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# Partial Differential Equations MTH 3326

Baylor University

GPA 3.85

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This 45 page Class Notes was uploaded by Dejon Bednar on Saturday October 3, 2015. The Class Notes belongs to MTH 3326 at Baylor University taught by Matthew Beauregard in Fall. Since its upload, it has received 34 views. For similar materials see /class/217912/mth-3326-baylor-university in Mathematics (M) at Baylor University.

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Date Created: 10/03/15

Problem 3 Parta llr DuXX au buJ XLAi tTz uUu U U U quotFU quot47 47 um DaU bU3 01 4 bk efga ur 113 11 If Problem 1 Part a Best Approximation Theorem Suppose fe L2a b has orthogonal expansion fX chXX a S XS 0 and let N 111 be fixed Among all possible choices of the coef cients cl 02 CN the choice that minimizes the L2 error between fX and the Nth partial sum SNX Z ch X ie minimizes HQ 2 HX H N f SN where this is the L2 norm is given by c n 1 K N ITI39I Part b Parseval s Theorem The following are equivalent a The Fourier series of f converges in the meansquare sense of a b b For every f e L2ab Parseval39s Equality holds 2cm W c fe L2ab Part c i 1101717HXHM1 07rde ltxgt 13W ii 0 r 0de 7T 2 1 cosXcosXgt Io coszXdX ltcos2XCOS2Xgt I0 cos22xdx 7 2 ltwslt3xgtgt i W IO COSZ3XdX 9 c3 Xm gjcoscx 0cos2 X cos3 X 0 lt Xlt 7r norm 1kg X 1 I deX1 Lig COSX COS3X dX 656 72393 7 7 when plugged into Mathematica 817239 16 MTH 3326 PDEs Exercises Spring 2011 Instructor Dr Beauregard Print Name The following questions are additional exercises on dimensional analysis of par tial differential equations Although these questions will not be collected it is stressed that it is important to complete them in due time 1 Consider the partial di erential equations speci ed over the real line Rewrite each problem in dimensionless form with all coef cients equal to one Discuss what the scalings on time and space suggests Utt Cum U Du Utaw 0 utuum 0 2 Consider the following nonlinear parabolic partial differential equation ut auerbuercuz abc gt 0 a ls it possible to write this equation in dimensionless form without any non dimensional parameters appearing Why or why not b Choose a scaling on uxt s and t such that no coef cients appear on the um and um What does the non dimensional parameter 6 represent Consider 6 ltlt 1 is it reasonable to neglect that term What does the e 0 solution represent c Choose a scaling on uxt s and t such that no coef cients appear on the um and uz What does the non dimensional parameter 6 represent Consider 6 ltlt 1 is it reasonable to neglect that term What does the E 0 solution represent d Choose a scaling on u t s and t such that no coef cients appear on the um and uzz What does the non dimensional parameter 6 represent Consider 6 ltlt 1 is it reasonable to neglect that term What does the E 0 solution represent What is the order in space of this partial di erential equation How is that different from the previous two cases e Show that this partial differential equation does not have a scaling symmetry 3 Consider the one dimensional Burger7s equation U uum 0 a Show that Burger7s equation has a scaling symmetry b Through an introduction a similarity variable 77 transform Burger7s equation into a di erential equation for 77 4 Transform the Korteweg de Vries KdV equation into a di erential equation through a suitable similarity transformation ut6uumui 0 a Show that Burger7s equation has a scaling symmetry b Through an introduction a variable transform Burger7s equation lHtO 777 a di erential equation fOI Selected Answers Chapter 1 11 1 uzt e lT W 2 a un 12z t2 and um 12z t2 so u um holds The initial condition satis ed is uz7 0 I 04 14 1 1 l urrT yue0uee080umur uee T T 20l 12 1 In the conservation law an extra rate out77 term appears of the form 7 f cuzt dz Where c gt 0 is a constant resulting in the PDE u i 170 Selected Answers kum 7 our 2 80301 uux 111 SO so u2um f 30 b 9 7111 so so 6 f E 0 c 3075 1322 so go arctanu39 f ltl 3 a u c211 7 g b um c211 7 dut 5 Compute the units of c Tpl 13 1 u 0t 0 2i utkum0ltzltitgt0 mu 7 mum t 7 Mm K gt 0 um f