Elliptic Functions. 1 Introduction. 2 Analysis of motion of a Pendulum. Mark Price. Spring 2001

Transcription

1 Elliptic Functions Mark Price Spring 00 Introction I have used the following notation in this essay: the set of all complex numbers is denoted by C and the set of all real numbers is denoted by R. Also where theorems have been taken from books, the proofs have been expanded and added to for the purpose of showing understanding. In this essay I will start by analysing a problem in classical mechanics which has a solution comprising of elliptic functions. I will then do some analysis on the elliptic functions which arise from the problem and will discuss briefly how this analysis can be applied to the problem and suggest how it can be used to predict other solutions. I will then look at the general theory of elliptic functions and then study another specific class before finally looking briefly at the algebraic theory of elliptic functions. Analysis of motion of a Penlum In this section I am going to analyse the motion of a simple penlum with no approximations. This should give motivation to the theory and also allow further analysis of the solution to be done after the theory. The following analysis is taken from [4] P Consider the following penlum with mass m, length a and angle from its rest position θ. In this case the total kinetic energy + total potential energy = the total energy. Hence ( ) dθ ma mga cos θ = E () dt

2 Now allow the motion to be oscillatory with amplitude α. So if θ = = 0 which implies that E = mga cos α. Putting this into () gives ±α, dθ dt ma( ) dθ dt mga cos θ = mga cos α a( ) dθ dt = g(cos θ cos α) ( ) dθ dt = g(cos θ cos α) ( dθ dt ) = p (cos θ cos α) where p = g a ( ) dθ dt = p ( sin θ + sin α ( ) dθ dt = 4p (sin α sin θ) () Now define φ by sin θ = sin α sin φ (3) Differentiating (3) with respect to time yields cos θ( dθ) = sin α dt Now multiply () by cos θ to get cos θ Substituting from (3) and (4) gives ( ) dθ = 4p cos θ sin dt ( α θ ) sin ) by trigonometric identities cos φ( dφ dt ) (4) 4 sin α cos φ ( ) dφ dt = 4p cos θ (sin α sin α sin φ) sin α cos φ ( ) dφ dt = p cos θ (sin α sin α + cos φ sin α) by trigonometric identities sin α cos φ ( ) dφ dt = p cos θ sin α cos φ cos φ ( ) dφ dt = p cos θ cos φ since sin α 0 cos φ ( ) dφ dt = p ( sin φ )( sin φ ) by trigonometric identities (5) cos φ ( ) dφ dt = p ( sin φ)( sin α sin φ) by (3) Now let y = sin φ dy dt So (5) becomes ( dy Now let x = pt So dx = p dt Then dy = cos φ( dφ dt ) and k = sin α dt ) = p ( y )( k y ) = ( dy dt )( ) = ( dy )( ) = ( y dx dt dx dt p )( k y ) So y = ( y )( k y )dy Since ( y )( k y ) is integrable. We may write the solution as y = sn(x + c).

3 hence the general motion of a simple penlum with no approximations is sin θ = sin α sn[p(t t 0)] where t 0 is a constant. sn is normally defined for 0 < k <. If we let k 0, we get the solution as a function of sin. This corresponds to the first approximation of the simple penlum. If we let k, then we get a solution which is a function of tan. This corresponds to the penlum doing a full revolution, i.e it rotates towards the unstable equilibrium point at the top but never reaches it in finite time. These are both extreme cases of the simple penlum. In reality it does something in between and so it is sensible to assume that 0 < k < gives a realistic solution. 3 Jacobian Elliptic Functions In this section, I will look at the sn function and the other functions closely related to it. These functions are called the Jacobian Elliptic Functions. I will now look at the Jacobian Elliptic Functions. I will start by defining sn and the other Jacobian Elliptic Functions. DEFINITIONS Define sn as the function which satisfies the following equation: u = y 0 dt ( t )( k t ) 0 < k < (6) and sn() = 0. Define cn as the function which satisfies the following equation: cn (u) = sn (u) (7) and cn(0) =. Define dn as the function which satisfies the following equation: dn (u) = k sn (u) (8) and dn(0) =. 3

