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The Green's function for the Laplacian in the plane is $$G(\mathbf{x},\mathbf{x_0}) = \frac{1}{4\pi}\log(||\mathbf{x}-\mathbf{x_0}||^2)$$

That is to say, $G$ satisfies $$\triangle_{\mathbf{x}}G = \delta(\mathbf{x}-\mathbf{x_0})$$

where $\triangle$ is the Laplacian operator, and $\delta$ is the Dirac function.

In an effort to convince myself of this, I computed by hand the Laplacian of $G$, and obtained $$\triangle_{\mathbf{x}}G = \frac{-1}{2\pi}\frac{2r^2-2r^2}{r^4}$$ where $r= ||\mathbf{x}-\mathbf{x_0}||^2$.

Two questions:

Is this really equal to the $\delta$ function? Certainly it's equal to $0$ for $r \neq 0$, but for $r=0$ can we really say that it blows up? I suppose since $r > 0$, the value tends to $\infty$.

The more important question: Why does the initial factor of $\frac{1}{4\pi}$ even matter? We could have started out with any arbitrary constant and obtained the same property for the Laplacian.

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  • $\begingroup$ The prefactor $\require{cancel}\cancel{1 \over 4\pi}$ must be, indeed $\color{#f00}{-\,{1 \over 2\pi}}$. Usually, the Green function is defined as $\ln\left\vert\,\vec{r} - \vec{R}\,\right\vert$ such that $\nabla^{2}\ln\left\vert\,\vec{r} - \vec{R}\,\right\vert = -2\pi\,\delta\left(\,\vec{r} - \vec{R}\,\right)$. $\endgroup$ – Felix Marin May 15 '17 at 3:18
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We can redefine the problem to circumvent use of the Dirac Delta. So, rather than describing the problem by the expression $\nabla^2 G(\vec \rho|\vec \rho')=\delta(\vec \rho-\vec \rho')$ we describe the function $G(\vec \rho|\vec \rho')$ to satisfy the equations

$$\begin{align} \nabla^2 G(\vec \rho|\vec \rho')&=0\,\,\ \dots\vec \rho\ne \vec \rho'\tag 1\\\\ \lim_{\epsilon\to 0}\oint_{|\vec \rho-\vec \rho'|=\epsilon}\nabla' G(\vec \rho|\vec \rho')\cdot \hat n'\,d\ell'&=1\tag2 \end{align}$$

As determined in the OP, $(1)$ is satisfied for $G(\vec \rho|\vec \rho')=\frac{1}{2\pi }\log(|\vec \rho-\vec \rho'|)$.

To see that this form of $G$ satisfies $(2)$ we see that on $|\vec \rho-\vec \rho'|=\epsilon$, we have $\color{blue}{\nabla' G(\vec \rho|\vec \rho')=\frac{1}{2\pi}\frac{\vec \rho'-\vec \rho}{|\vec \rho-\vec \rho'|^2}}$, $\color{red}{\hat n'=\frac{\vec \rho'-\vec \rho}{|\vec \rho-\vec \rho'|}}$, and $\color{green}{d\ell'=|\vec \rho-\vec \rho'|\,d\phi'}$, where $\phi'\in [0,2\pi]$. Putting everything together reveals

$$\oint_{|\vec \rho-\vec \rho'|=R} \color{blue}{\nabla'G(\vec \rho|\vec \rho')}\cdot \color{red}{\hat n'}\,\color{green}{d\ell'}=\int_0^{2\pi} \color{blue}{\left(\frac{1}{2\pi}\frac{\vec \rho'-\vec \rho}{|\vec \rho-\vec \rho'|^2}\right)}\cdot \color{red}{\left(\frac{\vec \rho'-\vec \rho}{|\vec \rho-\vec \rho'|}\right)}\,\color{green}{\left(|\vec \rho-\vec \rho'|\,d\phi'\right)}=1$$

as was to be shown! In fact, note that we didn't need to let $\epsilon\to 0$. We only require that $\epsilon$ is small enough so that the disk $|\vec \rho-\vec \rho'|=\epsilon$ is contained inside the open domain of interest.


Establishing that Solution to $(1)$ and $(2)$ is the Green (Green's) Function

We show now the motivation for describing the Green (or Green's) function to the problem $\nabla^2 \phi(\rho)=f(\rho)$, where $f$ is a smooth function with compact support. We write $\phi(\vec \rho)$ as the superposition integral

$$\phi(\vec \rho)=\int_S G(\vec\rho|\vec\rho')f(\vec \rho')\,dS'\tag 3$$

We have $\nabla^2 G(\vec \rho|\vec\rho')=0$ for $\vec\rho\ne\vec\rho'$. In addition, we require $\phi$ to satisfy $\nabla^2 \phi(\vec \rho)=f(\vec \rho)$.

