Elliptic functions and Elliptic Integrals

Transcription

1 Elliptic functions and Elliptic Integrals R. Herman Nonlinear Pendulum We motivate the need for elliptic integrals by looking for the solution of the nonlinear pendulum equation, θ + ω sin θ. () This models a mass m attached to a string of length L undergoing periodic motion. Pulling the mass to an angle of θ and releasing it, what is the resulting motion? We employ a technique that is useful for equations of the form θ + F(θ) θ Figure : A simple pendulum consists of a point mass m attached to a string of length L. It is released from an angle θ. L m when it is easy to integrate the function F(θ). Namely, we note that [ d θ(t) ] θ + F(φ) dφ ( θ + F(θ)) θ. For the nonlinear pendulum problem, we multiply Equation () by θ, θ θ + ω sin θ θ and note that the left side of this equation is a perfect derivative. Thus, [ ] d θ ω cos θ. Therefore, the quantity in the brackets is a constant. So, we can write θ ω cos θ c. () The constant in Equation () can be found using the initial conditions, θ() θ, θ(). Evaluating Equation () at t, we have c ω cos θ. Solving for θ, we obtain ω (cos θ cos θ ). This equation is a separable first order equation and we can rearrange and integrate the terms to find that θ ω cos θ ω cos θ. (3)

2 elliptic functions and elliptic integrals We can solve for θ and integrate the differential equation to obtain t ω (cos θ cos θ ). At this point one says that the problem has been solved by quadratures.namely, the solution is given in terms of some integral. We will proceed to rewrite this integral in the standard form of an elliptic integral. Using the half angle formula, sin θ ( cos θ), we can rewrite the argument in the radical as [ cos θ cos θ sin θ θ ] sin. Noting that a motion from θ to θ θ is a quarter of a cycle, we have that T θ. (4) ω sin θ sin θ This result can now be transformed into an elliptic integral. We define and Then, Equation (4) becomes z sin θ sin θ k sin θ. Elliptic integrals were first studied by Leonhard Euler and Giulio Carlo de Toschi di Fagnano (68-766), who studied the lengths of curves such as the ellipse and the lemniscate, (x + y ) x y. T 4 ω dz ( z )( k z ). (5) This is done by noting that dz k cos θ k ( k z ) / and that sin θ sin θ k ( z ). The integral in this result is called The complete elliptic integral of the first the complete elliptic integral of the first kind. kind.

3 elliptic functions and elliptic integrals 3 Elliptic Integrals of First and Second Kind There are several elliptic integrals. They are defined as F(φ, k) K(k) E(φ, k) E(k) sin φ π/ k sin θ ( t )( k t ) k sin θ ( t )( k t ) sin φ π/ (6) (7) (8) (9) k sin θ () k t t () k sin θ () k t t (3) (4) Elliptic Functions Elliptic functions result from the inversion of elliptic integrals. Consider u(sin φ, k) F(φ, k) sin φ k sin θ. (5) ( t )( k t ). (6) Note:F(φ, ) φ and F(φ, ) ln(sec φ + tan φ). In these cases F is obviously monotone increasing and thus there must be an inverse. The inverse of F(u, k) is sn (u, k) sin φ sin amu, where am(u, k) φ F (u, k) am is called the amplitude. Note that sn (u, ) sin u and sn (u, ) tanh u. Similarly, we have u u cn (u,k). (7) ( t )(k + k t ) dn (u,k). (8) ( t )(t k )

4 elliptic functions and elliptic integrals 4 Figure : Plots of the Jacobi elliptic functions for m sn(u) cn(u) dn(u) The Jacobi elliptic functions for m.75 are shown in Figure. We note that these functions are periodic. The Jacobi elliptic functions are related by sin φ sn (u, k) (9) cos φ sn (u, k) () k sin φ dn (u, k) () Furthermore, we have the identities sn u + cn u, k sn u + dn u. Derivatives Derivatives of the Jacobi elliptic functions are easily found. First, we note that d( sn u) du d( sn u) dφ dφ du cn u k sin φ cn u dn u, () where du dφ k sin φ results from integrating F(φ, k). Similarly, we have d d cn u sn u dn u, and du du dn u k sn u cn u. Differential Equations Let y sn u. Using we have or d( sn u) du cn u dn u, dy du y k y, ( ) dy ( y )( k y ). du

