center manifold | Pengfei's Notes

Let $X:\mathbb{R}^d\to \mathbb{R}^d$ be a $C^\infty$ vector field with $X(o)=0$ . Then the origin $o$ is a fixed point of the generated flow on $\mathbb{R}^d$ . Let $T_o\mathbb{R}^d=\mathbb{R}^s\oplus\mathbb{R}^c\oplus\mathbb{R}^u$ be the splitting into stable, center and unstable directions. Moreover, there are three invariant manifolds (at least locally) passing through $o$ and tangent to the corresponding subspaces at $o$ .

Theorem (Pliss). For any $n\ge 1$ , there exists a $C^n$ center manifold $C^n(o)=W^{c,n}(o)$ .

Generally speaking, the size of the center manifold given above depends on the pre-fixed regularity requirement. Theoretically, there may not be a $C^\infty$ center manifold, since $C^n(o)$ could shrink to $o$ as $n\to\infty$ . An explicit example was given by van Strien (here). He started with a family of vector fields $X_\mu(x,y)=(x^2-\mu^2, y+x^2-\mu^2)$ . It is easy to see that $(\mu,0)$ is a fixed point, with $\lambda_1=2\mu<\lambda_2=1$ . The center manifold can be represented (locally) as the graph of $y=f_\mu(x)$ .

Lemma. For $n\ge 3$ , $\mu=\frac{1}{2n}$ , $f_\mu$ is at most $C^{n-1}$ at $(\frac{1}{2n},0)$ .

Proof. Suppose $f_\mu$ is $C^{k}$ at $(\frac{1}{2n},0)$ , and let $\displaystyle f_\mu(x)=\sum_{i=1}^{k}a_i(x-\mu)^i+o(|x-\mu|^{k})$ be the finite Taylor expansion. The vector field direction $(x^2-\mu^2, y+x^2-\mu^2)$ always coincides with the tangent direction $(1,f'_\mu(x))$ along the graph $(x,f_\mu(x))$ , which leads to

$(x^2-\mu^2)f_\mu'(x)=y+x^2-\mu^2=f_\mu(x)+x^2-\mu^2$ .

Note that $x^2-\mu^2=(x-\mu)^2+2\mu(x-\mu)$ . Then up to an error term $o(|x-\mu|^{k})$ , the right-hand side in terms of $(x-\mu)$ : $(a_1+2\mu)(x-\mu)+(a_2+1)(x-\mu)^2+\sum_{i=3}^{k}a_i(x-\mu)^i$ ; while the left-hand side in terms of $(x-\mu)$ :

$(x-\mu)^2f_\mu'(x)+2\mu(x-\mu)f_\mu'(x)=\sum_{i=1}^{k}ia_i(x-\mu)^{i+1}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$

$=\sum_{i=2}^{k}(i-1)a_{i-1}(x-\mu)^{i}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$ .

So for $i=1$ : $2\mu a_1=a_1+2\mu$ , $a_1=\frac{-2\mu}{1-2\mu}\sim 0$ ;

$i=2$ : $a_2+1=a_1+4\mu a_2$ , $a_2=\frac{a_1-1}{1-4\mu}\sim -1$ ;

$i=3,\cdots,k$ : $a_i=(i-1)a_{i-1}+2i\mu a_i$ , $(1-2i\mu)a_i=(i-1)a_{i-1}$ .

Note that if $k\ge n$ , we evaluate the last equation at $i=n$ to conclude that $a_{n-1}=0$ . This will force $a_i=0$ for all $i=n-2,\cdots,2$ , which contradicts the second estimate that $a_2\sim -1$ . Q.E.D.

Consider the 3D vector field $X(x,y,z)=(x^2-z^2, y+x^2-z^2,0)$ . Note that the singular set $S$ are two lines $x=\pm z$ , $y=0$ (in particular it contains the origin $O=(0,0,0)$ ). Note that $D_OX=E_{22}$ . Hence a cener manifold $W^c(O)$ through $O$ is tangent to plane $y=0$ , and can be represented as $y=f(x,z)$ . We claim that $f(x,x)=0$ (at least locally).

Proof of the claim. Suppose on the contrary that $c_n=f(x_n,x_n)\neq0$ for some $x_n\to 0$ . Note that $p_n=(x_n,c_n,x_n)\in W^c(O)$ , and $W^c(O)$ is flow-invariant. However, there is exactly one flow line passing through $p_n$ : the line $L_n=\{(x_n,c_nt,x_n):t>0\}$ . Therefore $L_n\subset W^c(O)$ , which contradicts the fact that $W^c(O)$ is tangent to plane $y=0$ at $O$ . This completes the proof of the claim.

The planes $z=\mu$ are also invariant under the flow. Let’s take the intersection $W_\mu=\{z=\mu\}\cap W^c(O)=\{(x,f(x,\mu),\mu)\}$ . Then we check that $\{(x,f(x,\mu))\}$ is a (in fact the) center manifold of the restricted vector field in the plane $z=\mu$ . We already checked that $f(x,\mu)$ is not $C^\infty$ , so is $W^c(O)$ .

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