Non-Linear Dynamics Homework Solutions Week 2

Transcription

1 Non-Linea Dynamics Homewok Solutions Week Chis Small Mach, 7 Please me at smach9@evegeen.edu with any questions o concens eguading these solutions. Fo the ececises fom section., we sketch all qualitatively diffeent vecto fields that aise as is vaied, show that tanscitical bifucations occu at citical values of to be found and sketch the bifucation diagam of vs... ẋ = ( ) Fist we locate the fied points of ou system as a function of. To do this we facto ou epession fo ẋ. ẋ = ( ) = ( ( + )) If =, then ẋ = and = is the only fied point and is unstable. Othewise,, and so ẋ has zeos at = and = ( )/ (Note that we need to divide by ). Since ẋ is a quadatic polynomial when, its gaph takes the fom of a paabola which intesect the ais at the fied points we found above. Since is the leading tem of the polynomial ẋ, the sign of detemines whethe the paabola points up o down. Peicing togethe this infomation we obtain the vecto fields below Figue : Qualitatively Diffeent Vecto Fields To see that the fied point point = switches stability at a citical value (ie that a tanscitical bifucation occus), we could simply look at the diagams above, o we could see

2 that if we let ẋ = f(), then f () = + ( ). Evaluated at the fied point in question, this educes to f () = ( ). Thus we see that fo <, f () < indicating that the fied point is stable. Fo >, f () > indicating that the fied point is unstable. Hence, we have a tanscitical bifucation at c =. We give Figue fo a -D plot of ẋ vs and and the bifucation diagam Figue : -D plot of ẋ along with Bifucation Diagam.. ẋ = ( e ) As with.., we stat looking fo the fied points of the system. The point = will always be fied. We can also have ẋ = if e =, which occus iff ln =, so = ln is ou othe fied point. Note that ln is undefined if <. When is such that < <, ln <. When = ln = and when > ln >. Each of these egions coesponds to a qualitatively diffeent vecto field, each of which we plot in Figue Figue : Qualitatively Diffeent Vecto Fields Again, we could esot to a gaphical analysis to convince us that a tanscitical bifucation occus at c =, but as with.., we shall pesent a moe igoous aguement. Fist, we find that if ẋ = f(), the f () = ( + )e, which simplifies to f () = when =. This is positive fo > and negative fo < indicating that stability of the fied point switches at c =. Figue shows a -D plot of ẋ as a function of and, as well as a bifucation diagam of the system... This poblem is as those fom Section., only you must show that a pitchfok bifucation occus (instead of a tanscitical) and detemine whethe it is a supecitical o subcitical bifucation. ẋ = + +

3 Figue : -D Plot of ẋ and Bifucation Diagam Te begin we facto to get ẋ = ( + /( + )). Thus ẋ = = = o + /( + ) =. The latte condition implies that =. Fo <, the fied points ae not eal, fo = we get the same fied point we always had, and fo > we have two new fied points. Thus we have a bifucation occuing at c = and what s moe is that it must be some kind of pitchfok bifucation. When we plot these solutions, it is clea that we have a subcitical pitchfok bifucation. See Figues and Figue : Qualitatively Diffeent Vecto Fields Figue : -D Plot of ẋ and Bifucation Diagam Fo the emaining poblems of this section we ae to detemine what kind(s) of bifucation occus, whee they occu and sketch bifucation diagams of each systems fied points.

4 Figue 7: Plot of Fied Points vs Figue 8: Bifucation Diagam.. ẋ = + Factoing this equation, it is easy to see that fied points will lie along = and = (/) (when ). When we plot these two solution sets in Figue 7, we can see that thee apeas to be a tanscitical bifucation. To check that this is the case, note that if ẋ = f(), then f () = (+). When =, this simplifies to f () =. This implies that this constant fied point goes fom being unstable fo < to stable fo > (this follows fom the signs of the deivatives thee). This implies that at c = thee is a tanscitical bifucation. Figue 8 shows a complete bifucation diagam...7 ẋ = e To begin, we find that all fied points must satisfy = ± ln/ by solving fo in the equality ẋ =. Note that this epession is only defined when > since othewise we might not be able to divide by and cetainly wouldn t be able to take the natual log of divided by. Also note that unde these conditions ln / > making the squae oot well defined. Fo, ẋ since the maimum value that the tem e can take on is. By gaphing this solution cuve, and consideing ou discussion of the vlues of coesponding to fied points, we see that the bifucation occuing at the citical value of c = is of the blue-sky vaiety (see Figue 9 fo bifucation diagam and Figue fo a D plot of ẋ vs. and, which justifies the stabilities as given).

