Solution to Homework #4 Roy Malka

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1 1. Show that the map: Solution to Homework #4 Roy Malka F µ : x n+1 = x n + µ x 2 n x n R (1) undergoes a saddle-node bifurcation at (x, µ) = (0, 0); show that for µ < 0 it has no fixed points whereas for µ > 0 it has two fixed points, one stable and one unstable. Draw the bifurcation diagram for F µ. The FPs satisfies the following equation: x n = x n + µ x 2 n 0 = µ x 2 x R (2) the solutions are x = ± µ, since x R, there are two FPs for µ > 0, and no FP for µ < 0, and for µ = 0 there is one FP, thus this is a bifurcation point. D x F µ = 1 2x hence, x = µ is stable (λ < 1), and x = µ is unstable (λ > 1). Note that for µ 1 both solutions are unstable (and not as appears in the figure, this is from the condition for stability 1 < 1 2 µ < 1 µ > 0 and µ < 1 which is violated for µ 1) Figure 1: Saddle-Node bifurcation diagram 1

2 2. Consider a general one dimensional family of maps F µ (x) : R R. Find conditions under which F µ (x) undergoes a saddle-node bifurcation at some value (x, µ) = (x, µ ): define G(x, µ) = F µ (x) x, and using the implicit function theorem find conditions under which G(x, µ) = 0 has a unique parabola like solution G(x, µ(x)) for µ > µ (and µ µ and x x small). Verify that (1) indeed satisfies these conditions. the condition for FP is G(x, µ ) = 0. the condition for bifurcation point is x G(x, µ ) = 0. In order to have a parabola like solution we need to have G(x, µ(x)) = 0 at small neighborhood around (x, µ ), thus, if µ G(x, µ ) 0 then by the implicit function theorem there exists µ(x) s.t. G(x, µ(x)) = 0 holds in some small neighborhood of (x, µ ). In addition the bifurcation point, which is an extremum point of the curve G(x, µ(x)), should be a non-degenerate one, that is either minima or maxima, thus it should be d2 µ(x) dx 2 = 2 x 2 G(x,µ ) µ G(x,µ ) 0 (using implicit differentiation). Hence, the additional requirement is 2 G(x, µ ) 0. Specifically x 2 for the curve to exist for µ > µ, we need the extremum point to be a minimum, thus 2 G(x, µ )/ x 2 µ G(x, µ ) < 0. In the model form the previous question: µ G(x, µ ) = G(x, µ ) = 2 < 0, with a negative ratio between the two terms, x 2 as required. 3. Show that the flow: f µ : ẋ = µx x 2 x R undergoes a transcritical bifurcation at (x, µ) = (0, 0); show that the stability of the two fixed points is interchanged at the origin. Draw the bifurcation diagram for f µ. Such a bifurcation occurs when symmetry/modeling constraints imply that x = 0 is always a solution. Optional: You may want to verify that by breaking this symmetry assumption we are back to the saddle-node case. You may want to 2

3 derive general conditions for a flow f µ having this symmetry to have a transcritical bifurcation. x(µ x) = 0 x = 0, and x = µ are solutions. The linear stability is determined by D x f µ = µ 2x so x = 0 is stable for µ < 0 and unstable for µ > 0, while, x = µ (D x f µ x=µ = µ) is unstable for µ < 0 and stable for µ > 0, thus, it is clear that the FP interchange stability at the origin (for µ = 0 the two FPs unite). Figure 2: Transcritical bifurcation diagram breaking the symmetry: f µ : ẋ = µx x 2 + ε x R In this case x=0 is no longer a solution, and the FPs are: x 1,2 = µ 2 ± 1 2 µ2 + 4ε = µ 2 (1 ± 1 + 4ε/µ 2 ) for ε > 0 the solutions exist for all µ, so there is no bifurcation (Fig. 3 Left). For ε < 0 the solutions exist if µ > 4ε (saddle-node bifurcation, (Fig. 3 Right)). 3

4 Figure 3: perturbation of Transcritical bifurcation diagram: left ɛ > 0 and right: ɛ < 0 General Conditions for transcritical bifurcation: We require that ((x, µ )) to be a FP ( G(x, µ ) = 0 ) and that it is also a bifurcation point ( x G(x, µ ) = 0 i.e., non-hyperbolic FP). The difference between this bifurcation (transcritical bifurcation) and the saddle-node bifurcation is that x = 0 is a solution for all µ then µ G(x, µ ) = 0. Therefore, we will characterize the curve of the other solution, namely, the one defined by G(x, µ) = xh(x, µ) = 0, for x 0. Hence we get { G(x;µ) }, x 0, H(x, µ) = x (0, µ) x = 0. G x Since we want H(x, µ(x)) = 0 to be defined, we need µ H(x, µ ) 0 (Implicit function theorem) by the above definition of H(x, µ) we get that 2 µx G(x, µ ) 0. In addition, in order for the solution µ(x) to be locally linear dµ(x) dx = x H(x,µ ) µ H(x,µ ) 0 implies that 2 x 2 G(x, µ ) 0 4. Consider the tent map: x n+1 = F (x n ) { ax x 1/2 F (x) = a(1 x) x > 1/2 x [0, 1], a [0, 2]. (3) (a) Find the fixed points of F and their stability. 4

