Differential Equations on the two dimensional torus 1 Constant ...

November 1, 2010 9b-1<strong>Differential</strong> <strong>Equations</strong> on the two dimensionaltorus1 Constant vector fields on the torusWe will now study some simple differential equations on a two-dimensionaltorus. These will provide interesting minimal sets.The 2−torus T 2 is the product S 1 × S 1 of two circles.Writing S 1 aswe getS 1 = {z ∈ C : | z | = 1},S 1 × S 1 = {(z 1 , z 2 ) ∈ C 2 : | z 1 | = | z 2 | = 1.There is an alternative viewpoint which is useful.Consider the quotient group R/Z; i.e., the set of equivalence classes in Runder the equivalence relation x ∼ y iff x − y ∈ Z. There is a natural mapπ : R → Z assigning to each real number x its equivalence class [x].Notice that the map φ : t → e 2πit is a surjective map from R onto S 1such that each pre-image φ −1 (z) is an equivalence class in R/Z. Thus, themap ψ([x]) = φ(x) gives a well-defined bijection from R/Z to S 1 .We can use the map ψ to define a topology on R/Z, so we have notionsof continuity, etc.Note that we can also think of the circle as a closed interval [a, b] with itsendpoints identified.We can now apply these ideas to the product T 2 = S 1 × S 1 . There is abijection from R/Z × R/Z onto T 2 . This defines a topology on T 2 . There isanother natural bijection between R/Z × R/Z and R 2 /Z 2 , and we can thinkof this latter quotient set as T 2 as well. We will use all of these objects.In particular, we think of T 2 as either S 1 × S 1 or R 2 /Z 2 .Now we wish to define differential equations on T 2 . We will use therepresentation R 2 /Z 2 .For this purpose, let f(x, y), g(x, y) be two real-valued functions of tworeal variables such that they are periodic of period 1 in both variables. Thatis,

November 1, 2010 9b-2f(x + 1, y + 1) = f(x, y), g(x + 1, y + 1) = g(x, y), ∀x, y.The planar vector field X(x, y) = (f(x, y), g(x, y)) is a map from R 2 →R 2 . The periodicity conditions imply that it induces a well-defined map fromR 2 /Z 2 to R 2 . We call this a vector field on the torus T 2 . The associatedsystem of differential equationsẋ = f(x, y)ẏ = g(x, y)is called a differential equation on T 2 .The solutions (or orbits) are obtained by taking the orbits in R 2 andprojecting them down to T 2 .Let us take a simple example.Consider the constant vector field in R 2X(x, y) = ω 1 ∂ x + ω 2 ∂ y ,where ω 1 , ω 2 are positive real numbers.The associated system of differential equations isẋ = ω 1ẏ = ω 2(1)The periodicity conditions are obviously satisfied, so we get a differentialequation on T 2 . This is called the constant or linear vector field on T 2 .Let’s us consider its orbits.The solutions to (1) in R 2 are the linesx(t) = ω 1 t + c 1 , y(t) = ω 2 t + c 2 .Each such line has slope ω 2ω 1and passes through the point (c 1 , c 2 ).On the torus T 2 we simply take (x(t), y(t)) mod 1.We have two main cases:1. 2ω 1is rational. Write it as 2ω 1= p in lowest terms with p, q positiveqintegers.Consider the orbit γ 0 in R 2 through (0, 0) (i.e., c 1 = c 2 = 0).

