= Shared face flux cancellation
{title2=$M^{n+1}-M^n=k(F_{1/2}^n-F_{N+1/2}^n)$}
In a <finite volume method>, orient every face <numerical flux> to the right. For cells $1,\ldots,N$, write the discrete total as $M^n=\sum_{j=1}^N h_jU_j^n$. 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 contribute
$$
\begin{aligned}
h_j(U_j^{n+1}-U_j^n)&=kF_{j-1/2}^n-kF_{j+1/2}^n,\\
h_{j+1}(U_{j+1}^{n+1}-U_{j+1}^n)&=kF_{j+1/2}^n-kF_{j+3/2}^n.
\end{aligned}
$$
The interior face contributes $-kF_{j+1/2}^n+kF_{j+1/2}^n=0$. Summing all cells gives
$$
M^{n+1}-M^n=k\bigl(F_{1/2}^n-F_{N+1/2}^n\bigr).
$$
With <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 $\sum h_jU_j$, not the unweighted <sum> of <cell averages>.
\Image[/finite-volume-shared-face-bookkeeping.png]
{title=One oriented face transfer leaves the left cell and enters the right cell with opposite signs}
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 $k\cdot3=3/4$ cancels. If the receiving cell instead used $3+\varepsilon$ for that same face, the total would acquire the spurious term $k\varepsilon$. 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.
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