4 If k = 0 in the definition of sn, (6) reces to u = y dt and the 0 ( t ) solution is y = sin u. So sn is just a generalisation of sin. This leads to cn becoming cos when k = 0 and dn becoming. If in the definition of sn, t is replaced by t, we get the sign of y changed. So sn is an odd function. From our definition of cn it follows that cn( u) = ±cn(u). Put u = 0 to see that cn( u) = cn(u) and hence even. I will now consider some of the relationships between sn, cn and dn. From (6) we get ( d sn(u)) = ( sn (u))( k sn (u)) (9) Differentiating (7) gives cn(u) d so ( d sn(u)) = cn (u)dn (u) d sn(u) = cn(u)dn(u) d cn(u) = sn(u) sn(u) so cn(u) d cn(u) = sn(u)cn(u)dn(u) cn(u) = sn(u)dn(u) d Similarly (8) gives d dn(u) = k sn(u)cn(u) From the earlier analogy with sin and cos it would seem sensible to assume that there exist addition formulae. Here is the necessary theorem: THEOREM W e have sn(u + v) = cn(u + v) = sn(u)cn(v)dn(v)+sn(v)cn(u)dn(u) k sn (u)sn cn(u)cn(v) sn(u)sn(v)dn(u)dn(v) (v) k sn (u)sn (v) dn(u + v) = dn(u)dn(v) k sn(u)sn(v)dn(u)cn(v) k sn (u)sn (v) Proof. There are many alternate proofs to this theorem but this one comes from [6] P Let u + v = α where u and v vary but α is constant such that dv =. ) = ( sn (u))( k sn (u)) (0) From (9) we get ( dsn(u) Differentiating sn(v) and squaring gives ( dsn(v) k sn (v)) () ) ( dv ) = ( sn (v))( If you put k = 0 in the following relationships you will obtain the standard trigonometric formulae 4

5 Now differentiate (0) by u to get dsn(u). d sn(u) d sn(u) = sn(u)k dsn(u) sn(u) dsn(u) + 4k sn 3 (u) dsn(u) = k sn(u) sn(u) + k sn 3 (u) = ( + k )sn(u) + k sn 3 (u) () for cn(u) 0 and dn 0 Differentiate () by u to get ( dsn(v) dv )( d sn(v) )( dv d sn(v) ) = sn(v) dsn(v) ( dv ) k sn(v) dsn(v) ( dv ) + 4k sn 3 (v) dsn(v) = sn(v) k sn(v) + k sn 3 (v) = ( + k )sn(v) + k sn 3 (v) (3) for cn(u) 0 and dn 0 ( dv ) (0) sn (v) () sn (u) subtract () sn(v) (3) sn(u) subtract (5) (4) ( dsn(u) ( dsn(u) ) sn (v) = sn (v) ( + k )sn (u)sn (v) + k sn 4 (u)sn (u) ( dsn(v) ) sn (u) = sn (u) ( + k )sn (v)sn (u) + k sn 4 (v)sn (u) ) sn (v) ( ) dsn(v) sn (u) = sn (v) sn (u) + k sn 4 (u)sn (v) k sn 4 (v)sn (u) = (sn (v) sn (u))( k sn (u)sn (v)) (4) d sn(u) sn(v) = ( + k )sn(u)sn(v) + k sn 3 (u)sn(v) d sn(v) sn(u) = ( + k )sn(v)sn(u) + k sn 3 (v)sn(u) d sn(u) sn(v) d sn(v) d sn(v) sn(v) d sn(v) sn(u) ( dsn(u) ) sn(v) ( dsn(v) ( d dsn(u) ) dsn(v) dn(v) ( sn(u) dsn(u) dsn(v) sn(v) = k sn 3 (u)sn(v) sk sn 3 (v)sn(u) = k sn(u)sn(v)(sn (u) sn (v)) (5) ) = k sn(u)sn(v)(sn (u) sn (v)) ) = d (sn (v) sn (u))( k sn (u)sn (v)) ( ) k sn (u)sn (v) ( k sn (u)sn (v)) Integrating yields dsn(u) sn(v) dsn(v) sn(u) = C, a constant k sn (u)sn (v) but we already know dsn(u) and dsn(v), so cn(u)dn(u)sn(v) + cn(v)dn(v)sn(u) k sn (u)sn (v) = C This can be written as a function of two variables f(u + v) 5

