ELLIPTIC INTEGRALS AND FUNCTIONS

12. ELLIPTIC INTEGRALS AND FUNCTIONS

This is another collection of integrals and related functions which may arise in ap- plied problems and as computer answers (see problems). We shall merely summarize the basic definitions and properties—there are whole books on the subject—and you may find useful formulas and information in your computer program and in reference books and tables.

Section 12 Elliptic Integrals and Functions 555

Legendre Forms The Legendre forms of the elliptic integrals of the first and second kinds are:

E(φ, k) =

1−k 2 sin 2 θ dθ,

0 ≤ k ≤ 1.

There is also an elliptic integral of the third kind which occurs less frequently. In (12.1), φ is called the amplitude and k is called the modulus of the elliptic integral.

Jacobi Forms If we put t = sin θ, x = sin φ in the Legendre forms (12.1), we obtain the Jacobi forms of the elliptic integrals of the first and second kind:

2 cos θ . 1−t The limits θ = 0 to φ become t = 0 to x.

1−k 2 t 2 E(φ, k) =

Complete Elliptic Integrals The complete elliptic integrals of the first and second kind are the values of F and E when φ = π/2 or x = sin φ = 1:

1 dt K or K(k) = F

0 1−t 2 1−k 2 t 2 (12.3) π/2 1 π √

2 0 1−k sin θ

,k = 1−k 2 sin 2 1−k θ dθ = t 2 √

dt. Warning: The notation used for elliptic integrals is not uniform. Most references

E or E(k) = E

0 0 1−t 2

use F and E, but you may find φ replaced by x = sin φ, and instead of k you may find m = k 2 , or sin −1 k. Also (φ, k) may be written as (k, φ), and other variations exist. So check carefully the notation of any book or computer program you are using and reconcile the results with the notation used here.

Example 1.

1 − (1/2) sin 2 θ dθ = E(φ, k) = E(π/3, 1/

2 ) in our notation. Other

books or computer programs might give: E(φ, m) = E(π/3, 1/2), or E(x, k) = √ √ E( 3/2, 1/

2 ) or E(φ, sin −1 k) = E(π/3, π/4), etc. Of course, all of them will give the same numerical approximation 0.964951.

Many integrals can be written in the form of one of the integrals in (12.2).

556 Special Functions Chapter 11

Example 2.

16 − 8 sin 2 θ dθ becomes 4 times the integral in Example 1 if we divide

out a factor of 4 to get 4

1 − (1/2) sin 2 θ dθ.

5 , 2) in the notation of (12.2), except that we have previously required k < 1, and here k = 2. However, we can

= F (φ, k) = F (sin

put this integral in the standard form with k < 1 by making the change of variable 4t 2 =r 2 , or r = 2t. Substituting this into the given integral gives

which, by (12.2), is 1 2 F (φ, k) = 1 2 F (sin −1 4 5 , 1 2 ). (See Problem 24.) It is sometimes useful to note that the integrands in elliptic integrals are all

functions of sin 2 θ and so are even functions of θ. Thus an elliptic integral from −φ 1 to φ 2 (φ 1 and φ 2 both positive) is equal to the integral from 0 to φ 1 plus the

integral from 0 to φ 2 and we have φ 2

1−k 2 sin 2 θ dθ = E(φ 1 , k) + E(φ 2 , k)

and a similar formula for F (φ, k). Also we may note that a function of sin 2 θ has period π and is symmetric about θ = nπ + π/2 (look at a graph of sin 2 θ). Thus, using the complete elliptic integrals in (12.3), we can write (Problem 2)

F (nπ ± φ, k) = 2nK ± F (φ, k),

E(nπ ± φ, k) = 2nE ± E(φ, k).

Since k 2 sin 2 θ < 1 (for k 2 < 1), we get convergent infinite series for elliptic integrals by expanding their integrands using the binomial theorem, and then inte- grating term by term (Problem 1). For small k these series converge rapidly and provide a good method for approximating elliptic integrals when k ≪ 1.

