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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.