Przepływ indukowany różniczkowalnym polem prędkości jest różniczkowalny
Pozwolić $E$ być $\mathbb R$-Przestrzeń Banach, $\tau>0$ i $v:[0,\tau]\times E\to E$ takie że$^1$ $$x\mapsto t\mapsto v(t,x)\tag1$$ należy do $C^{0,\:1}(E,C^0([0,\tau],E))$. To wystarczy, aby zapewnić, że jest wyjątkowy$X^x\in C^0([0,\tau],E)$ z $$X^x(t)=x+\int_0^tv(s,X^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]\tag2$$ dla wszystkich $x\in E$. Teraz załóżmy$$v(t,\;\cdot\;)\in C^1(E,E)\;\;\;\text{for all }t\in[0,\tau]\tag3$$ i ${\rm D}_2v$jest (łącznie) ciągła. Znowu to wystarczy, aby zapewnić, że jest wyjątkowy$Y^x\in C^0([0,\tau],\mathfrak L(E))$ z $$Y^x(t)=\operatorname{id}_E+\int_0^tw_x(s,Y^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau],$$ gdzie$^2$ $$w_x(t,A):={\rm D}_2v(t,X^x(t))A\;\;\;\text{for }(t,A)\in[0,\tau]\times\mathfrak L(E),$$ dla wszystkich $x\in E$.
Chciałbym to pokazać $$E\to C^0([0,\tau],E)\;,\;\;\;x\mapsto X^x$$ jest różniczkowalna Frécheta i pochodna w $x$ jest dany przez $Y^x$ dla wszystkich $x\in E$.
Mogę pokazać to twierdzenie tylko przy założeniu, że $v(t,\;\cdot\;)\in C^2([0,\tau],E)$ i ${\rm D}_2^2v$ jest również (łącznie) ciągła, ponieważ wtedy ma zastosowanie twierdzenie Taylora.
Dla przypadku ogólnego: Niech $x,h\in E$i \ 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} dla$t\in[0,\tau]$. Możemy napisać \ 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} dla wszystkich$s\in[0,\tau]$. Pozwolić$$c_x:=\sup_{t\in[0,\:\tau]}\left\|{\rm D}_2v\left(X^x(t)\right)\right\|_{\mathfrak L(E)}<\infty\tag7$$ i $c_1$ oznaczają stałą Lipschitza dla $v$. Następnie \ 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} dla wszystkich$t\in[0,\tau]$. Teraz problemem jest znalezienie odpowiedniego wiązania$v\left(s,X^{x+h}(s)\right)-v\left(s,X^x(s)\right)-{\rm D}_2v\left(s,X^x(s)\right)Y^x(s)h$. Oczywiście, \ 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} dla wszystkich$t\in[0,\tau]$.
Ogólną wytyczną jest teraz odwołanie się do nierówności Gronwalla. Ale szacunek$(9)$ 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 $v$.
Odpowiedzi
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:
- $T_t$ 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 $$\sup_{t\in[0,\:\tau]}\left\|T_t^{-1}(x)-T_t^{-1}(y)\right\|_E\le e^{c_1}\tau\left\|x-y\right\|_E\tag{12}.$$
- $$[0,\tau]\times E\ni(t,x)\mapsto T_t(x)\tag{13}$$ is (jointly) continuous.
- $$\left\|X^x-X^y\right\|_t^\ast\le e^{c_1t}\left\|x-y\right\|_E\;\;\;\text{for all }t\in[0,\tau]\text{ and }x,y\in E\tag{14}.$$
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 $y\in\overline B_\varepsilon(x)$.