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0534552080_98858

Advanced Engineering Mathematics

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Unformatted text preview: 1716 Section 1 1.2 Chapter Eleven-n Qualitative Methods and. Systems I of Nonlinear; 1) ifferential Equations Section 11.1.1Nonlinea1 Systems and 1414219181108 of Solutions 1. For the 1101'1iil'1ear dammed pendulum we have :1:." =- F1(:13, :11) — 11:11“ 2 1432(13-31) — —w3 5111(1)— ‘JF; )F ")Fg 1 8F . 7y Caiculate ( =0, 3 1 =1 3 — --:113('.0s(:1; ) 3— c' Since 1?th and all four first 8:1: 81; 83: 8:1;— paitials ale continuous at all points of H3 theie exists a unique soiut ti011 for any given set of initial conditions :1:({}}' — 11,3;(0) = b. the value of (1. denotes initial angulaI displacement of the pendulum from the vertical, while the value of 1) denotes the time rate of change of this angular displacement. Mathematically, there are no restrictions on the values of a. and 11, even though there may be physical restrictions. ’2. 2.1101 the nonlineai spring we have Fl (:11 y)— -- y, F2(:1;,-)1 : ——:1¢ — £3; + 2:13. F}(:1:, y) F2(:1;, '1) m . 813 8F 8E 19 333817 ' and all four first ' 1 — —0, —1- c l, ‘3 2 I 33 33,3 2— M: L are continuous 8:1; 8:; 8:1: m 81 m 011 all of R3. Thus there exists a unique solution for any given set of initial conditions 33(0) : 11, “11(0) 2 b. The values of a and 11 represent initial displacement and initial velocity respectively of the mass on the end of the nonlinear spring. There are 110 restrictions on these values for mathematical reasons, although there may be restrictions for physical reasons. . . . h c o.- ’3 3. For the given dl‘lvmg force calculate F1(:1:,y} : y, F2(:r,y)— _- — ------- :1: — —y + —:1:3 + —:1:‘3 m m m F1(x,y),F2(:1:,y) and all first partials are (:oi'itinuous everywhere in R3, hence there 1s a unique soiution for any given initial conditions 33(0) — 11., 11(1)) 2 b, and such a solution will be defined on some interval ----h < t <: 11. Section 11.2 The Phase Plane, Phase Portraits, and Direction Fields 1. :1: = w~c1cos(t) — c2 si11(1‘.); y 2 (413] — C2) cos(t) + (cl + 4132) 5111(1). 2. :1: z 1: 1:33; 3,! = 8131 11:33 L1» ege33 3. 3:21:11? 33—1— 7c2833: 31:11:11 33—1—21221233 1.3:. to .' = (3"(131 cos(2't) + 13.2 si11(21)l; y 2 e"3[(2(31 .1. £22) 605(21) + (1:1 + 2132) 8111(‘20; cit c 5-) | | 11:11:33 -+ ZCQe“: :1} 2 (:21: 6. :1: 2 213112—331 —l— 313212—33: y 2 Cir?“ + 211219" '33 ,_ 11:1} 4.1. 1. ---— = —— 91131—4353 — 1:; 11110111111 curves ale ellipses centeied at (U, 0) ill-'1' Ely 1 (it! .113 “' 353 - - - 2 2 11 a 15. 1-1” : T, a homogeneous equi'1t11'1111vitl'1 general soiution :1: +1} : ks: or (:1:—1:) +y‘ : (:1: ”1:3; c3'integ1al cu1ves are ciicies tannent to :1; axis. 111 :1: -—-- 1 9 —--'i— ( 3' (:1: --- 3.)3—- (y —-— 2)3-—— 13:111teg1al cuives are liyperholas centered at (1, ~12). 11:51 (:1; + 2)’ Section 1 1.3 d 1 10. 11—: = cst(:1:) : y sin(;c); separable with solution y = ke‘ WSW) d :1: + 1 11. 8g : 7:31, a homogeneous equation with soiution 34— ~— :1:111 103:} —.