I am interested in the long-term behavior of the following OU type multivariate process: $$ dX(t) = \{-H(t)\cdot X(t) - \alpha(t)\mathbf{1}\}dt + \sigma(t)dB(t) $$ where $\alpha(t) = t^{-c}$ with $c>0$, and $\mathbf{1}$ is the vector of 1's with same dimension as $X(t)$. Suppose $H(t)$ and $\sigma(t)$ are deterministic positive definite and symmetric (for all $t$) matrices. $B(t)$ is a standard multivariate Brownian motion. As $t\to\infty$, is it true that $X(t)$ has the same stationary distribution as $$ d\tilde X(t) = -H(t)\cdot \tilde X(t)dt + \sigma(t)dB(t), $$ or do $X(t)$ and $\tilde X(t)$ converge to the same distribution as $t\to\infty$? Intuitively, this seems to be true, because $\alpha(t)\to 0$ as $t\to\infty$.
A major difficulty here is that the matrix $H(t)$ is time-varying in a general way. In particular, both its eigenvalues and eigenvectors are time-varying. It seems difficult to solve the SDEs (at least usual textbook methods won't apply), and compare the solutions.
I would appreciate if anybody could share an idea on how to rigorously prove this, or provides any reference for this. Thank you!
Edit (clarify the difficulty, and respond the comment of UBM): if $H(t)$ is a constant matrix $H(t)=H$, and $H$ is positive definite and symmetric, one can see that by solving the first and the second SDE (WLOG suppose $X(0)=\tilde X(0)=0$), $$ X(t) = -\int_0^t e^{-H(t-s)} t^{-c} \mathbf{1} ds + \int_0^t e^{-H(t-s)} \sigma(s) dB(s), $$ while the solution $\tilde X(t)$ does not have the first term on the right hand side. Therefore, one only has to show the first term on the right hand side vanishes as $t\to\infty$ whenever $c>0$, which can be done by careful estimation.
When $H(t)$ is not a constant matrix, I don't think one can solve $X(t)$ and $\tilde X(t)$ nicely.