We start with a discussion of the Sampling Theorem.

Transcription

1 Math 625 Lecture Notes - March 10, 2014 We start with a discussion of the Sampling Theorem. Discussion 0.1 Assume that the Fourier transform f of a function f satisfies f(w) = 0 for w > Ω, that is, f is compactly supported. Then the sampling theorem says that one has the following representation for f: ( ) kπ sin(ω(t πk/ω)) f(t) = f. Ω t πk/ω k= Here the function sin t/t is known as the sampling function. Sampling theorem is well known in the signal processing literature and has been important in the evolution of digital devices. Technical note : When f is compactly supported, f will be unbounded. This is something to keep in mind in applications, where the functions will be bounded, urging us to form a mathematical framework. Some historical notes about the sampling theorem: Nyquist (1927) Shannon (1949) Sampling theorem stated above is also known as Nyquist/Shannon sampling theorem. Some other names to remember: Kotelnikov (1933) Raabe (1939) Someya (1949) Back to the main discussion : Now we go back to our main discussion. So far we mainly dealt with two types of problems: (1) Finding roots and (2) Analyzing integrals (and summations as special cases). Former of these two was concerned with the perturbation theory: Given an equation that you can solve easily, perturb it to find the solutions for a more complex problem. General theory behind finding roots relied on the Inverse Function Theorem. 1

2 The second topic we studied so far, Integrals, was about asymptotics, and basically we developed the theory on Laplace s method. Now we move to a new topic, study of differential equations, which combines the techniques from perturbation theory and asymptotic analysis. Example 0.1 (i) Understanding the solutions of y + (1 + ɛ sin t)y = 0 with ɛ small using the simpler equation y + y = 0 corresponds to the relation with perturbation theory. (ii) Understanding the solutions of y = ty as t ± forms an example to demonstrate the relation to asymptotic theory. Now we review some basics about ordinary differential equations (ODE). Recall first that we can classify ODEs as nonlinear and linear equation. Focusing on the latter one, we have the further classifications as first order and higher order equations. In the case of linear first order equations, the ODE has the form y = f(t)y and one can check that the solution to the equation can be given as ( t ) y(t) = y(t 0 ) exp f(s)ds. t 0 We call such solutions with integrals as solutions by quadrature. Regarding the higher order linear equations, we note that a further classification as constant and nonconstant coefficient equations is necessary. For the constant coefficient case, the solutions turn out to be expressible in terms of elementary functions (i.e. finite combinations of polynomials, exponential function and trigonometric functions). On the contrary for the higher order differential equations with nonconstant coefficients, we obtain solutions by quadrature, and special functions emerge in this case. Discussion 0.2 Any function that is not an elementary function is considered as a special function. Special functions have representation via Power Series, Integrals or ODEs. For example Airy s function can be expressed in terms of each of these three. Next we move to today s main topic: Laplace s Method for ODEs. onsider the differential equation (a 0 z + a 1 )w + (b 0 z + b 1 )w + (c 0 z + c 1 )w = 0. (0.1) 2

3 Note that we are on the complex plane now. We are looking for solutions of the form w(z) = eyz v(y)dy. Once an appropriate contour is chosen (which we discuss later), we have w (z) = yeyz v(y)dy and w (z) = y2 e yz v(y)dy. Using this we have, for example, zw (z) = zye yz v(y)dy = yv(y)d(e zy ) = yv(y)e yz e yz (yv(y)) dy and similar computations can also be done for zw(z) and zw (z). Substituting all of these into (0.4), we obtain 0 = (a 0 y 2 + b 0 y + c 0 )v(y)e yz (0.2) + e yz ((a 1 y 2 + b 1 y + c 1 )v a 0 (y 2 v) b 0 (yv) c 0 v )dy. (0.3) In particular, for the second term on the right hand side to be zero, we need to have v v = a 1y 2 + (b 1 2a 0 )y + c 1 b 0 a 0 y 2 + b 0 y + c 0 which is a first order equation solvable by quadratures. Example 0.2 onsider Airy s equation: w zw = 0. For this special case, we have and so ( giving v = exp a 0 = 0, a 1 = 1, b 0 = b 1 = 0, c 0 = 1, c 1 = 0 y3 3 v v = y2 ). Hence w(z) = ezy y3 /3 dy. Now how do we choose? First note that should be unbounded and v should tend to 0 at the ends of. This will be the case when Re( y 3 ) at the ends of corresponding to cos(3θ + π) < 0 where y = re iθ. Thus the solutions turn out be w k (z) = k e yz y3 /3 dy, k = 1, 2, 3. These are linearly dependent with the relation among them being w 1 (z) + w 2 (z) + w 3 (z) = 0. 3

