Let $\Omega$ be a domain of a Riemannian manifold $M$. Let $\Omega_{D}$ be the diagonal of $\Omega$, that is,

$\Omega_{D} = \{(x,y)\in \Omega \times \Omega : x=y \}$.

According to my text, Green's function $G$ is defined as follows:

"A continuous function $G:(\bar{\Omega} \times \bar{\Omega}) \rightarrow \mathbb{R}$ is a Green's function of $\Omega$, if $G\mid \{(\Omega \times \Omega)\backslash\Omega_{D}\} \in C^{2}$,

$(\Delta _{y}G)(x,y)=0$

for all $x,y \in \Omega$.

$G\mid \Omega \times \Gamma =0$ (where $\Gamma=\partial \Omega$),

and, near $\Omega_{D}$, $G$ is given by

$G(x,y)=\psi(x,y) + h(x,y)$ ,

where $h\in C^{0}(\bar{\Omega}\times \bar{\Omega})\cap C^{2}(\Omega\times\Omega),$ and

$\psi(x,y)=\frac{1}{c_{n-1}}\biggl\{\begin{array}{11} d^{2-n}(x,y)/n-2,&n>2\\ -\log d(x,y),&n=2 \end{array} $

where $c_{n-1}$ is volume of $(n-1)$-sphere, $d(x,y)$ is distancefunction on $M$."

But according to many other books, Green's function is definde as


where $\delta$ is Dirac's delta function,

and satisfy appropriate boundary condition."

Is these same Green's function? I think the first one is defined as a standard function, the second one is defined as a distribution. But I have never heard that Green's function is a standard function, not a distribution.

I would appriciate your help.

  • $\begingroup$ It seems like there's a typo in the quoted definition. It says $G$ restricted to $\Omega \times \Omega \setminus \Omega_{D}$ is $C^{2}$ and then (mysteriously) $\Delta_{y} G(x,y) = 0$ if $(x,y) \in \Omega \times \Omega$, but presumably what was meant was $(x,y) \in \Omega \times \Omega \setminus \Omega_{D}$. $\endgroup$ – fourierwho Jul 11 '18 at 14:17
  • $\begingroup$ Thank you so much. $\endgroup$ – Manifolds Jul 21 '18 at 2:16

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