A Gaussian collocation defect has a high-order weighted endpoint bound

Let $y_h(s)=y(t+hs)$ and $e(s)=y_h(s)-u_h(s)$. Define the averaged Jacobian $A(s)={\int }_0^1{D}_{y}f(t+hs,u_h(s)+\theta e(s))d\theta$ and $B(s)=hA(s)$. For each unit vector $v\in {\mathbb{R}}^d$, the vector integral inequality [F8] gives $\mathrm{\parallel }A(s)v{\mathrm{\parallel }}_2\le {\int }_0^1\mathrm{\parallel }{D}_{y}f(t+hs,u_h(s)+\theta e(s))v{\mathrm{\parallel }}_{2}d\theta \le L$. Taking the supremum over unit $v$ proves $\mathrm{\parallel }A(s){\mathrm{\parallel }}_{op}\le L$ and hence $\mathrm{\parallel }B(s){\mathrm{\parallel }}_{op}\le hL$. Applying the scalar FTC to each coordinate along the segment gives $f(t+hs,y_h(s))-f(t+hs,u_h(s))=A(s)e(s)$. With $r_h=u_h^{\mathrm{\prime }}-h{g}_h$, [F13] and [F1] give $e^{\mathrm{\prime }}=Be-r_h$, $e(0)=0$.

For a continuous matrix function $X$ set $\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{\infty }}:={\sup }_{s\in [0,1]}\mathrm{\parallel }X(s){\mathrm{\parallel }}_{op}$. Define $(TX)(s)={\int }_s^{1}X(\tau )B(\tau )d\tau$ entrywise. For any unit vector $v$, the vector integral inequality [F8] and submultiplicativity in [A1] give $\mathrm{\parallel }(TX-TY)(s)v{\mathrm{\parallel }}_2\le {\int }_s^1\mathrm{\parallel }(X-Y)(\tau )B(\tau )v{\mathrm{\parallel }}_{2}d\tau \le hL\mathrm{\parallel }X-Y{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Taking the supremum over unit $v$ and $s$ proves $\mathrm{\parallel }TX-TY{\mathrm{\parallel }}_{\mathrm{\infty }}\le hL\mathrm{\parallel }X-Y{\mathrm{\parallel }}_{\mathrm{\infty }}$. For $hL\le 1\mathrm{/}2$, $T$ is a contraction. Starting with $P_0=I$, set $P_{n+1}=I+T{P}_n$. Successive differences are bounded by $2^{-n}\mathrm{\parallel }T{P}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$, so their entries are uniformly Cauchy; [F9] gives an entrywise limit and the geometric tail bound makes convergence uniform. Each iterate is continuous and [F7] makes the limit $P$ continuous. Passing to the limit in $P_{n+1}=I+T{P}_n$ gives $P=I+TP$, hence $P^{\mathrm{\prime }}=-PB$ and $P(1)=I$ by applying [F6] to each matrix entry. If $P$ and $\tilde{P}$ both solve it, $D=P-\tilde{P}=TD$ implies $\mathrm{\parallel }D{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{2}\mathrm{\parallel }D{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $D=0$; also $P=I+TP$ gives $\mathrm{\parallel }P{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2$.

The normalized exact path satisfies $y_h^{\mathrm{\prime }}=hf(t+hs,y_h(s))=O(h)$ by [F13] and [A1]. If $y_h^{(r)}=O(h^r)$ for $1\le r\le j$, the chain rule [F5] writes the $j$th derivative of $f(t+hs,y_h(s))$ as a finite sum whose argument-derivative orders total $j$; each term is $O(h^j)$. Differentiating the IVP identity gives $y_h^{(j+1)}=O(h^{j+1})$. Induction proves $y_h^{(j)}=O(h^j)$ through order $2M$.

The path $u_h+\theta (y_h-u_h)$ between the collocation polynomial and exact path has normalized derivatives of order $j\le M-1$ of size $O(h^j)$ by [F2] and step 2.2. Repeated chain and product rules for $D_{y}f$, then integration over $0\le \theta \le 1$ with uniform remainders on the compact tube, give $\mathrm{\parallel }{A}^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^j)$ and $\mathrm{\parallel }{B}^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^{j+1})$ for $0\le j<M$.

Differentiating $P^{\mathrm{\prime }}=-PB$ entrywise gives $P^{(j)}=-{\sum }_{k=0}^{j-1}\left(\genfrac{}{}{0px}{}{j-1}{k}\right)P^{(k)}{B}^{(j-1-k)}$ for $1\le j\le M$. Induction from $\mathrm{\parallel }P{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2$ and step 3.1 yields $\mathrm{\parallel }{P}^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_j{h}^j$. Apply the scalar Taylor formula [F4] to every matrix entry to get a matrix polynomial $q$ of degree at most $M-1$ with $\mathrm{\parallel }P-q{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{h}^M$ and $\mathrm{\parallel }{q}^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_j{h}^j$ for $j<M$. Finite dimensionality combines the entrywise bounds in the induced operator norm.

By [F1], $b_j={\int }_0^1{\mathrm{\ell }}_j(s)ds$; by [F11] the Gauss weights are $w_j=\int {\mathrm{\ell }}_{j}d\lambda$. The continuous-integral agreement [F12] gives $b_j=w_j$. Uniqueness of the monic orthogonal polynomial [F14] makes the order-$2M$ and order-$(2M-1)$ branch nodes identical, so [F3] applies to this tableau. By [F2], $\mathrm{\parallel }{g}_h-I_M{g}_h{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^M)$. The cardinal interpolation [F10] gives degree at most $2M-2$ for each coordinate of $q{I}_M{g}_h$, and [F3] therefore gives, coordinatewise, $\int q{I}_M{g}_h={\sum }_j{b}_{j}q(c_j)g_h(c_j)$. Split the weighted defect integral entrywise as ${\int }_0^{1}P(g_h-I_M{g}_h)={\int }_0^1(P-q)(g_h-I_M{g}_h)+({\int }_0^{1}q{g}_h-{\int }_0^{1}q{I}_M{g}_h)$.

The first integral in step 5.1 is $O(h^{2M})$ by step 4.1 and $\mathrm{\parallel }{g}_h-I_M{g}_h{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^M)$. For each scalar coordinate, the parenthesized term is the Gauss quadrature error for that coordinate of $q{g}_h$. Leibniz' rule [F5] and the bounds $q^{(j)}=O(h^j)$ and $g_h^{(j)}=O(h^j)$ show $(q{g}_h{)}^{(2M)}=O(h^{2M})$ coordinatewise, so the scalar remainder [F3] bounds every coordinate error by $O(h^{2M})$. Since there are finitely many coordinates, the vector integral is $O(h^{2M})$ in Euclidean norm. Hence ${\int }_0^{1}P(g_h-I_M{g}_h)=O(h^{2M})$.

By step 1.1, $e^{\mathrm{\prime }}=Be-r_h$, and step 2.1 gives $P^{\mathrm{\prime }}=-PB$, $P(1)=I$. Apply the product rule entrywise to obtain $(Pe{)}^{\mathrm{\prime }}=-P{r}_h$; the scalar FTC on each coordinate and $e(0)=0$ give $e(1)=-{\int }_0^{1}P{r}_h=h{\int }_0^{1}P(g_h-I_M{g}_h)$. Step 6.1 bounds this by $O(h^{2M+1})$, and [F1] identifies $u_h(1)$ as the collocation endpoint. This proves the local defect bound.