Smooth PEC edge finite element Maxwell error from compatible projections

If $T=0$, [F1] supplies the unique initialized map; Ritz best approximation [F3] and interpolation [F4] give the electric and curl error at zero, and [F5] gives the magnetic error and divergence compatibility. Norm equivalence [F6] proves the stated estimate on the singleton. Hence assume $T>0$. We first justify the closure and time derivatives; $E_t,E_{tt}$ denote the classical time derivatives. For each $u=E(t),E_t(t),E_{tt}(t)$, zero boundary tangential trace makes the boundary edge moments of $I_h^{1}u$ zero, so [F7] gives $I_h^{1}u\in V_h$. By [F4], $I_h^{1}u\to u$ in $H(\mathrm{curl})$ as $h\to 0$, and [F9] gives $u\in H_0(\mathrm{curl})$. Uniform bounds on $E_{tt}$ give $L^2$-Lipschitz continuity of $E_t$ by the pointwise mean-value bound [F8]. For a compact smooth vector test $\psi$, dominated differentiation gives $\frac{d}{dt}(\mathrm{curl}{E}_t,\psi )=\frac{d}{dt}(E_t,\mathrm{curl}\psi )=(E_{tt},\mathrm{curl}\psi )=(\mathrm{curl}{E}_{tt},\psi )$. The uniform bound on $\mathrm{curl}{E}_{tt}$ and [F8], followed by $L^2$ density, therefore give $\mathrm{\parallel }\mathrm{curl}{E}_t(t)-\mathrm{curl}{E}_t(s){\mathrm{\parallel }}_2\le K\mathrm{\mid }t-s\mathrm{\mid }$. The same test identities for $E$ show $E$ is graph-norm $C^1$ with derivative $E_t$: subtract the value of $E_t(s)$ from its difference quotient and use the just-proved Lipschitz bound, first in test pairings and then by density. Difference quotients of $E_t$ are uniformly graph bounded; their $L^2$ parts converge to $E_{tt}$ by dominated convergence, and their curls converge distributionally to $\mathrm{curl}{E}_{tt}$. Density and uniform boundedness upgrade the latter to weak $L^2$ convergence. Consequently every bounded graph-linear functional of $E_t$ is differentiable with derivative its value on $E_{tt}$. In particular the finite Ritz Gram equations give $(R_{h}E{)}_t=R_h{E}_t$ and $(R_{h}E{)}_{tt}=R_h{E}_{tt}$, with a bounded second derivative; continuity of $E_{tt}$ in the graph norm is unnecessary.

The initialization gives ${\theta }_h(0)=0$. Test the semidiscrete Ampere row at $t=0$ and subtract the exact weak row. By the curl orthogonality in [F5], $({\mu }^{-1}(B_h^0-B(0)),\mathrm{curl}{v}_h)=0$ for every $v_h$, so ${\dot{E}}_h(0)$ is the $ϵ$-orthogonal projection of $E_t(0)$ onto $V_h$. The zero tangential boundary trace gives zero boundary edge moments for $I_h^1{E}_t(0)$; [F7] therefore places this interpolant in $V_h$. Its best-approximation error is at most $\mathrm{\parallel }{E}_t(0)-I_h^1{E}_t(0){\mathrm{\parallel }}_{ϵ}\le Ch$ by [F4]. Also $\mathrm{\parallel }{R}_h{E}_t(0)-E_t(0){\mathrm{\parallel }}_{ϵ}\le Ch$ by [F3,F4]. Hence $\mathrm{\parallel }{\dot{\theta }}_h(0){\mathrm{\parallel }}_{ϵ}\le Ch$.

