Understanding the definition of linearization stability

Question

I've spent almost two hours understanding this definition. You can notice that there is also an example right after the definition, but still, I don't understand it. Could you help me understand this definition? Most importantly, could you explain what does the following phrase mean?

“infinitesimal variations which preserve F(·)” may be “integrated” to give honest curves preserving F(·) mean.

I don't see the motivation for defining such a thing. I would highly appreciate if you could briefly explain the concept with one or two examples. Thanks so much.

Answer 1

You mean

Definition 2.1 (Fischer–Marsden [FM75a, FM75b]). For $X, Y$ Banach manifolds and $F : X \to Y$ a differentiable mapping, then we say that $F$ is linearization stable at $x_0 \in X$ if for every $h \in T_{x_0}X$ so that $DF|_{x_0}\cdot h = 0$, there exists a continuously differentiable curve $x(t) \in X$ with $x(0) = x_0, F(x(t)) = F(x_0)$ and $x′(0) = h$.

I do not know what is confusing you here, so I am going to cover pretty much everything (though not in detail), which means I will undoubtedly explain parts that you understand fine. Please excuse this. Consider it fodder for less educated readers who may also come across this thread.

For $X, Y$ Banach manifolds

This just means $X, Y$ are spaces where differentiation can be defined. For understanding the definition you can assume that $X$ and $Y$ are open sets in $\Bbb R^n$ and $\Bbb R^m$ for some $n,m$. Banach manifolds can be infinite dimensional, but for the purposes here, that doesn't change anything.

and $F : X \to Y$ a differentiable mapping, then we say that $F$ is linearization stable at $x_0 \in X$ if for every $h \in T_{x_0}X$

That is, $h$ is a tangent vector at $x_0$, a vector indicating a speed and direction of travel that a curve passing through $x_0$ might possess.

so that $DF|_{x_0}\cdot h = 0$,

That is, the directional derivative of $F$ at $x_0$ in the direction $h$ is $0$. $F$ is flat in that direction at $x_0$.

there exists a continuously differentiable curve $x(t) \in X$ with $x(0) = x_0, F(x(t)) = F(x_0)$ and $x′(0) = h$.

That is, there is some curve in $X$ passing through $x_0$ in the direction of $h$ such that $F$ is constant along that curve.

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Let me give two even simpler examples than the one in the paper: let $$\begin{align}&F_1 : \Bbb R \to \Bbb R : t \mapsto t^2\\&F_2 : \Bbb R \to \Bbb R : t \mapsto \begin{cases}(t+1)^2& t<-1\\0& -1 \le t \le 1\\(t-1)^2&1 < t\end{cases}\end{align}$$

Let $h$ be the positive unit direction vector at $0$ on the real line, then $DF|_0\cdot h = F'(0)$. Clearly $F_1'(0) = F_2'(0)=0$. But if we leave $t = 0$ with $F_1$ the value changes. It is unavoidable. But with $F_2$ the value does not change until you are some distance from $t=0$. $F_2$ is linearization stable, while $F_1$ is not.

“infinitesimal variations which preserve $F(·)$” may be “integrated” to give honest curves preserving $F(·)$

"infinitesimal variations" refers to the behavior of the derivative of $F$. How does $F$ behave near $x_0$ in the limit as the size of "near" goes to $0$. If $F$ is constant, the derivative is $0$. So saying $DF|_{x_0}\cdot h = 0$ means that if you pass through $x_0$ in the direction of $h$, $F$ acts at $x_0$ as if it were constant - as if the value of $F$ is preserved in that direction. Both $F_1$ and $F_2$ have this property at $t=0$. But for $F_1$, it is misleading. As you leave $t = 0$, the value of $F_1$ is not preserved. For $F_2$, it is preserved - for awhile at least. There is a curve through $t=0$ in direction $h$, namely $x(t) = t$, for which $F_2(x(t))$ is constant in a neighborhood of $0$. ("Honest" is a fluff word in the sentence - the meaning does not change in the slightest if it is removed.)

With 2 or more dimensions, more possible behaviors emerge. The example given is

Let $\Phi : \Bbb R^2 \to \Bbb R, (x, y) \mapsto x(x^2 + y^2)$. The set $\Phi^{−1}(0) = \{(0, y) : y \in \Bbb R\}$ is one dimensional, and in particular $T_0\Phi^{−1}(0) = \text{span}\left\{\frac{\partial}{\partial y}\right\}$. However, $D\Phi|_0 = 0$, so $\ker(D\PhiΦ|_0) = T_0\Bbb R^2$. In particular, there is no curve $γ(t)$ through $\Phi^{−1}(0)$ with $γ′(0) = \frac{\partial}{\partial x}$

Once again here are the pieces. The example is actually saying some pretty simple things, but because it is being fit to the language of differential manifolds, it is being said in complicated ways.

$\Phi^{−1}(0) = \{(0, y) : y \in \Bbb R\}$

That is, $\Phi$ is $0$ on the $y$-axis, and non-zero everywhere else.

$T_0\Phi^{−1}(0) = \text{span}\left\{\frac{\partial}{\partial y}\right\}$

Because differential geometry defines tangent vectors as differential operators,$\frac{\partial}{\partial y}$ just means the canonical unit vector pointing along the positive $y$-axis. $\Phi^{−1}(0)$ (the $y$-axis) is a submanifold of $\Bbb R^2$. This is saying that all tangent vectors at $0$ (aka, $(0,0)$) are multiples of the unit $y$-axis vector. That is, they point along the $y$-axis.

However, $D\Phi|_0 = 0$, so $\ker(D\Phi|_0) = T_0\Bbb R^2$.

The derivative of $\Phi$ at $0$ is $0$ in every direction, not just along the $y$-axis.

In particular, there is no curve $γ(t)$ through $\Phi^{−1}(0)$ with $γ′(0) = \frac{\partial}{\partial x}$

(Okay, they've been referring to $\Phi^{−1}(0)$ enough that they've forgotten that here they just mean at $0 \equiv (0,0)$.)

It is only along the $y$-axis that $\Phi$ keeps the same value that it had at $0$. So if you pass through $0$ headed in any other direction, $\Phi$ will change in value. In particular if you travel in the direction of the $x$-axis, the value of $\Phi$ changes.

Since there are directions at $0$ where the derivative of $\Phi$ in those directions is $0$, but any curve running through $0$ in that direction will not have $\Phi$ constant along it, $\Phi$ is not linearization stable at $0$.

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One final example: $F : \Bbb R^2 \to \Bbb R^2 : (x,y) \to x^2 + y^2$, at $p = (1,0)$. If $\hat x = \frac{\partial}{\partial x}, \hat y = \frac{\partial}{\partial y}$ are the canonical basis vectors for $T_p\Bbb R^2$, then $$DF|_p\cdot \hat x = \frac{\partial F}{\partial x}(1,0) = 2\\DF|_p\cdot \hat y = \frac{\partial F}{\partial y}(1,0) = 0$$ And all other directions are linear combinations of these, so only vectors parallel to $\hat y$ give a derivative of $0$.

For any non-zero real number $a$, the curve $\phi(t) = (\cos at, \sin at)$ has $\phi(0) = p, \phi'(0) = a\hat y$ and $F\circ \phi$ constant. Thus every vector at $p$ for which $DF_p$ is $0$ has a curve in that direction along which $F$ is constant. Thus $F$ is linearization stable at $p$.