Theory of Invariant Manifold and Foliation and Uniqueness of Center Manifold Dynamics

Abstract

Here we prove that the dynamics on any two center-manifolds of a fixed point of any $C^{k,1}$ dynamical system of finite dimension with $k\ge 1$ are $C^k$-conjugate to each other. For pedagogical purpose, we also extend Perron’s method for differential equations to diffeomorphisms to construct the theory of invariant manifolds and invariant foliations at fixed points of dynamical systems of finite dimensions.

Appendix

A set of well-known results are collected here for references. Wherever proofs are omitted they can be filled by [5].

Theorem A.1

(Uniform Contraction Principle I) Let X, Y be two metric spaces with X being complete. Assume $f:X\times Y \rightarrow X$ is continuous and uniformly contractive with a contraction constant $0<\theta <1$. Then the unique fixed point ${\bar{x}}(y)$ is continuous and

$$\begin{aligned} d({\bar{x}}(z),{\bar{x}}(y))\le \frac{1}{1-\theta }d(f({\bar{x}}(y),z),f({\bar{x}}(y),y)). \end{aligned}$$

Theorem A.2

(Uniform Contraction Principle II) Let X, Y be two Banach spaces, and let $U\subset X,\ V\subset Y$ be open subsets. Let $f\in C^k({\bar{U}}\times V,{\bar{U}}), 1\le k<\infty$. Assume $f:{\bar{U}}\times V\rightarrow {\bar{U}}$ is a uniform contraction mapping, and $|D_xf(x,y)|$ is uniformly bounded by a constant $\theta <1$ in ${\bar{U}}\times V$. Let ${\bar{x}}(y)$ be the unique fixed point of $f(\cdot , y)$ in ${\bar{U}}$ for $y\in V$. Then ${\bar{x}}(\cdot )\in C^k(V, {\bar{U}})$ and the first derivative is

$$\begin{aligned} D{\bar{x}}(\cdot )=\sum _{n=0}^\infty [D_xf({\bar{x}}(\cdot ),\cdot )]^nD_yf({\bar{x}}(\cdot ),\cdot ). \end{aligned}$$

(A.1)

If f is $C^{k,1}$, then ${\bar{x}}(\cdot )$ is $C^{k,1}$, and if f is analytic in $U\times V$, then ${\bar{x}}(\cdot )$ is analytic from V to X.

Theorem A.3

(Implicit Function Theorem I) Let X, Y, Z be Banach spaces, $U\subset X,\ V\subset Y$ be open sets. Assume $F:U\times V\rightarrow Z$ is differentiable in $x\in U$ and both F and $D_xF$ are continuous in $(x,y)\in U\times V$. If there is a point $(x_0,y_0)\in U\times V$ such that $F(x_0,y_0)=0$ and $D_xF(x_0,y_0)$ has a bounded inverse, then there is a neighborhood $U_1\times V_1\subset U\times V$ of $(x_0,y_0)$ and a continuous function $f:V_1\rightarrow U_1$ with $f(y_0)=x_0$ such that $F(x,y)=0$ for $(x,y)\in U_1\times V_1$ iff $x=f(y)$.

Theorem A.4

(Implicit Function Theorem II) Let X, Y, Z be Banach spaces, $U\subset X,\ V\subset Y$ be open sets, and $F:U\times V\rightarrow Z$ be continuously differentiable in both variables. If there is a point $(x_0,y_0)\in U\times V$ such that $F(x_0,y_0)=0$ and $D_xF(x_0,y_0)$ has a bounded inverse, then there is a neighborhood $U_1\times V_1\subset U\times V$ of $(x_0,y_0)$ and a continuously differentiable function $f:V_1\rightarrow U_1$ with $f(y_0)=x_0$ such that $F(x,y)=0$ for $(x,y)\in U_1\times V_1$ iff $x=f(y)$. Also,

$$\begin{aligned} Df(y)=-[D_xF(f(y),y)]^{-1}D_yF(f(y),y). \end{aligned}$$

Moreover, if $F\in C^k(U\times V, Z),\ k\ge 1$ or $C^{k,1}$ or analytic in a neighborhood of $(x_0,y_0)$, then $f\in C^k(V_1,U_1)$ or $C^{k,1}$ or is analytic in a neighborhood of $y_0$.

