In a physics paper, I saw the following (weird) heuristic argument: Let $\theta,v>0$ be constants. Starting from the SDE

\begin{equation} dX_t=D(X_t)(U'(X_t) -\theta(X_t-vt))dt +\sqrt{2D(X_t)}dW_t \end{equation}

the authors claim that if $\theta$ (spring constant) is very large, the process $X_t$ will always be close enough to $vt$, such that they can "Taylor expand" the function U in such a way

\begin{equation} \begin{cases} U(X_t)\approx U(vt)+ U'(vt)\cdot(X_t-vt)\\ D(X_t)\approx D(vt) \end{cases} \end{equation}

so that they effectively replace the original SDE by

$$ d\tilde{X}_t=D(vt)(U'(vt)-\theta(\tilde{X}_t-vt))dt+\sqrt{2D(vt)}dW_t$$

which becomes essentially like a OU process. Is there a way this argument could be made precise (i.e. if $\theta$ is large, the distance between the two processes starting from the same initial condition will be small in expectation)? Furthermore, shouldn't any 'Taylor expansion' of $U(X_t)$ start from using Ito's lemma?

  • $\begingroup$ a simpler version of this question would be: if $\theta$ is large, is the solution of the SDE $ dX_t=-D(X_t)\theta(X_t-vt) +\sqrt{2D(X_t)}dW_t$ Close to the solution of $ dX_t= -D(vt)\theta(X_t-vt)+\sqrt{2D(vt)}dW_t$ in some sense? ($v$ is a constant) $\endgroup$
    – Asasuser
    Commented Dec 5, 2023 at 20:32
  • 1
    $\begingroup$ What are the functions $D,U$? Also add the physics reference by editing the post. $\endgroup$ Commented Dec 5, 2023 at 20:35


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