November 1, 2010 9b-3The solution in R 2 has the form( ) px(t) = ω 1 t, y(t) = ω 1 tqLet π : R 2 → T 2 denote the canonical projection.When t = q/ω 1 , we have x(t) = q, y(t) = p, so the projection π(γ 0 ) = γinto T 2 is a closed curve (topological circle) in T 2 .Since all the solutions are vertical tranlates of γ 0 we have that allsolutions in T 2 are periodic of the same period.2. ω 2ω 1is irrational.This case is more interesting. We will show that every orbit in T 2 isdense in T 2 . Hence, all of T 2 is a minimal set.Let us show that the orbit through (0, 0) is dense in T 2 . The analogousstatement for other orbits is similar.Consider the circle S 1 = π({x = 0}).There is a first return map F : S 1 → S 1 (also called Poincaré map)obtained by taking a point (0, y) and taking the point (1, F (y)) wherethe orbit through (0, y) hits the line {x = 1}.Let us compute F .The orbit through (0, y) is {(ω 1 t, ω 2 t + y), t ∈ R}.We getHence,ω 1 t = 1ω 2 t + y = F (y).t = 1/ω 1 , F (y) = y + ω 2 /ω 1So, letting α = ω 2ω 1we get that F (y) = y + α (mod 1).To show that the orbits of X are dense, it suffices to show that theiterates {F n (y), n ∈ Z} are dense mod 1

November 1, 2010 9b-4This is the number theoretic statement: for α irrational, andthenA = {nα + m : n ∈ Z, m ∈ Z},This elementary fact is proved as follows.(a) A is clearly an additive subgroup of R.A is dense in R. (2)(b) To show that A is dense, it suffices to show that 0 is an accumulationpoint of A.(c) For this latter fact, for each n ≥ 0, let m n ∈ Z be such thatx n = nα + m n ∈ [0, 1). Since α is irrational, the numbers x n areall distinct. Since [0, 1] is compact, there is a subsequence x ni of(x n ) which converges. In particular, if ɛ > 0 is arbitrary, there arepoints x n1 , x n2 such that| x n1 − x n2 | < ɛ.But then this difference x n1 − x n2 is in A ⋂ (−ɛ, ɛ). Since, ɛ isarbitrary, we have that 0 is an accumulation point of A as required.Let us now give an application of these ideas.Consider the mass spring system whose equation ismẍ + kx = 0where m is the mass of an object suspended vertically by a spring,k > 0 is the spring constant, and x = 0 represents the equilibriumposition.√Let ω = k. The associated first order planar system can be writtenmẋ = −ω yẏ = ω x(3)

November 1, 2010 9b-5Let us write this system (3) in polar coordinates.We havex = rcos(θ), y = rsin(θ)ẋ = ṙcos(θ) − rsin(θ) ˙θẏ = ṙsin(θ) + rcos(θ) ˙θ(4)So, we see that the simple polar coordinate systemṙ = 0˙θ = ω(5)is equivalent to (4).The solutions off (0, 0) arer = c > 0, θ(t) = ω t + θ 0 .Hence, the orbits in the plane off (0, 0) consist of the circles r = c andthe motion has constant rotational speed ω with period 2π ω .Now, consider two uncoupled springs whose equations in R 4 becomex˙1 = −ω 1 y 1y˙1 = ω x 1(6)x˙2 = −ω 2 y 2y˙2 = ω 2 x 2Using polar coordinates (r 1 , θ 1 , r 2 , θ 2 ) in R 4 in the obvious way, we getthe equivalent equationsr˙1 = 0θ˙1 = ω 1(7)r˙2 = 0θ˙2 = ω 2Hence all solutions in R 4 off the union of the two subspaces R 2 ×{(0, 0} ⋃ {0, 0}×R 2 lie on two dimensional tori r 1 = c 1 , r 2 = c 2 , and the motion on each is aconstant vector field. If ω 2 /ω 1 is rational, then all these orbits are periodic,and if ω 2 /ω 1 is irrational, then all these tori are minimal sets.

