Parameter-dependent ODE flows have the variational tangent on a common tube

If $T=0$, then $C([0,0];{\mathbb{R}}^d)$ is identified with ${\mathbb{R}}^d$ by evaluation at $0$, so $\theta \mapsto u_{\theta }$ is the given $C^1$ map $u_0$ and the initial condition determines its parameter tangent at the only time; this branch asserts no time derivative or IVP on the singleton. If $d=0$, then ${\mathbb{R}}^d=\{0\}$, so the state curve and tangent are the unique zero maps. If $m=0$, the parameter domain has at most one point; its map into $C([0,T];{\mathbb{R}}^d)$ has the unique derivative from ${\mathbb{R}}^0$, and the sensitivity matrix has no columns. In each case the asserted conclusion follows; hence assume $T>0$ and $d,m\ge 1$. Let ${\delta }_h(t)=u_{{\theta }_0+h}(t)-u_{{\theta }_0}(t)$. Differentiability of $u_0$ gives $\mathrm{\parallel }{u}_0({\theta }_0+h)-u_0({\theta }_0){\mathrm{\parallel }}_2\le C_0\mathrm{\parallel }h{\mathrm{\parallel }}_2$ for small $h$. Subtract the Volterra equations [F1] and apply [F2] on the segment between $(u_{{\theta }_0+h}(t),{\theta }_0+h)$ and $(u_{{\theta }_0}(t),{\theta }_0)$. Boundedness of $Df$ on $O$ gives $\mathrm{\parallel }{\delta }_h(t){\mathrm{\parallel }}_2\le C_0\mathrm{\parallel }h{\mathrm{\parallel }}_2+L{\int }_0^t(\mathrm{\parallel }{\delta }_h(s){\mathrm{\parallel }}_2+\mathrm{\parallel }h{\mathrm{\parallel }}_2)ds\mathrm{.}$ Since $t\le T$, [F3] yields ${\sup }_t\mathrm{\parallel }{\delta }_h(t){\mathrm{\parallel }}_2\le (C_0+LT)e^{LT}\mathrm{\parallel }h{\mathrm{\parallel }}_2$.

Put $A(t)=D_{u}f(t,u_{{\theta }_0}(t),{\theta }_0)$ and let $B_j(t)$ be column $j$ of $D_{\theta }f(t,u_{{\theta }_0}(t),{\theta }_0)$. For each parameter coordinate $j$, the augmented vector $Z_j=(S_{\cdot j},1)$ solves the homogeneous linear system $Z_j^{\mathrm{\prime }}(t)=\left(\begin{array}{cc}A(t)& B_j(t)\\ 0& 0\end{array}\right)Z_j(t),Z_j(0)=\left(\begin{array}{c}D{u}_0({\theta }_0{)}_{\cdot j}\\ 1\end{array}\right)\mathrm{.}$ Embed $Z_j(0)$ as one column of a square $(d+1)\times (d+1)$ initial matrix, with all other columns zero. The unique linear matrix solution supplied by [F5] has that vector solution as the selected column. Thus the tangent equation has a unique solution for each of the finitely many columns.

Put $A(t)=D_{u}f(t,u_{{\theta }_0}(t),{\theta }_0)$ and $B(t)=D_{\theta }f(t,u_{{\theta }_0}(t),{\theta }_0)$, and define ${\rho }_h(t)=f(t,u_{{\theta }_0}(t)+{\delta }_h(t),{\theta }_0+h)-f(t,u_{{\theta }_0}(t),{\theta }_0)-A(t){\delta }_h(t)-B(t)h$. Set $z_0(t)=(t,u_{{\theta }_0}(t),{\theta }_0)$ and $v_h(t)=(0,{\delta }_h(t),h)$, and let $\omega (r)=\sup \{\mathrm{\parallel }Df(z)-Df(w)\mathrm{\parallel }:z,w\in \bar{O},\mathrm{\parallel }z-w{\mathrm{\parallel }}_2\le r\}$. Uniform continuity of $Df$ on $\bar{O}$ gives $\omega (r)\to 0$ as $r\downarrow 0$. If $v_h(t)\mathrm{\ne }0$, the set $O\cap B(z_0(t),2\mathrm{\parallel }{v}_h(t){\mathrm{\parallel }}_2)$ is convex and open and contains both endpoints of the secant. For $R_t(z)=f(z)-Df(z_0(t))(z-z_0(t))$, its derivative on this set has norm at most $\omega (2\mathrm{\parallel }{v}_h(t){\mathrm{\parallel }}_2)$. Applying [F2] there gives $\mathrm{\parallel }{\rho }_h(t){\mathrm{\parallel }}_2\le \omega (2\mathrm{\parallel }{v}_h(t){\mathrm{\parallel }}_2)\mathrm{\parallel }{v}_h(t){\mathrm{\parallel }}_2$; if $v_h(t)=0$, then ${\rho }_h(t)=0$. Step 1.1 gives ${\sup }_t\mathrm{\parallel }{v}_h(t){\mathrm{\parallel }}_2=O(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$, so ${\sup }_t\mathrm{\parallel }{\rho }_h(t){\mathrm{\parallel }}_2=o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$. Also $u_0({\theta }_0+h)-u_0({\theta }_0)=D{u}_0({\theta }_0)h+o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$.

Let $E_h(t)={\delta }_h(t)-S(t)h$, and let $r_0(h)=u_0({\theta }_0+h)-u_0({\theta }_0)-D{u}_0({\theta }_0)h$. The Volterra equations and the field remainder ${\rho }_h$ from step 2.1 give $E_h(t)=r_0(h)+{\int }_0^t(A(s)E_h(s)+{\rho }_h(s))ds$, where $A(s)=D_{u}f(s,u_{{\theta }_0}(s),{\theta }_0)$. By [F4], $\mathrm{\parallel }{E}_h(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }{r}_0(h){\mathrm{\parallel }}_2+T{\sup }_s\mathrm{\parallel }{\rho }_h(s){\mathrm{\parallel }}_2+L{\int }_0^t\mathrm{\parallel }{E}_h(s){\mathrm{\parallel }}_{2}ds$. Both first terms are $o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$ by step 2.1 and differentiability of $u_0$. Applying [F3] yields ${\sup }_t\mathrm{\parallel }{E}_h(t){\mathrm{\parallel }}_2=o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$. Thus the Fréchet derivative at ${\theta }_0$ exists and equals $S$.

For any sequence ${\theta }_j\to {\theta }_0$ in $V_{\theta }$, step 1.1 gives $u_{{\theta }_j}\to u_{{\theta }_0}$ uniformly, and uniform continuity of $Df$ on $\bar{O}$ makes the tangent coefficients $A_j,B_j$ converge uniformly to $A,B$. Subtracting the tangent integral equations gives $S_j(t)-S(t)=D{u}_0({\theta }_j)-D{u}_0({\theta }_0)+{\int }_0^t(A_j(s)(S_j(s)-S(s))+(A_j(s)-A(s))S(s)+B_j(s)-B(s))ds$. The initial and coefficient differences tend to zero, while $A_j$ are uniformly bounded; [F3] then gives $\mathrm{\parallel }{S}_j-S{\mathrm{\parallel }}_{C([0,T])}\to 0$. Since ${\theta }_0$ is arbitrary in $V_{\theta }$, the derivative is continuous throughout that set, so the flow map is $C^1$.