A second-order elliptic differential operator $L$ is given by $$ Lu = \sum_{i,j = 1}^n a_{i,j}(x) \frac{\partial^2 u }{\partial x_i \partial x_j} + \sum_{j=1}^n b_j(x) \frac{\partial u}{\partial x_j}$$
where $x\in \mathbb{R}^n$, $b: \mathbb{R}^n \rightarrow \mathbb{R}^n$, and $a: \mathbb{R}^n \rightarrow \mathbb{R}^{n\times n}$.
The martingale problem is to find, for each $x_0 \in \mathbb{R}^n$, a probability measure $P^{x_0}$ over the space of all continuous functions (stochastic processes) $X: [0,\infty) \rightarrow \mathbb{R}^n$ such that $$ P^{x_0}(X_0 = x_0) = 1$$ and for any $f\in C^2(\mathbb{R}^n)$, $$f(X_t) - f(X_0) - \int_0^t L f(X_s)ds$$ is a local martingale under $P^{x_0}$.
Apparently, the martingale problem helps us solve the PDE $$ \begin{cases} u_t = Lu &\text{ in } \mathbb{R}^n \times [0,\infty)\\ u = f &\text{ for } t=0 \end{cases}$$
because the solution is $u(x,t) = \mathbb{E}_{P^x}[f(X_t)]$.
I'm having some trouble seeing the connection. I know the derivation involves showing that the infinitesimal operator on $\mathbb{E}_{P^x}[f(X_t)]$ at $t=0$ is equal to $L$, and that the infinitesimal operator on $u$ is "kind of like $u_t$"... but I don't have an intuition of what this stochastic setup is saying.
And how is this related to a diffusion SDE $dX_t = a(X_t)dt + b(X_t)dW_t$? Are $a,b$ in $L$ somehow related to the drift and variance coefficients in a diffusion equation?