Let $ E $ be a Banach space (for our purposes $ \mathbb R^n $ will suffice). Let $ U\subset E $ be open, and let $ X\colon U\to E $ be a smooth vector field on $ U $ (smooth meaning "of class $ \mathscr C^k $, for your favorite $ k = 1,\dots,\infty $").
For $ x_0\in U $ there exists a neighborhood $ V\subset U $ of $ x_0 $, an $ \alpha > 0 $ and a function $ \Phi\colon \left]-\alpha,\alpha\right[\times V\to U $ such that for each $ x\in V $ the curve $ \Phi({-},x)\colon \left]-\alpha,\alpha\right[\to U $ is the unique integral curve of $ X $ passing through $ x $ at $ t = 0 $. We'll call $ \Phi $ the local flow of $ X $ near $ x_0 $.
I want to show that the local flow is smooth. I'm reading Marsden&Ratiu's proof in Manifolds, Tensor Analysis and Applications, but I'm getting confused in a couple of places.
They start by defining a strange vector field on the set $ L(E,E) $ of all (bounded) linear maps of $ E $ to itself
$$
Y\colon L(E,E)\times \left]-\alpha,\alpha\right[\times V\to L(E,E)\text{,}\qquad Y(\psi,t,x) = \mathrm dX(\Phi_t(x))\circ \psi
$$
and they call the differential equation
$$
\dot\psi(t,x) = Y(\psi(t,x),t,x)\tag{*}\label{eq}
$$
the "linearized" or "first variation" differential equation. They claim that, if $ \psi\colon \left]-\beta,\beta\right[\times V\to L(E,E) $ is the solution to $ \eqref{eq} $ such that
$$
\psi(0,x) = \mathrm{id}_E
$$
then
$$
\psi(t,x) = \mathrm d{\Phi_t}(x)
$$
for all $ t\in \left]-\beta,\beta\right[ $ and $ x\in V $. At this point it's more or less clear where they are heading.
My question is why did they choose that strange vector field in the first place. What's the "intuition" behind this proof, if there is one?
Best Answer
For a fixed $x$, $\Phi(t, x)$ satisfies the ODE $$\Phi_t(t, x) = X(\Phi(t, x)), \hspace{20 pt} \Phi(0, x) = x.$$ From the existence proof, we know that $\Phi$ exists and is $C^1$ in the $t$ variable. Now the goal is to show smoothness of the solution $\Phi$ in the $x$ variable, which isn't very obvious from the ODE. So we first pretend that $D\Phi(t, x)$ exists and differentiate both sides of the ODE with respect to $x$ to get $$\partial_tD\Phi(t, x) = DX(\Phi(t, x))D\Phi(t,x),\hspace{20 pt} D\Phi(0, x) = I.$$ Thus $\psi(t, x) = D\Phi(t, x)$ satisfies the ODE $$\psi_t(t, x) = DX(\Phi(t, x))\psi(t, x), \hspace{20 pt} \psi(0, x) = I.$$ This is exactly the ODE that you wrote. Existence of the solution is a consequence of the existence theorem, so the hard work remains in showing that the solution $\psi$ is actually $D\Phi(t, x)$.