The Nyquist mode at and beyond the leapfrog CFL endpoint

For the Nyquist mode, $s_{N\mathrm{/}2}=1$, so the recurrence polynomial is $r^2-2(1-2{\nu }^2)r+1$. At $\nu =1$ this is $(r+1{)}^2$, with solution $w_n=(-1{)}^n(A+nB)$; the coefficient $B$ vanishes exactly when $w^1=-w^0$.

If $\nu >1$, then $q=1-2{\nu }^2<-1$. The roots of $r^2-2qr+1$ are real and negative because the discriminant is $4(q^2-1)>0$, their product is $1$, and their sum is $2q<-2$. Hence one root has modulus greater than one.

Define the periodic modified energy of adjacent states $X=U^n$, $Y=U^{n+1}$ by ${\mathcal{E}}_{\mathrm{p}\mathrm{e}\mathrm{r}}(Y,X)=\frac{1}{2k^2}\mathrm{\parallel }Y-X{\mathrm{\parallel }}_{2,h}^2+\frac{c^2}{2}\mathrm{Re}⟨D_{+}Y,D_{+}X{⟩}_{e,h}\mathrm{.}$ For one normalized Fourier mode ${\psi }_{\mathrm{\ell }}={\varphi }_{\mathrm{\ell }}\mathrm{/}\sqrt{L}$ with coefficients $x,y$, put $m=(x+y)\mathrm{/}2$ and $v=(y-x)\mathrm{/}k$. Reindexing the finite periodic sum gives $⟨D_{+}p,D_{+}q{⟩}_{e,h}=⟨-{\mathrm{\Delta }}_{h}p,q{⟩}_{2,h}$. Thus [F3] gives $\mathrm{\parallel }{D}_{+}{\psi }_{\mathrm{\ell }}{\mathrm{\parallel }}_{e,h}^2=4h^{-2}{s}_{\mathrm{\ell }}$, while [F4] gives orthogonality; the $N$ nonzero orthogonal vectors form a basis of the $N$-dimensional space ${\mathbb{C}}^{J_N}$. By [F5] they are normalized by $\sqrt{L}$. Since $x=m-kv\mathrm{/}2$ and $y=m+kv\mathrm{/}2$, $\mathrm{Re}(y\bar{x})=\mathrm{\mid }m{\mathrm{\mid }}^2-k^2\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}4$. Summing these diagonal contributions using [F6] gives the modal energy $\frac{1}{2}(1-{\nu }^2{s}_{\mathrm{\ell }})\mathrm{\mid }v{\mathrm{\mid }}^2+2c^2{h}^{-2}{s}_{\mathrm{\ell }}\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{.}$ At $\nu =1$, $0\le s_{\mathrm{\ell }}\le 1$, so every modal term is nonnegative, while the Nyquist velocity coefficient is zero. Orthogonal Fourier decomposition makes ${\mathcal{E}}_{\mathrm{p}\mathrm{e}\mathrm{r}}$ semidefinite.

Set the real amplitude $A=1$ and take $X=A{\varphi }_{N\mathrm{/}2}$ and $Y=-X$. Then $X\mathrm{\ne }0$, $\mathrm{\parallel }Y-X{\mathrm{\parallel }}^2=4\mathrm{\parallel }X{\mathrm{\parallel }}^2$, and $⟨D_{+}Y,D_{+}X{⟩}_{e,h}=-4h^{-2}\mathrm{\parallel }X{\mathrm{\parallel }}^2$, so ${\mathcal{E}}_{\mathrm{p}\mathrm{e}\mathrm{r}}(-X,X)=\frac{2}{k^2}(1-{\nu }^2)\mathrm{\parallel }X{\mathrm{\parallel }}_{2,h}^2\mathrm{.}$ It is zero at $\nu =1$ and negative at $\nu >1$.

The state $(Y,X)=(0,X)$ has ${\mathcal{E}}_{\mathrm{p}\mathrm{e}\mathrm{r}}(0,X)=\mathrm{\parallel }X{\mathrm{\parallel }}_{2,h}^2\mathrm{/}(2k^2)>0$. Together with step 1.4, this shows that the form is indefinite for $\nu >1$.