Accordingly, we apply the Laplacian to $(3)$ and write

$$\begin{align} \nabla^2 \phi(\vec \rho)&=\nabla \cdot \int_S \nabla G(\vec\rho|\vec\rho')f(\vec \rho')\,dS'\\\\ &=-\nabla \cdot \int_S \nabla' G(\vec\rho|\vec\rho')f(\vec \rho')\,dS'\\\\ &=-\nabla \cdot \int_S \left(\nabla'\left(G(\vec\rho|\vec\rho')f(\vec \rho')\right)-G(\vec\rho|\vec\rho')\nabla'f(\vec \rho')\right)\,dS'\\\\ &=-\nabla \cdot \underbrace{\oint_{\partial S} G(\vec\rho|\vec\rho')f(\vec \rho') \,\hat n'\,d\ell'}_{=0\,\,\text{since}\,f\,\text{is of compact support}}+\nabla \cdot \int_S G(\vec\rho|\vec\rho')\nabla'f(\vec \rho')\,dS'\\\\ &=-\int_S \nabla' G(\vec\rho|\vec\rho')\cdot \nabla'f(\vec \rho')\,dS'\\\\ &=-\int_S \nabla' \cdot \left(\nabla' G(\vec\rho|\vec\rho')f(\vec \rho')-f(\vec \rho')\underbrace{\nabla'^2 G(\vec \rho|\vec \rho')}_{=0}\right)\,dS'\\\\ &=-\underbrace{\oint_{\partial S}\hat n'\cdot\nabla' G(\vec\rho|\vec\rho')f(\vec \rho')\,d\ell'}_{=0,\,\text{since}\,f\,\text{is of compact support}}+\oint_{|\vec \rho-\vec \rho'|=\epsilon}\hat n'\cdot \nabla' G(\vec\rho|\vec\rho')f(\vec \rho')\,d\ell'\\\\ &=\underbrace{\oint_{|\vec \rho-\vec \rho'|=\epsilon}\hat n'\cdot \nabla' G(\vec\rho|\vec\rho')(f(\vec \rho')-f(\vec \rho))\,d\ell'}_{\to 0\,\,\text{as}\,\,\epsilon \to 0}+f(\vec \rho)\oint_{|\vec \rho-\vec \rho'|=\epsilon}\nabla' G(\vec\rho|\vec\rho')\cdot \hat n'\,d\ell' \end{align}$$

Therefore, we see that $\nabla^2 \phi(\vec \rho)=f(\vec \rho)$ if $\nabla^2 G(\vec\rho|\vec\rho')=0$ for $\vec \rho\ne \vec \rho'$ and

$$\lim_{\epsilon \to 0}\oint_{|\vec \rho-\vec \rho'|=\epsilon}\nabla' G(\vec\rho|\vec\rho')\cdot \hat n'\,d\ell'=1$$

This is precisely the description given by $(1)$ and $(2)$ and establishes a way forward without using distributions.

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  • $\begingroup$ There is still a bit of "formal integration by parts" going on in order to obtain your second identity, which is the whole idea of the distribution theory calculation, but still, nice. $\endgroup$ – Ian May 14 '17 at 22:14
  • $\begingroup$ @Ian No, there is not. I'm redefining the problem in a way that circumvents generalized functions. $\endgroup$ – Mark Viola May 14 '17 at 22:29
  • $\begingroup$ How do you arrive at (2) then? $\endgroup$ – Ian May 14 '17 at 22:40
  • $\begingroup$ @Ian I've edited to provide the answer to your question. ;-)) $\endgroup$ – Mark Viola May 14 '17 at 23:45
  • $\begingroup$ What a well-explained and thoughtful answer. Many thanks! $\endgroup$ – Alex May 15 '17 at 13:19
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You need distribution theory to really know what's going on. In the sense of distribution theory, the equation $\Delta G(x) = \delta(x)$ means that whenever $\phi$ is a smooth function with compact support, you have the "formal" equation

$$\int_{\mathbb{R}^2} (\Delta G)(x) \phi(x) dx = \int_{\mathbb{R}^2} \phi(x) \delta(x) dx.$$

The right side is understood by the definition of the delta function to be $\phi(0)$. The left side is understood through the definition of the distributional derivative as meaning

$$\int_{\mathbb{R}^2} G(x) (\Delta \phi)(x) dx.$$

This has meaning because the singularity in $G$ at $x=0$ is integrable.

This also tells you why the constant factor matters: if you replace $G$ with $cG$ for a constant $c \neq 1$ then you get $c \phi(0)$ out of the integration, which is not what you should have for the Green's function.

