E. I. KHARLAMOVA, P. V. KHARLAMOV

A NEW CASE OF INTEGRABILITY OF THE EQUATIONS OF MOTION OF A HEAVY RIGID BODY HAVING A FIXED POINT

(Presented by Academician P. Ya. Kochina on 28 II 1969)

The Euler–Poisson equations for the problem of the motion of a heavy rigid body in the case when the center of gravity lies on the first principal axis, and the gyrostatic moment is orthogonal to the third axis, have the form

\[ \begin{aligned} A\,dp/dt &= (B-C)qr+\lambda_2 r,\\ B\,dq/dt &= (C-A)rp-\lambda_1 r+\Gamma v_1,\\ C\,dr/dt &= (A-B)pq+\lambda_1 q-\lambda_2 p+\Gamma v_2,\\ dv_1/dt &= rv_2-qv_3,\quad dv_2/dt = pv_3-pv_1,\quad dv_3/dt = qv_1-pv_2. \end{aligned} \tag{1} \]

Here \(v_1, v_2, v_3\) are the components of the unit vector in the direction of the force of gravity; \(\Gamma\) is the product of the weight of the body and the distance from the fixed point to the center of gravity; \(\lambda_1, \lambda_2, 0\) are the gyrostatic moment; the remaining notation is standard.

The known integrals of these equations are:

\[ Ap^2+Bq^2+Cr^2-2\Gamma v_1=2E, \]

\[ (Ap+\lambda_1)v_1+(Bq+\lambda_2)v_2+Crv_3=k/\Gamma, \]

\[ v_1^2+v_2^2+v_3^2=1. \]

Let us subject the moments of inertia to the condition

\[ A=18C(B-C)/(10B-9C), \]

and, instead of \(\lambda_1,\lambda_2\), introduce the parameters \(u\) and \(\varepsilon\), putting

\[ \lambda_1=\frac{u}{9}\, \frac{25B^2-54BC+27C^2}{B^2C(10B-9C)}\,n\cos\varepsilon, \quad \lambda_2=-\frac{u}{3}\, \frac{(B-C)^2}{B^2C^2}\,n\sin\varepsilon. \]

Equations (1) then have, in this case, a solution in which the principal variables \(p,q,r,v_1,v_2,v_3\), expressed as functions of the new variable \(\sigma\), are as follows:

\[ p=-u\,\frac{B-C}{ABC} \left(\frac{1}{B}\frac{5B-3C}{10B-9C}n\cos\varepsilon+2\cos\sigma\right), \]

\[ q=\frac{u}{3BC} \left(\frac{B-C}{BC}n\sin\varepsilon-2\sin\sigma\right), \]

\[ r=\frac{u}{C} \left(r_0+r_1\cos\sigma+r'_1\sin\sigma+r_2\cos2\sigma+r_3\cos3\sigma+r'_3\sin3\sigma\right)^{1/2}, \]

\[ v_1=-\frac{u^2}{\Gamma}\frac{2B-3C}{36BC^2} \left(\chi_0+\chi_1\cos\sigma+\chi'_1\sin\sigma+\chi_2\cos2\sigma+\chi_3\cos3\sigma+\chi'_3\sin3\sigma\right), \]

\[ v_2=-\frac{u^2}{\Gamma}\frac{2B-3C}{36BC^2} \left(\chi_0+\chi_1\cos\sigma+\chi'_1\sin\sigma+\chi'_2\sin2\sigma+\chi_3\cos3\sigma+\chi'_3\sin3\sigma\right), \]

\[ v_3=\frac{u}{\Gamma}\frac{1}{18C}\left[-\frac{B-C}{B^2C}h\cos\varepsilon+ \frac{6(2B-3C)}{BC}\cos\sigma\right] \left(r_0+r_1\cos\sigma+r'_1\sin\sigma+ \right. \]
\[ \left. +r_2\cos2\sigma+r_3\cos3\sigma+r'_3\sin3\sigma\right)^{1/2}, \]

