The flow domain of a C¹ autonomous field is open and the flow is C¹ on it

Let $(t_0,a_0)\in \mathrm{\Omega }$ with $t_0>0$, and let $x:=x(\cdot ;a_0)$ on $[0,t_0]$. Its image $K:=x([0,t_0])$ is nonempty compact and lies in $U$. If ${\mathbb{R}}^n\setminus U$ is nonempty, [F3] gives a separation $d>0$; choose $0<\delta <d$, so every closed $\delta$-ball centred in $K$ lies in $U$. If $U={\mathbb{R}}^n$, choose any $\delta >0$. Define $d_K(z):={\min }_{k\in K}\mathrm{\parallel }z-k{\mathrm{\parallel }}_2$: for each fixed $z$, the reverse triangle inequality makes $k\mapsto \mathrm{\parallel }z-k{\mathrm{\parallel }}_2$ continuous, so the minimum is attained by [F4]. If $k_w$ attains the minimum for $w$, then $d_K(z)\le \mathrm{\parallel }z-k_w{\mathrm{\parallel }}_2\le \mathrm{\parallel }z-w{\mathrm{\parallel }}_2+d_K(w)$; interchanging $z,w$ gives $\mathrm{\mid }{d}_K(z)-d_K(w)\mathrm{\mid }\le \mathrm{\parallel }z-w{\mathrm{\parallel }}_2$. Thus the fixed tube $H:=\{z:d_K(z)\le \delta \}$ is closed. Since compact $K$ is bounded, there is $B\ge 0$ with $\mathrm{\parallel }k{\mathrm{\parallel }}_2\le B$ for $k\in K$; a minimizing $k$ for $z\in H$ gives $\mathrm{\parallel }z{\mathrm{\parallel }}_2\le B+\delta$, so $H$ is bounded and hence compact by [F4]. It lies in $U$: if the complement of $U$ is nonempty, every point in it has $d_K(z)\ge d>\delta$ by [F3]; when $U={\mathbb{R}}^n$ inclusion is automatic. The vector norm inequalities used here are those of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation. Put $M:={\sup }_H\mathrm{\parallel }F{\mathrm{\parallel }}_2$ and $L:={\sup }_H\mathrm{\parallel }DF\mathrm{\parallel }$, finite by [F4] and the norm continuity in [F1], before choosing any subdivision. Choose $N\ge 1$ with $h:=t_0\mathrm{/}N$ satisfying $hM\le \delta \mathrm{/}4$ and $hL\le 1\mathrm{/}2$, and set $x_j:=x(jh)$ for $0\le j\le N$. Every $\bar{B}(x_j,\delta )$ lies in $H$; thus its bounds $M_j,L_j$ satisfy $M_j\le M$, $L_j\le L$. Set $\rho :=\delta \mathrm{/}4$.

Small-step claim: for every $j$ and every $y\in \bar{B}(x_j,\rho )$, the maximal solution through $y$ is defined on $[0,h]$ with values in $\bar{B}(x_j,\delta )\subseteq U$, and $\mathrm{\Phi }$ is $C^1$ on $[0,h]\times B(x_j,\rho )$ with $\mathrm{\parallel }\mathrm{\Phi }(t,y)-\mathrm{\Phi }(t,x_j){\mathrm{\parallel }}_2\le e^{Lt}\mathrm{\parallel }y-x_j{\mathrm{\parallel }}_2$. Indeed $\rho +M_{j}h\le \delta \mathrm{/}4+\delta \mathrm{/}4<\delta$, so [F1] applies to $a_0:=x_j$, $T:=h$, $R:=\delta$ and the radius $\rho$: the bound on $[0,h]$ is $M_j\le M$, and the stated estimates are clauses 1 and 3 of [F1].

