Class #24 Monday, April 16, φ φ φ

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Blaze Harper
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1 lass #4 Moday, April 6, 08 haptr 3: Partial Diffrtial Equatios (PDE s First of all, this sctio is vry, vry difficult. But it s also supr cool. PDE s thr is mor tha o idpdt variabl. Exampl: φ φ φ φ = = 0 z Th dpdt variabl is φ. has idpdt variabls x, y, ad z. As with ODE s, th gral procdur is to hop that sombody ls ca tll you th aswr bfor you v start th problm. But, somtims you still hav to do it yourslf. I physics, thr ar a zillio rlvat PDE s. Exampl iclud hat trasfr, th quatios that dscrib th motio of wavs (.g., soud, or light, which is a wav of lctric fild, Schrodigr s quatio, tc. Most of ths quatios iclud a somwhr, which always rsults i drivativs with rspct to x, y, ad z. I additio, if you wat to kow th tmpratur (or whatvr, you ot oly hav to spcify whr you wat to kow φ (i.., at x, y, ad z, but oft you also hav to spcify wh you watd to kow φ (t. So, a lot of PDE s hav 4 idpdt variabls. Naturally, thr ar v mor kids of problms tha just fidig tmpratur as a fuctio of ths 4 variabls. You might istad hav a quatio that, if solvd, could tll you prssur as a fuctio of tmpratur ad dsity. Sparatio of Variabls Whvr possibl, w solv PDE s by a mthod calld sparatio of variabls, which is ufortuatly ot aythig lik th sparatio of variabls w usd to solv ODE s. For PDE s sparatio of variabls is a ickam for a mthod actually calld Eigfuctio dcompositio. Th Hat Equatio Lt s start with a simplifid form of th hat quatio. This quatio is about coductio: how th tmpratur i o part of a objct affcts th tmpratur i othr parts. Th basic quatio T is: k T =. This quatio assums that thr ar o sourcs of rgy mbddd i th t objct, so it is alrady somwhat simplifid. W will furthr simplify it by sayig that our objct is oly D, ad also by sayig that th tmpratur profil is stady (i.., that it is ot a T T fuctio of tim. With ths simplificatios, th quatio bcoms mrly + = 0. Notic that th coductivity of th matrial, k, caclld out altogthr as a rsult of makig it Pag of 6

Lookig at this quatio, I s two drivativs with rspct to x, ad two with rspct to y.

2 lass #4 Moday, April 6, 08 stady. As you imagi, this quatio is ot that itrstig yt. I fact, it has a ifiit umbr of aswrs. I ordr to v gt startd, w hav to rcogiz that thr must b som caus for th tmpratur to actually b o way istad of aothr. For most problms, that mas that w d to spcify boudary coditios. Lookig at this quatio, I s two drivativs with rspct to x, ad two with rspct to y. That mas that I d to spcify a total of four boudary coditios bfor this v bcoms a ral physics problm that I might wat to bothr with. So, hr is th ral problm I wat to solv: My objct is a rctagular sht of mtal (th kid of mtal appartly dos t mattr, sic k is go. I will hold th bottom dg agaist somthig hot, ad th othr thr dgs agaist somthig cold. I othr words, I will compltly spcify th tmpraturs aroud th dgs. As you might imagi, with ths boudaris, th tmpratur of parts of th plat ar th bottom dg will likly b hottr tha thos ar th othr dgs. That s our goal: to fid T(x, y. Lt s writ out th boudary coditios xplicitly: L y T = Tcold T = T cold T = Tcold T ( x = 0 = T T ( x = L = T T ( y = 0 = T T ( y = L = T H T = T hot L x It is customary, but ot strictly cssary, to algbraically maipulat th quatio ad all th boudary coditios i such a way that th rsult has o uits. This is do so that havig cratd th solutio i a gric way, w ca apply it to multipl similar situatios. It is aalogous to solvig Aalyt I problms symbolically istad of umrically. Hr is th usual way of rmovig uits from quatios. W crat w variabls that ar liarly rlatd to our origial idpdt ad dpdt variabls: x x = x = x L L y = = y y y L L L = = N L NL L T T ( T = TH T Θ + T TH T Pag of 6

3 lass #4 Moday, April 6, 08 Notic that o of th four w variabls has ay uits. Simpl substitutio dirctly ito our basic quatio ad also ito our boudary coditios rsults i: Θ Θ + = 0 ( 0 0 Θ x = = ( 0 Θ x = = = ( y 0 Θ y = N = ( 0 I othr words, w will rally solv for Θ(x, y. Oc do, w ca covrt it back ito T, if w car, usig th last quatio o th prvious pag. At this poit, it also customary to admit that w ar too lazy to bothr writig out all th stars. So, v though it s cofusig, w ll writ x wh w rally ma x. Now, w ca start th ral work. First, Sparatio of Variabls : this mas that w cross our figrs ad hop that th aswr is th product of two sparat aswrs, ach of which is itslf a fuctio of oly o of th idpdt variabls. Lt s call ths two fuctios X(x ad Y(y. So, w r hopig that X(x Y(y. Lt s substitut this bit of this hopfulss back ito th mai quatio: Θ X ( x Y ( y X ( x = = Y ( y, Θ X ( x Y ( y Y ( y = = X ( x So, Θ Θ X ( x Y ( y + = Y ( y + X ( x = 0 Dividig both sids by X ( x Y ( y ( ( X x Y y + = 0 X ( x Y ( y You might s why thy call this sparatio of variabls all th x s ar togthr i o group (both th variabl x ad th fuctio X, ad all th y s ar togthr, too. If som fuctio of oly x plus aothr fuctio of oly y add up to zro, always, th logically it must b tru that ach fuctio is a costat, ad o is th gativ of th othr. I ll writ it this way: Pag 3 of 6