Here are some examples where elliptic integrals occur. Example 4. Find the arc length of an ellipse. This is the problem that gave elliptic

integrals their name. We write the equation of the ellipse in the parametric form

x = a sin θ, y = b cos θ,

for the case a > b. (If b > a, use the form x = a cos θ, y = b sin θ; see Problem 15.) Then for a > b, we have

ds 2 = dx 2 + dy 2 = (a 2 cos 2 θ+b 2 sin 2 θ) dθ 2 . Since a 2 −b 2 > 0, we can write

a 2 2 −b − (a 2 2 −b 2 ) sin θ dθ = a 1−

ds =

sin θ dθ.

Section 12 Elliptic Integrals and Functions 557 This is an elliptic integral of the second kind where k 2 = (a 2 −b 2 )/a 2 =e 2 (e is the

eccentricity of the ellipse in analytic geometry). If we want the complete circumfer- ence, θ goes from 0 to 2π, and the answer is 4aE(π/2, k) = 4aE(k). For a smaller

arc, we use the appropriate limits φ 1 and φ 2 and obtain E(φ 2 , k) − E(φ 1 , k). For any given ellipse (that is, given a and b), we can find the numerical value of the desired arc length from computer or tables.

Example 5. Let a pendulum swing through large angles. We had in Section 8

2 (12.5) 2g ˙θ = cos θ + const.,

and we considered 180 ◦ swings, that is of amplitude 90 ◦ . Now we want to consider swings of any amplitude, say α; then ˙θ = 0 when θ = α, and (12.5) becomes

˙θ 2 = 2g (cos θ − cos α).

Integrating (12.6), we get

where T α is the period for swings from −α to +α and back. This integral can be written as an elliptic integral; its value (Problem 17) is

2K sin

Then (12.7) gives for the period

For α not too large (say α < 90 ◦ , 1 2 α < 45 ◦ , so that sin 2 (α/2) < 1 2 ), we can get a good approximation to T α by series (Problem 1):

For α small enough so that sin α/2 can be approximated by α/2, we can write

For very small α, we get the familiar formula for simple harmonic motion, T = 2π l/g independent of α. For somewhat larger α, say α = 1 radian (about 30 2 ◦ ), we get

This would mean that a pendulum started at 30 ◦ would get exactly out of phase with one of very small amplitude in about 32 periods.

For another physics problem giving rise to an elliptic integral, see Am. J. Phys. 55, 763 (1987).

558 Special Functions Chapter 11

Elliptic Functions Recall that

defines u as a function of x, or x as a function of u; in fact x = sin u. In a similar way, u = F (φ, k) in (12.2) defines u as a function of φ (or of x = sin φ) or it defines x or φ as functions of u (we are assuming k fixed). We write

or x = sn u. The function sn u (read ess-en of u) is an elliptic function. Since x = sin φ, we have

x = sn u = sin φ.

There are other elliptic functions, related to sn u; you will notice [in (12.14)] that they have some resemblance to the trigonometric functions. We define

cn u = cos φ = 2 1 − sin φ= 1 − sn 2 u= 1−x 2 , (12.14)

du/dφ

[The value of du/dφ is found from u = F (φ, k) in (12.2).] There are many formulas relating these functions—for example, addition formulas, integrals, derivatives, etc. These can be looked up or, in some cases, easily worked out. For example, since sn u = sin φ, we have

d d dφ

(sn u) =

(sin φ) = cos φ

For a physical problem using elliptic functions, see Am. J. Phys. 68, 888–895 (2000).

PROBLEMS, SECTION 12

1. Expand the integrands of K and E [see (12.3)] in power series in k 2 sin 2 θ (assuming small k), and integrate term by term to find power series approximations for the complete elliptic integrals K and E.

