Initial-data sensitivity of an ODE flow on a common compact tube

Compactness of $K$ and continuity of $D_{y}f$ give a finite constant $L$ bounding $\mathrm{\parallel }{D}_{y}f{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ and a modulus $\omega (\eta )\to 0$ such that $\mathrm{\parallel }{D}_{y}f(z)-D_{y}f(w){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \omega (\mathrm{\parallel }z-w{\mathrm{\parallel }}_2)$ for $z,w\in K$; this uses [F1] and [F4]. For fixed $s\in U$ and sufficiently small $h$ with $s+h\in U$, set ${\mathrm{\Delta }}_h(t)=\varphi (t;s+h)-\varphi (t;s)$. The two Volterra equations, convexity of $K$, and [F7] applied to the line segment between the two state values give $\mathrm{\parallel }f(t,\varphi (t;s+h))-f(t,\varphi (t;s)){\mathrm{\parallel }}_2\le L\mathrm{\parallel }{\mathrm{\Delta }}_h(t){\mathrm{\parallel }}_2$. Applying [F3] to the norm inequality from the integral equations yields ${\sup }_t\mathrm{\parallel }{\mathrm{\Delta }}_h(t){\mathrm{\parallel }}_2\le e^{L(b-a)}\mathrm{\parallel }h{\mathrm{\parallel }}_2$.

For each fixed $s$, the matrix equation $S^{\mathrm{\prime }}=A(t)S$, $S(a)=I_d$, with $A(t)=D_{y}f(t,\varphi (t;s))$ has a unique solution on $[a,b]$. View matrices as vectors in ${\mathbb{R}}^{d^2}$ with the Frobenius norm and extend $A$ continuously by its endpoint values to an open interval containing $[a,b]$. The extended linear vector field is locally Lipschitz in the matrix state, so the linear IVP has a unique maximal solution by [F8] and [F9]. Gronwall bounds $\mathrm{\parallel }S(t){\mathrm{\parallel }}_F$ on every finite subinterval. If its maximal interval ended at a finite time at or before $b$, then $S^{\mathrm{\prime }}=AS$ would be bounded there, making $S(t)$ Cauchy with a finite limit; local existence from that limit extends the solution, a contradiction. Its integral equation is unique by [F3].

Uniform continuity of $D_{y}f$ on $K$ implies, with $z=(t,\varphi (t;s))$ and $v={\mathrm{\Delta }}_h(t)$, that $\mathrm{\parallel }f(t,\varphi (t;s)+v)-f(t,\varphi (t;s))-D_{y}f(z)v{\mathrm{\parallel }}_2\le \omega (\mathrm{\parallel }v{\mathrm{\parallel }}_2)\mathrm{\parallel }v{\mathrm{\parallel }}_2$. The line segment remains in $K$ by convexity. For $g(\theta )=f(t,\varphi (t;s)+\theta v)-f(t,\varphi (t;s))-\theta D_{y}f(z)v$, the chain rule [F6] gives $g\mathrm{\prime }(\theta )=[D_{y}f(z+\theta (0,v))-D_{y}f(z)]v$, whose norm is at most $\omega (\mathrm{\parallel }v{\mathrm{\parallel }}_2)\mathrm{\parallel }v{\mathrm{\parallel }}_2$; applying [F7] on $[0,1]$ gives the claimed remainder bound.

Put $E_h(t)={\mathrm{\Delta }}_h(t)-S(t;s)h$. Subtract the integral equations for $\varphi (t;s+h)$, $\varphi (t;s)$, and $S(t;s)h$. The result is $E_h(t)={\int }_a^{t}A(r)E_h(r)dr+{\int }_a^t{q}_h(r)dr$, where $\mathrm{\parallel }{q}_h(r){\mathrm{\parallel }}_2\le \omega (e^{L(b-a)}\mathrm{\parallel }h{\mathrm{\parallel }}_2)e^{L(b-a)}\mathrm{\parallel }h{\mathrm{\parallel }}_2$ by steps 1.1 and 2.1. Applying [F3] gives ${\sup }_t\mathrm{\parallel }{E}_h(t){\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }h{\mathrm{\parallel }}_2\le (b-a)e^{2L(b-a)}\omega (e^{L(b-a)}\mathrm{\parallel }h{\mathrm{\parallel }}_2)\to 0$. Thus $D_s\varphi (t;s)=S(t;s)$ in the total-derivative definition [F1].

If $s_n\to s$ in $U$, step 1.1 gives uniform convergence $\varphi (\cdot ;s_n)\to \varphi (\cdot ;s)$. Uniform continuity of $D_{y}f$ on $K$ then gives uniform convergence $A(\cdot ;s_n)\to A(\cdot ;s)$. Subtract the two integral equations for $S(\cdot ;s_n)$ and $S(\cdot ;s)$; [F3], together with the uniform bound on $S$ obtained from its integral equation, implies ${\sup }_t\mathrm{\parallel }S(t;s_n)-S(t;s){\mathrm{\parallel }}_F\to 0$; equivalence of matrix norms gives the same convergence in operator norm. The derivative is continuous, so the flow is $C^1$ on $U$.