Deriving the Rocket Equation from Hamilton's Principle

Question

I've been trying to derive the rocket equation ($ m \ddot{x} = - \dot{m} v_e $) from Hamilton's Principle and I've hit a wall. The way I went about it was by extremizing the action $ S = \int L dt $, where $ L = K_{rocket} + K_{exhaust} $. That way the system (rocket + exhaust) is closed, and so in theory, Hamilton's Principle should apply. In particular, $ K_{rocket} = \frac{1}{2} m \dot{x}^2 $ and $$ K_{exhaust} = \int \frac{1}{2} (-dm) ( \dot{x} - v_e)^2 \implies K_{exhaust}(t) = - \frac{1}{2} \int_{t_0}^{t} \dot{m}(\tau) ( \dot{x}(\tau) - v_e)^2 d\tau .$$ This was more or less already addressed in the Stack Exchange (see The Tsiolkovsky equation derived via the Lagrange funtion), but inconclusively. In advance, thank you for helping or even just reading; I really appreciate it! (Please excuse any formatting issues — I’m new to Stack Exchange and still learning the ropes)

My issue comes down to this. Extremizing the action corresponding to $ L = K_{rocket} + K_{exhaust} $ does not result in the Rocket Equation as the EOM. However, extremizing the action corresponding to $ L = K_{rocket} - K_{exhaust} $ does result in the Rocket Equation as the EOM. I simply do not understand why. My work is as follows:

Suppose, in general, we have the following action:

$$ S = \int_{t_0}^{t_1} \left( G(t,x(t),\dot{x}(t))+\int_{t_0}^{t}f(\tau,x(\tau),\dot{x}(\tau))d\tau \right) dt $$

Now, $S$ can be rewritten as such:

$$ S = \int_{t_0}^{t_1} G(t,x(t),\dot{x}(t)) \; dt + \int_{t_0}^{t_1}\int_{t_0}^{t} f(\tau,x(\tau),\dot{x}(\tau)) \; d\tau \; dt \\ = \int_{t_0}^{t_1} G(t,x(t),\dot{x}(t)) \; dt + \int_{t_0}^{t_1}\int_{\tau}^{t_1} f(\tau,x(\tau),\dot{x}(\tau)) \; dt \; d\tau \\ = \int_{t_0}^{t_1} G(t,x(t),\dot{x}(t)) \; dt + \int_{t_0}^{t_1}\int_{t}^{t_1} f(t,x(t),\dot{x}(t)) d\tau \; dt \\ = \int_{t_0}^{t_1} G(t,x(t),\dot{x}(t)) \; dt + \int_{t_0}^{t_1} (t_1-t)f(t,x(t),\dot{x}(t)) \; dt \\ = \int_{t_0}^{t_1} G(t,x(t),\dot{x}(t)) + (t_1-t)f(t,x(t),\dot{x}(t)) \; dt $$

From here, the variation of the action will be taken with respect to $x$:

$$ \delta S = \delta \int_{t_0}^{t_1} G(t,x,\dot{x}) + (t_1-t)f(t,x,\dot{x}) \; dt \\ = \int_{t_0}^{t_1} \delta G(t,x,\dot{x}) + (t_1-t) \delta f(t,x,\dot{x}) \; dt \\ = \int_{t_0}^{t_1} \frac {\partial G}{\partial x} \delta x + \frac {\partial G}{\partial \dot{x}} \delta \dot{x} + (t_1-t) \left( \frac {\partial f}{\partial x} \delta x + \frac {\partial f}{\partial \dot{x}} \delta \dot{x} \right) dt \\ = \int_{t_0}^{t_1} \left( \left( \frac {\partial G}{\partial x} + (t_1-t) \frac {\partial f}{\partial x} \right) \delta x + \left( \frac {\partial G}{\partial \dot{x}} + (t_1-t) \frac {\partial f}{\partial \dot{x}} \right) \delta \dot{x} \right) dt $$

Then by using IBP and $ \delta x (t_0) = \delta x (t_1) = 0 $:

$$ \delta S = \int_{t_0}^{t_1} \left( \left( \frac {\partial G}{\partial x} + (t_1-t) \frac {\partial f}{\partial x} \right) \delta x - \frac{d}{dt} \left( \frac {\partial G}{\partial \dot{x}} + (t_1-t) \frac {\partial f}{\partial \dot{x}} \right) \delta x \right) dt \\ = \int_{t_0}^{t_1} \left( \frac {\partial G}{\partial x} + (t_1-t) \frac {\partial f}{\partial x} - \frac{d}{dt} \left( \frac {\partial G}{\partial \dot{x}} + (t_1-t) \frac {\partial f}{\partial \dot{x}} \right) \right) \delta x \; dt $$

