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# Class Note for MATH 254 at UA 32 pages

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MATH 254 Lectures 2230 For the next nine lectures7 we will study the second order autonomous system of two rst order ode s dz dy 8 1 3 7 Why a 7 CIyy and the connection of its solution trajectories z 1t7 y yt with the family of solution curves of the rst order ode dy 7 CIyy 82 g 7 FltIyyi We will learn about 0 phase plane analysis 0 equilibrium solutions and their stability properties 0 basins of attraction of asymptotically stable equilibrium points 0 stable and unstable manifolds of unstable saddle points Applications will include o mixing of salt solutions in two tanks 0 the pendulum o the motion of a particle in a quartic potential 0 predatorprey models 0 oscillators with nonlinear ampli cation and damping Applications Model 1 Mixing solutions in two tanks which are well stirred n T Hui1d 90 3143 a lt 8 leant Mm mlTol Col 390 MFA Ld c lfMRaA 09 58 mmLL comb i quot c l U l I 7 33 1 L C Let there be 3005 yt lbs of salt in tanks 111 dac y 10 2 F dt 10 320 5 83 dy 310 3y 20 10 may Volume in each tank remains constant at 20 85 10 gallons respectively 550 10 y0 20 Because 83 is a set of linear equations we can solve exactly a Exact solution Eliminate 10 from second equation t 20 6y Substitute in rst equation Find after multiplying across by 23 9 dy 3 3 dt2 20 dt 100g 1039 3 Complementary solution of y y y 0 Let y e W 9 2 9 3 9 x33 7 40 lt0 T2 0Tm0gtT 4 Oi 9 7 2 4 0lt0 yc Aer Bewt Particular solution By undetermined coef cients yp 10 Hence using ZEt 20 6y 83 i8 T T Determine A B from initial conditions 960 I Observe ZEt 20 0 Since r1r2 lt0 ya gt 10 ast gtoofor all AB 0 Since lrll lt lrgl 67quot c lmlt gtgt cmt e lmlt as t gets large E Thus N A 3 1 2 emf for large t and IEt approaches 20 10 along 20m 3x33 E the direction 3 2 6 N f 2 i n glamquot flip rlirnhfinn 71 m 27 2 b Hand drawn g Gm 330 we fke dfchCm 39 35 T l t m xLo C a 3 El dx dy G F G Where Fgt0Glt0 aStlvxlvyi Flt0Glt0 astTxiyi Flt0Ggt0 astTxiyT Fgt0GgtO astTxTyT On the next page we use SLOPES to draw direction eld and several solution curves Note the asymptot ically stable solution are 20 ye 10 corresponds to a concentration of llb gal in each tank 7 just What we expect We Will learn the name for are ye in this case later a stable node s19ng i3x233iyf183l3ylB3 x28 x ts 91 a xquot xx39x 3939quotx x39x 39a x i i x11 x w x x x a 2 a x x quotx x quotI 1 i X l 39s a 5 1 quota x x x Nquot quotx quotx quot quot x x x s K x x X i 1 391 g 192383 5 s x a E a s a x a a H X 1 t x 5 l I 919 511722 2 i x x x quotx x w a w x I i Fmvo XZ815734 a a R K R x 4 x a quota x x 3 i 39s l l l f 928M618 R x a R a x x X X x RR 1 i 5 i i 3quot ill xxx3939 39xxxxxxxx iii xxxquot 6 lamxxxxxxxxa xxxia x F39WGLw vo 5 x5 x s 2 s q R xx quot5 y 539 quot 5 XX 5 xx 2 xx x 3quot W 39 bil axxwwuxaxxxaa my E321 t l K a a a quotx a m 39x a a a w quota m m a a S L n a A 395 m 395 quota 5 quot5 395 quot a Mquot 539 SH 5 H quota in k K Rrref esh W39R39m39m r W 39 quot K x 13 use xllfxz Sr w w39w w w J i x x a quotx k N x a g use W er Mm 39 x i i 1 x x a k x g ggl l I m u My x g r I r a x x x x a a a a x 39 wN5Mx39 A x 155 3 pr a 35 39a x Starting from initial conditions 1020 2020 2520 205 55 we have drawn several solution trajectories Note that they all approach 20 10 the equilibrium point Which is an asymptotically stable node along the direction y 10 2 3az 20 2 20 Arrows on trajectories show how 97H 71H moves as time 7 increases Rinse H Mt Model 2 Pendl u 3 