# Vanishing viscosity method and random forcing

The vanishing viscosity method consists in viewing problem: $$(A) \hspace{1cm} u_t+g(u)_x = 0,\\[2ex]$$ as the limit of the problem: $$(B) \hspace{1cm} u_t+g(u)_x+\nu \varDelta u = 0,\\[2ex]$$ when $$\nu \rightarrow 0$$. So we could find solution of a problem (A) by finding the solution of the problem (B).

But could we do the same if we have random forcing in the system?

In other words, for a two problems given bellow, could we view the problem $$(A_1)$$ as a limit of the problem $$(B_1)$$, when $$\nu \rightarrow 0$$?

$$(A_1) \hspace{1cm} u_t+g(u)_x = f(u)\dfrac{dB}{dt},\\[2ex]$$ $$(B_1) \hspace{1cm} u_t+g(u)_x + \nu \varDelta u = f(u)\dfrac{dB}{dt},\\[2ex]$$

where B stands for Brownian motion and $$\nu$$ for vanishing viscosity coefficient.

What I have founded so far: In the paper [Kim2011], on the page 1651 (second page in the paper), the author comments a problem similar to ($$B_1$$). He says "By taking the limit as the viscosity coefficient $$\nu$$ tends to zero, we cannot directly obtain a strong solution because we have only weak convergence of viscosity solutions as random functions, and we do not have pointwise convergence over the probability space. This is in contrast to the case of deterministic equations."

My interpretation of this: Pointwise convergence in deterministic problems is equivalent to almost sure convergence in stochastic problems. In stochastic problem $$(B_1)$$ for every different $$\omega$$ we have different value of Brownian motion so we can't have a.s. convergence. But we may have convergence in the distribution now? Or I have missed the point completely?

More generally, I am very interested to find out more about taking limits when we have random function in the system, so if anyone know references in the literature that deals with this, write it down (or send me some other insights you have about this topic).