Flow induced by differentiable velocity field is differentiable

Question

Let $E$ be a $\mathbb R$-Banach space, $\tau>0$ and $v:[0,\tau]\times E\to E$ such that$^1$ $$x\mapsto t\mapsto v(t,x)\tag1$$ belongs to $C^{0,\:1}(E,C^0([0,\tau],E))$. This is enough to ensure that there is an unique $X^x\in C^0([0,\tau],E)$ with $$X^x(t)=x+\int_0^tv(s,X^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]\tag2$$ for all $x\in E$. Now assume $$v(t,\;\cdot\;)\in C^1(E,E)\;\;\;\text{for all }t\in[0,\tau]\tag3$$ and ${\rm D}_2v$ is (jointly) continuous. Again, this is enough to ensure that there is an unique $Y^x\in C^0([0,\tau],\mathfrak L(E))$ with $$Y^x(t)=\operatorname{id}_E+\int_0^tw_x(s,Y^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau],$$ where$^2$ $$w_x(t,A):={\rm D}_2v(t,X^x(t))A\;\;\;\text{for }(t,A)\in[0,\tau]\times\mathfrak L(E),$$ for all $x\in E$.

I would like to show that $$E\to C^0([0,\tau],E)\;,\;\;\;x\mapsto X^x$$ is Fréchet differentiable and the derivative at $x$ is given by $Y^x$ for all $x\in E$.

I'm only able to show this claim assuming that $v(t,\;\cdot\;)\in C^2([0,\tau],E)$ and ${\rm D}_2^2v$ is (jointly) continuous as well, since then Taylor's theorem is applicable.

For the general case: Let $x,h\in E$ and \begin{equation}\begin{split}Z(t)&:=X^{x+h}(t)-X^x(t)-Y^x(t)h\\&=\int_0^tv\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\:{\rm d}s\end{split}\tag5\end{equation} for $t\in[0,\tau]$. We may write \begin{equation}\begin{split}&v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\\&\;\;\;\;\;\;\;\;=v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-{\rm D}_2v\left(s,X^x(s)\right)\left(X^{x+h}(s)-X^x(s)\right)\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+{\rm D}_2v\left(s,X^x(s)\right)Z(s)\end{split}\tag6\end{equation} for all $s\in[0,\tau]$. Let $$c_x:=\sup_{t\in[0,\:\tau]}\left\|{\rm D}_2v\left(X^x(t)\right)\right\|_{\mathfrak L(E)}<\infty\tag7$$ and $c_1$ denote the Lipschitz constant of $v$. Then, \begin{equation}\begin{split}\sup_{s\in[0,\:t]}\left\|\left(X^{x+h}-X^x\right)'(s)\right\|_E&=\sup_{s\in[0,\:t]}\left\|v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)\right\|_E\\&\le c_1\sup_{s\in[0,\:t]}\left\|\left(X^{x+h}-X^x\right)(s)\right\|_E\le c_1e^{c_1t}\left\|h\right\|_E\end{split}\tag8\end{equation} for all $t\in[0,\tau]$. Now the problem is to find a suitable bound for $v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h$. Clearly, \begin{equation}\begin{split}&\sup_{s\in[0,\:t]}\left\|v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\right\|_E\\&\;\;\;\;\;\;\;\;\le\max(c,c_1)e^{c_1t}\left\|h\right\|_E+c\sup_{s\in[0,\:t]}\left\|Z(s)\right\|_E\end{split}\tag9\end{equation} for all $t\in[0,\tau]$.

The general guideline is now to invoke Gronwall's inequality. But the estimate $(9)$ is too weak to conclude the Fréchet differentiability from it, since on the right-hand side we would need to have $\left\|h\right\|_E^2$ instead of $\left\|h\right\|_E$ (which is the case, by Taylor's theorem, if we assume the aforementioned twice differentiability).

Can we do something to fix this problem?