169 2 quot3 A1th cut cu pvu clhr cf 11 btULcLuJuM and was is Cowcm ut ul quotA 393 MS a 1M gawk ar m4 bh ck v mam lna angle direction 1 dw were mgs1na m T T gravity friction 84 Note If a is small sing 2 a and 84 is linear and the same equation as that of a spring With bm 3 gl We can write 84 as a system y is angular velocity d1 dt y 85 y sinz dt Model 3 Particle rolling in quartic potential dzx 8V d3 3 d3 W a 90 d3 dt y 86 d d g I 63 3y Model 4 Predator prey Bt population of rabbits yt population of foxes d d C OzCC xy 87 d d i yz dazy On page 29 we discuss a model With competing foxes and rabbits and With saturation and hunting Examples Model 1 7 y 31 731 By Fltzygt7210 20Gltzygt720 10 83 18798 20710 Remark 1 Note that z 207 y 10 is the nal equilibrium state When both tanks have the brine With concentration 1 lbgali This is the point to Which the solutions 1t7 yt Will tend after a long time Model 2 Fwy y CIyy yi sinz 8 5 18 ye mr707 n 07i17i27 Remark 2 Note the equilibrium 1 0 or any multiple of 2 y 0 corresponds to the pendulum hanging vertically down With the mass below the hinge What do you think happens if you were to start 107 y0 near 00 The equilibrium 1 7r or any odd multiple of 7r7 y 0 corresponds to the pendulum being vertically up With the mass directly above the hinge What do you think Will happen if you start near I 7r7 y 0 Model 3 F079 y CIyy yi zr3 86 18798 17070707170 Model 4 FL y a6 1 7 y CIyy QW 51 8 7 7 a we 00 or W Model 5 Fwy y CIyy 6M1 7 12 I sis 18798 070 Exercises What are the equilibrium points of the following 2nd order autonomous systems dz 7 dy 7 a 7 17073 7 CIyy a Fzy y C17y I Is b Fzy ix 21y may y I2 y2 CF17y I i 12 7 my C17yyiiy2 Exy d Fzy y CM 1 e Fzy y C17y 7sinhz sinhz Stability of equilibrium points We are also interested however in What happens if we start near me ye namely if 10 7 re and y0 7 ye are small 18 ye is Lyapunov stable if7 for any 5 gt 07 there exists a 65 such that 65 A 0 as 5 A 0 and such that if 10 7 ve2 90 7 ye2 lt 55 then IO 7 ve W 7 ye2 lt 5 for all time t gt 0 Remark You start near7 you stay nearl lf rage is not Lyapunov stable7 we say it is unstablei rage is asymptotically stable if it is Lyapunov stable and if 103790 A 18792 as t A 00 The basin of attraction of the asymptotically stable equilibrium point 1e7ye is the set of all 107 y07 the trajectories 1t7 yt through which tend to 187 ye as t A 00 Examples Experiment and Explore We will do calculations later But for now7 I would like you to ask yourselves if the equilibrium points in models 1273475 are Lyapunov stable7 asymptotically stable or unstable Please use SLUPES and SYSTEMS to experiment and explore Model 1 We have already discussed this a little in class The equilibrium solution is 18 207 ye 10 Suppose you started with 10 21 and y0 117 would the resulting solutions 1tyt tend to 2010 as t A 00 What would be the set of107 y0 for which the solutions 1tyt would tend to 2010 This is the basin of attraction of 2010 You might want to run SYSTEMS and check this out experimentallyi Remember 1t7 yt should always be positive See page 4 of these notes Model 2 1e 07 ye 0 Suppose 10 ll radians and y0 ll radiansseci7 what does the pendulum do if the damping 07 ii if B gt 0 but small We will see how small when we do the calculations Model 3 Use SLUPES and SYSTEMS to explore the stability properties of the three equilibrium points 7107 07 07 10 in the case where 0 ii 6 02 Model 4 Again use SLUPES and SYSTEMS to explore the solution trajectories in the case where a 7 6 1 so that the equilibrium points are 07 0 and l7 1 Look