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}


\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}


\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?


2 Answers 2


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.


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 $0$?

  • $\begingroup$ I didn't choose the notation, but thank you for your answer :) Where did the delta functions go? $\endgroup$
    – user76284
    Commented Apr 19, 2016 at 20:36
  • $\begingroup$ Actually, in quite a few reasonable contexts, it is not at all obligatory to write $\ln$, nor to put arrows over names of vectors, and it is widely accepted to write $\Delta$ for Laplacians and Laplace-Beltrami operators... $\endgroup$ Commented Apr 19, 2016 at 20:53
  • $\begingroup$ I had some more time to show where the $\delta$-function still lives. @paul garrett when $\log$ is used its context must be established. Here it wasn't so bad because the reader could guess that $\log_{10}$ nor $\log_2$ wasn't intended, but when putting such a question to the world, every little bit of clarity helps. $\Delta$ for Laplacians is, however an abomination that may have come from a time when it was awkward to typeset $\nabla^2$. The notation of $x$ and $y$ for vectors which themselves have $x$ and $y$ components--don't have enough characters left for proper rant. [$\vec\nabla_x$!] $\endgroup$ Commented Apr 19, 2016 at 21:11
  • $\begingroup$ I think the notion that there are strict rules about usage, etc., is an artifact of lower-division math, and it might be good to anticipate that not many people actually adhere to whatever rules one may declare. E.g., in fact, $\Delta$ is standard: it is what mathematicians often write. If anything, the use of $\delta^2$ for $\delta$-on-$\mathbb R^2$ is pretty non-standard, and arguably un-necessary. Anyway, it's pretty hard to maintain a prescriptivist world-view in mathematical notation, and kind-of pointless to recommend it. $\endgroup$ Commented Apr 19, 2016 at 21:28
  • $\begingroup$ That this post sat undisturbed for $6$ hours before I touched it, and the fact that it was difficult for me to guess what it meant, and even so I'm not sure whether $\vec\nabla_x$ or $\vec\nabla_y$ was intended can be taken to demonstrate that the intent of the post was murky in the context of this forum. If you make people guess about what you mean, they will mostly just skip it. $\endgroup$ Commented Apr 19, 2016 at 22:15

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