4 lass #4 Moday, April 6, 08 X ( x X ( x = k Y ( y Y ( y = + k As you ca s, I calld th costat k, but it has othig to do with th coductivity k that w saw arlir. I othr words, if my hops work out, th what I rally d to do is solv ths two sparat ordiary diffrtial quatios, th multiply th two rsults togthr. Th dcisio to call th costat k istad of somthig mor ispiratioal such as coms from hidsight, from havig do th problm alrady ad oticig that if you calld it, th th aswr has a i it. So, by callig it k, w r hopig to mak th aswr look simplr i th log ru. Lt s ow solv ths two quatios sparatly: X Y X Y X ( x = k = + k Y = k X = + + X k X =0 Y k From chaptr 8, w us th auxiliary quatio "D" mthod to discovr: ky ky ( Acos( kx Bsi( kx ( D k Y ky X = Acos( kx + Bsi( kx Y = + D + + Y =0 ky W ow hav 5 ukows to solv for (A, B,, D, k, usig our 4 boudary coditios. I hop o of th ukows magically disappars! ky ky B # (lft dg: Θ(x = 0 = 0 thrfor A = 0 ( Bsi( kx ( + D ky ky B # (right dg: Θ(x = = 0 0 ( Bsi( k ( D = +. Th oly way for this to happ is if B is always zro, or if si(k is always zro. If B = 0, th w hav o solutio lft whatsovr, so lt s xami si(k = 0. I this cas, it must th b tru that k is som itgr multipl of π. I fact, it ca b ay ad vry itgr multipl of π : k = π πy πy So, our solutio so far is ( B si( π x ( + D kn kn B #3 (top dg: Θ(y = N = 0 0 ( D + πn = +, or D =. πn Pag 4 of 6

5 lass #4 Moday, April 6, 08 So, our total aswr so far is: ( B si[ πx] ( B si[ πx] πy πn πn πy πn πy πn If w rviw chaptr two a littl bit, w might rcogiz that this ca b simplifid. Spcifically, x x sih( x =. Pluggig this i rsults i this simplifid rsult so far: πn πy ( B si[ π x] E sih[ π ( N y ], whr E πn =. To sav myslf som troubl, I ll call th product B E to b som w fuctio F (this is what limiats my 5 th ukow, by th way: ( F si( π x sih ( π ( N y I hav o mor B: B #4 (bottom dg: Θ(y = 0 = ( F π x ( π ( N = si( sih 0 This is a lot lik th Fourir sris problms w v do rctly. Agai, to sav myslf som crampig i my had, I ll oc agai ivt a w lttr: b = F sih( π N. I othr words, I gt: = ( b π x si( If this wr a Fourir sris problm it would hav a /L i frot of it. So, this thig o th right had sid is th Fourir sris for f(x = usig L =. To complt th fial stps of this problm, w d to thik of this Fourir sris as rprstig a rpatig fuctio. Sic it oly has si trms i it, it bttr b a odd rpatig fuctio, as opposd to a v o: N Miimum that w d 0 3 Odd fuctio w ll gt. Pag 5 of 6

lass #4 Moday, April 6, 08 Solvig this for b as w did for all th othr Fourir sris last wk: b = L 0 πx L L ( si dx + ( + 0 πx si dx L = π ( cos( π Th fuctio cos(π is itrstig, ad might b simplifid if w

6 lass #4 Moday, April 6, 08 Solvig this for b as w did for all th othr Fourir sris last wk: b = L 0 πx L L ( si dx + ( + 0 πx si dx L = π ( cos( π Th fuctio cos(π is itrstig, ad might b simplifid if w r lucky: If is a v umbr, th cos(π =, ad so b = 0. 4 If is a odd umbr, th cos(π =, ad b =. π I ll tak this b, ad plug it back ito b = F sih( π N to fid FN, ad So, w hav arrivd at som vrsio of our complt aswr: N ( π ( N y ( π N 4 sih si( π x. odd π sih Our last stp would b to covrt ito som othr form m that isolats th odd valus for us, as w did last wk. I this cas, odd valus of ar gratd stps of m. So, for xampl, m = =, m = = 3, m = 3 = 5, tc. m= 4 ( m sih π sih That s b a lot of work hr. It rquirs us to combi parts from virtually vry sparat topic that w v studid so far this smstr. W usd sris, complx umbrs, chai ruls, itgrals, diffrtial quatios, ad Fourir sris. Th oly thig w did t us was matrics. S associatd Mathmatica sht for gratio of this D (cotour plot of th solutio: [( m π ( N y ] [( m πn ] si [( m πx] Pag 6 of 6

z 1+ 3 z = Π n =1 z f() z = n e - z = ( 1-z) e z e n z z 1- n = ( 1-z/2) 1+ 2n z e 2n e n -1 ( 1-z )/2 e 2n-1 1-2n -1 1 () z

z 1+ 3 z = Π n =1 z f() z = n e - z = ( 1-z) e z e n z z 1- n = ( 1-z/2) 1+ 2n z e 2n e n -1 ( 1-z )/2 e 2n-1 1-2n -1 1 () z Sris Expasio of Rciprocal of Gamma Fuctio. Fuctios with Itgrs as Roots Fuctio f with gativ itgrs as roots ca b dscribd as follows. f() Howvr, this ifiit product divrgs. That is, such a fuctio caot xist