Distributional Laplacian of Logarithmic function.

Question

Here is the exercise:

Compute the distributional Laplacian $\left(\text{in }\mathbb{R}^2\right)$ of $d(x,y)=\ln\left(\|(x,y)\|\right)=\ln\left(\sqrt{x^2+y^2}\right)$. Relate your answer to $\delta$ distributions.

Here is what I have tried so far:

We start by defining the tempered distribution

$$ \begin{aligned} d:S(\mathbb{R}^2)&\rightarrow \mathbb{R}\\ \phi(x,y)&\mapsto\int_{\mathbb{R}}\int_{\mathbb{R}}\ln\left(\sqrt{x^2+y^2}\right)\phi(x,y)dxdy. \end{aligned} $$

we know for a fact that $\Delta d[\phi]=d[\Delta\phi]$.

Let's start by writing the definition of distributional Laplacian using polar coordinates:

$$ \begin{aligned} \Delta d[\phi]&=d[\Delta\phi]=\lim_{\epsilon\rightarrow 0}\int_{\mathbb{R}^2-B(\bar{0},\epsilon)}\ln\left(\sqrt{x^2+y^2}\right)\Delta\phi(x,y)dxdy\\ &=\lim_{\epsilon\rightarrow 0}\int_{r>\epsilon}\int_{0}^{2\pi}\ln\left(r\right)\Delta\phi(r\cos(\theta),r\sin(\theta))rd\theta dr\\ &=\lim_{\epsilon\rightarrow 0}\int_{\epsilon}^{\infty}\int_{0}^{2\pi}\ln\left(r\right)\Delta\phi(r\cos(\theta),r\sin(\theta))rd\theta dr. \end{aligned} $$

Now, let's analyze the integrand. As $\epsilon$ approaches $0$, the integrand becomes highly oscillatory, and the contribution to the limit integral comes from an infinitesimally thin region around $r=0$ (i.e. the origin).

As $\epsilon\rightarrow 0$, we can approximate $\ln(r)$ with a re-scaled Dirac delta distribution:

$$ \ln(r)\approx -2\pi \delta(r), $$

where the factor of $-2\pi$ comes from integrating the Dirac delta distribution in polar coordinates. Therefore, we have:

$$ \Delta d[\phi]\approx -\lim_{\epsilon\rightarrow 0}\int_{\epsilon}^{\infty}\int_{0}^{2\pi}2\pi \delta(r)\Delta\phi(r\cos(\theta),r\sin(\theta))rd\theta dr. $$

Simplifying, we get:

$$ \Delta d[\phi]\approx -2\pi \int_{0}^{2\pi} \Delta\phi(0,0)d\theta=-4\pi^2\Delta\phi(0,0). $$

In other words, we have just proved that the distributional Laplacian of $d(x,y)=\ln\left(\|(x,y)\|\right)=\ln\left(\sqrt{x^2+y^2}\right)$ is given by

$$ \Delta d[\phi]=-4\pi^2\Delta\phi(0,0). $$

Could anyone please check if my reasoning is correct?

Answer 1

I'm not sure where you got that approximation from, but suffice to say it belongs in the home of a certain Oscar the Grouch. Your starting point was a good one, but you forgot the most elegant detail in this classic proof: Integration by Parts! Computing, we obtain

$$\int\limits_{\Bbb{R}^2-B_\epsilon(0)}\log r \cdot\Delta \phi(x) \:dx = \oint\limits_{\partial B_\epsilon(0)}\log r \:\nabla\phi(x)\cdot(-\hat{r})\:d\sigma - \int\limits_{\Bbb{R}^2-B_\epsilon(0)}\frac{\hat{r}\cdot\nabla\phi}{r}dx$$

$$\require{cancel}=\oint\limits_{\partial B_\epsilon(0)}\Bigr[\log r \:\nabla\phi(x)-\phi(x)\:\frac{\hat{r}}{r}\Bigr]\cdot(-\hat{r})\:d\sigma + \cancel{\int\limits_{\Bbb{R}^2-B_\epsilon(0)}0\cdot \phi(x)\:dx}$$

by using the formula for del in cylindrical coordinates to obtain the gradient and divergence, respectively, of $\log r$ and $\frac{\hat{r}}{r}$ (the polar terms of the $3$D cylindrical system give us exactly what the derivatives would be in $2$D polar coordinates). Since $\phi$ is smooth by assumption, plugging in $d\sigma = \epsilon \:d\theta$ we obtain

$$\Delta d[\phi] = \lim_{\epsilon\to0^+}\int_0^{2\pi}\phi(\epsilon,\theta)-\epsilon\log\epsilon\:\frac{\partial \phi}{\partial r}(\epsilon,\theta)\:d\theta = 2\pi\phi(0) \equiv 2\pi\delta[\phi]$$

by dominated convergence.