carefully at what happens if you start with 10 l2 y0 2 If you were an ecologist looking at the graphs 1t7 yt for this case7 might you be worried Model 5 Again use SLOPES and SYSTEMSi You will nd that instead of the solution 1tyt tending to a xed equilibrium point7 it keeps running around in a closed curve it is like a circle We call this asymptotic state a limit cyclei Methodology How do we analyze 81 Step 1 We rst look at the solution trajectory curves 1tyt in the 17y plane by solving dy CIyy m Fw a directly7 if we can7 and then drawing the curves or b by using SLUPES or SYSTEMS and drawing the curves one at a time Note The ratio 2 is not de ned at an equilibrium point 1e7ye where F1eye C187ye 0 Nevertheless we can analyze the behavior near 187 ye by simply looking at nearby trajectoriesl Step 2 We analyze mathematically the behavior of the solutions 1t7 yt near the point 187 ye by setting m 1 X w W and treating Xt Yt as small This will allow us to turn 81 into a pair of linear odels like model 1 which we can solve directlyi Using this idea we then write down general criteria on the partial derivatives 7 8FIevye BFIevye 7 6GIevye 6GIevye a 7 81 By 7 C 7 81 By which determine the kind of stable or unstable point me ye isi 811 b d Step 1 Case study li Apply to model 2 in the undamped casei Take 1 Then 85 is dz 812a a 7 y 8 12b i7 7sinz and as time t evolves the solution trajectories ztyt trace out curves in the z y plane which satisfy the rst order separable ode dy 7 sinz 8 13 37y Separating gives y dy 7 sin 1 dz lntegrating gives 2 K cosz when K is the constant of integration This is simply a statement of conservation of energy The sum of 2 the kinetic energy y with mass m l and potential energy with g l l l 7 cosz height above lowest point is K l which we will now call E the energy y2 8 14 il7coszE Observe since y2 gt 0 and l 7 cosz gt 0 E is always positive It is only zero at the equilibrium point 00i What do the curves 8 14 look like when I and y are close to zero Then E too is small If I is small cosz 217 g and then 814 is 2 2 y z 7 7 E 2 2 7 namely a family of circles about the origin with radius V 2Ei Remark If we had not taken 1 then 814 would read 3 27zl 7cos I E and then the trajectories for small E would be the ellipses 3 7 12 E with semimajor and minor axes 4 and x2Ei Remark 2 Since 814 is periodic in z with period 2 the same picture repeats if we shift I by i27ri 10 Solve 815 y d x 2cos E dt 2 Exercise Challenging I will ShOW you in class Solve 815 to get 816 t 4tar11 Note as t to gt oo7 t gt 4tar11 1 7T As If to gt 00 t gt 4tar111 7139 For E gt 2 the trajectory 814 yz 31 cosxE r cannot cross the Ie sinac Now let us look at SLOPES for dm y 12 Observe that any trajectory which begins near 7T7 0 will eventually leave this neighborhood We call W70 an unstable saddle point We call y 2cos the stable manifold of the unstable saddle point 7T7 0 because if a trajectory starts exactly on y 2cos it will tend to 7T7 0 as t gt 00 Of course7 if we start slightly off y 2cos 7 the trajectory will leave the neighborhood We call y 2 cos the unstable manifold of the unstable saddle point 7T7 0 because if we start a trajectory on it and run time backwards then 25t7 gt 7T7 0 as t gt 00 The unstable manifold of 7T7 0 is the trajectory which organizes the escape from 7T7 0 What happens if we look at the case of a weakly damped pendulum Now 85 gives cly i sina y cla i y and is not solvable by our usual methods We therefore must use SLOPES Nevertheless many of the original ideas remain Note that now 07 0 is asymptotically stable