t 3 a l b V c H d Vl e Vl f H g HI h IV 4 a The boundary condition ugc 075 gt 0 is an outward ux condition Heat is radiating from the left end of the bar to the surrounding rnediurnl However 1 t gt 0 is an inward ux condition so heat is being absorbed by the rod from the surrounding rnediurnl b absorbing c radiating 5 The displacement and slopes of the left endpoint and the right endpoint of the string must always be equall 6 No since the physical form is ugc 7u 7 7100 resulting in K lt 0 14 1 a linear b nonlinear c linear d nonlinear 2 a Eu 0 where 211 u u 7 u239 nonlinear b Eu 0 where Eu 2 u 7 ku linear hornogeneous c Eu sinz where 211 azu bzu czu linear nonhomo geneous d Eu z where Eu 2 uuquot nonlinear 171 01 on 5 0 50 H 5 b no a y Ce b y 26 16 a y 7 0712 b y 7 M27 g g a y 12 sinz C12 b y I2 sinz 712 a y 616 5 626 b y 5 26 5 5 C y Jeee 5 5866 a y 0164 Cgte g b y e 2t C y e732 7 7673 a y 62 cl cos it 52 sin t b y 62 cost 7 sin t C y gem cos t a y Ci Cos 4t 62 sin 4t 3 y cos 4t i sin 4t C y 7 cot4 cos 4t 7 sin 4t a y 011 3 02x2 b y 17451 COS1HI 62 sin1n a y 0165 62e 5 85 85 7 b y We 7 We 5 C y Cg cosh5t C4 sinh5t d y 5colsh5 Sinh5t e d1 172 Selected Answers Chapter 2 21 1 a lull W6 M W b ltuvgt 0 so not orthogonal c linearly independent d yes 2 a llfll NZ llyll 7r55 b sinmz for m i2i3 H works as does cosnz for n ili3i5 H C 00 3 a 1M b no 4 a Z J b x2323 5 a ut E 1 works b pt at2 5 works for any a c 6 R c for any pt from b 7 a A1 2 x1 10 A2 1 X2 731 b A is any real number u e 8 a Eu Au u0 0 ul 0 where 211 iu b A 0 results in E 0 so A 0 is not an eigenvalue c A lt 0 results in E 0 so there are no negative eigenvalues d A p2 gt 0 results in eigenvalues An mr2 n 12Hl with eigenfunctions up to a constant multiple sinAn z n 1 2 i H 9 a Eu Au uO 0 MG 0 where 211 iu b yes39 A0 0 uo 1 up to a constant multiple c none 1 An TmO2 cosAn z n 12 l H 10 a Eu Au u0 0 ul 0 where 211 u u b none c An W e E2 sinn7rz n 12Hl 1 none 173 11 H N 93 F 5 3 Eu Au u1 0 u5 0 Where 211 71211 7 11 b no 3 none 1 An mr 1n52 sinTn 1111 n 12H None at all a X AX 0 X0 0 XZ 0 2 b A WED Xnz mom I n 12 c Tnt Cexp7nkt n 12 H 00 d uz t Z on sin Ez exp7nkt T kT0 a X X 0 XO 0 XZ 039 b An lt2n2717rgt2 Xnz cos Ez n 12H c Tnt an cos Ect In sin Ect n 1 2 1 H d uz t 2 an cos Ect In sin Ect cos Ez 711 T Ac2T 0 a X X 0 X7Z XZ X 7Z X Z39 b A0 0 X0z E B An 7 Xnz ancosTnz 127 sin Ez n 12H c TOO C Tnt Cexp7nkt n 12 M d uz t co 2 an cos Ez In sin Ez exp7nkt T kT0 a X X0X00X1039 T T T0 b An mr2 sinn z n 12H c Tnt 642 an cos 4n7 1 t bnsin 4n71 0 n 12 1 uzt Zsinxn wait2 an cos 4A 71tgt 1 sin 4A 7175 1 174 Selected Answers 23 I 1 a 0 sinn7rzZsinm7rzZdz 7 ECW777123 ltT7gtJ dz 1 0 cos 7cos dz cos dz7 zcos dz 0 I 0 I 2nfm7r Sinltn7n7rzgt Sinltnn7rzgt sinn 7 m7r 7 sin0 7 Z 0 7 2nm7r Z Z 2n 7 m7r 2n myr 51117 m 51110 0 since mm are integers 5 a a0lzdz127 7171n n n27r2 7 1 an zcosmrzdz 1727 0 1 717 bn zsinnwzdzgn12pu 0 mr F b 71I ins n5 1n 71I ins n5 1n 71I ins n5 1n 2 I 7 a anZO fzsinEzdzn12pu 1 bn gzsinEzdzn12pu n5 0 3 OmoEm mm Emmma mg M N m Amv nSHNo xAHVmWSA 8Vamp83HFmiZ Genie 9w mtg oggiiggggh a We 5 595 Any Umomwm 0 Name M N o Amy DSHN0 lioo ik H mii N 5 H mAHVOOmAiviavkav EHFPZ Sn 0 3m 176 Selected Answers b c Oscillates inde nitely 1 10 a an 2 sinxn 1 dz n 1727 0 1 2 gzsinnz dz n 1727 H 0 2 bn7 JWA b c Dalnped oscillations to zero 11 The average value in the calculus sense of on the interval 0 lt z lt Zr 12 Do this and keep it handy You Will need it often in the future 24 1 Suppose X a cha 0 and Xb C2Xb 0 since every ho mogeneous Robin boundary condition can be Written in this form Then 177 Xd 751Xd and Xb 752Xbi Substitute these into the right hand side of 225 and simplify to get zero 4 b cl 0 02 7512 5 a yes b yes 6 a By direct computation show 7 P0zP1z dz 07 P0zP2z dz 07 P1zP2z dz 0 q i Use Theorem 23 00 i Rewrite the norms in terms of inner products and use the fact that this is an orthogonal family 9 a By direct computation show HoltzgtH1ltzgtwltzgtdz 0 HoltzgtH2ltzgtwltzgtdz 0 f H1ltzgtH2ltzgtwltzgt dz o where 6 12 o By direct computation show 0 L0zL1zwz dz 0 0 L0zL2zwz dz 0 0 L1zL2zwz dz 0 where 6 75 11 a co 0 on gem n 25 1 a Left convection with an outside