Geometric interpretation of this condition

Question

In a Differential Equations context we have the following theorem:

$\textbf{Theorem:}$ Let $D\subseteq\mathbb{R}^n$ be some open set and $f:D\to\mathbb{R}^n$ a $\mathscr{C}^1$ vector field. Then, for every $x_0\in D$ there are real numbers $\alpha,\beta>0$ and a $\mathscr{C}^1$ map $\varphi$ from $$I_\alpha\times B_\beta=\{(t,x)\mid \vert t\vert<\alpha\text{ and }\vert x-x_0\vert<\beta\}$$ to $D$ such that, for every $(t,x)\in I_\alpha\times B_\beta$ we have $$D_t\varphi(t,x)=f(\varphi(t,x))\quad\text{ and }\quad\varphi(0,x)=x$$ $$D_tD_x\varphi(t,x)=Df(\varphi(t,x))(D_x\varphi(t,x))\quad\text{ and }\quad D_x\varphi(0,x)=Id_{B_\beta}$$

$\textbf{Interpretation:}$ For $n=2$ I think of this theorem as follows: you have a plane $\mathbb{R}^2$ and a vector field on the plane. For $x_0$ in the plane we can take a cylinder "centered" at $x_0$ with base $B_\beta$ and height $2\alpha$. The cylinder is like a tube full of vertical spaghetti (one for every $x\in B_\beta$) and each spaghetti gets mapped by $\varphi$ to an integral curve of the field $f$ that at time $t=0$ passes through $x$. This is the same as saying $\varphi_x(t)=\varphi(t,x)$ is a solution of the differential equation $$\mathbf{u}'(t)=f(\mathbf{u}(t))\quad\quad \mathbf{u}(0)=x$$ So far so good. The second condition (which involves the $D_x$ and $Df$) must be saying something about how the solution $\varphi_x$ changes with $x$ but I don't know what exactly. I've tried interpreting it in many ways but don't find anything close to satisfactory.

Also, as I see it if $f(x_0)=0$ then the integral curves close to $x_0$ should be closed/periodic orbits and that doesn't seem to work nicely with my visualization

The closest thing I found is the rectification theorem but I can't connect both statements

Thanks!

Answer 1

The identity $$D_tD_x\varphi(t,x)=Df(\varphi(t,x))(D_x\varphi(t,x))$$ says that the linearization $D_x\varphi(t,x)$ of a solution along the initial conditions is a solution of the linearization $$y'=Df(\varphi(t,x))y$$ of the original equation along that solution. The latter is usually called a linear variational equation.

This means that the dependence of the solutions on the initial conditions can essentially be determined by studying the solutions of linear variational equations.

Let's take your example $f(\bar x_0)=0$.

In this case you get the equation $$y'=Ay\quad\text{with} \quad A=Df(\bar x_0).$$ This means that in order to understand how the solutions depend on the initial condition $x_0$ in a neighborhood of the original point $\bar x_0$ or, if you prefer, in a neighborhood of a critical point (but you need not have $f(x_0)=0$ for $x_0$ in the neighborhood), basically what matters are simply the eigenvalues of $A$ since the solutions of linear variational equation are obtained from $e^{At}$.

Something similar happens when the trajectory of $\bar x_0$ is periodic, although you need to study the much more complicated equation $y'=A(t)y$ with $A(t)$ periodic.