# Find the values of $A$, $B$, and $C$ such that the action is a minimum

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.

$$L=\frac{1}{2}m\dot{x}^2+Fx$$

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

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.

$$\dot{x}=\frac{F}{m}$$

Differentiate $$x(t)$$ once.

$$B+2Ct=\frac{F}{m}$$

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

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.

$$x=\frac{Ft^2}{2m}$$

$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$

$$A=\frac{Ft^2+Ft-F}{2m}$$

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?

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 $$B=\frac{a - Ct_0^2}{t_0}$$.

• Where the capital A had gone when finding equation for B. $$x(t_0)=a \\\\ => A+Bt_0+Ct_0^2=a \\\\ => B=\frac{a-A-Ct_0^2}{t_0}$$
– user313317
Sep 8, 2021 at 12:07
• 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. 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.

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