Mesh-uniform reaction-diffusion commutator bound on a periodic grid

Let $w$ be a twice differentiable scalar function on a coordinate lift with bounded $w^{\mathrm{\prime }\mathrm{\prime }}$. By [F6], $w^{\mathrm{\prime }}$ is continuous; hence $s\mapsto w^{\mathrm{\prime }}(x+s)-w^{\mathrm{\prime }}(x-s)$ is continuous and integrable on $[0,h]$ by [F4]. Apply [F3] to the differentiable function $s\mapsto w(x+s)+w(x-s)-2w(x)$, whose endpoint derivatives are the corresponding one-sided restrictions. This gives $w(x+h)-2w(x)+w(x-h)={\int }_0^h(w^{\mathrm{\prime }}(x+s)-w^{\mathrm{\prime }}(x-s))ds\mathrm{.}$ For $s>0$, [F5] applied to $w^{\mathrm{\prime }}$ on $[x-s,x+s]$ gives $\mathrm{\mid }{w}^{\mathrm{\prime }}(x+s)-w^{\mathrm{\prime }}(x-s)\mathrm{\mid }\le 2s\mathrm{\parallel }{w}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$; at $s=0$ this also holds. Integrating yields $\mathrm{\mid }w(x+h)-2w(x)+w(x-h)\mathrm{\mid }\le h^2\mathrm{\parallel }{w}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus the centered second difference divided by $h^2$ has the same mesh-independent bound. Only the continuous first derivative was integrated; integrability or continuity of $w^{\mathrm{\prime }\mathrm{\prime }}$ is not assumed.

Fix a component $i$ and a spatial direction $j$. The chain rule first gives ${\mathrm{\partial }}_j(F_i(u))={\sum }_k{\mathrm{\partial }}_k{F}_i(u){\mathrm{\partial }}_j{u}_k$. Applying the product rule to each summand and the chain rule to ${\mathrm{\partial }}_k{F}_i(u)$ gives ${\mathrm{\partial }}_{jj}(F_i(u))={\sum }_k{\mathrm{\partial }}_k{F}_i(u){\mathrm{\partial }}_{jj}{u}_k+{\sum }_{k,\mathrm{\ell }}{\mathrm{\partial }}_{k\mathrm{\ell }}{F}_i(u)({\mathrm{\partial }}_j{u}_k)({\mathrm{\partial }}_j{u}_{\mathrm{\ell }})\mathrm{.}$ These are pointwise identities under the stated second-order spatial differentiability of $u$ and $F\in C^2$ near its range. To express the bound using only the quantities listed in the Statement, let $w_k={\sup }_R{z}_k-{\inf }_R{z}_k$ and $H_{kj}=\mathrm{\parallel }{\mathrm{\partial }}_{jj}{u}_k{\mathrm{\parallel }}_{\mathrm{\infty }}$. For $g(s)=u_k(x+s{e}_j,t)$ on the periodic lift, $g^{\mathrm{\prime }}$ is continuous by [F6], and $r\mapsto g^{\mathrm{\prime }}(r)-g^{\mathrm{\prime }}(0)$ is integrable on $[0,s]$ by [F4]. Apply [F3] to $r\mapsto g(r)-g(0)-r{g}^{\mathrm{\prime }}(0)$ to get $g(s)-g(0)-s{g}^{\mathrm{\prime }}(0)={\int }_0^s(g^{\mathrm{\prime }}(r)-g^{\mathrm{\prime }}(0))dr$. By [F5] on $g^{\mathrm{\prime }}$, $\mathrm{\mid }{g}^{\mathrm{\prime }}(r)-g^{\mathrm{\prime }}(0)\mathrm{\mid }\le r{H}_{kj}$ for $0\le r\le s$. The remainder is therefore bounded by $s^2{H}_{kj}\mathrm{/}2$. Since $\mathrm{\mid }g(s)-g(0)\mathrm{\mid }\le w_k$, this yields, for every $s>0$, $s\mathrm{\mid }{\mathrm{\partial }}_j{u}_k(x,t)\mathrm{\mid }\le w_k+\frac{s^2}{2}{H}_{kj}\mathrm{.}$ If $w_k{H}_{kj}>0$, choose $s=\sqrt{2w_k\mathrm{/}{H}_{kj}}$ to obtain $\mathrm{\mid }{\mathrm{\partial }}_j{u}_k\mathrm{\mid }\le \sqrt{2w_k{H}_{kj}}$. If $H_{kj}=0$, the same inequality for every $s$ gives ${\mathrm{\partial }}_j{u}_k=0$; if $w_k=0$, the component is constant and the same conclusion holds. Therefore the displayed chain-rule terms are uniformly bounded in terms of $R$, $D^{2}u$, $DF$, and $D^{2}F$, as asserted.

In each component, ${\mathcal{A}}_h{F}_h(P_{h}u)$ is a finite sum of $d_i$ times the centered second differences of $F_i\circ u$, while $DF(P_{h}u){\mathcal{A}}_h{P}_{h}u$ is the finite sum of ${\mathrm{\partial }}_k{F}_i(u)d_k$ times the centered second differences of $u_k$. Step 1.1 bounds each difference uniformly, and step 2.1 bounds every coefficient. There are only $d{m}^2$ terms and $D$ is fixed, so their sum is bounded by a constant independent of all $h_j$.