A diagonal stiff diffusion step with linear reaction

For coordinate $i$ with ${\lambda }_i\in \{1,1000\}$, put $a={\lambda }_i+1>0$. The exact equation is $y_i^{\mathrm{\prime }}=-a{y}_i+1$, $y_i(0)=0$. The product rule [F6] and chain rule [F7] show $(e^{at}{y}_i(t){)}^{\mathrm{\prime }}=e^{at}$. Integrating by [F4] gives $e^{ah}{y}_i(h)={\int }_0^h{e}^{as}ds=(e^{ah}-1)\mathrm{/}a$, since [F3], [F7] give antiderivative $e^{as}\mathrm{/}a$. Dividing by the positive exponential using [F5] yields $y_i(h)=(1-e^{-ah})\mathrm{/}a$.

By [F8], powers of $A=\mathrm{diag}(-{\lambda }_i)$ are diagonal with entries $(-{\lambda }_i{)}^j$, so the series [F2] evaluates the matrix phi functions coordinatewise. The exact exponential-Euler map has $e^{-{\lambda }_{i}h}{y}_i(0)=0$ and reaction value $g_i(0)=1$, so its coordinate is $h{\phi }_1(-{\lambda }_{i}h)$. The ETD2 predictor has that same coordinate; hence $g_i(a)=1-h{\phi }_1(-{\lambda }_{i}h)$ and the ETD2 coordinate is $h[({\phi }_1(-{\lambda }_{i}h)-{\phi }_2(-{\lambda }_{i}h))+{\phi }_2(-{\lambda }_{i}h)(1-h{\phi }_1(-{\lambda }_{i}h))]=h({\phi }_1(-{\lambda }_{i}h)-h{\phi }_1(-{\lambda }_{i}h){\phi }_2(-{\lambda }_{i}h))$.

In particular the stiff coordinate uses ${\lambda }_{2}h=1$: its exact, exponential-Euler, and ETD2 values are respectively $(1-e^{-1.001})\mathrm{/}1001$, ${10}^{-3}{\phi }_1(-1)$, and ${10}^{-3}({\phi }_1(-1)-{10}^{-3}{\phi }_1(-1){\phi }_2(-1))$. All exponential and phi actions are at explicit matrix/scalar arguments; any numerical implementation must return the certificates specified by Certified-action implementation of exponential Euler and ETD2.