MPRK22(1) has a uniform second-order defect and one-step bound on positive tubes

Define $A(y)$ as in [F2]. On $U$, the coordinate reciprocals $y_j^{-1}$ have continuous partials ${\mathrm{\partial }}_k(y_j^{-1})=-{\delta }_{jk}{y}_j^{-2}$ and ${\mathrm{\partial }}_{\mathrm{\ell }}{\mathrm{\partial }}_k(y_j^{-1})=2{\delta }_{jk}{\delta }_{\mathrm{\ell }j}{y}_j^{-3}$, as follows from the scalar reciprocal difference quotients on $\mathbb{R}\setminus \{0\}$; the product rule [F7] then makes every entry of $A$ $C^2$ from the rate data [F6]. By the continuous-partials theorem [F6], these entries have continuous total derivatives through order two. Direct multiplication gives $A(y)y=F(y)$. The matrix $I-hA(y)$ is invertible for $h>0$ by [F4] and equals $I$ at $h=0$; its determinant and cofactors are $C^2$ finite sums of products, and its determinant is nonzero, so the adjugate formula and the scalar reciprocal calculation make its inverse $C^2$ on the compact set of admitted $(h,y)$. It is uniformly bounded there for small $h$. Since $y>0$, [F4] gives $z=(I-hA(y){)}^{-1}y>0$. The identity $(I-hA{)}^{-1}=I+hA+h^2{A}^2(I-hA{)}^{-1}$ now gives $z=y+hF(y)+O(h^2)$ uniformly; in particular $z-y=O(h)$, so after decreasing $h_0$ we have $z\in V\subset \mathrm{\Omega }$ uniformly for $y\in U$. Thus all stage denominators and the final rate evaluations are valid.

Write $B=B(y,z)$ for the final coefficient matrix in [F3]. Its entries are $C^2$ on the compact region of pairs produced inside $V$ by step 1.1: the rates are $C^2$, the stage map is $C^2$, and each reciprocal $z_j^{-1}$ has the same coordinate derivative formulas as in step 1.1. Its columns sum to zero, and direct multiplication using $p_{ij}=d_{ji}$ gives $B(y,z)z=\frac{1}{2}(F(y)+F(z))$. For small $h$, step 1.1 keeps every $z_j$ bounded away from zero, so $B$ and its first derivatives are uniformly bounded. Also $B(y,y)=A(y)$; the mean-value theorem on the compact tube and $z-y=O(h)$ give $B(y,z)=A(y)+O(h)$. The final matrix $I-hB$ is invertible by [F4] and its inverse is uniformly bounded by the adjugate formula. Expanding its inverse yields ${\mathrm{\Psi }}_h(y)=y+hB(y,z)y+O(h^2)=y+hF(y)+O(h^2)$. Comparing with step 1.1 gives ${\mathrm{\Psi }}_h(y)-z=O(h^2)$.

The stage and final maps are $C^1$ in $y$ near the compact positive tube: their rates are $C^2$, their denominators stay bounded away from zero, and their matrix inverses are $C^2$ by the adjugate formula and the reciprocal calculation in steps 1.1–2.1. For any direction $v$, differentiating $z=y+hA(y)z$ gives $(I-hA)D_{y}z[v]=v+h(DA(y)[v])z$; boundedness from step 1.1 therefore gives $D_{y}z[v]=v+O(h)$ uniformly for $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le 1$.

Using ${\mathrm{\Psi }}_h-y=hB{\mathrm{\Psi }}_h$, substitute $z$ and use step 2.1 to obtain ${\mathrm{\Psi }}_h-y=hBz+hB({\mathrm{\Psi }}_h-z)=\frac{h}{2}(F(y)+F(z))+O(h^3)$. Taylor expansion of $F$ at $y$, together with $z-y=hF(y)+O(h^2)$ from step 1.1, gives $F(z)=F(y)+hDF(y)F(y)+O(h^2)$. Thus ${\mathrm{\Psi }}_h(y)=y+hF(y)+\frac{h^2}{2}DF(y)F(y)+O(h^3)$. By [F5]–[F7], the exact solution has the same expansion: $y^{\mathrm{\prime }\mathrm{\prime }}=DF(y)F(y)$ and its third derivative is bounded, so the one-variable Taylor remainder applied coordinatewise is $O(h^3)$. Their difference is the asserted uniform exact-start defect in [F9].

Differentiating ${\mathrm{\Psi }}_h=y+hB(y,z){\mathrm{\Psi }}_h$ in direction $v$ gives $D_y{\mathrm{\Psi }}_h[v]=(I-hB(y,z){)}^{-1}(v+h(D_{1}B(y,z)[v]+D_{2}B(y,z)[D_{y}z[v]]){\mathrm{\Psi }}_h)$, with the inverse defined by step 2.1.

Steps 1.1–2.2 give $z=y+O(h)$, ${\mathrm{\Psi }}_h=y+O(h)$, $B(y,z)=A(y)+O(h)$, and $(I-hB{)}^{-1}=I+hB+O(h^2)$ uniformly. Substitution yields $D_y{\mathrm{\Psi }}_h[v]=v+h(A(y)v+(D_{1}B(y,y)[v]+D_{2}B(y,y)[v])y)+o(h)$. Differentiating $B(y,y)y=F(y)$ with the chain and product rules identifies the coefficient as $DF(y)v$; compactness makes the remainder uniform, so $D_y{\mathrm{\Psi }}_h(y)=I+hDF(y)+hE(h,y)$ with ${\sup }_{y\in U}\mathrm{\parallel }E(h,y){\mathrm{\parallel }}_2\to 0$.

Compactness bounds $\mathrm{\parallel }DF(y){\mathrm{\parallel }}_2$ uniformly, and step 4.1 makes the $E$ contribution at most $h$ for sufficiently small $h$. The norm properties in [F10] then give $\mathrm{\parallel }{D}_y{\mathrm{\Psi }}_h(y){\mathrm{\parallel }}_2\le 1+Lh$ after decreasing $h_0$ and increasing $L$. Applying [F11] on convex $U$ proves $\mathrm{\parallel }{\mathrm{\Psi }}_h(u)-{\mathrm{\Psi }}_h(v){\mathrm{\parallel }}_2\le (1+Lh)\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$.