medium of 0 but disobeys Newton s Law of Cooling iiei nonphysical Robin BC Right xed tempera ture o i b X0z B1 7 z 178 Selected Answers P c Results from tanhp p having no nonzero solution d tanp p e Graph y tanp and y p on the same coordinate plane f An p3 where 10 is the nth positive solution of the eigenvalue equation in A1 201907 A2 596795 A3 1189 A4 197858 A5 296554 With corresponding eigenfunctions Xnz sinpnz 7 pn cospnz n H a Left convection With an outside medium of 0 Which obeys Newton s Law of Cooling ie physical Robin BC Right convection With an outside medium of0 Which obeys Newton s Law of Cooling ie phys ical Robin BC b X AX 0 X 07X0 0 X 1X1 039 T AT0 c Show A 0 produces a trivial solution To rule out A lt 0 either do the standard argument or apply Theorem 23 The positive eigenval ues come from tanp 1572 Let An p3 n 1 2 Where pn is the nth positive solution of that equation The associated eigenfunc tions are cospnz 17 sinpnz n 12 d Tnt Cexp7Ant n 12 e uzt Zan cospnz pinsin nn exp7Ant 7 01 cospnz pin sinpnz dz 0 a ltXnltzgt7Xnltzgtgt 1 1 2d 0 cospnzp7s1npnz 1 12 a Left insulated Right convection With an outside medium of 0 Which obeys Newton s Law of Cooling ie physical Robin BC b X AX 0 X 0 0 X 1X1 039 T AT 0 c Show A 0 produces a trivial solution To rule out A lt 0 either do the standard argument or apply Theorem 23 The positive eigenval ues come from tanp i Let An p3 n 12 Where pn is the nth positive solution of that equation The associated eigenfunctions are Xnz cospnz n d Tnt Cexp7Ant n 12 H e uzt Zan cospnz exp7Ant ltfltzgtXnltzgtgt folfrcosPnIdI n ltXnx7 fol cos2pnz dz f an 12 179 26 a Left convection With an outside medium of 0 Which disobeys New ton s Law of Cooling iiei nonphysical Robin BC Right convection With an outside medium of 0 Which obeys Newton s Law of Cooling iiei physical Robin BC b X X 0 X 0 7X0 X 1 7X1 T T 0 c Show A 0 produces a trivial solutioni d A1 71 X1z coshz 7 sinhz e The positive eigenvalues come from sinp 0 Then An pi n7r2 n 1727 and Xnz cosTnz7 ksin Ez n 1727Hi f T1t Cexpt Tnt Dexp7nt n 1727Hi g uz7 t a1 coshz 7 sinh 1 exp t 00 1 gun cos z m sm Ez exp Ant ltfx7X1 fol coshz 7 sinh 1 dz ltX71I7X71Igt f01coshz7sinhz2dz 7 ltfzXnzgt ful l mama 7 jgsinw z dz h a1 an fol cos Ainz 7 7 sinTnz 2 dz a Note that a0 is the proportionality constant from the convection boundary condition at z 0 Similarly ag is the proportionality con stant from the convection boundary condition at z Zr So there is a zero eigenvalue if and only if E that is When the combina tion of the proportionality constants on the lefthand side matches the interval lengt i b X0z Baoz 1 or you could say X0z aoz 1 up to an arbitrary multiplicative constant a uzkum 0ltzlt tgt0 u0t 100 tgt 0 Mm 0 tgt 0 1410 fz 0 lt z lt A b v z0 0ltzlt 7 v0 100 vZ 0 Which has solution vz E 100 12m 180 Selected Answers 3 wtkwm 0ltzlt 7 tgt07 w0t 0 tgt 0 1075Lt07 tgt07 wz0 fz7vz7 0ltIltZi wzt Zen simmz exp7nkt Where An W 721 C fjltfltzgt 7 vltzgtgt sinltmzgt dz n 1 2 n fjsin2Tnzdz 7 7 d uzt vz wz t for vz and wzt above 6 yes 2 a utkum 0ltIltLtgt07 117507 t u0t 7 5007 t gt 07 1 t 07 tgt 07 11170fz7 0ltIltZi b v z07 0ltzlt 7 vO Kv0 7 5007 vZ 0 Which has solution vz E 500 c wtkwm 0ltzlt 7 tgt07 14ch07i110t707 tgt07 1075M t 07 t gt 07 wz0fz7vz 0ltIltZi oo 1 wzt 7 gun cospnz 07 s1npnz exp7nkt Where An 7 pi pn is the nth positive solution of tanp 1177 and KW 7 vltzgtgt comm 177 sinltpnzgt dz an n fog cospnz 17 sinpnz2 dz 1 uzt vz wz t for vz and wzt above 6 yes 3 a utkum 0ltIltLtgt07 u0t 100 tgt 0 7u Lt u t 7 5007 t gt 07 uz0 0 lt z lt i b v z07 0ltzlt 7 v0 100 4a Kv 7 500 181 400K Which has solution vz Kprlz 100 C wtkw117 0ltzlt tgt0 w0t 0 tgt0 iwgcanf wZt70 tgt0 wz0 fz7vz 0ltIltZi oo wz t Z In sinn z exp7nkt Where An p3 pn is the nth 711 fjltfltzgt 7 M smltmzgtdz fog sin2Tnz dz positive solution oftanp 7p and In n 1 2 i i uz t vz wzt for vz and wzt above yes Mr C2uxm7 u0t 1 um 07 11170 161 714170 91 v z 0 110 1 vZ 0 Which has solution vz 0ltzlt tgt0 tgt0 tgt0 0ltzlt 0ltzltZi 0ltzlt 1 0ltzlt tgt0 tgt0 tgt0 wn 027002027 7 w0t 0 mm 07 wz0fz7vz 0ltIltL wzz0gz 0ltzlt i d uzt vz wzt for 01 and wzt above H f9z so fg is an odd a f91 09990 f0 171 