rather than Lyapunov stable Note also that whereas the stable and unstable manifolds of the unstable point 7T7 0 are slightly shifted7 they do not go away no matter how strong the damping may become Note that the basins of attraction of 07 0 and 27T7 0 are separated by the stable manifold of 7T7 0 13 039 81 e 1 39 l 39 5 x1 isemsse y i12639647 N 32 51186813 I 2 156399 4M2ka a g Mumi i39bquot PB neo ei 39 com MM 311 Ii integral 39 W WA Es slopes Re refresh 3 12 use 11le r B use my 2 some 01 C Help y UNLOM K l3w LWLM I 914k MQMPFLc k a icx 1 0 Nohi 0o H l f 8 20 83 r 140 L awgv S F m h 7 m CLka a as Rte15 ew d39o x 7 bur W a looks in 1 Emme k OW a w Lk ou 3W J r 5 w r 14 15 i r quot quot 3 139 139 z I I f 52quot 539 K quot5 3 393 1 39139 quot quot x 397 395 KW W H 1 x quot 5 395 391 I 3 i I I nquot E i I i a 1 39 4quot xquot 1quot rquot 1quot I 339 X quot quot39639 9 6 239 If f f 3 x W I s I 5 a h I 39x 391 s x 39s g 1 r v39 15 x x 5 I f i 5quot quot 393 r x mxmwxxxxx x Ammxax1xx A a I r 39 x I I 1 I l a x I 1 3 55x x x x39mw vH39xquot r 3 r r x I Jquot quotK N N 3 Mu r39w wquot jquot 3quot f f fa xxxxxxxmhwxxxrxxxx 5 ca A r A a 1 xT s I 5 55 i m m 335 xmn win39311 quot5 quot N IN 5quot m mquot s a sh mmmxa m mam 9 3 Ira a quotamm quot 395 393 3 a 3 5 im vial is v I c a E m an jmm s l HrIISGI tvIH39 m 5quot 5 A j q cmm nanmun In Sun NC 5 2 ma 4 mucus94 mac 2 312 UmevImmv Arrows ShOW direction of x05 yt as 75 increases dydx sintxhiyy x 32 501 xi1 11 1 81 X163314535 Y11834785 X218Bl76491 Y2623645785 s h 2 09 eel h1 e11 Ii integral quot391 5 slopes Re refresh 12 use x1x22 9 use orig lemmeh Help F Dawea More 39 Nok o LMW39 M W jaoLJJts L c mtgN owe W2 TiL1 16 L39 quotLWA 31m eam h I WNW w CeyCCQA asvv H T w l DEL K haw haw 5A4 Om 39 as 010 mm 3vtv1A 0 W L 430 w 3 03 Step 2 Local analysis near the equilibrium point It is useful to analyze the behavior of the solutions near the equilibrium point Set 310 13 zeXt W yeYt substitute into the equations 81 and treat XY as small This will allow us approximate 81 by a pair of linear and homogeneous no forcing terms equations in X Y which are easy to solve Before giving you the general formula let us work out a few cases by hand Example 1 Consider 83 Because this is linear already we will not have to neglect any terms The approximation is exact Let zt 20 Xt yt 10 Yt substitute into 83 and nd dX 72 10Y 320X 7 3X Y dt 7 10 20 7 20 10 317 dY 3 3 3X 33 E lt20Xgt lt10Ygt Solve just like we did in obtaining the exact solution earlier X20dY 818 E E W and substitute to nd d2Y 9 dY 3 7 if 730 32 20 dt 100 819 Note the similarity with y y Wgoy from earlier If we set y 10 Y then we get 819 The solution is Yt Aer Be where 7 1 w lt 0 7 2 lt 0 So using 818 we have 820 lt i8 gt lt gt Alt 2092 gtemBlt 2092 gtemri Just as we did earlier the behavior of the solution ztyt as it approaches zeye 2010 can be ascertained from 820 We will give the de nitions later but the equilibrium point 20 10 in this case is a stable node 17 Example 2 The damped pendulumi Consider 8 5 and the equilibrium point me 07ye 0 Set 1t 0Xt7 yt 0Y in 85 and nd dX E Y 821 dY g E 7 763 7 jsinXi Since X is small7 we can approximate sinX by the rst and linear term in its Taylor expansion sinXX7 7375i So near X Y 07 we approximate 821 by g Y 822 Suppose we look in the neighborhood of the other equilibrium point me 7r7 ye 0 Set zt TV X7 Mt 0 Y7 substitute and nd dX w Y 823 dY g E 7 763 7 js1n7r X But sin7r X sin7rcosX cos7rsinX 7 sinX which is approximated by 7Xi Hence 823 is dX w Y 824 dY g E iY BY Let us now solve 8 22 and 824 Substitute Y