---

$^1$ So, $v$ is Lipschitz continuous with respect to the second argument uniformly with respect to the first, has at most linear growth with respect to the second argument uniformly with respect to the first and is (jointly) continuous.

$^2$ For every $x\in E$, $w_x$ has the same Lipschitz and linear growth properties as $v$.

Answer 1

Let $$\left\|f\right\|_t^\ast:=\sup_{s\in[0,\:t]}\left\|f(s)\right\|_E\;\;\;\text{for }f:[0,\tau]\to E\text{ and }t\in[0,\tau],$$ $c_1\ge0$ with $$\left\|v(\;\cdot\;,x)-v(\;\cdot\;,y)\right\|_\tau^\ast\le c_1\left\|x-y\right\|_E\tag{10}$$ and $$T_t(x):=X^x(t)\;\;\;\text{for }(t,x)\in[0,\tau]\times E.$$

We will need the following easy-to-verify results:

1. $T_t$ is bijective for all $t\in[0,\tau]$, $$[0,\tau]\ni t\mapsto T_t^{-1}(x)\tag{11}$$ is continuous for all $x\in E$ and $$\sup_{t\in[0,\:\tau]}\left\|T_t^{-1}(x)-T_t^{-1}(y)\right\|_E\le e^{c_1}\tau\left\|x-y\right\|_E\tag{12}.$$
2. $$[0,\tau]\times E\ni(t,x)\mapsto T_t(x)\tag{13}$$ is (jointly) continuous.
3. $$\left\|X^x-X^y\right\|_t^\ast\le e^{c_1t}\left\|x-y\right\|_E\;\;\;\text{for all }t\in[0,\tau]\text{ and }x,y\in E\tag{14}.$$

Now let $x\in E$. I claim that $$\frac{\left\|X^{x+h}-X^x-Y^xh\right\|_\tau^\ast}{\left\|h\right\|_E}\xrightarrow{h\to0}0\tag{15}.$$

