sample-midterm-exam-f08-sols

sample-midterm-exam-f08-sols - Math 550:391 ‘ I +, Sample...

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Unformatted text preview: Math 550:391 ‘ I +, Sample Exam Name: 0 U» lows III-Class Part Print your name neatly. If you forget to write your name, or write so sloppy that I can’t read it, you can lose all possible points on this exam! Answer all the questions that you can. Circle your answer. WARNING: This is a closed—book, closed—note, no—caicutator exam. Exam format: The exam has been divided into two equally weighted parts: Part 1 is elementary concepts problems. Part 2 is intermediate level calculation-based problems. Part 1 consists of 16 elementary level problems that test your understanding of the basic concepts. Each problem is worth 1 point. There wiil be no partial credit on any of these elementary problems. NO EXCEPTIONS! Do any 15 of the 16 problems. You may do all 16, but i will stop grading after you get the first 15 correct. in other words, you cannot earn more than 100% on this exam. Part 2 consists of two calculationwbased questions with multiple parts. The value of each part is displayed next to the probiem number. ' You must Show your work. You will not receive credit for lucky guesses. Show your work as clearly as you can: if i can’t understand how you got an answer, I will not give you credit for it. Grading Scheme for 3 point problems Each problem wiil be graded according to the foliowing scheme: Minor algebra/ calculus mistake with a correct approach: 2 out of 3 points given Major algebra/caicuius mistake with a correct approach: 1 out of 3 points given Wrong approach: 0 points given The diiierences between major and minor mistakes wiil be determined by the instructor, not the student! This algorithm will be strictiy enforced. No exceptionsl DO NOT WRITE ON THIS PAGE IN THE SPACE BELOW 1...__ 2......”— 3_.__ 4W 5—« 6~__.. 7..._.~ 8_____H. 9m...— 10_.___ 11....— 12W 13...._ 14% 15,—. 16 _._._ 17a m...— 17’b __..._ lSa m 18b m_._ 18c _...__.. Solid cl 0+5 0 :: 5415,15 fixed Phj Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 2 { Le+ Aer/ow one o —-= “whisk-H'er pm In problems 1-5 you are given the graph of :i: m f on the mowplane, Where v m :f: is the velocity (rate of change of Your job is to draw the flow field superimposed along the x-axis. Be sure to label all fixed points as stable, unstable, or halfustable with a solid dot, hollow dot, or half—hollow/half—solid dot. Use three arrows close together >>> to denote the flow direction in each region bounded by fixed points, and / or a fixed point and infinity. Problem 1: Figure 1: The Slope field for the ODE is f(:r) : ~:r(1 - ' Problem 2: 1.0 “1.5 —1.0 Figure 2: The slope field for the ODE is Hm) = —(:13 + 1):r(:c —— l). Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 3 Problem 3: Figure 3: The Siope field for the ODE is f = 562(93 + 1). Problem 4: TM‘J'ArmN +56. ‘F’ow “GEM I'gr Cm?— ferlrork cmA repeai‘ “1.0 ~15 Figure 4: The Siope field for the ODE is f = COS 3:. Math 550.391, fall 08 Sample Exam Wayne Hacker 2008. 4 f’roblem 5: In this graph be sure and label where the flow slows down. V c. z Valet-Hy 1.2 1.0 / “(flew “Win-j 0.8 0.6 0.4 slower mow'n 0.2 J 5 0.5 1.0 1.5' 2.0 mecr _0_2 Flow {$63M .I'I ainyd‘ mow'n “fb’fhenf law!» Her/<3ng w ~H.e Mtrouk H’ 3a+s.:fn (ff; I+Legamés GXPMen‘hkfly £3093? Figure 5: The slope field for the ODE is f = 8“”. i he HM" b““s émifcahe unfimhle fawn“ cm 44'“: hf” :‘fi'w‘a fixed FG‘H“ Solid be”: indi'cgd-a s’f’abte fad—Hon: on “H161 MH£>$+4ME “G‘XE’d path-\- In problems. 6-8 you are given the graph of the potential function V(x) corresponding to an ODE of the form: z'c = f (:13), Where f m ~g. Use this graph to draw the flow field. Problem 6: 1.0 'The. «air-cs 6F +58 fowg‘q‘ fin —0.5 —1.0 ' 1 1 Figure 6: The potential for the flow fieid for the ODE is V(m) m —&x4 + 5:132. Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 5 Problem 7: ‘yE' ;“°“(\ Figure 7: The potential for the flow field for the ODE is V(:c) = —:c —— cc. Problem 8: ~10 1 Figure 8: The potenfiiai for the flow fieid for the ODE is V(3:) m 13:4 - $272 — Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 6 In problems $41 find all of the fixed points to the firstworder autonomous ODE and do a local analysis to determine the stability of the fixed points. Recall: Let 32* fixed point to the ODE 9': m f Then, provided that f’(a:*) ;£ 0, if f’ > O, 33* is an unstable fixed point if f’(a:*) < 0, $* is a stable fixed point *2...“ 13roblem 9: $3241.93), 5.3+ fix) a: x(:——x) z x-wx‘“ (1w mm- x .... of) Fixed Points: 0” 'ac/{OJ :1 > o :27 unrl'dtle Stable Points: l Unstable Points: Halfmstable Points: Nah (no double refit?) 523' 5300 :3le 55—0 :22) )(.-"-‘~‘lffi’l‘J nnéZ,n d2” (Mm-0M “ f’(n1f)‘:: can: [m0 2: Pr0blemll,{}:w:i15"sin :3. Fixed Points: fin»an nel— Steble Points: mi’J n (Mel ‘ Unstable Points: HT?) V1 €03" Half—stable Points: norm SeffCKl": "'3 a w» Problem 11: :2: =1— €32. a, . o :1..([v-x + laylal’flcfométflbo‘d J 5300 .— X1+8Cxqj Fixed Points: 0' Stable Points: None Unstable Points: Mane, ‘ Halli-stable Points: 0 Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 7 Problem 12a: :‘c m 0053: 2:53“) V00 2 —- Scans! AK : “jelshnx = «smx +V° (can 5+0? hare ‘1 3 59+ WE’efif'fi-‘ficfi Pghen‘I-x‘o,‘ 5: O 7': > VD $0 (+1455 9+3? 1'! 100+ haterrary) ’ :V(x):-—5|'nx Problem 121): $ = — sinh :c-::§(K)(Note: / sinh :1: dcc w coshzz: + constant) V(x):: *jfbfiMx .7. ~S~s£nkx<ix a: mam :- coshx +vo. Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 8 Problems 1344: (Impossibility of oscillations) Recall: In class we showed that the solutions to a first-order autonomous equation of the form :is = f are always mono— tonic on the real line. That is, if we take any initial condition between two consecutive fixed points, then the solution will always move to the left or to the right. AnsWer the following true / false questions. Problem 13: The solutions to the systemgzt w cosa: cannot oscillate in time. . Thai: 4'5” 6. "fi'rffvonlr-J" mutton omoux xys+emwt+h .Srwo‘Ha ‘53“, False (encie one) ‘ Problem 14: The general solution to the harmonic oscillator i3 + (.0233 = O‘is given by cc(t) = (:1 cos(wt) + Cg sin(wt). This Solution cannot oscillate in time because the system is autonomous. :[3— Aces use; “1+3 Bu'l‘ ‘HM'K [3“ at 73‘“ order qtil‘bmumou ODE True (Circie one) QWX our Free-F {5” only Valitbl‘lhrfi'gm (filer “Whmmws ODE. Problems 15-16 (Existence and uniqueness) Consider the first»order autonomous IVP: m' = f(m) with 17(0) m 330. Theorem 1 (Existence and uniqueness theorem for a first-order autonomous ODE Let I g R be an open interval on the m—axis. If f’ is continuous on I (a subset of the Lit—axis), then there is an interval on the t-axis (t9 —— T, tg+fl centered at to for some 7’ > 0 and a unique function a: = 33(t) defined on (to — T,t0 + ’r) (the domain of the solution and contained in the range of 1c, denoted Rm g I satisfying the ODE. Using the theorem, determine if the foilowing ODES have unique solutions. i’robiems 15: 51': m 36%; 33(0) = 0. Tl“: 'H‘ec’f'fim "all: +0 halal he“? becau-‘e 3:60 :: x ’3 :3”; fix) :- % xng’fi has “Sm‘hgwlarifl oil" x z: 0 J w Mali Es ad— +53 . I t Paint" 0-? ow!" I_C , 3:“) Sol“ neecl he‘i‘ be unique. Anti J we dofi‘ilnw uumquesolr,‘ ((Non uniquesomrl No’h'ca ‘l‘ho‘l' X150 (5 :1 Sci“ +o WIVP above. 73%“. 0k 2“ morn:an Sol“ we {win m’ie— “Hua ODE Lulu S‘Cfdrn'lflcono't: Variables. item A” M x“ Vaax ran Mesa ext? == be #1231; % 017520” to :> f" 30‘ 3/ m (‘Zfld Sol") $>X££Lzz Problems 16: True or False: Since f = 3:2 is a smooth function on the real iine, it i 0 follows from the theorem that the solution of the IV? 9': m 9:2 with $(G) = ace exists and is continuous for all time t and for every initial condition. You must justify your answer to receive any credit. Hint: It’s not hard to solve this equation using separation of variables. In fact, you’ve done in your homework. I ' I t I __ n I “A 7711‘: Sol“ has gyros-fun ecu: swam (LPU'QnH-e—‘hme He)qu )jm) flfzfiaefifxg m ) '77“? clear no‘l' whining ‘Hie “fheurem Jbefiausehi‘l' only guarantee; q So‘hon Some :nmd 0‘“, time. 32+ Mates to frame: eta “Hie Sol .5 mm vs >0. 