Why Must the Transversality Condition Be Less Than or Equal to Zero?

[Avatrin]

Hey, I have a theorem I cannot prove.

We have a function x^* that maximizes or minimizes the integral:
\int^{t_1}_{t_0} F(t,x(t),\dot{x}(t))dt

Our end point conditions are:
x(t_0) = x_0, x(t_1) \geq x_1

I am told that x^* has to satisfy the Euler equation. That I can fully understand since x^*(t_1) can be equal to x_1. However, then it gives me the transversality condition:
\left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1} \leq 0 \text{ ( = 0 if $x^*(t_1) > x_1$)}

I can understand the statement in the parentheses. However, I do not understand why \left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1} must be less than or equal to zero if x^*(t_1) = x_1. Why can it not be more than zero?

 

[Orodruin]

The original variation of the functional (call the functional $\mathcal F$) is given by
$$
\delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} \delta x + \frac{\partial F}{\partial \dot x} \delta \dot x\right) dt.
$$
Integration by parts of the second term now leads to
$$
\delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} - \frac{d}{dt} \frac{\partial F}{\partial \dot x} \right) \delta x \, dt
+ \left[\frac{\partial F}{\partial \dot x} \delta x\right]_{t=t_0}^{t_1}.
$$
Since $\delta x$ is arbitrary, the Euler-Lagrange equation has to hold if $x$ is a stationary function of the functional. This, together with $\delta x(t_0) = 0$ leads to
$$
\delta \mathcal F = \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \delta x(t_1).
$$
It follows that to have a local stationary function, you must have
$$
\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} = 0.
$$
However, there is also the option of having a function at the boundary $x(t_1) = x_1$. If you are at the boundary, then only variations $\delta x(t_1) \geq 0$ are allowed and therefore $\delta \mathcal F$ can only be positive if
$$
\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} > 0
$$
implying that the value of $\mathcal F$ would increase with the allowed variation. Your condition
$$
\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \leq 0
$$
is therefore equivalent to requiring that $\delta \mathcal F \leq 0$, i.e., that $x^*$ maximises the functional (at least locally). For a minimisation, you would get the opposite inequality.