Consider a PDE,
$$\partial_t u -a \nabla u - ru (1-u) = 0$$
at a given snapshot in time. The inverse problem is to find the diffusion coefficient $a \in L^{\infty}$ from a noisy measurement $$Y = \Phi(x) + E$$
Where $Y$ is the observations, $E ∼ N (0, \Gamma)$ for an SPD matrix $\Gamma \in R^{m×m}$ is the observational noise, and $\Phi(x) = Bu \in R^m$ is the forward operator. It can be shown that $\Phi$ is continuous and measurable.
$a : \Omega \to L^{\infty}(D)$ is the variable to be inferred. It is modelled as a RV for a probability space $(\Omega, \mathcal{A}, \mathbb{P})$. i.e
$$ a(\omega, x) ∈ R,\quad ω ∈ Ω, x ∈ D$$
Furthermore, let $u(x,y)$ be the finite element solution (i.e weak solution) of the aforementioned PDE. For a fixed $\omega \in \Omega$, by appropriate conditioning and Lax-Milgram, we can say that there is a unique weak solution for the diffusion coefficient $a(\omega)$ or $u(\omega, .)$ - by abuse of notation since the solution depends on each realization of $a$.
By Bayes theorem we proove that the posterior distribution is defined to be:
$$ \pi_{X|Y} \propto \pi_{Y|X} \pi_X $$
Here's where things get a little confusing:
To perform Bayesian inverse, first have to choose a prior, $\pi_x$. The prior measure can then be discretized by the KL expansion.
For computational purposes, it is much more convenient, to formulate the inverse problem in terms of the RV $(\xi_j)_{j \in N} : \Omega \to [−1, 1]^N$ : With the uniform prior $\mu$ on $[−1, 1]\mathbb{N}$ , and the measurement $\Phi(u(\omega, .)) + E \in R^m$ The goal is to determine $(\xi_j )_{j \in N}$. The diffusion coefficient can then be obtained as $a(\omega, .) = T((\xi_j )_{j \in N})$.
Where
$T((\xi_j)_{j\in \mathbb{N}} = m + \sum_{j\in \mathbb{N}} \xi_j l_j \phi_j(x)$.
$T:[-1,1]^{\mathbb{N}} \to L^{\infty}(D)$
Questions
What exactly is $\Phi(x)$ referred to as quantity of interest, even when we're searching for the joint distribution of $a$ and $y$?
How can KL expansion of the prior encapsulates the posterior as well?
I looked at a similar 1D example $-(a(t)u'(t))' = f(t), t \in D:=[0,1]$:
If we take $a(\omega, t) = 1 + \sum_{j=1}^{\infty} z_j(\omega) \sqrt{l_j}\phi_j(t)$ then the goal would be to find coefficients: $(z_j)_{j=1}^{\infty}$.
Prior is then given in terms of z ($\pi_z$) and likelihood becomes: $\pi_{Y|z} \propto (y - u(z,t_j))$ - for the additive noise model. But this confused me even more.- 2.1: Why is it being formulated in terms of $z$?
- 2.2: What does $u(z, t_j)$ mean in terms of solution of PDE
- 2.3: For the finite element solution, does this mean that the solution needs to go through all points in the mesh - or just the points for which we have observed data? How would it look like for my sample problem, $u(z, [x, y])$?
Why is $a(\omega, \cdot)$ considered to be a random field? The value of a does not depend spatially, it's constant for the entire domain.
