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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.