Consider the stochastic differential equation
$$d Z_t = M Z_t dt + dB_t,$$
where $M \in \mathbb{R}^{n\times n}$ is assumed to have eigenvalues that have negative real parts, $B_t$ is $n$-dimensional spherical brownian motion.
The solution of this SDE is given by
$$ Z_t = \exp(Mt)Z_0 + \int_0^t \exp(M(t-s)) dB_s$$
and has the distribution $$Z_t \sim \mathcal{N}\left(\exp(Mt)Z_0,\int_0^t \exp(M(t-s))\exp(M^T(t-s))ds\right).$$
Now, consider the function $$f(z) = z^T\Sigma z.$$ I am trying to understand for what choice of $\Sigma$ positive (semi-)definite we have that $\mathbb{E}[Z_t^T\Sigma Z_t]$ is decreasing.
In the case of a deterministic equation, this yields to the Lyapunov equation, such that all $\Sigma$ that solve $$M\Sigma + \Sigma M^T = - Q,$$ for some positive definite $Q$, yield a Lyapunov function for the system But in the stochastic case, we obtain an additional term that results from the fact that
$$ \mathbb{E}[Z_t^T\Sigma Z_t] = \mathbb{E}[Z_t]^T\Sigma \mathbb{E}[Z_t] + \operatorname{tr}[\Sigma C(t)]$$
with $C(t) = \operatorname{Cov}[Z_t]$.
Differentiating yields
$$ 2\mathbb{E}[Z_t] \Sigma M \mathbb{E}[Z_t] + \operatorname{tr}[\Sigma (M C(t) + C(t)M^T + I_n)]$$
And then I'm stuck. Theoretically, $C(\infty)^{-1}$ should work, but I also don't know how to show that. So, any help on how to determine the general form of the solutions would be appreciated!
Edit: After some thoughts, and also considering the answer below, it seems that one has to consider the initial conditions carefully.
As an example, consider $Z_0=0$. Then we have $\frac{d}{dt} \mathbb{E}[f(Z_t)]\vert_0 = \operatorname{tr}[\Sigma]$, so this should also hold in some neighborhood of $0$. Intuitively, the noise has to 'spread out' first from the initial delta distribution.
As I noted in the comments, I am unsure of the answer as I am unsure of my multivariate Ito calculus, but here it goes:
Here, we can use Ito's formula, namely, for any $f$, we can find the change in the value of the function to be
$$df = (\mu_t \frac{df}{dx} + \frac{\sigma^2_t}{2}\frac{d^2f}{dx^2})dt + ...dB_t $$
Now, in expectation, the terms containing $dB_t$ go away, and we are now left with
$$\mathbb{E}[\mu_t \frac{df}{dx} + \frac{\sigma^2_t}{2}\frac{d^2f}{dx^2}]$$
The first term is the same as in the deterministic case, which gives us
$ \mathbb{E}[Z_t^\top (\Sigma + \Sigma^\top) M Z_t ]$ for the first term
We note that the Hessian of $f$ is constant. As such, the second term simply reduces to trace of $\Sigma$. As such, a sufficient condition should be to ensure that
$x^\top (\Sigma + \Sigma^\top)M x + tr(\Sigma) < 0$.