# behavior of SDE as parameter goes to infinity (Ornstein-Uhlembeck?)

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

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

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

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

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?

• 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) Commented Dec 5, 2023 at 20:32
• What are the functions $D,U$? Also add the physics reference by editing the post. Commented Dec 5, 2023 at 20:35