I am reading Whittaker's Analytical Dynamics. This is chapter 9 Principle of least action and least curvature. The paragraph is 99 Hamilton’s principle for conservative holonomial systems.
Let us consider a conservative holonomial system with $n$ degrees of freedom $(q_1,\dots ,q_n)$. Let $L$ be the kinetic potential. Let $AB$ be an arc in the $n$-dimensional space that is part of the trajectory of the system. Let $CD$ be part of a neighboring arc that is not necessarily the trajectory. Let $t$ be the moment in which the representative point $(q_1,\dots ,q_n)$ takes the position $P$ on $AB$. We are going to assume that each point on $CD$ is assigned to some value of time $t$, such that there is a point $Q$ on $CD$, which corresponds to the same value of $t$ that $P$ corresponds to. As we describe the arc $CD$ we assume that the corresponding values of $t$ change in one and the same direction. Thus if a point is describing $CD$ it will go thorough states corresponding to a continuous sequence of values of $q_1,\dots ,q_n,t$, and therefore to each point in CD will correspond a system of values for $\dot q_1,\dots ,\dot q_n$.
We will denote by $\delta$ the variation by which we go from a point in $AB$ to the point in $CD$ that corresponds to the same value in time, and with $t_0,t_0+\Delta t_0,t_1,t_1+\Delta t_1$ the moments in time corresponding to the endpoints $A,C,B,D$ respectively. Let $L_R$ denote the value of $L$ at any point in each arc.
Let us now take the difference of the integral $\int L(\dot q_1,\dots,\dot q_n,q_1,\dots, q_n,t)dt$ taken along the arcs $AB$ and $CD$ respectively.
Therefore, $$\int_{CD}Ldt-\int_{AB}Ldt=L_B\Delta t_1-L_A\Delta t_0+\int_{t_0}^{t_1}\delta Ldt$$
Would anyone shed some light on this formula. Does it stay true for any integrable function on the phase space? Which members on the right side correspond to which ones on the left? How can one derive the formula?
(I have little to no background in calculus of variations and was not the most dedicated student when I took calculus.)
...Let us now assume that that $C$ coincides with $A$ and $B$ coincides with $D$ and that the moments corresponding to $C$ and $D$ are $t_0$ and $t_1$ respectively, so that $\Delta q_1,\dots,\Delta q_n,\Delta t$ are all zero in $A$ and $B$...
$(\Delta q_i)_B$ here is the change of $q_i$ in going from point $B$ to point $D$. An analogous definition is in force for point $A$ and $C$.
... Therefore $$\int_{CD}Ldt-\int_{AB}Ldt=0$$ which shows that the integral $\int Ldt$ has a stationary value for an arbitrary part of the actual trajectory for as long as comparative lines are considered the trajectories $CD$, that have the same end points as the actual trajectory and for which the time has the same end values.
What does it mean to take any trajectories for "comparative?" How does this equation speak for a stationary point?
The least action principle is trying to state that a real trajectory of particle is one that minimize the action. So here AB is true trajectory (which needs to be shown to have the least principle) and CD is possible trajectory. Computing the difference of actions of both trajectories directly produces $$\begin{align} &\int_{CD}L(t)dt-\int_{AB}L(t)dt\\ =&\int_{t_0+\Delta t_0}^{t_1+\Delta t_1}L_{CD}(t)dt-\int_{t_0}^{t_1}L_{AB}(t)dt\\ =&\int_{t_1}^{t_1+\Delta t_1}L_{CD}(t)dt+\int_{t_0}^{t_1}[L_{CD}(t)-L_{AB}(t)]dt-\int_{t_0}^{t_0+\Delta t_0}L_{AB}(t)dt\\ \approx&L_B\Delta t_1+\int_{t_0}^{t_1}\delta Ldt-L_A\Delta t_0 &(\geq 0) \end{align}$$ The last line is an approximation due to the fact that CD is a neighbor of AB.
In order to find the true trajectory according to the principle adding that A coincides C, B coincides D as well as $\Delta t_0=\Delta t_1=0$, we need $$\delta\int_{t_0}^{t_1}Ldt=\int_{CD}L(t)dt-\int_{AB}L(t)dt=\int_{t_0}^{t_1}\delta Ldt=0$$ to assure that $CD$ is exactly what we believe the true trajectory. We say the trajectory we solve is the stationary trajectory of the variation of action $\int_{t_0}^{t_1}Ldt$. Comparative lines are defined in the first paragraph in your quotation.