Hamilton's principal function

Remarks

Assume that $L$ is hyperregular, with Hamiltonian $H$. Let $p$ be the momentum of $\gamma_{q, t}$ at time $t$. The first variation of the action with free endpoint gives $\nabla_q S(q, t) = p$ and $\partial_t S(q, t) = -H(q, p, t)$. Hence $S$ solves the Hamilton–Jacobi equation $\partial_t S + H(q, \nabla_q S, t) = 0$.

Example: for $L = \frac{1}{2} m \dot q^2$ with $n = 1$, the path is $\gamma(s) = q_0 + (q - q_0)(s - t_0)/(t - t_0)$, and $$S(q, t) = \frac{m (q - q_0)^2}{2 (t - t_0)}.$$ Check: $\partial_t S = -\frac{m (q - q_0)^2}{2 (t - t_0)^2}$ and $\frac{1}{2m} (\partial_q S)^2 = \frac{m (q - q_0)^2}{2 (t - t_0)^2}$, so $\partial_t S + \frac{1}{2m} (\partial_q S)^2 = 0$.

Non-example: for the oscillator $L = \frac{1}{2} \dot q^2 - \frac{1}{2} q^2$, every solution starting at $q_0$ at $t_0$ is at $-q_0$ at time $t_0 + \pi$. So for $t = t_0 + \pi$ there is no solution ending at $q \ne -q_0$ and infinitely many ending at $-q_0$, and $S$ is not defined near such points.