1
$\begingroup$

Recently, I have been trying to use Jacobian Elliptic Functions to solve for angular velocities in the Euler's Equations of Motion, which look like this: $$ I_1\dot\omega_1-\omega_2\omega_3(I_2-I_3)=0\\ I_2\dot\omega_2-\omega_3\omega_1(I_3-I_1)=0\\ I_3\dot\omega_3-\omega_1\omega_2(I_1-I_2)=0 $$ Where $I_n$'s are constants. According to this paper, if we assume that $I_1<I_2<I_3$, we can express the angular velocities $\omega_n$ as: $$ \omega_1=\sqrt{ \frac{L^2 − 2T I_1}{ I_1(I_3−I_1)}} sn(\tau, k)\\ \omega_2=\sqrt{ \frac{L^2 − 2T I_1}{ I_2(I_3−I_2)}} sn(\tau, k)\\ \omega_3=\sqrt{ \frac{L^2 − 2T I_1}{ I_3(I_3−I_1)}} dn(\tau, k) $$ Where $L$ and $T$ are expressed by: $$ L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2\\ 2T=I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2 $$ However both are constants, so the only thing needed to express them are the three $\omega_n$'s at some point in time. $\tau$ and $k$ are the following: $$ \tau=\sqrt{\dfrac{(L^2−2TI_1)(I_3−I_2)}{\sqrt{I_1I_2I_3}}} dt \\ k=\sqrt{\dfrac{(I_2−I_1)(2TI_3−L^2)}{(I_3−I_2)(L^2−2TI_1)}} $$ My question is the following: If I want to use the Jacobian functions to find $\omega_n$, assuming that I know all the $I_n$'s and the initial $\omega_n$'s, do I just input all the values into the function, substituting RHS of the ninth equation for $\tau$ and treating $dt$ as $t$? I have tried that, but the results were different from when I used Mathematica to numerically integrate the first three equations. Also, sometimes a hole in the graph appeared when using the Jacobian function - the graph wasn't continuous. What am I doing wrong?

Also, one of the assumptions stated in the article was that $t=0$ when $\omega_2=0$. Obviously, I followed this assumption, but what should I do if I want to integrate for a case when $\omega_2$ does never equal? Or if it does at some point in time, but that point is unknown and only some other non-zero initial conditions are given?

$\endgroup$
  • 1
    $\begingroup$ In the paper, the $d$ in equation(16) is a typo. The expression for equation(22) also have typos, the one appear in the formula of $\omega_1$ should be $cn(\tau,k)$. I'm not sure about the correctness of that scaling constant, you should check it again. Another thing you should know is Mathematica uses a different convention for the arguments in Jacobic elliptic function. For example, the $m$ in JacobiSN[z, m] is equal to $k^2$ in the paper. $\endgroup$ – achille hui Oct 16 '15 at 20:27
  • $\begingroup$ Thank you, this helps a lot. If you also know the answer to the last question "What if I want to choose a different starting point that $\omega_2=0$", than you should post your comment as an answer $\endgroup$ – Bruno KM Oct 17 '15 at 9:17
0
$\begingroup$

HINT:

Since the derivatives of the three basic Jacobi elliptic functions are, (Hille,Zwillinger)

$$ \frac{\mathrm{d}}{\mathrm{d}z}\, \mathrm{sn}\,(z) = \mathrm{cn}\,(z)\, \mathrm{dn}\,(z),$$

$$ \frac{\mathrm{d}}{\mathrm{d}z}\, \mathrm{cn}\,(z) = -\mathrm{sn}\,(z)\, \mathrm{dn}\,(z),$$

$$ \frac{\mathrm{d}}{\mathrm{d}z}\, \mathrm{dn}\,(z) = - k^2 \mathrm{sn}\,(z)\, \mathrm{cn}\,(z). $$

with some re-arrangements they could be be directly expressed as required.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.