- :1: in lx|+ d1 1.2. ti: — y?) separable with solution y:ke"1/m l7? Problems 1 through 6 have phase portraits which illustrate all possibilities. Because of the repetition of appearance in the phase portraits of the remaining exercises, oniy these six are shown here. 1.. The characteristic equation A2 + M + 4 2 0 has equal roots A; : Ag : ——2 with a single eigenvector. Thus the origin is an improper node. Solution is :1: = ole—gi—E—Mc} —02)te_2t; y : ogre—2‘ + 5(c1 # cake—2‘ - . . . 4 2. A; = 4,)12 : m3; (0, 0) 15 a saddie pomt. Solution 13 :1: 2 51:11:24! — (3263:; y = 01841.4“ age31 f/fff/f /////// /// /:"///// / [z //////x’ ///////-zw\1 /////// ~w}?! fifififi'ffié 311:3? ///,-’///- 7/// /Z//// f/f/ / Z'Q-‘i/ X I’D-4f / ff/Z/ [/7/ {ff/t ff/r' Z!{£\ //// 1; A,» If}! [\Jf //// m/x / l/l/ // // f/f!’ x/rr-r'rrflfifl-d “HRH-«N 2' /// /,r// ////(~4’+~‘ ' mauve. KN /// - //// / «(arr-Mr MWHH‘H‘H N Problem 1 - Improper Node Problem 2 — Saddle Point 3. A1 :— 21', A2 2 —-21l; {0, 0) is a center. Solution is :1: : (C1—262)Sln(23)+(201+82) cos(2t);y 2 c1 si11(2t) + c2 cos(2t) 4 A;- u 2 Ag— m 3; (U U) is a node} source Solution 1s 3— * 70113:” + c2 (32 :31: 618183: + [:2 m Hard-”Hm ////////// fwd—«sew /////////.r ///////zzfl.d. Wm; ///////./.// . ..__ ///////zz. . . Her. ///////!!J’___ - HEM“ ///////z../ -' ' ///////z- M/H/rwewrw HKWK/erww (wwxawxwrflfl w/WWMJKWWW -_;/////// ak//////// “////////// -’)//////// -///X////// -/2)/////// //’//////// ffiJfl#%f/Efl ////////// Problem 3 - Center Problem 4 - Nodal Source 178 Section 1 l .4 5. A1 2 4 + 521, Ag 2 4 W St; (0,0) is a spiral point. Solution is a: z (361 — 5-22)eq si1i(5t) —— (Sc; + 309034: cos(5t); y : 2c1e‘llsiii(5t) —l— 216:26‘“ cos(5t) 6- A1 I *51A2 = —3; (0, 0) is a social sink. Solution is :1: : 7c1e 3t + (325: ”fly 2 5,318 36 __ (:28 5t x’ x z / / / / 1" f 1 r / / / / z / / Mi??? 1 Z X f I X f f' 1’ / 1' / / Z I .r’ X f :z i 5 :i :i :1 I m,” ( ‘ \‘w—eM/x , ; L’s/2,); / f/X/f/ - ,r ///f// , - / ”‘MHMHMMHWA . x , , ____ ////// {L . I! HHMH‘I—I-A‘u-figk ...____., hfi“, //////_ / / Problem 5 - Spiral Point Problem 6 — Nodal Sink 7. A1—3’i2—3 with a single eigeiivcctor; (0:0) is 3“ improper node. Solution is it ; 016:3: + Cat-93¢; y ‘ (317' (3)8“ '1' 03mm 8. A1 = 31'i,)\2 = —\/3—1i; (0,0) is a center. Solution is a: -—- (3.91 — mCQ)SlIi(x/3—1t)+ (Mc1+1302)cos(\/Zfit);y = 8:21 Sl11(\/§lt)+ 8c2 cosh/fit) 9. A1 2 #2 + x/I’li, A2 = —2 — \r/gl; (0,0) is a spiral point. Solution is :1: = ole—El cosh/5t) — ege_2‘sin(\/§t); y = c16_2lsin(\/§t) + 3026—2: cosh/gt) 10. A1 2 A2 = —13 with a single eigeiivectoi‘; (0,0) is an improper node. Solution is {I} Z c-1e_13l+ 7(C1 — Cgflfl _1‘lt ,y— w (126' _13t -l‘ INC} — c2)te_l3l Section 11.4 Critical Points and Stability 1. The characteristic equation A2 + 4A +4 : 0 has equal roots A1 2 A2 = —2 with a single eigenvector. Since both eigenvalues are negative, the origin is an impi‘Oper node and is both stable and asymptotically stable. 2. A1 = 4, A2 = —3; (0,0) is a saddle point which is unstable. 3. A} 2 2i, A2 : —2i, (0,0) is a center which is stable but not asymptotically stable. 4. A1—. 2, A2— a 3, (0,0) is a nodal source and is unstable. 5. A1 2 4 + 51', A2 = 4 — 5i; (0,0) is a spiral point. Since eigenvalues have positive real part, the spiral point is unstable. 6. A1 = --—5, A2 = —3; (0,0) is a nodal sink and is both stable and asymptotically stable. l. A1 ~—- A2 w- 3 “’ltll a Sill Ile Ell rel'lVE‘ClZOT; 0 0 is an im HT) ')01' IlOClB. Since the Cl "GIIVELer l8 is i l pOSlthG this node is unstable. 