Solving a Lagrangian using an Ansatz

[bantalon]

We were working on some Lagrangian and trying to solve it using an ansatz. Due to some problems in the results we got, we started to doubt the correctness of the method in use. Here is a very simple Lagrangian which shows the problem very easily:
L=\frac{1}{2}\dot{x}^{2}-gx
We took m=1 for the mere sake of simplicity. Solving this using the Euler-Lagrange equation gives \ddot{x}=-g as expected, and finally (taking initial values to zero):
x=-\frac{1}{2}gt^2
So far so good. Now, let's assume the following ansatz on the form of the solution:
x=\alpha{t}^{2}.
Where alpha is a constant. Now the fun begins. We wish to put the ansatz for x into the Lagrangian, vary it by alpha this time, and get the solution:
\alpha=-g
We believe this should work since this ansatz suits the solution we got for x. Putting the ansatz into the Lagrangian we get:
L=2\alpha^{2}t^{2}-g\alpha{t}^{2}
Varying by alpha, we get the equation:
4\alpha{t}^{2}-gt^2=0
And finally:
\alpha=\frac{g}{4}
not quite as expected.
Needless to mention, if we allow alpha to be a general function of t instead of simply a constant, everything works seamlessly.
We would appreciate your thoughts regarding why this method does not work in this case. Can you specify the cases in which it should work?

 

[Bill_K]

Setting x = αt² is imposing a constraint, not just an ansatz. In other words, you're solving a different problem, which has a different minimum. Effectively you're adding a term λ(x-αt²) to the Lagrangian, where λ is a Lagrange multiplier. There will be a constraining force acting on the particle, forcing it to follow a parabolic path, and among this family of paths, α = -g is no longer the minimum.