function b In ff favequot sinmrzZ dz 0 since the integral of an odd odd function functlon over a syrnrnetrlc 1nterval about the orlgin 1s zero 00 Fourier cosine series39 Fourier sine series39 full Fourier series 9 a It is actually a Fourier cosine seriesi y 182 Selected Answers 32 H E0 93 H N 93 b It is actually a Fourier sine seriesl C 1 E 0 a 02932020715413 136744 b 492485 c 158925 a 105813067 105173717 010784891 13 1 c 01191686 a Show that 11 7 IN 7 11 7gt 0 as N 7gt 00 To do this use the fact that 0 lt z lt 1 in this probleml b Show that max03 31117 IN 711 74gt 0 as N 7gt 00 Do this by computing maxoggcgl 117IN711 and then taking the limit as N 7gt 00 Againy keep in mind that 0 g I g 1 here c Show that 01117IN7 012 dz 7 0 as N 7gt 00 Do this by computing the integral and then taking the limit as N 7gt 00 This exercise is a refresher on p integrals from calculusl calculus book or nd one in the libraryl Dig up your a 040 Elan cosn7rzZ In sinn7rzZ an fig cosmrzZ dz n 071271 In fffzsinn7rzZdz n 1727111 Where Lisa and Z i c need not be de ned f0 3 f0 so the Fourier series converges pointWise to w 3 at z 0 d Since f does not have a continuous periodic extension the Uniform Convergence Theorem does not app y e Since fjl dz lt 0 then f 6 L2711 and hence its Fourier series converges in the L2 sense on 717 11 a Recall that the full Fourier series corresponds to the periodic shift extension 0 i b Recall that the Fourier cosine series corresponds to the periodic even extension of c Recall that the Fourier sine series corresponds to the periodic odd extension of 4 a 040 Elan cosmrzZ In sinn7rzZ an ff cosn7rzZ dz 01 N 93 183 n 012 H In fffz sinmrzZ dz n 12Hi where is the given piecewise function and Z 1 c Yesi Since f is piecewise smooth on 71 g z 1 by Theorem 31 the Fourier series converges pointwise d No Since f does not have a continuous periodic extension by The orem 32b the Fourier series does not converge uniform y e Since fil dz lt X then f 6 L27l1 and hence its Fourier series converges in the L2 sense on 711 by Theorem 33 PCT Pointwise Convergence Theorem UCT Uniform Convergence Theorem LQCT L2 Convergence Theorem DT Differentiation The orem a PCT applies so yes39 UCT applies so yes39 L2CT applies so yes39 DT applies so yes for all points in 71 lt z lt 1 b PCT applies so yes39 by UCTb no L2CT applies so yes39 DT doesn t apply c PCT doesn t apply since f is not piecewise smooth at z 039 UCT doesn t apply39 LQCT applies so yes39 DT doesn t apply 1 PCT doesn t since f is not piecewise smooth at z 039 UCT doesn t apply39 LQCT applies so yes39 DT doesn t apply e PCT applies so yes39 UCT applies so yes39 L2CT applies so yes39 DT applies so yes for all z f 0 f PCT doesn t apply since f is not piecewise smooth at z 039 by UCTb no by L2CT no since since 1 z 4 dz diverges39 DT doesn t apply g PCT applies so yes39 UCT applies so yes39 L2CT applies so yes39 DT applies so yes for all points in 727T lt z lt 27f except z 7w07r since f does not exist at these points a yes b no c no 0 yes a do 7r d17 d2 0 0437947 W3 656 b ME 7 E 00770668 184 Selected Answers 4 a on W n 012 where Pnz is the nth Legendre poly norn1all 2 1 2 b 2513435 7 161 dz M1102 1 n0 8 no 9 a energy 0 power 12 b energy sinh 239 power 0 c Any signal with nite energy will have zero powerl Any signal with nonzero power will have in nite energyl Any in niteduration in creasing function eigl 6 will not be either 10 a 20 terms Hint You can do this by trial and error or you can create a table of values of integralsl b 2 terms Same hint here 12 no Chapter 4 41 1 ln R3 write V I ltgtl Then for f I fzyz and F I F1F2F3gt we have a a a gag Wiltaiy gtfiltafayazy 7 8 8 8 iaFl 8F2 BFS V FiltE7E7Egt ltF17F27F3gtEEEl 2 a Vf lt21y3ycoszy312y2zcoszygt Af 2y37y2 sinzy 612y 7 12 sinzy 3 a yes 3 yes c yes x I f 39 f a l x 4i a l 185 7 00 50 H H H 93 H 01 H on H H b 2 c A positive divergence indicates expansion at every point b 0 c A zero divergence indicates neither expansion nor compression at any point a Show by direct calculation that V gtltF 0 a Direct calculation b no i Calculate each part directlyi b meaningful equals the zero vector always d meaningful equals the zero scalar