Smooth consistency orders for the five periodic advection stencils

If $a<0$, set $A=-a>0$ and $\tilde{u}(x,t)=u(-x,t)$. For fixed $x$, its time difference quotient is that of $u$ at $(-x,t)$, so ${\tilde{u}}_t(x,t)=u_t(-x,t)$. In its spatial difference quotient, substitute $s=-h$ to get ${\tilde{u}}_x(x,t)=-u_x(-x,t)$. Repeating these coordinate calculations gives ${\mathrm{\partial }}_x^p{\mathrm{\partial }}_t^q\tilde{u}(x,t)=(-1{)}^p({\mathrm{\partial }}_x^p{\mathrm{\partial }}_t^{q}u)(-x,t)$, so the period and derivative bounds are preserved. The PDE gives ${\tilde{u}}_t+A{\tilde{u}}_x=u_t(-x,t)-A{u}_x(-x,t)=0$. The definition's reflection $j\mapsto -j$ identifies each negative-speed stencil residual with the corresponding positive-speed residual of $\tilde{u}$ at the reflected point. If $a>0$, take $A=a$ and $\tilde{u}=u$. It is therefore enough to derive the estimates for positive speed $A$.

For this positive-speed solution, differentiating ${\tilde{u}}_t=-A{\tilde{u}}_x$ in $t$ and $x$, then using [F4] to commute the order-two mixed partials, gives ${\tilde{u}}_{tt}=-A{\tilde{u}}_{xt}=-A{\tilde{u}}_{tx}=A^2{\tilde{u}}_{xx}$. Differentiating this identity in $t$ and the PDE twice in $x$, then using [F4] to commute the order-three partials, gives ${\tilde{u}}_{ttt}=A^2{\tilde{u}}_{xxt}=A^2{\tilde{u}}_{txx}=-A^3{\tilde{u}}_{xxx}$. By [F3] these derivatives exist, and the bounded derivatives in the hypotheses make the Taylor remainders uniform. Taylor's formula from [F2] yields $\frac{\tilde{u}(x,t+k)-\tilde{u}(x,t)}{k}={\tilde{u}}_t+\frac{k}{2}{\tilde{u}}_{tt}+\frac{k^2}{6}{\tilde{u}}_{ttt}+O(k^3),$ $\frac{\tilde{u}(x+h,t)-\tilde{u}(x-h,t)}{2h}={\tilde{u}}_x+\frac{h^2}{6}{\tilde{u}}_{xxx}+O(h^4),\frac{\tilde{u}(x+h,t)-2\tilde{u}(x,t)+\tilde{u}(x-h,t)}{h^2}={\tilde{u}}_{xx}+\frac{h^2}{12}{\tilde{u}}_{xxxx}+O(h^4)\mathrm{.}$ The one-sided expansions are $(\tilde{u}(x,t)-\tilde{u}(x-h,t))\mathrm{/}h={\tilde{u}}_x-\frac{h}{2}{\tilde{u}}_{xx}+O(h^2)$ and $(\tilde{u}(x+h,t)-\tilde{u}(x,t))\mathrm{/}h={\tilde{u}}_x+\frac{h}{2}{\tilde{u}}_{xx}+O(h^2)$.

Substituting the step 2.1 expansions into [F1] with speed $A=\mathrm{\mid }a\mathrm{\mid }$ and using ${\tilde{u}}_t+A{\tilde{u}}_x=0$ gives ${\tau }_{\mathrm{u}\mathrm{p}}=\frac{k}{2}{\tilde{u}}_{tt}-\frac{Ah}{2}{\tilde{u}}_{xx}+O(k^2+h^2),{\tau }_{\mathrm{d}\mathrm{o}\mathrm{w}\mathrm{n}}=\frac{k}{2}{\tilde{u}}_{tt}+\frac{Ah}{2}{\tilde{u}}_{xx}+O(k^2+h^2),{\tau }_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{d}}=\frac{k}{2}{\tilde{u}}_{tt}+O(k^2+h^2)\mathrm{.}$ For Lax--Friedrichs the symmetric average contributes $-h^2{\tilde{u}}_{xx}\mathrm{/}(2k)$ to the residual, so ${\tau }_{\mathrm{L}\mathrm{F}}=\frac{k}{2}{\tilde{u}}_{tt}-\frac{h^2}{2k}{\tilde{u}}_{xx}+O(k^2+h^2+h^4\mathrm{/}k)$. For Lax--Wendroff the $k{\tilde{u}}_{tt}\mathrm{/}2$ term cancels the second-difference term by ${\tilde{u}}_{tt}=A^2{\tilde{u}}_{xx}$, leaving ${\tau }_{\mathrm{L}\mathrm{W}}=\frac{k^2}{6}{\tilde{u}}_{ttt}+\frac{A{h}^2}{6}{\tilde{u}}_{xxx}+O(k^3+h^4+k{h}^2)$. Reflection in step 1.1 transfers these residual bounds to the negative-speed branch.

The mesh ratio bounds imply $k=O(h)$ and $h^2\mathrm{/}k=O(h)$. Thus the first four residuals are uniformly $O(h)$, and every displayed Lax--Wendroff term is $O(h^2)$ or smaller. By step 1.1 the same estimates hold for either velocity sign. The conclusion concerns only the sampled smooth solution in the stated periodic setting.