# Proper linearization of ODEs of the form $\dot{x}(t) + f(x(t)) + \sigma(t) = 0$?

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 $$\varepsilon$$...

• 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). Aug 8, 2023 at 20:24
• @peek-a-boo can you elaborate your comment into a complete answer? I am not sure why you mention the term "variational ODE" Aug 8, 2023 at 22:32
• 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 $x^*$. Aug 8, 2023 at 23:10
• 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). Aug 8, 2023 at 23:49
• @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 $x^*$. Aug 9, 2023 at 2:26

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.