Variational equations and commuting local flows

For $m=0$, define ${\mathbb{R}}^0$ as the singleton consisting of the empty coordinate tuple, denoted $0$, with its unique vector-space structure and norm $\mathrm{\parallel }0\mathrm{\parallel }=0$. Its open subsets are empty or the singleton; the empty domain has no initial point and all assertions there are vacuous. On the singleton both fields necessarily equal zero. Define ${\varphi }_X^t(0)={\varphi }_Y^t(0)=0$ for every real $t$. These are the unique solution curves and satisfy the flow law. Every map into this zero vector space has the unique zero linear map as total derivative, since its differentiability remainder is identically zero; iterating gives continuous derivatives through order two. Thus the flows are jointly $C^2$, $J$ is the unique endomorphism of ${\mathbb{R}}^0$, which is also its identity $I_0$, and ${\mathrm{\partial }}_{t}J=DX({\varphi }_X)J=0$. The bracket and both transport expressions are zero, and both flow compositions are the identity. This proves every assertion for $m=0$ without invoking the positive-dimensional suppliers. In the remaining steps assume $m\ge 1$.

For $p$ near $q$ put ${\mathrm{\Delta }}_h(t):={\varphi }_X^t(p+h)-{\varphi }_X^t(p)$. The integral equations from [F1] and the one-variable fundamental theorem applied to $\theta \mapsto X({\varphi }_X^s(p)+\theta {\mathrm{\Delta }}_h(s))$ give ${\mathrm{\Delta }}_h(t)=h+{\int }_0^t{A}_h(s){\mathrm{\Delta }}_h(s)ds,A_h(s):={\int }_0^{1}DX({\varphi }_X^s(p)+\theta {\mathrm{\Delta }}_h(s))d\theta \mathrm{.}$ The segments lie in $K$ after shrinking the initial ball and time interval. The local Lipschitz estimate in [F1] gives ${\sup }_t\mathrm{\parallel }{\mathrm{\Delta }}_h(t)\mathrm{\parallel }=O(\mathrm{\parallel }h\mathrm{\parallel })$. Uniform continuity of $DX$ on $K$ from [F4] makes $A_h(s)$ converge uniformly to $DX({\varphi }_X^s(p))$ as $h\to 0$.

Let $\mathcal{I}=[-\delta ,\delta ]$ and let ${\mathcal{C}}_1=C(\mathcal{I},M_m(\mathbb{R}))$ with the supremum operator norm. This space is complete: for a supremum-Cauchy sequence $(J_k)$, each matrix entry at each $s$ is Cauchy because $\mathrm{\mid }{A}_{ij}\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$; [F5] applied to the $m^2$ entries gives a pointwise matrix limit $J(s)$, and the uniform Cauchy estimate passed to each entry gives uniform convergence of all entries. Each limit entry is continuous by the $\epsilon \mathrm{/}3$ argument for uniform limits of continuous functions. The finite-coordinate bound $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\sum }_{i,j}\mathrm{\mid }{A}_{ij}\mathrm{\mid }$ then gives uniform operator-norm convergence, so $J\in {\mathcal{C}}_1$. The space is nonempty because it contains the constant zero matrix curve. Put $A_0(s):=DX({\varphi }_X^s(p))$ and define $T(J)(t):=I_m+{\int }_0^t{A}_0(s)J(s)ds$. It maps ${\mathcal{C}}_1$ into itself and is a contraction with constant at most $L\delta <1$. By [F6], it has a unique fixed point $J$. Define $R_h(t):={\mathrm{\Delta }}_h(t)-J(t)h$. Subtracting the integral equations and writing the integrand difference as $(A_h-A_0){\mathrm{\Delta }}_h+A_0{R}_h$ gives $\mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le L\delta \mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}+\delta \mathrm{\parallel }{A}_h-A_0{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathrm{\Delta }}_h{\mathrm{\parallel }}_{\mathrm{\infty }}=o(\mathrm{\parallel }h\mathrm{\parallel })$, by step 1.2 and uniform continuity of $DX$ on $K$. Thus $D_q{\varphi }_X^t(p)=J(t,p)$. The continuous integrand in the equation for $J$ has average over $[t,t+h]$ tending to its value at $t$: the quotient error is at most ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\parallel }DX({\varphi }_X^s(p))J(s,p)-DX({\varphi }_X^t(p))J(t,p)\mathrm{\parallel }$, which tends to zero. Thus ${\mathrm{\partial }}_{t}J=DX({\varphi }_X^t(p))J$ and $J(0,p)=I_m$.

