Derivative of complete Elliptical integral and show steps of solution.

Questions
1

Hello,

Please I use below code to calculate partial derivative of U with respect to r and y. I have couple of questions.

I know the derivative of E(k) =(E(k)-K(k))/k. However, the result shows that the answer is E(k) =(E(k)-K(k))/2*k. I do not know why there is 2 in denominator.

Where K(k) and E(k) are complete elliptical integral kind one and two

Clear[r, y, Y, k, U, Ur, Uy]
k = r^3 + y^3;
U := EllipticE[k];
Ur = D[U, y] // FullSimplify
Uy = D[U, r] // FullSimplify

Result from Mathematica is

(3 y^2 (EllipticE[r^3 + y^3] - EllipticK[r^3 + y^3]))/(2 (r^3 + y^3))
(3 r^2 (EllipticE[r^3 + y^3] - EllipticK[r^3 + y^3]))/(2 (r^3 + y^3))

The answer should be

    (3 y^2 (EllipticE[r^3 + y^3] - EllipticK[r^3 + y^3]))/((r^3 + y^3))
    (3 r^2 (EllipticE[r^3 + y^3] - EllipticK[r^3 + y^3]))/((r^3 + y^3))

Thanks

2

Is there anyway to show the solution of derivative step by step?

This has been asked many times before. You can check this question

get-a-step-by-step-evaluation-in-mathematica and How to show steps in Wolfram Mathematica?

3

Hello,

Thank you for your reply. It is not the same result.

Thanks

4

The factor of 2 in the denominator appears to be correct:

http://functions.wolfram.com/EllipticIntegrals/EllipticE2/20/01/02/

It seems that the result without the factor of 2 in the denominator, that one finds in a number of places (e.g., http://mathworld.wolfram.com/CompleteEllipticIntegraloftheSecondKind.html and https://en.wikipedia.org/wiki/Elliptic_integral#Derivative_and_differential_equation and http://dlmf.nist.gov/19.4) is not correct. (But, I am going to sit down and re-derive it in the old fashioned way by hand later today…)
[[Edit: see my comment below–both are correct, but the definitions of the integral are slightly different and that accounts for the confusion.]]

Consider the derivative of the argument of the Elliptic integral

D[(1 - z Sin[\[Theta]]^2)^(1/2), z]

which gives

-(Sin[\[Theta]]^2/(2 Sqrt[1 - z Sin[\[Theta]]^2]))

Now numerically integrate that (as a test case) at the value of z=1/4:

NIntegrate[
 Evaluate[-(Sin[\[Theta]]^2/(2 Sqrt[1 - z Sin[\[Theta]]^2])) /. 
   z -> 1/4], {\[Theta], 0, \[Pi]/2}]

which gives

-0.436576

and this is the same as the value

(EllipticE[z] - EllipticK[z])/(2 z) /. z -> 1/4.

I do not suspect that this is a branch cut issue though perhaps someone can advise on that.

5

Hello,

Thank you for your reply. Really I do not know if the correct answer is with 2 in denominator or no. I am confused now.

I will wait to your reply.

Regards

6

I just returned home, pulled out my ancient copy of Whittaker and Watson, and now see what the issue is. In the documentation for EllipticE in Mathematica one sees that it is defined as

Integrate[(1 - m Sin[t]^2)^(1/2), {t 0, Pi/2}]

whereas other definitions use m^2 as in

Integrate[(1 - m^2 Sin[t]^2)^(1/2), {t 0, Pi/2}]

and this will account for the factor of 1/2 difference between the usages.

So everyone is correct but the definitions of how the modulus parameter is used in the definition differ.

7

Of course, my rational self told me that Whittaker and Watson could not possibly have been wrong!

8

Hello,

Thank you so much for your reply.

For this code. I am trying to find partial derivative of U with respect to y. I have k^2 = ((4*R*r)/((y - Y)^2 + (r + R)^2)).

My question is.

Do I have to define k= ((4Rr)/((y - Y)^2 + (r + R)^2)) or k=((4Rr)/((y - Y)^2 + (r + R)^2))^1/2`

In order to get denominator without 2? In other words how I can define k to get answer without 2? I still got answer with 2 even when I define k in both cases.

Clear[r, R, y, Y, k, U, Ur, Uy]
k = ((4*R*r)/((y - Y)^2 + (r + R)^2))^1/2;
U = EllipticE[k];
Uy = D[U, y]

The result

-(((y - Y) (EllipticE[(2 r R)/((r + R)^2 + (y - Y)^2)] - 
    EllipticK[(2 r R)/((r + R)^2 + (y - Y)^2)]))/((r + R)^2 + (y - 
    Y)^2))
9

I was going to mention this. There are indeed two common conventions for the modulus parameter, and that does give rise to confusion.