Now, after applying Hamilton's Principle ($ \delta S = 0$):

$$ \frac {\partial G}{\partial x} + (t_1-t) \frac {\partial f}{\partial x} - \frac{d}{dt} \left( \frac {\partial G}{\partial \dot{x}} + (t_1-t) \frac {\partial f}{\partial \dot{x}} \right) = 0 \\ \implies \frac {\partial G}{\partial x} - \frac {d}{dt} \frac {\partial G}{\partial \dot{x}} + (t_1-t) \left( \frac {\partial f}{\partial x} - \frac {d}{dt} \frac {\partial f}{\partial \dot{x}} \right) + \frac {\partial f}{\partial \dot{x}} = 0 $$

Here's where I start making sketchy decisions... Let $t_1 = t$, effectively restricting the action to the time interval containing the initial and current time:

$$ \implies \frac {\partial G}{\partial x} - \frac {d}{dt} \frac {\partial G}{\partial \dot{x}} + \frac {\partial f}{\partial \dot{x}} = 0 $$

As a gut check, plugging in $ G(t,x,\dot{x}) = \frac{1}{2} m \dot{x}^2 $ and $ f(t,x,\dot{x}) = -kx \dot{x} = \frac{d}{dt} \left( - \frac{1}{2} kx^2 \right) $, which should be the same as choosing $ L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} kx^2 $, results in the EOM of $ \; m \ddot{x} + kx = 0 \; $ as expected.

Now, for the rocket-exhaust system, choosing $ G(t,x,\dot{x}) = \frac{1}{2} m \dot{x}^2 $ and $ f(t,x,\dot{x}) = - \frac{1}{2} \dot{m}(t) ( \dot{x} - v_e )^2 $, corresponding with $ L = K_{rocket} + K_{exhaust} $, results in an EOM of $ \; m \ddot{x} + 2 \dot{m} \dot{x} - \dot{m} v_e = 0 $, which is definitely not the rocket equation! What's bizarre is that choosing $ f(t,x,\dot{x}) = \frac{1}{2} \dot{m}(t) ( \dot{x} - v_e )^2 $ (keeping $G$ the same), corresponding with $ L = K_{rocket} - K_{exhaust} $, results in an EOM of $ \; m \ddot{x} + \dot{m} v_e = 0 $, which is the rocket equation! But how? What on Earth is going on?!

Answer 1

This is not an answer, it is a comment. (It's just that this comment is larger than the comment space limit.)

As background I will first present general discussion of differential equations and variational equations, and then I will discuss how that motivates a recommendation.

It seems to me that an effort to involve Hamilton's stationary action in deriving the Tsiolkovsky equation is a mathematical instance of what is known as a Rube Goldberg machine.

I notice that in the answer to the 2020 question you link to it is pointed out that:

[...] 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 $\dot{\xi}(t)$.

As I understand that:
If there is a Lagrangian for this problem it must feature that integral-over-the-history-of-the-function.

General consideration about application of Calculus of Variations:
The following is essential: the Euler-Lagrange equation is a differential equation.

I'd like to point out the properties that make differential equations so powerful.

Comparison:
There is the class of equations that has a number as its solution, an example of that is an equation to find the roots of a function; the solution space is a space of numbers.
By contrast: we have that the solution to a differential equation is a function; the solution space of a differential equation is a space of functions.

In the case of application of differential calculus in mechanics:
There is a trajectory, and the constraint is that the differential relation must be satisfied for all points along the trajectory concurrently.

The differential relation specifies a local relation, but of equal importance: the differential relation is to be satisfied for all points along the trajectory concurrently. That must-be-satisfied-everywhere-concurrently makes the differential equation a global equation.

About variational equations: As we know: the solution space of a variational equation is a space of functions.

The derivation of the Euler-Lagrange equation

The derivation of the Euler-Lagrange equation starts with a variational relation: over the course of the derivation all non-essential elements are dismissed.

The variational equation is stated in terms of a start point and an end point.
The EL-equation does not involve the start/end point, the boundaries have been dismissed.
The EL-equation does not feature integration: the integration has been dismissed.

The point is:
The Euler-Lagrange equation - a differential equation - has the same expressive power as the variational equation.

(More precisely: we can think of the Euler-Lagrange expression as a differential operator that is populated with a Lagrangian.)

Reminder:
The solution space of a differential equation is a space of functions.
As pointed out earlier: with a variational equation is it also the case that the solution space is a space of functions.

The point that I'm getting at: The equation you want to arrive at: $ \; m \ddot{x} + \dot{m} v_e = 0 $; a differential equation.

So your aim is to arrive at that equation using energy considerations (whereas the usual approach is to derive that equation using momentum considerations).

Given that the aimed for result is a differential equation it should be possible to arrive at it using differential reasoning only.