Put ${\rho }_0:=\rho e^{-L{t}_0}$ and, for $0\le j\le N$, ${\rho }_j:=\rho e^{-L(t_0-jh)}$, so that ${\rho }_N=\rho$, ${\rho }_j\le \rho$ for $j<N$, and ${\rho }_{j+1}=e^{Lh}{\rho }_j$. I claim, by induction on $j$, that $\mathrm{\Phi }(jh,\cdot )$ is $C^1$ on $W:=B(a_0,{\rho }_0)$ and $\mathrm{\Phi }(jh,W)\subseteq B(x_j,{\rho }_j)$. For $j=0$ this says $\mathrm{\Phi }(0,a)=a\in B(a_0,{\rho }_0)$ and is [F2]. If it holds for $j$, then for $a\in W$ the point $y:=\mathrm{\Phi }(jh,a)$ lies in $B(x_j,{\rho }_j)\subseteq B(x_j,\rho )$, so the small-step claim of step 2.1 gives that $\mathrm{\Phi }(h,y)$ is defined and $\mathrm{\Phi }((j+1)h,a)=\mathrm{\Phi }(h,\mathrm{\Phi }(jh,a))$; moreover $\mathrm{\parallel }\mathrm{\Phi }(h,y)-x_{j+1}{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\Phi }(h,y)-\mathrm{\Phi }(h,x_j){\mathrm{\parallel }}_2\le e^{Lh}\mathrm{\parallel }y-x_j{\mathrm{\parallel }}_2<e^{Lh}{\rho }_j={\rho }_{j+1}$, so $\mathrm{\Phi }((j+1)h,W)\subseteq B(x_{j+1},{\rho }_{j+1})$, and $\mathrm{\Phi }((j+1)h,\cdot )$ is a composite of $C^1$ maps on the relevant open sets by [F5]. This proves the claim; in particular $\mathrm{\Phi }(t_0,\cdot )$ is $C^1$ on $W$.

The endpoint regularity in [F1] means regularity of a local extension to an open time neighbourhood, as its proof establishes. For $(t_0,a_0)$ choose the last step, writing $t_0=(N-1)h+h$. By step 3.1, $g(a):=\mathrm{\Phi }((N-1)h,a)$ is $C^1$ on $W$ and $g(a_0)=x_{N-1}$. By step 2.1 and [F1], the small-step flow is defined and $C^1$ on an open neighbourhood $V$ of $(h,x_{N-1})$. Continuity of $(t,a)\mapsto (t-(N-1)h,g(a))$ gives a product neighbourhood $J\times W^{\mathrm{\prime }}$ of $(t_0,a_0)$, with $W^{\mathrm{\prime }}\subseteq W$, whose image lies in $V$. On it [F2] gives $\mathrm{\Phi }(t,a)=\mathrm{\Phi }(t-(N-1)h,g(a))$, so the flow is defined and $C^1$ by [F5]. This proves openness and regularity near every positive-time point without requiring any seam assertion.

For a point $(0,a_0)$ choose $R>0$ with $\bar{B}(a_0,R)\subseteq U$, put $M:={\sup }_{\bar{B}(a_0,R)}\mathrm{\parallel }F{\mathrm{\parallel }}_2$, and choose $\rho =R\mathrm{/}2$ and $h>0$ with $Mh<R\mathrm{/}2$. Apply [F1] to both $F$ and $-F$ on $[0,h]\times B(a_0,\rho )$. Time reversal and uniqueness [F2] identify the second flow with $\mathrm{\Phi }(-s,a)$, so $(-h,h)\times B(a_0,\rho )\subseteq \mathrm{\Omega }$. The state derivatives are continuous on both sides and equal $I_n$ at $0$, while the time derivative on the whole interval is $F(\mathrm{\Phi }(t,a))$ by the maximal-solution equation. The two continuous pieces agree at $0$, so [F5] gives $C^1$ regularity on this open product. For $t_0<0$, apply steps 1.1-4.1 to $-F$ at $(-t_0,a_0)$ and reverse time; uniqueness identifies its maximal interval with $-I_a$ (a proper extension would reverse to an extension of the original maximal solution). Thus every point of $\mathrm{\Omega }$ has an open neighbourhood on which $\mathrm{\Phi }$ is defined and $C^1$. On all these neighbourhoods ${\mathrm{\partial }}_t\mathrm{\Phi }(t,a)=F(\mathrm{\Phi }(t,a))$, proving the statement.