6 Let v = 0 cn(u)dn(u)sn(0) + cn(0)dn(0)sn(u) k sn (u)sn (0) So f is sn. Hence sn(u + v) = cn(u)dn(u)sn(v)+cn(v)dn(v)sn(u) k sn (u)sn (v) By definition cn (u) = sn (u), so = f(u) cn (u + v) = sn (u + v) ( k sn (u)sn (v))sn ( (u + v) = ( k sn ( u)sn (v)) ( sn (u + v)) = ( k sn (u)sn (v)) ( ) ) sn(u)cn(v)dn(v)+sn(v)cn(u)dn(u) k sn (u)sn (v) = ( k sn (u)sn (v)) (sn(u)cn(v)dn(v) + sn(v)cn(u)dn(u)) = k sn (u)sn (v) + k 4 sn 4 (u)sn 4 (v) sn (u)cn (v)dn (v) sn(u)cn(v)dn(v)sn(u)cn(u)dn(u) sn (v)cn (u)dn (u) = k sn (u)sn (v) + k 4 sn 4 (u)sn 4 (v) sn (u)( sn (v))( k sn (u)) sn (v)( sn (u))( sn (v)) sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) = k sn (u)sn (v) + k 4 sn 4 (u)sn 4 (v) sn (u) + k sn (u)sn (v) +sn (v)sn (u) k sn 4 (v)sn 4 (u) sn (v) + k sn (u)sn (v) +sn (v)sn (u) k sn 4 (u)sn (v) sn(u)sn(v)cn(u)cn(v)dn(u)dn(v) = + k 4 sn 4 (u)sn 4 (v) k sn 4 (u)sn (v) k sn 4 (v)sn (u) sn (u) sn (v) +sn (v)sn (u) sn(u)cn(u)dn(u)sn(v)cn(v)dn(v) = sn (u)sn (v)( k sn (u))( k sn (u)) + ( sn (u))( sn (v)) sn(u)cn(u)dn(u)sn(v)cn(v)dn(v) = cn (u)cn (v) + sn (u)sn (v)dn (u)dn (v) sn(u)sn(v)cn(u)cn(v)dn(u)dn(v) = (cn(u)cn(v) sn(u)sn(v)dn(u)dn(v)) cn(u + v) = ± cn(u)cn(v) sn(u)sn(v)dn(u)dn(v) k sn (u)sn (v) Let u = 0 cn(v) = ± cn(0)cn(v) sn(0)sn(v)dn(0)dn(v) k sn (0)sn (v) = ±cn(v) So have to take the positive term. Hence cn(u + v) = cn(u)cn(v) sn(u)sn(v)dn(u)dn(v) k sn (u)sn (v) The same method is used to obtain the remaining formula. 6

