How to prove that there is only trajectory given fixed boundary conditions, if you know the Lagrangian of the system? In my particular problem, the Lagrangian of the system is:
$$
    L = \frac{m(\dot r^2 + r^2\dot \varphi^2)}{2}
    + \frac{m\omega^2 (r\sin \varphi)^2}{2}
$$
From there, we can derive the equations of motion:
$$
    \begin{cases}
\frac{d}{dt} \frac{\partial L}{\partial \dot r}
- \frac{\partial L}{\partial r} = 0 \\
\frac{d}{dt} \frac{\partial L}{\partial \dot \varphi}
- \frac{\partial L}{\partial \varphi} = 0
    \end{cases}
$$
or
$$
    \begin{cases}
    m\ddot r - mr\dot \varphi^2
    - m\omega^2 r\sin^2\varphi = 0\\
    mr^2\ddot \varphi
    - m\omega^2 r^2 \sin \varphi \cos \varphi = 0 
    \end{cases}
$$
The action functional:
$$
    S = \int_{t_0}^{t_1} L dt
    = \int_{t_0}^{t_1} \left( \frac{m(\dot r^2 
        + r^2\dot \varphi^2)}{2}
    + \frac{m\omega^2 (r\sin \varphi)^2}{2} \right) dt
$$
So, how do you prove that there is only one trajectory that the system can travel from point $q_0$ to point $q_1$ during a fixed time interval $T = t_1 - t_0 > 0$.
I don't understand how to tackle these type of problems. Do I need to solve the equations of motion and show that given the boundary conditions, there is only one solution? Or should I somehow use the action functional?
 A: In general you can't, because it isn't true.  Consider the simple pendulum with boundary conditions $\phi(0)=0$ and $\phi(T)=0$ for some $T$.  There are clearly an infinity of trajectories which obey the equations of motion and obey these fixed boundary conditions, including the trivial solution $\phi(t)=\phi'(t)=0$, because the pendulum could go around the loop any integer number of times.
If you carefully follow the derivation of the Euler-Lagrange equations, the statement is that if there exists an extremal trajectory $\phi$ with fixed boundary conditions which extremizes the action $S[\phi]$, then
$$\delta S[\phi]  = 0 \iff \frac{d}{dt}\left(\frac{\partial L}{\partial \dot \phi}\right) = \frac{\partial L}{\partial \phi}$$
This does not guarantee the existence of such a trajectory, and if it does exist, it is not guaranteed to be unique.  The pendulum problem has the latter feature; for an example of the former, consider the Lagrangian $L(x,\dot x) = \frac{1}{2}\dot x^2 + \frac{1}{2}x^2$ and boundary conditions $x(0)=0$, $x(2\pi)=1$.

The system you describe also features non-unique solutions.  In Cartesian coordinates, that's just a particle which is free in the $x$-direction and subject to a harmonic potential in the $y$-direction.  If our boundary conditions are chosen correctly, we could have unique solutions or non-unique ones.  For example, $x(0)=0,y(0)=0$ and $x(2\pi/\omega)=0,y(2\pi/\omega)=0$ has the trivial solution of a particle sitting at the origin as well as the particle oscillating once in the y-direction.
That being said, for almost all choices of boundary condition, the action-extremizing solutions are unique.  In the $x$-direction, fixing $x(0)$ and $x(T)$  fixes the velocity uniquely, because it's a free particle.  In the $y$-direction, fixing $T$ (along with the initial conditions) tells you what part of the oscillation you end in; from there, you can show with a bit of algebra that only one choice of initial $y$-velocity will work unless $T=2\pi/\omega$ and $y(0)=y(T)=0$.
In mechanics, however, we often don't talk about trajectories with fixed boundary conditions, but rather trajectories with fixed initial conditions.  The uniqueness of these trajectories (and possible conditions for non-uniqueness) follows from the theory of second order ODEs.