2. Use a graph of sin 2 θ and the text discussion just before (12.4) to verify the equations (12.4). Note that the area under the sin 2 θ graph from 0 to π/2 and the area from π/2 to π are mirror images of each other, and this will be true also for any function of sin 2 θ.

3. Computer plot graphs of K(k) and E(k) in (12.3) for k from 0 to 1. Also plot 3D graphs of F (φ, k) and E(φ, k) in (12.1) for k from 0 to 1 and φ from 0 to π/2 and also from 0 to 2π. Warning: Be sure you understand the notation used by your computer program; see text discussion just after (12.3) and Example 1.

Section 12 Elliptic Integrals and Functions 559

In Problems 4 to 13, identify each of the integrals as an elliptic integral (see Examples 1 and 2). Learn the notation of your computer program (see Problem 3) and then evaluate the integral by computer.

Z 5π/4 p

Z π/4 Z dθ 3π/8 dθ 10. p

14. Find the circumference of the ellipse 4x 2 + 9y 2 = 36.

√ 15. Find the length of arc of the ellipse x 2 + (y 2 /4) = 1 between (0, 2) and ( 1 2 ,

3 ). (Note that here b > a; see Example 4.)

16. Find the arc length of one arch of y = sin x. 17. Write the integral in equation (12.7) as an elliptic integral and show that (12.8)

gives its value. Hints: Write cos θ = 1 − 2 sin 2 (θ/2) and a similar equation for cos α. Then make the change of variable x = sin(θ/2)/ sin(α/2).

18. Computer plot graphs of sn u, cn u, and dn u, for several values of k, say, for example, k = 1/4, 1/2, 3/4, 0.9, 0.99. Also plot 3D graphs of sn, cn, and dn as functions of u and k.

19. If u = ln(sec φ + tan φ), then φ is a function of u called the Gudermannian of u, φ = gd u. Prove that:

„π φ « d u = ln tan

+ , tan gd u = sinh u, sin gd u = tanh u, gd u = sech u. 4 2 du

20. Show that for k = 0: u = F (φ, 0) = φ, sn u = sin u, cn u = cos u, dn u = 1; and for k = 1: u = F (φ, 1) = ln(sec φ + tan φ) or φ = gd u (Problem 19),

sn u = tanh u, cn u = dn u = sech u.

R π/2

√ 21. Show that the four answers given in Section 1 for 0 dθ/ cos θ are all correct.

Hints: For the beta function result, use (6.4). Then get the gamma function results by using (7.1) and the various Γ function formulas. For the elliptic integral, use the hint of Problem 17 with α = π/2.

560 Special Functions Chapter 11

22. In the pendulum problem, θ = α sin pg/l t is an approximate solution when the amplitude α is small enough for the motion to be considered simple harmonic. Show that the corresponding exact solution when α is not small is

where k = sin(α/2) is the modulus of the elliptic function. Show that this reduces to the simple harmonic motion solution for small amplitude α.

23. A uniform solid sphere of density 1 2 is floating in water. (Compare Chapter 8, Prob- lem 5.37.) It is pushed down just under water and released. Write the differential equation of motion (neglecting friction) and solve it to obtain the period in terms of K(5 −1/2 ). Show that this period is approximately 1.16 times the period for small oscillations.

24. Sometimes you may find the notation F (φ, k) in (12.2) used when k > 1. Allowing this notation, show that 1 F (sin −1 3 , 4 )= 1 F (sin −1 4 , 3 3 5 3 4 5 4 ). Hints: Using the Jacobi form of F in (12.2), write the integral which is equal to 1 3 F (sin −1 3 5 , 4 3 ). Follow Example 3 to make a change of variable, write the corresponding integral, and

verify that it is equal to 1 4 F (sin −1 4 5 , 3 4 ).

25. As in Problem 24, show that 1 2 2 F (sin −1 4 15 , 5 2 )= 1 5 F (sin −1 2 3 , 5 ).