Theorem A.5

(Global Inverse Function Theorem) Let $A_{n\times n}$ be nonsingular and $h\in C^1({\mathbb {R}}^n)$. Then there is a small number $\delta >0$ so that $\sup _{x\in {\mathbb {R}}^n} (|h(x)|+|Dh(x)|)<\delta$ implies $f(x)=Ax+h(x)$ is invertible and the inverse $f^{-1}$ is as smooth as f. Moreover, $f^{-1}$ can be expressed as $f^{-1}=A^{-1}+g$ with $g=-A^{-1}\circ h\circ f^{-1}$, $\sup _{x\in {\mathbb {R}}^n} (|g(x)|+|Dg(x)|)\le \epsilon$ and $\lim _{\delta \rightarrow 0}\epsilon =0$. Furthermore, if f is $C^k$, or $C^{k,1}$, for $k\ge 1$, or analytic then $f^{-1}$ is also $C^k$, or $C^{k,1}$, or analytic, respectively.

Proof

Let $X=C^1({\mathbb {R}}^n)$ be the Banach space of functions from ${\mathbb {R}}^n$ to itself for which they and their derivatives are uniformly continuous and uniformly bounded with norm

$$\begin{aligned} ||h||_1=\sup _{x\in {{\mathbb {R}}}^n}(|h(x)|+|Dh(x)|). \end{aligned}$$

We look for inverse of the form $\phi =A^{-1}+g$ with $g\in X$

$$\begin{aligned} {\textrm{id}}=\phi \circ f=(A^{-1}+g)\circ (A+h)={\textrm{id}}+A^{-1}\circ h+g\circ (A+h) \end{aligned}$$

equivalent to

$$\begin{aligned} F(g,h):=A^{-1}\circ h+g\circ (A+h)=0. \end{aligned}$$

Obviously, $F(g,h)\in X$, showing $F:X\times X\rightarrow X$. Also, F is differentiable in g, h with $D_gF(g,h)v=v\circ (A+h)$ and $D_hF(g,h)v=Dg\circ (A+h)v$ for any $v\in X$, showing $F\in C^1(X\times X,X)$. Moreover, $D_gF(0,0)v=v\circ A=w$ for any $w\in X$ iff $v=w\circ A^{-1}$. This shows $D_gF(0,0)\in L(X,X)$ is invertible with a bounded inverse since $v=[D_gF(0,0)]^{-1}w=w\circ A^{-1}$ and $|[D_gF(0,0)]^{-1}|=1$. Since in addition $F(0,0)=0$, therefore, by IFT there are open neighborhood $V=N_{\delta _1}(0),U=N_{\delta _2}(0)\subset X$ for some small numbers $\delta _1,\delta _2>0$ and a $u\in C^1(V,U)$ so that $F(g,h)=0$ for $(g,h)\in U\times V$ iff $g=u(h)$. So, the left-inverse $\phi (h)=A^{-1}+u(h)$ exists and is of $C^1$.

To show $\phi$ is also the right-inverse, consider similarly the right-inverse of the form $\psi =A^{-1}+g$ with

$$\begin{aligned} {\textrm{id}}=f\circ \psi =(A+h)\circ (A^{-1}+g)={\textrm{id}}+A\circ g+h\circ (A^{-1}+g) \end{aligned}$$

equivalent to

$$\begin{aligned} G(g,h):=A\circ g+h\circ (A^{-1}+g)=0. \end{aligned}$$

It is similar to show $G\in C^1(X\times X,X)$ and $G(0,0)=0$. It is slightly different to show $D_gG(0,0)$ has a bounded inverse. Specifically, for any $v\in X$,

$$\begin{aligned} D_gG(0,0)v=[A+Dh(A^{-1}\cdot )]v=A[\textrm{id}+A^{-1}Dh(A^{-1}\cdot )]v, \end{aligned}$$

which means

$$\begin{aligned} {[}D_gG(0,0)v](x)=[A+Dh(A^{-1}x)]v(x)=A[\textrm{id}+A^{-1}Dh(A^{-1}x)]v(x). \end{aligned}$$

