Consider a simple SPDE as follows:

$\partial_t u(t,x)=\partial_x^2 u(t,x)+V(u(t,x))+\dot{W}(t,x)$, $t>0$, $x\in(0,1)$,



where $V$ is a bounded, smooth potential function, $\{W_t, \mathcal{F}_t\}$ is a cylindrical Brownian motion and $\dot{W}$ stands for the standard space-time white noise. $u(t,x)$ is called its solution in the sense of generalized function (or weak solution) if and only if $\forall \varphi \in C^2[0,1]$ satisfying $\varphi(0)=\varphi(1)=0$,

$\langle u(t),\varphi \rangle-\langle v,\varphi \rangle=\int_0^t\langle u(s), \partial_x^2\varphi \rangle ds+\int_0^t\langle V(u(s)), \varphi \rangle ds+W_t(\varphi)$ a.s..

Under my settings this coincides with the mild solution, and can be proved jointly continuous in $t$ and $x$.

I am wondering if this equation still holds when $\varphi$ is exchanged by an $\mathcal{F}_0$-measurable random function:

Given an $\mathcal{F}_0$-measurable r.v. $\Phi$ which takes value in $\{\varphi \in C^2[0, 1]; \varphi(0)=\varphi(1)=0\}$, would the following holds for a weak solution $u$:

$\langle u(t),\Phi \rangle-\langle v,\Phi \rangle=\int_0^t\langle u(s), \partial_x^2\Phi \rangle ds+\int_0^t\langle V(u(s)), \Phi \rangle ds+\int_0^t\Phi dW_s$, a.s..

Thanks a lot.

  • $\begingroup$ looks like a monotone class argument. start with $\Phi=\sum_i \varphi_i 1_{A_i}$ $\endgroup$ – marcoromito Mar 12 '14 at 10:23

The $\sigma-$algebra $\mathcal{F}_0$ is trivial, i.e.$\forall A\in\mathcal{F}_0$, $P(A)=0$ or $1$. So the answer is true. I think the answer is also true if $\varphi$ is a $\mathcal{G}-$measurable function when $\mathcal{G}$ is dependent of the filtration generated by the Brownian motion.


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