Uniform inverse bound for the sampled midpoint linearized BVP

Write $A(t)=D_{y}f(t,y^{\ast }(t))$ and let $X(t)$ solve $X^{\mathrm{\prime }}=A(t)X$, $X(a)=I_d$. Since $A$ is continuous and bounded on $[a,b]$, extend it continuously by its endpoint values to an open interval containing $[a,b]$. Local existence follows from [F3], and the integral equation [F8] gives the Gronwall bound for the maximal solution from [F7]; if it stopped at or before $b$, boundedness of $X^{\mathrm{\prime }}=AX$ would give a finite endpoint limit from which local existence extends it, a contradiction. Thus $X$ exists on $[a,b]$. Define $Q=B_1+B_{2}X(b)$. If $Qc=0$, then $v(t)=X(t)c$ satisfies $\mathcal{L}v=(0,0)$; [F2] forces $v=0$, and $v(a)=c$ then gives $c=0$. The square matrix $Q$ is therefore invertible.

For each mesh let $C_i^h=D_{y}f(t_i+h\mathrm{/}2,(Y_i^{\ast }+Y_{i+1}^{\ast })\mathrm{/}2)$. Uniform continuity of $y^{\ast }$ and $D_{y}f$ on a compact neighborhood implies ${\max }_i\mathrm{\parallel }{C}_i^h-A(t_i+h\mathrm{/}2){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\to 0$. Also ${\sup }_{h,i}\mathrm{\parallel }{C}_i^h{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<\mathrm{\infty }$. Applying the chain rule [F5] to the rows in [F1], the interior equation $J_{h}z=(r_0,\dots ,r_{N-1},q)$ is $(I-h{C}_i^h\mathrm{/}2)z_{i+1}=(I+h{C}_i^h\mathrm{/}2)z_i+h{r}_i$. For small $h$, $I-h{C}_i^h\mathrm{/}2$ is invertible by its convergent geometric series; its inverse norm is at most $2$, and the induced transition matrices $M_i^h=(I-h{C}_i^h\mathrm{/}2{)}^{-1}(I+h{C}_i^h\mathrm{/}2)$ satisfy $\mathrm{\parallel }{M}_i^h\mathrm{\parallel }\le 1+C_{0}h$ for a constant independent of $i,h$.

Let $M_h=M_{N-1}^h\cdots M_0^h$, the discrete homogeneous transfer from $z_0$ to $z_N$. By [F8] on each column, the exact fundamental matrix $X$ satisfies $X(t_{i+1})-X(t_i)={\int }_{t_i}^{t_{i+1}}A(t)X(t)dt$. The integrand is uniformly continuous, so its average over each cell differs from its midpoint value by a quantity tending uniformly to zero as $h\to 0$. Explicitly, put $m_i=t_i+h\mathrm{/}2$, $H(t)=A(t)X(t)$ and ${\omega }_G(\rho )={\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \rho }\mathrm{\parallel }G(s)-G(t){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ for $s,t\in [a,b]$ and $G=H,X$. Write $S_X={\sup }_t\mathrm{\parallel }X(t){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$, $S_A={\sup }_t\mathrm{\parallel }A(t){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$, and ${\delta }_h={\max }_i\mathrm{\parallel }{C}_i^h-A(m_i){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\to 0$ by step 1.2. By [F9], $\mathrm{\parallel }{h}^{-1}{\int }_{t_i}^{t_{i+1}}H(t)dt-H(m_i){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\omega }_H(h\mathrm{/}2)$. Define the normalized midpoint defect by ${\eta }_i^h=(X(t_{i+1})-X(t_i))\mathrm{/}h-C_i^h(X(t_i)+X(t_{i+1}))\mathrm{/}2$. Adding and subtracting $H(m_i)$ and $A(m_i)(X(t_i)+X(t_{i+1}))\mathrm{/}2$ therefore gives ${\max }_i\mathrm{\parallel }{\eta }_i^h{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\omega }_H(h\mathrm{/}2)+{\delta }_h{S}_X+S_A{\omega }_X(h\mathrm{/}2)\to 0$, since $H$ and $X$ are uniformly continuous on $[a,b]$. The transition bound in step 1.2 propagates the increment defects $h{\eta }_i^h$ with total bound at most $C_1(b-a){\max }_i\mathrm{\parallel }{\eta }_i^h\mathrm{\parallel }$; hence $M_h\to X(b)$.

Put $Q_h=B_1+B_2{M}_h$. By step 2.1, $Q_h\to Q$. Since $Q$ is invertible, determinant continuity makes $Q_h$ invertible for all sufficiently small $h$, and the adjugate formula gives a uniform bound on $\mathrm{\parallel }{Q}_h^{-1}\mathrm{\parallel }$. For arbitrary forcing blocks $r_i$, the recurrence in step 1.2 and its transition bound give $z_N=M_h{z}_0+w_h$ and ${\max }_i\mathrm{\parallel }{z}_i{\mathrm{\parallel }}_2\le C_2(\mathrm{\parallel }{z}_0{\mathrm{\parallel }}_2+{\max }_i\mathrm{\parallel }{r}_i{\mathrm{\parallel }}_2)$, with $\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_2\le C_2{\max }_i\mathrm{\parallel }{r}_i{\mathrm{\parallel }}_2$. The boundary row $B_1{z}_0+B_2{z}_N=q$ becomes $Q_h{z}_0=q-B_2{w}_h$, so $z_0$ and then every $z_i$ are bounded by $C_3\max ({\max }_i\mathrm{\parallel }{r}_i{\mathrm{\parallel }}_2,\mathrm{\parallel }q{\mathrm{\parallel }}_2)$ uniformly in $h$.

For every right-hand side $(r_0,\dots ,r_{N-1},q)$, step 3.1 constructs a unique solution of the square linear system $J_{h}z=(r_0,\dots ,r_{N-1},q)$ and gives the asserted uniform operator bound. Therefore $J_h$ is onto and one-to-one, hence invertible, with the stated estimate.