$C^1$ dependence of solutions on initial data

Write $x_{\ast }(t):=\mathrm{\Phi }(t,x_0)$. Its graph $G:=\{(t,x_{\ast }(t)):t\in I\}$ is compact and lies in the open ODE domain $D$. Hence it has a positive interior margin: there is $r>0$ such that the compact tube $K:=\{(t,x):t\in I,\mathrm{\parallel }x-x_{\ast }(t){\mathrm{\parallel }}_2\le r\}$ lies in $D$. Indeed, a finite cover of $G$ by open balls whose doubled balls lie in $D$ gives a positive uniform distance from $G$ to the complement of $D$; choose $r$ smaller than that distance. The tube is closed and bounded because $x_{\ast }$ is continuous. On $K$, the continuous state Jacobian is bounded by $M:={\sup }_K\mathrm{\parallel }{D}_{x}F\mathrm{\parallel }<\mathrm{\infty }$ and uniformly continuous. Each time slice of $K$ is a convex state ball, so integrating $D_{x}F$ on its state segments gives the state-Lipschitz bound $M$ on that slice.

Shrink the given initial-state neighbourhood to an open ball $U_0$ about $x_0$ with $e^{Mh}\mathrm{\parallel }y-x_0{\mathrm{\parallel }}_2<r\mathrm{/}2$ for every $y\in U_0$. Every given solution through $y\in U_0$ stays within distance $r\mathrm{/}2$ of $x_{\ast }$ on all of $I$: up to a hypothetical first time when its distance from $x_{\ast }$ reaches $r\mathrm{/}2$, both paths and their connecting state segment lie in $K$. Subtracting their Volterra equations and applying the norm inequality and Gronwall gives $\mathrm{\parallel }\mathrm{\Phi }(t,y)-x_{\ast }(t){\mathrm{\parallel }}_2\le e^{M\mathrm{\mid }t-t_0\mathrm{\mid }}\mathrm{\parallel }y-x_0{\mathrm{\parallel }}_2<r\mathrm{/}2$, contradicting that first time. The reflected estimate gives the identical contradiction backward from $t_0$. Thus any two such paths and every same-time state segment joining them lie in $K$. Subtraction and the same estimate now give ${\sup }_{t\in I}\mathrm{\parallel }\mathrm{\Phi }(t,y^{\mathrm{\prime }})-\mathrm{\Phi }(t,y){\mathrm{\parallel }}_2\le e^{Mh}\mathrm{\parallel }{y}^{\mathrm{\prime }}-y{\mathrm{\parallel }}_2$ for $y,y^{\mathrm{\prime }}\in U_0$. This proves uniform-in-time continuous dependence on the entire given interval, without assuming that a single product cylinder covers the base graph.

Fix $y\in U_0$ and set $C_y(t):=D_{x}F(t,\mathrm{\Phi }(t,y))$. This is continuous on $I$ with $\mathrm{\parallel }{C}_y(t)\mathrm{\parallel }\le M$. By [L2] the variational equation $A_y^{\mathrm{\prime }}(t)=C_y(t)A_y(t)$, $A_y(t_0)=I_n$, has a unique solution on $I$. For each unit vector $v$, apply the vector norm inequality and Gronwall to the integral equation of $A_y(t)v$, whose initial norm is $1$; taking the supremum over $v$ gives ${\sup }_{t\in I}\mathrm{\parallel }{A}_y(t)\mathrm{\parallel }\le e^{Mh}$ in the Euclidean operator norm.

Let $u\in {\mathbb{R}}^n$ with $y+u\in U_0$ and put $z_u(t):=\mathrm{\Phi }(t,y+u)-\mathrm{\Phi }(t,y)$. By [F2], $z_u(t)=u+{\int }_{t_0}^t{B}_u(s)z_u(s)ds$, where $B_u(s):={\int }_0^1{D}_{x}F(s,\mathrm{\Phi }(s,y)+\theta z_u(s))d\theta$. This state-variable fundamental-theorem-of-calculus formula is legitimate because step 2.1 puts every displayed secant point in $K\subseteq D$. In particular $\mathrm{\parallel }{B}_u(s)\mathrm{\parallel }\le M$. Also step 2.1 gives ${\sup }_I\mathrm{\parallel }{z}_u{\mathrm{\parallel }}_2\le e^{Mh}\mathrm{\parallel }u{\mathrm{\parallel }}_2\to 0$, so uniform continuity of $D_{x}F$ on $K$ gives ${\epsilon }_y(u):={\sup }_{s\in I}\mathrm{\parallel }{B}_u(s)-C_y(s)\mathrm{\parallel }\to 0$ as $u\to 0$.

Put ${\rho }_u(t):=z_u(t)-A_y(t)u$. Subtracting the integral equations gives ${\rho }_u(t)={\int }_{t_0}^t{B}_u(s){\rho }_u(s)ds+{\int }_{t_0}^t(B_u(s)-C_y(s))A_y(s)uds$. On either oriented half of $I$, the norm inequality and Gronwall, with forcing bounded by $h{e}^{Mh}{\epsilon }_y(u)\mathrm{\parallel }u{\mathrm{\parallel }}_2$, yield ${\sup }_{t\in I}\mathrm{\parallel }{\rho }_u(t){\mathrm{\parallel }}_2\le h{e}^{2Mh}{\epsilon }_y(u)\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Dividing by $\mathrm{\parallel }u{\mathrm{\parallel }}_2$ for $u\mathrm{\ne }0$ and using step 3.2 shows that the remainder tends to zero uniformly in $t$. Hence $D_y\mathrm{\Phi }(t,y)=A_y(t)$ for every $t\in I$.

For $y,y^{\mathrm{\prime }}\in U_0$, set $E(t):=A_{y^{\mathrm{\prime }}}(t)-A_y(t)$. Subtracting the variational integral equations gives $E(t)={\int }_{t_0}^t{C}_{y^{\mathrm{\prime }}}(s)E(s)ds+{\int }_{t_0}^t(C_{y^{\mathrm{\prime }}}(s)-C_y(s))A_y(s)ds$. Steps 1.1 and 2.1 imply $\delta (y^{\mathrm{\prime }},y):={\sup }_{s\in I}\mathrm{\parallel }{C}_{y^{\mathrm{\prime }}}(s)-C_y(s)\mathrm{\parallel }\to 0$ as $y^{\mathrm{\prime }}\to y$. The same two-sided Gronwall estimate gives ${\sup }_{t\in I}\mathrm{\parallel }E(t)\mathrm{\parallel }\le h{e}^{2Mh}\delta (y^{\mathrm{\prime }},y)\to 0$. Thus $y\mapsto A_y$ is continuous in the uniform norm; continuity of each $A_y$ in time also makes $(t,y)\mapsto A_y(t)$ jointly continuous.

Steps 4.1 and 4.2 prove that $\mathrm{\Phi }$ is $C^1$ in the initial-state variable and that its derivative is the unique solution of the variational equation along each trajectory.