5 (b) Show graphically convergence to a stable fixed point and divergence from an unstable fixed point (choose apropriate a values). (c) Draw the graph of F 2 (x) for a = 2. How many fixed points it has? How about F n (x)? (a) for a (0, 1): the only FP is x = 0, since this is the only intersection of F with the identity map. (for x = ax it is clear, for x = a(1 x) we get that there is a unique FP at x = a/(1 a), which is less than 1/2 for a < 1). The stability is D x F = a < 1 x = 0 is stable. for a (1, 2] the FP at the origin is unstable since D x F = a > 1. The second FP x = a/(1 a) linear stability gives D x F = a > 1, hence it is also unstable. For a = 0 the map is identically zero, so still one stable FP, x = 0, However, for a = 1 there is an interval of FPs, x [0, 1/2] are all FP, which are non-hyperbolic, and non is stable, and we can see that at the edge of the domain, for x = 1/2 + ɛ we get F (x) = 1 x = 1/2 ɛ, thus it is not even stable from one direction. (b) See Fig. 4 Left for convergence and Fig. 4 Right for divergence. (c) In Fig. 5 we can see that F 2 (x) has 4 fixed points. Form numerical experimentation F n (x) has 2 n fixed points. But this is clear when we notice that the tent map F (x) is a double cover of the unit interval, thus, it has two crossing of identity, and that F F generates 4 covers of the interval, thus it has 4 fixed points, etc. Bonus questions: 1. (20pts) Solve numerically the following equation at λ = 0.1, 0.5 : x 5 + λx 3 + 2λx = 0 5

6 Figure 4: Iteration of Tent map: (Left) a < 1 the orbit converge to zero. (Right) a > 1 the orbit diverge from x = x/(1 + a) Figure 5: F 2 (x) (compare: perturbation method, continuation method and brute force numerical root finding) Using brute force numerical root finding, that find zeros based on zero crossing (zero knowledge procedure that only assumes that the function is continuous). For λ = 0.1, x 0 = while for λ = 0.5, x 0 = , this is quit robust in this case, since the other roots are complex, but for each λ, only one solution was found, while there are 5 roots (4 are complex). 6

7 I am focusing on the real root, finding the complex roots is just some more algebra (simply solving it twice when x = a + ib etc ). Perturbation method: lets do an expansion in λ λ 0 : the equation is x = 0 x 0 = 1 λ 1 : the equation is x 5 + λx 3 + 2λx = 0, using x = 1 + aλ, 0 = 1 + 5aλ + O(λ 2 ) + λ( 1 + aλ) 3 + 2λ( 1 + aλ) = 5aλ λ+2λ+o(λ 2 ) a = 1/5 x 0 = 1 1/5λ+O(λ 2 ) λ= λ 2 : the equation is x 5 +λx 3 +2λx 2 +1 = 0, using x = 1 1/5λ+bλ 2, so ( 1 + aλ + bλ 2 ) 5 + λ( 1 + aλ + bλ 2 ) 3 + 2λ( 1 + aλ + bλ 2 ) = 0, where we need only terms of order λ 2 5bλ 2 10a 2 λ 2 + 3aλ 2 4aλ 2 = 0 5b 10a 2 a = 0 b = (10a 1)a/5 = 3/25 so x 0 = 1 1/5λ 3/25λ 2 + O(λ 3 )) λ= We see that we are getting closer and closer to the solution obtained by the brute-force numerical computation. The continuation is a way to utilize the perturbation method to not so small perturbation, by using the expansion in λ but for increasing values of λ, where at each step, we use the the result from the previous step for initialization. 2. (5pts) Find a 1 d map which cannot be the return map of any smooth flow on an orientable smooth 2 d manifold. The basic requirement, regardless of dimension is continuity of the map, since it is a time integral of a continuous vector field. In the case of 2 d flow, the return map must be monotonic (continuous and 1-1), this results from the uniqueness of solutions of the ODE and the Jordan curve theorem. Thus a map like the tent map can not be a return map of a 2 d flow. However, the Jordan Curve theorem is a topological property of the plane. So this is just a partial answer, since the question is about any orientable smooth 2 d manifold and not just on R 2. But while Jordan 7

8 curve does not divide orientable smooth manifold to interior (bounded) domain and exterior domain, (thus, the theorem does not hold on the tours for example) it does divide the manifold, that is, we can not connect two points on the two sides of the line without crossing the line (notice that we lose this property in a non-orientable manifold like Mobius band). 3. (5pts) Give an example of a diffeomorphism f : R R such that f(0) = 0, every orbit of the differential f (0) (i.e. of the linearized map) is bounded, and every orbit of f different then the origin is unbounded. The requirements are the following: Diffeomorphism: f(x) is a C r (R) (r 1) function, and f 1 (x) is also a C r (R) function. f(0) = 0 orbits of linearized (x = 0) system are bounded: n, x R, (f (0)) n x < M <. non-zero orbits of system are unbounded: x R \ {0}, M < n 0 (M) < n, s.t. f n (x) > M or f n (x) > M We can choose f(x) = arctan(x) (choosing the ( π/2, π/2) branch): (f : R ( π/2, π/2), f 1 : ( π/2, π/2) R) Thus f is diffeomorphism: f (x) = 1/(1 + x 2 ) is continuous on R, and (f 1 ) (x) = 1/cos 2 (x) is continuous on ( π/2, π/2). f(0) = arctan(0) = 0 orbits of linearized (x = 0) system are bounded: f (x) = 1/(1 + x 2 ) x=0 = 1 thus, n, x R, (f (0)) n x = x < M <, holds taking M = x + ɛ, (ɛ > 0). non-zero orbits of system are unbounded: Clearly, this holds: x R\{0}, M < n 0 (M) < n, s.t. f n (x) > M or f n (x) > M, since the α-limit set of ( π/2, π/2) \ {0} is, that can be seen for example by the linearized system: (f 1 ) (x) = 1/cos 2 (x) > 1 for all x ( π/2, π/2) \ {0}. 8