8. A1 a \/31i,)\2:-~\/31i; (,0 0) )is a cent e1 “which is stable but not asymptotically stable. Section 1 1.5 179 9_ A1 2 —2 + J3}, A2 = ——2 ._ V61; (0,0) is a spiral point. Since eigenvalues have negative real part, the spiral point is both stable and asymptotically stable. 10. A} = A2 : —13 with a single cigenvector; (0,0) is an improper node. The eigenvalue is negative so the node is both stable and asymptotically stable. 11. (a) For e = U the eigenvalues are A] : fit and A2 : —\/5i so (0,0) is a center which is stable but not asymptotically stable. (h) For e > 0 the eigenvalues are A1 2 §mé~flc — 2P — 24 and Ag = g—% (J; 2)2 ~-- 24 which have positive real part. Thus (0, U) is an unstable spiral point whenever {J c s < 2\/E+ 2. For e > 2\/E+ 2 the origin becomes an unstable saddle point. For e = Aft—5+ 2 the origin is an unstable improper node. 12. (a) For a = U the characteristic equation is A2 4— 6A + 9 I 0 which has repeated room A] 3 A2 = —3. 'This repeated eigenvalue has only one eigenvector? namely V1 2 —1 improper node. (b) For 6 > U the eigenvalues are A1 = ““6 + %\/(E'i‘10)2 — 100 and A2 2 % — %i/(c "iv 10)'2:mfi5. Thus for E > 0 these eigenvalues are real and distinct. Ii‘urthermorc for [l -< E < 9/ 8 these eigenvalues are both negative, so (0, 0) is a stable and asyn'iptot— ically stable nodal sink. For 6 >- 9/8 the eigenvalues have opposite sign, so (0,0) is an unstable saddle point. For e 2 9/8 the origin is not an isolated singularity. ( 1 ). Since the eigenvalue is negative, (0,0) is stable and asymptotically stable Section 11.5 Almost Linear Systems 1. (a) We have f(:r,y) = a: — y + :52 and 903,1) 2 £1? + 2y, with first partial derivatives f9; = 1 + 23;, fy = “1,93, m 1,513, — 2. Since the first partials are continuous everywhere, the system is almost linear. (13) Critical points are the simultaneous solutions of flay) = a; — y + 3:2 = 0 and May) 2 a: + 21; = 0 which are easily found to be (0,0) and (—3/21 3/4). 1—1 (c) We easily identify Aw‘n) : ( ), which has eigen— 1 2 l .. -- as 3 is. 3 x/fi. . i m values A1 : §+ TI and A2 = E — 72. Since these are 2! ' $3} 1 . . . . . FF x 1\Y'\ \ r complex conjugate eigenvalues with posit1ve real part} the “—2: a, i‘] ,5, j, . . . . . . ‘H . .5 I I f I origin of the nonlinear system is an unstable spiral point. 22 i __ l55555 _ —2 —1 . . Qt 5‘; ’" The matrix Ai-W? 3X4) 2 _ , which has eigenn Q: h ' l 2 ' it h \\ \ values A1 : V’E and A2 = —\/3. Since these are of opposite sign? the point (—3/2,3/4) is an unstable saddle point of the nonlinear system. 180 2. (a) (b) (C) 3. We have f(a:,y) : :c + 3y — :52 sin(y) and Many) = 2m ~1— y — Iy2, with first partial derivatives f3; := 1 —# 23? sin(y),fy : 3—332 cos(y), 9,, : 2—3,:2, 9y : 1—2:cy. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of f(3;,y) : 354—33; —m2 sin(y) x O and g(3:,y) : 2$+y w my? a 0 one of which is easily found to be (0,0). The other is determined numerically to be R: (—2.7533, —1.5074-) 13 21 Hex/5 and A2 : 1—\/5. Since these are of opposite sign, the origin of the nonlinear system is an unstable saddle point. “4.5029 3.2774 .5837 “7.8513 ’ which has eigenvalues approximately ~7.7386 and —4.8156, so (—2.7533, ~1.6074) is an asymptotically stable nodal sink. We identify Amp) : ( ), which has eigenvalues A1 = We also find A{~2.7533,~1.6074) “ We have flay) 2 ~23: + 2y and g(3:,y) = m + 4y + 3,12, with first partial derivatives f5, = #2,}?! = 2,9x m 1,9,, 2 41 + 23;. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of f(m,y) = ~2$+ 2y 2 0 and 9(55, y) = :s +4y+y2 : O which are easily found to be [0, 0) and (—5, —5). —2 2 1 4 Lies A1 2 1 + x/ll and A2 : 1 ~— \/11. Since these are of op posite sign, the origin of the nonlinear system is an unstable saddle point. The matrix A(_5!_5) = ( _f m: ), which has eigenvalues A1 = "4 ~i~ x/E and A2 = _4 — J5. Since these are uneclual and both negative, the point (m5, m5) is an asymptotically stable nodal sink of the nonlinear system. We easily identify Awe] : ( ) , which has eigenval~ N ‘I‘ N _\. '\ x x x 'L. \- ‘u. ,r Section 1 1.5 \ wzzxxxz—n. \ 1r Wlf/Ifl-"H \ 'I, W/f/I/Jh\ 1 W)/////r"‘~\, l WJ///} //;///f H 5' x n x 1 H I I . x‘ x x \ x x \. a \, “a, Section 1 1.5 (a) (b) (C) We have f(a‘,y) : —2:i: eSywy2 and g(:t,y) : m-Hiy, with first partial derivatives fI : w2,,1’y ,—, —3 — 219,93, 2 1,9,, 2 4. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of f(:I:,y) = —22: e 33; —— y2 z D and g(:r:,y) = :1: —I— 4y 2 0 which are easily found to be (0, U) and (~20, 5). ”2 m3 1 4 values A1 2 1 + «6 and A2 : 1 — x/(fii. Since these are of opposite sign, the origin of the nonlinear system is an un— We easily identify Amt!) z ( ), which has eigen- 1 4 which has eigenvalues A1 = 1 + 23’ and A2 2 1 w 2i. Since these are complex conjugates with positive reai part, the point (—20,5) is an unstable spiral point of the nonlinear system. stable saddle point. The matrix AGES) : _2 _13 )1 We have f($,y) 2 313+ 12;; and g(:i:,y) z m$-3y+$3, with first partial derivatives fa; : 3, fy 2 l2,gz z —1 + 3:52,eg = —3. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of f(:i:,y) : 33: + 123,: m 0 and g(a:,y) = mm m 33,! + m3 z 0 which are easily found to be (0,0), (1/2, —1/8) and (ml/2, 1/8). 3 12 —1- —3 values A1 = 3i and A2 2 --3i. Since these are pure imagi- nary, the origin of the linear system is a center, whereas the origin of the nonlinear System could be either a center or a spiral point, and could be unstable or asymptotically stew . 3 12 ble. The matrices Auygrlyg) = AFUQ‘HS) ( *1/4 _3 ), which has eigenvalues A1 2 x/é and A2 = mx/(i. Since these are of opposite sign, the points (1/2, —1/8) and (—1/‘2, 1/8) are both unstable saddle points of the nonlinear system. We easily identify Ame) : ), which has eigenw \NNK '-\'\\ '\\\\ \'\\ \'\‘\\ \'\\ . \'\\\ \\\. '.\‘\\\ _‘\.?\.'\ Vii KW: \\_'=.'\ \K\ \"\'\ i,\'\'\ \\ '\'\'\ \'\\\_ \'\'\ ) ‘\ ' \NN , 1* m .\\\ \\\ '1 N \ ‘\ \\ \ \ HHMHM‘E-‘I—n‘H-‘h-W‘ mummeHHH—u H-ca-u—H—m—H-Hu‘u—Mhn “‘x‘H-x‘H‘I-q'h—i-‘HH‘H‘h- "KH‘IQHHHHHH'U- .Hhhhhhhhhh. 182 (a) (a) (b) We have flay) 2 2:1: _ 4g; + 33:3} and {Kay} 2 s: + y + $2, with first partial derivatives fI : 2 +337, fy : —4+393,g$ 2 1 + 2.1:, gy : 1. Since the first partials are continuous every- where, the systei'n is almost linear. Critical points are the simultaneous solutions of f(:r,y) = 2.1: ----- 43} + 33:3; : U and g(s:,y) = :L' + y + {E2 : O which are found to be (e, n) and ((1 :l: five, —(20 : 2mm). 2 —4 3. 