I browsed through many similar questions about the Initial Condition Problem (ICP) and Boundary Value Problem (BVP) for Euler-Lagrange equations, here some interesting but (in my opinion) incomplete examples: E.g.1 or E.g.2 or E.g.3. This question is within the subject but slightly different.
We know that studying the first variation of the action functional between two positions $q(t_{1})=q_{1}$ and $q(t_{2})=q_{2}$, where a particle is found at the instants $t_{1}$ and $t_{2}$, with $t_{1}<t_{2}$, is equivalent to study the BVP with the Euler-Lagrange equation \begin{equation} \frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=0 \end{equation} and the boundary conditions $q(t_{1})=q_{1}$ and $q(t_{2})=q_{2}$.
My question is, if we look at the matter "backwards": from an ICP with the Euler-Lagrange equation above and an initial condition $q(t_{3})=q_{3}$ and $\dot{q}(t_{3})=\dot{v}_{3}$, with $t_{1}<t_{3}<t_{2}$, assuming all the regularity conditions for the Picard–Lindelöf theorem are satisfied and the solution exists and it's unique, what can we say in terms of the action functional? Will the first variation of the action functional be zero if we plug the solution that we found from the ICP?
