Newton potential for Neumann problem on unit disk

Question

Problem: Show that

$$G(x, y) = \frac{1}{2\pi} \left( \log \lvert x - y \rvert + \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \right)$$

is a Green's function for the Neumann problem on the unit disk.

Attempt at solution: I was able to show that

\begin{equation} \begin{split} \Delta \log \lvert x - y \rvert &= \nabla \cdot \nabla \log \lvert x - y \rvert \\ &= \nabla \cdot \frac{\nabla \lvert x - y \rvert}{\lvert x - y \rvert} \\ &= \nabla \cdot \frac{(x - y) \cdot \nabla (x - y)}{\lvert x - y \rvert^2} \\ &= \nabla \cdot \frac{x - y}{\lvert x - y \rvert^2} \\ &= \nabla \cdot \frac{\widehat{(x - y)}}{\lvert x - y \rvert} \\ &= 2 \pi \nabla \cdot \frac{\widehat{(x - y)}}{2 \pi \lvert x - y \rvert} \\ &= 2 \pi \delta^2 (x - y) \end{split} \end{equation}

and

\begin{equation} \begin{split} \Delta \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert &= \nabla \cdot \nabla \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \\ &= \nabla \cdot \frac{\nabla \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert}{\left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert} \\ &= \nabla \cdot \frac{\left( \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right) \cdot \nabla \left( \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right)}{\left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert^{2}} \end{split} \end{equation}

Hence

\begin{equation} \begin{split} \Delta K(x, y) &= \Delta \frac{1}{2 \pi} \left( \log \lvert x - y \rvert + \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \right) \\ &= \frac{1}{2 \pi} \Delta \log \lvert x - y \rvert + \frac{1}{2 \pi} \Delta \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \\ &= \frac{1}{2 \pi} 2 \pi \delta^2 (x - y) + \frac{1}{2 \pi} \Delta \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \\ &= \delta^2 (x - y) + \frac{1}{2 \pi} \Delta \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert \end{split} \end{equation}

How should I proceed? Is the second term zero?

Answer 1

You want to show that for a fixed $y$ in the unit disk, the function $$u(x) = \log \left\lvert \frac{x}{\lvert x \rvert} - \lvert x \rvert y \right\rvert $$ is harmonic in the unit disk. Consider the change of variable $z = x/|x|^2$, which is a conformal map of the punctured unit disk onto its exterior. This transforms the function into $$\log|z-y| - \log|z|$$ which is harmonic for $|z|>1$, per your first computation.

Since conformal maps preserve harmonicity, we conclude that $u$ is harmonic for $0<|x|<1$. And since $u(x) \to u(0)=0$ as $x\to 0$, the singularity at $0$ is removable.

Answer 2

Your problem statement suffer from atrocious notation. $\log$ instead of $\ln$ is ambiguous, using $x$ and $y$ to represent vectors $\vec x$ and $\vec y$ which themselves have $x$ and $y$ components, and later using $\nabla$ instead of $\vec\nabla_x$ make it ambiguous (the reader might think you mean $\vec\nabla_y$) not to mention the horrible $\Delta$ for $\nabla_x^2$.