1 use A1 : g + élet and A2 2 g 4— % 13d. Since these are complex with positive real part? the origin of the nonlinear system is an unstable spiral point. The other two critical points are both unstable saddie points of the nonlinear sys- tern. We easily identify AND} : ( j , which has eigenvai— We have flay) = —3:r — 43,! + 3:2 —— y2 and g(a:,y) : :1: + y, with first partial derivatives fa; = —3+2:t, fy = flew-2y, gI : 1, 9y = 1. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of flay) '2 —3:i: — 43,! + 2:2 — y2 I 0 and g(:r:,y) : .1: + y z D which is easily found to he only (0, 0). W3 mi 1 1 values A1 2 +1 and A2 : ~1. Since these are equal and negative, the origin of the linear. approximation is an as- ymptotically stable improper node, whereas the nonlinear system could have an asymptotically stable node or spiral point. we easily identify A(U,0) = ( ), which has eigen~ Section 1 i.5 Section 11.5 183 (a) We have f(:t,y) = —3:E — 4y arid.g(:t,y} : —a‘ 4 y — 3:21, with first partial derivatives ffig 2 —--3, f1, = 44,93; 2 —1 — 2351;}gy : 1 — x2. Since the first partials are eontimlous everywhere, the system is almost linear. (1)] Critical points are the simultaneous solutitms of f{:i:,y) = 43:3—43; : 0 and 9(35, y) : —:t +y—x2y : D which is found to be Univ (0, O). —3 —4 —1 1 values A1 2 ——1 + 2% and A2 = —1 --— 2V5. Since these are of opposite Sign, the origin of the nonlinear system is an unstable saddle point. (e) We easily identify Ame) : ( )1 which has eigen— [a) We have f(a:,y) : —23:—y+y2 and Many) : —-4:i:—l—y, with first partiai derivatives fa: = —2, fy 2 ---1+2y, 9n: = —4, 99 x 1. Since the first partials are continuous everywhere, the system is almost iinear. (b) Critical points are the simultaneous solutions of f($,y) = 423—194—332 = U and 9(33, y) : —4:1:+'y : U which are easiiy found to be (0, 0) and (3/8, 3/2). . . . H2 —--1 . . (c) We eas11y identlfy AU) 9) x , which has elgen- ' —’-1 1 Ill/ll ......... - . 55:54: values A} = 3 and A2 a —2. Since these are of opposrte 555??? sign , the origin of the nonlinear system is an unstable 5%??? 1' f .r' f I I _ _ __2 2 ‘ z r I I I I saddie pomt. The matrix Awg 3’12) : _ winch i5}??? ' —4 1 I 55 :5 5 i i __ I I / 1’ I" .r“ . 1 23, 1 $73 _ 415.4444 . has eigenvalues A; = —— + ——t and A2 : ——— w —3. NH 2 2 '2 2 5'55”; Since these are complex with negative real part, the point (3/8, 3/2) is an asymptotically stahio spiral point of the non— linear system. 184 10.‘ 11.1 (8») OJ) We have f($,y) a 2:1: — y — 9:3 sin(m) and g(:t,y) 2 ~23: + y + rayg, with first partial derivatives fa, : 2 4 3:122 sin(m) — m3 cos(a:), fy = —1,gI z w-2+y2,gy : 1+2my. Since the first partials are continuous everywhere, the system is almost linear. Critical points are the simultaneous solutions of f(:i:,y) — 22: ‘—- y — 3:3 sin(m) = E} and g(a:,y) = ”2:1: + y + :ty2 : (Lot which there are seven. (0, 0) is easily found by inspection, the others are determined numerically to be approximately (—33.3617,1.5707),(43.2912,~1.2704),(1.0918,1.0285), (1334-28, —1.7299), (2.8100, --~-1.6033),(2.9608,1.2554). 