4 Figure 1 : urves of Integration Now focusing on w 1 and substituting y = it, we have w 1 (z) = i This in particular implies Ai(z) = 1 2πi e itz+it3 /3 dt = 2πiAi(z). 1 e yz y3 /3 dy. Also letting ω = e 2πi/3, we have 2 = 1 /ω and 3 = ω 1. These give w 2 (z) = 1 ω w 1 ( z ω ) and w 3 (z) = ωw 1 (ωz). Question : What is Bi(z) in terms of w k? 4

5 Math 625 Lecture Notes - March 12, 2014 We started the discussion with a physical experiment where we have a VHS videocassette (and we had one), and we observe its rotation once we toss it into the air. onclusion was that the behavior of the cassette in the air was different once we changed the axis of rotation. Also we mentioned that this would only work for rectangular prisms, and not for, for example, an American football or a soccer ball due to their symmetry. We concluded this discussion by briefly talking over the mathematical theory of our experiment using vector calculus. Then we moved to the following two comments related to previous lecture: First we clarify the following statement: If the Fourier transform f of f is compactly supported than f(t) is not compactly supported. Let a be large enough so that f is 0 outside the interval [ a, a]. Also let g(t) = sin c 1t c 2 where c t 1, c 2 are nonzero constants chosen so that the Fourier transform is given by ĝ(w) = χ [ a,a] (w). Then we have f = f ĝ, which implies via the convolution theorem that f = f g is satisfied. Now since g is not compactly supported, we conclude that f is also not compactly supported. Second note is on the Airy function and deformation of paths. Last time we showed that Airy function satisfies Ai(z) = 1 e yz y3 /3 dy 2πi 1 where 1 is as in Figure 1 from previous lecture (pink curve). By using a deformation of paths, we also have the following representation of the Airy function Ai(z) = 1 e i(yt+t3 /3) dt. 2π We also discussed deformation of our paths into other infinite contours, for which it might be non-trivial to check that the contours are indeed equivalent in terms of integration. 5

6 Back to the main discussion : Going back to the main discussion, last time we showed that the equation (a 0 z + a 1 )w + (b 0 z + b 1 )w + (c 0 z + c 1 )w = 0. (0.4) is solvable in quadratures. Indeed, one can also solve the following generalized higher order version using the same strategy: n (s k z + r k )w (k) (z) = 0. k=0 That is, using w(z) = eyz v(y)dy again reduces the problem to a first order equation for v. Moreover, similar techniques can also be applied to finite difference equations. Let s first consider the following A n y n+2 + B n y n+1 + n y n = 0. For this case, letting y n = r n, A n = a, B n = b, n = c, we have ar 2 +br+c = 0. For example, letting A n = 1, B n = n = 1, we have y n+2 = y n+1 + y n which is the recursive relation for Fibonacci sequence. For this special case, we have r 2 = r + 1. Next as a finite difference variant of (0.4), we consider (a 0 n + a 1 )y n+2 + (b 0 n + b 1 )y n+1 + (c 0 n + c 1 )y n = 0. Letting w(z) = n 0 y nz n, we have n 0 ny nz n 1 which implies and we have zp 1 (z)w (z) + p 2 (z)w(z) = f(z, y 1, y 2 ) y n = 1 2πi z =ρ provided that w is analytic at z = 0. Perturbations of linear ODES w(z) dz zn+1 We divide perturbations of ODES into two categories: regular and singular. Starting with the regular case, and considering the equation Y = A(t, ɛ)y where A(t,.) is analytic 6

7 we have So A(t, ɛ) = Y = A k (t)ɛ k. k=0 y k (t)ɛ k. k=0 Then we need to have y 0 = A(t, 0)y to be solvable. Singular perturbations are further classified as poles and essential singularities. WKB (Wentzel-Kramers-Brillouin) approximation method is used for the poles case to find approximate solutions to partial differential equations (This is also known as LG or Liouville-Green method). Not much is known about the essential case averaging is one techniques that applies in some cases. 7