Mesh-uniform first-order Lie error for a smooth periodic reaction-diffusion solution

Let $Y_h(t)=P_{h}u(t)$ and $E_h(s)=e^{s{\mathcal{A}}_h}$. By [F1]–[F3], the sampled equation is $Y_h^{\mathrm{\prime }}={\mathcal{A}}_h{Y}_h+F_h(Y_h)+r_h$, where $F_h$ applies $F$ at each node and $\mathrm{\parallel }{r}_h(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\mathrm{s}\mathrm{p}}{h}_{\max }^{}$. Differentiate $E_h(\tau -s)Y_h(t+s)$ using [F9], then integrate from $s=0$ to $s=\tau$ using [F10]; this gives $Y_h(t+\tau )=E_h(\tau )Y_h(t)+{\int }_0^{\tau }{E}_h(\tau -s)(F_h(Y_h(t+s))+r_h(t+s))ds\mathrm{.}$ By [F5], $\mathrm{\parallel }{E}_h(s){\mathrm{\parallel }}_{\mathrm{\infty }\to \mathrm{\infty }}\le 1$.

For any two grid states $X,Z$ nodewise in $R$, the reaction flows satisfy $\mathrm{\parallel }{\mathrm{\Phi }}_{F_h}^{\tau }(X)-{\mathrm{\Phi }}_{F_h}^{\tau }(Z){\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{L_F\tau }\mathrm{\parallel }X-Z{\mathrm{\parallel }}_{\mathrm{\infty }}$, where $L_F={\sup }_R\mathrm{\parallel }DF\mathrm{\parallel }$; this follows by subtracting their nodewise integral equations and applying [F7]. Composing with the diffusion contraction [F5] gives $\mathrm{\parallel }{S}_{h,\tau }(X)-S_{h,\tau }(Z){\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{L_F\tau }\mathrm{\parallel }X-Z{\mathrm{\parallel }}_{\mathrm{\infty }}$ for the exact Lie map $S_{h,\tau }$.

Write $V_h(s)={\mathrm{\Phi }}_{F_h}^s(Y_h(t))$ for the nodewise exact reaction flow. By [F8], [F11], and integration of its reaction ODE using [F10], the exact Lie step from $Y_h(t)$ equals $E_h(\tau )Y_h(t)+{\int }_0^{\tau }{E}_h(\tau )F_h(V_h(s))ds$. Subtract the identity in step 1.1 and add and subtract $E_h(\tau )F_h(Y_h(t+s))$ inside the integrand. The defect is therefore the sum of ${\int }_0^{\tau }[E_h(\tau )-E_h(\tau -s)]F_h(Y_h(t+s))ds,{\int }_0^{\tau }{E}_h(\tau )[F_h(V_h(s))-F_h(Y_h(t+s))]ds,-{\int }_0^{\tau }{E}_h(\tau -s)r_h(t+s)ds\mathrm{.}$

For a scalar $C^2$ component $g$, two integrations of $g^{\mathrm{\prime }\mathrm{\prime }}$ give $g(x+h)-2g(x)+g(x-h)={\int }_0^h(h-s)(g^{\mathrm{\prime }\mathrm{\prime }}(x+s)+g^{\mathrm{\prime }\mathrm{\prime }}(x-s))ds$, so the absolute centered second difference divided by $h^2$ is at most $\mathrm{\parallel }{g}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Applying this in each coordinate to $u$ and using [A1] gives $\mathrm{\parallel }{\mathcal{A}}_h{Y}_h(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_u$ independently of the mesh. By [F4], ${\mathcal{A}}_h{F}_h(Y_h)-DF(Y_h){\mathcal{A}}_h{Y}_h$ is uniformly bounded; [A1] bounds $DF$, so $\mathrm{\parallel }{\mathcal{A}}_h{F}_h(Y_h(t+s)){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_A$ for all $s$ and grids. Since $E_h^{\mathrm{\prime }}(r)=E_h(r){\mathcal{A}}_h$ by [F9] and $E_h$ contracts by [F5], $\mathrm{\parallel }[E_h(\tau )-E_h(\tau -s)]F_h(Y_h(t+s)){\mathrm{\parallel }}_{\mathrm{\infty }}\le {\int }_{\tau -s}^{\tau }\mathrm{\parallel }{E}_h(r){\mathcal{A}}_h{F}_h(Y_h(t+s)){\mathrm{\parallel }}_{\mathrm{\infty }}dr\le C_{A}s\mathrm{.}$ Thus the first integral in step 2.1 has norm at most $C_A{\tau }^2\mathrm{/}2$.

The reaction path $V_h(s)$ stays in $R$ by [F11], so its time derivative is uniformly bounded by ${\sup }_R\mathrm{\parallel }F\mathrm{\parallel }$. Also $\mathrm{\parallel }{Y}_h(t+s)-Y_h(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le s\sup \mathrm{\parallel }{u}_t{\mathrm{\parallel }}_{\mathrm{\infty }}$. Hence, integrating both time derivatives by [F10], $\mathrm{\parallel }{V}_h(s)-Y_h(t+s){\mathrm{\parallel }}_{\mathrm{\infty }}\le Cs$. The mean-value bound for $F$ on the convex rectangle $R$ and contraction of $E_h$ show that the second integral in step 2.1 has norm at most $C{\tau }^2\mathrm{/}2$. The residual term has norm at most $C_{\mathrm{s}\mathrm{p}}\tau h_{\max }^{}$ by [F3] and [F5]. Consequently the one-step defect is at most $C({\tau }^2+\tau h_{\max }^{})$ with $C$ independent of grid and step.

Set $e_n=\mathrm{\parallel }{U}^n-Y_h(t_n){\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F6] and the assumed PDE range, both inputs in step 1.2 lie in $R$. Applying that stability estimate and the defect bound of step 3.2 gives $e_{n+1}\le e^{L_F\tau }{e}_n+C\tau (\tau +h_{\max }^{})$, with $e_0=0$. Iterating, $e_n\le C\tau (\tau +h_{\max }^{}){\sum }_{k=0}^{n-1}{e}^{L_{F}k\tau }\le CT{e}^{L_{F}T}(\tau +h_{\max }^{})$ for $n\tau \le T$. This proves the stated estimate after enlarging $C_T$.