Help needed with Elliptic Integrals

[TerryW]

Homework Statement

See Attempt at a Solution

Relevant Equations

See Attempt at a Solution

I want to find the solution to the integral

$\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}$

I can see that $\frac{d^2u}{d\theta^2} = A +Bu+Cu^2$ is a Weierstrass elliptic function, which can be generated from $\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3$ (A = 0, B=-1, C=3)

So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck.

TerryW

 

[renormalize]

I want to find the solution to the integral

$\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}$

Mathematica can evaluate the indefinite integral on the right side:

Here, $\text{EllipticF}\left[\phi,m\right]$ is the elliptic integral of the 1rst kind $F\left(\phi\mid m\right)$ and the $r_k$ are the 3 roots of the cubic polynomial equation $c-r^2+2\,r^3=0$. According to the documentation, the roots are indexed as follows:
"The root indexing representation $\text{Root}\left[f,k\right]$ applies to polynomial functions $f$ only. The indexing of roots takes the real roots first, in increasing order. For polynomials with rational coefficients, the complex conjugate pairs of roots have consecutive indices."

 

[TerryW]

Many thanks for this.

I think I can console myself that I didn't miss an easy route to a solution! It's interesting that the elliptic integral of the first kind appears in the solution, but where the rest of it comes from may remain a mystery to me.

I'm still puzzled by the fact that on the one hand there are statements like:

"Modern mathematics defines an 'elliptic integral' as any function $f$ which can be expressed in the form $f(x) = \int_c^xR(t,\sqrt{P(t)})dt$" (from Wikipedia)

but then Wikipedia does not go beyond the elliptic integral of the third kind.

Do you know of any reference for elliptic integrals in general?

TerryW

 

[renormalize]

Do you know of any reference for elliptic integrals in general?

There are doubtless entire monographs devoted to elliptic integrals, but one concise reference is:
Handbook of Mathematical Functions by Abramowitz and Stegun, chap. 17
I happen to have a print copy (from 1968!) but you can find PDF versions floating around on the web.

 

[TerryW]

I've found a pdf of Abromowitz and Stegun. I'll see if it clarifies things for me.

Looking at the result from Mathematica, I wonder if it is actually correct? For example, the term $(r2-r3)^2$ on the denominator of the second square root cancels out the term $(r2-r3)$ at the beginning of the expression. Also, if r1, r2, and r3, are the roots of my cubic, doesn't $\sqrt{(r1-u)}\sqrt{(r2-u)(-r3+u)}$ cancel out $\sqrt{(c+u^2(-1+2u)}$?

Or am I just misunderstanding the formulation?

Maybe I should try and find the derivative of the expression WRT u.

Regards and thanks for the reference.

TerryW

 

[renormalize]

Looking at the result from Mathematica, I wonder if it is actually correct? For example, the term $(r2-r3)^2$ on the denominator of the second square root cancels out the term $(r2-r3)$ at the beginning of the expression. Also, if r1, r2, and r3, are the roots of my cubic, doesn't $\sqrt{(r1-u)}\sqrt{(r2-u)(-r3+u)}$ cancel out $\sqrt{(c+u^2(-1+2u)}$?

Good observation! You're absolutely correct, up to a sign: depending on the specific values of the roots $r_2,r_3\,$, the two ratios you cite reduce to:$$\frac{r_{2}-r_{3}}{\sqrt{\left(r_{2}-r_{3}\right)^2}}=\pm1,\quad\frac{\sqrt{(r_1-u)}\sqrt{(r_2-u)(-r_3+u)}}{\sqrt{(c+u^{2}(-1+2u)}}=\frac{1}{\sqrt{2}}$$so the Mathematica expression for the integral becomes simply:$$\pm\,\frac{\sqrt{2}\,F\left(\sin^{-1}\left(\frac{u-r_{3}}{r_{2}-r_{3}}\right)\mid\frac{r_{2}-r_{3}}{r_{1}-r_{3}}\right)}{\sqrt{r_{1}-r_{3}}}$$You'll have to determine the overall sign by examining the roots of your particular cubic.

 

[TerryW]

Good observation! You're absolutely correct, up to a sign: depending on the specific values of the roots $r_2,r_3\,$, the two ratios you cite reduce to:$$\frac{r_{2}-r_{3}}{\sqrt{\left(r_{2}-r_{3}\right)^2}}=\pm1,\quad\frac{\sqrt{(r_1-u)}\sqrt{(r_2-u)(-r_3+u)}}{\sqrt{(c+u^{2}(-1+2u)}}=\frac{1}{\sqrt{2}}$$so the Mathematica expression for the integral becomes simply:$$\pm\,\frac{\sqrt{2}\,F\left(\sin^{-1}\left(\frac{u-r_{3}}{r_{2}-r_{3}}\right)\mid\frac{r_{2}-r_{3}}{r_{1}-r_{3}}\right)}{\sqrt{r_{1}-r_{3}}}$$You'll have to determine the overall sign by examining the roots of your particular cubic.

Which brings me back to what was my original problem, which is to work out how to get my cubic expression ($\int\frac{du}{\sqrt{((c+u^2(-1+2u))}}$) into one of the forms of elliptic integral which has a solution in the form of $F(sin^{-1}\theta|\alpha)$. I'll see if Abromowitz and Stegun can help with this. I'll post my result if I manage to do this. One more observation - why should r3 be specially selected as part of the $sin^{-1}$ expression?

 

[renormalize]

I'll see if Abromowitz and Stegun can help with this.

Or, assuming the roots $r_k$ of your cubic are all real, you can perhaps simply look up your integral in Gradshteyn and Ryzhik, Table of Integrals, Series, and Products:

 

[TerryW]

That's great!

Thanks for all your help with this.

TerryW

 

[JimWhoKnew]

assuming the roots $r_k$ of your cubic are all real,

It is easy to see that all roots will be real if and only if $~0\leq c \leq 1/27~$.

(The case $~u=1/3~,~c=1/27~$ corresponds to the circular orbit of light at $~r=3M~$ around a Schwarzschild BH, where $~\frac{du}{d\phi}=0~$)