The Tsiolkovsky equation derived via the Lagrange funtion I wanted to derive the Tsiolkovsky rocket equation using the Lagrange formalism.
$$ T = \frac{m(t)}{2}\dot{\xi(t)}^2 + \frac{\mu(t)}{2}(\dot{\xi(t)} - u)^2$$
$m$ - mass of the rocket
$\xi$ - some coordinate along which the rocket is moving
$\mu$ - mass of the thrown out fuel
$u$ - fuel velocity (in the rocket's reference frame).
The standard procedure yields the following equation of motion:
$$\ddot{\xi} (m+\mu) + \dot{\xi} (\dot{m} + \dot{\mu}) = \dot{\mu} u.$$
Since $m(t) = m_0 - \eta t$, and $\mu = \eta t$, there follow $\dot{m} + \dot{\mu} = 0$, $m+\mu = m_0$, and $\dot{\mu} = - \dot{m}$.
Therefore we have
$$m_0 \frac{d}{dt}\dot{\xi} = m_0 \ddot{\xi} = -u \eta.$$
This is not the anticipated result, the Tsiolkovsky equation is
$$\frac{d}{dt}\dot{\xi} = - u \frac{dm}{m},$$
where $m = m(t)$. Specifically, my issue is the fact that $m_0 \neq m(t)$.
Where is my mistake? Is the lagrangian wrong, or am I missing something else here?
 A: Your Lagrangian doesn't accurately describe the energy of the system.  You have a term in your kinetic energy
$$
\frac{\mu(t)}{2}(\dot{\xi(t)} - u)^2
$$
which is intended (I assume) to describe the kinetic energy $T_\text{fuel}(t)$ of all of the fuel that has been thrown out of the rocket up to time $t$.   But the fuel that's thrown out at a time $t_e < t$ will not be moving at speed $\dot{\xi}(t) - u$;  it will be moving at speed $\dot{\xi}(t_e) - u$.
We can try to write down an expression for $T_\text{fuel}(t)$ by integrating over the rocket's history.  The kinetic energy of the fuel that's thrown out between $t_e$ and $t_e + \Delta t_e$ will be
$$
\frac{\Delta \mu}{2}(\dot{\xi(t_e)} - u)^2 
$$
where $\Delta \mu \equiv \dot{\mu}(t_e) \Delta t_e$.   Thus, the total kinetic energy of the ejected fuel  will be
$$
T_\text{fuel}(t) = \frac{1}{2} \int_0^t \dot{\mu}(t_e) (\dot{\xi}(t_e) - u)^2 \, dt_e.
$$
Note that since your Lagrangian $\mathcal{L}$ will now involve an integral over the past history of the function $\xi(t)$, rather than simply depending on $\xi(t)$ and $\dot{\xi}(t)$.  This means that the plain-vanilla Euler-Lagrange equations $d/dt(\partial\mathcal{L}/\partial \dot{\xi}) = \partial\mathcal{L}/\xi$ cannot straightforwardly be applied.  You will instead have to return to first principles in the calculus of variations to derive the equations of motion.