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    $\begingroup$ Another way to understand what's going on is to consider instead some nascent delta function $\delta_t(\mathbf{x})$ which converges (in an appropriate weak sense) to $\delta(\mathbf{x})$ as $t\to 0^+$. In that case, the requirement that $\int_{\mathbb{R}^2}\delta(\mathbf{x})\,d\mathbf{x}=1$ again fixes the overall normalization constant. $\endgroup$ – Semiclassical May 14 '17 at 21:50
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Here is a non-rigorous argument (along the lines of SemiClassical's comment), which I hope you find satisfying nonetheless:

Without loss of generality, set $\mathbf x_0 = 0$. So $$ G(\mathbf x, \mathbf 0) = \frac 1 {4\pi} \ln r^2 = \frac 1 {2\pi} \ln r,$$ where $r$ is the radial polar coordinate.

Let's consider the integral $$ \int_{B(0, \epsilon)} \nabla^2_{\mathbf x} G(\mathbf x, \mathbf 0) \ dA,$$ where $B(0, \epsilon)$ is the disk of radius $\epsilon$ centred at the origin. Since $$ \nabla^2_{\mathbf x} G(\mathbf x, \mathbf 0) = \delta (\mathbf x),$$ we expect the integral to be equal to one.

Now, let's check this, by "computing" the integral using the divergence theorem: \begin{multline} \int_{B(0, \epsilon)} \nabla^2_{\mathbf x} G(\mathbf x, \mathbf 0) \ dA = \int_{\partial B(0, \epsilon)} \nabla_{\mathbf x} G(\mathbf x, \mathbf 0) . \mathbf n \ dl \\ = \int_{\phi = 0}^{\phi = 2\pi} \frac{\partial }{\partial r} \left( \frac{1}{2\pi} \ln r \right) \big\rvert_{r = \epsilon} .\epsilon d \phi = \frac{1}{2\pi\epsilon}.(2\pi \epsilon) = 1,\end{multline} which is what we want.

The reason why this argument is non-rigorous is that the "function" we're integrating is not a genuine function: it takes "values" of zero and infinity. As Ian explained, we need to think about the delta function as a distribution in order to do this properly. (See also Section 2.2.4 of Evans's PDEs book.) Nonetheless, I hope my intuitive argument manages to convince you that we need the factor of $\frac 1 {4\pi}$ in the Green's function.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Integrate, for example, along a circle in the $\ds{xy}$ plane which encloses the origin. Namely,

\begin{align} \int\nabla^{2}\ln\pars{r}\,\dd^{2}\vec{r} & = \int\nabla\cdot\nabla\ln\pars{r}\,\dd^{2}\vec{r} = \int\braces{\partiald{}{x}\bracks{\partiald{\ln\pars{r}}{x}} - \partiald{}{y}\bracks{-\,\partiald{\ln\pars{r}}{y}}}\,\dd^{2}\vec{r} \\[5mm] & = \int\braces{\nabla \times \bracks{-\,\partiald{\ln\pars{r}}{y}\,\hat{x} + \partiald{\ln\pars{r}}{x}\,\hat{y}}}_{\ z}\,\dd^{2}\vec{r} \\[5mm] &= \int\nabla \times \bracks{-\,\partiald{\ln\pars{r}}{y}\,\hat{x} + \partiald{\ln\pars{r}}{x}\,\hat{y}}\cdot \dd\vec{S} \\[5mm] & = \int\bracks{-\,\partiald{\ln\pars{r}}{y}\,\hat{x} + \partiald{\ln\pars{r}}{x}\,\hat{y}}\cdot\dd\vec{r}\qquad\qquad\pars{~Stokes\ Theorem~} \\[5mm] & = \int\bracks{-\,{y \over r^{2}}\,\hat{x} + {x \over r^{2}}\,\hat{y}}\cdot\dd\vec{r} = -\int{ x\,\dd y - y\,\dd x \over r^{2}} = -\int{\pars{\vec{r} \times \,\dd\vec{r}}_{z} \over r^{2}} \\[5mm] & = -\int{r\verts{\dd\vec{r}} \over r^{2}} = -\,{1 \over r}\ \overbrace{\int\verts{\dd\vec{r}}}^{\ds{2\pi r}} = \bbx{-2\pi}\,,\qquad\quad \nabla^{2}\ln\pars{r} = -2\pi\,\delta\pars{\vec{r}} \end{align}

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Another approach, similar in spirit to my comment on Ian's answer, is to modify the Green's function to $G_a(\mathbf{x}-\mathbf{x}_0)=\dfrac{1}{4\pi}\ln[a^2+(\mathbf{x}-\mathbf{x}_0)^2]$. Then $$\Delta G_a(\mathbf{x}-\mathbf{x}_0)=\left(\partial_x^2+\partial_y^2\right)\left(\frac{1}{4\pi}\ln[a^2+(x-x_0)^2+(y-y_0)^2]\right)=\frac{1}{\pi}\frac{a^2}{(a^2+(\mathbf{x}-\mathbf{x_0}^2)^2}.$$ One can then verify that $\int_{\mathbb{R}^2} \Delta G_a(\mathbf{x}-\mathbf{x}_0)\,dA=1$ for all $a>0$; if the constant factor wasn't there, this integral wouldn't equal 1.

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