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Integral Transforms Appendix A Integral Transforms In this appendix a brief review is given of some integral transform methods. These are techniques used to reduce a differential equation to an algebraic equation. The main transforms are the Laplace transform, the Fourier transform and the Hankel transform. These will be presented here, together with some of their main properties. Derivations of the theorems will be given in condensed form, or not at all. Complete derivations are given by Titchmarsh (1948), Sneddon (1951) and Churchill (1972). Extensive tables of transforms have been published by the staff of the Bateman project (Erdélyi et al., 1954). Short tables of Laplace transforms, Fourier transforms and Hankel transforms are presented, with references to their derivation, and some numerical illustrations and verifications. Finally, an elegant and effective method is described for the determination of the inverse Fourier-Laplace transform for certain problems, in particular problems of elastodynamics (De Hoop, 1960). A.1 Laplace Transforms A.1.1 Definitions The Laplace transform is particularly useful for problems in which the variables are defined in a semi-infinite domain 0 <t<∞, where t may, for instance, be the time, and t = 0 indicates the initial value of time. The Laplace transform of a function f(t)is defined as ∞ F(s)= f(t)exp(−st)dt, (A.1) 0 where s is a parameter, which is assumed to be sufficiently large for the integral to exist. By the integration over the time domain, for various values of s, the func- tion f(t) is transformed into a function F(s). For various functions the Laplace transform can be calculated, sometimes very easily, sometimes with considerable A. Verruijt, An Introduction to Soil Dynamics, 395 Theory and Applications of Transport in Porous Media 24, © Springer Science+Business Media B.V. 2010 396 A Integral Transforms Table A.1 Some Laplace = ∞ − transforms No. f(t) F(s) 0 f(t)exp( st)dt 1 11 s 2 t 1 s2 ! 3 tn n sn+1 1 4exp(at) s−a 5sin(at) a s2+a2 6cos(at) s s2+a2 sin(at) a 7 t arctan( s ) effort. Tables of such transforms are widely available (Churchill, 1972; Erdélyi et al., 1954). A short table is given in Table A.1. The integrals in this table can all be evaluated with little effort, using techniques such as partial integration. The fundamental property of the Laplace transform appears when considering the transform of the time derivative. Using partial integration this is found to be ∞ df (t ) exp(−st)dt = sF(s) − f(0). (A.2) 0 dt Thus differentiation with respect to time is transformed into multiplication by s, and subtraction of the initial value f(0). A.1.2 Example In order to illustrate the application of the Laplace transform technique consider the differential equation df (t ) + 2f = 0, (A.3) dt with the initial condition f(0) = 5. Using the property (A.2) the differential equa- tion (A.3) is transformed into the algebraic equation (s + 2)F (s) − 5 = 0, (A.4) the solution of which is 5 F(s)= . (A.5) s + 2 Inverse transformation now gives, using transform no. 4 from Table A.1, f(t)= 5exp(−2t). (A.6) A.1 Laplace Transforms 397 Substitution into the original differential equation (A.3) will show that this is indeed the correct solution, satisfying the given initial condition. This example shows that the solution of the problem can be performed in a straightforward way. The main problem is the inverse transformation of the so- lution (A.5), which depends upon the availability of a sufficiently wide range of Laplace transforms. If the inverse transformation can not be found in a table of transforms it may be possible to use the general inverse transformation theorem (Churchill, 1972), but this requires considerable mathematical skill. A.1.3 Heaviside’s Expansion Theorem A powerful inversion method is provided by the expansion theorem developed by Heaviside, one of the pioneers of the Laplace transform method. This applies to functions that can be written as a quotient of two polynomials, p(s) F(s)= , (A.7) q(s) where q(s) must be a polynomial of higher order than p(s). It is assumed that the function q(s) possesses single zeroes only, so that it may be written as q(s) = (s − s1)(s − s2) ···(s − sn). (A.8) One may now write p(s) a a a F(s)= = 1 + 2 +···+ n . (A.9) q(s) s − s1 s − s2 s − sn The coefficient ai can be determined by multiplication of both sides of (A.9)by (s − si), and then passing into the limit s → si . This gives (s − si)p(s) ai = lim . (A.10) s→si q(s) Because q(si) = 0 the limit may be evaluated using L’Hôpital’s rule, giving = p(si) ai . (A.11) q (si) Inverse transformation of the expression (A.9) now gives, using formula no. 4 from Table A.1, n = p(si) f(t) exp(sit). (A.12) q (si) i=1 This is Heaviside’s expansion theorem. It provides a useful method to determine the inverse Laplace transform of functions of the form (A.7). It can also be used to 398 A Integral Transforms determine the inverse transform of functions of a more general form, although such inverse transforms can usually be found in a more general way by application of the complex inversion integral (Churchill, 1972). A.2 Fourier Transforms A.2.1 Fourier Series For certain partial differential equations the Fourier transform method can be used to derive solutions. These include problems of potential flow, and elasticity prob- lems, especially in the case of problems for infinite regions, semi-infinite regions, or infinite strips. The main principles of the method will be presented in this sec- tion. The main property of the Fourier transform can most easily be derived by first considering a Fourier series expansion. For this purpose let there be given a function g(θ), which is periodic with a period 2π, such that g(θ + 2π)= g(θ). This function can be written as ∞ 1 g(θ) = A + Ak cos(kθ) + Bk sin(kθ) , (A.13) 2 0 k=1 where 1 +π Ak = g(t)cos(kt) dt, (A.14) π −π and 1 +π Bk = g(t)sin(kt) dt, (A.15) π −π These formulas can be derived by multiplication of (A.13) by cos(jθ) or sin(jθ), and then integrating the result from θ =−π to θ =+π. It will then appear that from the infinite series only one term is unequal to zero, namely for k = j.This leads to (A.14) and (A.15). For a function with period 2πl the Fourier expansion can be obtained from (A.13) by replacing θ with x/l, t by t/l and then renaming g(x/l) as f(x). The result is ∞ 1 f(x)= A + Ak cos(kx/l) + Bk sin(kx/l) , (A.16) 2 0 k=1 where now 1 +πl Ak = f(t)cos(kt/l) dt, (A.17) πl −πl A.2 Fourier Transforms 399 and 1 +πl Bk = f(t)sin(kt/l) dt. (A.18) πl −πl Example As an example consider the block function defined by 0, |x| >πl/2, f(x)= (A.19) 1, |x| <πl/2. For this case the coefficients Ak and Bk can easily be calculated, using the expres- sions (A.17) and (A.18). The factors Bk are all zero, which is a consequence of the fact that the function f(x)is even, f(−x) = f(x). The factors Ak are equal to zero when k is even, and the uneven terms are proportional to 1/k. The series (A.16) finally can be written as 1 2 x 1 3x 1 5x 1 7x f(x)= + cos − cos + cos − cos +··· . 2 π l 3 l 5 l 7 l (A.20) The first term of this series represents the average value of the function, the sec- ond term causes the main fluctuation, and the remaining terms together modify this first sinusoidal fluctuation into the block function. Figure A.1 shows the approximation of the series (A.20) by its first 40 terms. It appears that the approximation is reasonably good, except very close to the discon- tinuities. This is a well known effect, often referred to as the Gibbs phenomenon (Weisstein, 1999). The approximation becomes better, of course, when more terms are taken into account, but the overshoot near the discontinuities remains. Fig. A.1 Fourier series, 40 terms 400 A Integral Transforms A.2.2 From Fourier Series to Fourier Integral Substitution of (A.17) and (A.18)into(A.16)gives ∞ 1 +πl 1 +πl f(x)= f(t)dt+ f(t) f(t)cos[k(t − x)/l] dt. (A.21) 2πl − πl − πl k=1 πl The interval can be made very large by writing 1/l = ξ. Then (A.21) becomes ∞ ξ +π/ξ ξ +π/ξ f(x)= f(t)dt+ f(t) f(t)cos[kξ(t − x)] dt. 2π − π − π/ξ k=1 π/ξ (A.22) Writing kξ = ξ and letting ξ → 0 this reduces to 1 ∞ +∞ f(x)= dξ f(t)cos[ξ(x − t)] dt. (A.23) π 0 −∞ This can also be written as 1 ∞ f(x)= A(ξ) cos(xξ) + B(ξ)sin(xξ) dξ, (A.24) 2π 0 where ∞ A(ξ) = 2 f(t)cos(ξt) dt, (A.25) −∞ and ∞ B(ξ) = 2 f(t)sin(ξt) dt.
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