Calculation of the Laplacian of a function in $\mathbb{R}^3$.

Question

I have to calculate the Laplacian distributional sense) of the following function $$\frac{e^{i\sqrt{\lambda}|x|}-1}{4\pi|x|}$$ with $\lambda>0$, $x\in\mathbb{R}^3$. I've procedded in the following way $$\Delta\bigg(\frac{e^{i\sqrt{\lambda}|x|}-1}{4\pi|x|}\bigg)=\Delta\bigg(\frac{e^{i\sqrt{\lambda}|x|}}{4\pi|x|}\bigg)+\Delta\bigg(-\frac{1}{4\pi|x|}\bigg)=-\lambda\frac{e^{i\sqrt{\lambda}|x|}}{4\pi|x|}-\delta_0+\delta_0=-\lambda\frac{e^{i\sqrt{\lambda}|x|}}{4\pi|x|}$$ where $\delta_0$ is the distribution of Dirac centred in $0$. Is it right?

Answer 1

The answer is correct. Justification of the formula for $\Delta\bigg(\dfrac{e^{i\sqrt{\lambda}|x|}}{4\pi|x|}\bigg)$ is lacking. One way to make this rigorous is to avoid $\delta_0$ altogether by arguing that the given function is in $W^{2,1}_{\rm loc}(\mathbb R^3)$, and therefore its distributional Laplacian is a locally integrable function given by the pointwise Laplacian. To begin with, the function $$u(x)=\frac{e^{i\sqrt{\lambda}|x|}-1}{4\pi|x|} = \frac{1}{4\pi}(i\sqrt{\lambda}+(i\sqrt{\lambda})^2 |x|+ (i\sqrt{\lambda})^3 |x|^2+\dots)$$ is locally Lipschitz continuous, which means $u\in W^{1,\infty}_{\rm loc}$. Therefore, its distributional gradient is represented by the pointwise gradient. Since $\nabla |x|^p=p|x|^{p-2}x$, we have $$\nabla u(x) = \frac{1}{4\pi}\left((i\sqrt{\lambda})^2 \frac{x}{|x|}+ (i\sqrt{\lambda})^3 (2x)+(i\sqrt{\lambda})^4 (3x)|x|+ \dots\right)$$ Inspecting this function, we find it locally Lipschitz in $\mathbb R^3\setminus \{0\}$. Therefore, it is absolutely continuous on almost every segment parallel to coordinate axes. The pointwise Hessian $D^2u $ consists of two terms: one comes from $\dfrac{x}{|x|}$ and the rest is bounded near the origin. The term coming from $\dfrac{x}{|x|}$ can be computed explicitly: $\dfrac{|x|^2I-x\otimes x}{|x|^3}$, which is in $L^1_{\rm loc}(\mathbb R^3)$. Alternatively, use homogeneity: since $\dfrac{x}{|x|}$ is homogeneous of degree $0$, its derivatives are homogeneous of degree $-1$, hence $\|D^2u(x)\|\le C_1|x|^{-1}+C_2$. Either way, the combination of absolute continuity on lines and the local integrability of partial derivatives implies that $\nabla u\in W^{1,1}_{\rm loc}(\mathbb R^3)$. Therefore, the distributional divergence $\operatorname{div}\nabla u$ is a locally integrable function which is represented by the pointwise divergence of $\nabla u$.