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For a scalar ODE of the form $$\dot{x}(t) + f\left(x(t)\right) = 0 \label{1}\tag{1}$$ where $f \colon \mathbb R \to \mathbb R$ is some smooth function admitting a unique root $x^*$ such that $f(x^*) = 0$. The linearization of \eqref{1} is straightforward as we can insert $$ x(t) = x^* + \varepsilon g(t) \label{2}\tag{2}$$ into \eqref{1}, where $0 < |\varepsilon| \ll 1$, to see that the equation for $g$ (neglecting $\mathcal{O}(\varepsilon^2)$ terms) will be $$ \dot{g}(t) + f'(x^*) g(t) = 0 \label{3} \tag{3}.$$ However, suppose that we want to perturb the ODE \eqref{1} by some external signal modelled by another time-dependent function $\sigma(t)$ (whose precise expression is not available) such that $\sigma(t) \to 0$ as $ t \to \infty$ (we can also impose the condition that $|\sigma(t)|$ is bounded by some exponentially decaying function $\mathrm{e}^{-\lambda t}$), i.e., we consider the perturbed ODE $$\dot{x}(t) + f\left(x(t)\right) + \sigma(t) = 0 \label{4}\tag{4}$$ such that $x^*$ remains to be a long-time equilibrium state. May I know how can we can "linearize" the equation \eqref{4}? Apparently, employing the ansatz \eqref{2} will not give us a equation for $g$ at the order of $\varepsilon$...

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  • $\begingroup$ Since $\sigma$ doesnt depend on $x$ the variational ODE remains the same (almost by definition). But in this case you should be linearizing about an actual solution of the original ODE, which for nontrivial $\sigma$, cannot be a constant $x^*$ (I guess you could linearize about the constant $x^*$, but then you’d have to interpret the solution of the variational equation as telling you information in an asymptotic sense). $\endgroup$
    – peek-a-boo
    Aug 8, 2023 at 20:24
  • $\begingroup$ @peek-a-boo can you elaborate your comment into a complete answer? I am not sure why you mention the term "variational ODE" $\endgroup$
    – Fei Cao
    Aug 8, 2023 at 22:32
  • $\begingroup$ variational ODE is the name for your “linearized ODE” (3). More properly, (3) is called the (linear) variational ODE associated to (1) along the (constant) solution $x^*$. $\endgroup$
    – peek-a-boo
    Aug 8, 2023 at 23:10
  • $\begingroup$ I don’t know how much help it’s going to be but here is a PhySE answer of mine about linearizing. There I happened to discuss the autonomous case, but the discussion extends almost verbatim to the non-autonomous case (simply replace the full Frechet derivative $D$ by the derivative in the spatial variables only). $\endgroup$
    – peek-a-boo
    Aug 8, 2023 at 23:49
  • $\begingroup$ @peek-a-boo Thank you for your comment, although I am not sure whether that is super-related to what I am trying to ask. Regarding your very first comment, I am trying to linearize about a long-time equilibrium of the ODE (4), which contains at least the point/state $x^*$. $\endgroup$
    – Fei Cao
    Aug 9, 2023 at 2:26

1 Answer 1

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This is not an answer, but an elaboration on my comment as requested by the OP.

The linearized version of this equation is given by $$ \dot g(t) + f'(x^*)\,g(t) + \frac1\epsilon\sigma(t) = 0. $$ This equation holds as long as $g(t)$ is $\mathcal{O}(1)$, which is a problem because of the $1/\epsilon$ term in the equation. Let us write $\sigma(t) = e^{-\lambda t}\mu(t)$ for some $\mu(t)$ with $|\mu(t)|<1$. We can rewrite the above equation as $$ \dot g(t) + f'(x^*)\,g(t) + \frac{e^{-\lambda t}}{\epsilon}\mu(t) = 0. $$ There is a time $t^* = \ln(1/\epsilon)/\lambda$ after which the prefactor $e^{-\lambda t}/\epsilon$ becomes smaller than one. For $t>t^*$, this equation provides a perturbative solution to the original equation up to $\mathcal{O}(\epsilon)$.

There is no guarantee that a perturbative solution exists for an early time. Let me try to explain what could go wrong. Near $x^*$, the term $f(x(t))$ is small in the unperturbed equation. If the $\sigma(t)$ term is not equally small, the equation would be approximately $\dot x(t) = -\sigma(t)$. Now imagine if $\sigma(t)$ does not decay to zero near $t=0$. In this case, even if we start near $x^*$, $x(t)$ grows away from $x^*$ at a macroscopic rate, and therefore, no perturbative solution around $x^*$ could exist.

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