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To Be Identified. Then the Flow Line at the Origin Under the ODE System Is 4.2. LINEAR FLOWS ON THE TORUS 89 to be identified. Then the flow line at the origin under the ODE system is a straight line of slope γ emanating from the origin and meeting the right edge of the unit square at the point (1; γ). But by the identification, we can restart the graph of the line at the same height on the left side of the square (at the point (0; γ). Continuing to do this, we will eventually reach the top of the square. But by the identification again, we will drop to the bottom point and continue the line as before. See Figure 6 In essence, we are graphing the flow line as it would appear on the unit square. When we pull this square out of the plane and bend it to create our torus T, the flow line will come with it. Suppose γ ∈~ Q. What can we say about the positive flow line? Figure 6. Wrapping the flow onto T, the unit square with oppo- site sides identified. Proposition 4.16. if γ !2 is irrational, then the flow is minimal. If γ , = !1 ∈ Q then every orbit is closed. Proof. Choose a point x = (x1; x2) and let Sx be a Poincare section (a first- return map for the flow.) Then, along Sx, the Poincare map is precisely Rγ (see the left side of Figure 7). Since γ ∈~ Q, Ox ∩ Sx is dense in Sx. 2 So let y be some arbitrary point in T . We claim that arbitrarily close to y is a point in Ox. Indeed, choose > 0 and construct an open -neighborhood of y in 2 T , N(y). Then ∃z ∈ Sx and a t > 0, such that y = !t + z But then ∃u ∈ Ox ∩ Sx, where d(u; z) < . And then v = !t + u ∈ N(y) and v ∈ Ox. See the right side of Figure 7 Remark 4.17. For a Poincare first return map, if every orbit intersects the Poincare section, we call the section a global section. Otherwise, it is a local section. !2 Now let γ . Then Rγ is a rational rotation on Sx. Then t0 0 = !1 ∈ Q ∃ > such that x = !t0 + x mod 1. Thus !t0 = 0 mod 1 and Ox is closed. But then 2 for any other point y ∈ T , we have y − x = z mod 1. Thus x = y − z mod 1 and y − z = !t0 + y − z so that y = !t0 + y mod 1. Hence Oy is closed. Exercise 108. Draw enough of the flow lines passing through the origin to indicate general behavior for the following values of γ: √ √ 1 1 5 1 5 (a) γ 1; (b) γ 2; (c) γ ; (d) γ + ; (e) γ − : = = = 2 = 2 = 2 90 4. RECURRENCE 2 Figure 7. The orbit of x is dense in T . Exercise 109. Show that a linear map on the real line f(x) = kx corresponds to a continuous map on S1 iff k is an integer. Graph the circle map on the unit interval (with the endpoints identified) that corresponds to f(x) = 3x. Identify all fixed points. Can you find a period-3 orbit? Hint: Remember that two points in the real line correspond to the same point on the circle is their distance apart is an integer. This is the parameterization that we will always refer to by default. 2 2 Exercise 110. Let h ∶ R → R be a linear map so that h(x) = Ax, where a b A and a; b; c; d . Do the following: = c d ∈ Z 1 1 (a) Show h induces a map on the standard two torus T = S × S . Hint: Two vectors in the plane are in the same equivalence class on the torus (correspond to the same point on T if they differ by a vector with integer entries.) (b) Describe, as best as you can, the dynamics of the linear maps on T correspond- ing to 0 1 1 1 2 1 i:A = ; ii:A = ; iii:A = : −1 0 0 1 1 1 4.2.1. Application: Lissajous Figures. We can look at toral flows in an- 4 other way: Via an embedding of a torus into R with simultaneous rotations in the 1 2 two orthogonal coordinate planes. To see this, think of S ∈ R as a circle of radius r centered at the origin. Then we can represent T as the set 4 2 2 2 2 2 2 T = (x1; x2; x3; x4) ∈ R V x1 + x2 = r1; x3 + x4 = r2 : 2 Now recall a continuous rotation in R is given by the linear ODE system x_ = Bαx, 0 α where B is the matrix whose eigenvalues ±αi are purely imaginary. Do −α 0 2 this for each pair of coordinates (each of two copies of R ) to get the partially 4 uncoupled system of ODEs on R , ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ x_1 ⎥ ⎢ 0 α1 0 0 ⎥ ⎢ x1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x_2 ⎥ ⎢ −α1 0 0 0 ⎥ ⎢ x2 ⎥ x_ = Ax; ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ : ⎢ x_3 ⎥ ⎢ 0 0 0 α2 ⎥ ⎢ x3 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x_ ⎥ ⎢ 0 0 α 0 ⎥ ⎢ x ⎥ ⎣ 4 ⎦ ⎣ − 2 ⎦ ⎣ 4 ⎦ 4.2. LINEAR FLOWS ON THE TORUS 91 We will eventually see that this is the model for the spherical pendulum.For now, Figure 8 shows two planes, which under uncoupled rotations would leave concentric 4 circles invariant. Joining these two planes only at the origin comprises R (hard to 2 2 visualize, no?) The 2-torus T is a subspace of R invariant under this flow. 4 Figure 8. The x1x2 and x3x4-planes as subspaces of R , Some notes: 2 2 2 2 2 2 ● The two circles x1 + x2 = r1 and x3 + x4 = r2 are invariant under this flow. ● We can define angular coordinates on T via the equations x1 = r1 cos 2π'1 x2 = r1 sin 2π'1 x3 = r2 cos 2π'2 x4 = r2 sin 2π'2: αi Then, restricted to these angular coordinates and with !i = 2π , i = 1; 2, we recover '_ 1 = −!1; '_ 2 = −!2: Motion is independent along each circle, and the solutions are 'i(t) = !i(t − t0): If α2 !2 , then the flow is minimal. ● α1 = !1 ∈~ Q Exercise 111. Do the change of coordinates explicitly to show that these two interpretations of linear toral flows are the same. Now, for a choice of ! and r1 = r2 = 1, project a solution onto either the (x1; x3) or the (x2; x4)-planes. The resulting figure is a plot of a parameterized curve whose two coordinate functions are cosine (resp. sine) functions of periods which are rationally dependent iff ! is rational. In this case, the figure is closed, and is called a Lissajous figure. See the figure below for the case of two sine functions (projection onto the (x2; x4)-plane, in this case), where !1 = 2 and !2 = 3. Q. What would the figure look like if !1 and !2 were not rational multiples of each other? Exercise 112. Draw the Lissajous figure corresponding to the x2x4-planar projection of the toral flow when !1 = 2 and !2 = 5. For these same values, draw 2 the orbit also on the torus using the unit square in R representation (with sides identified appropriately), and as well on the \surface of a doughnut" representation 3 in R . 92 4. RECURRENCE A nice physical interpretation of this curve is as the trajectory of a pair of uncoupled harmonic oscillators, given by x¨1 = −!1x1 x¨2 = −!2x2: Toral flows also show up as a means to study a completely different class of dynamical system; a billiard. We will eventually focus on some more general fea- tures of this class of dynamical systems. For now, we will introduce a particularly interesting example, where a linear flow on a torus allows us to answer a rather deep question. 4.2.2. Application: A polygonal Billiard. Consider the unit interval I = [0; 1] with two point masses x1 and x2, with respective masses m1 and m2 respec- tively, free to move along I but confined to stay on I. Without outside influence, these point masses will move at a constant, initial velocity. Eventually, they will collide with each other and with the walls (see left side of Figure 9). Assume also that theses collisions are elastic, with no energy absorption or loss due to friction. Here, elastic means that, upon a wall collision, a point mass' velocity will only switch sign. And upon a point mass collision, the two point masses will exchange velocities. For now, assume that m1 = m2 = 1. Figure 9. The state space (right) of two point masses in the unit interval (left). 4.2. LINEAR FLOWS ON THE TORUS 93 2 The state space in R is 2 T = (x1; x2) ∈ R V 0 ≤ x1 ≤ x2 ≤ 1 : Here, T is the region in the unit square above the diagonal line which is the graph of the identity map on I (right side of Figure 9). The edges of the region T are included; since the point masses have no size, they can occupy the same position at the point of contact. An interesting question to ask yourself is: How does the state space change if the point masses had size to them? Now, given an initial set of data, with initial positions and velocities v1 and v2, respectively, what is the evolution of the system? The answer lies in the study of these types of dynamical systems called billiards. Evolution will look like movement in T . A point in T comprises the simultaneous positions of the two particles, and movement in T will consist of a curve parameterized by time t.
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