Periodic PIC particle and force errors under a prescribed initial coupling

If particles $i$ and $j$ start at the same phase state, subtract their grid equations. The terms with common second argument satisfy $\mathrm{\mid }{z}_i(t)-z_j(t)\mathrm{\mid }\le L_{\epsilon ,h}{\int }_0^t\mathrm{\mid }{z}_i(s)-z_j(s)\mathrm{\mid }ds,$ because the two-variable Lipschitz bound in [F2] applies with only the first argument changed. By [F5], the two paths agree. Thus the finite particle flow defines a single path $X_t(a)$ for each distinct atom $a$ of ${\nu }_0^h$, independent of the label representing $a$.

Extend $X_t$ by the identity off the finite support of ${\nu }_0^h$. This is a jointly Borel map, its paths are absolutely continuous on that support, and their path suprema are integrable because the support is finite and each path is continuous on $[0,T]$. Also ${\nu }_t^h=(X_t{)}_{\mathrm{\#}}{\nu }_0^h$. For a support point $a$, the finite system equation can be written as $r_h(t,a)=\int ({\mathbb{K}}_{\epsilon ,h}(X_t(a),X_t(b))-{\mathbb{K}}_{\epsilon }(X_t(a),X_t(b)))d{\nu }_0^h(b)\mathrm{.}$ Thus ${\dot{X}}_t(a)=\int {\mathbb{K}}_{\epsilon }(X_t(a),X_t(b))d{\nu }_0^h(b)+r_h(t,a)$. This measurable residual has norm at most ${\delta }_{\epsilon ,h}$ by the uniform defect hypothesis. The continuum characteristic flow has residual zero.

Fix any prescribed initial coupling ${\pi }_0$ of ${\nu }_0^h$ and ${\mu }_0$ with cost $D_0$. Apply [F6] to the two characteristic flows in steps 1.1 and 2.1, with ${\eta }_{\mu }=0$ and ${\eta }_{\nu }={\delta }_{\epsilon ,h}$. The transported coupling has cost at most $e^{2L_{\epsilon }t}(D_0+t{\delta }_{\epsilon ,h})$, so by [F3] the same expression bounds $W_1({\nu }_t^h,{\mu }_t)$.

Step 3.1 holds for every initial coupling. Taking the infimum over its initial cost $D_0$ gives ${\delta }_0$ by [F3], yielding the claimed inequality. No coupling attaining the infimum is selected.