Consider a simple SPDE as follows: 


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


$u(t,0)=u(t,1)=0$, 


$u(0,x)=v(x)$, 


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. 



On the solution of a stochastic partial differential equation
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.