5 elliptic functions and elliptic integrals 5 Differentiating with respect to u again, we have the nonlinear second order differential equation y ( + k )y + k y 3. We note that this differential equation is amenable to solution using Simulink. Such a model is shown in Figure 3. y'' k y 3 s Integrator ( + k )y y' Product k s Integrator +k y Scope One u.9 Figure 3: Simulink model for solving y ( + k )y + k y 3. Product y'' - ( + k )y + k y 3 3 y Gain u()^3 Math Function k Fcn Periodicity Consider F(φ + π, k) +π k sin θ. φ+π + k sin θ φ π k sin θ F(φ, k) + k sin θ F(φ, k) + 4K(k). (3) Since F(φ + π, k) u + 4K, we have sn (u + 4K) sin(am(u + 4K)) sin(am(u) + π) sin am(u) sn u. In general, we have sn (u + K, k) sn (u, k) (4) cn (u + K, k) cn (u, k) (5) dn (u + K, k) dn (u, k). (6) The plots of sn (u), cn (u), and dn(u), are shown in Figures 4-6.

6 elliptic functions and elliptic integrals 6.5 Figure 4: Plots of sn (u, k) for m,.5,.5,.75,...5 sn(u) m m.5 m.5 m.75 m u.5 Figure 5: Plots of cn (u, k) for m,.5,.5,.75,...5 cn(u) m m.5 m.5 m.75 m u Complex Arguments Values of the Jacobi elliptic functions for complex arguments can be found using Jacobi s imaginary transformations, sn (iu, k) i sc (u, k ) (7) cn (iu, k) nc (u, k ) (8) dn (iu, k) dc (u, k ). (9).5 Figure 6: Plots of dn (u, k) for m,.5,.5,.75,.. dn(u).5 m m.5 m.5 m.75 m u

7 elliptic functions and elliptic integrals 7 These results are found by rewriting the elliptic integral. We show this for the first result by considering u F(φ, k) in the form F(φ, k) We introduce the transformation This gives sin θ cos θ k sin θ. t + t, ( ) t + t t + t. (3) cos θ ( + t ) 4t ( + t ) ( t ) ( + t ), or +t Applying this variable substitution to the elliptic integral, we have u s k sin θ ( ) ( + t ) k t +t s ( + t ) 4k t s + ( k )t + t 4. (3) Inserting t ix, and noting that the integrand is an even function of x, we obtain is dx u i ( k )x + x. 4 i is Introducing k k, leads to is u i iu is i is dx ( k )x + x 4. (3) dx ( ( k ))x + x 4 dx ( + k )x + x 4 dx + ( k )x + x 4. (33)

8 elliptic functions and elliptic integrals 8 Therefore, we have Equation (33) is the same as Equation (3) and the inverse function is sn (iu, k ). Using the transformation, we find that sn (iu, k ) is pure imaginary: sn (iu, k ) is s i sin φ cos φ sn (u, k) i cn (u, k) i sc (u, k). (34) We can exchange k with k to obtain the final result sn (iu, k) i sc (u, k ). There is a problem when cn (u, k ). Noting that and sn (, k), cn (, k), dn (, k), sn (K, k), cn (K, k), dn (K, k) k, and that cn (u, k) has period 4K, then cn (u, k ) for u (n + )K. Thus, sn (iu, k) has imaginary period of ik. Plots of the Jacobi elliptic functions in the complex plane using domain coloring for k.7 are shown in Figures 7-9. In this case we have K(.7).8457 and K (.7) K(.7 ).866. This gives the periods for sn(u) as and 3.753i, which can be seen in Figure Figure 7: Domain coloring plot of sn (u, k) for u x + iy and k

9 elliptic functions and elliptic integrals Figure 8: Domain coloring plot of cn (u, k) for u x + iy and k Figure 9: Domain coloring plot of dn (u, k) for u x + iy and k

10 elliptic functions and elliptic integrals Addition Formulae Letting s i sn (u i ), for i,, etc., we have sn (u + v) cn (u + v) sn u cn v dn v + sn v cn u dn u k sn x sn. y (35) cn u cn v sn u sn v dn u dn v k sn x sn. y (36) dn (u + v) dn u dn v k sn u sn v cn u cn v k sn x sn. (37) y From these formulae and the Jacobi imaginary transformation, one can derive formula for complex arguments. Arithmetic-Geometric Mean The Arithmetic-Geometric Mean (AGM) iteration of Gauss is given by a two-term recursion a n+ a n + b n, b n+ a n b n. (38) These sequences converge to a common limit, In 799 Gauss saw that lim a n lim b n M(a, b ). n n M(, ) π t up to eleven decimal places. This is an example of M(, x) π/. π ( x ) sin θ Letting x sin α, we can write K(cos α) π M(, sin α).