Final Exam - 2009 Solutions

Final Exam - 2009 Solutions - Seabird/L3 2007 9M1(Egd’tf...

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Unformatted text preview: Seabird/L3 2007 9M1 (Egd’tf W Problem ] [lsntsl Let us solve a system of equations with the Gauss-Seidel iterative matrix solver S (a) Explain how the method works ’5 (b) Why is the Gauss-Seidel iterative scheme attractive for very large sparse systems of simultaneous equa- tions? 3 (c) What is a necessary condition for the solution to converge 2 (d) Suppose we Were solving a differential equation such as = 2.1:. Assume we diserefiZe this equation dxz using a standard second order accurate central approximation. Would standard Gauss-Seidel solve the resulting system of discrete equations and converge? - 5- 09.9%” a, L/Awer . «r-qurwee a Ava-rm E muff VEDA “Lu—5 9 3 i; on»? amps-mm— om iratr' Nu~2{£v TEENS, 0(MJ6WIWUNJ @ DlPéOHAL" 1‘2“ "' L y derflANj— 3 c) DIAGWQLJJ‘FRM @ ‘ ) Em {Mp EQUATflJN i Zi j W media-(LU CLMUMHAL-f Um CW L————.._-—-“‘l @ (9 [Q \up Problem 2 (lSptsl Let’s consider Lagrange interpolation of any order. 7, (a) What is the interpolation fiinction g(x) equal to at the interpolation points or nodes? 1 (b) What is the error, e(x)=f(x)—g(x) at the interpolation points or nodes? 5 (c) Explain the general form of the error function, commenting on the polynomial portion of the error fimc- tion and the derivatiVe portion of the error functionCie. Specifically when can you space your interpofating points more widely?) 1W5 (3 (d) What are the three advantages of using Newton forward interpolation to derive the interpolating poly— nomial1 g(:c_)? {uTEEPDLA'nm-f Per/U71 Vflf/C‘flrv f’y'f 73187 femurs LC!) Assuaes Em) = 0 AT ALL Mir-tn! 77+: DERIVATIVE f5 SMALL 542 77/12 4 {:14 £1}; 1cm 1 m1? ' (DOMan 0mm Ame Easy 70 ADD \5 SIMPLE w CALCULATE Will-Emu, wee {JARWMLY sump DASED 0'” :9 Emma. 3) NEW Tgfam’ W P65;:Lr:{1'-’1;L, wow/0+ m +0 Sow: Wadi-e25 5) DO {genial-LL: oh" O—LWC. FEEVNU'S 71;“? (115 Problem 3 (252$) Let’s consider the convection—diffiision equation with initial and boundary conditions. 5_C+V‘3_C=D§_C. 0$x510 a: 6x 31.3 with intial conditions C(x,0) = 5x and boundary conditions C(0,t) a o VC+D§§ = 50V+ 5D x mm V and D are constant coefiicients. Assume Ax = 2 spacing and Ar = 1 time step. Use a Crank-Nicolson time discretization and a second order accurate spatial discretization to set up (but not solve) the system of simultaneous equations which time steps from r a 0 to r = I . ___,___.___....._._-————--—-—J 1 (9C Vflé _. 3) BC. I f. 01 .1 "I as P? ‘ éfi 4: 673‘ de o, 6 e- ! '5 'i 2%" "Eu D I w n “W 0‘3 a NEH: 9‘ a 0 t I a , 3 H, V +Ci+L?"'Ci—f "Cf—a JD *2 .' ¢C:€”—+Ci»?—zc,-LD+C;:?) ytzu xrfl xr'i Yrg «V35 xbw 1 "pi-M M 2. if A Zflx __\L_-_ D 9"” —l“~+l .“E” (AL—ll) 5’“: “V 9.. FL” I 1 “‘0 V D “‘13 WAX 241711)?“ ‘+ A"? 4M2 C’ 4' ‘Mx 2m?- C"" if 2422’" C’” +(Rfl-AX‘ C; +Gfix+fi§ba:—E? - g- '0 NEW v b 5‘13 D m .3— D "50 L‘é-fificfifh (1 +396 “’+ *3?" E m = ("i-“256% *0‘ JC; “L (‘3 ““E‘JCH K g: - HEW: Cf“) = O [39”): 10 C59»: (1.0 CI '— 0 c5” = (0 cw°b=ao cem= 50 m ' ’ anrsw D ";‘.3'C}"-€"-;"WZ;W+ : SUV-t- r. + C Elg LI (3?, ) %C$NEW9DC;W+(V+ Egan“: 30V+SD Ss—r u? Em Cimwzo (1m ‘ a 21—36:“ +(«+%)c;"‘ *(-%—%)c,"“ = CL%+%)(2® + (rm + (~¥~+—2)@ l I —-°)c,~€~ +(~-®¢s~¢~ *C-%-%)c=~w = <-%4%X30> a(v%3(z®~+(%+~%¥wv r '3 ‘ . D C MW +0” "staff-gm +Gga'3é) Cam—H : ('%'T%3(Lfo~) “(Maya 4” (%+%)Cm) I) a MW!” “3: S a; .. +0+%)csw +('%-%) = c-‘én 13603 WW9 ‘* W *9 / a—cf“, D Cf” + (w 33- a “W :: so V+ SD Problem 4 (25 pts[ Consider the equation: 5'?