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TehFinalReport2 - Dispersive Quantization of the linear...

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Dispersive Quantization of the linear Schr¨odinger equation Sudarshan Balakrishnan September 18, 2011 Abstract . The study of linear dispersive partial differential equations with piecewise constant periodic initial data leads to quantized structures at times which are rational multiples of π and fractal profiles at irrational multiples of π . We will present an overview of these results for the Schrodinger equation on the torus. § Introduction The purpose of this paper is to study the behavior of the solution to the Schr¨ odinger wave equation with specific initial conditions. First, we go over the methods to analyze the linearly dispersive wave equation established in [ ? ]. Next, we proceed to carry out the same analysis for the Schr¨ odinger wave equation and write programs in MATLAB to plot its solution. We also present an overview of the Fast Fourier Algorithim(FFT) and plot the error involved in the solution to the Schr¨ odinger wave equation. § 1 Linearly dispersive wave equation The Linearly dispersive wave equation is a specific periodic initial-boundary value prob- lem on the interval 0 x 2 π given by ∂u ∂t = 3 u ∂x 3 (1) With initial conditions: u ( t, 0) = f ( x ) , (2) Satisfying: u ( t, 0) = u ( t, 2 π ) , (3) ∂u ∂x ( t, 0) = ∂u ∂x ( t, 2 π ) , (4) 1
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2 u ∂x 2 ( t, 0) = 2 u ∂x 2 ( t, 2 π ) . (5) First, we take the initial data as a unit step f ( x ) = σ ( x ) = - 1 , 0 < x < π ; 1 , π < x < 2 π (6) The precise values assigned at its discontinuities are not important, although choosing f ( x ) = 1 2 at x = 0 , π, 2 π is consistent with Fourier analysis. The boundary conditions allow us to extend the initial data and solution to be 2 π periodic functions in x. In order to construct the solution, let us consider a time-dependent Fourier Series u ( t, x ) X k = -∞ b k ( t ) e ikx . (7) As we substitute (7) in equation (1) we get b k ( t ) will satisfy the elementary linear ordinary differential equation: db k dt = ik 3 b k ( t ) . (8) Then, we get: b k ( t ) = b k (0) e ik 3 t . (9) Re-substituting (9) in (7), we get : u ( t, x ) X k = -∞ b k (0) e i ( kx - k 3 t ) . (10) Observe that the solution (10) is 2 π periodic in both t and x . Moreover, since the k -th summand is a function of , periodic waves of frequency k move with wave speed k 2 ie., ω ( k ) = - k 2 , thereby justifying the dispersive nature of the system. As we solve for the initial value problem, we note the initial condition for the Fourier series given by: f ( x ) X k = -∞ c k e ikx . (11) Where we have : c k = 1 2 π Z 2 π 0 f ( x ) e - ikx dx. (12) Now, when we equate u (0 , x ) = f ( x ), we see that b k (0) = c k which will be the fourier coefficients of the initial data. We observe the step function (6), we notice that the fourier coefficients are given by: 2
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b k (0) = i πk k odd ; 1 2 π k= 0 (13) Note, here that when k = even, we have that b k (0) equals zero. Now, when we combine this with (10), we get that the solution to the Linearly dispersive wave equation is given by : u ( t, x ) 1 2 - 2 π X j =0 sin((2 j + 1) x - (2 j + 1) 3 t ) 2 j + 1 . (14) t § 1.1 Plots We plot the solution (14) at time t = . 5 π and t = 2 π . The X-axis represents the spatial coordinates x [0 , 2 π
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