Write $J_h(s)=D_q{\varphi }_X^s(p+h)$ and $J_0(s)=D_q{\varphi }_X^s(p)$. Subtracting their integral equations gives $J_h-J_0={\int }_0^t{A}_0(J_h-J_0)ds+{\int }_0^t(A_h-A_0)J_{h}ds$, where $A_h=DX({\varphi }_h)$, $A_0=DX({\varphi }_0)$, ${\varphi }_h={\varphi }_X^s(p+h)$, and ${\varphi }_0={\varphi }_X^s(p)$. The line-segment chain rule and Newton–Leibniz supplier claims in [F2]–[F3], applied entrywise to $DX$ along the segment from ${\varphi }_0$ to ${\varphi }_h$, give the exact formula $A_h-A_0={\int }_0^1{D}^{2}X({\varphi }_0+\theta ({\varphi }_h-{\varphi }_0))[{\varphi }_h-{\varphi }_0,\cdot ]d\theta$ and therefore $\mathrm{\parallel }{A}_h-A_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_2\mathrm{\parallel }{\varphi }_h-{\varphi }_0{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F1], $\mathrm{\parallel }{\varphi }_h-{\varphi }_0{\mathrm{\parallel }}_{\mathrm{\infty }}=O(\mathrm{\parallel }h\mathrm{\parallel })$; the integral equation gives $\mathrm{\parallel }{J}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le (1-L\delta {)}^{-1}$. Absorbing the first integral using $L\delta <1$ yields $J_h-J_0=O(\mathrm{\parallel }h\mathrm{\parallel })$ uniformly in time. Uniform continuity of $D^{2}X$ on $K$, the same line-segment formula, and ${\varphi }_h-{\varphi }_0=J_{0}h+o(\mathrm{\parallel }h\mathrm{\parallel })$ from step 2.1 now give $A_h-A_0=D^{2}X({\varphi }_0)[J_{0}h,\cdot ]+o(\mathrm{\parallel }h\mathrm{\parallel })$ uniformly, and hence $(A_h-A_0)J_h=D^{2}X({\varphi }_0)[J_{0}h,J_0(\cdot )]+o(\mathrm{\parallel }h\mathrm{\parallel })$. For each base point $p$, define ${\mathcal{C}}_2=C(\mathcal{I},\mathrm{Bil}({\mathbb{R}}^m\times {\mathbb{R}}^m,{\mathbb{R}}^m))$ with the supremum bilinear-operator norm. The coordinatewise completeness proof of step 2.1 applies with $m^3$ coefficients, so ${\mathcal{C}}_2$ is complete; it is nonempty because it contains the zero bilinear-map curve. The map $(\mathcal{T}K)(t)[u,v]:={\int }_0^t(A_0(s)K(s)[u,v]+D^{2}X({\varphi }_0(s))[J_0(s)u,J_0(s)v])ds$ is a contraction with constant at most $L\delta$ on ${\mathcal{C}}_2$. By [F6] it has a unique fixed point $K_p$. Subtracting its equation evaluated at $(h,v)$ from the equation for $J_h-J_0$ gives $R_h:=J_h-J_0-K_p[h,\cdot ]$ with $\mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le L\delta \mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}+o(\mathrm{\parallel }h\mathrm{\parallel })$, so $R_h=o(\mathrm{\parallel }h\mathrm{\parallel })$. Therefore $D_{p}J[h](v)=K_p[h,v]$, and $K_p$ is the second derivative $D_q^2{\varphi }_X$. Its equation makes it bilinear. The coefficients, forcing, ${\varphi }_p$, and $J_p$ vary continuously in $p$ uniformly for $s\in \mathcal{I}$; for fixed-point maps $T_p$ with contraction constant $L\delta$, $\mathrm{\parallel }{K}_p-K_{p^{\mathrm{\prime }}}{\mathrm{\parallel }}_{\mathrm{\infty }}\le (1-L\delta {)}^{-1}\mathrm{\parallel }{T}_p(K_{p^{\mathrm{\prime }}})-T_{p^{\mathrm{\prime }}}(K_{p^{\mathrm{\prime }}}){\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. The integral equation gives continuity in $t$ as well. Thus the second derivatives in $q$ are continuous.

Let $V(t):=Y({\varphi }_X^t(q))$ and $W(t):=D_q{\varphi }_X^t(q)Y(q)$. By [F2] and the variational equation, $V^{\mathrm{\prime }}=DY({\varphi }_X^t)X({\varphi }_X^t)$ and $W^{\mathrm{\prime }}=DX({\varphi }_X^t)W$. The identity $[X,Y]=0$ makes $V$ solve this same linear equation with $V(0)=W(0)=Y(q)$; uniqueness from the contraction argument gives $D_q{\varphi }_X^t(q)Y(q)=Y({\varphi }_X^t(q))$.

On the open interior of the chosen time and initial-point neighborhood, step 2.1 gives $D_q{\varphi }_X=J$ and ${\mathrm{\partial }}_{t}J=DX({\varphi }_X)J$, while step 3.1 gives the continuous second derivative $D_{q}J=K$. The chain rule [F2] applied to ${\mathrm{\partial }}_t{\varphi }_X=X({\varphi }_X)$ gives ${\mathrm{\partial }}_{q_i}{\mathrm{\partial }}_t{\varphi }_X=DX({\varphi }_X)J{e}_i$, and the variational equation gives ${\mathrm{\partial }}_t{\mathrm{\partial }}_{q_i}{\varphi }_X=DX({\varphi }_X)J{e}_i$; also ${\mathrm{\partial }}_t^2{\varphi }_X=DX({\varphi }_X)X({\varphi }_X)$. Hence all first- and second-order coordinate partial derivatives of ${\varphi }_X$ exist and are continuous there. Apply [F7] pointwise first to ${\varphi }_X$ and then to its first-derivative map: its Jacobian is the matrix of first partials, and the Jacobian of that map is the continuous array of second partials. Thus ${\varphi }_X$ is jointly $C^2$.

For fixed $t$, the curve $s\mapsto {\varphi }_X^t({\varphi }_Y^s(q))$ has derivative $D{\varphi }_X^{t}Y({\varphi }_Y^s(q))=Y({\varphi }_X^t({\varphi }_Y^s(q)))$ by step 3.2, and starts at ${\varphi }_X^t(q)$. It is therefore the $Y$-flow from that initial point; IVP uniqueness from [F1] yields ${\varphi }_X^t{\varphi }_Y^s(q)={\varphi }_Y^s{\varphi }_X^t(q)$.