So $D_gG(0,0)$ is invertible if $T\in L(X,X)$ with $T(x)=A^{-1}Dh(A^{-1}x)$ is bounded by $\sup _{x\in {\mathbb {R}}^n}|T(x)|<1$ which holds if $\sup _{x\in {\mathbb {R}}^n}|Dh(x)| < 1/|A^{-1}|:=r$. So if we let $Y={\bar{N}}_r(0)\subset X$, then for $G\in C^1(X\times Y,X)$, $D_gG(0,0)\in L(X,X)$ has a bounded inverse. Therefore, by IFT there are open neighborhood $V=N_{\delta _1'}(0)\subset Y,U=N_{\delta _2'}(0)\subset X$ for some small numbers $\delta _1',\delta _2'>0$ and a $w\in C^1(V,U)$ so that $G(g,h)=0$ for $(g,h)\in U\times V$ iff $g=w(h)$. That is, the right-inverse $\psi (h)=A^{-1}+w(h)$ exists.

Next, to show $\phi$ and $\psi$ are the same function, let $\delta =\min \{\delta _1,\delta _1'\}$, and $\gamma =\max \{\delta _2,\delta _2'\}$, then both u and w map $V=N_\delta (0)\subset X$ to $U=N_\gamma (0)\subset X$. Because of the continuity, $\lim _{\delta \rightarrow 0}\epsilon =0$ where $\epsilon =\max \{{\Vert u\Vert }_1,{\Vert w\Vert }_1\}$. As a result, both $\phi (h)=A^{-1}+u(h)$ and $\psi (h)=A^{-1}+w(h)$ are defined for $h\in V$ so that $\phi (h)\circ f=\textrm{id}$ and $f\circ \psi (h)=\textrm{id}$ imply

$$\begin{aligned} \phi (h)=\phi (h)\circ \textrm{id}=\phi (h)\circ (f\circ \psi (h))=\psi (h) \end{aligned}$$

by the associative law of composition. By definition, we have $\phi (h)=f^{-1}$.

Finally, if h is $C^k$, or $C^{k,1}$, for $k\ge 1$, or analytic, then both F and G have the same smoothness, and by IFT both $\phi$ and $\psi$ have the same smoothness as well. As a consequence, $f^{-1}$ is as smooth as h is. $\square$

Lemma A.1

(Cut-off Function) For each $r>0$ there exists a $C^\infty$ function $\rho _r:{\mathbb {R}}^n\rightarrow [0,1]$ so that $\rho _r|_{N_r}\equiv 1$ and $\textrm{supp}\{\rho \}\subset N_{2r}$, where $N_r$ is the Euclidean ball of radius r in ${\mathbb {R}}^n$ centered at 0.

Theorem A.6

(Local Inverse Function Theorem) Let $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ be a $C^k$ function for $k\ge 1$. Assume at a point $x_0$, $Df(x_0)$ is invertible. Then there is a small open neighborhood U of $x_0$, a small open neighborhood V of $y_0=f(x_0)$ so that $f:U\rightarrow V$ is 1-1, onto, and the inverse $f^{-1}$ is also $C^k$.

Proof

First we claim that $f:U\rightarrow V$ is invertible iff $g: U'=U\oplus \{-x_0\}\rightarrow V'=V\oplus \{-y_0\}$ is invertible where ${\bar{y}}=g({\bar{x}})=f({\bar{x}}+x_0)-y_0$, ${\bar{x}}=x-x_0\in U'$. This can be checked directly as follows. Specifically, if f is invertible with inverse $f^{-1}$, then $g^{-1}({\bar{y}})=f^{-1}({\bar{y}}+y_0)-x_0$ because

$$\begin{aligned} g\circ g^{-1}({\bar{y}})=f(g^{-1}({\bar{y}})+x_0)-y_0=({\bar{y}}+y_0)-y_0={\bar{y}}, \end{aligned}$$

and similarly $g^{-1}\circ g({\bar{x}})={\bar{x}}$. If g is invertible with inverse $g^{-1}$, then $f^{-1}(y)=g^{-1}(y-y_0)+x_0$ because

$$\begin{aligned} f\circ f^{-1}(y)=[f(g^{-1}(y-y_0)+x_0)-y_0]+y_0=g\circ g^{-1}({\bar{y}})+y_0=y, \end{aligned}$$

and similarly $f^{-1}\circ f(x)=x$.

