Lagrangian of the Euler equations Do the Euler equations (where $I_1,I_2,I_3$ are principal moments of inertia):
$$I_1\dot{\omega}_1+(I_3-I_2)\omega_2\omega_3=M_1$$
$$I_2\dot{\omega}_2+(I_1-I_3)\omega_3\omega_1=M_2$$
$$I_3\dot{\omega}_3+(I_2-I_1)\omega_1\omega_2=M_3$$
in their above general form have a Lagrangian? If not, does a specific case of $\omega_1=\omega_2=0$ (and so $M_1=M_2=0$) have a general Lagrangian? ($M_3$ is the torque coming from a central gravitational potential - a planet - keeping the body (a satellite) on an elliptic orbit.)
 A: It's some time since I worked this, but I believe that the Eqs  of motion  for $\omega$ ${\bf p}$, ${\bf r}$ from the action functional
$$
S[\omega, {\bf p}, {\bf r}]= \int \left(\frac 12 I_1\omega_1^2+\frac 12 I_2\omega_2^2+\frac 12 I_3\omega_3^2+ {\bf p}\cdot (\dot {\bf r}+
\omega \times {\bf r})\right)dt,
$$ 
after the elimination of ${\bf p}$ and $ \dot {\bf r}$ in favour of the angular velocity vector $\omega$ gives the Euler equations for $\omega$ in the case $M_1=M_2=M_3=0$. A check strongly recommended  as  I posted a totally stupid wrong ``answer'' this morning, and am doubting my comptetence.   
OK a check:  The equations of motion that come from varying the vectors $\omega$, ${\bf p}$, ${\bf r}$, are (in that order)
$$
{\bf L} \equiv  {\bf I}\omega= -({\bf r}\times {\bf p})\\
\dot {\bf r}= - (\omega\times {\bf r})\\
\dot {\bf p}= - (\omega \times {\bf p})
$$
Then 
$$
\dot {\bf L} = -(\dot {\bf r}\times {\bf p})-({\bf r}\times\dot {\bf p})\\
= -{\bf p}\times(\omega\times {\bf r})- {\bf r}\times({\bf p}\times \omega)\\
= - \omega\times ({\bf p}\times {\bf r})\\
= - \omega\times {\bf L}.
$$ 
In the last-but-one step I have use the vector triple-product Jacobi identity
$$
{\bf a}\times ({\bf b}\times {\bf c}) +{\bf b}\times ({\bf c}\times {\bf a})+{\bf c}\times ({\bf a}\times {\bf b})=0.
$$
Writing  out
$$ 
\dot{\bf  L}+\omega\times {\bf L}=0
$$
with ${\bf L}= (I\omega_1, I_2\omega_2,I_3\omega_3)$ yields the torque-free Euler equations. 
The idea of using Lin constraints to derive  Euler's equations  isquite modern.  I think I saw this done in  Cendra and Marsden "Lin Constraints Clebsh potentials  and Variational Principles" Physica 27D (1987) 63-89. 
A: They do.  In the case of the symmetric top with $I_1=I_2\ne I_3$ for instance, the kinetic energy is given by  (if my algebra is right)
\begin{align}
T_{\hbox{rot}}&=\textstyle\frac{1}{2}\sum_k I_k\omega_k^2\, ,\\
&=\textstyle\frac{1}{2}I_1\left((\dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi)^2+
(\dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi)^2\right)
+\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, ,\\
&=\textstyle\frac{1}{2}I_1\left(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2\right)
+\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, .
\end{align}
where the expressions
\begin{align}
\omega_1&= \dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi   \\
\omega_2&= \dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi \\
\omega_3&= \dot{\psi}+\dot{\phi}\cos(\theta)\, ,
\end{align}
transition from the the angular frequencies to the time-derivatives of Euler angles.
