Distributional Laplacian of Logarithmic function.

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?

• How do you justify $\ln(r)\approx -2\pi \delta(r)$? Commented Aug 28, 2023 at 19:06
• Since $\Delta d[\phi]=-4\pi^2\Delta\phi(0,0)=-4\pi^2(\Delta\delta)[\phi]$ your result would imply that $d=-4\pi^2\delta+h,$ where $\Delta h=0.$ Since that is not the case, your result is incorrect. Commented Aug 28, 2023 at 19:10
• I think that the result should be $\Delta d=-2\pi\delta.$ Commented Aug 28, 2023 at 19:11
• You forgot the most critical step in this classic proof: Before taking the limit, you must integrate by parts. Commented Aug 28, 2023 at 20:18

$$\int\limits_{\Bbb{R}^2-B_\epsilon(0)}\log r \cdot\Delta \phi(x) \:dx = \oint\limits_{\partial B_\epsilon(0)}\log r \:\nabla\phi(x)\cdot(-\hat{r})\:d\sigma - \int\limits_{\Bbb{R}^2-B_\epsilon(0)}\frac{\hat{r}\cdot\nabla\phi}{r}dx$$
$$\require{cancel}=\oint\limits_{\partial B_\epsilon(0)}\Bigr[\log r \:\nabla\phi(x)-\phi(x)\:\frac{\hat{r}}{r}\Bigr]\cdot(-\hat{r})\:d\sigma + \cancel{\int\limits_{\Bbb{R}^2-B_\epsilon(0)}0\cdot \phi(x)\:dx}$$
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
$$\Delta d[\phi] = \lim_{\epsilon\to0^+}\int_0^{2\pi}\phi(\epsilon,\theta)-\epsilon\log\epsilon\:\frac{\partial \phi}{\partial r}(\epsilon,\theta)\:d\theta = 2\pi\phi(0) \equiv 2\pi\delta[\phi]$$