Without loss of generality, we can assume $x_0=y_0=0$ for $f\in C^k({\mathbb {R}}^n,{\mathbb {R}}^n)$. Now, let $A=Df(0)$, $k(x)=f(x)-Ax$. Then $k(0)=0$, $Dk(x)=Df(x)-A$ and $Dk(0)=0$. So by the continuous differentiability of f for any $\delta _1>0$ there is a small r-ball $N_{r}$ of $x=0$ so that

$$\begin{aligned} \sup _{x\in N_r}(|k(x)|+|Dk(x)|)\le \delta _1. \end{aligned}$$

Let $\rho _r$ be a cut-off function from the previous lemma. Define

$$\begin{aligned} h(x)=\rho _{r/2}(x)k(x). \end{aligned}$$

Then the support of h is inside $N_r$, and

$$\begin{aligned} |Dh(x)|=|D\rho _{r/2}(x)k(x)+\rho _{r/2}(x)Dk(x)|\le K\delta _1 \end{aligned}$$

for a constant K and all $x\in {\mathbb {R}}^n$. Hence,

$$\begin{aligned} \sup _{x\in {\mathbb {R}}^n}(|h(x)|+|Dh(x)|)\le (K+1)\delta _1:=\delta \end{aligned}$$

Therefore, by the Global Inverse Function Theorem, for sufficiently small $r>0$, $F(x)=Ax+h(x)$ is $C^k$ invertible in ${\mathbb {R}}^n$. For $x\in N_{r/2}$, since $\rho _{r/2}(x)\equiv 1$, we have $F(x)=Ax+h(x) =Ax+k(x)=f(x)$. Hence f is locally invertible from $U= N_{r/2}$ to $V=F(U)$, and the inverse, $f^{-1}=F^{-1}|_V$, is also $C^k$. $\square$

Adapted Norm: For any nonsingular square matrix A, there is an invertible matrix consisting of generalized eigenvectors P so that A is similarly to a Jordan canonical form J, $A\sim J$ with $AP=PJ$. Using the column vectors of P as a new basis for a new coordinate system, the matrix A becomes its Jordan form. That is, without loss of generality we can assume a coordinate system is chosen so that A is in its Jordan canonical form, $A=\textrm{diag}(D_1,\dots , D_k)$, a block-diagonal matrix. Here each $D_i$ is one of the following forms:

$$\begin{aligned} \left[ \begin{array}{cccccc} D &{} N &{} 0 &{} \dots &{} 0\\ 0 &{} D &{} N &{}\dots &{} 0\\ \vdots &{} \vdots &{} \ddots &{} &{} \vdots \\ 0 &{} 0 &{} \ldots &{} D &{} N \\ 0 &{} 0 &{} \ldots &{} 0 &{} D \\ \end{array}\right] \end{aligned}$$

where either $D=\lambda , N=0$, or $D=\lambda , N=\epsilon$, or $D=\left[ \begin{array}{cc} a &{} -b\\ b &{} a \end{array}\right]$, $N=\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} 0 \end{array}\right]$, or $D=\left[ \begin{array}{cc} a &{} -b\\ b &{} a \end{array}\right]$, $N=\left[ \begin{array}{cc} \epsilon &{} 0\\ 0 &{} \epsilon \end{array}\right]$, with $\lambda$ or $a+ib$ the eigenvalues of $D_i$. The parameter $\epsilon >0$ can be made arbitrarily small a priori with respect to any parameter, say $\max \{|\sigma (A)|\}<\alpha$, by only re-adjusting the (non-zero) length of each column vector of the similarity matrix P. For the resulting coordinate x use the maximum norm, ${\Vert x\Vert }=\min _{1\le i\le d}{|x_i|}$, which in turn gives rise to the matrix norm satisfying

$$\begin{aligned} {\Vert A\Vert }<\alpha , \end{aligned}$$

and for $0<\beta <\min \{|\sigma (A)|\}$,

$$\begin{aligned} {\Vert A^{-1}\Vert }<1/\beta \ \hbox { and } \beta < {\Vert A\Vert }. \end{aligned}$$

Usually, such norms can be chosen in combinations for pseudo-hyperbolic splits of matrices. They are referred to as adapted norms.