Function for which action is the minimum

Question

On page 2 of "Mechanics" Landau & Lifshitz say that $q=q(t)$ is a function for which action is a minimum. Before this they say that at times $t_1$ and $t_2$, the system occupies coordinates $q^{(1)}$ and $q^{(2)}$, respectively. However the function for which action is a minimum, or $q$ is $\int\limits_{t_1}^{t_2} L(q,\dot{q},t)\,dt$, and not just the coordinates. Why do they use the same symbol $q$ twice? Another thing is that I don't understand when they say:

"Since, for $t_1$ and $t_2$, all the functions $q+{\delta}q$ must take the values $q^{(1)}$ and $q^{(2)}$ respectively, it follows that ${\delta}q(t_1)={\delta}q(t_2)=0$"

This would be understandable if $q$ was $\int\limits_{t_1}^{t_2} L(q,\dot{q},t)\,dt$, but later on they justify the fact that $[\frac{\partial{L}}{{\partial{\dot{q}}}}{\delta}q]^{t_2}_{t_1}=0$ by the above, once again treating $q$ as if it were coordinates. So what do the $q$ refer to after all?

Answer 1

This is a statement of the principle of minimal action. Let $$\int\limits_{t_0}^{t_f}\text{d}t L (q(t),\dot{q}(t),t) $$ be the energy functional ($L = E_{kin} - E_{pot}$). First note that the energy is function of coordinate ($q$) and speed ($\dot{q}$). Then consider a infinitesimal variation of the trajectory from point $q(t_0)$ and $q(t_f)$: $$q(t) = q_0(t) + \delta q(t)$$, the action becomes: $$S(q_0(t) + \delta q(t)) = S(q_0(t)) + \delta S = \int\limits_{t_0}^{t_f} \text{d}t L (q_0(t),\dot{q}_0(t),t) + \bigg (\frac{\delta L }{\delta q} - \partial_t \frac{\delta L }{\delta \dot{q}}\bigg ) \delta q $$ $q_0(t)$ is the trajectory with minimal action, if the variation of the action with respect to $\delta q(t)$ vanishes.

Now about your questions: the symbol $\dot{q}$ is the speed, and not the coordinate. It is the time derivative of the position. About the second question, you want to vary infinitesimally the trajectory between $q^{(1)}$ and $q^{(2)}$, but not $q^{(1)}$ and $q^{(2)}$ themselves. So you require $\delta q(t_1) = \delta q(t_2) = 0$.