2 —1 —2 1 values A] 3 and A2 2 0. Since one of these is zero, the matrix ACLU) is singular and the critical points of the lin- ear approximation are not isolated; in fact every point on the line y 2 22; is a critical point of the linear approxima~ tion. r[‘he current theory does not provide a classification of the stability of the origin of the almost linear system. The remaining six critical points are found to be respectively asymptotically stable nodal sink, unstable saddle point, un~ stable saddle point, asymptotioaiiy stable nodal sink, un— stable saddle point, asymptotically unstable nodal source. Enlarge phase portraits of these points are shown below. We easily identify Awifl) = , which has eigen— Section 1 1.5 Local Enlargements of Phase Portrait Near Each Critical Point. fW/n/f ,waa_ ,zz/z 13 ' i? $55 wk“; 4..... .. ’ ' _...-...~..-..~.:I\\. { i z 1 x ’ =HHWHHWM4HHH Mamnxxxx rrrzx La . aaa\\\\\ {XXIX Hum \\\\ tiff! I}. ‘u \. \ \ I .I‘ I f \\\\ III L2 RH ‘ " i \. r a it all. . . 5 i.) I (1.0918, 1.0285) ("3.361715707) . . Saddle Pomt Nodal Smk ##fi/fxltlfl \\NHM&MMM _(/{//r 1/ ._-._-._-.—a~/—‘-//// I _\ \‘x-«ammhenc-n / (K, '//r #HFAP-rrr—bzzz / ‘KN‘H‘HHHEH. Ij/I _ zrr-r trazizifi k 22:22:22 4”,” mumwauggz '\ “.44.. ..... H r:;:;:-” r -/fimfia-w -¢rr4w4WWfr “flHHHNHHmaxx “WHHHH'HHHM “x. \ mac-awn \ ! {/rra-f‘.» {fr/xwxr ff/JffJV 13731 «3.3 3723 abs (—3.2912,-—1.2704) Saddle Point ea (1.7423_,~1.7299) Nodal Sink 2.94 2 ‘ (2.9608, 1.2554) Nodal Source {/K.'//z////f l\\\\.‘u‘m r//////z1/{f \NNNHM (ff////////f \ \\‘\.\M“H (err/xxxz/l / \ \ \M‘xwa fFf////f{f/f \\\NHHM I} \NxHHHM / / \, HEM-Imam. r \MH-u-mmn—n / w( H ‘n—. / ._..-_.-..»..\ J.» K '\ // HHHHHHMHHR. \ ; I/fr-J'Q-l.» Hmmmham‘mkc‘x \ ////»)—*fi “name-emuhau'x k x I 22/ 22,1» aammwaxx\\\ IJ/l/zxr EHHHH\N\\\\ XIII/z»; Hamam\\\H\t !/////// HHHHNR\\\\\ fix/xxx; . . 44 v" 2.?3 2.?8 23? 2.62 2. {2.8300, —1.6033) . Saddle Pornt Section 11.6 ' 135 11. (a) By using Tayior series we find that both systems have the iinearization X" = Y, Y’ = wX, for which the eigenvalues are A1 2 i,)\;; 2 wt, so (0, U) is a center for the associated linear system of both systems. H: —J_2'!: +\/—3}2 (b) Using polar coordinates we get it cos( (6)! < r —> {3 as 7‘ —> 0. Similarly were ' (c) Since r2 = $2 —l—- y2 we easily get by differentiating with respect to t, that "r?" : 3:3," +193)“: d 0?," (ti) For the first system compute 5' OT: -— a: c +yy— -" —(:1:2 +y2)\!:r2 + y? 2 “-‘T'3, or hit 2 ----- T2. -< r —w 0 as r —> 0 so both Systems are almost linear. 1 .. Thus r’(t) < 0 for all t. Solve to: 1(t ) subject to i' (to): To to get r(t)= Eng—T1 and _ 0 4_ _0 note as t —> 00 r'(t) —> 0 so the origin is asymptotically stable for the first system. (J? i . (e) For the second system compute ri— — 3:3: + yy— — r3 or ;—:—— — r2. Soive for r(t) subject 1 I 1 -' I to rt : r to et 5" t 2 -————r———. As t w) (t -~l— —) , note that 1" t “4 00, so the (e) 0 g () (ewe—t e a o origin is unstable for the second system. Section 11.6 Lyapunov’s Stability Criteria 1. Compute the orbital derivaztive V : ——-4{ta:2y2 + b(—— 2my3 — 33 3y) — 2czsgy 2. Choosing b = 0,0: 2 c I 1 gives V = 332 + y2 which rs positive definite and V— — —6:r 23,: 2 which is negative selnidefinite. By Theorem 10.4, (0,0) is stable. 2. The orbital derivative is V 2 “20,352 c082(y) + b( easy cos2 y + (6* )a‘y ‘2) + 2c(6 — 9:)y3. This expression is neither positive definite nor negative definite for any choice of constants a, b, c. Thus it seems that Lya...
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