Differentiate the exact weak Ampere equation in [F2] and use $B_t=-\mathrm{curl}E$. Since $\mu$ is time-independent, this gives, for every $v\in H_0(\mathrm{curl};\mathrm{\Omega })$, $(ϵE_{tt},v)+({\mu }^{-1}\mathrm{curl}E,\mathrm{curl}v)=-({\mathrm{\partial }}_t{J}_{\mathrm{s}\mathrm{r}\mathrm{c}},v)\mathrm{.}$ Differentiating the finite-dimensional Ampere row and using $\dot{b}+Ce=0$ gives the identical Galerkin equation for $E_h$: $(ϵ{\ddot{E}}_h,v_h)+({\mu }^{-1}\mathrm{curl}{E}_h,\mathrm{curl}{v}_h)=-({\mathrm{\partial }}_t{J}_{\mathrm{s}\mathrm{r}\mathrm{c}},v_h)\mathrm{.}$ The solution exists by [F1]; the exact time pairings are differentiable by step 1.1, and the discrete differentiation is valid because $J_{\mathrm{s}\mathrm{r}\mathrm{c}}$ is $C^1$.

Let $R_h$ be fixed in time and put ${\theta }_h=E_h-R_{h}E$. Step 1.1 justifies $(R_{h}E{)}_{tt}=R_h{E}_{tt}$ for its time-independent Gram matrix. Subtract the two equations in step 2.1 and use Ritz orthogonality [F3], namely $a(E-R_{h}E,v_h)=-m(E-R_{h}E,v_h)$. For every $v_h\in V_h$ this yields $m({\ddot{\theta }}_h,v_h)+a({\theta }_h,v_h)=m((I-R_h)(E_{tt}-E),v_h)\mathrm{.}$ The zero tangential boundary trace gives zero boundary edge moments for $I_h^1(E_{tt}-E)$, so [F7] places this interpolant in $V_h$. Best approximation [F3] and the canonical estimate [F4] then show uniformly in time that $\mathrm{\parallel }(I-R_h)(E_{tt}-E){\mathrm{\parallel }}_{ϵ}\le Ch$.

Define $Q_h(t)=\frac{1}{2}(\mathrm{\parallel }{\dot{\theta }}_h(t){\mathrm{\parallel }}_{ϵ}^2+\mathrm{\parallel }\mathrm{curl}{\theta }_h(t){\mathrm{\parallel }}_{{\mu }^{-1}}^2)$. Test the equation in step 3.1 with $v_h={\dot{\theta }}_h$. Symmetry gives $Q_h^{\mathrm{\prime }}(t)=m((I-R_h)(E_{tt}-E),{\dot{\theta }}_h),\mathrm{\mid }{Q}_h^{\mathrm{\prime }}(t)\mathrm{\mid }\le \mathrm{\parallel }(I-R_h)(E_{tt}-E){\mathrm{\parallel }}_{ϵ}\sqrt{2Q_h(t)}\mathrm{.}$ For $\delta >0$, the derivative of $\sqrt{Q_h+\delta }$ has absolute value at most $Ch\mathrm{/}\sqrt{2}$ by step 3.1. The scalar mean-value bound [F8], the initial estimate in step 1.2 and then $\delta \downarrow 0$ give $\sqrt{Q_h(t)}\le Ch$ on $[0,T]$. The finite-dimensional mean-value bound and ${\theta }_h(0)=0$ also give ${\sup }_t\mathrm{\parallel }{\theta }_h(t){\mathrm{\parallel }}_{ϵ}\le T{\sup }_t\mathrm{\parallel }{\dot{\theta }}_h(t){\mathrm{\parallel }}_{ϵ}\le Ch$.

By step 4.1, ${\theta }_h$ is $O(h)$ in the electric $L^2$ norm and its curl norm. The Ritz best-approximation estimate [F3] and interpolation estimate [F4] give $\mathrm{\parallel }{R}_{h}E-E{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }\mathrm{curl}(R_{h}E-E){\mathrm{\parallel }}_{L^2}\le Ch$. The triangle inequality therefore gives the asserted electric and curl error bound.

Subtract the exact Faraday equation from the semidiscrete magnetic row and integrate in time: $B_h(t)-B(t)=B_h^0-B(0)-{\int }_0^t\mathrm{curl}(E_h(s)-E(s))ds\mathrm{.}$ The initial difference is $O(h)$ by [F5], and the curl error is $O(h)$ uniformly by step 5.1, so the magnetic $L^2$ error is at most $Ch+TCh$. Also $D{B}_h(t)=D{B}_h^0-{\int }_0^{t}DC{e}_h(s)ds=0$ by [F5] and the discrete identity $DC=0$ in [F2]. Norm equivalence [F6] completes the result.