How is the first-order variation derived for an endpoint that is free to vary in both x and t?

Question

When solving a calculus of variations problem with an endpoint that is free to vary in two dimensions (e.g. in t and x), it is necessary to relate the variation in $x_f$ with the variations in $x(t_f)$, and $t_f$, as seen here. According to the textbook, Optimal Control Theory: An Introduction by Donald E. Kirk, this is found to be the following:

$$\delta{x_f} \doteq \delta{x(t_f)} + \dot{x}^*(t_f)\delta{t_f}$$

Where the $\doteq$ sign means "equals on the first order" and we are assuming $\delta{x(t_f)}$ and $\delta{t_f}$ are unrelated and arbitrary. However, when I try solving for this relationship, I get an extra first-order term in my expression. My method is as follows:

1. Start with the following equation based on the geometry shown in the graph linked above: $$\delta{x_f} = x(t_f + \delta{t_f}) - x^*(t_f)$$
2. Focus, momentarily, on the first part of the right-hand expression: $$x(t_f + \delta{t_f}) = x^*(t_f + \delta{t_f}) + \delta{x^*(t_f + \delta{t_f})}$$
3. Find the first-order taylor approximation of $x(t)$ about $t=t_f$: $$x(t) \doteq x^*(t_f) + \delta{x^*(t_f)} + \dot{x}^*(t_f)(t-t_f) + \delta\dot{x}^*(tf)(t-t_f)$$
4. Plug this back into (1) and get the following: $$\delta{x_f} \doteq x^*(t_f) + \delta{x^*(t_f)} + \dot{x}^*(tf)\delta{t_f} + \delta\dot{x}(t_f)\delta{t_f} - x^*(t_f)$$
5. Finally, cancel out the $x^*(t_f)$ terms and get: $$\delta{x_f} \doteq \delta{x(t_f)} + \dot{x}^*(t_f)\delta{t_f} + \delta\dot{x}^*(t_f)\delta{t_f}$$

Why does the textbook solution not include this extra $\delta\dot{x}^*(t_f)\delta{t_f}$ term? Did I mess up somewhere in my solution? Any help would be greatly appreciated.

Answer 1

I have found that, a lot of times, in the book by Donald E. Kirk the product of two variations is dropped implicitly, and one is left wondering where the terms went. I do not see any mistakes in your deduction so this must be one of those times.

However, I do not love your notation or your approach. And my answer will focus on showing a different approach with a notation that is less confusing in my opinion.

Start with the following assumptions \begin{align} t_f &= t_f^* + \delta t_f \,, \tag{1}\label{1} \\ x(t_f^*) &= x^*(t_f^*) + \delta x(t_f^*) \,, \tag{2}\label{2} \\ \delta x_f &= x(t_f) - x^*(t_f^*) \,. \tag{3}\label{3} \end{align} From \eqref{1} $$ x(t_f) = x(t_f^* + \delta t_f) \tag{4}\label{4} \,.$$ Apply Taylor series to \eqref{4} keeping only the first order term, $$ x(t_f) = x(t_f^*) + \dot{x}(t_f^*) \delta t_f \,, \tag{5}\label{5}$$ and solve for $x^*(t_f^*)$ in \eqref{2}, $$ x^*(t_f^*) = x(t_f^*) - \delta x(t_f^*) \,. \tag{6}\label{6}$$ Finally, \eqref{3} suggests to find the difference between \eqref{5} and \eqref{6}: $$ \delta x_f = x(t_f) - x^*(t_f^*) = \delta x(t_f^*) + \dot{x}(t_f^*) \delta t_f \tag{7}\label{7}$$ Below you may find a picture illustrating this notation. In this picture, imagine that the two functions are very very simmilar and that the plot shows the functions zoomed-in to the point that they look linear in $t$.