Subtracting analytic Cauchy jets

For $0\le j<m$, differentiation gives ${\mathrm{\partial }}_t^{j}P={\sum }_{\mathrm{\ell }=j}^{m-1}{t}^{\mathrm{\ell }-j}{g}_{\mathrm{\ell }}(x)\mathrm{/}(\mathrm{\ell }-j)!$. Thus ${\mathrm{\partial }}_t^{j}P(0,x)=g_j(x)$ and ${\mathrm{\partial }}_t^{m}P=0$. Its tangential derivatives are obtained by replacing $g_{\mathrm{\ell }}$ with ${\mathrm{\partial }}_x^{\alpha }{g}_{\mathrm{\ell }}$ and are analytic by F1.

For each allowed slot $(\alpha ,j)$ put $P_{\alpha ,j}={\mathrm{\partial }}_x^{\alpha }{\mathrm{\partial }}_t^{j}P$. The transformed right-hand side is $\tilde{F}(t,x,(V_{\alpha ,j}))=F(t,x,(V_{\alpha ,j}+P_{\alpha ,j}(t,x)))$. This finite analytic substitution is made near the actual initial jet, so after translation of that centre F1 applies to zero-constant increments. It is analytic near $(0,0,0)$ and introduces no new derivative slot. Since ${\mathrm{\partial }}_t^{m}P=0$, ${\mathrm{\partial }}_t^{m}u=F$ is exactly ${\mathrm{\partial }}_t^{m}v=\tilde{F}$.

Step 1.1 gives ${\mathrm{\partial }}_t^{j}v(0,x)={\mathrm{\partial }}_t^{j}u(0,x)-g_j(x)$, so the old data hold precisely when all the new data vanish. Conversely adding P to any zero-data solution reverses step 2.1 and restores every old data function. Addition and subtraction of P are inverse maps on the solution germs.