Fundamental solution of Laplace's equation

Question

For $n=2$ the fundamental solution of the Laplace's equation is given by $P=-\frac{\log(|x|)}{2\pi}$. But how do I show that $\int_{\mathbb{R}^2}\Delta P \, f \,dx=f(0)$ for all $f\in C^\infty_c(\mathbb{R}^2)$. My Idea is \begin{align} & \int_{\mathbb{R}^2}\Delta Pf \, dx =\lim\limits_{\varepsilon\to 0} \int_{\mathbb{R}^2\setminus B_\varepsilon} \Delta Pf \, dx \\[8pt] = {} &-\lim_{\varepsilon\to 0} \left(\int_{\partial B_\varepsilon}\nabla Pf\cdot n \, dS+\int_{\mathbb{R}^2\setminus B_\varepsilon}\nabla P\cdot \nabla f \, dx \right) \\[8pt] = {} &-\lim_{\varepsilon\to 0} \left( \int_0^{2\pi}\frac{1}{\varepsilon^2}f \, d\varphi+\int_{\mathbb{R}^2\setminus B_\varepsilon}\nabla P\cdot \nabla f \, dx \right) \end{align}

Answer 1

You're idea is on the right track. Here is a way forward if $f$ is differentiable and of compact support.

It can be shown that we can differentiate once (but not twice) under the integral to find

$$\begin{align} \Delta \int_{\mathbb{R}^2}P(\vec x, \vec x')f(\vec x')\,d\vec x'&=\nabla \cdot \int_{\mathbb{R}^2}\nabla P(\vec x, \vec x')f(\vec x')\,d\vec x'\\\\ &=-\nabla \cdot \int_{\mathbb{R}^2}\nabla' P(\vec x, \vec x')f(\vec x')\,d\vec x'\\\\ &=\nabla \cdot \int_{\mathbb{R}^2} P(\vec x, \vec x')\nabla'f(\vec x')\,d\vec x'\\\\ &=- \int_{\mathbb{R}^2} \nabla ' P(\vec x, \vec x')\cdot \nabla'f(\vec x')\,d\vec x'\\\\ &=-\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}^2\setminus B_\varepsilon} \nabla ' P(\vec x, \vec x')\cdot \nabla'f(\vec x')\,d\vec x'\tag1 \end{align}$$

Now, we integrate by parts the integral on the right-hand side of $(1)$ to obtain

$$\begin{align} \Delta \int_{\mathbb{R}^2}P(\vec x, \vec x')f(\vec x')\,d\vec x'&=\lim_{\varepsilon\to 0^+} \oint_{ B_\varepsilon} \hat n'\cdot \nabla ' P(\vec x, \vec x')f(\vec x')\,dS'\\\\ &=\frac1{2\pi }\lim_{\varepsilon\to 0^+} \int_0^{2\pi}\frac1\varepsilon f(\vec x'=\vec x+\varepsilon \hat n')\,\varepsilon\,d\phi'\\\\ &=f(\vec x) \end{align}$$

as was to be shown! And we are done.

Answer 2

You have $$ \int_{\mathbb{R}^2}\Delta Pf\,dx=\int_{\mathbb{R}^2\setminus B_\varepsilon}\Delta Pf\,dx+\int_{B_\varepsilon}\Delta Pf\,dx. $$ Since $\Delta P=0$ outside of $B_\varepsilon$, the first integral is zero. Therefore, integrating by parts, $$ \int_{\mathbb{R}^2}f\Delta P \, dx=\int_{B_\varepsilon}f\Delta P \, dx=-\int_{B_\varepsilon}\nabla P\nabla f \, dx+\int_{\partial B_\varepsilon}f\nabla P\cdot n \, dS_\varepsilon= $$ $$ =-\frac{1}{2\pi}\int_{B_\varepsilon}\frac{x}{\|x\|^2}\cdot \nabla f\,dx-\frac{1}{2\pi}\int_{\partial B_\varepsilon}f\frac{x}{\|x\|^2}\cdot n dS_\varepsilon $$ Now, the first integral approaches zero as $\varepsilon\to 0$ because $\nabla f$ is bounded and $x/\|x\|^2\sim 1/\varepsilon$ while $Area (B_\varepsilon)\sim\varepsilon^2$. As for the second, $x\cdot n=-\|x\|$ and $x\cdot n/\|x\|^2=-1/\varepsilon$. Therefore, $$ -\frac{1}{2\pi}\int_{\partial B_\varepsilon}f\frac{x}{\|x\|^2}\cdot n dS_\varepsilon=\frac{1}{2\pi\varepsilon}\int_{\partial B_\varepsilon}fdS_\varepsilon=f(\xi) $$ for some $\xi\in\partial B_\varepsilon$. In the limit $\varepsilon\to 0$, $f(\xi)\to f(0)$, as desired.