7 From the definition, dn (u + v) = k sn (u + v) ( k sn (u)sn (v)) dn (u + v) = ( k sn (u)sn (v)) ( k sn (u + v)) = ( k sn (u)sn (v)) k (sn(u)cn(v)dn(v) + sn(v)cn(u)dn(u)) = k sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) k sn (u)cn (u)dn (u) k sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) k sn (v)cn (u)dn (u) = k sn (u)sn (v) + k 4 sn 4 (u)sn 4 (v) k sn (u)( sn (v))( k sn (v)) k sn (v)( sn (u))( k sn (u)) k sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) = k sn (u)sn (v) + k 4 sn 4 (u)sn 4 (v) k sn (u) + k sn (u)sn (v) +k 4 sn (v)sn (u) k 4 sn 4 (v)sn (u) k 4 sn (v) + k sn (v)sn (u) +k 4 sn (u)sn (v) k 4 sn 4 (u)sn (v) sk sn(u)cn(v)dn(v)sn(v)sn(u)cn(u)dn(u) = ( k sn (u))( k sn (v)) + k 4 sn (v)sn (u) k 4 sn 4 (v)sn (u) +k 4 sn (u)sn (v) k 4 sn 4 (u)sn (v) k sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) = ( k sn (u))( k sn (v)) + k 4 sn (u)sn (v)( sn (u))( sn (v)) k sn(u)cn(v)dn(v)sn(v)cn(u)dn(u) = dn(u)dn(v) + k 4 sn (u)sn (v)cn (u)cn (v) k 3 sn (u)cn (v)dn (v)sn (v)cn (u)dn (u) = (dn(u)dn(v) k sn(u)sn(v)cn(u)cn(v)) dn(u + v) = ± dn(u)dn(v) k sn(u)sn(v)cn(u)cn(v) k sn (u)sn (v) Let u = 0 dn(v) = ± dn(0)dn(v) k sn(0)sm(v)cn(0)cn(v) k sn (0)sn (v) = ±dn(v) So again take the positive term. Hence dn(u + v) = dn(u)dn(v) k sn(u)sn(v)cn(u)cn(v) k sn (u)sn (v) This concludes the theory on Jacobian Elliptic Functions. In the next section, I will rigourously define elliptic functions and generalise the theory. In this section I will prove some of the main results in the general theory of elliptic functions. The first thing I need to do is define an elliptic function. DEFINITION A function, f, which is holomorphic except at poles and has singularities in a finite part of the plane and satisfies the following equations: 7

8 is called elliptic. f(z + ω ) = f(z) and f(z + ω ) = f(z) for some ω and ω C Firstly I shall prove some important theorems in the theory of elliptic functions and then I will draw some conclusions on the nature of elliptic functions based on these results before looking at some specific examples. LEMMA An elliptic function without poles is constant Proof. Let f(z) be an elliptic function with periods ω and ω. f(z) can be enclosed by a parallelogram, P α, with vertices α, α+ω, α+ω and α+ω +ω. This idea and why it works will be discussed later. As we have no poles and f is bounded, Liouville s Theorem says that f is constant. LEMMA 3 T he sum of the resies of an elliptic function is zero. Proof. Let P α be the same as in the previous proof. By the Resie Theorem we have n j= (f, z j) = πi = πi ( P α f(z)dz where there is a pole at every z j α+ω f(z)dz + α+ω +ω α α+ω f(z)dz + α+ω α+ω +ω f(z)dz + α As f is periodic f(z) = f(z + ω ) and f(z) = f(z + ω ), so changing the variables on the second and third integral gives ( n j= (f, z j) = α+ω πi = πi + α+ω f(z + ω α α )dz + α α+ω f(z + ω )dz + α α+ω f(z) f(z + ω α )dz α+ω f(z) f(z + ω πi α )dz ) α+ω f(z)dz ) α+ω f(z)dz As f is periodic both of these integrals are zero thus n j= (f, z j) = 0 8