and nd 82 becomes d2X dX g 7 7 7 8 22 dtQ 6 dt X 0 and 824 becomes 2 4X 75X0i 7 8 24 dt2 dt 1 We will see that the solutions to these equations are very different Let X equot in 822 Find 01 If the damping B gt 0 is weak so that gt 372 then and the general solution of 822 is Xt e 2 ltAcoslt 7 gttBsinlt 7 gttgt 77 6 5 5 7 E E 55 Yt E76 tlt72AB 7 4gtcos 7 4te tlt72B7A 7 4gts1n 7 4t If the solution begins near 00 then it slowly spirals into 0 0 as t 7gt 00 For obvious reasons the equilibrium point 00 is called a stable spiral see graph on pg 16 2 If 37 gt B gt 0 then the two roots are real and negative In this case 00 is a stable node The trajectory does not circle the equilibrium point 00 as it approaches it The second case 824 is very different Now if we set X equot we nd T2BT70 or 7 9 T1775 jjgt0 and 7 9 T277 7 jjlt0l The two roots are real but of opposite sign Let us begin to examine the behavior near I 7r y 0 or X 0 Y 0 in the frictionless case B 0 For simplicity we also take 1 Then Xt Ae Bequot zt 7 7r 8 25 Yt Ae 7 Bequot yt If 10 y0 are the initial conditions then ABz077r AI0 Wy0 826 or A 7 B y0 B 7 7r 7 Note that as long as A f 0 ltzt77r 7 00 as t 7gt 00 W gt In other words the solution trajectory zt yt which starts near 7r0 does not stay near it The equilibrium point 7r 0 is unstable Because the roots are of opposite sign it is called a saddle point The reason for the name saddle will be clear shortlyl 19 If A was zero namely if 10 y0 lay on the line y0 7z0 7 7r then gt7gt0 as t7gtooi The line y0 7z0 7 7r is the approximation to what we previously de ned as the stable manifold of the unstable saddle point 7r 0 If B was zero namely if 10 y0 lay on the line y0 10 7 7r then mmw as So the line y0 10 7 7r is the linear approximation to the unstable manifold of the unstable saddle point 7r 0 Recall from pg 13 that the stable manifold for 7r0 was I 2 7i y cos 2 LetyY z7rXand nd X X Y 2cos 72sin 2 7X or y I i 7r The unstable manifold for 7r 0 was I 72 7i y cos 2 Again set yY z 7rX and nd YX or yz77ri General Classi cations of the stability properties of equilibrium points Begin with 85 dz dy 827 E 7 FIy7 a 7 Czyy and the equilibrium point me ye Set 828 m me Xt yt ye Yti Substitute and expand by Taylor series 6F 8F Fm Xye Y Fzeye XEreye Ya7yreye hioiti 829 8G 8G Cace X ye Y Gltzeye X7ze ye Y7zeye hioiti 81 By where hot means higher order terms containing products such as X2 XY Y2 X3 i We ignore them because we take X Y small 20 Call the partial derivatives a b c d namely 8F 6F BC as 830 a EweyyeL b aiy eyyeh C EweyyeL d aiy eyyel Examples 1 8 3 Fzy 7 7 CM 3 7 zeye 2010 a 7 b 7amp7 C 27307 d 2 8A5 FIyy y CIyy yi sinz reyye 00 I a 0 b 1 c 70081 xi 7 d 73 187987T0104 0 b 1 c 7Cosz mgr d 75 3 8 6 Fwy y Czyy ix 13 7 y Ievye 7170 I a 0 b 1 c71312171 2 d 75 reyye 070 I a 07 b 1 c 71 312x0 71 d in awe 10a0 b1 c 71312T 2 d7 i 4 8 7 Fwy M 7 My CIyy 77y 61y Eeyye 00aa 2751 0c6y 0d77i y0 10 awe w maxim a 0 b a c4782 dw6z 0 yg 7 up 5 8 8 Fwy y CL y 6M1 7 12 I 187980701a07b175 17d Exercise Calculate a b c d at each of the equilibrium points for the examples given in Exercise 8 In the small X Y linear approximation 831 g aX bY 832 g 5X dY Observe The original system of equations 81 is in general nonlineari Near the equilibrium point me ye Xt zt ire Yt yt eye we get a system of linear equations 831 832 for Xt Yti We now solve 831 832 in the various cases 21 Case 1 b c 0 Now the equations are each rst order The solutions are Xt X0e quot7 Yt Y0ed i Conclusion The equilibrium point me ye is asymptotically stable if a lt 0 and d lt 0 It is unstable