always e not meaningful i Apply Stokes7 Theoremi C fIyy12yzy2y2C Afryy yzC Note that V 9F V lt9F179F279F3gt 9F1 395392 9F3 i Use the Divergence Theorem in Rgi i Use Exercise 8 i If the minus were not present the time derivative of the total heat energy would be positive when heat is owing outward from the boundary 186 H N 93 P H Selected Answers a Left 7V1 n Bottom 7V1 n b Left ugc Ku7 TK Right iu 7R Top u T39 Bottom uy R Ku 7 TK Right 7Vu n 7R Top u T39 7R a If N gt 0 737 lt 0 Which is absorbing If N lt 0 then 7 gt 0 Which is radiating c Rewrite as 727 Ku 7 lf K gt 0 this is a physical Robin condition lfK lt 0 it is a nonphysical Robin condition a left xed temperature of 0 right insulated bottom xed temper ature of 0 top insulated b left radiating at N units39 right physical Robin condition ie con vection obeying Newton s Law of Cooling With an outside medium of R1 degrees39 bottom nonphysical Robin condition ie convection disobeying Newton s Law of Cooling With an outside medium of 7R2 degrees39 top xed temperature of T degrees top and right xed displacement of zero bottom and left edge maintains a zero slope but the displacement may vary a BC 239 1C 1 b BC 439 1C 0 c BC 439 lC 2 d BC 439 1C 2 e BC 1239 1C 0 f BC 639 lC 0 g BC 639 1C 0 h BC 639 1C 1 i BC 239 1C 2 Viewing Green s First Identity in the rearranged form VvVudA vd37 vAudA D 3D an D should help Use Green s First ldentity With v E 1 No For any solution u to this problem u K is also a solution for any constant K 187 45 1 a uz7 y Z sin Ey an cosh Ez In Where An 2 1 211 iysanltmygtdyian coshwm m1n 7 2 f0 951n n 9 d9 bn 7 0W n 7 b Fixed temperature of 0 on the top and bottom edges speci ed tem perature of y along the left edge speci ed temperature of 7y along the right edge u umusm 1 2 a uz y Z sin Ez an cosh Ey In sinh Ey Where An 2 1 2 71 7 7 2f 10051nm1d1 7 773 7r an anAn bn sinh2xO n7xcosh2x n 7 17 27 b Left edge xed temperature of 0 Right edge insulated Top xed temperature of 100 Bottom convection With an outside medium of 0 but disobeys Newton s Law of Cooling u DA as m 1 3 a uz7 y aoz 120 Zcosxn y an coshn z bn sinhn 711 Where An n2 0 0 an 0 n 172 a0 fgcosy2dy bn 7rsinh2m7r 07F cosy2 cos n y dy n 172 b Top and bottom edges insulated Left edge temperature xed at 0 Right edge temperature is speci ed by cosy2 188 Selected Answers 4 The associated 1 problem has the solution vz7 y 2 an cosh Aily In sinh E sinEz7 Where An TNTa2 n 12H and an 3011 pz sin Ez dz7 a b f qz sinn z dz 7 an coshn b n 1 2 sinhnb On the other hand wz7 y Z en cosh Mn z dn sil1hTnz sinQ y7 711 Where Mn Tm102 n 12 t t t and b on fy sinmm dy dn f0bgy sinTny dy 7 an coshTn a n Sinhlt Ma 12m Fina11y 110079 v17yw17y 46 1 a uz7 y t Z Z sin Wz sin Zy amn COSC Amn t bmn sincx Amn t where W mix un mrW Am W um and f5 In m y sinwm sinwm dz dy amn fob f0 sin2t mm z sin2 dz dy b a bum 1 f0 f0 gzy s1n Mmzs1n ydzdy myn1727 C mn f 61 sin2 mm z sin2 y dz dy 189 2 a uz y t Z Z cmn cos Mm z cosMy exp7mnkt Where Mm nlm 1 LTgm V 7anmi Am Mm and f f fz y cosWz cosMy dz dy fob foa COS2 Mm 1 COS217nydx dy 7 Cn Chapter 5 51 1 a u39rt9 ZanrmsinEt9 Where An Tm902 711 2 7 Ancm l o an 90 f 9sin n 0 d6 n 12 H 0 2 a uT t9 a 0i T7 an cosm9 In sinm9 Where a0 i if d9 711 an if cosm9 d9 12 if sinm9 d9 n 1 2 i H 52 2 a 4 b 72 C 1 3 Show 2 satis es A2 0 in D and 2 0 on BDi By the Maximum Principle 2 E 0 Why Hence 1 E 1w 53 1 a uT 9 t Z Z J2nnm 7 sinQEO anm cosnm ct bnm sinnm cw mil n1 7 f0 OW2 m whomquot T sinTnt9r 4 b 7 0 f0quot 0 3an T gnawing 494T Anm znmp2 Where znm is the mth zero of J2nz and Mn 2702 nm 12Hi Where anm 2 a 1mm ZaomJomEr Z Z Jnnm 7 anm cosm9 bum sinm9 exp7nmkt 190 Where Selected Answers fo fir f 0J0m7 7 dOdT fop fir Ty 0Jnm7 cosn0r dOdT b 7 f0quot f3 fn 0Jn ATmr sinm9rdt9dr m I I Jam mane m 1 nm gumm2 aOm anm Where znm is the mth zero of JAE mm 1727 H Index L2 norm 86 Au 94 curlF 95 V x F 95 grad f 93 area integral 95 basis 27 Bessel function rst kind 133 second kind 133 Bessel s Inequality 88 Bessel s ODE 131 boundary condition 8 absorbing 13 Dirichlet 8 106 homogenizing 61 116 mathematical 124 Neumann 8 106 nonhomogeneous 60 periodic 8 physical 124 radiating 13 Robin 8 55 106 symmetric 49 boundary