9 Now I can define the order of an elliptic function. Let c be a constant. The order of f is the number of roots of the equation f(z) = c which is equal to the number of poles in P α. THEOREM 4 Elliptic functions have order Proof. If an elliptic function has order, it must have a non-zero resie. However this contradicts lemma 3. So elliptic functions must have order >. In this essay I will just examine the case where the order of the elliptic function is. When the order is two, there are two subcases. One when there is a single irrecible double pole with zero resie and another when there are two poles with equal but opposite resies. I have already looked at the elliptic functions with two poles, as these are the Jacobian Elliptic Functions. In the next section, I will consider the other type of order. 4 Weierstraß-ϕ Function In this section, I will start off by constructing an elliptic function of order two with a single irrecible pole with zero resie. Firstly, I will define the function: ϕ(z) = z + ( (z ω) ) where ω = n ω ω + n ω ω 0 This is the Weierstraß-ϕ function. This function has its pole at the origin. THEOREM 5 T he above sum converges, has a double period and hence is an elliptic function. Proof. This argument is taken from [6] P Firstly estimate, 9

10 Let ω > z so z ω) ω = ω (z ω) ω (z ω) = ω z +ωz ω ω (z ω) = z(ω z) ω (z ω) ωz z ω z ω 3 z+ω ωz z by triangle inequality 4 ω z ω 3 z + ω 4 4 z z ω ω 4 ω3 ω + ω4 z ω 4 z z ω ω 3 z ω 3 6 z ω 3 6 z So the series converges if ω 0 converges. ω 3 This is true since ω ω is non-real, so K > 0 such that n ω + n ω K( n + n ) (n, n ) R. Consider only integer pairs of (n, n ). There are 4n which have the property n + n = n. So ω 0 4 ω 3 k 3 which converges. n 3 Now al that is left to prove that ϕ is an elliptic function is to prove that it is doubly periodic. Let f(z) = + ( ) z ω 0 (z ω) ω then f (z) = z 3 ω 0 (z omega) 3 f (z) = ω (z ω) 3 This sum has two distinct periods, ω, ω, which implies that f(z + ω ) f(z) = a constant and f(z + ω ) f(z) = a constant. As f is even by definition, let z = ω f(ω ω ) f( ω ) = f(ω ) f(ω ) = 0 0

11 and now let z = ω f(ω ω ) f( ω ) = f(ω ) f(ω ) = 0 From our definition of ϕ and f it is clear that ϕ(z) f(z) is a constant and by the construction of ϕ at the origin, ϕ(z) = f(z) and so ϕ is an elliptic function. ϕ has zero reside and so can be integrated to obtain F (z) = z ω 0 z by convention F (z) is called ζ(z) so ζ(z) = + z ω ω ω z ω z z ω ω ω Once again convergence has to be shown. Firstly estimate Let ω > z + + z z ω ω ω = ω +(z ω)ω+z(z ω) ω (z ω) = ω +zω ω +z zω ω z ω 3 = z ω z ω 3 z z by triangle inequality ω z ω 3 ω 3 z ω 3 z z ω ω ω ω z ( ω ω ) z ω Now only have to show that ω 0 converges. ω This series converges since ω ω is non-real and so M > 0 st n ω + n ω M( n + n ) (n, n ) R. Hence ζ converges. As ϕ is periodic it follows that ζ is periodic and thus ζ(z + ω ) = ζ(z) + η (6) and ζ(z + ω ) = ζ(z) + η (7). There is a very important relationship between η, η, ω and ω. THEOREM 6 T he following relationship holds: πi = ω η ω η

12 Proof. Recall the parallelogram, P α, used earlier. Now use the Resie Theorem. = n j=0 (f, z j) = πi = πi ( P α ζ(z)dz α+ω ζ(z)dz + α+ω +ω α α+ω ζ(z)dz + α+ω α+ω +ω ζ(z)dz + α α+ω ζ(z)dz Use (6) and (7) to change the variables in the second and third integrals to get: πi = α+ω ζ(z)dz + α+ω ζ(z + ω α α ) + η dz + α α+ω ζ(z + ω ) + η dz + α = α+ω ζ(z) ζ(z + ω α )dz α+ω ζ(z) ζ(z ω α )dz + α+ω α = (αη + ω η αη ) + (αη αη ω η ) πi = ω η ω η ω ζ(z)dz η dz + α α+ω η dz This is called Legendre s Relation. From the study of the Jacobian Elliptic Functions it seems sensible to assume that ϕ has an addition formula and here it is: THEOREM 7 W e have ϕ(z + y) = 4 ( ) ϕ (z) ϕ (y) ϕ(z) ϕ(y) ϕ(z) ϕ(y) ) Proof. It is important to note that if y = ±z(mod ω.ω ) then the formula does not hold. Let y ±z(mod ω, ω ). The following equations are needed: ϕ (z) = αϕ(z) + β (8) ϕ (y) = αϕ(y) + β (9) Firstly look at ϕ (γ) αϕ(γ) β. It has a triple pole when γ = 0 and so has three irrecible zeros by Theorem and hence γ = z, γ = y and γ = z y are the poles.