So I am going to assume that you mean $$\begin{align}G(\vec x,\vec y)&=\frac1{2\pi}\left(\ln||\vec x-\vec y||+\ln\left|\left|\frac{\vec x}{||\vec x||}-||\vec x||\vec y\right|\right|\right)\\ &=\frac1{4\pi}\left(\ln\left(x_1^2+x_2^2-2x_1y_1-2x_2y_2+y_1^2+y_2^2\right)+\ln\left(1-2x_1y_1-2x_2y_2+\left(x_1^2+x_2^2\right)\left(y_1^2+y_2^2\right)\right)\right)\\ &=\frac1{4\pi}\left(\ln(x^2-2\vec x\cdot\vec y+y^2)+\ln(1-2\vec x\cdot\vec y+x^2y^2)\right)\end{align}$$ Where we have written $||\vec x||^2=x_1^2+x_2^2=x^2$ and $||\vec y||^2=y_1^2+y_2^2=y^2$. Then $$\vec\nabla_xG=\frac1{4\pi}\left(\frac{2\vec x-2\vec y}{x^2-2\vec x\cdot\vec y+y^2}+\frac{-2\vec y+2y^2\vec x}{1-2\vec x\cdot\vec y+x^2y^2}\right)$$ On the unit circle, $||x||^2=x^2=1$ so $\hat n=\frac{\vec x}{||\vec x||}=\vec x$ and $$\begin{align}\vec\nabla_xG\cdot\hat n&=\frac1{2\pi}\left(\frac{(1+y^2)\vec x-2\vec y}{x^2-2\vec x\cdot\vec y+y^2}\right)\cdot\vec x\\ &=\frac1{2\pi}\left(\frac{x^2-2\vec x\cdot\vec y+x^2y^2}{x^2-2\vec x\cdot\vec y+y^2}\right)\\ &=\frac1{2\pi}\end{align}$$ Then $$\begin{align}\nabla_x^2G=\vec\nabla_x\cdot\vec\nabla_xG&=\frac1{2\pi}\left(\frac2{x^2-2\vec x\cdot\vec y+y^2}-\frac{(\vec x-\vec y)\cdot(2\vec x-2\vec y)}{(x^2-2\vec x\cdot\vec y+y^2)^2}+\frac{2y^2}{1-2\vec x\cdot\vec y+x^2y^2}-\frac{(y^2\vec x-\vec y)\cdot(2y^2\vec x-2\vec y)}{(1-2\vec x\cdot\vec y+x^2y^2)^2}\right)\\ &=\frac1{2\pi}\left(\frac{2x^2-4\vec x\cdot\vec y+2y^2-2x^2+4\vec x\cdot\vec y-2y^2}{(x^2-2\vec x\cdot\vec y+y^2)^2}+\frac{2y^2-4y^2\vec x\cdot\vec y+x^2y^4-2y^4x^2+4y^4x^2-2y^2}{(1-2\vec x\cdot\vec y+x^2y^2)^2}\right)\\ &=0\end{align}$$ Where it is defined. While $\ln\left|\left|\frac{\vec x}{||\vec x||}-||\vec x||\vec y\right|\right|$ is regular in the unit disk, $\ln||\vec x-\vec y||$ is undefined when $\vec x=\vec y$ so we can use the divergence theorem to find $$\begin{align}\int\int_{||\vec x||<1}\nabla_x^2\frac1{2\pi}\ln||\vec x-\vec y||d^2\vec x&=\int_{||\vec x||=1}\vec\nabla_x\frac1{2\pi}\ln||\vec x-\vec y||\cdot\hat n\,ds\\ &=\frac1{2\pi}\int_0^{2\pi}\frac{\vec x-\vec y}{x^2-2\vec x\cdot\vec y+y^2}\cdot\vec x\,d\theta\\ &=\frac1{2\pi}\int_0^{2\pi}\frac{1-y_1\cos\theta-y_2\sin\theta}{1-2y_1\cos\theta-2y_2\sin\theta+y_1^2+y^2_2}\,d\theta\end{align}$$ Ran out of time, but hopefully the reader can show that this last is equal to $1$.

EDIT: Just got some time back. Let $y_1=r\cos\phi$ and $y_2=r\sin\phi$. Then $y_1^2+y_2^2=r^2$ and $y1\cos\theta+y_2\sin\theta=r\cos(\theta-\phi)$ and $$\begin{align}\frac1{2\pi}\int_0^{2\pi}\frac{1-y_1\cos\theta-y_2\sin\theta}{1-2y_1\cos\theta-y_2\sin\theta+y_1^2+y_2^2}d\theta&=\frac1{2\pi}\int_0^{2\pi}\frac{1-r\cos(\theta-\phi)}{1-2r\cos(\theta-\phi)+r^2}d\theta\\ &=\frac1{2\pi}\int_0^{2\pi}\frac{1-r\cos\nu}{1+2r\cos\nu+r^2}d\nu\\ &=\frac1{2\pi}\int_0^{2\pi}\left(\frac12+\frac{\frac12(1-r^2)}{1+2r\cos\nu+r^2}\right)d\nu\\ &=\frac12+\frac1{2\pi}\frac{(1-r^2)}{2(1+r^2)}\int_0^{2\pi}\frac{d\nu}{1+e\cos\nu}\end{align}$$ Where we have used the substitutions $\nu=\theta-\phi-\pi$ and $e=\frac{2r}{1+r^2}$ and used the periodicity of the integrand to remap its limits. Now we have showed that $$\begin{align}\int_0^{2\pi}\frac{d\nu}{1+e\cos\nu}&=\int_0^{2\pi}\frac{dE}{\sqrt{1-e^2}}=\left.\frac E{\sqrt{1-e^2}}\right|_0^{2\pi}\\&=\frac{2\pi}{\sqrt{1-e^2}}=\frac{2\pi}{\sqrt{1-\left(\frac{2r}{1+r^2}\right)^2}}=\frac{2\pi(1+r^2)}{1-r^2}\end{align}$$ Where $$\sin E=\frac{\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$ is the eccentric anomaly. So that shows that in fact $$\frac1{2\pi}\int_0^{2\pi}\frac{1-y_1\cos\theta-y_2\sin\theta}{1-2y_1\cos\theta-y_2\sin\theta+y_1^2+y_2^2}d\theta=1$$ Since $\nabla_x^2G(\vec x,\vec y)=0$ for $\vec x\ne\vec y$ and $$\int\int_{||\vec x||<1}\nabla_x^2G(\vec x,\vec y)d^2\vec x=1$$ it follows that $\nabla_x^2G(\vec x,\vec y)=\delta^2(\vec x,\vec y)$. Since we also showed that the normal derivative of $G(\vec x,\vec y)$ is $\frac1{2\pi}$ at the boundary of the unit disk, I think that should do it. But shouldn't the normal derivative be $0$?