if either a gt 0 or d gt 0 Case 2 At least one of b or c is non zero Suppose I f 0 Then from 8 317 833 w g 7 aXgt Substitute 833 in 832 and get d2X 834 W 7adad7bcX0 Exercise Suppose c f 0 and we choose to eliminate X from 831 and thereby get a second order ode for Yti Show that you get the same equation as 8347 namely7 d2Y dY W 7 adE ad7bcY 0 Therefore the behaVior of the solution does not depend on which variable you eliminate Let us rewrite 834 in a familiar form by setting pad7 qad7bc to obtain 8 35 10 4X 0 Let Xt equot and obtain 8 36 T2prq0 The roots are 837 Tgi q We now discuss all the possibilities Case 2a qlt0i Roots T17g p4727qgt0 T2717277 1727qlt0are real andofoppositesigni Xt Aer Ber21 YO 2146711 H b b rage is an unstable saddle pointi lts stable manifold is the set of points X0 7 re and Y0 7 ye for which A 0 in which case Y0 X0i lts unstable manifold is the set of points X0 7 re Y0 7 ye for which E 0 in which case Y0 bi X0i BeTgt Example Undamped pendulum at me 7r7 ye 0 with 1 Then a 07 b 17 c 17 d 07 p 07 q 71 lt 0 The roots 7 1 and 7 2 are l and 71 respectively The stable manifold is the set of 107 y0 siti y0 7 0 7z0 7 7r since I 71 The unstable manifold is y0 10 7 TL 22 Example Damped pendulum at me 7r ye 0 with g 1 Then a 0 b 1 c 1 d 7B p 6 q 71 and 7 1 7 12 T2 7g7u1 jgi The stable manifold is y0 lt7 7 1 372 7 7r Which reduces to y0 7z0 7 7r when 6 0 The unstable manifold is gm 7 7 ltzltogt 7 2 Case 2b pf gt L gt 0 Roots 7 1 71277 M17472 7q T2 75 7 MT 74 are both real and negative if p gt 0 Then 18 ye is an asymptotically stable node If p lt 0 the roots are both positive and we have an unstable nodei Example Two tank mixingi zeye 2010 a 72730 b 7 c 2730 d 7 p 7ad 2 1 q W301 4 81150 i The trajectory approaches 18 ye along the d1rectlon lt Thu gt i b 2 Case 2c pf lt q Roots 7 1 71277 2 q 7 1172 T2 7g 7 iuq 7 1472 are complex conjugatesi The solution 2 2 Xte Acosx 47t4rBsinx q7pZtgt spirals in out to from X 0Y 0 I may ye ifp gt 0 p lt 0 Accordingly for q gt 17 p gt 0 re ye is a stable spirali For 4 gt 1172 p lt 0 re ye is an unstable spirali Example Weakly damped pendulum with 1 and B2 lt 4 at me 0 ye 0 Then a 0 b 1 c 71 d 7B p and q 11 See page 16 Case 2dr 4 gt 0 p 0 Then Xt Acos tJrBsin t Yt 7 aA cos t 7 A 7 aB sin t and the trajectory Xt Yt circles 00 in an ellipse We call me ye a center Examplei Undamped pendulum with if near zeye 00i Then a0b1c7d0p0q and zt70AcostBsint yt70 Bcos t7 Asin ti Exercise Show directly that 2 2 z y 1 A2B2 A2B2 In case 2d the analysis does not prove de nitively that z 18 y ye is Lyapunov stablei More needs to be done and we will discuss this in class 23 a at a 0th L h t b 0 quot 9 Summary wwo 3 ng ubi o bc S SarLoUC L act ha yoiht l9quot 1 dx dy g F7y7 a ways Feaye Geaye 0 8F 8F 8G 8G a EtvevyE b 8 y 7y 7 6 E7 d 8 y 7ye p ad qad bc Note Case 1 with b C 0 and q ad p a d is consistent with this Classi cation because if q ad lt 0 we have a saddle and if q gt 0 then we have a stable node if p gt 0 and an unstable node if p lt 0 24 Phase plane for ball rolling between over two hills agdx xxxxb9fg gncdec x Ra incidec a incfdec a Eh incidec h incdec I Cc incage g GIRL9LT xlyxh1 RFD azh1c1 I Integral SS slnpes Ru refresh 12 use x1522 Nn New lnit B use only Dd Show 1n1t 22 zoom ni ESCAPE quit N W o f Mhhn 4mwmmpww m S Av r quot d nraquotzt vm 1 11 da L sax L Lr dt y we dy describes the motion of a puck on g 1 x3 by a frictionless surface I 0 Take case b 0 Solution curves 1 2375 y yt of 1 trace out curves in 13 y plane and these curves are solutions of ode of 1st order d 3 2 2 4 2 2 i 