Fourier-mode stability of the periodic leapfrog benchmark

The period and zero-set facts in [F3, F5] make $e_{\mathrm{\ell }}$ well-defined on $\mathbb{Z}\mathrm{/}N\mathbb{Z}$. Put ${\theta }_{\mathrm{\ell }}=2\pi \mathrm{\ell }\mathrm{/}N$. Euler's formula and the addition formulas [F3, F4], together with the complex multiplication formulas [F2], give $e_{\mathrm{\ell }}(j+1)=e^{i{\theta }_{\mathrm{\ell }}}{e}_{\mathrm{\ell }}(j)$ and $e_{\mathrm{\ell }}(j-1)=e^{-i{\theta }_{\mathrm{\ell }}}{e}_{\mathrm{\ell }}(j)$. Therefore $(D_h{e}_{\mathrm{\ell }})(j)=h^{-2}(e^{i{\theta }_{\mathrm{\ell }}}-2+e^{-i{\theta }_{\mathrm{\ell }}})e_{\mathrm{\ell }}(j)=-4h^{-2}{\sin }^2(\pi \mathrm{\ell }\mathrm{/}N)e_{\mathrm{\ell }}(j),$ where the last equality uses [F3, F4]. Thus a mode sequence $u_j^n=a_n{e}_{\mathrm{\ell }}(j)$ obeys $a_{n+1}-2c_{\mathrm{\ell }}{a}_n+a_{n-1}=0$ for every update step.

For the constant mode, define ${\bar{u}}^n=N^{-1}{\sum }_{j=0}^{N-1}{u}_j^n$. By [F8], a cyclic shift preserves a finite sum, so ${\sum }_j(D_h{u}^n{)}_j=0$. Averaging the update [F1] therefore gives ${\bar{u}}^{n+1}=2{\bar{u}}^n-{\bar{u}}^{n-1}$, and averaging the startup gives ${\bar{u}}^1={\bar{u}}^0+k{\bar{v}}^0$. Induction yields ${\bar{u}}^n={\bar{u}}^0+nk{\bar{v}}^0$. If $0\le nk\le T$, then [F12, F13] give $\mathrm{\mid }nk{\bar{v}}^0\mathrm{\mid }=nk\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }\le T\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }$, and the real triangle inequality [F15] gives $\mathrm{\mid }{\bar{u}}^n\mathrm{\mid }\le \mathrm{\mid }{\bar{u}}^0\mathrm{\mid }+T\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }$. If ${\bar{v}}^0\mathrm{\ne }0$, then [F13] gives $\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }>0$. Given any $M>0$, [F16] supplies a natural $n\ge 1$ with $n>(M+\mathrm{\mid }{\bar{u}}^0\mathrm{\mid })\mathrm{/}(k\mathrm{\mid }{\bar{v}}^0\mathrm{\mid })$; this quotient is defined because $k\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }>0$. Since $nk\ge 0$, [F12, F13] give $\mathrm{\mid }nk{\bar{v}}^0\mathrm{\mid }=nk\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }$, and [F15] applied to $nk{\bar{v}}^0=({\bar{u}}^0+nk{\bar{v}}^0)-{\bar{u}}^0$ yields $\mathrm{\mid }{\bar{u}}^0+nk{\bar{v}}^0\mathrm{\mid }\ge nk\mathrm{\mid }{\bar{v}}^0\mathrm{\mid }-\mathrm{\mid }{\bar{u}}^0\mathrm{\mid }>M$, proving unboundedness.

The derivative identities in [F3] give ${\sin }^{\mathrm{\prime }}=\cos$ and ${\cos }^{\mathrm{\prime }}=-\sin$ everywhere, so both functions are continuous and differentiable on the closed subintervals used here. By [F18], ${\cos }^{\mathrm{\prime }}<0$ on $(0,\pi \mathrm{/}2)$; applying [F17] to cosine on any $[a,b]\subseteq [0,\pi \mathrm{/}2]$ with $a<b$ shows $\cos b<\cos a$. For $0<x<\pi \mathrm{/}2$, apply [F17] to sine on $[0,x]$ and $[x,\pi \mathrm{/}2]$. Since [F3, F9] give $\sin 0=0$ and $\sin (\pi \mathrm{/}2)=1$, there are $c_1\in (0,x)$ and $c_2\in (x,\pi \mathrm{/}2)$ with $\sin x\mathrm{/}x=\cos c_1$ and $(1-\sin x)\mathrm{/}(\pi \mathrm{/}2-x)=\cos c_2$. The just-proved decrease gives $\sin x\mathrm{/}x\ge (1-\sin x)\mathrm{/}(\pi \mathrm{/}2-x)$. Multiplying by the positive denominators and rearranging gives $\sin x\ge 2x\mathrm{/}\pi$. This also holds at $x=0,\pi \mathrm{/}2$ by their stated values. For $1\le \mathrm{\ell }<N$, let $m=\min \{\mathrm{\ell },N-\mathrm{\ell }\}$, so $1\le m\le N\mathrm{/}2$. Parity and the $\pi$ shift in [F4, F9] give $\sin (\pi \mathrm{\ell }\mathrm{/}N)=\sin (\pi m\mathrm{/}N)>0$. Applying the chord bound at $x=\pi m\mathrm{/}N$ gives $\mathrm{\mid }\sin (\pi \mathrm{\ell }\mathrm{/}N)\mathrm{\mid }\ge 2m\mathrm{/}N$. Consequently, since $h=2\pi \mathrm{/}N$, one has ${\omega }_{\mathrm{\ell }}=2h^{-1}\mathrm{\mid }\sin (\pi \mathrm{\ell }\mathrm{/}N)\mathrm{\mid }\ge 2m\mathrm{/}\pi \ge 2\mathrm{/}\pi$, uniformly in $N$. Taking successive minima over the finite list still gives an attained positive minimum on each grid. This proof is a frequency bound only and does not bound the initial-data-dependent coefficients of the two-root representation uniformly over arbitrary unscaled pairs.

