This is from page 17-18 of Trudinger and Gilbarg

Let $\Omega$ be a domain for which the divergence theorem holds. Let $\Gamma(x-y)$ be the normalised fundamental solution of the Laplace's equation, then Green's representation formula

$$u(y)=\int_{\partial\Omega}\bigg(u\frac{\partial\Gamma}{\partial\nu}(x-y)-\Gamma(x-y)\frac{\partial u}{\partial v}\bigg)\text{d}s+\int_\Omega\Gamma(x-y)\Delta u\text{d}x$$

If $u$ has compact support in $\mathbb{R}^n$, then this yields

$$u(y) = \int \Gamma(x-y)\Delta u(x)\text{d}x$$

I do not really understand this. I assume the first term disappear by taking a domain, say $\Sigma$ is large enough so that $u$ is $0$ on $\partial\Sigma$, but then $\Delta u$ may not necessarily be well defined on $\partial\Omega$, so why does it hold?

  • $\begingroup$ I think the reasonable assumption is that $u$ is compactly supported in $\Omega$. The discussion is a little vague in that regard; I think there's reason to be confused. $\endgroup$ – DisintegratingByParts Jul 8 '14 at 0:42
  • $\begingroup$ @T.A.E. I have trouble with (maybe) a similar argument later in the book too. It makes me grit my teeth. $\endgroup$ – Lost1 Jul 8 '14 at 10:06

If $u$ has compact support in $\mathbb R^n$...

This is what happens when people try to pack too much information into too few words. They meant to say

If $\Omega=\mathbb R^n$ and $u$ has compact support...

Notice that there is no $\Omega$ under the integral on the next line, indicating they are thinking of $\mathbb R^n$.

Why $\mathbb R^n$, and not general $\Omega$? We have that same representation whenever $u$ is compactly supported in $\Omega$. But then it does not quality as frequently useful (the words they use to describe (2.17)). When working within a domain, one would integrate $\Delta u$ against Green's function of that domain, not against $\Gamma$. Green's function enforces the Dirichlet boundary condition, not matter what is plugged in instead of $\Delta u$. Convolution with $\Gamma$ does not; it returns a function with zero boundary values only when given that very special input $\Delta u$.


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