conditions 19 boundary value problem 19 Burgers7 equation 5 CauchyEuler equation 20 characteristic equation 17 compatibility condition 114 constitutive equation 5 convergence L2 76 82 meansquare 82 pointWise 76 81 uniform 76 81 curl 95 del 93 diffusion equation see heat equation 5 dimension 27 directional derivative 110 divergence 93 Divergence Theorem 97 dot product 27 eigenfunction 29 complex 50 expansion 75 eigenvalue 29 complex 50 eigenvalue problem 29 eigenvector 29 existence 109 extension even 68 odd 68 periodic 70 periodic even 70 periodic odd 70 periodic shift 70 shift 69 Fick s Law 5 101 Fisher s equation 6 ux 4 100 107 Fourier series 192 INDEX L2 convergence 82 best approximation 86 classical 80 complex 54 cosine 42 differentiation of 83 full 44 generalized 75 integration of 83 meansquare convergence 82 pointWise convergence 81 sine 40 uniform convergence 81 Fourier s Law 5 101 Fundamental Conservation Law 5 general solution 18 gradient 93 Green s Identity First 110 Second 110 harmonic function 104 heat energy 101 heat equation 1D 5 2D 101 Hermite polynomials 53 homogeneous 15 hyperbolic trig functions 21 initial condition 18 initial value problem 18 inner product 27 integrating factor 17 Laguerre polynomials 53 Laplace s equation 104 Laplacian operator 94 Legendre polynomials 52 line integral 96 linear differential equation 15 operator 14 linearity 14 linearly dependent 27 linearly independent 27 Maximum Principle 128 Mean Value Property 128 nonhomogeneous 15 nonlinear differential equation 15 operator 14 norm 27 31 normal derivative 107 108 110 operator 14 linear 14 nonlinear 14 orthogonal 27 family 47 orthogonality 40 eigenfunctions 49 weight function 53 Parseval s Equality 88 Parseval s Theorem 88 particular solution 18 Poisson s Integral Formula 127 potential equation 104 potential function 93 separation of variables 33 spans 2 stability 110 steady state solution 60 Stokes7 Theorem 97 Superposition Principle 15 surface integral 96 transient solution 61 transport equation 5 transportdiffusion equation 5 uniqueness 109 vector eld 93 conservative 93 divergence free 94 gradient 93 incompressible 94 INDEX 193 vector space 27 volume integral 95 wave equation 2Dy 103 wave speed 7 103 wellposed 110 Problem 6 a uV 11 represents the convective acceleration of a given ow This affects the acceleration of the uid with respect to space and is independent of time The expression is nonlinear b Au represents the Laplacian of u the divergence of the gradient of the vector 11 This affects the net ux through the boundary c The incompressibility condition V u 0 represents no divergence or outward ux In addition the density of the given uid is constant First we start with the mass of 32 oz of water mass HIpdV V EdV JJJ V JdV J mass LIX V dt dp i VJ0 J V dt p VpV0 amp iy z Dt t X t y t z t 03 5 i7V VV V0 Dr 3 p p The compressibility condition asserts that the density of a uid is constant In relation to mass it is shown that the change in mass of a uid is zero when dependent on time Therefore mass is not time dependent MTH 3326 PDEs Wrap Up Assignment Spring 2011 Instructor Dr Beauregard Print Name The following question are to be completed and turned in prior to the nal exam scheduled on May 6th at 430pm Due to the number depth and com plexity of each questions I urge each one of you to give adequate enough time to complete the assignment Mathematica can be used to calculate any of the in tegrals or derivatives All solutions should be typewritten using Mathematica Microsoft Word or LaTeX If you elect to handwrite your solutions it will need to be submitted on white lined paper in a single column onesided format 1 Consider the space of real functions de ned by 7 Wm f szdvc lt co which is equipped with the inner product 17 ltuv gt Thus the L2 norm is de ned as lt uu gt Let Xn3 be a set of orthogonal functions in L2 a b Given f E L2ab Let N be a positive integer We de ne N lt f X gt fN gchnQ where on ltXn 7 ngt n 012 a State the Best Appmmz39matz39on Theorem b State Parseval s Theorem c Suppose we want to make the approximation 1 z ice 01 cosx 02 cos2x 03 cos3 0 lt z lt 7T in such a way that the L2 norm of the error