• $d\in\left\{2,3\right\}$
  • $\mathcal V_t\subseteq\mathbb R^d$ be the bounded domain occupied by an incompressible Newtonian fluid at time $t\ge 0$
  • $\Phi_t:\mathcal V_0\to\mathcal V_t$ such that $\Phi_t(x_0)$ is the position at time $t\ge 0$ of a particle which started at position $x_0\in\mathcal V_0$ at time $0$

Then, the velocity field $u_t:\mathcal V_t\to\mathbb R^d$ at time $t\ge 0$ is defined by $${\rm d}\Phi_t(x_0)=\underbrace{u_t\left(\Phi_t\left(x_0\right)\right)}_{\displaystyle =:v_t(x_0)}{\rm d}t\;\;\;\text{for all }x_0\in\mathcal V_0\;.\tag 1$$ Using $(1)$ and the chain rule, we obtain $$\frac{\partial v}{\partial t}(t,x_0)=\left[\frac{\partial u}{\partial t}+(u\cdot\nabla)u\right]\left(t,\Phi_t\left(x_0\right)\right)\;\;\;\text{for all }t\ge 0\text{ and }x_0\in\mathcal V_0\;.\tag 2$$ By the momentum conservation equation $$\frac{\partial v}{\partial t}(t,x_0)=\underbrace{\left[\nu\Delta u-\nabla p\right]}_{\displaystyle =:f}\left(t,\Phi_t\left(x_0\right)\right)\;\;\;\text{for all }t\ge 0\text{ and }x_0\in\mathcal V_0\;,\tag 3$$ we obtain $$\left[\frac{\partial u}{\partial t}+(u\cdot\nabla)u\right](t,x)=\left[\nu\Delta u-\nabla p\right](t,x)\;\;\;\text{for all }t\ge 0\text{ and }x\in\mathcal V_t\;.\tag 4$$ The crucial point in $(4)$ is that we can replace $\Phi_t(x_0)$ of $(2)$ and $(3)$ by a uniquely determined $x\in\mathcal V_t$, since by definition $\mathcal V_t=\left\{\Phi_t(x_0):x_0\in\mathcal V_0\right\}$. Thus, we can easily recast $(4)$ into an equation $$\left[\frac{\partial u}{\partial t}+(u\cdot\nabla)u\right](t)=\left[\nu\Delta u-\nabla p\right](t)\;\;\;\text{for all }t\ge 0\tag 5$$ in a Hilbert space $H$ of "functions in space", e.g. $H=L^2(\mathcal V;\mathbb R^d)$ where $\mathcal V=\bigcup_{t\ge 0}\mathcal V_t$.

Now, I want to add a stochastic forcing to $(1)$ and obtain a stochastic PDE similar to $(5)$ using an Itō formula in place of the chain rule in the derivation. I've tried to consider $${\rm d}\Phi_t(x_0)=u_t\left(\Phi_t\left(x_0\right)\right){\rm d}t+\xi_t\left(\Phi_t\left(x_0\right)\right){\rm d}W_t\;\;\;\text{for all }x_0\in\mathcal V_0\;.\tag 6$$ instead of $(1)$. Therefor, let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $U$ be a separable Hilbert space
  • $Q$ be a bounded, linear, nonnegative and symmetric operator on $U$ with finite trace
  • $(W_t)_{t\ge 0}$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$
  • $\xi:\Omega\times[0,\infty)\times\mathbb R^d\to\operatorname{HS}(U_0,\mathbb R^d)$, where $\operatorname{HS}(U_0,\mathbb R^d)$ is the space of Hilbert-Schmidt operators from $U_0:=Q^{1/2}U$ to $\mathbb R^d$

Using the Itō formula (see Da Prato, Theorem 4.32) we obtain

\begin{equation} \begin{split} {\rm d}u^i(t,x_t)&=\left(\xi_t(x_t){\rm d}W_t\cdot\nabla\right)u^i(t,x_t)\\ &+\left[\frac{\partial u^i}{\partial t}(t,x_t)+\left(u_t(x_t)\cdot\nabla\right)u^i(t,x_t)+\operatorname{tr}\left[\nabla^2u^i(t,x_t)\left(\tilde\xi_t(x_t)\right)\left(\tilde\xi_t(x_t)\right)^\ast\right]\right]{\rm d}t \end{split}\tag 7 \end{equation}

where $x_t:=\Phi_t(x_0)$, $u^i$ is the $i$-th component of $u=(u^1,\ldots,u^d)$, $\nabla^2u^i(t,x_t)$ denotes the Hessian of $u^i$ at $(t,x_t)$ and $\tilde\xi_t(x_t):=\xi_t(x_t)Q^{\frac 12}$.

My goal is to obtain a SPDE as considered by Da Prato. I don't know if my approach is the correct approach to obtain a stochastic Navier-Stokes equation. In any case, I need help in rewriting $(7)$ as an equation in a Hilbert space of functions in $x$. Maybe the trace term can be simplified. And maybe we need to rewrite $(7)$ in a coordinate-free form.

I've asked many questions on MSE (1, 2, 3, 4, 5, 6), but didn't find a solution.

  • $\begingroup$ In eq (1) you are using differentials, but shouldn't the right hand side then also contain the additional terms with $dx_i$ wher the $x_i$ are coordinates on the initial configuration? I don't know if this is relevant to your problem since I'm not familiar with SPDEs, but it seems strange from the classical perspective. $\endgroup$ – Michael Bächtold May 22 '16 at 9:12
  • $\begingroup$ @MichaelBächtold Sorry, I don't understand what you mean. $(1)$ is simply $$\Phi_t(X_0)=\Phi_0(X_0)+\int_0^tv_t(x_0)\;{\rm d}t.$$ $\endgroup$ – 0xbadf00d May 22 '16 at 9:17
  • $\begingroup$ I guess you are only considering time as independent variable, while the initial position $x_0$ is considered constant in that equation. You can do that, while I was considering space coordinates also as independent variables. Btw is there a difference between big X and small x in your previous eq? $\endgroup$ – Michael Bächtold May 22 '16 at 9:34
  • $\begingroup$ @MichaelBächtold No, there should be no big $X$. I've intended to write that $(1)$ is simply $$\Phi_t(x_0)=\Phi_0(x_0)+\int_0^tv_t(x_0)\;{\rm d}t.$$ I've deleted my previous comment in order to prevent confusion. $\endgroup$ – 0xbadf00d May 22 '16 at 10:05

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