EE222: Solutions to Homework 2

Transcription

1 Spring 7 EE: Solutions to Homework 1. Draw the phase portrait of a reaction-diffusion sstem ẋ 1 = ( 1 )+ 1 (1 1 ) ẋ = ( 1 )+ (1 ). List the equilibria and their tpes. Does the sstem have limit ccles? (Hint: Show that this is a gradient sstem.) Solution. B solving ẋ 1 =,ẋ =, we get that the sstem has three equilibria: p =(, ), p 1 =(1, 1) and p =( 1, 1). Linearizing the sstem at the equilibria, it is eas to see that p is a saddle, while p 1 and p are stable nodes. The unstable and stable spaces of p are given b: 1 = ( 1 < 1) and 1 + =, respectivel. In fact, if we denote the vector field of the sstem b f =(f 1,f ) T,then f 1 == f, 1 which implies that f is a gradient vector field. B integrating f 1 relative to 1, differentiating the obtained function with respect to and equating the result with f,wegetthatf = grad V,where V ( 1, )= 1 ( 1 + )+1 ( 1 + ) 1. = ( ) + (1 ) = ( ) + (1 ) Figure 1: The phase portrait of f in problem 1. Therefore, b eercise 5. from Homework 1, the sstem has no closed orbits. (Alternativel, one could observe that the divergence of f is 3( 1 + ) and use Bendison s theorem.) 1

2 The phase portrait of f is shown in Fig. 1.. Shown below are phase portraits of some vector fields in the plane. Determine which of them are correct and which are incorrect. Modif the incorect ones, not b deleting an orbits, but b changing their stabilit tpe or adding new orbits. Solution. Phase portrait (a) is correct, but here s how the given trajectories can be etended. (b) (c) Let P be the region bounded b four saddle connections. We claim that P must contain an eqilibrium. Since P is invariant, for an p P, ω(p) andα(p) can be: (i) an equilibrium, or: (ii) a limit ccle or: (iii) a finite connected union of saddle connections. If (ii), then the region bunded b the limit ccle must contain an equilibrium. If (iii), appl the same reasoning to the region Q bounded b these saddle connections (if Q = P, reverse the time).

3 3. Draw phase portraits of the following 1-dimensional sstems as μ changes: (a) ẋ = μ 3 (b) ẋ = μ α +μ 3 5,fordifferentα. Solution. (a) Let f μ () =μ 3. The equilibria are (μ) =, 1 (μ) =μ and (μ) = μ. B looking at f μ ( i(μ)) for i =, 1,, we see that μ = is the bifurcation value and obtain the following bifurcation diagram (Fig. ). 1 Figure : Problem 3.(a) (b) Let f μ () =αμ +μ 3 5.Thesolutionstof μ () =are: 1 (μ) =, (μ), 3 (μ), (μ) = (μ) and 5 (μ) = 3 (μ), where: (μ) = μ + μ 1+α, 3 (μ) = μ μ 1+α. We also have: f μ() =αμ +6μ 5. To investigate stabilit of equilibria of ẋ = f μ (), we need to consider the following cases: Case 1: α< 1 (Fig. 3). The onl equilibrium is 1 (μ) and since f μ ( 1(μ)) = αμ,itisstableforμ. Ifμ =, then f μ () = 5, which also has a stable equilibrium at. Case : 1 <α<. The equilibria are 1 (μ),..., 5 (μ), if μ, and onl 1 (μ) ifμ<. B checking the sign of f μ ( i(μ)), for i =1,..., we obtain: if μ, the onl equilibrium 1 (μ) =isstable; if μ>, then: 1 (μ), (μ), (μ) are all stable, and 3 (μ), 5 (μ) are unstable. Case 3: α = 1 orα = (Fig. 5). 3

4 μ Figure 3: Problem (b), Case μ 5 Figure : Problem (b), Case. = 3 (μ) = μ 1 μ 1 μ = 5 = Figure 5: Problem (b), Case 3, cases α = 1 (left)andα = (right).

5 Case : α>. The onl equilibria are 1 (μ), (μ) and (μ) = (μ), for all μ R. B checking the sign of f μ at these points we obtain that 1(μ) is unstable, while the other ones are stable for all μ (Fig. 6). 1 μ Figure 6: Problem (b), Case.. Let f be a smooth vector field on the annulus A = {(, ) R :1 + }. Assume f points inward along the boundar of A, andthatforever α π, the radial segment (in polar coordinates) S α = {(r, θ) :1 r, θ = α} is a local cross section in the sense defined in class; that is, at ever point p S α, the angle between f(p) and S α is not zero. (a) Let p S be an arbitrar point. Show that the orbit of f starting at p returns to S after some positive time. (b) Show that ever continuous function φ :[1, ] [1, ] has a fied point. (c) For p S,letψ(p) bethepointoffirstreturntos of the orbit starting at p. Show that ψ : S S has a fied point. (d) Show that there eists a closed orbit of f in A. Solution. (a) First of all, observe that since f points inward along the boundar of A, A is positivel invariant relative to the flow Φ t of f. Let Θ be the angular coordinate in the polar coordinate sstem on A. Let p =(r, θ) A be an arbitrar point and assume it belongs to S α,forsomeα. Then b assumption that S α is a local cross section for the flow Φ t,weobtain d dt Θ(Φ t(r, θ)). Since both Θ and Φ t are smooth on A, the left hand side is either alwas > oralwas<. d Assume the former. Since A is compact, Θ(Φ dt t(r, θ)) has a minimum m>ona. B the Fundamental Theorem of Calculus, Θ(Φ t (r, θ)) Θ(r, θ) mt 5

