L6b Solving PDEs with Fourier Series

Transcription

1 6b Solvig PDEs with Fourier Series Rev 1 Feb 13 Alice thought she had ever see such a curious croquet-groud i her life; it was all ridges ad furrows; the balls were live hedgehogs, the mallets live flamigoes, ad the soldiers had to double themselves up ad to stad o their hads ad feet, to make the arches. The chief difficulty Alice foud at first was i maagig her flamigo: she succeeded i gettig its body tucked away, comfortably eough, uder her arm, with its legs hagig dow, but geerally, just as she had got its eck icely straighteed out, ad was goig to give the hedgehog a blow with its head, it would twist itself roud ad look up i her face, with such a puzzled expressio that she could ot help burstig out laughig. Refer to Dawkis PDE chapter, startig o p. 34. If you have ot already looked at Examples Heat Equatio, do so before cotiuig. It is o coicidece that the proportioal relatioship betwee the time rate of chage i the temperature of a solid object ad its iteral thermal gradiet is kow as Fourier s aw. Joseph Fourier (our hero from the last lab) solved this problem usig (o surprise) the trigoometric series which came to bear his ame. We used Fourier s aw to derive the simple form of the oe dimesioal Heat Equatio, u t = -k u xx. et s see how we fid the Fourier Series solutio for this PDE. Heat equatio agai We assume the variables-separable solutio, uxt (, ) FxGt ( ) ( ), obtaiig the ODEs dg dt d F kg()ad t F( x), with the separatio costat. dx k t For the time equatio, Gt () Ge, plai old expoetial coolig: Iitially G = G ; evetually G. For the x equatio, the secod derivative yields F( x) Acos x Bsi x. kt Put em all together: uxt (, ) FxGt ( ) ( ) Ge ( Acos x Bsi x). If you do t like takig si ad cos of square roots, you could call the separatio costat.

2 et s take simple boudary coditios: u(, t) = u(, t) = ad the iitial coditio u(x,) = f(x); arbitrarily settig k = 1. We rule out the trivial A = B =. Note that f() = f() = is a boudary coditio o the iitial coditio. At x =, we lear that A =. At x =, we fid. Ivokig the iitial coditio, f ( x) Csi x, where we combied costats ito C. But ca be ay positive iteger eigevalues! Ad that meas lots of eigefuctio solutios; plus the sum (ay, liear combiatio) of those solutios is also a solutio. Hece there is a geeral solutio give by k t (, ) si uxt e C x 1 ad that sum looks like a Fourier Series. If t=, it is the Fourier Series for the iitial coditio fuctio f(x). That meas we ca calculate the Fourier coefficiets C from ay iitial coditio fuctio f(x)! Souds like fu. Or maybe a lot of work. Here is a clever trick that helps: For ay iteger m, mx x f( x) si dx C si 1 mx si dx I that step, all we did was replace f(x) with its Fourier Series. The ext step, x mx x mx ( C si si ) dx Csi si dx, 1 1 just says that the itegral of a sum is the sum of itegrals. But the itegral of si x si mx terms evaluates to either (whe m ) or (si x ) dx whe m =, a happy circumstace which you should verify. Hit: Try fidig all possible values of The look at the square terms, where m.

3 If all is well, you obtaied the coefficiets C f( x)si xdx, aka the Fourier sie series for f(x). Is it a problem that this solutio oly icludes sies? Hit: look back at the boudary coditios! Example et f(x) = 4 x ( x), which is at both x = ad x =. The Fourier Series solutio to the Heat ( ) t Equatio, takig k = 1, is uxt (, ) ae si x, where the coefficiets 8 a x( x)si xdx, which you ca easily evaluate. What happes to the eve terms? Pluggig i a few values for, we obtai coefficiets which have a costat umerator ad deomiators icreasig as 3 for odd values of. That is great ews, because it meas we do t eed very may terms before they become very small., Plot of the temperature fuctio with = 1 for 1, 3, 5 ad 7 Fourier Series terms. The differece betwee 5 terms ad 7 terms is egligible ideed. The time dimesio is domiated by that expoetial; after some time has elapsed, differeces betwee the Fourier Series solutios do ot eve show. 3

