Study the action in a one-dimensional movement (Hamilton Principle)

[alex23]

Homework Statement

Study the action (in the sense of Hamilton) for the one-dimensional movement of a particle of mass 1 in a field of constant forces and prove that it reaches a minimum for the real trajectory of the particle.

Relevant Equations

$L=T-V$
$T=\frac{1}{2}m\dot x^2$
$V=-mgx$
Action: $S=\int{L(t,x,\dot x)dt}$

Using the Lagrangian of the system I reached that $x(t)=\frac{1}{2} gt^2+ut$ is the real trajectory of the particle.
After that, I consider different trajectories: $x(\alpha,t) = x(t) + \alpha(t)$ with $\alpha(t)$ being an arbitrary function of t expect for the conditions $\alpha(0)=\alpha(T)=0$, where $T$ is the final time of the movement. This way $x(\alpha,t)$ shows a family of trajectories that start and end in the same points as the real trajectory $x(t)$.
Now, the problem asks me to study the action and prove that the real trajectory makes the action an extremal (in this case a minimum). For that, I calculated the variance of $S: \delta S= S(\alpha) - S(0)$.
To do that I calculated the Lagrangian of both $x(\alpha,t)$ and $x(0,t)$ and introduced them both to calculate $S(\alpha)$ and $S(0)$ and introduce them in the formula above. Doing that I reached this:
$$ \displaystyle\int_{0}^{T}{[\frac{1}{2}(gt+u+\dot \alpha(t))^2 -g(\frac{1}{2}gt^2+ut+\alpha(t))]dt} - \displaystyle\int_{0}^{T}{[\frac{1}{2}(g^2t^2+u^2+2ugt)-(\frac{1}{2}g^2t^2+ugt)]dt} $$
Where $\dot\alpha(t)$ is used to denote the total time derivate of $\alpha(t)$, so $\dot\alpha(t)=\frac{d\alpha(t)}{dt}$.
After expanding the integral and combining them I reached to this:
$$ \displaystyle\int_{0}^{T}{(\frac{1}{2}\dot\alpha^2(t)+gt\dot\alpha(t)+u\dot\alpha(t)-g\dot\alpha(t))dt} $$
This leaves me with 3 integrals, the first:
$\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)dt}$, which is always positive, which would make it so it had a minimum in $\alpha=0$ and hence proving that the real trajectory makes the action S an extremal.
The second integral:
$u\displaystyle\int_{0}^{T}{\dot\alpha(t)dt}=u[\alpha(T)-\alpha(0)]=0$, which would make it so it didn't contribute.
And the third integral and the reason for my problems:
$g\displaystyle\int_{0}^{T}{(t\dot\alpha(t)-\alpha(t))dt}$, which I can't seem to make 0 since I think it would not be 0 in the general case and I can't see where I was mistaken before to make this integral appear.
I know for a fact that the objectives is to reach that $\delta S=S(\alpha)-S(0)=\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)}$. And making the argument I made above would be sufficient to solve the problem.
I don't think the mistakes is after expanding and simplifying the integrals, so the error should come from before that, but I can't seem to locate where I was mistaken. Any help pointing where the mistake is would be greatly appreciated.
Thanks in advance!

Edit: I edited the post to properly format everything in LaTeX since I didn't know it was supported by the forum, hopefully it makes it easier to read! And sorry for the inconvenience.

 

[TSny]

I think you just have a careless sign error (easy to do). There are two negative signs that contribute to the $mgx$ term in the Lagrangian. One negative sign is the one in $L = T-V$ and the other is in $V = -mgx$.

Otherwise, your work looks good to me.

You might think about doing integration by parts for $\int t \dot \alpha dt$

Thanks for using LaTex!

 

[alex23]

I think you just have a careless sign error (easy to do). There are two negative signs that contribute to the $mgx$ term in the Lagrangian. One negative sign is the one in $L = T-V$ and the other is in $V = -mgx$.

Otherwise, your work looks good to me.

You might think about doing integration by parts for $\int t \dot \alpha dt$

Thanks for using LaTex!

You are absolutely right, I had a sign error in the Lagrangian. After fixing that and doing the integration by parts as you suggested I managed to get $\delta S=\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)dt}$ as I as supposed to.
Thanks!