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/Part 1: Diffusion Processes/Chapter 2: Integral Manifolds/Shenanigans

2.1 Integrable Bundles

Definition 2.1.1 (Integrable).label Let $X$ be a $C^{p}$ ($p \ge 2$) manifold, $E \subset TX$ be a subbundle, and $x_{0} \in X$, then $E$ is integrable at $x_{0}$ if there exists a submanifold $x_{0} \in Y \subset X$ such that

\[T\iota: TY \to E \subset TX\]

is an isomorphism. If $E$ is integrable at every $x_{0} \in X$, then $E$ is integrable.

Lemma 2.1.2.label Let $E, F$ be Banach spaces, $U \subset E$, $V \subset F$ be open sets, $\eps > 0$, $g \in C^{k}((-\eps, \eps) \times U \times V; F)$ ($k \ge 2$), and $(x_{0}, y_{0}) \in U \times V$, then there exists $\delta > 0$, $U_{0} \in \cn_{E}(x_{0})$, $V_{0} \in \cn_{F}(x_{0})$ and a unique $C^{k}$-mapping

\[\beta: (-\delta, \delta) \times U_{0} \times V_{0} \to V\]

such that

  1. (1)

    For all $(x, y) \in U_{0} \times V_{0}$, $\beta(0, x, y) = y$.

  2. (2)

    For all $(t, x, y) \in (-\delta, \delta) \times U_{0} \times V_{0}$,

    \[\frac{d}{dt}\beta(t, x, y) = g(t, x, \beta(t, x, y))\]

  3. (3)

    For each fixed $y \in V_{0}$, let $\beta(t, x) = \beta(t, x, y)$, then for each $(t, x) \in (-\delta, \delta) \times U_{0}$,

    \[(\partial_{t}\partial_{x}\beta)(t, x) = (\partial_{x}g)(t, x, \beta(t, x)) + (\partial_{y}g)(t, x, \beta(t, x)) \circ (\partial_{x}\beta)(t, x)\]

Proof, [Proposition VI.2.1, Lan12].Let

\[G: (-\eps, \eps) \times U \times V \to E \times F \quad (t, x, y) \mapsto (0, g(t, x, y))\]

then by the existence and uniqueness of ODEs, there exists $\delta > 0$, $U_{0} \in \cn_{E}(x_{0})$, $V_{0} \in \cn_{F}(y_{0})$, and a unique $C^{k}$-mapping

\[B: (-\delta, \delta) \times U_{0} \times V_{0} \to E \times F\]

such that

  1. (a)

    For each $(x, y) \in U_{0} \times V_{0}$, $B(0, x, y) = (x, y)$.

  2. (b)

    For each $(t, x, y) \in (-\delta, \delta) \times U_{0} \times V_{0}$,

    \[\frac{d}{dt}B(t, x, y) = (0, g(t, B(t, x, y)))\]

Let $\beta(t, x, y) = \pi_{2}(t, x, y)$, then

  1. (1)

    For each $(x, y) \in U_{0} \times V_{0}$, $\beta(0, x, y) = \pi_{2}(x, y) = y$.

  2. (2)

    For each $(t, x, y) \in (-\delta, \delta) \times U_{0} \times V_{0}$,

    \[\frac{d}{dt}\beta(t, x, y) = \pi_{2}(g(t, B(t, x, y))) = g(t, x, \beta(t, x, y))\]

  3. (3)

    For each fixed $y \in V_{0}$, let $\beta(t, x) = \beta(t, x, y)$, then for each $(t, x) \in (-\delta, \delta) \times U_{0}$, by the chain rule,

    \[(\partial_{t}\partial_{x}\beta)(t, x) = (\partial_{x}g)(t, x, \beta(t, x)) + (\partial_{y}g)(t, x, \beta(t, x)) \circ (\partial_{x}\beta)(t, x)\]

$\square$

Theorem 2.1.3.label Let $E, F$ be Banach spaces, $U \subset E$ and $V \subset F$ be open subsets, and $f \in C^{k}(U \times V; L(E; F))$ ($k \ge 2$).

