Green's functions/fundamental solution to a non-constant coefficients pde

Question

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.

Answer 1

First, unless I misinterpret what you wrote, your integral representation for $u(t,x)$ is not correct. At the very least, it's missing boundary terms. Without boundary terms the formula obviously fails for any $u(t,x)\ne 0$ with $Lu(t,x)=0$. Whatever the right boundary term is, that will give you the $L$-analog of the Poisson kernel.

Here's how to find it. First, you need the formal adjoint $L^*$ of the operator $L$. Everybody knows how to find the formal adjoint, by using integration by parts. But what that really implies is that there exists a pair of bidifferential operators $V[-,-]$ and $W[-,-]$, linear in each argument such that $$ v L[u] - L^*[v] u = \partial_t V[v,u] + \operatorname{div}_x W[v,u] . $$ Basically, $V$ and $W$ collect all the boundary terms that you get by moving all the derivatives from $u$ to $v$ using integration by parts. If that's confusing, see this recent answer for some explanation. In your case, \begin{align*} L&=\frac{\partial}{\partial t}+\operatorname{div}_x\big(A(x,t)\nabla_x\big) , \\ L^*&=-\frac{\partial}{\partial t}+\operatorname{div}_x\big(A(x,t)\nabla_x\big) , \\ V[v,u] &= v u , \\ W[v,u] &= v A(t,x) \nabla_x u - u A(t,x) \nabla_x v . \end{align*}

Next, you actually need $G(s,y;t,x)$ to satisfy $$ L_{s,y}^*[G](s,y;t,x) = \delta(s-t) \delta(y-x) . $$ Then, for $(t,x) \in \Omega = [0,T] \times D$, and assuming that $G$ and $u$ are sufficiently regular for all the differentiations to be well defined (perhaps in a weak sense), \begin{align} u(t,x) &= \int_\Omega \delta(s-t) \delta(y-x) u(s,y) \, ds dy \\ &= \int_\Omega L_{s,y}^*[G](s,y; t,x) u(s,y) \, ds dy \\ &= \int_\Omega \left( \partial_s V[u,v] + \operatorname{div}_y W[u,v] \right) \, ds dy + \int_\Omega G(s,y; t,x) L_{s,y}[u](s,y) \, ds dy \\ &= \left. \int_D V[G,u] \right|_{s=0}^T \, dy + \int_{\partial D}\left( \int_0^T n(y)\cdot W[G,u] \, ds \right) dy_{\|} + \int_\Omega G(s,y; t,x) L_{s,y}[u](s,y) \, ds dy , \end{align} where $n(y)$ is the outer normal vector to the boundary $\partial D$ and $dy_{\|}$ is the surface area element of $\partial D$.

So, in your case, the final integral representation of $u(t,x)$ inside $\Omega$ is \begin{multline*} u(t,x) = \left. \int_D G(s,y; t,x) u(s,y) \right|_{s=0}^T \, dy \\ + \int_{\partial D} \int_0^T \left( G(s,y; t,x) \, n(y)\cdot A(s,y)\nabla_y u(s,y) - u(s,y)\, n(y)\cdot A(s,y) \nabla_y G(s,y; t,x) \right) \, ds dy_{\|} \\ + \int_\Omega G(s,y; t,x) L[u](s,y) \, ds dy . \end{multline*} Depending on what properties you require of $u$ and $G$ in the interior and on the boundary of $\Omega$, some of the terms on the right hand side of that identity will vanish. For instance, if $L[u] = 0$, then you are left with a boundary integral representation of the form $$ u(t,x) = \int_{\partial\Omega} p(s,y; t,x) \, (ds dy)_{\|} , $$ where again the $\|$ subscript denotes the appropriate surface area element on $\partial \Omega$. I guess $p(s,y; t,x)$ would be your $L$-analog of a Poisson kernel, which you can read off from the preceding identity.