# Sampling strategies to solve a stochastic partial differential equation

Suppose I had a stochastic partial differential equation of the form:

$\nabla^2U=F(x,D)$, where $x\in\Omega\equiv [0,1]$ and $F(x,D)$ is a function which depends on position $x$ and a uniform random coefficient D.

An approximate the stochastic solution can be easily obtained by monte-carlo simulations. That is, I can choosing different values of D randomly, obtaining the approximate solution of the PDE by some numerical scheme, then accumulate these solutions in histogram bins in the $X-U$ space.

Now then, let's suppose that the function $F$ depends not on a single random parameter, but on a vector of random parameters $\vec{D}$. For simplicity, let's assume that each component of $\vec{D}$ has the same probability distribution. Let's further assume that each component of F is a function of a single, unique random parameter only. Then,

1. Would a monte-carlo approach require substantially more samples than in the 1 random parameter case? If so, how can I quantify the rate of convergence in terms of the sample size and number of random parameters?

2. Is there a more efficient approach to approximate the solution of the stochastic pde which requires fewer random samples in the case of the vector of random parameters?

Update:

I found this presentation (slide #50), which proposes a similar problem to the one I proposed above. Slide #52 proposes a sparse grid approach to resolving this problem, but does not give too much detail about this approach. I'm really curious about the details of its implementation and whether it can be applied to the problem I proposed above.

• Is the solution to your PDE separable in terms of any of the random variables in the vector? If so, you can likely solve it for each component and then combine the results at the end. Is each variable in the vector independent from the rest? – Costis May 28 '12 at 11:38
• I'm working with the case where each component of the RHS vector is a function of a unique random variable which does not appear in any other component. – Paul May 28 '12 at 13:05
• But can you separate any of the variables in the solution U? i.e. write $U(x,D_1,D_2,...)=A(x)B(D_1)C(D_2)$ or similar? – Costis May 29 '12 at 1:32
• In general, no. But I'm curious to know how it would work in the variable separable case. – Paul May 29 '12 at 14:19
• I'm not certain as I haven't really done this before, but in the separable variable case, I would think that you could run the stochastic simulation taking only one random variable into consideration at a time and combine the results. – Costis May 30 '12 at 9:32