November 1, 2010 9b-62 Gradient vector fields on the torusTo specify a gradient vector field on the two dimensional torus, we need toconsider smooth potential functions on the torus T 2 = R 2 /Z 2 . These aresmooth functions ψ(x, y) such thatψ(x + 1, y + 1) = ψ(x, y) for all (x, y) ∈ R 2 .From the theory of Fourier series, we can write such functions as sums∞∑ψ(x, y) = a k cos(2πkx) + b k sin(2πk)k=0The simplest such examples are the trigonometric polynomialsN∑ψ N (x, y) = a k cos(2πkx) + b k sin(2πk)k=0for a positive integer N.Givena function ψ(x, y) with ψ(x + 1, y + 1) = ψ(x, y), we obtain itsgradient vector field byX(x, y) = (∂ x ψ(x, y), ∂ y ψ(x, y))We wish to give a simple geometric interpretation of this in a special case.First, we recall some concepts about surfaces of revolution in R 3 .Let (X, Y, Z) denote the coordinates of a point in R 3 .Let I = (a, b) be an open interval in R with a < b.Consider a C r parametrized curve η : I → R 3 in the XY -plane given bysuch thatandη(t) = (X(t), Y (t), 0)inf | X(t) |) > 0t∈Iinf |t∈I X′ (t) 2 + Y ′ (t) 2 | > 0

November 1, 2010 9b-7with r ≥ 1.Thus, the image of η is disjoint from the Y − axis, and the tangent vectorsto η never vanish.The surface obtained by revolving η about the Y −axis can be parametrizedbyγ(t, θ) = (X(t)cos(2πθ), Y (t), X(t)sin(2πθ))with 0 < θ < 1).Thus, the map γ : (a, b)×(0, 1) → R 3 is a C r embedding from (a, b)×(0, 1)into R 3 . Its image is the surface of revolution determined by η and the Y −axis.Next, consider the open unit box in R 2 given by (0, 1) × (0, 1). We usethis to parametrize an open dense set of a torus of revolution in R 3 as follows.Let 0 < r < R be two positive real numbers.Consider the embedding η from (0, 1) into the X − Y −plane given byη(t) = (R + r cos(2πt), r sin(2πt), 0) = (X(t), Y (t), 0)This maps (0, 1) onto the circle{(X, Y, 0) : (X − R) 2 + Y 2 = r 2 }with a point removed.Using this to parametrize the surface obtained by revolving η about theY −axis, we get the map Φ : (0, 1) × (0, 1) → R 3 defined byΦ(t, θ) = ((R + rcos(2πt)cos(2πθ), r sin(2πt), (R + rcos(2πt)sin(2πθ)).The image is an open torus of revolution with two circles removed. Theimage is dense in the torus of revolution.Consider the height function h(X, Y, Z) = Z from R 3 to R. This functionis defined as a C ∞ function on all of R 3 , and, in particular on the closure ofthe image of Φ.This gives a function ψ(t, θ) from (0, 1) × (0, 1) to R defined byψ(t, θ) = (R + rcos(2πt))sin(2πθ)Replacing t by x and θ by y, and still using ψ to denote the function, wehave

November 1, 2010 9b-8ψ(x, y) = (R + rcos(2πx))sin(2πy). (8)This formula can be extended to a doubly periodic (periodic in both xand y) function from R 2 to R which induces a function ˜ψ : T 2 → R.Let Σ denote the closure of the image of Φ. This is a representation of atorus of revolution in R 3 . The function h restricted to Σ is (by definition) aC ∞ function on Σ. Consider the level sets of h in Σ; i.e., the intersections of{h(X, Y, Z) = c} and Σ. These sets are one of the following types.1. isolated points2. disjoint union of two topological circles3. a topological figure eight4. topological circlesFor p ∈ Σ, let T p Σ denote the tangent plane to Σ at p.One can define a vector field on Σ by assigning to each point p ∈ Σ thevector in T p Σ obtained by projecting the vector (0, 0, 1) at p into T p Σ. Thisis defined to by the gradient vector field on Σ induced by the function h | Σ.Using the parametrization Φ : (0, 1) × (0, 1) → Σ this vector field can bepulled back to (0, 1) × (0, 1). This latter vector field can be extended to allof R 2 , and this is the gradient vector field of the function ψ above.Remark The function ψ is a real-valued function from R 2 to R withinfinitely many critical points. Its level sets in (0, 1) × (0, 1) are the preimagesby Φ of the level sets of h on the image of Φ.