Two stage right-hand sides reuse one shifted-Jacobian factorization

Use the autonomous extension $f(t,y)=Jy$ on $D=\mathbb{R}\times {\mathbb{R}}^2$, with in_domain returning true and exact field evaluation and storage checks. The W approximation is $\tilde{J}=0$, so $W=I_2$. For the first active column $x=(1,0{)}^T$, [F3] has $s=1$, $\widehat{r}=1$, $\widehat{\alpha }=-1$, $v=(2,0{)}^T$, $w=4$, and $\widehat{\tau }=1\mathrm{/}2$. Its trailing-block dot is zero, so the output block is $\mathrm{diag}(-1,1)$. The second one-dimensional active column similarly has $v=2,w=4,\widehat{\tau }=1\mathrm{/}2$ and produces $R=-I_2$. By [F5] every exact representable intermediate is unchanged; all nonzero intermediates are among the listed stored normal values, no range branch fails, and both pivots are nonzero. Thus the one factorization succeeds.

The first RHS is $f(y)=(-1,-4{)}^T$. In the first reflector replay the dot is $-2$ and the active vector $p=\widehat{\tau }v$ is $(1,0{)}^T$, so the update sends $(-1,-4{)}^T$ to $(1,-4{)}^T$. The second replay has dot $-8$ and scalar $p=1$, so it sends the second coordinate to $-4-(-8)=4$. Every replay product, accumulated sum and subtraction is zero or one of the stipulated normal values, hence is represented exactly by [F5]. Thus the transformed RHS is $(1,4{)}^T$, and descending substitution with $R=-I_2$ has zero off-diagonal products and exact divisions by $-1$, giving $k_1=(-1,-4{)}^T$. The midpoint is $y+k_1\mathrm{/}2=(1\mathrm{/}2,0{)}^T$, so $F_1=J(y+k_1\mathrm{/}2)=(-1\mathrm{/}2,0{)}^T$; all these exact values are stored.

The second RHS is $F_1-k_1=(1\mathrm{/}2,4{)}^T$. The first stored reflector has dot $1$ and active $p=(1,0{)}^T$, so it sends the first coordinate to $1\mathrm{/}2-1=-1\mathrm{/}2$ and leaves the second at $4$. The second reflector has dot $8$ and scalar $p=1$, so it sends the second coordinate to $4-8=-4$. All replay products, sums and subtractions are again zero or stipulated normal represented values. Descending substitution with $R=-I_2$ is exact as in step 2.1 and gives $q_2=(1\mathrm{/}2,4{)}^T$. Hence $k_2=q_2+k_1=(-1\mathrm{/}2,0{)}^T$.

For either solve, $W=I_2$ makes $\widehat{v}=x$ and the computed residual zero, so the evaluator returns $\bar{r}=0$. The other norm upper bounds and ${\gamma }_2=2u\mathrm{/}(1-2u)<0.021$ show the first certificate expression is at most $0.021\cdot 2\cdot 5+0.01\cdot 10=0.31<1\mathrm{/}2$, so the evaluator can soundly return $\bar{W}=1\mathrm{/}2$ and then $\rho =1\mathrm{/}2$ for $\bar{r}+\bar{W}$. Thus both solves meet tolerance one. By [F1]–[F2], this successful call uses one factorization and two triangular solves.