on the interval 07T is minimized i What is the norm of x ii Determine the values of CO 01 02 03 What is the norm ofthe approximation iii Plot z and the approximation on the set of axes 2 9 Consider the following linear parabolic partial differential equation with hornogenous Dirichlet boundary conditions utumau agt0 E01 tgt0 1 u 0 z17 2 a Show that this can be written in the form u where is a linear operator Determine the operator and show that it is linear b Use separation of variables to determine the exact solution Clearly state the eigenfunctions eigenvalues and the fourier coef cients a c Let t 0 well show that if you only consider the rst M functions in the sum that the error can be small Find an M such that 00 Z lama lt 01 nM1 Hints Consider the range of nx and to use the integral test d Consider the lirnHoout For what values of a is the limit zero For what values of a is the limit in nite For what values of a is the limit of nite function of x e Let a 7T2 1 Use Mathematica to generate a 2 D plot of ut at t 0 1 and 2 using M number of terms in the series as found in part f Describe how the additional term on effects the solution Consider the following nonlinear parabolic partial differential equation over the in nite real line utDumauibu3 abD gt 0 z ER tgt 0 MW 0 f96 a lntroduce new dirnensionless variables ls it possible to restate the problem with all coe icients equal to one Why or why not b Determine the steady state solutions c This is an emploratory question but a very important question that arises in many eonteaits What does it mean if there are multiple steady state solutions How could we determine in some sense where the solution may tend to 4 U 03 RECON Consider the eigenvalue problem f x A2fx 0 z E 06 Determine the eigenvalues and eigenfunctions for each set of boundary conditions That7s right you can simply copy your tablel a Dirichlet Dirichlet boundary conditions Robin Robin boundary conditions The above conditions are call symmetric boundary conditions State the de nition for symmetric boundary conditions and show that for each case this condition is satis ed Consider the 1 D wave equation with a source term over the unit interval utt um17 z E 01 tgt0 u0at 07 u17t 17 u0 x3 utz0 2x a Find the general solution Hint Let utvrvt we where Uet and are chosen in a special way b Use Mathematica to create a plot of the solution at t 0 1 and 2 c What effect does the source term have on the solution Consider the Navier Stokes equations for an incompressible ow 6 plt uVugt 7VPMAu Vu 0 uyzt ltuzy2twyztwy2tgt a Describe What the term u V u represents In other words what effect will this have on the solution b Describe what the term Au represents In other words what e ect will this have on the solution c What does the incompressibility condition V u 0 represent d Consider a 32 ounce Nalgene body lled with water which is sealed Show that the total ux out of the Nalgene must be zero by the incompressibility condition How is the incompressibility condition related to the mass Complete all parts of section 46 1 on page 122 Complete all parts of section 46 2 on page 122 Complete all parts of section 53 2 on page 137 Problem 5 Part a uquot uxx1 X 4Q00 u1t1 mxm uX02X La mxowxomo ygl X 81 81 EMMH WXQVA t WX1 X Vquot VXX WXX1 X WXX1 XX 1 mxowxomw maowaowm Qo 0 Qo mLowLnMD VLt01 gt VLt1 mxmwxmmm xmaxf a xm MgtU uX0 VX00 Is02X gt Wamp02X V VXX W X 1 050 WxX 12 v0 t 1 W X 13 xm w x4f Mx g 2X Now we transform VX t into the Transient and SteadyState solutions let m t UX tMX Transient Steady State Uquot Um quotX0 00 t 0 M0 00 t 0 u1 1 ULt fXMX 1 1 vLtgX W 1 6 Separation of Variables AX 2 X let Uampt Xpt Xpquotr quotXpr wt we 2 4 P0 X 0 gp 0 BC S using the same A W 1 0 A 11702 in 117702 X a cosZI b sine171 mx sing2X Using the superposition principle we generate r 39mn at 1mm U X Esm rX a cosW s1ngt mo fIXHXZansinwZX 111 J X tXSin1ZXdX an 1112 J sin2 1 X dX Ux t isin1ZXb Zcos1Zt anJZsin1Zt mm gm ibn sin1ZX L J0 gXsinJZXdX J7 j sin2Vl XdX 0 b 1112 Now we collect all parts together mo mm M where vgt 01 t uX uamp0 U tMX WX ug t X X 13 isinQZX an cosGZt b sin 7 t Problem 8 Part a urkA11 0ltXlta 0ltyltb tgt0 11X0ytuayt0 