m257_316FinalDec06_Solutions

m257_316FinalDec06_Solutions - M92 57/f/K $2: 72005 L...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: M92 57/f/K $2: 72005 L “a” . mar-nyfiz—mw (I) (a) Mumosumummmmmammm (b) findmvalmofrmchduduammofmmflc) - "gum". (c) GunmapanfiminmbmhwflulHMOfll). (d) Dammineihcndiuufmoffieuiuhm. no 1 (a) X: o 1.: 4 [tat/M: swan/ac finer x > o M: An Olalflfl‘l Pour: run" \ °° nrf—I “ 0° )6 yam. (K) fix) :7 "’2 an X a :9 ! anherh’ iy51annfl- m, n '0 find 06 '0 7", nvf-rl — - + 2 a x "257,4, 0’47, fig anl-JHHI’NYHT’) *é’i” (Ix nan " nzm-I flcoaf’l” 00 o: a, [aw-(mwr— 13Xr+ ,5 @n [@¢r)[2n+2r,2w_7—{} + amjxm" mu [m/ awe (WP/(2W ya 5:”: “Zara = (2r+:)(rrl)=o 7": 4,3 7w. (c) 7791 14'than m 4,, 4r.- 4,., :2 -- 4.4—: 07*r)(2m2r-I)—l (7'. ,1 an s - m.-. . ~ * (n+')73"*’)'l (2n +3310 0"";4’ “13$! =::";9 “"49"” " '7: 9753-0 @nv3)(2nn)...7.6.-77! 1” 47,03 = X "Z: £2223... ' @nfi)w7.§. a T'sol/z: '0» = —- Qn—l = ’4’”, = —an a n # ’ ("4/2 )Zim'z)” an—Wn—n—I 2 rf‘is’mlL I -/ '1 Ian's) V} a’ g "La-L‘A” 42-: ’6" =2?» “3‘ 141.“ at"! 4n=é0 an /.(—I .21! .2. 3.6" I: (Zn—3).-2.(—:).n! 4/2 (x) = X affilWX" s x 4’(/+x.52__:_z_3+...) "‘Oén—3J.-2.(»Jnl 2 ’8 fliémtm S‘odaf/M/ Is a; 77% Fat” yd!) = C, .5; (x) I t, a: (X) [4) .9“: Aim/an :7: A»; m o V/x/(ao ryrrd/m/g I'D-9‘0 ""“ ln-I 7"” éms) ’1 \l Aw xlvnnrtrt Ila/o: a; cwmma' 2. Solve the Ina: conduction problem u; = dang—u, 0<z<1, t>0 u(0,t) = 1andu(l,t)=0 142,0) = ain(31rz). 0<z<1 1 malain(nxz)ainh(l;z)dz=i_raac-Ln-m—)zginh(é) [15ml zoo/r HA I 971297 6777/ swat/Mr Vtx)! 0 : “2‘60: ’Y V(0)-'-=/ Yll): O Ann/(x):- e“ «a? o: xzb’il =0 1; :- ’/x 7.1M: v0): A s/‘nA (/’x)/,( . I I 5: \ I a ’_ :____.- W) o 6/ (Marga/fly No). A Sin/70x)- / :17 A fill/A) was £53511(I’XVKj/Serfl/K) 417 61/293) s Vtx) + W092!) 73M W‘- =’ szxx “W wlo,z{)==o W(/,f)=o wag 0):..- sin (3171:) -' Vtx) 4;»? Walt): th) 71*) I 2:," g Xxxg‘A; " .‘Z\\+/\"IZ= 0 2.10):- o s 20) ab 1(1):: lasflx¢€6§h Ax it»): Asa XII);- 6’5z‘rzAso Ansnfi‘ 149/, 2,.)szsl'n/nrx) . 90 WM 0) -= 570 (3)7») - sin/1 {I’XD/K s f 3,, 5"}:(nix) 5I'fllt( Mi.) ’1‘, I 5’ = zj s;n(an)—5M(zvx)/a< sinéznwx -_~ /-.-.Zo(777n n J W] WI. 00 .— 1 xzfnr" £- . an“): 62nh(I—x)/o< + f E-Ifiilz’; :‘e {f 'H 5’»: (mm) I saw/.0 M “‘3 +' .. ..._—...__ -..--~ 3. Consider the wave equation with the following boundary and initial conditions: - u“ — c’un, 0 < a: < 1 u(0, t) = 0 and 110, t) = 0 . u(:z:,0) f(:e) and u¢(:c,0) = 0 (a) Determine the solution to this equation. (13) If e = 1/1r and if the function f (x) is given by _ a: if0<:c<% f(“)‘{1—zir§sz<1 determine the shape of the string at t = 1r/2. [15 marks] (a) i Y Z Q‘maaré Fox/#4441 “0‘2”: 31 [ {300—015) + ¥o(x+c~kfl want 430 (x) Immm 7-42- .z—romxoz/c 02.29 fX’Z’VS/M at f; (L) 45:0 .marfotx—g)= Cx-VZ) :04 x6[0,/.7- m gong): I—(x+v2)= v24 XGCOJ‘J Mr 60/1/6403! 77th [LUST/2): Jian/ZH (Ural =0. 4. Use separation of variables to solve the following boundary value problem: ufl+éur+$ugg = 0. 0<r<a, 0<0<1r u(r,0) = Oandu9(r,1r)==0 “(010) = Assume that the solutionuis bouudodasr —) 0. [lSmarks] arcs) : 2m 9(9) 2- “ I \ \\ r(,€+7__k)=‘Q Ké'f) 9(9) (9“ + A2 (9 r— 0 Q = A (05/1 9 +3.99: A9 . 