6 which becomes greater than π for sufficientl large t >. Therefore, the orbit of f starting at p S returns to S after some positive time. (b) If φ(1) = 1 or φ() =, there is nothing to prove. So assume φ(1) > 1andφ() <, and define g(s) =φ(s) s. Theng is continuous, g(1) >, and g() <. Therefore, b the Intermediate Value Theorem, there eists a point s (1, ) such that g(s )=. Clearl,s is a fied point for φ. (c) Observe that b the Implicit Function Theorem ψ is a smooth, hence continuous function. Therefore, since S =[1, ] {} [1, ], we can take ψ = φ and use part (b). (d) Let p S be a fied point of ψ from (c). Then, b definition of ψ, p = ψ(p) =Φ τ (p) for some τ>, i.e., the orbit of p is closed. 5. Consider the following parametrized famil of differential equations: ẋ = ẏ = μ 1 + μ + 3. (a) What does the flow near (, ) look like when μ 1 = μ =? (b) What bifurcations occur at μ 1 =andatμ =(forμ 1 < )? (c) Use Bendison s theorem and inde theor to rule out parameter regions where there are no periodic orbits. (d) Conjecture the full bifurcation diagram. Solution. (a) For the phase portrait when μ 1 = μ =, see Fig. 7. (b) The equilibria of the sstem are: p (μ) =(, ), p 1 (μ) =( μ 1, ), p (μ) =( μ 1, ), where μ =(μ 1,μ ). At p (μ), the eigenvalues of the linearization are [μ ± μ +μ 1]/, so: μ 1 < μ :ifμ >, then p (μ) is a source (Fig. 8), and if μ <, it is a sink (Fig. 9); μ <μ 1 < : if μ <, then p (μ) is a stable node (Fig. 1), and if μ >, then it is an unstable node (Fig. 11); if μ 1 > andμ R, thenp (μ) is a saddle. At p 1, (μ), the eigenvalues of the linearization are: μ 1 + μ ± (μ 1 + μ ) 8μ 1. 6

7 = = Figure 7: The phase portrait in problem 5 for μ 1 = μ =. = m1 = 16 = m1 + m + 3 m = Figure 8: The phase portrait in problem 5(b) for μ 1 < μ / andμ >. 7

8 = m1 = 16 = m1 + m + 3 m = Figure 9: The phase portrait in problem 5(b) for μ 1 < μ / andμ <. = m1 = 16 = m1 + m + 3 m = Figure 1: The phase portrait in problem 5(b) for μ / <μ 1 < andμ <. 8

9 = m1 = 16 = m1 + m + 3 m = Figure 11: The phase portrait in problem 5(b) for μ / <μ 1 < andμ >. Since μ 1 <, both p 1 (μ) andp (μ) are saddles. To summerize, at μ 1 =,whenμ >, two saddles (± μ, ) and the unstable equlibirium point (, ) merges into the unstable equlibrium (, ). At μ 1 =,whenμ <, two saddles (± μ, ) and the stable equlibirium point (, ) merges into the unstable equlibrium (, ). i.e. a pitchfork bifurcation appears. At μ =,whenμ 1 <, the eigenvalue of the linearization at (, ) cross the jw-ais (i.e. from C o when μ < toc o + when μ > ) and not through the origin. In fact, a stable limit ccle and an unstable equlibrium arise from a stable equilibrium, i.e. a (supercritical) Hopf bifurcation appears. (c) Since the divergence of f is div f = μ,wegetthatdivf forμ, so f has no closed orbits when μ. It is not difficult to check that if μ 1, the onl equilibrium, namel,, is a saddle, hence has inde 1, which guarantees noneistence of closed orbits (recall that the inde of a smooth vector field on the disk bounded b a closed orbit is 1). Therefore, f has no closed orbits when μ 1 orμ. 9

10 6. Let f : R R be a diffeomorphism; in other words, f is invertible, and both f and f 1 are smooth maps. Let p,p 1,...,p N 1 be a period N ccle of f; thatis,p k = f k (p )for1 k N 1andf N (p )=p. Prove that the matrices Df N (p k ) have the same spectrum for k =,...,N 1. Solution. B the Chain Rule, Df N (p k )=A k 1 A k...a A N 1 A N...A k+1 A k, where A i = Df(p i ). Since f is a diffeomorphism, each matri A i is invertible, so we get Df N (p k 1 )=A 1 k 1 DfN (p k )A k 1. Therefore, for all k =1,,...,N, Df N (p k 1 )anddf N (p k ) are similar, thus have the same spectrum. 1