3
$\begingroup$

Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a vector field given by $f_i(x)=\sum_{k=1}^n \sin(x_i-x_k), \forall x\in \mathbb{R}^n $. Now consider the ODE $\dot x=f(x)$. We observe that the Jacobian of $f(x)$ is singular (each row sums to zero). The intuition behind this singularity is that the vector field is invariant under shift, i.e., if $f(x)=f(x+a\mathbf{1}^n)$, where $\mathbf{1}^n$ is a vector of all ones and $a$ is any constant. What can we say about the center manifold of this ODE? Is it true that the center manifold in this case is unique and is equal to the one-dimensional subspace (i.e., the null space of the Jacobian)?

Another question is that how can I handle this kind of singularity? I am asking because I cannot apply many theorems (like the inverse function theorem) because of this singularity.

Thank you, in advance, for your comments.

$\endgroup$
  • 1
    $\begingroup$ The way you have defined $f$ makes it a scalar function; hence for $x \in \Bbb R^n$, $n \ge 2$, the equation $\dot x = f(x)$ can't be quite right. Any suggestions? Cheers! $\endgroup$ – Robert Lewis Feb 5 at 1:49
  • 1
    $\begingroup$ I revised the question. Sorry for the typo. $\endgroup$ – Arthur Feb 5 at 3:04
  • 1
    $\begingroup$ Usually you go for the differences $\gamma_i = x_i - x_1$, for example. New system will have equation $\dot{\gamma}_1 = 0$. Since it is reasonable to assume that $\gamma_1 \equiv 0$, just discard this equation and put zero instead of $\gamma_1$ in other equations. This is often done in papers about phase oscillators with RHS depending on phase differences only. $\endgroup$ – Evgeny Feb 5 at 16:17
  • $\begingroup$ Thanks for the comment. I agree, but sometimes, it is not reasonable to discard $\dot \gamma_1=0$. For example, if the vector field is more complicated and $\gamma_1$ appears in some other terms. I was wondering if there is another way to handle such situations. Thanks! $\endgroup$ – Arthur Feb 5 at 20:25

Your Answer

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

Browse other questions tagged or ask your own question.