Adjoint pairing computes the first variation of the terminal cost

Consider the fixed-horizon problem of BertsekasCTModel with $C^1$ data $f,g,h$, an admissible pair $(u,x)$ on $[0,T]$, and a costate $p$ that is continuous on $[0,T]$, satisfies the terminal condition $p(T)=\nabla h(x(T))$, and solves the adjoint equation

off a finite set of times.

Fix $\tau\in(0,T)$ and let $\varepsilon\mapsto y_\varepsilon$ be a family of perturbed trajectories which, for all small $\varepsilon>0$, is continuous on $[\tau,T]$ and satisfies the same state equation $\dot y_\varepsilon=f(y_\varepsilon,u)$ with the same control $u$ off a finite set. Assume the initial deviation at $\tau$ has a first-order expansion,

Then the tail cost accumulated on $[\tau,T]$ has first variation given by the adjoint pairing at $\tau$:

The mechanism is the classical pairing identity. Writing $\Delta_\varepsilon=y_\varepsilon-x$, the deviation obeys the variational equation up to a remainder that is $o(\lVert\Delta_\varepsilon\rVert)$ uniformly in $t$, because $f$ is $C^1$ and both trajectories remain in a compact tube; a Gronwall estimate gives $\lVert\Delta_\varepsilon\rVert=O(\varepsilon)$ on $[\tau,T]$. Differentiating $t\mapsto\langle p(t),\Delta_\varepsilon(t)\rangle$ off the finite exceptional set and using the adjoint equation shows that its increment cancels the first-order running-cost increment $\int_\tau^T\langle\nabla_x g(x,u),\Delta_\varepsilon\rangle$, while the terminal condition $p(T)=\nabla h(x(T))$ converts the terminal-cost increment into $\langle p(T),\Delta_\varepsilon(T)\rangle$. What survives is the pairing at the left endpoint, $\langle p(\tau),\Delta_\varepsilon(\tau)\rangle$, whose $\varepsilon$-quotient converges to $\langle p(\tau),w\rangle$.