Second-order consistency of the homogeneous periodic Yee scheme

Let $u_x=e_x\mathrm{/}{h}_x$ denote the line average of the $x$ component on an $x$-edge; define $u_y,u_z$ cyclically. For values on the corresponding staggered locations, define ${\delta }_i{\delta }_{j}f(r)$ by the four-corner formula $[f(r+\frac{h_i}{2}{e}_i+\frac{h_j}{2}{e}_j)-f(r+\frac{h_i}{2}{e}_i-\frac{h_j}{2}{e}_j)-f(r-\frac{h_i}{2}{e}_i+\frac{h_j}{2}{e}_j)+f(r-\frac{h_i}{2}{e}_i-\frac{h_j}{2}{e}_j)]\mathrm{/}(h_i{h}_j)$. Expanding the four signed face incidences around an $x$-edge and inserting [F1] gives the normalized electric operator $h_x^{-1}(M_e^{-1}{C}^T{M}_{b}Ce{)}_x=c^2(-{\delta }_y^2{u}_x-{\delta }_z^2{u}_x+{\delta }_x{\delta }_y{u}_y+{\delta }_x{\delta }_z{u}_z),c^2=(ϵ\mu {)}^{-1},$ where ${\delta }_j^{2}f=(f(x+h_j{e}_j)-2f(x)+f(x-h_j{e}_j))\mathrm{/}{h}_j^2$ and ${\delta }_i{\delta }_j$ is the corresponding centered staggered mixed difference. The $y$ and $z$ components follow by the same cyclic incidence expansion.

For every face, the exact circulation cochain $C{e}_{\ast }^n$ equals the face flux of $\mathrm{curl}E(t_n)$ by [F4]. Taylor’s formula in time around $t_n$ gives ${\int }_{t_n-\mathrm{\Delta }t\mathrm{/}2}^{t_n+\mathrm{\Delta }t\mathrm{/}2}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_F\mathrm{curl}E(t)\cdot ndSdt=\mathrm{\Delta }t{\int }_F\mathrm{curl}E(t_n)\cdot ndS+O(\mathrm{\mid }F\mathrm{\mid }\mathrm{\Delta }{t}^3\mathrm{\parallel }{\mathrm{\partial }}_t^2\mathrm{curl}E{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{.}$ Faraday’s equation in [F3] identifies the left side with $-{\int }_F(B(t_n+\mathrm{\Delta }t\mathrm{/}2)-B(t_n-\mathrm{\Delta }t\mathrm{/}2))\cdot ndS$. Thus each face entry of $s_h^n$ is bounded by $C\mathrm{\mid }F\mathrm{\mid }\mathrm{\Delta }{t}^3$ times the derivative bound. The face weights in [F1] turn the sum of their squared entries into at most $C_T\mathrm{\Delta }{t}^6$ times the fixed box volume, so $\mathrm{\parallel }{s}_h^n{\mathrm{\parallel }}_{M_b}\le C_T\mathrm{\Delta }{t}^3\le C_T^{\mathrm{\prime }}\mathrm{\Delta }{t}^2$ on the stated time range.

Taylor’s formula with integral remainder gives ${\delta }_j^{2}f={\mathrm{\partial }}_j^{2}f+O(h_j^2\mathrm{\parallel }{\mathrm{\partial }}_j^{4}f{\mathrm{\parallel }}_{\mathrm{\infty }})$ and ${\delta }_i{\delta }_{j}f={\mathrm{\partial }}_i{\mathrm{\partial }}_{j}f+O((h_i^2+h_j^2)\mathrm{\parallel }{D}^{4}f{\mathrm{\parallel }}_{\mathrm{\infty }})$. Averaging along the edge changes the midpoint value of each needed derivative by at most $O(h^2)$ by the same symmetric Taylor estimate. Therefore the stencil in step 1.1 differs from $\mathrm{curl}\mathrm{curl}E$ by at most $C{h}^2$ pointwise. The centered time second difference of the exact edge integrals differs from the edge integral of $E_{tt}$ by at most $C\mathrm{\Delta }{t}^2$ times the stated fourth time-derivative bound. The continuous equation [F3] cancels the leading terms. For an $x$-edge, the resulting electric covector entry is bounded by $Cϵh_y{h}_z(h^2+\mathrm{\Delta }{t}^2)$, with cyclic area factors for the other orientations. Summing squared entries with the reciprocal edge masses in [F1] contributes a factor proportional to $h_x{h}_y{h}_z$ per edge, so $\mathrm{\parallel }{r}_h^n{\mathrm{\parallel }}_{M_e^{-1}}\le C_T(h^2+\mathrm{\Delta }{t}^2)$, since the dual cell volumes sum to the fixed box volume.

The electric bound from step 2.1 and the separate face bound from step 1.2 give the two asserted residual estimates.