5 fied Figue 9: Bifucation Diagam Figue : D Plot of ẋ vs. and.. ẋ = ( + + ) By factoing ou epession fo ẋ, we see that fied points occu iff = o + ( )/( + ) =. If the latte holds then = ± /( + ). This implies that must be such that <, since othewise /( + ) <, in which case the squae oot will not poduce eal outputs. As fom the left the solutions = ± /( + ). This indicates that the numbe of solutions goes fom to as is vaied continuously fom just above zeo to just below it. We conclude that a subcitical pitchfok bifucation occus at the the citial value c =. We also get some kind of bizzae bifucation at =, which might not have a name yet. I m gonna call it a ω- ski-slope bifucation, since two infitely high and steep ski slopes come out of nowhee as becomes geate than. See Figues and fo bifucation diagam and D plot of ẋ vs and...7 We nondimensionalize the system given by the logistic equation Ṅ = N( N/K), N() = N. a) Find the dimension of the paametes, k and N. The paamete K must be in tems of population elements (such as people o tiges o whateve population is in question) so that the N/K tem is unitless (this is necessay fo N/K to make any sense. The paamete must make the dimensions of Ṅ be in tems of population elements pe unit time, so must have units of pe unit time since N( N/K) has units of populaton elements.

6 Figue : Bifucation Diagam fied Figue : D Plot of ẋ vs. and

7 b) Rewite the system in the fom d dτ = ( ), () =. To do this we ae going to define = N/K, so that the tem ( N/K) = ( ), which is pat of what we need. Net, we want to come up with a definition fo τ that gets id of the need fo in ou epession. Note that whateve choice we make, by the chain ule we must have that d dτ = d dn dn dt dt dτ, which evaluates to d/dτ = ( ) dt/dτ. We want to have dt/dτ =. Setting τ = t we get pecisely that. Since at τ = we have t = (we ae assuming hee that > ), = τ= = N/K τ= = N/K τ= = N /K. c) Find a diffeent nondimensionalization in tems of u and τ such that u = always. To do this we assume that N so that we can set u = N/N and τ = t. Using the chain ule as in pat b) we find that du/dτ = u( N /(Ku)). Note that this is indeed nondimensionalized, since N /K is unitless (since both K and N have the same units). Net we find that u = u τ= = N/N τ= = N/N τ= = N /N =, as it should. d) What ae the advantages of one nondimensionalization ove the othe? The st nondimensionalization is easie in that = and = ae the only fied points and don t vay given specific paametes. The nd nondimensionalization is easie in that the initial condition is always, which is only a possible tanslation if N... Conside the system given by ẋ = + a, a, R. a) Sketch all qualitatively diffeent bifucation diagams that can be obtained by vaying a. Since ẋ = ( a ), we get that fied points occu at = and = a ± a +. Thus when =, Thee ae two fied points, one at = and the othe at = a. Also, the vete of the paabolic shape fomed by the paametic plot of = a ± a + occus when the tem afte the ± sign is zeo, which occus when = a. This cuve in (, a) coesponds to blue sky bifucations, ecept when = a = which coesponds to a supecitical pitchfok bifucation. All othe places in the phase space whee = ae tanscitical bifucations. Finally we see that the qualitively diffeent bifucation diagams of vs can be classified accoding to whethe a <, a = o a >. Figue shows thee dimensional plots of ẋ vs and fo a values in each of these egions, and Figue shows the coesponding bifucation diagams in each of those egions. b) In the (, a) plane plot the egions which coespond to qualitatively diffeent classes of vecto fields. Identify and classify these egions. We ve aleady done the gunt wok, all we need to do is make the gaph. See Figue. 7

8 Figue : D Plot of ẋ vs. and fo a <, a = and a > (fom left to ight, and then down) Figue : Bifucation Diagams fo a <, a = and a > (fom left to ight) 8

9 Tanscitical Bifs... Pitchfok Bif a Blue Sky Bifs. a Figue : (, a) Phase Plane 9