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Elliptic Functions: Introduction Course

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Elliptic Functions: Introduction Course Elliptic functions: Introduction course Vladimir G. TKACHEV Department of Mathematics, Royal Institute of Technology Lindstedtsv¨agen 25, 10044 Stockholm, Sweden email: [email protected] URL: http://www.math.kth.se/˜tkatchev Contents Chapter 1. Elliptic integrals and Jacobi’s theta functions 5 1.1. Elliptic integrals and the AGM: real case 5 1.2. Lemniscates and elastic curves 11 1.3. Euler’s addition theorem 18 1.4. Theta functions: preliminaries 24 Chapter 2. General theory of doubly periodic functions 31 2.1. Preliminaries 31 2.2. Periods of analytic functions 33 2.3. Existence of doubly periodic functions 36 2.4. Liouville’s theorems 38 2.5. The Weierstrass function ℘(z) 43 2.6. Modular forms 51 Bibliography 61 3 CHAPTER 1 Elliptic integrals and Jacobi’s theta functions 1.1. Elliptic integrals and the AGM: real case 1.1.1. Arclength of ellipses. Consider an ellipse with major and minor arcs 2a and 2b and eccentricity e := (a2 b2)/a2 [0, 1), e.g., − ∈ x2 y2 + = 1. a2 b2 What is the arclength `(a; b) of the ellipse, as a function of a and b? There are two easy observations to be made: (1) `(ra; rb) = r`(a; b), because rescaling by a factor r increases the arclength by the same factor; (2) `(a; a) = 2πa, because we know the circumference of a circle. Of course, π is transcendental so it is debatable how well we understand it! 1 0.5 ±2 ±1 0 1 2 ±0.5 ±1 2 y2 Figure 1. Ellipse x + 4 = 1 The total arclength is four times the length of the piece in the first quadrant, where we have the relations 2 xb 1 y = b 1 (x/a) , y0(x)= . − − a2 1 (x/a)2 p − 5 p Thus we obtain a 2 `(a, b) = 4 1+ y0 (x) dx = Z0 substitutingp z = x/a 1 1 ez2 = 4a − dx = 1 z2 Z0 r − 1 1 ez2 = 4a − dx. 2 2 0 (1 ez )(1 z ) Z − − This is an example of an elliptic integral ofp the second kind. 1.1.2. The simple pendulum. How do we compute the period of motion of a simple pendulum? Suppose the length of the pendulum is L and the gravitational constant is g. Let θ be the angle of the displacement of the pendulum from the vertical. The motion of the pendulum is governed by a differential equation g θ00(t)= sin θ(t). −L In basic calculus and physics classes, this is traditionally linearized to g θ00(t)= θ(t), θ 0, −L ≈ so that the solutions take the form g θ(t)= A cos ωt + B sin ωt, ω = . L r We obtain simple harmonic motion with frequency ω and period 2π/ω. We shall consider the nonlinear equation, using a series of substitutions. First, note that our equation integrates to 1 2 2 θ0 ω cos θ = const 2 − Assume that the pendulum has a maximal displacement of angle θ = α; then θ0(α) = 0 so we have 1 2 2 θ0 = ω (cos θ cos α), 2 − and thus, θ0 = ω 2(cos θ cos α). ± − We take positive square root before the maximalp displacement is achieved. Integrating again, we obtain θ dφ 1 θ dφ ωt = = . 0 2(cos φ cos α) 2 0 2 α 2 φ Z Z sin sin − 2 − 2 Substituting p q φ θ sin 2 sin 2 2 α z = α , ρ = α , e = sin [0, 1), sin 2 sin 2 2 ∈ 6 we obtain ρ dz ωt = . 2 2 0 (1 z )(1 ez ) Z − − At maximal displacement θ = α we have ρ = 1, so the first time where maximal displacement p occurs is given by T 1 1 dz = , 2 2 4 ω 0 (1 z )(1 ez ) Z − − where T is the period of the oscillation (whichp is four times the time needed to achieve the maximal displacement). These are examples of elliptic integrals of the first kind. Finally, we should point out that actually computing the function θ(t) involves inverting the function ρ dz ρ . 