Fourier Series and the Wave Equation

Transcription

1 Fourier Series ad the Wave Equatio We start with the oe-dimesioal wave equatio u u =, x u(, t) = u(, t) =, ux (,) = f( x), u ( x,) = This represets a vibratig strig, where u is the displacemet of the strig at poit x at time t The coditios u(, x) = u(, x) = reflect the requiremet that the strig is fixed at both edpoits The coditio ux (,) = f( x) says that the iitial positio (at time t=) u of the strig is give by the fuctio f(x) Fially, the coditio ( x,) = says that the iitial velocity is Solvig this problem takes several steps We bega by guessig we could fid some basic solutios of the form uxt (, ) = XxTt ( ) ( ) I the homework, we foud this guess alog with coditios u(, x) = u(, x) = led to solutios of the form u( x, t) = ( acos( t) + bsi( t))si( x) Next, we compute u = ( asi( t) + bcos( t) ) si( x), ad so applyig the last coditio we get u ( x,) = b si( x ) =, from which we obtai b = for all So our basic solutios take the form u ( x, t) = a cos( t)si( x) Fially, we ote that our equatio is liear, so we ca add up our basic solutios to build ew solutios i the form uxt (, ) = u( xt, ) So i order to satisfy the remaiig costrait, ux (,) = f( x), we will wat to pick values for the a i order to satisfy ux (,) = u( x,) = = = a si( x) = f( x) = a cos()si( x) This last equatio leads to what is called a Fourier sie series = We have see that (ice) fuctios ca be represeted by power series (called Taylor series) We ca also represet (ice) fuctios by sums of sies ad/or cosies (ad ice here does t have to be early as ice as for Taylor series) Such series are called Fourier series To fid the Fourier series whe we are give the fuctio, we start with the assumptio that the fuctio ca be represeted by a series of sies, which will be a valid =

2 assumptio for ay of the fuctios we ecouter i this course defied betwee ad p which have the property that f() = f(p) = (ad eve more fuctios tha that it will tur out) So we will assume that f ( x) = a si( x) ad the try to work out what are the = values of the coefficiets a Two crucial facts that we will use i workig out the values of the a are si( x)si( mx) = if m, si ( x) = Checkig these itegrals is part of the homework assigmet for this week We ca the use these facts to compute f ( x)si( mx) = a si( x) si( mx) = = = a si( x)si( mx) am =, where the ifiite sum collapses dow to a sigle value because all but oe of the itegrals are The we solve this equatio for am = f( x)si( mx) This tells us how to fid the coefficiets of the Fourier sie series for a fuctio Let s work out a example Suppose we pull our strig back at the ceter, with the strig stretched liearly from the edpoits to the ceter This is represeted by usig the fuctio x, x f( x) = x, x The we compute the Fourier coefficiets, which requires that we split the itegral ito two parts (correspodig to the two parts of the fuctio) ad the evaluate each part usig itegratio by parts

3 a = f( x)si( x) / = xsi( x) ( x)si( x) + / / u = x dv = si( x) = xsi( x ) si( x ) xsi( x ) + / / du = v = cos( x ) / / x x = cos( x) + cos( x) + c os( x) cos( x) cos( x) / + / / / / x x = cos( x) + si( x) cos( x) cos( x) si( x) + / + / /, divisible by 4 4, lea ves remaider whe divided by 4 =, leaves remaider whe divided by 4 4, leaves remaider 3 whe divided by 4 Just as we have to split the itegral ito two pieces, sice we have a fuctio defied i two parts, we split the aswer ito 4 pieces, sice si ad cos take differet values depedig o the remaider leaves o divisio by 4 Note that we could have used the symmetry of the fuctios si(x) ad the fuctio f(x) to recogize that all the eve coefficiets would have to be from the start I ay evet, oce we have these differet values for a, we observe we ca simplify them to, eve a = ( )/ 4 ( ), odd Recallig that + will cout through all the odd itegers, we ca the write the Fourier series for f(x) as 4 f ( x) = ( ) si ((+ ) x) = ( + ) If we plot the first 5 terms of this series, we see that the series is quickly covergig to the fuctio

4 Fially, we ca write our solutio to the wave equatio as 4 uxt (, ) = ( ) si ((+ ) x) cos ((+ ) t) ( + ) = Next we will work out aother example of fidig a Fourier series This example is simpler to compute, though it is a bit more complicated from a theoretical perspective Suppose we have the fuctio f(x) = x The we compute the Fourier coefficiets usig itegratio by parts u = x dv = si( x) a = xsi( x) du = v = cos( x ) x = cos( x) + cos( x) x = cos( x) + si( x) ( ) = + ( ) + = ( )

5 The calculatios are much easier for this example, sice we do t have to cut the itegral ito separate pieces As before, oce we have the Fourier coefficiets we ca write out the fuctio as a Fourier series + x = ( ) si ( x) = We plot the first terms of the series to see how the series is covergig We ote that the series is covergig more slowly this time, but it will coverge, at least betwee x= ad x=p The difficulty comes at the right edpoit, x = p Sice we are buildig a series of sie fuctios, ad si(p) = for all values of, our series must have value at x = p, yet the series is workig reasoably well eve with a fuctio which is ot at x = p Furthermore, outside the rage from to p, the Fourier series does t come close to the fuctio f(x) = x The behavior of the Fourier series outside the basic iterval is govered by the fact that si(x) is p-periodic ad odd, so the graph of the Fourier series flips over rather tha followig the origial fuctio These difficulties will affect us i some problems but ot others If our strig is ot exactly p log, we will eed to scale our work to get the Fourier series to match up correctly Ad while our otio of a wave o a strig requires that the iitial positio be cotiuous, that wo t always be the case For example, studyig shock waves may lead to discotiuities I other differetial equatios, discotiuities may arise eve more aturally Jea Fourier was iitially iterested i questios of heat coductio, so he was tryig to solve the heat

6 u u equatio, = κ Whe he tried to model placig a hot bar ext to a cold bar, his x iitial temperature distributio was discotiuous Fourier igored this problem ad solved the problem usig his trigoometric series Whe he submitted his work to the Frech Academy however, other mathematicias were t sure that this was valid A Taylor series, for example, ca t hadle a discotiuity, so why should a Fourier series, which is a sum of cotiuous fuctios, ot break dow as well? It took years to carefully verify mathematically that Fourier s iitial work was ideed correct ad early a cetury to work out all the questios that arose about how Fourier series coverge We will look at some other examples i lab Tuesday ad try to get a sese of how Fourier series coverge ad how their covergece differs from that of Taylor series