David Randall. ( )e ikx. k = u x,t. u( x,t)e ikx dx L. x L /2. Recall that the proof of (1) and (2) involves use of the orthogonality condition.

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1 ! Revised April 21, :27 P! 1 Fourier Series David Radall Assume ha u( x,) is real ad iegrable If he domai is periodic, wih period L, we ca express u( x,) exacly by a Fourier series expasio: ( ) = u x, ( )e ikx k = The complex coefficies ( ) ca be evaluaed usig (1) ( ) = 1 x+ L /2 u( x,)e ikx dx L x L /2 Recall ha he proof of (1) ad (2) ivolves use of he orhogoaliy codiio (2) where is he Kroecker dela 1 L x+ L /2 e ikx e ilx dx = δ kl, x L /2 δ kl 1, k = l 0, k l (3) (4) From (1), we see ha he x -derivaive of u saisfies x x, ( )e ikx ( ) = ik k = (5) Copyrigh David Radall, 2010

2 ! Revised April 21, :27 P! 2 Ispecio of (5) shows ha x does o have a coribuio from û 0 ; he reaso for his should be clear A umerical model uses equaios similar o (1), (2), ad (5), bu wih a fiie se of wave umbers, ad wih x defied o a fiie mesh: ( ) u x, ( )e ikx, k = ( ) 1 u ( x, )e ikx, k, ( ) ik x x, ( )e ikx k = (8) Noe ha we have used approximaely equal sigs i (6) - (8) I (7) we sum over a grid wih pois I he followig discussio, we assume ha he value of is chose by he user The value of, correspodig o a give value of, is discussed below Subsiuio of (6) io (7) gives ( ) = 1 ( )e ilx e ikx, k l= (9) This is of course a raher circular subsiuio, bu he resul serves o clarify some basic ideas If expaded, each erm o he righ-had side of (9) ivolves he produc of wo wave umbers, l ad k, each of which lies i he rage o The rage for wave umber l is explicily spelled ou i he ier sum o he righ-had side of (9); he rage for wave umber k is udersood because, as idicaed, we wish o evaluae he lef-had side of (9) for k i he rage o Because each erm o he righ-had side of (9) ivolves he produc of wo Fourier modes wih wave umbers i he rage o, each erm icludes wave umbers up o ±2 We herefore eed complex coefficies, ie values of he ( ) (6) (7) Because u is real, i mus be rue ha û k = û * k, where he * deoes he cougae To see why his is so, cosider he +k ad k coribuios o he sum i (6): Copyrigh David Radall, 2010

3 ! Revised April 21, :27 P! 3 Τ k where R k e iθ ( ) ad R k e iµ û k idepedece, our assumpio ha u x, Τ k ( x ) mus be zero, for all x I follows ha ( x ) ( )e ikx + û k ( )e ikx R k e iθ e ikx + R k e iµ e ikx, (10) ( ), ad R k ad R k are real ad o-egaive By liear ( ) for all x is real implies ha he imagiary par of The oly way o saisfy his for all x is o se R k si ( θ + kx ) + R k si ( µ kx ) = 0 for all x (11) ad θ + kx = ( µ kx ) = µ + kx, which meas ha θ = µ, (12) Eqs (12) ad (13) imply ha as was o be demosraed R k = R k û k = *, (13) (14) Eq (14) implies ha ad û k ogeher ivolve oly wo disic real umbers I addiio, i follows from (14) ha û 0 is real Therefore, he complex values of embody he equivale of oly disic real umbers The Fourier represeaio up o wave umber is hus equivale o represeig he real fucio u x, ( ) o grid pois, i he sese ha he iformaio coe is he same We coclude ha, i order o use a grid of pois o represe he ampliudes ad phases of all waves up o k = ±, we eed ; we ca use more ha pois, bu o fewer As a very simple example, a highly rucaed Fourier represeaio of u icludig us wave umbers zero ad oe is equivale o a grid-poi represeaio of u usig 3 grid pois The real values of u assiged a he hree grid pois suffice o compue he coefficie of wave umber zero (he mea value of u ) ad he phase ad ampliude (or sie ad cosie coefficies ) of wave umber oe Subsiuig (7) io (8) gives Copyrigh David Radall, 2010

4 ! Revised April 21, :27 P! 4 x x l, u ( x, )e ikx eikx l k = ( ) ik Reversig he order of summaio leads o (15) where x x l, ( ) α ( ) l u x,, (16) α l i ke ik ( x l x ) k = (17) The poi of his lile derivaio is ha (16) ca be ierpreed as a fiie-differece approximaio - a special oe ivolvig all grid pois i he domai From his poi of view, specral models ca be regarded as a class of fiie-differece models Eq (9) ca be rewrie as ( ) = 1 The coribuio of wave umber l o is e i( l k)x l = (18) The we ca wrie For l = k, we recover ( ) l 1 ( )x e i l k = ( ) l l = = ûl ( ei l k )x (19) (20) For l k, we ge ( ) k = u k 1 = (21) Copyrigh David Radall, 2010

5 ! Revised April 21, :27 P! 5 ( ) l = ûl { ( ) x ( ) x } cos l k + isi l k From (22) we ca see ha, provided ha l k fis o he grid, we will have ( ) l = 0 for l k The resuls (21) ad (23) are as would be expeced (22) (23) The The coribuio of grid poi o he Fourier coefficie is ( ) 1 e i( l k)x l = (24) = ( ), k We mus ge idepede complex values of These are (25) Comparig wih (7), we see ha as expeced ( ) = e ikx e ilx l = ( ) = u( x )e ikx, k, (26) (27) Copyrigh David Radall, 2010