A particle is subjected to the potential $V (x) = −F x$, where $F$ is a constant. The particle travels from $x = 0$ to $x = a$ in a time interval $t_0 $. Assume the motion of the particle can be expressed in the form $x(t) = A + B t + C t^2$. Find the values of $A, B$ and $C$ such that the action is a minimum.

I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.


$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0\implies m\ddot{x}=F\implies \ddot{x}=\frac{F}{m}$$

Differentiate $x(t)$ twice. $$2C=\frac{F}{m}\implies C=\frac{F}{2m}$$

For finding B I was thinking to integrate $\ddot{x}$ once. $$\dot{x}=\int \ddot{x} \mathrm dt =\ddot{x}t$$

initial position is 0 so, not writing constant.


Differentiate $x(t)$ once.


$$\implies B=\frac{F}{m}-\frac{2Ft}{2m}=-\frac{Ft}{2m}$$

Again, going to integrate $\ddot{x}$ twice.

$$x=\iint \ddot{x} dt dt=\frac{\ddot{x}t^2}{2}$$

initial velocity and initial position is 0.




According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.

A person were saying that "It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory."

In my work where should I put the interval?


2 Answers 2


The Euler-lagrangian equation gives the equations of motion that once solved give you a family of solutions that minimize the action. A unique solution is given by specifying boundary conditions. It is just a case of inputing those boundary conditions.

Wlog let $ x(0)=0 $ and $x(t_0)=a $. Integrating $\ddot{x} = \frac{F}{m}$ gives the general solution $x(t)=\frac{F}{2m}t^2 +Bt + A$, fixing C. Subbing in $x(0)=0$ gives $A=0$ and subbing $x(t_0)=a$ gives $B$ as $B=\frac{a - Ct_0^2}{t_0}$.

  • $\begingroup$ Where the capital A had gone when finding equation for B. $$x(t_0)=a \\\\ => A+Bt_0+Ct_0^2=a \\\\ => B=\frac{a-A-Ct_0^2}{t_0}$$ $\endgroup$
    – user313317
    Sep 8, 2021 at 12:07
  • $\begingroup$ The A vanishes by setting wlog $x(0)=0$. You state in your question that the particle is at $0$ initially, and at $x=a$ after a time $t_0$. This is not sufficient information to get a solution mathematically, you must state a time for each boundary condition ie if $x(t_i)=0$ for the initial time $t_i$ then the other condition is $x(t_i+t_0)=a$. Since this labelling along the time axis is somewhat arbitrary (ie the physics is the same regardless of the value of $t_i$ you choose but mathematically the solution looks different) then is convenient to choose $t_i=0$ hence A=0. $\endgroup$ Sep 8, 2021 at 12:22

I suspect that this is not the approach that the problem author wants you to take. Were I confronted with this problem, I would do the following:

  • Determine what the conditions $x(0) = 0$ and $x(t_0) =a$ tell me about the parameters $A$, $B$, and $C$. I should be able to eliminate two of them in favor of the third using these two equations.
  • Plug the resulting form for $x(t)$ into the action and integrate it from $0$ to $t_0$.
  • Differentiate the resulting quantity with respect to the remaining parameter to find the value of the parameter which minimizes the action.

One should, of course, find that $C = F/2m$ after all this.

  • $\begingroup$ +1 but, I don't have reputation that's why I can't upvote... $\endgroup$
    – user313317
    Sep 8, 2021 at 15:38

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