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W09_201B_HW4sol thomases

W09_201B_HW4sol thomases - LL H0007€\N(21LQ N0 4 It I A...

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Unformatted text preview: LL} H0007€\N(21LQ N0 4 It I A?) 17 ' ‘ {.5} M w ‘ :72 XWJTEH: X7”) T“) 29‘ 1151:5122 2* TM) XML) T;(‘):-/\/"’(f) :‘> X8» = —\ X”) Iy/MU; _W JIM B+C=0 w a 101x L K ’7 é't' X{) () D ”L -HL \ngxCZO 2» M ,Wza‘ have: >\>o: A.) SW6): A G’Af 8X00 2 g) M“ (fixfitc QMUT’X) fig”? HLSMTZD X gnfi Vnzi L 400 2 9 ”W -nth ”2! L L ADO M750): fox): £0n/J’lmnzx QDO/Ww Claus/wt; 73) 1> fwd) 4%, [(5 Wm “jg-L ”QM/QM j§> N V. //.n‘ 2 4 Q ),flnafic> 2 0,, 11 Mill/l2 ~> On: / g (“’4 ”VJ—134% L 0 L an if “UL/é) : E On 4"?"17‘1) ,fl MI T” 7.9 Suppose that u(x, t) is a smooth solution of the one-dimensional wave equation, u” — czuxx = 0. Prove that (142+ c2143,» — (Zczutuxh = 0. If u(0, t) = u(1,‘t) = o for all t, deduce that 1 / lut(x, t)!2 + czlux(x, t)l2 dx = constant. 0 Expanding the quantity (u? + c2u§)t + (2c2utux)x gives (u? + c2143,» 2 21mm + 2c2uxuxt, and —(202utux)x = —2c2(utxux + utuxx). Combining these, we have (u? + c2u§)t — (Zczutuxfi = 2mm: + 2c2uxuxt ~— 2c2utxux — Zczutuxx = Zututt — 2c2utuxx Notice that here we used the fact that u is smooth, so we can change the order of mixed derivatives. Now, by the fact that u solves the PDE, we have 2 Zututt — 2C utuxx = Zut(utt — uxx) = 0. Now, consider the time derivative d 1 1d aft) lut(x,t)|2+c2|ux(x,t)|2dx«=/0 fi< §+c2u§)dx. Again, we have used the fact that u is smooth to interchange integration and differentia— tion. By the previous part of the problem, we have 1 d 1 /0 E( %+czu§)dx=/O (2c2utux)xdx. Which is easily integrated. However, because of the Dirichlet boundary conditions, ut (0, t) = u(1, t) = 0 for all t. Thus, the integral is zero. Summing up, we have d 1 32/0 lut(x,t)|2 + C2|ux(x,t)[2dx = 0, and the result is proved. I Prolo‘em 3 Ex. 7‘ l0 Show ”WM \MX/C): f<x+ct> + 3(x—ct) v‘s 0x golwéim af d2 one ‘oéimem‘om( wave Gail/Nib” / Wt —— em = 6’, for arbifmrj functions f am/ y w ( J’A/mEe/t '5 So/uf'lbn) Proof. We have Mt=Cf'——Cg’, WP Cl{"+c2g", w = w 92 l/(xy: f'I—F g”, hence ) aft : len+ Cl?“ 2: C7-( {'(‘F 9“) t C2 Uxx , 5 t PY OUQM A?! L—J ,7 ‘ 7H [,th £1: {(r,e)‘r<[j EQ The (AH/{E 0U“ [m the (ruse. W PlOd/IG/ where (Y, 63) are WPOZQV WOCCm‘fiZg‘ TA? 5%»de sf” L is we omit ctrcle 7‘. Let car, 9) be a Solwé'ion 9f (oflaaan's inwon M Q , -,—’;Crt¢)r + 7/:uee =0 r< 1, We define {,3 6 HT) 1? f(&) = cam/,6», 0(9) zoom, 62) Show fixov‘t 8 = Hf where H (‘5 {k2 Pei/(belie Hin/i- W5 arm/ H - Car) .9 Um Ht ( 3: f 9‘“) = 3:1‘ (sgnmfi e“ ”twee 1‘ \ MW]? The golwfioh of Hue Laplcwe's HIM/Tim :3 Lang) 2 i an YM’QMG rah-w i :>‘,' Mg" — in an VJ”! ang MI:- fm: Men 6): I mane h:-e., 2; m: :4» am (VMHEW ()0 , \ gtmgww was»: :2 MWWG ~ «770 00 Heme/ 8(8) - :00 (~ OKs—3,190) \ CV1 anal/19 Lad Pro Uem g EX 7 I? Show HWf 14‘6” i5 I‘Wl‘ffcv/ dam f G (“DY—W) for wLAoL {he ivww Proflem fay Laszls aim/oh “H + “xx = O / Mme) -fo<) L“ (X/ 0) t m M gomom WM um) e L <7“) r‘n cm 2mm! Hid“, Mambo ' ”(‘0 ” 1le .J' " 873M H. Lei 71W) -H::mfh Q * €‘7‘8r/27/29a( Ur) U”) i 5 0L Cowffm‘ HWY» (Aft :‘(jxx/ bufi _0(9MV\ V W" '1: V11 Uh , Wk‘d/‘v (MA/Pug) Mn“) 2 {n em: Heme, ._ E f VAX/t) \ 2‘ f” 9V1 em)‘ nt-Oo ~o +—J‘ (109%) 2 ma, 8“ «m #6! x :) :3 WWW: 1 , . mun “Note I @700 {W any t7 0‘ WM) (‘5; END (m cm {NBA/M H14 8/ Mm 570‘ LEA Pmbkem? (54. 