6‘2-74/1 sin/{19 ¢KJ645219 \ 0(a) = o 9‘07): 0 (QM-4:0 607) = EA (as (M): o AnséZfl—I) V’s/J'ZJ... 62; 0,169): m {Mal-)9) 2“ \ 2 7’1?" 7* an?” _. Axe”.— I?" (r) g.- Cnv‘ + 0],, 7’ Own: 6459) <90 A5 1’2» 0 mr Ira/Aer n’nzo 90 a . .'. 4/549) = C” 7'” 2) sm [67305) Hz, Dd y)_/ ' fakamla): Z 6;, d[-é)$/n[fl-l)§) V15] ‘ ; ZY Fox/Maw 522746 tmey/M/ ( —/ . 6,, a" w- 3 ffm) 5Jn/év—l)a)d¢9 77‘ o z 0/ oz 7’ ' f d9 C = 8m «L 6‘ ” 77737—14) g 7c”) ) ) 5. Considet the following boundary value problem: (9/: — (3-1 ’)' = Az‘3¢, 1 < a: < 2 (2) ' «1) = o and ¢(2)=o (3) Determine the eigenvalues and eigenfunctions of the boundary value problem (2). (b) Use these eigenvalues and eigenfunctions to solve the heat conduction prob]: m = zzuu—zuz.l<z<2,t>0 u(1,t) = 0mdu(2,t)=0 u(a:,0) = z, 1<z<2 [20marks] (a) 15¢: —(x"¢')\=/\x'3¢ /<:(42 {KP/xv: Xz¢"—-X¢|+ /\¢=0 ¢{/)=0 =¢/{z) f3) 741/; xx 47/ 42/442 552 $0 447' Q00? x7» {ff—I)—b’+/\=0 Yr- +bi 4’44 s /://--2\1 £ 0 - Fm mum wt; to; («f/om: mr (flax/f /\‘>/ =29 b’ = / 2': I M-I = / i‘lfl. l‘f/fl l-vf 37/00: C, X +62 x = X[ACos(/Ax) i 359164940] gig): A=o 55(2): .chmggAz);-o =z> f9”: 2477 ids/22,... 2 07 2 15 . -°- A”: law?" = /7‘{Z;) 10:52,... 7752' flédalzwrcf/oa/J first ff, (X): X 67/1 {/r, Xx [Al Mam: aux/x) .—= 10:) 720:» = xZX“-— xX\ —.- _./\ clmsr 72’) Xlx) 7'7») = .A7 a» 725/: 55”” .—-- 2 f" v BX“—XZ\+/\.X=O =l> An=/+{fl.’r) r1292“ KH=A=15M¢AAQ A; " JIM-=0 sz'a) 00 ca :- ,_ ~5— ‘E” n J‘ nf/M (ab/,3) =— Z 6,, e. A" na). = g5)“: (22;) Xsm( 2 fl:/ no ’ X: (Hat/o) r.— C"1 anKX) 14:1 .. 5' art! 62) A! A .2»; 2:36;”;2‘2'2r' lav/7’ Wflffz’. Fwd/020W TO!) :: X 3 WI 26074327 x v- w ‘ Mm dam: 2 [fix 3) x ¢m60dx -.— fl cn fix 3) an) $00 A I ’ 2 C j 5’- 5m mm dx —_- cm j x15/n2mrAx)dx JV 01W¢zi¥¢z¢ 0:10 ’2 x3. 2 ’{2 i 73 Z 5,»: 2mm mag smpnru)“ = Cm f [1-— (‘as Zmfi'mfldlosfgltu- x 0 A12 '2— 0 42 AL 2 (Wm-10d 0 . z ‘___ .. S mfll. m 0. cm AZECO (fi)0=o?;[/”(”)j (ruff/A; (M77). 6. Solve the inhomogeneous heat conduction problem: at = un+8e"oos(3x), 0<x<1r,t>0 u,(0, t) = 0 and u,(1r,t) = 0 u(:z,0) = 1 [15 marks] 73?! f7/91071/M [/fZFaA/Jflo/VS a: 77/4 PWM 67779:?" ¢II +/\¢50 ¢‘[0)=:0-; ¢'[}i) sAfoyx-fgS/h X l ¢EOL¢ .élso - .. . .. 2. /l,,. v» 34-0)”... A”, n J N WA’ wow 051M fiéMva/Cf/a/I/ 4771mm» pm 544x) 5‘) = 5 ("a DO .. lit 2 45,1 (057m!) 51” = Z (-402) 6,, CoSan) flag a”. I1—‘o 147’ A78: fi‘fif}: “1/4 (SM/(79' 0 n 00 v C 2 -“' -- §[1"+”Cn' ‘3 5n3]605nx:o =0 45- t ‘ if» ,4 r1241 :=‘ 36 (n3 0“ 73‘ ' 873 .— ns‘?’ (8 C5) 5: ‘76 Q 9359:: €t+ A 6—91; +2173: - 3 2 -—V21§' méz: e c”) = o 00 9417:) =: Ana .. 2 at LLZxJa‘) == "5 Ame "teas/ma) 4- 8 Cosé’x) =0 00 5 “(’90) = "2,0 A“ (0501):) + 60563!) / Ida/flaw COfiF/C/MZY oz (arm/5- F5 2 at wag” ,(2/1’7’ ¢‘:a1f/¢Ji//rflx+é%6a$ K @405 cosnx. )Cosémz ...
View Full Document

This note was uploaded on 04/06/2012 for the course MATH 257 taught by Professor Peirce during the Fall '08 term at The University of British Columbia.

Page1 / 6

m257_316FinalDec06_Solutions - M92 57/f/K $2: 72005 L...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online