For any $\xi, \eta \in C^{k}(U \times V; E)$, denote

\[\Xi(x, y) = (\xi(x, y), f(x, y) \cdot \xi(x, y)) \quad H(x, y) = (\eta(x, y), f(x, y) \cdot \eta(x, y))\]

If for every $\xi, \eta \in C^{k}(U \times V; E)$ and $(x, y) \in U \times V$,

\[[Df(x, y) \cdot \Xi(x, y)] \cdot \eta(x, y) = [Df(x, y) \cdot H(x, y)] \cdot \xi(x, y)\]

then for every $(x_{0}, y_{0}) \in U \times V$, there exists $U_{0} \in \cn_{E}(x_{0})$, $V_{0} \in \cn_{F}(y_{0})$, and a unique $\alpha \in C^{k}(U_{0} \times V_{0}; V)$ such that

  1. (1)

    For every $y \in V_{0}$, $\alpha(x_{0}, y) = y$.

  2. (2)

    For every $(x, y) \in U_{0} \times V_{0}$,

    \[(\partial_{x} \alpha)(x, y) = f(x, \alpha(x, y))\]

Proof. Using translation, assume without loss of generality that $x_{0} = 0$ and $y_{0} = 0$. Let $B \in \cn_{E}(0)$ and

\[g: (-\eps, \eps) \times B \times V \to F \quad (t, z, y) \mapsto f(tz, y) \cdot z\]

then by Lemma 2.1.2, there exists $\delta > 0$, $B_{0} \in \cn_{E}(0)$, and $\beta: (-\delta, \delta) \times B_{0} \times V_{0}$ such that

  1. (a)

    For each $(z, y) \in B_{0} \times V_{0}$, $\beta(0, z, y) = y$.

  2. (b)

    For each $(t, z, y) \in (-\delta, \delta) \times B_{0} \times V_{0}$,

    \[\frac{d}{dt}\beta(t, z, y) = f(tz, \beta(t, z, y)) \cdot z\]

  3. (c)

    For each fixed $y \in V_{0}$, if $\beta(t, z) = \beta(t, z, y)$, then for each $(t, x) \in (-\delta, \delta) \times B_{0}$ and $h \in E$,

    \begin{align*}(\partial_{t}\partial_{z}\beta)(t, z)\cdot h&= t(\partial_{x}f)(tz, \beta(t, z))\cdot h \cdot z + f(tz, \beta(t, z))\cdot h \\&+ (\partial_{y}f)(tz, \beta(t, z)) \circ (\partial_{z} \beta)(t, z)\cdot h \cdot z\end{align*}

Following (c), let $k(t, z) = (\partial_{z}\beta)(t, z)\cdot h - tf(tz, \beta(t, z))\cdot h$, then

\begin{align*}\frac{d}{dt}k(t)&= (\partial_{t}\partial_{z}\beta)(t, z)\cdot h - f(tz, \beta)\cdot h \\&- t [(\partial_{x}f)(tz, \beta)]\cdot z \cdot h - t[(\partial_{y} f)(tz, \beta)]f(tz, \beta) \cdot z\cdot h\end{align*}

where by assumption,

\begin{align*}&[(\partial_{x}f)(tz, \beta)]\cdot z \cdot h - [(\partial_{y} f)(tz, \beta)]f(tz, \beta) \cdot z \cdot h \\&= Df(tz, \beta) \cdot (z, f(tz, \beta) \cdot z) \cdot h \\&= Df(tz, \beta) \cdot (h, f(tz, \beta) \cdot h) \cdot z \\&= [(\partial_{x}f)(tz, \beta)]\cdot h \cdot z - [(\partial_{y} f)(tz, \beta)]f(tz, \beta) \cdot h\cdot z\end{align*}

Therefore by (c),

\begin{align*}\frac{d}{dt}k(t)&= (\partial_{y}f)(tz, \beta) \circ (\partial_{z} \beta)(t, z)\cdot h \cdot z - t[(\partial_{y} f)(tz, \beta)]f(tz, \beta) \cdot h\cdot z \\&= (\partial_{y} f)(tz, \beta)[\partial_{z} \beta(t, z) - tf(tz, \beta)] \cdot h \cdot z \\&= (\partial_{y} f)(tz, \beta) \cdot k(t) \cdot z\end{align*}

Since $0$ is a solution to the above equation, $k(t) = 0$ by the uniqueness of solutions to ODEs. Hence for every $t \in (-\delta, \delta)$ and $z \in B_{0}$,

\[\partial_{z} \beta(t, z) = tf(tz, \beta(t, z))\]

By adjusting $\delta$ and $B_{0}$, assume without loss of generality that $\delta > 1$. In which case, if $\alpha(x) = \beta(1, x, y)$, then

\[\partial_{x}\alpha(x, y) = \partial_{z}\beta(1, x, y) = f(z, \beta(1, x, y)) = f(x, \alpha(x, y))\]

$\square$

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