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Consider a continuous-time dynamical system defined by \begin{equation} \dot{x}=f(x),\ \ \ \ \ \ x\in\mathbb R^{n} \ \ \ \ \ \ \ (1) \end{equation} where f is sufficiently smooth, $f(0) = 0$. Let the eigenvalues of the Jacobian matrix A evaluated at the equilibrium point $x_0 = 0$ be $\lambda_1, ... , \lambda_n$. Suppose the equilibrium is not hyperbolic and that there are thus eigenvalues with zero real part. Assume that there are $n_+$ eigenvalues (counting multiplicities) with $\Re (\lambda)>0$, $n_0$ eigenvalues with $\Re (\lambda)=0$, and $n_-$ eigenvalues with $\Re (\lambda)<0$. Let $T^c$ denote the linear eigenspace of A corresponding to the union of the $n_0$ eigenvalues on the imaginary axis. Let $\varphi^t$ denote the flow associated with (1).

Center Manifold Theorem:

There is a locally defined smooth $n_0$-dimensional invariant manifold $W_{loc}^c (0)$ of (1) that is tangent to $T^c$ at $x=0$. The manifold $W_{loc}^c$ is called the center manifold. Moreover, there is a neighborhood U of $x_0 =0$, such that if $\varphi^t x\in U$ for all $t\geq 0$ ($t\leq 0$), then $\varphi^t x\rightarrow W_{loc}^c (0)$ for $t\rightarrow +\infty$ ($t\rightarrow -\infty$).

My question relates to smooth continuous-time system that depends smoothly on a parameter: \begin{equation} \dot{x}=f(x, \alpha),\ \ \ \ \ \ x\in\mathbb R^{n},\alpha\in\mathbb R\ \ \ \ \ \ \ (2) \end{equation} Is still valid the Center Manifold Theorem for system (2)? If the system (2) hasn't imaginary eigenvalues ($\Re (\lambda)=0$) can we say that there isn't center manifold?

thank you very much

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up vote 2 down vote accepted

No, there is no center manifold if there is no eigenvalue with zero real part.

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