SOLUTIONS - 1 Physics 384 2016 Assignment...

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1 Physics 384 2016 Assignment 8 SOLUTIONS Problem I Consider the inhomogeneous ODE d 2 y ( x ) dx 2 - k 2 y ( x ) = f ( x ) where the boundary conditions are that y ( ±∞ ) = finite. Use the Fourier transform method to show that the Green’s function G ( x ; x 0 ) for this problem, which satisfies d 2 G ( x ; x 0 ) dx 2 - k 2 G ( x ; x 0 ) = δ ( x - x 0 ) (1) can be expressed as G ( x ; x 0 ) = - 1 2 π ˆ -∞ e ik 0 ( x - x 0 ) k 0 2 + k 2 dk 0 Evaluate the integral and show that G ( x ; x 0 ) = - 1 2 k e - k | x - x 0 | x Solution To get started, it’s handy to use the fact that the Green’s function should depend only on the separation x - x 0 , since there is no preferred orgin in Eq. (1), to write the ODE as d 2 G ( x ) dx 2 - k 2 G ( x ) = δ ( x ) (2) We write G ( x ) as a Fourier transform G ( x ) = 1 2 π ˆ -∞ ˜ G ( k 0 ) e ik 0 x dk 0 apply the differential operator, and identify the result with the delta-function in terms of its own Fourier transform d 2 dx 2 - k 2 G ( x ) = 1 2 π ˆ -∞ ( - k 0 2 - k 2 ) ˜ G ( k 0 ) e ik 0 x dk 0 = 1 2 π ˆ -∞ e ik 0 x dk 0
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2 Equating the integrands gives ˜ G ( k 0 ) = - 1 k 0 2 + k 2 Now it’s time to use complex analysis to explicitly evalute the F.T. for G ( x ) G ( x ) = - 1 2 π ˆ -∞ 1 k 0 2 + k 2 e ik 0 x dk 0 The integrand has poles at k 0 = ± ik As usual, to get damped exponentials at complex infinity, we close the contour above the real line for x > 0 , and below the real line for x < 0 , hence G ( x ) = - 1 2 π × 2 πi × e i ( k 0 = ik ) x 2 ik
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