31 33 In this case you can integrate y x BI E y VI T T T energy kinetic potential Which can also be written y2 132 12 1 4E 302 1 xl 4E For E lt intersects with X aXis y 0 are x2 1 i V1 4E 132 1 xl 4E For ooltElt0 132 lt0 andxi 1x1 4E ForE0xi For0ltElt 2331 x1 4E mi1x1 4E For E gt i No intersects 47 25 ForE yx2 1 0r y x2 1 T T unstable stable manifold manifold of 10 of 10 dydx xxixxbigfg x B a 1 6 b 1 L 81 c B 3 xl gill 1 1 a E i 91 ci 1 3 X18264346 911857864 X2 43536924 1 922 97814 Xx incden x ha incden a Incden a B5 incfdec b dec Cc Inpdgc 9 GIRLALT xxg1h KPD a1b1cl 11 Integral SS slopes 1 Hr refresh 12 use x15x2 Mn New lnit 6 use 091g Dd show inlt 3 22 mean ufi ESCRPE quit Same problem with b 01 NOW 0 0 is an asymptotically stable equilibrium it is a stable spiral Again i1 0 are unstable saddle points 26 u IE 5 demand Vodka W o n L Ron vouMA mm E10 inc aill39llelluulfll u LIlII LI Il lllllll IllilllllalllruIJ 1Illitllvl ll 1l Ilrllvlfflllliil italah IIIfl till glutluf vtn wt if it lit quota u x 131411337 x x quota xx xxwiffrz a 4 u t arrathxFET 2 r s 39t 1 t quotp z xxfflqucxx a u p u u I it 1 n 1 I a f I R new 4 x f ff t sxk x a u T J i xxx44 3 3 a x xx Inf IIIIIT Illllllllllln t xti T a x 334 4 quot5 a N s Jrr If r e an raw 5 W as tstI llll 2a A J m U K a uIll lllflJPf I I 4 quot a t x 1 L 1 a J a ast ixa 3 a a w I r flirrulllh xx fllfJJJitla39JInIIIIn uIIIIIII Iill l 51 P r IIrlnIIIIIII II ll In P I Il l lulnlliu l u t TIILIJIIlIII l l mulhlllliil1lt fume Ms ni nnhlntii Mums dt Example m dz dt 31 dt Intc quottwe atl hP vms quotq F 37 q aha x amp4 I g5 21 1313 K J 4 3 C 39l F I I r l I 2 u llll 1 HI hcl ltii n 9 Iilrvlfltloltl III silt I39llJlll i c n r 6 1 c r5 1 rn Cm C I I I n l n I I l I i Illrllrll39l y w m b c r e m m Lam e r 1 a d C m g m Tu p m y t dd o r e y M m 3 t a dd m r f n m a m W n u 2 S 6 1 M 4m W 2 d m S 2 y 1 m22 l 0 that the solution equation is E Stable and unstable manifolds are the same very unusual note that this away e g point A on gure may loop the loop before settling in on one or other of the asymptotically Note how trajectories x05 starting near 1 0 spiral in to 1 0 trajectories which start far stable spiral points i1 27 6x oi EXfAM t It A gt X X eta 003 net GHQ WA ck 3w 24 auon moot m M we cf TN atvm dLli liet bktwL 1 Mn NotL How Fuicdawix LinMilk A 6 Lilo 3 hajcckovlg DJ 33 km A 0A admvij M quot OM or away 39l d Solut1on curves for d i y a 23 01g There are asymptotically stable equilibria at i10 and an unstable saddle at 00 Man 4 a m 0101 1 SW 1903 Qwes39 bn NJ ml MLW 3 L Jrlt LO39D Question What role does the stable manifold of the unstable saddle 0 0 play 28 Rabbits and foxes Goal To construct a model about how two species say rabbits and foxes might interact and to learn how to design and make parameter choices to achieve certain goalsl Let XT YT be populations of rabbitsfoxes measured in units of 1000 say and let T be time measured in years Left to themselves and in an environment of unlimited resources the rabbits will grow in population via Malthus7 lawi dX 1 E kX where you can determine k by measuring the time Tgr needed to double the population recall kTgT ln2i But if there are nite resources the habitat can only sustain a nite population say K st 1 is replaced by 2 game If we add the effects of culling or hunting which say removes a certain amount per year then the logistic equation 2 is modi ed to 3 gawkhltx gti lt1 gtm Please go back and study what happens 4h 4h 7 1a Wgt 1 3 cases lt l W 7TH equilibrium points Now we add a predator the fox Rabbits get eaten by foxes at a rate