+xfl"+yly+2a"f=o 05x56 05.1956 5 Bx“ 6" 6f with boundary conditions H=10y along x=0and 053136 H=4y along x=6and 0$ysfi 6H ___= z < < 6y 0 along y 0 and O..x-6 %=0 along yfifi and 05x56 Assume that Ax = 2 and Ay = 2 , develop and write the system of simultaneous equations that you need to solve this problem. Show all steps along the way including the computational mesh, defining the unknowns, and developing the necessary algebraic equations to solve for the unknowns. $£\‘“9Ial fdul :0 AI, __— .: \50 ‘ffi‘fl / Q] 5 *l[ HZ 2 = 7'" *1; a H3= my = 90 :1 01 ch . :1,- -'= n H w 0 W5 21:" |H__I v/H'Za ‘0 '3 = y .— I: #0 n1 “7;”: 1-6 HI? :- = % K M MDES 3J9) 3,] '1 Hrs : W = [6 8 6H HR: :' {fY “2", x _.———-_; o 537’ _ v , 000000000000 0 EL: '37Cn"qJ[;'-;'+£‘—2 so - - ' ' ' ' . . - ' u 20 Y 2L, ' v ' ' » r l ' j c :0 M _ ' ’ ' ' t ' ° T F“3#‘:‘IH¢J'-+ ll; ;=O " 3 “EH ' ' ** *' “ ' * ~ C9 77 .51 'JJ-l ’01 ZZZ‘*7""" 0 (Stay Dal-1510"?" MAma FUR. ' 2 So 2 r . 7 r ' - o FMWMW 0: Bflcmmarzbs) - [-1 3 = a r - , o .‘___‘ " " I S‘I‘I' I f ' ’ I O 3#2”q#7“’#6:0 III 122*.9” o i: r 1 a I r '2_ 5'0 2 l a q v o Burl—WI,” *#w :0 1g?!) . , . . l-If 3 . .. . O n \i' ' a . v ' I! i O o BHS’W#L+#?‘° ‘ r , . 0'00 8 .. r . I I r . n ' ' [G LL _ +H ~0 A . o o: o 3‘! LN!” ” 00000000000000! 21 :01 NuDE s 'CJflJIUJH ; HF—HS—ZH‘ZE- Ali‘l' i-H" #:3- ' - v u r » 5( J hi " 4- x Lag.) +\/'Z_I-__+2 rJJ+I"'ZH1_,J +l/iJJq :0 2h ’4 h: 5* EON-w -+ 40+ 1112—3 #;- ah i 1 “J C 7’ ) “(5 2)’v’:—»3 +2y';5+a+2#051' =0 ‘l""2"PT§ 7H", + 2H; + 3H2 + 2H7+ 2H5 :0 7H,, + Sal—l7 +3113 + 2/47, +21% 1-: 0 CHIN—r 2H”, —+ Hg-F 2H" +2147 =0 q#;5+50Hu +Hy-r2llu +2.41,” :0 Problem 5 (20 points) You compute a piezometric head at a point using a numerical solution to the Laplace equation using spatial grid size hl = 20:71. Your solution equals H,1| «1 9.373. You now compute another solution using a much finer grid 11?. = 2m and you get HJul = 9.265 . Assume that your spatial discretization algorithm is second order accurate. (a) Derive the appropriate formula that estimates the error to both the coarse and fine grid solutions. (b) Compute error estimates to both the coarse and fine grid solutions. 10 Paoawm '3’ 0-) (21:14)”? D50“) 8 CK") 75ft“ $2.1? (*3 (my ' + m: 7 M "a: ((0955) = 1”“: '1" d“ + HOT 3 E 541114 PaMTraN: ’ 1 ‘ é - 1 1 _ Sr'mmFY' #52:? 01’”; “Alma-Dd,” E— ~F 16 H H ' Ll " (9WD: ”‘ 0sz 1 (415, "##1 d _-_- h“ — ha in: h“ —. I 1 ;_4_ .13.. 2 (0-1:;er ACCUMTE £41. “L ) _ 2 .. ’L P H E'Gmbi '2 CK“! " th'Hh‘L 1": .3 p p (‘7. 1'1, ’20 hz_h'a. }”r‘ '3 w: “'1’0 7. a H 1 2. EC .' .. EX 7' :1 H '# 2 ) a D; in :1 :1. [)2 3 I PF 8 £=1uhfiu51= E{1373"£13265:(EME'JES’ b) ’“2' ' " " > E :P t 10" 7 CF”; 6mm, ’ Ian” 9 - 1—— 101. 040% : ‘- 0J0? 1] ...
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This note was uploaded on 03/02/2012 for the course CE 30125 taught by Professor Westerink,j during the Fall '08 term at Notre Dame.

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Final Exam - 2009 Solutions - Seabird/L3 2007 9M1(Egd’tf...

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