Consider 
$$ Au:=\operatorname{div}\left(y^{\beta}\nabla u\right) \text{ for } (x,y)\in \mathbb{H} $$
and $u|\_{\mathbb{R}}(x,0)=\phi(x)$ and some $\beta\in (0,1)$. For $\phi\in L^{2}(\mathbb{R},dx)$ (in fact less is required) we have a unique solution with
$$\int\_{0}^{\infty}\|\nabla\_{(x,y)}u\|^{2}y dy<\infty$$
(see [Boundary value problems for degenerate elliptic equations and systems](https://arxiv.org/abs/1307.5056)).


I am interested to know if there is a corresponding diffusion $X\_{t}=(X\_{1,t},X\_{2,t})$ s.t.
$$u(z)=E\_{z}[\phi(X\_{\tau\_{\partial \mathbb{H}}})].$$



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> Q: Is my following approach correct? If not or there is a simpler approach, please only provide hints to help me learn. 
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1. The corresponding process is $(B\_{t}, R\_{t}(\beta))=(\mbox{Brownian motion}, \mbox{Bessel}(\beta))$ because it has the same generator
$$\Delta\_{(x,y)}u(x,y)+\frac{\beta}{y}u\_{y}=0.$$
2. In ["On the martingale problem for degenerate"](https://www.ams.org/journals/tran/2015-367-11/S0002-9947-2015-06243-6/), they show that for the degenerate pde $$y\Delta\_{(x,y)}u(x,y)+\beta u\_{y}=0,$$  

there is a corresponding unique-in-law process that solves the martingale problee. Up to a time change, we get the above process.
3. Since the semi-elliptic $Au=0$ has a unique solution, we get that $u(z)=E\_{z}[\phi(X\_{\tau\_{\partial \mathbb{H}}})]$ solves because it solves the martingale problem and taking expectation of Dynkin's formula (see Oksendal pg. 181 "stochastic Dirichlet problem" ).



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> Q2: The main issue here is whether the L2-boundary data causes any issues with uniqueness.
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The Bessel diffusion is not an Ito diffusion but it is equal in law to an Ito diffusion process (see Oksendal on Bessel processes). So the most uniqueness we can hope for is in law.



Divergence form degenerate pde and Feynman Kac
This is just an extended comment, not an actual answer. In fact, I am not sure if I understand your question. Apparently what you are attempting to do is quite standard and rather general.


1. Given any Hunt process $X(t)$, any non-negative or bounded Borel function $\phi$ and any open (or even Borel) set $D$, the function $u(x) = \mathbb{E}^x \phi(X(\tau\_D))$ is $X(t)$-harmonic in $D$; that is, for any pre-compact $U$ with closure contained in $D$ one has $u(x) = \mathbb{E}^x u(X(\tau\_U))$. Here $\tau\_U$ stands for the first exit time of $U$. The same holds true whenever $u$ is well-defined (in the sense that the expectation converges absolutely). This is a simple consequence of the strong Markov property of $X(t)$ (plus loads of technicalities, such as measurability issues), and any textbook on Hunt processes or probabilistic potential theory should be a good reference (I did not check it, but I am virtually sure that both Blumenthal–Getoor and Bliedtner–Hansen contain this statement).
2. If $X(t)$ is actually a "nice" diffusion process, then harmonicity with respect to $X(t)$ is equivalent to harmonicity in the PDE sense, with respect to the differential generator of $X(t)$. This is discussed in a systematic way in a lovely two-volume book *Markov processes* by Dynkin. Alternatively, one can refer to the theory of Itô's diffusions.
3. Existence of a diffusion corresponding to the operator $A$ in your question is quite clear: as you point out, one can simply take a pair of independent processes: the Bessel process and the Brownian motion. No need to employ more advanced tools, such as $A\_2$ weights.
4. For the above process $X(t)$ and $D$ the interior of the half-space, the distribution of $X(\tau\_D)$ is known explicitly: if I am not mistaken, for the starting point is $(x, y)$ the hitting position has density $c(\beta) y^{2\beta} / (y^2 + (x - z)^2)^{d + 2 \beta}$ with respect to the Lebesgue measure $dz$ on the boundary. This is certainly proved by Caffarelli–Silvestre in their famous paper; I am not sure if earlier references also mention this. Hence, $L^2$ (or $L^1 + L^\infty$, or even yet more general) boundary data are perfectly OK.
5. The only non-obvious problem here is continuity of $u$ up to the boundary. In the general case, this is a very delicate question. Here, however, $u$ is given by a convolution of the boundary data $\phi$ with a kernel that is an approximate identity with respect to the $y$ variable. Therefore, $u(\cdot, y)$ converges to $\phi$ in $L^2$ as $y \to 0^+$.


