I'm considering the following (stochastic) PDE: $${\rm d}U_t=\kappa(t)\Delta U_t{\rm d}t+\sigma W_t\tag1$$ on $[0,1)^2$ with Neumann boundary conditions, where $\kappa:[0,T]\to(0,\infty)$ is linear and $W$ is a cylindrical Wiener process on $H:=L^2([0,1)^2)$. I somehow need to compute/estimate (or at least find a sharp upper bound on) $\operatorname E\left[\|g(U_t)\|_H^2\right]$, where $g:H\to H$ is nice enough.
Is there anything in the literature on this?
I'm particularly interested in $g(t,\;\cdot\;)=\nabla\ln p_t$, where $p_t$ is the "density" of the projections of $U_t$ with respect to the Lebesgue measure. That is, if $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$ and $$H_d:=\operatorname{span}\{e_1,\ldots,e_d\}\;\;\;\text{for }d\in\mathbb N,$$ then $p_t$ is a Borel measurable function $H\to[0,\infty)$ such that $$\operatorname E\left[\varphi(\pi_dU_t)\right]=\int\lambda^{\otimes d}({\rm d}x)p_t(\iota_dx)\varphi(\iota_dx)\tag2$$ for all bounded Borel measurable $\varphi:H_d\to\mathbb R$ and $d\in\mathbb N$, where $$\iota_d:\mathbb R^d\to H_d\;,\;\;\;a\mapsto\sum_{i=1}^da_ie_i$$ and $\pi_d$ is the canonical projection of $H$ onto $H_d$. However, I don't think that this precise shape of $g$ simplifies the question.
EDIT: Let be elaborate on what we might can try. Assume that $X$ is an Ito process on $H$ satisfying $${\rm d}X_t=\varphi_t{\rm d}t+\Phi_t{\rm d}W_t\tag3,$$ where $W$ is any $Q$-Wiener process on another Hilbert space $U$ and $\varphi:[0,T]\times H\to H$ and $\Phi:[0,T]\times H\to\operatorname{HS}\left(Q^{\frac12}U,H\right)$ are nice enough. Let $g:[0,T]\times H\to H$ be smooth enough for the following and define $f(t,x):=\|g(t,x)\|_H^2$. By the Ito formula, \begin{equation}\begin{split}\|g(t,X_t)\|_H^2&=\|g(0,X_0)\|_H^2\\&\;\;\;\;+2\int_0^t\langle g(s,X_s),\partial_tg(s,X_s)\rangle_H\:{\rm d}s\\&\;\;\;\;+2\int_0^t\langle g(s,X_s),{\rm D}_xg(s,X_s)\varphi_s\rangle_H\:{\rm d}s\\&\;\;\;\;+\frac12\int_0^t\operatorname{tr}\left[Q_sQ_s^\ast{\rm D}_x^2f(s,X_s)\right]\:{\rm d}s\\&\;\;\;\;+2\int_0^t\langle g(s,X_s),{\rm D}_xg(s,X_s)\Phi_s\rangle_H\;{\rm d}W_s,\end{split}\tag4\end{equation} where I defined $Q_s:=\Phi_sQ^{\frac12}$. This expression seems to be tremendously complicated to compute/estimate numerically. Especially the trace term is bothering me. We can write \begin{equation}\begin{split}&\operatorname{tr}\left[Q_sQ_s^\ast{\rm D}_x^2f(s,X_s)\right]=\\&\;\;\;\;2\sum_{n\in\mathbb N}\left\langle Q_s^\ast\left({\rm D}_x^2g(s,X_s)Q_se_n\right)^\ast g(s,X_s),e_n\right\rangle_H\\&\;\;\;\;+2\left\|{\rm D}_xg(s,X_s)Q_s\right\|_{\operatorname{HS}(U,\:H)}^2,\end{split}\end{equation} where $(e_n)_{n\in\mathbb N}$ is any orthonormal basis of $U$. I have no idea how I can simplify the first summand on the right-hand side. But even if I could, it seems still be hard to compute/estimate this in practice.
As written in the question, I only need the expectation of this expression. Maybe it simplifies in a useful manner?
