Strict spectral CFL positivity and Yee modified energy

If $N_e=0$, the electric cochain is necessarily zero and $C$ is the zero map from the zero space. The Faraday row in [F2] makes $b^{n+1\mathrm{/}2}=b^{n-1\mathrm{/}2}$ at every step, while the electric row is an equality in the zero space. Thus $b_c^n=b^{n+1\mathrm{/}2}$ is constant and ${\mathcal{E}}^n=\frac{1}{2}\mathrm{\parallel }{b}_c^n{\mathrm{\parallel }}_{M_b}^2$. With the stated convention ${\lambda }_{\max }=0$, this proves conservation, the lower bound, and positivity on the remaining magnetic state for every $\mathrm{\Delta }t>0$; there is no positive-eigenvalue endpoint case. For the rest of the proof assume $N_e>0$. The quotient defining ${\lambda }_{\max }$ is continuous on the compact set $\{e:e^T{M}_{e}e=1\}$, so the maximum exists and is finite. Also $e^{T}Ke=(Ce{)}^T{M}_b(Ce)\ge 0$, hence ${\lambda }_{\max }\ge 0$. From the first row of [F2], $b^{n-1\mathrm{/}2}=b_c^n+\mathrm{\Delta }tC{e}^n\mathrm{/}2$ and $b^{n+1\mathrm{/}2}=b_c^n-\mathrm{\Delta }tC{e}^n\mathrm{/}2$. Expanding their $M_b$ inner product gives ${\mathcal{E}}^n=\frac{1}{2}\mathrm{\parallel }{e}^n{\mathrm{\parallel }}_{M_e}^2+\frac{1}{2}\mathrm{\parallel }{b}_c^n{\mathrm{\parallel }}_{M_b}^2-\frac{\mathrm{\Delta }{t}^2}{8}\mathrm{\parallel }C{e}^n{\mathrm{\parallel }}_{M_b}^2\mathrm{.}$

Using [F2], the electric-energy increment is $\frac{1}{2}\mathrm{\parallel }{e}^{n+1}{\mathrm{\parallel }}_{M_e}^2-\frac{1}{2}\mathrm{\parallel }{e}^n{\mathrm{\parallel }}_{M_e}^2={\left(\frac{e^{n+1}+e^n}{2}\right)}^T{M}_e(e^{n+1}-e^n)=\mathrm{\Delta }t{\left(\frac{e^{n+1}+e^n}{2}\right)}^T{C}^T{M}_b{b}^{n+1\mathrm{/}2}\mathrm{.}$ The magnetic cross-energy increment is $\frac{1}{2}(b^{n+1\mathrm{/}2}{)}^T{M}_b(b^{n+3\mathrm{/}2}-b^{n-1\mathrm{/}2})=-\mathrm{\Delta }t{\left(\frac{e^{n+1}+e^n}{2}\right)}^T{C}^T{M}_b{b}^{n+1\mathrm{/}2},$ by applying the Faraday row at times $n$ and $n+1$. The two increments cancel, proving conservation.

By the definition of ${\lambda }_{\max }$, $\mathrm{\parallel }C{e}^n{\mathrm{\parallel }}_{M_b}^2=(e^n{)}^{T}K{e}^n\le {\lambda }_{\max }\mathrm{\parallel }{e}^n{\mathrm{\parallel }}_{M_e}^2$. Substitution into step 1.1 gives the stated lower bound. When $\mathrm{\Delta }{t}^2{\lambda }_{\max }<4$, its electric coefficient is positive. If an admissible staggered state has zero energy, the bound forces $e^n=0$ and $b_c^n=0$; the Faraday row in [F2] then gives $b^{n+1\mathrm{/}2}=b^{n-1\mathrm{/}2}$, so $b_c^n=b^{n-1\mathrm{/}2}=b^{n+1\mathrm{/}2}=0$. Thus the energy is positive definite on update states.

Subtract the electric row in [F2] at indices $n$ and $n-1$, then use the magnetic row to replace $b^{n+1\mathrm{/}2}-b^{n-1\mathrm{/}2}$; this gives $M_e(e^{n+1}-2e^n+e^{n-1})+\mathrm{\Delta }{t}^{2}K{e}^n=0$. If $\mathrm{\Delta }{t}^2{\lambda }_{\max }=4$, a maximizing vector $v$ satisfies $Kv={\lambda }_{\max }{M}_{e}v$ by the constrained Rayleigh-quotient maximum. Thus $e^n=q_{n}v$ obeys $q_{n+1}-2q_n+q_{n-1}+\mathrm{\Delta }{t}^2{\lambda }_{\max }{q}_n=0$, whose characteristic polynomial is $(r+1{)}^2$. Hence $q_n=(-1{)}^n(a+bn)$ is a solution and grows linearly when $b\mathrm{\ne }0$. It is compatible with the full first-order rows by setting $b^{n+1\mathrm{/}2}=(q_{n+1}-q_n)Cv\mathrm{/}(\mathrm{\Delta }t{\lambda }_{\max })$: the Ampere row follows from $Kv={\lambda }_{\max }{M}_{e}v$, and the Faraday row follows from the scalar recurrence. For $\mathrm{\Delta }{t}^2{\lambda }_{\max }>4$, the same recurrence has real reciprocal roots, one with modulus greater than one. This proves the endpoint caveat and instability above it.