Analytic Verlet retains order-two energy accuracy on exponential times

Mechanical identification with a uniform smaller step. Decrease the existing positive $h_0$, independently of $n$, so that $h_0\le \delta \mathrm{/}(4M)$ and $h_0\le \rho \mathrm{/}(64d\sqrt{2d}M)$. Starting at any $z\in K$, construct the three subflows successively using polydiscs of radius $\rho \mathrm{/}(4\sqrt{2d})$ at their initial points. By [F3] their accumulated displacement, including every partial subflow time, is at most $(1\mathrm{/}2+1+1\mathrm{/}2)Mh=2Mh\le \delta \mathrm{/}2$. These polydiscs lie in $U_{\rho }$, as in the distance-domain construction of the flow supplier, after further decreasing $h_0$ to ensure $2M{h}_0\le \rho \mathrm{/}16$. For real mechanical initial data each subflow is real: on the real neighbourhood inside $U$ its vector field is real, and uniqueness makes its local solution equal to its complex conjugate. If this real continuation reached the boundary of $U$ before a factor ended, the accumulated displacement bound would put that boundary point within $\delta \mathrm{/}2$ of the starting point in $K$, contradicting the definition of $\delta$. Local uniqueness therefore continues the real solution through every factor inside $U$. The mechanical fields integrate to affine drift and kick paths there, so [F3] identifies every complete internal segment and its endpoint with the corresponding real factor. Thus ${\mathrm{\Psi }}_h=K_{h\mathrm{/}2}\circ D_h\circ K_{h\mathrm{/}2}$ in the potential-first order, and equals the position-Verlet variant in the other order, uniformly for every starting point in $K$. Retention of step endpoints in $K$ lets this same argument apply at every step. All original energy estimates remain valid after decreasing $h_0$.

Modified conservation and local comparison. Along the modified flow, differentiation gives $d{H}_h(X_{H_h})=\mathrm{\nabla }{H}_h^{\mathsf{T}}J\mathrm{\nabla }{H}_h=0$ by skew symmetry. For $z_j\in K$, the splitting displacement is at most $2Mh$ by [F3], and the modified-flow displacement is at most $C_{G}h$ by [F2], since $J$ is an isometry and integration of $\dot{Y}=X_{H_h}(Y)$ bounds displacement by the supremum of $\parallel X_{H_h}\parallel$ times elapsed time. Put $C_{\ast }:=\max \{2M,C_G\}$ and decrease $h_0$ so that $C_{\ast }{h}_0\le \rho \mathrm{/}16$. Every point of the segment joining the two endpoints is then within $\rho \mathrm{/}16$ of $z_j\in U$, hence in $D(\rho \mathrm{/}8)$. Integrating the bounded gradient along this segment and using [F1] gives $\mathrm{\mid }{H}_h(z_{j+1})-H_h(z_j)\mathrm{\mid }\le Ch{e}^{-c\mathrm{/}h}$.

Telescoping and original energy. Summing step 1.2 over the finite $n$ steps gives $\mathrm{\mid }{H}_h(z_n)-H_h(z_0)\mathrm{\mid }\le Cnh{e}^{-c\mathrm{/}h}$. By [F2], $\mathrm{\mid }H(z_n)-H(z_0)\mathrm{\mid }\le 2C_E{h}^2+Cnh{e}^{-c\mathrm{/}h}$. If $nh\le e^{c\mathrm{/}(2h)}$, the last term is at most $C{e}^{-c\mathrm{/}(2h)}$. Applying [F4] to $x=c\mathrm{/}(2h)>0$ gives $e^{-c\mathrm{/}(2h)}\le 8h^2\mathrm{/}{c}^2$. Enlarging $C$ proves the asserted order-two estimate for both splitting orders.