13 So ϕ(z), ϕ(y) and ϕ(z + y) are the roots of the following cubic 4ϕ 3 (γ) α ϕ (γ) (αβ + g )ϕ(γ) (β + g ) = 0 From the theory of polynomials, it follows that ϕ(z) + ϕ(y) + ϕ(z + y) = 4 α4 Solving (8) and (9) for α gives α = ϕ (z) ϕ (y) ϕ(z) ϕ(y) ( ) Hence ϕ(z + y) = ϕ (z) ϕ (y) ϕ(z) ϕ(y) 4 ϕ(z) ϕ(y) This has proved the addition formula however as noted earlier that it does not hold if y = ±z(mod ω, ω ). It is possible to derive the formula when y = z, by taking the limit as y goes to z in the addition formula. Here is the necessary theorem. THEOREM 8 W e have ϕ(z) = 4 ( ) ϕ (z) ϕ(z) ϕ (z) Proof. Taking limits gives lim y z ϕ(z+y) = lim ( ϕ ) (z) ϕ (y) ϕ(z) limy z 4 y z ϕ(y) ϕ(z) ϕ(y) If z is not a period then we get ϕ(z) = lim ( ϕ (z) ϕ (z+h)) 4 h 0 ϕ(z) ϕ(z+h) ϕ(z) Apply Taylor s Theorem to get ϕ(z + h) and ϕ (z + h) to get ϕ(z) = ( ϕ ) (z) 4 ϕ (z) ϕ(z) This has shown some of the properties of ϕ. This concludes the work on elliptic functions in my essay, in the next section I will consider some more abstract theory. 5 Algebraic Theory of Elliptic Functions I am now going to go back and say something about the parallelogram, P α, discussed earlier and comment on the field of elliptic functions. P α was formed by taking a point in the complex plane and using the periods of the elliptic function to calculate the sides of the parallelogram. 3

14 If this parallelogram is translated back to the origin, and then copied and placed end to end it can cover the whole of the complex plane. This is a lattice, L = ω Z ω Z, where in each parallelogram the behaviour of the elliptic function is identical e to its periodicity. Now quotient C by L to obtain a parallelogram with α = 0 but this is on the complex torus. It can be proved that all elliptic functions belong to the field of functions of rational character over the complex torus, i.e. as all elliptic functions are quotients of the form C then it must be rational. This gives some idea of how to look L at elliptic functions algebraically. 6 Conclusion To conclude I have given some motivation for the study of elliptic functions by looking at an example in classical mechanics which gives rise to them. There are of course many other systems which give rise to elliptic functions, but there is not enough space in this essay to include them all. I then looked briefly at the Jacobian Elliptic Functions before looking at the general theory of elliptic functions before analysing the Weierstraß-ϕ function and proving some of its basic properties. I then finally considered some of the more abstract theory. References [] Alhfors Complex Analysis McGraw-Hill (979) [] Lang elliptic Functions Springer-Verlag [3] McKean and Moll Elliptic Curves Cambridge University Press (997) [4] Synge and Griffith Principles of Mechanics McGraw-Hill [5] Roquette Theory of Elliptic Functions [6] Whittaker and Watson Course of Modern Analysis Cambridge University Press (963) 4