2 2 → 0 (1 z )(1 ez ) Z − − 1.1.3. The arithmetic-geometricp mean iteration. The arithmetic-geometric mean of two numbers a and b is defined to be the common limit of the two sequences a ∞ and { n}n=0 b ∞ determined by the algorithm { n}n=0 a0 = a, b0 = b a + b (1.1) a = n n , b = a b , n = 0, 1, 2 ..., n+1 2 n+1 n n where bn + 1 is always the positive square rootp of anbn. Note that a1 and b1 are the respective arithmetic and geometric means of a and b, a2 and b2 the corresponding means of a1 and b1, etc. Thus the limit M(a, b) := lim an = lim bn (1.2) n n →∞ →∞ really does deserve to be called the arithmetic-geometric mean (AGM) of a and b. This algorithm first appeared in papers of Euler and Lagrange (sometime before 1785), but it was Gauss who really discovered (in the 1790s at the age of 14) the amazing depth of this subject. Unfortunately, Gauss published little on the AGM during his lifetime. 1 Theorem 1.1. Let a and b be positive real numbers. Then the limits in (1.2) do exist and coincide. Proof. We will assume that a b> 0, and we let an n∞=0 and bn n∞=0 be as in (1.1). The usual inequality between arithmetic≥ and geometric{ means,} { } a + b n n a b 2 ≥ n n p 1By May 30th, 1799, Gauss had observed, purely computationally, that 1 1 2 dt and 4 M(1, √2) π 0 √1 t Z − agreed to at least eleven (!) decimal places. He commented in his diary that this result ”will surely open up a whole new field of analysis” — a claim vindicated by the subsequent directions of nineteenth-century mathematics. The inverse of the above (indefinite) integral is the lemniscate sine, a function Gauss studied in some detail. He had recognized it as a doubly periodc function by the year 1800 and hence had anticipated one of the most important developments of Abel and Jacobi: the inverse of algebraic integrals. 7 Figure 2. GAUSS Carl Friedrich (1777-1855) immediately implies that a b for all n 0. Actually, much more is true: we have n ≥ n ≥ a a . a a . b b . b b (1.3) 1 ≥ 2 ≥ ≥ n ≥ n+1 ≥ ≥ n+1 ≥ n ≥ ≥ 1 ≥ 0 and n 0 a b 2− (a b). (1.4) ≤ n − n ≤ − To prove (1.3), note that a b and a b imply n ≥ n n+1 ≥ n+1 a + b a n n = a b = a b b , n ≥ 2 n+1 ≥ n+1 n n ≥ n and (1.3) follows. From b b we obtain p n+1 ≥ n 1 a b a b = 2− (a b ), n+1 − n+1 ≤ n+1 − n n − n and (1.4) follows by induction. From (1.3) we see immediately that limn an and limn bn exist, and (1.4) implies that the limits are equal. →∞ →∞ Thus, we can use (1.2) to define the arithmetic-geometric mean M(a, b) of a and b. Below we list the simple properties of the AGM. Fact 1: M(a,a)= a; Fact 2: M(a, b)= M(b, a); Fact 3: M(a, 0) = 0; Fact 4: M(a, b)= M(a1, b1)= M(a2, b2)= . .; Fact 5: M(λa, λb)= λM(a, b); a+b √ Fact 6: M(a, b)= M 2 , ab . In particular, the latter relation leads us to 1+ x M(1, x)= M , √x , 2 which shows that the AGM f(x) := M(1, x) is a solution to the following functional equation 1+ x 2√x f(x)= f . 2 1+ x Our next result shows that the AGM is not as simple as indicated by what we have done so far. We now get our first glimpse of the depth of this subject. 8 Theorem 1.2 (Gauss, 1799). Let a and b are positive reals. Then 1 2 π/2 dφ = M(a, b) π 2 2 2 2 Z0 a cos φ + b sin φ Proof 1. As before, we assume that a pb > 0. Let I(a, b) denote the above integral, and set µ = M(a, b). Thus we need to prove≥ π I(a, b)= . 2µ The key step is to show that I(a, b)= I(a1, b1). (1.5) Let us introduce a new variable φ0 such that 2a sin φ sin φ = 0 . (1.6) a + b +(a b) sin2 φ − 0 π π Note that 0 φ0 corresponds to 0 φ . To see this we consider the function ≤ ≤ 2 ≤ ≤ 2 2at f(t) := . a + b +(a b)t2 − Then (a + b (a b)t2) 2ab f 0(t) = 2a − − > 0 (a + b +(a b)t2)2 ≥ (a + b +(a b)t2)2 − − which means that f(t) increasing in [0, 1]. On the other hand, f(0) = 0, f(1) = 1, which yields our claim. Now, we note that dφ dφ = 0 . (1.7) 2 2 2 2 2 2 2 2 a cos φ + b sin φ a1 cos φ0 + b1 sin φ0 Indeed, one can find fromp (1.6) p 2 cos φ a2 cos2 φ + b2 sin2 φ cos φ = 0 1 0 1 0 (1.8) a + b +(a b) sin2 φ p − 0 and it follows (by straightforward manipulations) that 2 a + b (a b) sin φ0 a2 cos2 φ + b2 sin2 φ = a − − .
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