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I have been studying Green's functions for Laplace/Poisson's equation and have been having some trouble on a few things. In Strauss's book he claims the solution to the Dirichlet problem is: $$u(\bf x_0)= \iint_{bdy D} u(\bf{x})\frac{\partial G(\bf{x},\bf{x_0})}{\partial n} \,dS\label{1}\tag{1}$$ But in other texts I have seen it defined as $$u(x) = \int_\Omega G(x,\xi)\Delta u \,dy + \int_{\partial \Omega} u(y) \frac{\partial G(x,\xi)}{\partial n} \,dS\label{2}\tag{2}$$ I must be missing something here, but I cannot pinpoint what exactly. Which definition is the correct one? Are they equivalent? I suspect the first equation is assuming a homogenous PDE.

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  • $\begingroup$ Your suspicion is correct, otherwise they are "equivalent". $\endgroup$ May 13, 2020 at 6:04

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Formula \eqref{2} you give includes the solution to the Dirichlet problem for both equations, i.e. for Laplace's and Poisson's equations.

Precisely, the Dirichlet problem for Poisson's equation reads as $$ \begin{cases} \Delta u(x)=f(x) & x\in \Omega\\ u|_{\partial\Omega}=U(x) & x\in \partial\Omega \end{cases}, $$ while the same problem for Laplace’s equation looks identical except for the fact that in this case $f\equiv0$.
If $f(x)\not\equiv 0$ then $\Delta u(x)\not\equiv 0$ and you have the proper Poisson's equation, thus $$ \int_\Omega G(x,\xi)\Delta u \,dy\neq 0, $$ and formula \eqref{2} holds in full. On the other hand, if $f(x)\equiv 0$ then $\Delta u(x)\equiv 0$ thus you have the proper Laplace's equation and $$ \int_\Omega G(x,\xi)\Delta u \,dy= 0, $$ thus formula \eqref{2} reduces to formula \eqref{1}.

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  • $\begingroup$ Thank you! So in that case, is using the Green's function method as simple as computing the integral, assuming the Green's function is known? $\endgroup$
    – WBSS
    May 13, 2020 at 7:48
  • $\begingroup$ @WBSS yes, the true real difficulty is computing the Green ‘s function for a given domain. And as you may have guessed from \eqref{2}, the directional derivative of Green’s function for a given domain on its boundary is nothing less that the Poisson kernel for that domain. Thus the information provided by Green’s function is truly complete. $\endgroup$ May 13, 2020 at 7:54
  • $\begingroup$ Ah I see, things are a lot clearer now! Thank you! $\endgroup$
    – WBSS
    May 13, 2020 at 8:06

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