Prove that in $\mathbb R^2$, we have the following representation formula for harmonic function $u$:$$u(\vec x_0)=\frac{1}{2\pi}\int_{\partial D}\left(u(\vec x)\frac{\partial}{\partial \vec n}(\ln|\vec x-\vec x_0|)-\ln|\vec x-\vec x_0 |\frac{\partial u}{\partial \vec n}(\vec x)\right)\,\mathrm ds.$$
In class we found the representation formula in $\mathbb R^3$, which is $$u(\vec x_0)=\frac{1}{4\pi}\iint_{\partial D}\left(-u(\vec x)\frac{\partial}{\partial \vec n}\left(\frac{1}{| \vec x-\vec x_0|}\right)+\frac{1}{|\vec x-\vec x_0 |}\frac{\partial u}{\partial \vec n}(\vec x)\right)\,\mathrm dA.$$
In the process we did not use at all the Dirac delta function. I have seen a proof* for the representation formula in $\mathbb R^2$ using that function and Heavyside-function as well, but I did not understand this part: $$v(x)=\frac{1}{2\pi}\log| x- x_0|$$ satisfies the identity $$\Delta v=\delta(x-x_0).$$
Could someone explain with details that part?
Alternatively, how to use the representation formula in $\mathbb R^3$ to prove in $\mathbb R^2$? If you see Chee Han comment, maybe could help although I did not understand his comment.
Please help me with the proof.
Proof*
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Without getting into technical details of generalized functions, here is a classical method to verify the identity.
Assume that $D$ is an open and bounded area in $\mathbb{R}^2$ with $C^1$ boundary. Suppose $B(x_0, 2r_0) \subseteq D$. Since $u \in C^1(\overline{B(x_0, r_0)})$, then there exists $M > 0$ such that $|∇u| \leqslant M$ on $\overline{B(x_0, r_0)}$. Note that $∆v = 0$ on $D \setminus \{x_0\}$. For $0 < ε < \min(r_0, 1)$, applying Green's second identity on $D \setminus B(x_0, ε)$,\begin{align*} 0 &= \int\limits_{D \setminus B(x_0, ε)} (u∆v - v∆u) \,\d x = \int\limits_{\partial (D \setminus B(x_0, ε))} \left( u \frac{\partial v}{\partial \vec{n}} - v \frac{\partial u}{\partial \vec{n}} \right) \,\d s\\ &= \int\limits_{\partial D} \left( u \frac{\partial v}{\partial \vec{n}} - v \frac{\partial u}{\partial \vec{n}} \right) \,\d s - \int\limits_{\partial B(x_0, ε)} \left( u \frac{\partial v}{\partial \vec{n}} - v \frac{\partial u}{\partial \vec{n}} \right) \,\d s, \end{align*} thus$$ \int\limits_{\partial D} \left( u \frac{\partial v}{\partial \vec{n}} - v \frac{\partial u}{\partial \vec{n}} \right) \,\d s = \int\limits_{\partial B(x_0, ε)} u \frac{\partial v}{\partial \vec{n}} \,\d s - \int\limits_{\partial B(x_0, ε)} v \frac{\partial u}{\partial \vec{n}} \,\d s. \tag{1} $$
Because$$ \left. \frac{\partial v}{\partial \vec{n}} \right|_{\partial B(x_0, ε)} = \left. \frac{\partial v}{\partial r} \right|_{\partial B(x_0, ε)} = \left. \frac{\partial}{\partial r} \left( \frac{1}{2π} \ln r \right) \right|_{r = ε} = \frac{1}{2πε} $$ and $u$ is continuous at $x_0$, then$$ \lim_{ε → 0^+} \int\limits_{\partial B(x_0, ε)} u \frac{\partial v}{\partial \vec{n}} \,\d s = \lim_{ε → 0^+} \frac{1}{2πε} \int\limits_{\partial B(x_0, ε)} u \,\d s = u(x_0). $$ Because $|∇u| \leqslant M$ on $B(x_0, ε) \subseteq \overline{B(x_0, r_0)}$, then\begin{align*} &\peq \left| \ \int\limits_{\smash{\partial B(x_0, ε)}} v \frac{\partial u}{\partial \vec{n}} \,\d s \right| = \left| \ \int\limits_{\smash{\partial B(x_0, ε)}} v ∇u · \vec{n} \,\d s \right|\\ &\leqslant \int\limits_{\smash{\partial B(x_0, ε)}} |v∇u| · |\vec{n}| \,\d s \leqslant M \int\limits_{\smash{\partial B(x_0, ε)}} |v| \,\d s\\ &= M \int_{-π}^π \left( -\frac{\ln ε}{2π} \right) · ε \,\d θ = -M ε \ln ε, \end{align*} which implies$$ \lim_{ε → 0^+} \int\limits_{\smash{\partial B(x_0, ε)}} v \frac{\partial u}{\partial \vec{n}} \,\d s = 0. $$ Making $ε → 0^+$ in (1),$$ \int\limits_{\partial D} \left( u \frac{\partial v}{\partial \vec{n}} - v \frac{\partial u}{\partial \vec{n}} \right) \,\d s = u(x_0). $$