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I am following these notes on Green's function for Poisson's equation, which are based on Evan's PDE book.

Let $\Omega \subset \mathbb{R}^n$ be open and bounded. Let $u \in C^2(\overline{\Omega})$ be arbitrary. Fix $x \in \Omega$ and choose $\epsilon > 0$ such that $\textrm{dist}(x, \partial \Omega) < \epsilon$. Define $V_\epsilon \equiv \Omega - B(x, \epsilon)$. Let $\Phi$ be the fundamental solution to Laplace's equation.

$\Phi$ has a singularity at $y = x$ and the domain $V_\epsilon$ was chosen to avoid this point. On page 2 of the notes I linked they claim that $$\lim_{\epsilon \rightarrow 0^+} \int_{V_\epsilon} \Phi(y-x)\Delta u(y) dy = \int_\Omega \Phi(y-x) \Delta u(y) dy.$$

I am unclear on how the above equality is established. How does the integral on the right make sense despite this singularity?

CBBAM
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1 Answers1

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Do you agree that $\int_0^1\frac{1}{\sqrt{r}}\,dr$ makes sense even though the integrand approaches $\infty$ at the origin?

Same story here. The fundamental solution $\Phi$ is locally integrable on $\Bbb{R}^n$, i.e for all $r>0$, we have $\int_{B_r(0)}|\Phi(x)|\,dx<\infty$; this is a trivial check to make by changing to polar coordinates, and I leave this to you to verify. So, with this, the function $y\mapsto |\Phi(x-y)\Delta u(y)|\cdot \mathbf{1}_{V_{\epsilon}}(y)$ is bounded independently of $\epsilon$ by $y\mapsto |\Phi(x-y)\Delta u(y)|$ on $\Omega$, which is a bounded set, so by the above remark on $\Phi$ and since $u\in C^2(\overline{\Omega})$, this function is in $L^1(\Omega)$. Hence, we can apply dominated convergence to take the limit $\epsilon\to 0^+$.

peek-a-boo
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  • Thank you very much! Since $\int_\Omega \Phi(y-x) \Delta u(y) dy$ exists because $\Phi$ is locally integrable and $\Omega$ is bounded, why bother isolating the singularity and introducing $V_\epsilon$ in the first place? – CBBAM May 01 '24 at 18:55
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    @CBBAM so that you can integrate by parts/apply the divergence theorem (which requires a $C^1$ integrand atleast each time you use it) integrability alone is not enough. Hence we need to stay strictly away from the point $x$. Once you’re done with all the IBP, you look at both sides of your equation and see what limits you can take (at this stage when taking limits, you only need whatever is required for DCT). – peek-a-boo May 01 '24 at 19:15
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    See also the middle of my answer here, for a related problem on how to calculate the distributional derivatives the classical way of removing a ball around the singularity and then taking limits appropriately (along with a few other approaches). – peek-a-boo May 01 '24 at 19:22
  • Your comment helped clear up all my confusion, thanks again for all the help! – CBBAM May 01 '24 at 20:04