Doubt in the expression of Lagrangian of a system [duplicate]

Question

There is a problem given in Goldstein's Classical Mechanics Chapter-1 as

20. A particle of mass $\,m\,$ moves in one dimension such that it has the Lagrangian \begin{equation} L\boldsymbol{=}\dfrac{m^2\dot{x}^4}{12}\boldsymbol{+}m\dot{x}^2V\!\left(x\right)\boldsymbol{-}V^2\!\left(x\right) \nonumber \end{equation} where $\,V\,$ is some differentiable function of $\,x$. Find the equation of motion for $\,x\left(t\right)\,$ and describe the physical nature of the system on the basis of this equation.

$L=T-U$ is only for the cases when the all the forces acting on the system is conservative and the potential is dependent on coordinates only.
In the above case, the potential is dependent on coordinates only.
So, our Lagrangian has conservative forces and there may or may not be some non-conservative forces included in the Lagrangian.
After solving the Lagrange's equation I get the following equation of motion,
$$\Big(m\ddot{x}+\frac{dV}{dx}\Big)\Big(m\dot{x}^2+2V(x)\Big)=0.$$

I have the following doubts

1. We know that when we use generalized coordinates the kinetic energy term involves
   $T_o$(independent of $\dot{x}$) + $T_1$ (linear in $\dot{x}$) + $T_2$ (quadratic in $\dot{x}$).
   But in the Lagrangian how we get the term $(\dot{x})^4$ (biquadratic term in generalized in velocity)? How is this possible?

2. The equation which we get in the bold what is its physical implications?
   Like we can say that
   Either $\Big(m\ddot{x}+\frac{dV}{dx}\Big)=0$ or $\Big(m\dot{x}^2+2V(x)\Big)=0$ or both
   $\implies$ Either $m\ddot{x}=-\frac{dV}{dx}=F_{conservative}$ or $\frac{1}{2}m\dot{x}^2+V(x)=0$ or both.
   $m\ddot{x}=F_{conservative}$ means that the only force acting in the system is a conservative force.
   But what does the second term $\frac{1}{2}m\dot{x}^2+V(x)=0$ implies? If we interpret $\frac{1}{2}m\dot{x}^2$ as kinetic energy then we can say total energy is 0. But I think we can't interpret it as kinetic energy because then in Lagrangian there can't be a term biquadratic in $\dot{q}$.

I am able to solve the Lagrange's equation but I am not able to resolve my above two doubts which in short are 1) structure of Lagrangian and 2) the physical interpretation of motion.

Please help me in clarifying the doubt.

Addendum -
I have one more doubt which is very closely related to this one, so I don't want to ask a new question for that.
We know that if $L'=L+\frac{dF(q,t)}{dt}$, then $L'$ gives same equation of motion as that given by the $L$ after solving the Lagrange's equation.
In the above example, $L\boldsymbol{=}\dfrac{m^2\dot{x}^4}{12}\boldsymbol{+}m\dot{x}^2V\!\left(x\right)\boldsymbol{-}V^2\!\left(x\right)$ gives same equation as that of $L=\frac{m\dot{x}^2}{2}-V(x)$.

So, I want to know is there some specific class of functions on the Lagrangian which gives same equation of motion as that of the original or simplest Lagrangian $(L=T-V)$ after solving the Lagrange's equation?

Answer 1

One cool thing about Lagrangian mechanics that Goldstein seems to want to point out with this problem is: there is no uniqueness in the Lagrangian. You can get different Lagrangians which, at least classically, lead to the very same results. In other words, some equations of motion can be derived from a Lagrangian that looks nothing like $L = T - U$.

The amazing problem you solved is an example. As you pointed out, there are a few terms that one wouldn't find in a conservative potential or in the kinetic energy, but still the equations of motion are (I'm trusting your calculations) $$\left(m\ddot{x} + \frac{dV}{dx}\right)\left(\frac{1}{2}m\dot{x}^2+V(x)\right) = 0.$$

Suppose the second factor vanishes. Differentiating $\frac{1}{2}m\dot{x}^2+V(x)=0$ with respect to time leads to $m\ddot{x} + \frac{dV}{dx} = 0$, and hence the other one is satisfied. If we try to do the inverse, we'll simply get $\frac{1}{2}m\dot{x}^2+V(x)=E$ for some constant $E$. Therefore, in general your equations of motion tell us that $$\frac{1}{2}m\dot{x}^2+V(x)=E,$$ for some constant $E$. Taking a time derivative leads to $$m\ddot{x} + \frac{dV}{dx} = 0,$$ which are the exact same conclusions we'd get provided we considered the Lagrangian $$L = \frac{m \dot{x}^2}{2} - V(x)$$ in first place.

I should also point out that in this sense we can consider a wider class of Lagrangians than the usual ones, and sometimes this will allow us to describe different phenomena or simply will make calculations a bit easier. While in Classical Mechanics you might find no reason to work with this weird Lagrangian instead of the more simple one which leads to the same equations of motion, in other contexts one of them might be much easier to manipulate.