1L :3 k ~ — Jae-M 6- Check. gézxz' w’x”Z‘AK:A-h«i~—>-x‘=£ to xfoirzxo Xe ° 0 Xe , "-1 ....._ l r ' I x0e): rib-nan (ems-mm new “PL Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 9 Probiems 17 (6 points total) (a) (3 points) Consider the system :i: m "m: + :53, Where 7" > 0. Show that 03(t) we 00 in finite time, starting from any initial condition 339 > 0. Hint: Notice that :3: > O and m > 0 (Vt 2 0). You shouid be able to find a simple ” comparison” system for which this is true. (b) (3 points) Suppose a stabiiizing nonlinearity is added. In particular, consiéer the system :2: = rm ~i- 9:3 w m5 . Is it still true that 513(t) —> $00? To receive any credit, you must justify your answer with a calculation. A yes or no answer Wilt receive a zero. No. X Lt} -=> {‘53 L251 a: {Taco Warning: Do not try to solve the equation exactly! ‘9’" My X0 >°' Hint: What’s the large a: behavior? Try using an approximation. Also, you may find the graph below to be helpfui. )2 :3 fix; fir [We x a“) I"? X" >> Ifi‘hen XL—U Jams“ mmofim‘mllyqn Kira, L-gL V-voxL‘f 2 543.5% ‘Q‘KJ 9+ we: K-cucr} 2.0 Figure 9: f(:c)m:v+cc3——:I:5, forr=1. 3 3 Shae 2(0):)! F?OJCU,LAJ‘3 K>0JWC haJ/‘e— c o 3 \ Re I fly mu” Coumh“ 4+!eommj [41 but cmwhow has «can *‘thflb oqujJrhen Jo aloe: ).(::\f‘“>< +X3- ' 5 wa— éi.” Y3 JeP-V‘“ -*3ciymci‘b W"? Y ~— t"”Z£o A4: \ ‘ I m \t: Lia ¢ “y sures I: —-Z_ to fiffry‘ffifzc' / "10 w (70 W i m 222. 65> if H: w; y w mi") 4‘35?" ‘E W Jy‘z“ 2* Inuit £9 70“ =_ e» as: time aw cf ° Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 11 Problems 18:(9 points) Consider the equation a": =r33—~ atln(1 +33) with parameter r, on the interval (—00, 00), Where f (LII, r) = m: — a: 1n(1 + 93):: x (f—— ln(l+X)) F :e xcr-wx) (22W ) (lam We a. «amass imam ) (a) (3 points) _ Step I: For each real value of r, find all the fixed points of this system. You should be able to find explicit expressions for these. Step 2: Find the critical values (mare) where the bifurcation occurs. Hint: set 22(1) (’1‘) = 33(2) (b) (3 points) Step 3: After finding the point (we, re), do a local analysis of the ODE to determine the normal form for the bifurcation. lise this analysis to determine the stability of all of the fixed points for both cases: r < re and r > re. Recall: The normal forms are: X :: R :I: X 2 (saddle-point bifurcation) - X = RX (1 - X) (transcritical bifurcation) (c) (3 points) Step 4: Now put together all of the information that you have: :rmfi'), for 2' = 1,..11, (n roots), and the stability information about each branch, to sketch the bifurcation diagram (is, the plot of the fixed points as a function of r with the stability indicated by a dashed line for an unstable branch and a solid line for a stabie branch). same em31 with Seesaw) $0. ' X*:O (indepmzlgefi‘l' 69 V3 x(r~1n(‘+*llfio 7“? : [no-w) a.» xra an} ° (I) ,_ r These-QnJure our branches“ 1 X (r) 3: O I QYlOl X (Z) (r) 3 E. "“ l (In 'er hffimcm-Hon cilagmm p G we, . r 0 EN} ‘Hne. C(\+|CA.\ fair (Kara) where. ‘Hue hréurm‘l’mn (scum, Sig-fix W O achcl S“ K "’“Kw: 1:) “nae c,“ w clear ’Hla'l’ H19 ODE hafamnscrl‘l'n’ml ° (km 4010). {s hhéfenaml’al: r) H13? Pally )m ! ¢ New ‘. Since )3“ =3 O ‘4 flat: by exfanAi—nj fpg‘r) m afiylerseneiahodl” 09:ch . 5; when. We now van 3 :0 I? Ina“) a: X mix 4.." Math 550:391, fall 08 Sample Exam Wayne Hacker 2008. 12 Problems 18 Scratch space) ' .531“ (La) 54398) Loan “$175.3 «Lava» xcso Max R=r m 11:0. )2 2.; X (r... |n(I+x)) a: x Cur-“(x +6080) :: x(r~——x)-+ my) Let+ GV‘A R s: r J +113“ k 3X anfl. OW [00nd normal-(Bryn {I I ' (cal lax'firm+t'bh> 0 flu; 13F rzRCOJ-J-hen‘fi‘ICOn'g I V — ‘ K “H I—F r2K7OJ "HRH: +he ortgfin £5 “V‘SM‘CJ «ml 4kg brunch " K“c~xk " V\ Draw ‘er LII'me‘Cmfl'on cl‘kgmm . Be Sure +0 [0456‘ where «Jew +5681)! CW3! ‘Hle. 'G‘XBA WW are Shame and Im fickle . - X~axfls ...
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This note was uploaded on 02/28/2011 for the course MATH 550.391 taught by Professor Dr.hacker during the Fall '08 term at Johns Hopkins.

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