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,

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

$\square$