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.


1 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.

  • $\begingroup$ The above are quite useful to me. But still one thing confuses me. In the paper,$$p(x,t;y,s)=\frac{\partial G}{\partial N(y,s)}(x,t;y,s)$$ where $N(y,s)=A(y,s)n(y)$ with $n(y)$ is the unit inner normal to $\partial D$ at $y$. However, I can not get the same answer from the above result. $\endgroup$ Jan 4, 2018 at 9:53
  • $\begingroup$ @chengchengling: I saw this notation in your question and also in the paper you linked to. I can only say one thing about it: I don't understand it, meaning that I don't understand the notation. Can you rewrite it using normal partial derivatives of $G$ with respect to its arguments? $\endgroup$ Jan 4, 2018 at 11:04
  • $\begingroup$ Also, you need to be clear about any hidden assumptions that you are (or the linked paper is) relying on. For example, the only way to get the Poisson kernel in the first equation of your question is to choose $G$ that would vanish on $\partial \Omega$ in the appropriate argument. Otherwise, the general formula that I derived shows that extra terms would need to be taken into account. $\endgroup$ Jan 4, 2018 at 11:07
  • $\begingroup$ OK, I see. I would read the paper and your answer carefully to figure out the relationship. Firstly I need to check if I fully understand or not. Thank you sincerely! $\endgroup$ Jan 4, 2018 at 11:36

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