Local second-order midpoint collocation error with an algebraic residual budget

Let $Y_i^{\ast }=y^{\ast }(t_i)$. Applying scalar Taylor's formula coordinatewise to the $d$ components of the $C^3$ curve $y^{\ast }$ on each cell gives $(Y_{i+1}^{\ast }-Y_i^{\ast })\mathrm{/}h=y^{\ast \mathrm{\prime }}(t_i+h\mathrm{/}2)+O(h^2)$ and $(Y_i^{\ast }+Y_{i+1}^{\ast })\mathrm{/}2=y^{\ast }(t_i+h\mathrm{/}2)+O(h^2)$. Since $y^{\ast \mathrm{\prime }}=f(t,y^{\ast }(t))$, bounded first derivatives of $f$ make each normalized interior residual $O(h^2)$ uniformly. The exact endpoint pair makes the boundary block zero; [F2] names the boundedly invertible continuous linearized operator assumed here, and thus $\mathrm{\parallel }{G}_h(Y^{\ast }){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_0{h}^2$. The Taylor remainder bounds are uniform because the derivatives are bounded on a compact neighborhood of the trajectory.

On a fixed-radius ball about $Y^{\ast }$, all midpoint states and endpoint pairs stay in compact subsets of the domains for sufficiently small radius. Bounded second derivatives of $f$ and $B$ imply a constant $L_G$, independent of $h,N$, such that $\mathrm{\parallel }D{G}_h(Y)-D{G}_h(Z){\mathrm{\parallel }}_{\mathrm{\infty }\to \mathrm{\infty }}\le L_G\mathrm{\parallel }Y-Z{\mathrm{\parallel }}_{\mathrm{\infty }}$ on this ball: every interior row depends on only two node blocks and its nonlinear part is $f$ evaluated at their average, while the linear increment term has constant derivative; the boundary row depends only on the two endpoint blocks. The bounds on each row therefore do not grow with the number of mesh cells.

Choose $r>0$ small enough that the closed ball ${\bar{B}}_{\mathrm{\infty }}(Y^{\ast },r)$ lies in the nodal domain and $K{L}_{G}r\le 1\mathrm{/}2$. Define $T_h(Y)=Y-J_h^{-1}{G}_h(Y)$. The mean-value bound from step 1.2 gives $\mathrm{\parallel }{T}_h(Y)-T_h(Z){\mathrm{\parallel }}_{\mathrm{\infty }}\le K{L}_{G}r\mathrm{\parallel }Y-Z{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{2}\mathrm{\parallel }Y-Z{\mathrm{\parallel }}_{\mathrm{\infty }}$ on the ball. Also $\mathrm{\parallel }{T}_h(Y^{\ast })-Y^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}\le K{C}_0{h}^2$, so for small enough $h$ this is at most $r\mathrm{/}2$; then $T_h$ maps the ball into itself. It is a contraction on a nonempty complete ball by [F4] and [F5].

The contraction theorem gives a unique fixed point $Y_h$ in the ball. Since $J_h$ is invertible, $T_h(Y)=Y$ is equivalent to $G_h(Y)=0$, so this is the unique residual zero there. The fixed-point estimate gives $\mathrm{\parallel }{Y}_h-Y^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2K{C}_0{h}^2$. Further, $\mathrm{\parallel }D{G}_h(Y_h)-J_h\mathrm{\parallel }\le L_G\mathrm{\parallel }{Y}_h-Y^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^2)$. For small $h$, $\mathrm{\parallel }{J}_h^{-1}(D{G}_h(Y_h)-J_h)\mathrm{\parallel }<1\mathrm{/}2$; the geometric-series inverse of $I+J_h^{-1}(D{G}_h(Y_h)-J_h)$ shows $D{G}_h(Y_h)$ invertible with inverse norm at most $2K$.

For any candidate $Y$ in the ball, put $e=Y-Y_h$. Each interior residual block depends only on two node blocks, and the boundary block only on the two endpoint blocks. Apply [F6] to each scalar component of $G_h(Y_h+te)$ for $0\le t\le 1$; the segment lies in the open nodal domain. The uniform second-derivative bounds from step 1.2, row locality, and fixed $d$ give a constant $C_G$, independent of $h$ and $N$, such that $G_h(Y)=D{G}_h(Y_h)e+q$ with $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_G\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. Since both points lie in the radius-$r$ ball, $\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2r$; reduce $r$ initially if needed so $4K{C}_{G}r\le 1\mathrm{/}2$. Using [step 3.1], $\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2K\mathrm{\parallel }{G}_h(Y){\mathrm{\parallel }}_{\mathrm{\infty }}+2K{C}_G\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}^2\le 2K{\epsilon }_{\mathrm{a}\mathrm{l}\mathrm{g}}+\frac{1}{2}\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $\mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}\le 4K{\epsilon }_{\mathrm{a}\mathrm{l}\mathrm{g}}$. The piecewise-linear interpolant is a convex combination of adjacent nodes, hence $\mathrm{\parallel }{P}_h(Y-Y_h){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }e{\mathrm{\parallel }}_{\mathrm{\infty }}$. The linear interpolant of $Y^{\ast }$ differs from $y^{\ast }$ by at most $h^2\mathrm{\parallel }{y}^{\ast \mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}8$ on each cell, and $\mathrm{\parallel }{P}_h(Y_h-Y^{\ast }){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{Y}_h-Y^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^2)$ by step 3.1. Adding these bounds proves the stated estimate.