EDIT2
Let's get back to (1). Consider the slightly more general form $${\rm d}U_t=\kappa(t)AU_t{\rm d}t+{\rm d}W_t\tag5,$$ where $A=\Delta u$ and $W$ is a $Q$-Wiener process for some nonegative self-adjoint $Q\in\mathfrak L(H)$. Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ with $$-Ae_n=\lambda_ne_n\tag6$$ and $$\frac\partial{\partial\nu_{\partial_{[0,\;1)^2}}}e_n=0$$ for all $n\in\mathbb N$ for some nondecreasing $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$ with $\lambda_n\xrightarrow{n\to\infty}\infty$. Assume $$\langle Qe_i,e_j\rangle_H=\delta_{ij}\sigma_i^2\;\;\;\text{for all }i,j\in\mathbb N\tag7$$ for some $(\sigma_n)_{n\in\mathbb N}\subseteq(0,\infty)$. Then, by the Ito formula, we obtain \begin{equation}\begin{split}\|g(t,U_t)\|_H^2&=\|g(0,U_0)\|_H^2\\&\;\;\;\;+2\int_0^t\langle g(s,U_s),\partial_tg(s,U_s)\rangle_H\:{\rm d}s\\&\;\;\;\;+2\int_0^t\kappa(s)\langle g(s,X_s),{\rm D}_xg(s,U_s)AU_s\rangle_H\:{\rm d}s\\&\;\;\;\;+\int_0^t\sum_{n\in\mathbb N}\sigma_n^2\left\langle g(t,x),\left({\rm D}_x^2g(s,U_s)e_n\right)e_n\right\rangle_H\;{\rm d}s\\&\;\;\;\;+\int_0^t\sum_{n\in\mathbb N}\sigma_n^2\left\|{\rm D}_xg(t,x)e_n\right\|_H^2\;{\rm d}s\\&\;\;\;\;+2\int_0^t\langle g(s,U_s),{\rm D}_xg(s,U_s)\rangle_H\;{\rm d}W_s.\end{split}\tag8\end{equation} The question is: If $g(t,\;\cdot\;)=\nabla\ln p_t$, can $(8)$ be simplified further? Maybe using the Fokker-Planck equation as suggested in an answer below?
I think your question is essentially finite dimensional. To make my life easier I am going to assume that $\kappa(t)=at +b$ for strictly positive $a,b$. You can adapt the argument for another set of assumptions. Let $e_i, i=1,2 \dots \,$ denote the eigenfunctions of the Neumann Laplacian $-\Delta$. Then, $X_t=\pi_d U_t$ solves the following SDE: \begin{align} dX_t = -\kappa(t) \Lambda_d X_t \, dt + \sqrt{2}dB_t^d \,, \end{align} where $B_t^d$ is standard d-dimensional Brownian motion and $\Lambda_d$ is diagonal matrix with entries $(\Lambda_d)_i=\lambda_i$, with $\lambda_i\geq0, i=1, \dots,d$ the first $d$ eigenvalues of the $-\Delta$ on your domain with Neumann boundary conditions. Note that these are explicitly computable. I have used the fact that $W_t=\sqrt{2}\sum_{i=1}^\infty e_i B_t^i$ (I have added the $\sqrt{2}$ so that I have a better normalisation) where the $B_t^i$'s are independent one dimensional Brownian motions (this is a pretty standard representation of cylindrical Wiener processes).
For convenience, I will write $p_t$ for both the measure and its density. Note now that $p_t= \mathrm{Law}(X_t)$ and since $X_t$is just a diffusion process (or more specifically an Ornstein--Uhlenbeck process with time-dependent drift), $p_t$ is the solution of the following Fokker--Planck equation \begin{align} \partial_t p_t = \Delta p_t + \kappa(t)\nabla \cdot ((\Lambda_d \cdot x) p_t) \, . \end{align} Furthermore, the observable whose evolution you are interested in also has a simple representation, i.e. \begin{align} \mathbb{E}[|\nabla \ln p_t (\pi_d U_t)|^2] = & \mathbb{E}[|\nabla \ln p_t (X_t)] \\ =&\int_{\mathbb{R}^d}|\nabla \ln p_t|^2 \, \mathrm{d}p_t \, . \end{align} The above term is the Fisher information of the law with respect to the Lebesgue measure. It is more convenient to compute the evolution of the following quantity \begin{align} \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t \end{align} The evolution of the above quantity along solutions of Fokker--Planck equations is relatively well-understood. The only real difference is that the potential depends on time through the presence of the function $\kappa(t)$. After a rather long and tedious calculation one can see that \begin{align} &\frac{d}{dt} \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t\\=& -2 \int_{\mathbb{R}^d}\mathrm{tr}\left(D^2 \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\cdot\left[ D^2 \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right]^T\right) \, \mathrm{d}p_t \\ &- 2 \int_{\mathbb{R}^d} \left \langle \nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}, \kappa(t)\Lambda_d \nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)} \right\rangle \, \mathrm{d}p_t \\ &+2 \kappa'(t)\sum_{i=1}^d\int_{\mathbb{R}^d}\lambda_i\partial_{x_i} \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)} \, \mathrm{d}p_t \, . \end{align} The only novel part of this calculation is the third term, the first two show up in the time-independent case as well. If you want a reference look at equation (21) from the Markowich and Villani (see https://cedricvillani.org/sites/dev/files/old_images/2012/07/P01.MV-FPReview.pdf). Let us now try and bound the terms. The first term is lesser than or equal to $0$ while for the second term we apply the bound \begin{align} &- 2 \int_{\mathbb{R}^d} \left \langle \nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x ,\Lambda_d x \rangle}{2} \right)}, \kappa(t) \Lambda_d \nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)} \right\rangle \, \mathrm{d}p_t \\ \leq & -2 \lambda_* b \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t \, , \end{align} where $\lambda_* = \min_{i=1,\dots,d} \lambda_i$. For the last term, we apply Young's inequality to obtain \begin{align} &2 \kappa'(t)\sum_{i=1}^d\int_{\mathbb{R}^d}\lambda_i\partial_{x_i} \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)} \, \mathrm{d}p_t \\ \leq & + \lambda_* b \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t + \frac{ a^2}{\lambda_*^2 b^2}\sum_{i=1}^d \lambda_i^2\, . \end{align} I have assumed $\lambda_*>0$ which is true because the Neumann Laplacian has a spectral gap. We can now close the estimate using Gronwall's inequality to obtain the bound \begin{align} \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t \leq C \, , \end{align} for some explicit constant which is independent of time but depends on the eigenvalues $\lambda_i$. Now to treat the original quantity, we note that (using $|a+b|^2 \leq 2a^2 + 2 b^2$) \begin{align} \int_{\mathbb{R}^d}\left|\nabla \ln p_t\right|^2 \, \mathrm{d}p_t \leq 2 \int_{\mathbb{R}^d}\left|\nabla \ln \frac{p_t}{\exp\left(-\kappa(t)\frac{\langle x , \Lambda_d x \rangle}{2} \right)}\right|^2 \, \mathrm{d}p_t +\frac14 |\kappa(t)|^2 \sum_{i=1}^d \lambda_i^2 \int_{\mathbb{R}^d}|x|^2 \, \mathrm{d}\rho_t \, . \end{align} We are almost done except for the second moment term that shows up in the second term whose behaviour can explicitly controlled by applying Itos formula to for the observable $|X_t|^2$ and taking expectation. I will leave you to do this. I think if you bound stuff brutally you end up with a bound which scales like $\kappa(t)t\sim t^3$.
I think all of these bounds can be improved. As a starting point, the bound of $\kappa(t)\Lambda_d \geq b \lambda_* \mathrm{id}$ seems inefficient and it may make sense to retain the time dependence of $\kappa(t)$.
Edit: Define the function $F: H \to \mathbb{R}$ as $F(u)=\ln p_t( \pi_d \cdot u)$ where I think of $\pi_d \cdot u$ as an element of $\mathbb{R}^d$. Then, its Frechet $D_u F(u)\in H'$ can be (formally) defined for any $\eta \in H$ as follows \begin{align} \langle D_u F(u) , \eta\rangle_{H',H} = (\nabla \ln p_t (\pi_d \cdot u))\cdot(\pi_d\cdot \eta) \, . \end{align}
We would now like to explicitly identify it with an element of $H$. We can do this by noting that \begin{align} \langle D_u F(u) , \eta\rangle_{H',H} = \int_{[0,1)^2} \frac{\delta F}{\delta u}[u](y)\eta(y) \, \mathrm{d}y \, , \end{align} where \begin{align} \frac{\delta F}{\delta u}[u](y) =\sum_{i=1}^d \partial_{x_i} p_t(\pi_d \cdot u) e_i(y) \, . \end{align} Now if we compute the norm of the object above in $H$ evaluate it at $u=U_t$, and take expectation, we obtain \begin{align} &\mathbb{E}\left[\int_{[0,1)^2}\left|\frac{\delta F}{\delta u}[U_t](y)\right|^2 \, \mathrm{d}y\right] \\ =&\int_{\mathbb{R}^d}|\nabla \ln p_t|^2 \, \mathrm{d}p_t \,. \end{align} Note that I have used the fact that the $e_i$'s are orthonormal which implies that \begin{align} \int_{[0,1)^2}\left|\frac{\delta F}{\delta u}[U_t](y)\right|^2 \, \mathrm{d}y = |\nabla \ln p_t (\pi_d\cdot U_t)|^2 \, . \end{align}