Let $\varepsilon>0$. Since $(13)$ is continuous, $$K:=\left\{\left(t,X^y(t)\right):(t,y)\in[0,\tau]\times\overline B_\varepsilon(x)\right\}$$ is compact. Let $$\omega(\delta):=\sup_{\substack{(t,\:y_1),\:(t,\:y_2)\:\in\:K\\\left\|y_1-y_2\right\|_E\:<\:\delta}}\left\|{\rm D}_2v(t,y_1)-{\rm D}_2v(t,y_2)\right\|_{\mathfrak L(E)}\;\;\;\text{for }\delta>0.$$ Note that $\omega$ is nondecreasing. Since ${\rm D}_2v$ is (jointly) continuous, it is uniformly continuous on $K$ and hence $$\omega(\delta)\xrightarrow{\delta\to0+}0\tag{16}.$$ By the fundamental theorem of calculus, $$v(t,y_2)-v(t,y_1)=\int_0^1{\rm D}_2v\left(t,y_1+r(y_2-y_1)\right)(y_2-y_1)\:{\rm d}r\tag{17}$$ for all $t\in[0,\tau]$ and $y_1,y_2\in E$ and hence \begin{equation}\begin{split}&\left\|v(t,y_2)-v(t,y_1)-{\rm D}_2v(t,y_1)(y_2-y_1)\right\|_E\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\left\|y_1-y_2\right\|_E\int_0^1\left\|{\rm D}_2v(t,y_1+r(y_2-y_1))-{\rm D}_2v(t,y_1)\right\|_{\mathfrak L(E)}{\rm d}r\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\left\|y_1-y_2\right\|_E\omega\left(\left\|y_1-y_2\right\|_E\right)\end{split}\tag{18}\end{equation} for all $t\in[0,\tau]$ and $y_1,y_2\in E$ with $$(t,y_1+r(y_2-y_1))\in K\;\;\;\text{for all }r\in[0,1)\tag{19}.$$ Now let $h\in B_\varepsilon(x)\setminus\{0\}$ and \begin{equation}\begin{split}Z(t)&:=X^{x+h}(t)-X^x(t)-Y^x(t)h\\&=\int_0^tv\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\:{\rm d}s\end{split}\tag{20}\end{equation} for $t\in[0,\tau]$. Observe that$^1$ $$\left(t,X^x(t)+r\left(X^{x+h}(t)-X^x(t)\right)\right)\in K\;\;\;\text{for all }t\in[0,\tau]\text{ and }r\in[0,1)\tag{21}$$ and hence \begin{equation}\begin{split}&\left\|v\left(t,X^{x+h}(t)\right)-v\left(t,X^x(t)\right)-{\rm D}_2v\left(t,X^x(t)\right)\left(X^{x+h}(t)-X^x(t)\right)\right\|_E\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\left\|X^{x+h}(t)-X^x(t)\right\|_E\omega\left(\left\|X^{x+h}(t)-X^x(t)\right\|_E\right)\\&\;\;\;\;\;\;\;\;\;\;\;\;\le e^{c_1t}\left\|h\right\|_E\omega\left(e^{c_1t}\left\|h\right\|_E\right)\end{split}\tag{24}\end{equation} by $(18)$ and $(14)$ for all $t\in[0,\tau]$. Let $$a:=e^{c_1\tau}\omega\left(e^{c_1\tau}\left\|h\right\|_E\right).$$ By $(6)$ and $(24)$, \begin{equation}\begin{split}&\left\|v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\right\|_E\\&\;\;\;\;\;\;\;\;\;\;\;\;e^{c_1s}\left\|h\right\|_E\omega\left(e^{c_1s}\left\|h\right\|_E\right)+c_x\left\|Z\right\|_s^\ast\le a\left\|h\right\|_E+c_x+\left\|Z\right\|_s^\ast\end{split}\tag{25}\end{equation} for all $s\in[0,\tau]$ and hence \begin{equation}\begin{split}\left\|Z\right\|_t^\ast&\le\int_0t^t\left\|v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h\right\|_E{\rm d}s\\&\le a\left\|h\right\|_Et+c_x\int_0^t\left\|Z\right\|_s^\ast{\rm d}s\end{split}\tag{26}\end{equation} for all $t\in[0,\tau]$. Thus, by Gronwall's inequality, $$\left\|Z\right\|_t^\ast\le a\left\|h\right\|_Ete^{c_xt}\;\;\;\text{for all }t\in[0,\tau]\tag{27}$$ and hence $$\frac{\left\|Z\right\|_\tau^\ast}{\left\|h\right\|_E}\le a\tau e^{c_x\tau}\xrightarrow{h\to0}0\tag{28}.$$

This finishes the proof and we have shown that the map $$E\to C^0([0,\tau],E)\;,\;\;\;x\mapsto X^x$$ is Fréchet differentiable at $x$ with derivative equal to $Y^x$ for all $x\in E$.

---

$^1$ Let $t\in[0,\tau]$, $r\in[0,1)$, $$z:=(1-r)X^x(t)+rX^{x+h}(t)$$ and $$y:=T_t^{-1}(z).$$ By construction $$X^y(t)=z\tag{22}$$ and hence $$(t,z)\in K\Leftrightarrow y\in\overline B_\varepsilon(x).$$ By $(12)$ and $(14)$, $$\left\|x-y\right\|_E=\left\|T_t^{-1}(T_t(x))-T_t^{-1}(z)\right\|_E\le e^{c_1t}\left\|T_t(x)-z\right\|_E\le e^{2c_1t}\left\|h\right\|_E\tag{23}.$$ Since $\left\|h\right\|_E<\varepsilon$ and $e^{2c_1t}\le 1$, we obtain $\left\|x-y\right\|_E<\varepsilon$ and hence $y\in\overline B_\varepsilon(x)$.