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Let

  • $U$ and $H$ be separable Hilbert spaces
  • $Q\in\mathfrak L(U)$ be nonnegative and symmetric with finite trace
  • $U_0:=Q^{1/2}U$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(W_t)_{t\ge 0}$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$
  • $X_0$ be a $H$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$
  • $v:[0,\infty)\times H\to H$ be continuously Fréchet differentiable with respect to the first argument and twice continuously Fréchet differentiable with respect to the second argument
  • $\xi:\Omega\times[0,\infty)\times H\to\operatorname{HS}(U_0,H)$ suitable for the following

Assuming that $$X_t=X_0+\int_0^tv_s(X_s)\;{\rm d}s+\int_0^t\xi_s(X_s)\;{\rm d}W_s\;\;\;\text{for }t>0\;,\tag 1$$ I would like to use an Itō-like formula to find an expression for $$Y_t:=v_t(X_t)\;\;\;\text{for }t\ge 0\;.\tag 2$$ However, the Itō formula (see Da Prato, Theorem 4.32) for $(1)$ can only be used to find an expression for $f_t(X_t)$, if $f$ is a continuously partially Fréchet differentiable mapping $[0,\infty)\times H\to\color{red}{\mathbb R}$.

Since I want to derive a SPDE for $Y$, I'm unsure what I need to do. Maybe we can do the following: Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $$v^{(n)}_t(x):=\langle v_t(x),e_n\rangle_H\;\;\;\text{for }n\in\mathbb N\;.$$ Then, each $v^{(n)}$ is continuously partially Fréchet differentiable and we can apply the Itō formula to find an expression for $$Y^{(n)}_t:=v_t^{(n)}(X_t)\;\;\;\text{for }t\ge 0\;.$$ Is this a good idea? We should be able to obtain $(2)$ by $$Y=\sum_{n\in\mathbb N}Y^{(n)}e_n\;.$$ Later, I'm interested in numerically obtaining $Y$.


EDIT: By the referenced version of the Itō formula, we obtain

\begin{equation} \begin{split} \langle v(t,X_t),e_n\rangle_H&=\langle v(0,X_0),e_n\rangle_H+\int_0^t\langle{\rm D}u(s,X_s)(\xi_s(X_s){\rm d}W_s),e_n\rangle_H\\ &+\int_0^t\langle\frac{\partial u}{\partial t}(s,X_s),e_n\rangle_H+\langle{\rm D}u(s,X_s)(u_s(X_s)),e_n\rangle_H\\ &+\frac 12\operatorname{tr}\langle{\rm D}^2u(s,X_s)\left(\tilde\xi_s(X_s)\tilde\xi_s^\ast(X_s)\right),e_n\rangle_H{\rm d}s \end{split}\tag 3 \end{equation}

where $\tilde\xi:=\xi Q^{1/2}$. I've got two questions:

  1. Can we write the infinite system of equations $(3)$ as one equation, as it is possible in the case $H=\mathbb R^d$ and $e_n=n\text{-th standard basis vector}$? It seems like that's the case, cause almost all terms are simply projections to the $n$-th basis vector, but I don't know how I need to deal with the trace term.
  2. In my real application, I have another known expression for $v(t,X_t)$. Thus, my SPDE would be obtained by equating the SPDE of the question with that other expression. Now the question is: I want to solve that SPDE numerically. Is there any recommended method for such a type of equation? [I should note that I will consider something like $H=[L^2(\mathcal V)]^3$ or $H=[H_0^1(\mathcal V)]^3$ for some bounded domain $\mathcal V\subseteq\mathbb R^3$].
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    $\begingroup$ Another book to check is Michel Metivier, Semimartingales. $\endgroup$ – Nate Eldredge Apr 27 '16 at 17:03
  • $\begingroup$ @NateEldredge Thanks for your comment, but the book you've mentioned is related to SDEs not SPDEs. $\endgroup$ – 0xbadf00d Apr 27 '16 at 18:28
  • $\begingroup$ Well, I don't have a copy handy, but I have a note that it discusses stochastic calculus of Hilbert-space valued processes, so I thought it might have an appropriate version of Ito's formula. $\endgroup$ – Nate Eldredge Apr 27 '16 at 18:38
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    $\begingroup$ You are correct, this can be written coordinatewise without any problems. It can also be written in $H$ exactly the same way it is written in $\mathbb{R}$ (except some obvious notation change). There is absolutely no problem with that if the equation coefficients are good. There would be a problem only if there were some unbounded operators in the coefficients, in which case some mild Ito formula could have been used. $\endgroup$ – zhoraster May 2 '16 at 14:37
  • $\begingroup$ @zhoraster Thank you very much for your comment. Please take note of my edit. If $H=\mathbb R^d$, we're in the convenient case where we can choose $(e_1,\ldots,e_d)$ to be the standard basis of the euclidean space. In that case $\langle\text{something},e_n\rangle_H$ is considered to be the $n$-th component of $\text{something}$ and we can write all $d$ equations $(3)$ as one equation (where the derivatives are understood to be taken componentwise). $\endgroup$ – 0xbadf00d May 5 '16 at 12:17
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This longer version of my comment addresses only the first question.

Yes, you can do that. Let, for simplicity, $v$ be independent of $t$, i.e. $v\colon H\to H$, and let it be twice continuously differentiable with bounded derivatives. Then $D^2 v(x)$ can be thought as a bilinear map $H\times H\mapsto H$. On the other hand, $\xi\xi^*(x)$ can be regarded as a trace class operator on $H$. In turn, the space of trace class operators is the closure of $H\otimes H$ with respect to the trace norm. Hence, the map $H\otimes H\mapsto H$ which sends $h\otimes g$ into $D^2v(x) (h,g)$ can be extended to the space of trace class operators by continuity; denote this by $\operatorname{tr}_H(D^2v(x)\cdot)$. This makes the writing $\operatorname{tr}_H\big(D^2v(X_s) \xi Q\xi^*(X_s)\big)$ meaningful as an element of $H$.

So you can write $$ v(X_t) = v(X_0) + \int_0^t Dv(X_s)\Big(v(X_s)\mathrm{d}s + \xi_s(X_s) \mathrm{d}W_s\Big) \\ + \frac12 \int_0^t \operatorname{tr}_H\big(D^2v(X_s) \xi Q\xi^*(X_s)\big)\mathrm{d}s. $$

Long story short, it is just a matter of making some notation to write this in a coordinate-free form on $H$.

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  • $\begingroup$ Did you intend to write "Let [...] $v$ be independent of $[0,\infty)$" (instead of $H$)? $\endgroup$ – 0xbadf00d May 6 '16 at 17:53
  • $\begingroup$ @0xbadf00d, independent of $t$, sure. $\endgroup$ – zhoraster May 6 '16 at 17:56
  • $\begingroup$ You say, that "the space of trace class operators is the closure of $H\otimes H$ with respect to the trace norm". Do you have a reference for that statement? $\endgroup$ – 0xbadf00d May 7 '16 at 17:34
  • $\begingroup$ @0xbadf00d, this seems pretty obvious to me. I will nevertheless give you a reference if I find this. $\endgroup$ – zhoraster May 8 '16 at 6:36
  • $\begingroup$ I'm not so familiar with operator theory. If $H\otimes H$ denotes the tensor product, how can its closure (why is the trace norm even applicable?) be a space of operators? $\endgroup$ – 0xbadf00d May 12 '16 at 14:37

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