The nearest-neighbor CFL condition is necessary but not sufficient

Put $R=aT\epsilon \mathrm{/}(4(1+\epsilon ))$ and take $\chi$ from [F4]. Set $u_0(x)=\chi (3x\mathrm{/}R)$. Then $u_0$ is smooth, equals $1$ at zero, and is supported in $[-2R\mathrm{/}3,2R\mathrm{/}3]$, a compact subset of $(-R,R)$; by [F8] it is an admissible real-valued datum.

By finite induction [F3], the value at index $j$ after zero steps depends only on $j$. If every value after $n$ steps depends only on initial indices $j-n,\dots ,j+n$, then the next update reads indices $j-1,j,j+1$ at time $n$, whose initial dependence intervals have union $[j-(n+1),j+(n+1)]$. Thus after $n$ steps only initial indices $j-n,\dots ,j+n$ can affect the value.

Along the violating subsequence put $n_m=\lfloor T\mathrm{/}{k}_m\rfloor$ and $j_m=\lfloor aT\mathrm{/}{h}_m+1\mathrm{/}2\rfloor$. By [F7], $j_m\le aT\mathrm{/}{h}_m+1\mathrm{/}2<j_m+1$, so $\mathrm{\mid }{j}_m{h}_m-aT\mathrm{\mid }\le h_m\mathrm{/}2$. Since $n_m{h}_m\le T{h}_m\mathrm{/}{k}_m\le aT\mathrm{/}(1+\epsilon )$, the left edge of the numerical dependence interval obeys $j_m{h}_m-n_m{h}_m\ge \frac{aT\epsilon }{1+\epsilon }-\frac{h_m}{2}\mathrm{.}$ For all sufficiently large $m$, this exceeds $R$. Hence every sampled initial value on which $U_{j_m}^{n_m}$ depends is zero, so $U_{j_m}^{n_m}=0$.

The constant-velocity characteristic is $X(s;t,x)=x+a(s-t)$: it has derivative $a$ and terminal value $x$, and any other solution agrees by [F6] applied to its difference from $X$. The explicit function $u^{\ast }(y,s)=u_0(y-as)$ has partial derivatives $u_y^{\ast }=u_0^{\mathrm{\prime }}(y-as)$ and $u_s^{\ast }=-a{u}_0^{\mathrm{\prime }}(y-as)$ by one-variable difference quotients, so it solves $u_t+a{u}_x=0$ with initial data $u_0$. To verify uniqueness, let $u$ be any classical solution on an open neighborhood of this characteristic segment with the same initial data, fix $(x,t)$, and set $v(s)=u(x+a(s-t),s)$ on $[0,t]$. For interior $s$ and small nonzero $\delta$ whose coordinate segments remain in the solution domain, applying [F6] separately in the first and second coordinates gives $\frac{v(s+\delta )-v(s)}{\delta }=a{u}_x({\xi }_{\delta },s+\delta )+u_t(x+a(s-t),{\eta }_{\delta }),$ where ${\xi }_{\delta }$ lies between $x+a(s-t)$ and $x+a(s+\delta -t)$, and ${\eta }_{\delta }$ lies between $s$ and $s+\delta$. As $\delta \to 0$, these intermediate points tend to $(x+a(s-t),s)$, so continuity of the partials in [F2] yields $v^{\mathrm{\prime }}(s)=a{u}_x(x+a(s-t),s)+u_t(x+a(s-t),s)=0$. Then [F6] makes $v$ constant, hence $u(x,t)=u_0(x-at)$. Since $0\le T-n_m{k}_m<k_m$ and $k_m\to 0$, the exact characteristic foot $j_m{h}_m-a{n}_m{k}_m$ tends to $aT-aT=0$. By continuity and $u_0(0)=1$, the exact sample tends to $1$, while step 2.1 gives numerical value zero. Thus the sampled maximum error does not tend to zero.

For the separate centered-family claim, fix $P>0$ and use $N_q=4q$, $h_q=P\mathrm{/}(4q)$, $k_q=h_q\mathrm{/}(2a)$, and $r_q=\lfloor T\mathrm{/}{k}_q\rfloor$. Then [F1] gives the centered update with $a{k}_q\mathrm{/}{h}_q=1\mathrm{/}2$, and the Fourier index $\mathrm{\ell }=q$ is admissible with $z_{\mathrm{\ell }}=\exp (2\pi iq\mathrm{/}(4q))=i$. By [F9] its multiplier is $g=1-i\mathrm{/}2$, so $\mathrm{\mid }g\mathrm{\mid }=\sqrt{5}\mathrm{/}2>1$. Repeated application gives $Q_q^{r_q}{\varphi }_q=g^{r_q}{\varphi }_q$. The mode is nonzero, and the finite Fourier basis in [F9] diagonalizes $Q_q^{r_q}$ with finitely many finite multipliers, so this operator is bounded. By [F10] its norm is at least $\mathrm{\mid }g{\mathrm{\mid }}^{r_q}$. Since [F7] gives $r_q\ge T\mathrm{/}{k}_q-1=8aTq\mathrm{/}P-1\to \mathrm{\infty }$, these norms are unbounded while all time levels satisfy $r_q{k}_q\le T$. Thus the domain-of-dependence bound is not sufficient for uniform finite-time power stability.