So the general solution is n k t n 0 k e dk 2 t 11241

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, so the general solution is ˜ n ( k, t ) = ˜ n 0 ( k ) e - Dk 2 t . (11.241) where ˜ n 0 ( k ) ˜ n ( k, t = 0), to be determined by the initial conditions. The answer (i.e. general solution) comes via an inverse FT: n ( x, t ) = Z -∞ dk 2 π ˜ n ( k, t ) e ikx = Z -∞ dk 2 π ˜ n 0 ( k ) e ikx - Dk 2 t . (11.242) 71
n(x,t) x increasing time Figure 11.18: Variation of concentration with distance x at various diffusion times. SPECIFIC EXAMPLE: We add a small drop of ink to a large tank of water (assumed 1- dimensional). We want to find the density of ink as a function of space and time, n ( x, t ). Initially, all the ink ( S particles) is concentrated at one point (call it the origin): n ( x, t = 0) = S δ ( x ) (11.243) implying (using the sifting property of the Dirac delta function), ˜ n 0 ( k ) ˜ n ( k, 0) = Z -∞ dx n ( x, t = 0) e - ikx = Z -∞ dx δ ( x ) e - ikx = S. (11.244) Putting this into Eqn. (11.242) we get: n ( x, t ) = Z -∞ dk 2 π ˜ n ( k, t ) e ikx = Z -∞ dk 2 π ˜ n 0 ( k ) e ikx - Dk 2 t = Z -∞ dk 2 π S e ikx - Dk 2 t = S 2 π 2 Dt e - x 2 / (4 Dt ) . (11.245) (we used the ‘completing the square’ trick that we previously used to FT the Gaussian). Compare this with the usual expression for a Gaussian, 1 2 πσ exp - x 2 2 σ 2 (11.246) and identify the width σ with 2 Dt . So, the ink spreads out with concentration described by a normalized Gaussian centred on the origin with width σ = 2 Dt . The important features are: normalized: there are always S particles in total at every value of t centred on the origin: where we placed the initial drop width σ = 2 Dt : gets broader as time increases 72
σ t : characteristic of random walk (‘stochastic’) process σ D : if we increase the diffusion constant D , the ink spreads out more quickly. The solution n ( x, t ) is sketched for various t in Fig. 11.18. FOURIER ANALYSIS: LECTURE 18 11.3 Fourier solution of the wave equation One is used to thinking of solutions to the wave equation being sinusoidal, but they don’t have to be. We can use Fourier Transforms to show this rather elegantly, applying a partial FT ( x k , but keeping t as is). The wave equation is c 2 2 u ( x, t ) ∂x 2 = 2 u ( x, t ) ∂t 2 (11.247) where c is the wave speed. We Fourier Transform w.r.t. x to get ˜ u ( k, t ) (note the arguments), remembering that the FT of 2 /∂x 2 is - k 2 : - c 2 k 2 ˜ u ( k, t ) = 2 ˜ u ( k, t ) ∂t 2 . (11.248) This is a harmonic equation for ˜ u ( k, t ), with solution ˜ u ( k, t ) = Ae - ikct + Be ikct (11.249) However, because the derivatives are partial derivatives, the ‘constants’ A and B can be functions of k . Let us write these arbitrary functions as ˜ f ( k ) and ˜ g ( k ), i.e. ˜ u ( k, t ) = ˜ f ( k ) e - ikct + ˜ g ( k ) e ikct . (11.250) We now invert the transform, to give u ( x, t ) = Z -∞ dk 2 π h ˜ f ( k ) e - ikct + ˜ g ( k ) e ikct i e ikx = Z -∞ dk 2 π ˜ f ( k ) e ik ( x - ct ) + Z -∞ dk 2 π ˜ g ( k ) e ik ( x + ct ) = f ( x - ct ) + g ( x + ct ) and f and g are arbitrary functions.

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