where

\[ r_0=2E_*-\frac{2B-3C}{27B^2C^3}(16B^2+13BC-18C^2)+ \]
\[ +\frac{1}{162B^3C^2}\left[18\frac{(B-C)^2}{C} -\frac{(5B-3C)(35B-33C)}{10B-9C}\right]n^2\cos^2\varepsilon, \]

\[ r_1=-\frac{1}{36}\frac{(B-C)(50B-27C)}{B^3C^2}\,n\cos\varepsilon, \]

\[ r'_1=\frac{1}{36}\frac{(B-C)(10B+9C)}{B^3C^2}\,n\sin\varepsilon, \]

\[ r_2=-\frac{1}{9}\frac{(2B-3C)(11B-6C)}{B^2C^2},\qquad n^2=\frac{4}{3}\frac{(2B-3C)(4B-3C)B^2}{(B-C)^2}, \]

\[ r_3=-\frac{1}{12}\frac{(B-C)(2B-3C)}{B^3C^2}\,n\cos\varepsilon, \]

\[ r'_3=-\frac{1}{12}\frac{(B-C)(2B-3C)}{B^3C^2}\,n\sin\varepsilon, \]

\[ \varkappa_0=\frac{3}{2}\frac{10B-9C}{BC} -\frac{(B-3C)(5B-3C)}{9B^3C(2B-3C)}\,n^2\cos^2\varepsilon, \]

\[ \varkappa_1=\frac{1}{3B^2C}\left(\frac{5B+3C}{2} +\frac{10B^2-15BC+9C^2}{2B-3C}\right)n\cos\varepsilon, \qquad \varkappa'_1=\frac{3}{2}\frac{B-C}{B^2C}\,n\sin\varepsilon, \]

\[ \varkappa_2=\frac{6(2B-C)}{BC},\qquad \varkappa_3=\frac{3(B-C)}{2B^2C}\,n\cos\varepsilon,\qquad \varkappa'_3=\frac{3(B-C)}{2B^2C}\,n\sin\varepsilon, \]

\[ \chi_0=-\frac{2(B-C)(B-3C)}{B^3C(2B-3C)}\,n^2\cos\varepsilon\sin\varepsilon,\qquad \chi_1=-\frac{9}{2}\frac{B-C}{B^2C}\,n\sin\varepsilon, \]

\[ \chi'_1=\frac{1}{3B^2C}\left(-\frac{5B+3C}{2} +\frac{10B^2-15BC+9C^2}{2B-3C}\right)n\cos\varepsilon, \qquad \chi'_2=\frac{8}{C}, \]

\[ \chi_3=-\frac{3(B-C)}{2B^2C}\,n\sin\varepsilon,\qquad \chi'_3=\frac{3(B-C)}{2B^2C}\,n\cos\varepsilon, \]

\[ E_*=\frac{4}{27}\frac{(B-2C)(4B-3C)}{BC^3} +\frac{1}{5184B^2}\left[ 3\frac{(B-3C)^2}{B^2C^2}\frac{10B-7C}{2B-3C} +9\frac{5B-3C}{B^2C^2}\times \right. \]
\[ \left. \times\frac{6B-7C}{10B-9C} -2(34B-51C)(5B-3C)\frac{B-3C}{2B-3C} -36\frac{(B-C)^2(8B-9C)}{B^2C^3} \right]n^2\cos^2\varepsilon \]

The dependence between \(\sigma\) and \(t\) is established by inverting the integral

\[ t-t_0=\frac{3(B-C)}{uB}\int_0^\sigma \left(r_0+r_1\cos\sigma+r'_1\sin\sigma+r_2\cos2\sigma+ \right. \]
\[ \left. +r_3\cos3\sigma+r'_3\sin3\sigma\right)^{-1/2}\,d\sigma \]

The parameters \(B,C,\Gamma,\varepsilon\) in this solution are independent.

Donetsk Computing Center
of the Academy of Sciences of the Ukrainian SSR

Received
21 II 1969