Periodic leapfrog amplification, finite-time bounds, and the Nyquist endpoint

Substituting one mode from [F2] into [F1] uses $-{\mathrm{\Delta }}_h{\varphi }_{\mathrm{\ell }}=4h^{-2}{s}_{\mathrm{\ell }}{\varphi }_{\mathrm{\ell }}$ and gives the displayed scalar recurrence, whose characteristic polynomial is $p_{\mathrm{\ell }}(r)=r^2-2(1-2{\nu }^2{s}_{\mathrm{\ell }})r+1.$

For $\mathrm{\ell }=0$, $s_0=0$, so successive first differences are constant and $w_0^n=w_0^0+n(w_0^1-w_0^0)=w_0^0+t_n{\widehat{V}}_0^0$. For $\mathrm{\ell }\mathrm{\ne }0$ and $\nu <1$, put $q=1-2{\nu }^2{s}_{\mathrm{\ell }}$; then $-1<q<1$, and $\rho =q+i\sqrt{1-q^2}$ and $\bar{\rho }$ are distinct roots with modulus one.

For a nonconstant mode the recurrence solution matching $w_{\mathrm{\ell }}^0,w_{\mathrm{\ell }}^1$ is $w_{\mathrm{\ell }}^n=w_{\mathrm{\ell }}^0\frac{{\rho }^n+{\bar{\rho }}^n}{2}+(w_{\mathrm{\ell }}^1-q{w}_{\mathrm{\ell }}^0)\frac{{\rho }^n-{\bar{\rho }}^n}{\rho -\bar{\rho }}\mathrm{.}$ Indeed both sequences solve the recurrence and the right side equals $w_{\mathrm{\ell }}^0,w_{\mathrm{\ell }}^1$ at $n=0,1$. The quotient is ${\sum }_{r=0}^{n-1}{\rho }^{n-1-r}{\bar{\rho }}^r$, so its modulus is at most $n$. Also $\mathrm{\mid }({\rho }^n-{\bar{\rho }}^n)\mathrm{/}(\rho -\bar{\rho })\mathrm{\mid }\le 1\mathrm{/}\sqrt{1-q^2}$ since $\mathrm{\mid }{\rho }^n-{\bar{\rho }}^n\mathrm{\mid }\le 2$ and $\mathrm{\mid }\rho -\bar{\rho }\mathrm{\mid }=2\sqrt{1-q^2}$.

Since $w_{\mathrm{\ell }}^1-q{w}_{\mathrm{\ell }}^0=k{\widehat{V}}_{\mathrm{\ell }}^0+(1-q)w_{\mathrm{\ell }}^0$ and $\frac{1-q}{\sqrt{1-q^2}}=\frac{\nu \sqrt{s_{\mathrm{\ell }}}}{\sqrt{1-{\nu }^2{s}_{\mathrm{\ell }}}}\le \frac{{\nu }_0}{\sqrt{1-{\nu }_0^2}},$ step 2.2 gives $\mathrm{\mid }{w}_{\mathrm{\ell }}^n\mathrm{\mid }\le (1+{\nu }_0\mathrm{/}\sqrt{1-{\nu }_0^2})\mathrm{\mid }{w}_{\mathrm{\ell }}^0\mathrm{\mid }+t_n\mathrm{\mid }{\widehat{V}}_{\mathrm{\ell }}^0\mathrm{\mid }$. The same bound holds for the mean mode.

The modes are a basis by [F3] and orthogonal by [F4]. By [F6], each has squared norm $Nh$, so Parseval's identity holds in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{2,h}$. Squaring the coefficient bounds, summing, and using ${\left({\sum }_{\mathrm{\ell }}{a}_{\mathrm{\ell }}{b}_{\mathrm{\ell }}\right)}^2\le \left({\sum }_{\mathrm{\ell }}{a}_{\mathrm{\ell }}^2\right)\left({\sum }_{\mathrm{\ell }}{b}_{\mathrm{\ell }}^2\right),$ which follows by expanding ${\sum }_{\mathrm{\ell }<r}(a_{\mathrm{\ell }}{b}_r-a_r{b}_{\mathrm{\ell }}{)}^2\ge 0$, yields the stated weighted-grid norm estimate.

If $\nu =1$, $N$ is even, and $\mathrm{\ell }=N\mathrm{/}2$, then $s_{\mathrm{\ell }}=1$ and $p_{\mathrm{\ell }}(r)=(r+1{)}^2$. Its solutions are $(-1{)}^n(A+nB)$; the initial values give $B=-w_{\mathrm{\ell }}^1-w_{\mathrm{\ell }}^0$, so $B=0$ exactly when $w_{\mathrm{\ell }}^1=-w_{\mathrm{\ell }}^0$. For $B\mathrm{\ne }0$ the mode grows linearly in $n$, proving the endpoint caveat.