 Shared face flux cancellation

In a finite volume method, orient every face numerical flux to the right. For cells $1,\dots ,N$, write the discrete total as $M^n={\sum }_{j=1}^N{h}_j{U}_j^n$. If $u$ is a mass per unit length, this total approximates mass; for a different conserved field, it measures that field's integral. Multiplying the update by the cell width gives an amount balance, so adjacent cells contributeThe interior face contributes $-k{F}_{j+1/2}^n+k{F}_{j+1/2}^n=0$. Summing all cells givesWith periodic boundary conditions, the two boundary labels denote the same physical face and must use the same numerical flux; then $M^{n+1}=M^n$. On an open interval the total changes by the boundary transfers. Unequal widths require the weighted total $\sum h_j{U}_j$, not the unweighted sum of cell averages.

The diagram uses two unit-width cells, $k=1/4$, cell averages $(2,1)$ and right-oriented numerical fluxes $(1,3,2)$ at their three faces. The updated cell averages are $(3/2,5/4)$: the total changes from $3$ to $11/4$, exactly $k(1-2)=-1/4$. The shared interior transfer $k\cdot 3=3/4$ cancels. If the receiving cell instead used $3+\epsilon$ for that same face, the total would acquire the spurious term $k\epsilon$. Conservation is therefore an algebraic property of sharing the transfer, even when the numerical flux is a poor approximation. In floating-point arithmetic, independently rounded cell updates may still leave small roundoff errors in the computed total.