Suppose $N$ is even and $\lambda =1$. Repeated $\pi$-shift use of [F9] gives $\sin (\pi j)=0$ and $\cos (\pi j)=(-1{)}^j$ for every grid index $0\le j<N$; Euler's formula [F3] identifies the Fourier character $e_{N\mathrm{/}2}(j)$ with the alternating array $(-1{)}^j$. Its periodic neighbors are equal to its negative. By [F1], $(D_h((-1{)}^{\cdot }){)}_j=-4h^{-2}(-1{)}^j$, so the mode recurrence has coefficient $c=1-2{\lambda }^2=-1$ and characteristic polynomial $(z+1{)}^2$.

If $\mathrm{\ell }\mathrm{\ne }0$, [F5] gives $s:=\sin (\pi \mathrm{\ell }\mathrm{/}N)\mathrm{\ne }0$, so $s^2>0$ by [F14]. Also ${\cos }^2(\pi \mathrm{\ell }\mathrm{/}N)\ge 0$ by [F14] and the zero case, hence [F6] gives $s^2\le 1$. As $0<\lambda <1$, it follows that $0<2{\lambda }^2{s}^2\le 2{\lambda }^2<2$, so $-1<c_{\mathrm{\ell }}<1$. Then $r_{\mathrm{\ell }}:=\sqrt{1-c_{\mathrm{\ell }}^2}$ exists and is positive by [F7]. Direct substitution shows that $z^2-2c_{\mathrm{\ell }}z+1=0$ has roots $c_{\mathrm{\ell }}\pm i{r}_{\mathrm{\ell }}$; they are distinct. By [F10], their complex moduli are $\sqrt{c_{\mathrm{\ell }}^2+r_{\mathrm{\ell }}^2}=\sqrt{1}=1$, where the last equality uses uniqueness of the nonnegative square root [F7].

Since the roots are distinct, for any initial pair $a_0,a_1$ there are unique complex constants $A,B$ with $a_0=A+B$ and $a_1=A{z}_{\mathrm{\ell },+}+B{z}_{\mathrm{\ell },-}$. The sequence $A{z}_{\mathrm{\ell },+}^n+B{z}_{\mathrm{\ell },-}^n$ obeys the recurrence in step 1.1 and has that initial pair; induction on $n$ therefore identifies it with $a_n$. By multiplicativity in [F11], induction gives $\mathrm{\mid }{z}_{\mathrm{\ell },\pm }^n\mathrm{\mid }=\mathrm{\mid }{z}_{\mathrm{\ell },\pm }{\mathrm{\mid }}^n=1$. The complex triangle inequality and multiplicativity in [F11] then give $\mathrm{\mid }A{z}_{\mathrm{\ell },+}^n+B{z}_{\mathrm{\ell },-}^n\mathrm{\mid }\le \mathrm{\mid }A\mathrm{\mid }\mathrm{\mid }{z}_{\mathrm{\ell },+}^n\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }\mathrm{\mid }{z}_{\mathrm{\ell },-}^n\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }$ for all $n$, so the mode is bounded on every finite slab.

Write $u_j^n=a_n(-1{)}^j$ and put $q_n:=(-1{)}^n{a}_n$. The repeated-root recurrence from step 1.4 gives $q_{n+1}-2q_n+q_{n-1}=0$, so its first differences are constant and $q_n=a_0+n(-a_1-a_0)$. The startup in [F1] with $u_j^0=a_0(-1{)}^j$ and $v_j^0={\beta }_0(-1{)}^j$ gives $a_1=-a_0+k{\beta }_0$, hence $a_n=(-1{)}^n(a_0-nk{\beta }_0)$. For ${\beta }_0\mathrm{\ne }0$ its magnitude grows linearly in $n$; for ${\beta }_0=0$ it is bounded.