0ltyltb tgt0 uyx0tuXbt0 0ltXlt a tgt0 uxy0 fXy 0ltXlt a 0ltyltb Separation of Variables 16f UXy t XP 7t XPY T0 k quotXy 7U XPquotY7t Ta ra K 1 gt Au u MU X PU WNW 0 quot 0 TMT0 305 BC r M 0 0 a 0 ltp 0ltpb0 fM 212 my Um 2m 172392 1 n Um 2a 2b X oweEX 11 12 pm cosrEy 112 12 m Using superposition we generate the general solution my r cmcosgExyosg yym u2zy0 aw gcmcoswaxyoswy J mocoswzxyoswmyxo 0 cm 11m12 Z Z 0 0 cos MXOS Mydey Part b 1140 y l 0 The ux at the left wall side is 0 neither absorbing nor radiating heat at time t lbw5 t 0 The temperature at the right wall side is 0 at time t IMXQ00 The ux at the bottom wall side is 0 neither absorbing nor radiating heat at time t mxao0 The temperature at the top wall side is 0 at time t Problem 2 Parta Lu uXX ur all L i a Proof for determining that L is linear Lfg L1 Hg Lltfgfg gfgafg aquot 0quot 5 g0gg Et Eg af ag 0quot 82 g E afgg Egag 52 52 Lf Lggf Efafj g Etgag LCV cLV Lcv CV tcv acv wagm c mmm cLv gm gm am cvc tvcav Partb Separation of Variables 11 109 1 XPt XP39t XquotPt a XP t m w a 1 P0 X SteadyState Solution Temporal Solution X1a X0 p39t1pt0 1542 a 5 Homogenous Dirichlet BC s produce quotX5 X0 r10 Homogeneous Dirichlet BC39s produce r 1 5 Zlmlz 01 a an 6W 1 A 5 a Mr sineEX mm sing2 aX 1112 Using superposition we generate the general solution ug t Z a sine1 aXe A 39 111 UX0 X1 x Ea sing2 aX 111 IX1 mine2 aXdX 4 1 1 lgsinz meX quot3 3 mathematica Part c 2 2 N 2 2 EN llfXll ZanllsmmrXH 111 2 2 1 N 2 EN llfXll 52 111 E l a 2 N 211 I APPIYing the Integral Test we produce WW N I 7239 16 1 1 1quot64 he N EN1 16 4 lt01 117239 Z 1 64 H 0001 116 is 665703 00017r 11 295493 11 M1 M 11 1 295493 1 M 195493 m 2 Part d lim ux t 0 when a lt 1172392 tgtoo ltim ux t is in nite when a gt 1172392 ltim ux t is a nite lnction of X when a 1172392 Part 2 see next page Part f The term am is a source term The source term reduces the discrepancy between the function uxt and the approximation Problem 4 A a b V c V e V 8 1 fltogt fa sing2X f0 M 20 0 0 B in 11 sz 008EX F 1 fan fa LJ singTEX f0 f a cosGZx b sine17X f0 fa f0 c1110 47 ctan Z 1 1quot1 02 f0 By de nition boundary conditions are symmetric if 1quotxjg X fXg 0 a b c V d DI39rI39clzlet Dirichlet foogoo mglt2o13 0 1quot10 0g f00 0g0 0 0 0 0 V Neumazm Neumazm foogoo mgoo 0 0g 1700 0g0 f00 0 0 0 0 V DI39rI39clzlet Neumazm fxgx mgoo 0 0g 1700 f 0X0 0 0 0 0 0 0 Periodic foogoo wagon 0 1quot0g0 f0g0 1quot0g0 f0g0 0 0 0 0 e Robz39n Robin mmm 21mm 0 P1g1 mm mom og0 0 r 1 r 1 mi f1P1JLc 0 j 0P0J 0 C C 0 0 0 V Problem 7 Part a u czAu 0ltXlta 0ltyltb tgt0 u0ytuayt0 0ltyltb tgt0 u2g0tuXbt0 0ltXlta tgt0 ugy0fgy 0ltXlta 0ltyltb urgy0ggy 0ltXlta 0ltyltb Separation of Variables 16f HUMJP Xlt0Y7U X y7t 02 quotXlt0Y7t Xlt0quoty7t Tm 0271 X 00 l gt lyu NA 0 HU 0 T 021T0 L35 305 11czl0 0 a0 p0 pb 0 r0ijc 2 2 lun U lum num a quot b 7 amsin c t bmcos c t XsinLT X n12 msinEy m 12 Hamngam i Using superposition we generate the general solution u y t iisin EXsin uimyxam sin c t b wigE0 mm fXy if hm sinExin y I I fg y sin1LTnXsinEydXdy b 0 11m12 I I sin2 Xsin2 ydXdy 0 0 0 HUG0 amino in 1 j I fgysinEXSin EydXdy 11 111 L2 0 0 a 0 at El sin2 1 an Mydey exProblem 9 Parta urku ull7u99 0ltrltp 7rlt639lt7r up6t0 7rlt6lt7r tgt0 ur60fr6 0ltrltp 7rlt6lt7r Separation of Variables let ur 639 t Rf19 6391 RM kR3919T lIR19T 17R19 quotT T R R 8 7 7 77 1 T R fR 119 r2 r M iu R R 9 IJR39rR t LLR0 RP0 let r A39X1 13XA 112X0 yX01nXCZYX RU 011100 02 YA Bessel s Eqn 12ar gt oo as r gt 0 so 02 0 R0 011104 0 let am Zm mm zero of fnX ZIIIII p 2 Am 01quot gt 0 Applying the superposition principle we form mam momma mao imaniimmzm 111 111 m 3 0 L35 19 7r 197139 x939 Ir 83900 0 0 190X 1 u 112 19quotX an cos116 b cos116 11 L2 ur 6 t i JO 10m rc0me39l kr 0 i In Mrlam cos116 bm cosn6e39w 1 111 ln1 Apply initial condition ur 60 fr 639 i Amrym In MrIam cos116 b 005016 1 111 ln1 p lt m 6 JO Mr I I fa 6JO Aomr drda c 2 0 m12 lt10 10 200 MM dr 0 lt f 6 I mryosoww I j ff t9 rgtosn6rdrdz9 0 a 4 11m12 m ltJnmr05 g 111mr0116gt I j J 2 ryosz m9rdrdt9 lt f 6 In rgin ggt It I 1006 Mrsinn6rdrd6 b 11111 L2 Iquot lt1 rsinn6 Jnmrsinn6gt I j anmrsinz grdrd6

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