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Given a differential equation on a Banach space $\mathcal{X}$ of the form $\frac{d u}{d t} = F(u)$, it is often the case that $F$ is equivariant under translations, i.e. that $T_\alpha F(u) = F(T_\alpha u)$ where $(T_\alpha f)(x) = f(x+\alpha)$. The set of translation operators {$T_\alpha, \alpha \in \mathbb{R}$} is a (unitary ?) representation of the Lie group $\mathbb{R}$; its infinitesimal generator is $\tau = \frac{d}{d x}$. Assume that $u_* $ is an equilibrium of the differential equation, so $F(u_*)=0$.

My question is if and when it is possible to write a solution to the above differential equation as $u(t) = T_{\alpha(t)} (u_* + v(t))$, where $v$ is transversal to $\tau u_* $ in the following way: take the linear form $\phi_* \in \mathcal{X}^* $ for which $\phi_* (\tau u_* ) = 1 $, then $v \in \mathcal{H} = $ { $v \in \mathcal{X}: \phi_*(v) = 0$ }.

Loosely speaking, this means that we can split the action of the semiflow generated by the differential equation in a part transversal to the direction of the translation group action at $ u_* $, followed by a translation. This specific form of the solution was stated as an Ansatz in Haragus&Iooss (Springer, 2011), 'Local bifurcations, center manifolds and normal forms in infinite-dimensional dynamical systems', p. 56.

Thanks in advance for any (useful) reply!

Frits Veerman, Universiteit Leiden, The Netherlands

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Just a small marginal comment: Unitary makes only sense if you have a Hilbert space... – András Bátkai Nov 29 '12 at 13:38
Very true. I have to admit that the applications I have in mind are all in a Hilbert space setting; I'm just looking for the most general context in which the above might be true. Thanks for your remark. – Frits Veerman Nov 30 '12 at 14:15

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