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 Sneddon (1951) and Churchill (1972). Extensive tables of transforms have been published by the staff of the Bateman project (Bateman, 1954). A.l Laplace transforms The Laplace transform is particularly useful for problems in which the variables are defined in a semi-infinite domain, say for 0 < t < oo, 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) = 100 F(t) exp( -st) dt, (A.1) 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 function F(t) is transformed into a function f(s). For various functions the Laplace transform can be calculated, sometimes very easily, sometimes with considerable effort. Tables of such transforms are widely available (Churchill, 1972; Bateman, 1954). A short table is given in table A.l. The integrals in this table can all be evaluated with little effort. The fundamental property of the Laplace transform appears when considering the transform of the time derivative. Using partial integration this is found to be oo dF(t) 1 -- exp(-st)dt = sf(s)- F(O). (A.2) 0 dt Thus differentiation with respect to time is transformed into multiplication by s, apart from the subtraction of the initial value F(O). EXAMPLE In order to illustrate the application of the Laplace transform technique let us consider the differential equation 356 INTEGRAL TRANSFORMS 357 No. F(t) f(s) = Jt F(t) exp(-st)dt 1 1 1 - s 1 2 t s2 n! 3 tn 8n+l 1 4 exp(at) -- s-a a 5 sin( at) s2 + a2 s 6 cos( at) s2 + a2 Table A.l. Some Laplace transforms. dF(t) 2F = 0 dt + , (A.3) with the initial condition F(O) = 5. Using the property (A.2) the differential equation (A.3) is transformed into the algebraic equation (s+2)/-5=0, (A.4) the solution of which is 5 !=-. (A.5) s+2 Inverse transformation now gives, using transform no. 3 from table A.1, F = 5 exp( -2t). (A.6) 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 solution (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. 358 APPENDIX A 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) (A.7) s 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- sl)(s- s2) · · · (s- sn)· (A.8) One may now write p(s) a1 a2 an f(s) = -() = --+ --+···+ --. (A.9) q S S - St S - S2 S - Sn The coefficient a; can be determined by multiplication of both sides of eq. (A.9) by (s- s;), and then passing into the limit s- s;. This gives 1. ( s - s;) p( s) a;= ,_,;tm q ( s ) . (A.lO) Because q(s;) = 0 the limit may be evaluated using L'Hopital's rule, giving p(s;) a; = --. (A.ll) q'(si) Inverse transformation of the expression (A.9) now gives, using formula no. 4 from table A.l, ~ p(s;) F(t) = L...J '(s·) exp(s;t). (A.l2) i=l q I This is Heaviside's expansion. It gives the inverse transform of the function (A.7). It has been derived here for the case of the quotient of two polynomials, but it can be used equally well for more general cases of the quotient of two functions, provided that the denominator is of higher order than the numerator, and that the denominator has zeroes of the first order only. A.2 Fourier transforms 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 section. 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( 8), which is periodic with a period 211", such that g( 8 + 211") = g( 8). This function can be written as INTEGRAL TRANSFORMS 359 1 00 g(O) = 2Ao + L {Ak cos(kO) + Bk sin(kO)}, (A.13) k=l where Ak = -11+.,.. g(t) cos(kt) dt, (A.14) 7r _.,.. and 11+.,.. Bk = ; _.,.. g(t) sin(kt) dt, (A.15) These formulas can be derived by multiplication of eq. (A.13) by cos(jO) or sin(jO), and then integrating the result from (} = -1r to (} = +1r. It will then appear that from the infinite series only one term is unequal to zero, namely for k = j. This leads to eqs. (A.14) and (A.15). For a function with period 21rl the Fourier expansion can be obtained from (A.13) by replacing(} with x/1, t by t/1 and then renaming g(x/1) as f(x). The result is 1 00 f(x) = 2Ao + L {Ak cos(kx/1) + Bk sin(kx/1)}, (A.16) k=l where now 11+'11'1 Ak = l f(t) cos(kt/1) dt, (A.17) 7r -'11'1 and 11+71'1 Bk = - 1 f(t) sin(kt/1) dt. (A.18) 7r -71'1 EXAMPLE As an example consider the block function defined by !( ) _ { 0, if lxl > 7rl/2, (A.19) x - 1, if lxl < 7rl/2. For this case the coefficients Ak and Bk can easily be calculated, using the expres­ sions (A.17) and (A.I8). 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 I/k. The series (A.I6) finally can be written as I 2 { x I 3x 1 5x 1 ( 7x ) } f(x) = 2" +; cos( l)- acos( T) + S cos( T)- 7 cos T + .... (A.20) 360 APPENDIX A f(x) /[\ 2 ···················:···················:···················:···················:·················· ~ ~ 1 1 ~ ~ ~ ~ ·· · · · · · ·· ··· ·······I·················· ·I·················· ·I·················· ~ ·· ··· ··· ·· ·· · · · · · · ~ ~ l 1 f..- ......... ~-· ... .... : ........ -~····· ... ·Jro....-+--1114 .................. : .................. : .................. : ................................... ~ ~ 1 ~ ~ l l X OL-~~-=--~--~=--~---=0 1 2 3 ~~~------~--~4 5,.. 11"1 Figure A.l. Fourier series, 40 terms. The first term of this series represents the average value of the function, the second 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 discontinuities. The approximation becomes better, of course, when more terms are taken into account. FROM FOURIER SERIES TO FOURIER TRANSFORM Substitution of (A.17) and (A.18) into (A.16) gives 1 J+..-1 f(x) = -2 I f(t) dt + 7r -..-1 00 1 J+..-1 {; f(t) 1rl -..-I f(t) cos[k(t- x)f~ dt. (A.21) The interval can be made very large by writing 1/1 = aa. Then the formula (A.21) becomes aa J+..-/Aa f(x) =- f(t) dt + 27r -..-/A a oo a a !+..-/A a {; f(t) 7 -..-/Aa f(t) cos[kaa(t- x)] dt. (A.22) If aa ~ 0 this reduces to f(x) = -11oo da J+oo f(t) cos[a(x- t)] dt. (A.23) 7r 0 -oo INTEGRAL TRANSFORMS 361 This can also be written as l(x) = 100 {A(o:) cos(o:x) + B(o:) sin(o:x)} do:, (A.24) where A(o:) = ;11 -oo00 l(t) cos(o:t) dt, (A.25) and 00 B(o:) = ;11 -oo l(t) sin(o:t) dt. (A.26) It can be seen from (A.25) that A(o:) is an even function, A(-o:) = A(o:), and from (A.26) it can be seen that B(o:) is uneven, B(-o:) = -B(o:). Therefore, if G(o:) =A+ iB, it follows that 00 I: G(o:) do:= 21 { A(o:) cos(o:x) + B(o:) sin(o:x)} do:, (A.27) and thus eq. (A.24) may also be written as l(x) = 211 -oo00 G(o:) exp(-io:x)do:, (A.28) where G(o:) =;11 -oo00 l(t) exp(io:t)dt, (A.29) Finally, by writing F(o:) = ~G(o:) the factor ~can be eliminated from the expres­ sion (A.28), l(x) =I: F(o:) exp(-io:x)do:, (A.30) where now 00 F(o:) = 217r 1-oo l(t) exp(io:t) dt, (A.31) This is the basic formula of the Fourier transform method.