4 Work through the problems at If ecessary, ru through their heat equatio module ( first. For review, see Wave equatio with Fourier Series If you have ot already worked through Examples- Wave Equatio, do so ow. We ve already see that the oedimesioal wave equatio, u tt (x, t) = c u xx (x, t), yields very icely to separatio of variables. Applyig boudary coditios to the resultig x fuctio ad the iitial coditio to the t fuctio gives us the solutio we seek. Suppose we have the boudary coditios u(, t) = = u(, t). M. Dirichlet, say hello to M. Fourier. Separatio of variables produces the wave fuctio solutio u(x,t) = F(x)G(t), where F( x) C coskx C sikx,i the usual way. 1 the boudary coditio at x = requires C 1 =. Thus C must be ozero to avoid a trivial solutio. The eigefuctios of the wave equatio come from settig si k =, which requires eigevalues k ad thus ( ) si F x C, where is ay positive iteger. Sice we will have costats i the time fuctios, we are free to set C = 1 at this stage. 4

5 The time fuctios are solutios to G G, such as G ( t ) B cos t C si t with ck c. Thus the most geeral wave fuctios satisfyig these BCs are give by (, ) [ cos c si c u x t B t C t]si x. Note that the eigefuctios have a set of frequecies ad a correspodig set of modes kow as the fudametal ( = 1) ad overtoes or harmoics ( > ). The odal poits (also 1 kow simply as odes) x,,..., have o displacemet as u (x, t) = at those poits. These are stadig waves. Iitial coditios: ux (,) f( x) ad u t (x,) = We kow that ay sum of the eigefuctios u (x, t) as defied above will also solve the wave equatio. Our Fourier Series solutio will take the geeral form uxt (, ) ( BcostCsi t)si x, for 1 c although we will usually be happy with a approximatio with some fiite umber of terms. Settig t =, the coefficiets must satisfy ux (,) B si xad thus B f( x)si xdx 1. Sice the iitial velocity is, C =. Refer to Dawkis PDE, p Example Suppose ux (,) f( x) si for = ad c = 1. x Verify the Fourier Series solutio by determiig the coefficiets ad plottig for several values of. 5

6 Plot of the first 5 terms for t = Plot of the sum of just the first 3 terms (t = ) with the iitial coditio superimposed. The Fourier Series with two ozero terms is Note that this is real world stuff, as all physical systems are iheretly bad-limited. We ca either geerate, trasmit, receive, amplify or detect a ifiite umber of frequecies. Ad all bad-limited Fourier Series have some of that edge-effect amed for Mr. Gibbs. Go back to 6a to see the Gibbs effect i actio. Plot a aimatio with time of a 5 term series (it s a thig of beauty!) 6

7 Neuma coditio Suppose the ICs are based o u(x,) = (iitial positio at equilibrium) with some iitial motio, u gx ( ). et s differetiate the geeral solutio to obtai t ut( x, t) ( BsitCcos t)si x, 1 which we evaluate for t = : ut( x,) Csi x g( x). The coefficiets must satisfy C g( x)si xdx g( x)si xdx c The B are all, so the solutio becomes 1. uxt (, ) Ccostsi x 1 Example et = 1 ad c = 1; 1.1 x, x u (,) t x 1.1(1 x), x 1 At t =, the first 5 terms i the series are Plot of the first two ozero terms at t =. The 5 term series solutio is :. 7

8 Fially, we recall M. D Alembert (from Examples: Wave Equatio) ad ote that our Fourier Series wave fuctios also satisfy travelig wave relatioships xvt uxt (, ) f( xvt) f( xvt) gsds ( ) v xvt Now let us proclaim from o high: Stadig or travelig, heatig or wavig; these PDEs have solutios composed of eigefuctios from the boudary coditios ad the Fourier Series of the iitial coditios. See for additioal examples. 8