Show that the solution of $\frac{\rm d}{{\rm d}t}X^x(t)=v(t,X^x(t))$, $X^x(0)=x$, is differentiable in $x$

Question

Let $X^x$ be the solution of$^1$ \begin{align}\frac{\rm d}{{\rm d}t}X^x(t)&=v(t,X^x(t))\\ X^x(0)&=x\end{align} and $$T_t(x):=X^x(t).$$

Assuming that $v$ is differentiable in the second argument, can we show that $T_t$ is differentiable?

I'm only able to prove this when $v$ is twice differentiable in the second argument, since then Taylor's theorem is applicable.

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$^1$ Assume $v:[0,T]\times\mathbb R^d\to\mathbb R^d$ is Lipschitz continuous in the second argument uniformly with respect to $t$ and continuous in the first variable.

Answer 1

Define an augmentation of the ODE system via $$ \frac{d}{dt}U^x(t)=\frac{\partial v}{\partial x}(t,X^x(t))\,U^x(t), ~~~ U^x(0)=I. $$ Then use Grönwall or similar to find a bound for $E^{x,Δx}(t)=X^{x+\Delta x}(t)-X^x(t)-U^x(t)\Delta x$.

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Assume that the following considerations are restricted to a compact domain so that $X^x(s)$, $X^{x+Δx}(s)$, and the connecting segment are inside the domain for all $s\in[0,t]$.

Obviously, $E^x(0)=0$ by construction. Then \begin{align} E^{x,Δx}(t)&=\int_0^t\frac{d}{dt}E^{x,Δx}(s)\,ds \\ &=\int_0^t\left(v(s,X^{x+Δx}(s))-v(s,X^{x}(s))-∂_xv(s,X^{x}(s))U^x(s)Δx\right)\,ds \\ &=\int_0^t\left[v(s,X^{x+Δx}(s))-v(s,X^{x}(s))-∂_xv(s,X^{x}(s))\left(X^{x+Δx}(s)-X^{x}(s)\right)\right]\,ds \\&\qquad +\int_0^t∂_xv(s,X^{x}(s))\left[X^{x+Δx}(s)-X^{x}(s)-U^x(s)Δx\right]\,ds \\ &=\int_0^t\rho_v\left(X^{x}(s),X^{x+Δx}(s)-X^{x}(s)\right)\,ds + \int_0^t∂_xv(s,X^{x}(s))E^{x,Δx}(s)\,ds \end{align} Now one could argue that the first integrand is uniformly $o(Δx)$ by continuity of $X^x$ in $x$ and the definition of differentiability of $v$, while $∂_xv$ in the second integrand is bounded by the Lipschitz constant $L$ of $v$. This allows to find $δ>0$ for some given $ε>0$ so that for all $\|Δx\|<δ$ one gets the integral inequality with its solution per Grönwall $$ \|E^{x,Δx}(t)\|\le ε\|Δx\|t+L\int_0^t\|E^{x,Δx}(s)\|\,ds \implies \|E^{x,Δx}(t)\|\le ε\|Δx\|\frac{e^{Lt}-1}{L} $$ from where $E^{x,Δx}(t)=o(Δx)\implies\dfrac{∂X^x(t)}{∂x}=U^x(t)$ follows.

Answer 2

Let me add the assumption that there is some uniformity in the differentiabilty of $v(t,x)$ in that if we write $$ f(t,x,\Delta x) = v(t,x+\Delta x) - v(t,x) - \frac{\partial v}{\partial x}(t,x), $$ then given $\epsilon>0$, there exists $\delta>0$ such that $$ | f(t,x,\Delta x)| \le \epsilon |\Delta x| \quad \text{ for $|\Delta x| < \delta$} $$ (that is to say, the choice of $\delta$ is uniform).

Since $v$ is Lipschitz in the second variable, the partial derivatives of $v$ are uniformly bounded: $$ \left|\frac{\partial v}{\partial x}\right| \le M $$

Also, by an application of Grönwall's inequality, we can see that $T_t(x)$ is Lipschitz with constant at worst $e^{tM}$.

It seems to me that we only need to show $T_t(x)$ is differentiable at $x = 0$. And then we can assume without loss of generality that $v(t,0) = 0$.

Consider the ODE $$ \frac{d}{dt} Y^x = \frac{\partial v}{\partial x} v(t,0) Y^x , \quad Y^x(0) = x ,$$ and write $U_t(x)$ for $Y^x(t)$. This ODE is linear, so $U_t(x)$ is clearly differentiable with respect to $x$, with derivative $U_t$.

Now look at $X^x(t) - Y^x(t)$: \begin{align} \frac d{dt} (X^x(t) - Y^x(t)) &= v(t,X^x) - \frac{\partial v}{\partial x} v(t,0) X^x + \frac{\partial v}{\partial x} v(t,0) (Y^x(t) - X^x(t)) \\&= f(t,0,X^x) + \frac{\partial v}{\partial x} v(t,0) (Y^x(t) - X^x(t)) . \end{align} So as long as $X^x(t)$ and $Y^x(t)$ are less than $\delta$, which is the case if $|x| \le e^{-tM} \delta$, then \begin{align} \left|\frac d{dt} (X^x(t) - Y^x(t)) \right| &\le \epsilon |X^x(t)| + M|X^x(t) - Y^x(t)| \\ &\le \epsilon e^{tM}|x| + M|X^x(t) - Y^x(t)| \end{align} So by using a version of Grönwall's inequality that looks like the variations of parameters formula, we see that if $|x|$ is sufficiently small, then $$ T_t(x) - U_t(x) = o(|x|) .$$ Thus $$ \nabla T_t(0) = U_t .$$

Answer 3

I've asked for a stronger result (but with the additional assumption that ${\rm D}_2v$ is (jointly) continuous) on mathoverflow. I was able to provide an answer to that question. For convenience, I post this answer here as well, but please note that the equation numbers are taken from the other question:

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 $x\in E$.

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$^1$ 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 $y\in\overline B_\varepsilon(x)$.