Fourier Series and the Wave Equation Part 2


 Willis Thornton
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1 Fourier Series ad the Wave Equatio Part There are two big ideas i our work this week. The first is the use of liearity to break complicated problems ito simple pieces. The secod is the use of the symmetries of the sie fuctio to help aalyze complicated fuctios as sums of simple fuctios. O Moday, we used the first idea to show how we could take the very simple solutios of the wave equatio foud usig separatio of variables ad put them together to build more complicated solutios. We used the secod idea whe we used the orthogoality of the fuctios si(x) over the rage from to p to write a geeral fuctio as a Fourier series. We will develop both these geeral themes further today. O Moday, we cosidered the problem u u u(, t) = u(, t) () ux (,) = f( x), u ( x,) =. I this problem, we had a vibratig strig whose edpoits were fixed at ad whose iitial positio was give by the fuctio f(x), startig with iitial velocity. This last coditio is ot particularly reasoable from a physical stadpoit. If you pluck a guitar strig ad let it vibrate, the pluckig ot oly chages the iitial positio but also imparts a iitial velocity. So we should istead look at a problem of the form w w w(, t) = w(, t) () wx (,) = f( x), w ( x,) = gx ( ). where g(x) deotes the iitial velocity. The problem is that we used the fact that the iitial velocity was as part of our simplifyig our large collectio of solutios geerated by separatio of variables dow to a small set which we could use to build a Fourier series. The key is to apply our basic idea of liearity agai. We will split problem () ito two parts ad solve each part separately. This should remid you of how we solved liear equatios i chapter by lookig for the homogeeous ad particular solutios separately. Both approaches are based o the key advatage of liear equatios, that you ca split large problems ito small problems, solve the small problems separately, ad the add the solutios to the small problems together to solve the large problem. Sice we already kow how to solve problem (), where the iitial positio is give ad the iitial velocity is, what we have to do to fiish solvig the full problem () is to work with a give velocity ad a iitial positio. This leads to the problem
2 v v v(, t) = v(, t) (3) vx (,) v ( x,) = gx ( ). We solve this problem i a maer almost idetical to our solutio to problem () o v v Moday. From the differetial equatio = ad the boudary coditios v(, t) = v(, t) = we coclude that v( x, t) = ( acos( t) + bsi( t))si( x). The we use the requiremet that v(x,) = for all x to coclude a = ad hece that v( x, t) = bsi( t)si( x). We ow form a more geeral solutio by addig these simple solutios to get vxt (, ) = bsi( t)si( x). Next we compute v ( x,) = b si( x ) = g ( x ) to see that all we eed to do is fid aother Fourier x, x series expasio. For example, if gx ( ) =, the from our work x, x 4 Moday we kow that gx ( ) = ( ) si ((+ ) x), so ( + ), if eve; b = 4 Note the fact that we had b as the coefficiet of our ( ), if odd. 3 series for g(x) led to the extra power of i the deomiator of b, compared to our result from Moday. Usig the fact that + rus through all the odd itegers, we ca the write our fial solutio to problem () as 4 vxt (, ) = ( ) si 3 ((+ ) x) si ((+ ) t). ( + ) Fially, we put our solutios to () ad (3) together to solve the full problem (). Let wxt (, ) = uxt (, ) + vxt (, ) where u ad v are the solutios we foud to problems () ad (3) respectively. The
3 w ( u+ v) u v u v ( u+ v) w = = + = + = w(, t) = u(, t) + v(, t) = + w(, t) = u(, t) + v(, t) = + =, wx (,) = ux (,) + vx (,) = f( x) + = f( x), w ( u+ v) u v ( x,) = ( x,) = ( x,) + ( x,) = + g( x) = gx ( ), where we have used the fact that both our equatio ad our coditios are liear to show x, x that the solutio to the full problem () with f( x) = g( x) = is x, x 4 wxt (, ) = ( ) si ((+ ) t) + cos ((+ ) t) si ((+ ) x). (+ ) Note that the coditio that the iitial positio ad iitial velocity are equal is ot quite as ulikely as you might thik. After all, the poits that have bee pulled farthest from equilibrium whe the strig is plucked are goig to be movig faster tha the parts that have t bee moved as far from equilibrium. O the homework you have a slightly more geeral problem, i which the edpoits are fixed at poits a ad b rather tha at. This just adds oe more simple problem to be solved, ad the you add thigs up just as we did here to fid the full solutio. The other big idea we ve worked with is usig the symmetry properties of the sie fuctio to simplify the process of writig a complicated fuctio as a sum of sie terms. You should have further developed a sese of how these symmetry properties work rig the lab o Tuesday. I particular, you should have oticed that if the fuctio f(x) is symmetric about x = p/, the the eve coefficiets of the Fourier series are all, while if f(x) is atisymmetric about p/ the the odd coefficiets are. This, ad the other facts we eed about sie fuctios, ca be proved with the aid of some trig idetities. si( θ )si( ϕ) = ( cos( θ ϕ) cos( θ + ϕ) ) si ( θ) = cos( θ) si( θ) = si( + θ) si( + θ) = si( θ), if eve si( + θ) = si( θ), if odd
4 For example, suppose we wat to show that if f(x) is atisymmetric about, i.e. f ( x) = f( + x), the all the odd coefficiets i the Fourier series for f(x) are. I will write out the steps ad the explai them at the ed. am = f( x)si( mx) / = f( x)si( mx) f( x)si( mx) + / x= u x= + u = = / m m = f ( u)si( mu)( ) f( u)si( mu) / / / m m = f ( u)si( mu) + f ( + u)si( + mu) / / m m = f( u)si( mu) f( u)si( mu) + / m m = f ( u) si( mu) si( mu) + / = f( u) [ ] =. We start with the defiitio of the Fourier coefficiets am = f( x)si( mx) ad split the itegral ito two pieces, oe from to ad oe from to p i the secod lie. We the make the idicated chage of variables i the third lie, where the sig o the switches the upper ad lower limits of the first itegral i the fourth equatio. The we use the atisymmetry coditio we are give for f(x) i the fifth equatio. We the write the differece of the itegrals as the itegral of the differece, where we factor out the commo term f ( u). Fially, sice m is odd, the last trig idetity from the previous page shows the differece i the brackets is, ad hece the whole itegral is.
5 There are a lot of tricks like this you ca develop. It should t be surprisig that si( x)si( mx) = for m (which is oe of the key facts we used i fidig the Fourier series), sice si(x) ad si(mx) will have differet symmetries. You are asked to show several results like this o the homework. You are also asked to fid the Fourier series for oe particular fuctio ad the solve a problem similar to problem () where you use that fuctio for the iitial positio, which is t hard oce you ve got the Fourier series. While we have cocetrated o applicatios of Fourier series to differetial equatios (sice this is a differetial equatios course after all), it should be oted they have applicatios i may differet areas. Fourier series, ad especially the discrete aalogue, the Fast Fourier Trasform, play a crucial role i sigal aalysis ad data compressio. Furthermore, may of the tricks we developed this week will work for other sets of fuctios that will be better adapted to differet problems. The extra credit assigmet takes you through a stadard approach to buildig such series for SturmLiouville problems, which use some of the deeper applicatios of liearity to see how the orthogoality coditio we use for Fourier series will work i a wide variety of fuctios that arise i variable coefficiet problems. We will look more closely at a specific example, the wave equatio for a circular drum, o Moday, ad show how we ca explai mathematically the toal differeces betwee drums (vibratig circles) ad violis (vibratig strigs).
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