7‘12: B=§encx>};:, is an errngoygfl 56W f~ MW). Rode: CO) V 0 UR, ’er sm‘ Ba: {QnCX-(ng is an OMB 1% L1( Maw», (L) Ham, 1% se+ 13°: Hagan)”: n G» 0N8 )5” L4 [0, 0]), (C3 WH—k er Cmvewfiovx tkm- 9mm; are Manda)? to (X (Q’jer vimh Hm flair dig/hm Mn 67 my +Aem equal to 0 35 U a m an OMB 14,“ Liam). d) Ukesz \s OM 0GB fw EUR)- Ravf, (a) OMmrr/wb’y: ff” 8mm) - €n(X‘a) nyif’ any) any) 4y 2 f l r {ft/17:14 0 / (5f {Mil/1 . CMpWew v 36 e L2([a,a+z]), ’ ‘ flx) & , pwe 14M Q; /\ m A fW-Jrox) 3h; fled?) 2»; gem—a). n 3514» {A f ' ‘ “ ”I. I f, e "Gura maffium/ 0f /-—(y)if(j+a). (b) grfiombty: f: [ammo JEQnCCYHxL‘ where ya fa c&(cx)€n('cx)dx fix / ** f 6:40))6xéflol)‘ \__ ~ I f f mm 0 / ff Mk”. Wtemess; V f E €L1([0,I/C]), V ‘;C)( ’0 “5‘ A W fsz<>L“-——-f<%> :rme»zj>~:f are fi is #2 I-Mier oefi’iawr of [3(7) é 'fflyc). L8.) (C) Hider Hm Cmvem‘fon, I‘f’j “IV/via! 44M 8 U B, @35 (UL OMWM {617. We wrtl glm gUB, (‘5 0:»?ch V f e LZMOQJ) ' {00: 2 fiuwikx) + :f find“) Ah vx/LGIE fwm are #2 [runner coeffm‘H-MS 9% %& 7f”)£/ ff )f(l)2 [0 I] 2] , ,2] (d) Lu «m M U; BK is QM WWW fie?- m WQ w (A Sim» s‘f'W CaMPLn/tgk mum/2) / def/m fm/kEZO/s (n) : 7C 7[/EK, HI] / WWW; {m C- L2( fk,[<fl] )_ 1+ follows M {on = 3— if“) QM K€Z “u“ /' gnu) QEXS' efi’m: g j 0 / Wham (k) are 1%: Hana aszuMS f 7Cut) I<C Z 8::(%>~ {6m k), stskn 0 , elst . 1. Exercise 7 .13: Use Fourier series to solve the following initial—boundary value problem for th Schrodin- der equation, that describes a quantum mechanical particle in a box: 7hit : “um“ u(0,t) = u(1,t) for all t, ”(DMD = f(1‘) Derive the following two conservation laws from your Fourier series solution and directly from the PDE: d 1 2 d 1 2 a 0 new dab-=0, a /) an dwzo. Soln: To start we need an orthonormal basis for the space we are in, which happens to be L2([0, 1]) (Schrodinder’s equation always implies we are on L2([a, b]) for some a, b). We look for functions of the form en 2 cei’lk'“. We want these to satisfy, (en, em) 2 6mm. A simple computation shows we want 0 = 1 and kn : 27r ,Vn. So our orthonormal basis is {en}°_°oo s.t. en : 627””. Now write f in terms of the ONE, f = Z::_m fnez’m” and assume u(x,t) is seperable in the form u(m, t) : ZZZwO un(t)e2“"“. Then u; = Z::_oo an(t)e2“‘m and um = Z::_x —(27rn)2un(t)ez"lm. Equating coefficients of in; : —um we get mu 2 (27rn)2un or an : —i(27rin)2un. The solution of this with the intial condition given by u(x, 0) = fix) or un(0) = fn is, un(t) : fne‘m’mlit. So our general solution is u(x, t) = an=—x fne'i(2"")2teg"i"” = 21:01—00 f,,e’“2"")2‘(e"im)2. Its easy to see/check that this satisfies all intial conditions. In order to show both, g; fol |u(:r, t)|2dm = 0,and%f01 luxlzda: = 0, multiply each side of wt 2 —u” by 2n, and massage to get (m2); = ——2(uuz)m + 21%. Integrating and using the periodicity of u we get i f; fiugdc = 2 [J uidm or since we have the unstated fact that u(:r, t) is real valued, we have 2‘ 1 9— u (E, It) 2dr 2 2 1 um 2dr. Being a physics problem we play like optimistic physicists and blindly 0 dt 0 1 1 switch the order of integration and differentiation and get, if; f0 |u(x,t)|2dx = 2 f0 |u$l2dam The right side of this equation is purely imaginary and the left is purely real, hence each side is zero. And since 1 2 . . d 1 2 f” [url d1 = 0 clearly 3? f0 [um] d3: = 0. ...
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