proportional to the probability they meet olt XY so we add to 4 km dT gt7h7DXY Foxes on the other hand will die out if they have no food source and we will suppose their only food source is rabbitsl They will be sustained if they meet enough rabbitsl Their population equation is 5 ika 7 EX 29 E is different from D because a fox may need several rabbits to sustain him for a given period Now equations 4 5 have lots of parameters k K h D kf and E all of which are either measurable or controllable We can measure 16 kf and for a given habitat K E and D can be estimated h and K are controllablei We can decide on how much hunting we allow and we can always increase K by providing more food for rabbits We do not want however to deal with 6 parameters And we don7t have to By choosing the units in which we measure rabbits the units in which we measure the fox population and the unit of time we can reduce the number of parameters in 45 from 6 to 31 Here is how we do it Let X X01 Y Yoy and T Toti 6 Put these changes of variables in the equations X0 dz 7 X01 To a 7 16on lt17 7 h7DX0YOIy Yo dy 7 To E 7 kfy17 Eon or multiplying across by 3 and respectively d kToz lt17 Ex 7 7 DToYozy dt K X0 7ka0y 1 7 Eon1 Now choose kTo 1 7 st t TLC kT is a dimensionless time unit When T 1 year t k year 373 yearsi Whent 1 T yearsi Next choose EXO 1 8 st when X 1 namely 1 thousand z Xio 1 E thousandi Conversely when I 1X thousandi Finally choose DTOYO DYDk 1 9 This simpli es the equation to 30 10 Iiwr2i 1yyv y d k 11 dig 7ay171 1fi We are going to explore how solnsi behave when we change the parameters 76 oar Example 1 Take 7 0 In nite resources for rabbits no huntingi Examine phase plane trajectories for 10 11 with a l 23 passing through the points 5 3 and 8 8 Show your results by printing screen with all 6 trajectoriesi Is there much qualitative difference between a l 2 3 What are the equilibrium points 18798 M Classify their stability properties by calculating 3F 3F 3G 3G 12 a7 biTy CE diTy 15112 062m 062m 062w a17y b71 cay d7a171 062m 12w 15115 15115 Find 17 7a d L ad 7 be at each equilibrium point and from LN 2229 pg 24 classify the stability properties of each equilibrium point Example 2 Let us introduce nite saturation 7 gt 0 and hunting and remove foxes altogether by setting y0 0 If y0 0 exactly yt 0 Then d j 171275 wltzezigtltzezgt l l 1 EmmyaEF um As lO 14 1740 We will assume 1 gt 457 What happens to the population of rabbits in this case if 10 lt 1 1 lt 10 lt 1 10 gt 1 Example 3 Reintroduce foxesi y0 gt 0 Take however the case of no hunting 6 0 d1 dt di dt 1771271y iay I 31 Take a 2 and examine what happens as you increase 7 Note that the equilibrium points are 1 0 y 0 1 l y 0 and 1 17 7 What happens if we increase 7 from 0 4 say to 0 6 to 1 Explore Al For 0 lt 7 lt 1 what are the stability properties of the 3 equilibrium points Important For 0 lt 7 lt 1 you will nd 11 7 7 is an asymptotically stable spiral What is its basin of attraction Draw in direction elds Hit S Example 4 Include everything d17 2 dyi 7 Eziyz i izy E777y171i Takea72 The equilibrium points are 1 0 1 0 1 ye What are 1 1 and ye What happens ye if 1 lt B 7 For 1 gt B 7 determine the stability properties of each of the equilibrium points Are they stable spirals stable nodes unstable saddle points etc Discuss the cases for 7 0 4 0 3 0 4 0 5 Determine the stability properties of all equilibrium points in the three cases Draw the stable manifold of 1 0 and the unstable manifold of 1 0 for each case What happens when 6 gt 7 32

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