Imagine there is a mechanical system described in unitary units by the equation: $$\dot{x} = -\text{sgn}(x)\sqrt{|x|},\quad x(0)=1 \tag{Eq. 1}$$ such it has a finite duration solution: $$x(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2 \tag{Eq. 2}$$

Is this enough information to reconstruct its Kinetic and Potential Energies to obtain its Lagrangian of this System and its Least Action Principle's integral? What are these values in terms of $x(t)$?


Recently I have learned about the existence of finite duration solutions of differential equations on these papers: Finite Time Differential Equations and Finite Time Controllers by Vardia T. Haimo, and since everyday phenomena are of finite duration, I want to know how will behave the Energy and the Least Action Principle on this kind of system with finite duration solutions, and this is the only example I have so far of an autonomous system that stands finite duration solutions (maybe $\dot{y} = -\sqrt{y},\,y(0)=1$ also works if the solutions is restrained to the reals, since after $(y,\,\dot{y})=(0,\,0)$ the derivative is never going to rise up again since the square root is positive).

I am trying to make a mechanical system with $x(t)$ the solution to their equation of motions, not in the other way.

  • $\begingroup$ Mechanical system is described by second order differential equation $\endgroup$
    – Eli
    Mar 27 at 7:41
  • $\begingroup$ @Eli I left the explicit solution $x(t)$: It is not possible to obtain the equations of motion by differentiated it twice? $\endgroup$
    – Joako
    Mar 28 at 19:32
  • $\begingroup$ the equations of motion are described by the Newton second law. $\endgroup$
    – Eli
    Mar 28 at 20:40

1 Answer 1

  1. A 1st-order ODE $$\dot{x}~=~f(x,t)\tag{A}$$ cannot be a Euler-Lagrange (EL) equation if we are only allowed to use a single real variable $x(t)$, cf. e.g. this Math.SE post.

  2. With several variables it is easy. We can e.g. use a Lagrange multiplier $$ S[x,\lambda]~=~ \int_{t_i}^{t_f}\!dt~ \lambda(\dot{x}-f(x,t)) \tag{B}$$

  3. Care should be taken wrt. boundary conditions, so that they are compatible with the physical system at hand, i.e. one might need to add boundary terms to the action (B).

  • $\begingroup$ Thanks for answering. Since the explicit solution is given, I was expecting to form the kinetic energy as $K=\frac{1}{2}\dot{x}$ ($m=1$, and $\dot{x}$ from the equation), use a generic potential $V$ and since the $\int\,K - V\,dt$ should give the same equation for $\dot{x}$ one could determine $V$ (again, this because I know the explicit solution of $x(t)$). If this approach is mistaken, It could be possible to obtain the equation of motion from the Work-Energy Theorem? (since I have the initial Kinetic energy, and the final Kinetic Energy is zero - the solution $x(t)$ is of finite-duration) $\endgroup$
    – Joako
    Mar 28 at 19:25
  • 2
    $\begingroup$ Hi Joako, For starters, your proposed kinetic term $K=\frac{1}{2}\dot{x}$ is a total time derivative, and hence does not contribute to the EL equation. $\endgroup$
    – Qmechanic
    Mar 28 at 19:47
  • $\begingroup$ Let me see If I am following you: From mechanics principles (like Newton´s laws, Lagrangian/Hamiltonian, others) it is possible to obtain the equations of motion, but from the equation of motion, is not always possible to reconstruct these principle's equations back? $\endgroup$
    – Joako
    Mar 28 at 20:01
  • $\begingroup$ Forget variational principles for a moment. How did you/Can you derive eq. (1) from Newton's laws? $\endgroup$
    – Qmechanic
    Mar 28 at 20:08
  • $\begingroup$ Don´t really know... I am trying to form a mechanical system that can stand a solution of finite duration from the only example with explicit solution I know, so I am working backwards.... an alternative idea could be differentiating one time the equation $\dot{y} = -\sqrt{y}$ which also stand the same solution (since is also non Lipschitz at zero)... the second derivative will be the $F\cdot a$ term and the other side a potential, something similar to the Norton's Dome but with convergent instead of divergent solutions. $\endgroup$
    – Joako
    Mar 28 at 20:16

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