We already know the relationship between Green's function and solution to elliptic partial differential equation, i.e $$u(y)=\int_{\partial \Omega}u\frac{\partial G}{\partial n} ds+\int_\Omega G\Delta u dx $$ where $n$ is the unit outward normal , $G$ is Green's function on $\Omega$, and $p=\frac{\partial G}{\partial n}$ is called the Poisson kernel.
So now I'm wondering the case if we have a partial differential operator $$Lu(t,x)=\partial_tu(t,x)+\sum_{i,j}^da_{i,j}(t,x)\frac{\partial^2u(t,x)}{\partial x_i\partial x_j}$$ I know that in this case we call $G$ the fundamental solution if $$LG(s,y;t,x)=\delta(s-t)\delta(x-y)$$ Then we can find that $$u(t,x)=\int G(s,y;t,x)Lu(s,y)dsdy$$ without boundary conditions, if my understanding is correct.
Are there the similar relationship between fundamental solution and Poisson kernel (if this notion is correct) on the boundary? In one paper, $$L=\frac{\partial}{\partial t}+\operatorname{div}\big(A(x,t)\nabla_x\big)$$ where $A$ is a $d\times d$ real symmetric matrix, uniformly elliptic on $\Omega=D\times(0,\infty)$. They have the corresponding Poisson kernel $$p(x,t;y,s)=\frac{\partial G}{\partial N(y,s)}(x,t;y,s)$$ where $N(y,s)=A(y,s)n(y)$ with $n(y)$ is the unit inner normal to $\partial D$ at $y$. It confused me for several days. I really want to figure out how it was calculated.
