Laplacian in 4-dimensions

Question

I know for 3-D $$\nabla^2 \left(\frac1r\right)=-4\pi\, \delta(\vec{r})\,.$$ I would like to know, what is $$\text{Div}\cdot\text{Grad}\left(\frac{1}{r^2}\right)$$ in 4-Dimensions ($r^2=x_1^2+x_2^2+x_3^2+x_4^2$)?

Answer 1

Let $\sigma_k$ denote the hypersurface area measure on the $k$-sphere. For example, $$\text{d}\sigma_1(\varphi_1)=\text{d}\phi_1\,,\,\,\text{d}\sigma_2(\varphi_1,\varphi_2)=\sin(\varphi_2)\,\text{d}\varphi_1\,\text{d}\varphi_2\,,$$ and $$\text{d}\sigma_3(\varphi_1,\varphi_2,\varphi_3)=\sin(\varphi_2)\,\sin^2(\varphi_3)\,\text{d}\varphi_1\,\text{d}\varphi_2\,\text{d}\varphi_3\,.$$ In general, $$\text{d}\sigma_k(\varphi_1,\varphi_2,\ldots,\varphi_k)=\prod_{j=1}^k\,\left(\sin^{j-1}(\varphi_j)\,\text{d}\varphi_j\right)\text{ for all }k=1,2,3,\ldots\,.$$ For simplicity, write $\Phi_k$ for the angular tuple $\left(\varphi_1,\varphi_2,\ldots,\varphi_k\right)$. The volume element in the polar coordinates of $\mathbb{R}^n$ is given by $$\text{d}\lambda_n(\mathbf{x})=r^{n-1}\,\text{d}r\,\text{d}\sigma_{n-1}\left(\Phi_{n-1}\right)\,,$$ if $\mathbf{x}\in\mathbb{R}^n$ is represented by the polar coordinates $(r,\varphi_1,\varphi_2,\ldots,\varphi_{n-1})$. Define $$\Psi_n(\mathbf{x}):=\frac{1}{\|\mathbf{x}\|_2^{n-2}}\text{ for all }\mathbf{x}\in\mathbb{R}^n\text{ for }n>2\,,$$ where $\|\_\|_2$ is the usual Euclidean norm on $\mathbb{R}^n$.

For a differentiable function $f:\Omega\to\mathbb{R}$, where $\Omega$ is an open region in $\mathbb{R}^n$, the gradient $\boldsymbol{\nabla}f:\Omega\to\mathbb{R}^n$ is given by $$(\boldsymbol{\nabla}f)(\mathbf{x}):=\sum_{j=1}^n\,\left(\frac{\partial f}{\partial x_j}(\mathbf{x})\right)\,\mathbf{e}_j\text{ for all }\mathbf{x}\in\Omega\,,$$ where $\mathbf{x}=\left(x_1,x_2,\ldots,x_n\right)$ and $\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n$ are the usual standard basis vectors of $\mathbb{R}^n$. For a differentiable vector function $\mathbf{v}:\Omega\to\mathbb{R}^n$, the divergence $\boldsymbol{\nabla}\cdot\mathbf{v}:\Omega\to\mathbb{R}$ is the function $$(\boldsymbol{\nabla}\cdot\mathbf{v})(\mathbf{x}):=\sum_{j=1}^n\,\left(\frac{\partial v_j}{\partial x_j}(\mathbf{x})\right)\,,$$ where $\mathbf{v}=\left(v_1,v_2,\ldots,v_n\right)$. The Laplacian operator $\nabla^2$ is defined by $$\nabla^2f:=\boldsymbol{\nabla}\cdot(\boldsymbol{\nabla}f)\,.$$

We want to evaluate $$L(f):=\int_{\mathbb{R}^n}\,\Psi_n(\mathbf{x})\,(\nabla^2f)(\mathbf{x})\,\text{d}\lambda_n(\mathbf{x})$$ for a well behaved function $f$ (say, $f$ is sufficiently fast decaying at large distances from the origin $\boldsymbol{0}_n$ of $\mathbb{R}^n$, at least equipped with second weak derivatives, and with bounded first weak derivatives). We can see that the distribution $\nabla^2\Psi_n$ satisfies $$\int_{\mathbb{R}^n}\,f(\mathbf{x})\,(\nabla^2\Psi_n)(\mathbf{x})\,\text{d}\lambda_n(\mathbf{x})=(-1)^2\,\int_{\mathbb{R}^n}\,\Psi_n(\mathbf{x})\,(\nabla^2f)(\mathbf{x})\,\text{d}\lambda_n(\mathbf{x})=L(f)\,.$$ (The equality above is where the assumption that $f$ is fast decaying at large distances comes into play.) Note that $$\left(\boldsymbol{\nabla}\Psi_n\right)(\mathbf{x}) = -(n-2)\,\left(\frac{\mathbf{x}}{\|\mathbf{x}\|_2^{n}}\right)\text{ for all }\mathbf{x}\neq \boldsymbol{0}_n\,,$$ and so $$(\nabla^2\Psi_n)(\mathbf{x})=-(n-2)\,\left(\frac{n}{\|\mathbf{x}\|_2^n}-n\,\sum_{j=1}^n\,\frac{x_j^2}{\|\mathbf{x}\|_2^{n+2}}\right)=0\text{ for every }\mathbf{x}\neq \boldsymbol{0}_n\,.$$ That is, $$L(f)=\lim_{R\to0^+}\,\int_{\mathbb{R}\setminus B_R(\boldsymbol{0}_n)}\,\Psi_n(\mathbf{x})\,\big(\nabla^2 f\big)(\mathbf{x})\,\text{d}\lambda_n(\mathbf{x})\,,\tag{*}$$ where $B_\rho(\mathbf{y})$ is the open ball centered at $\mathbf{y}\in\mathbb{R}^n$ with radius $\rho>0$.

We can write $$f(\mathbf{x})\,(\nabla^2\Psi_n)(\mathbf{x})=\big(\boldsymbol{\nabla}\cdot(f\,\boldsymbol{\nabla}\Psi_n)\big)(\mathbf{x})-\big((\boldsymbol{\nabla}f)(\mathbf{x})\big)\cdot\big((\boldsymbol{\nabla}\Psi_n)(\mathbf{x})\big)$$ and $$\Psi_n(\mathbf{x})\,(\nabla^2f)(\mathbf{x})=\big(\boldsymbol{\nabla}\cdot(\Psi_n\,\boldsymbol{\nabla}f)\big)(\mathbf{x})-\big((\boldsymbol{\nabla}\Psi_n)(\mathbf{x})\big)\cdot\big((\boldsymbol{\nabla}f)(\mathbf{x})\big)\,.$$ Thus, $$f(\mathbf{x})\,(\nabla^2\Psi_n)(\mathbf{x})=\Psi_n(\mathbf{x})\,(\nabla^2f)(\mathbf{x})+(\boldsymbol{\nabla}\cdot \mathbf{v})(\mathbf{x})\,,$$ where $$\mathbf{v}:=f\,(\boldsymbol{\nabla}\Psi_n)-\Psi_n\,(\boldsymbol{\nabla}f)\,.$$ For $\mathbf{x}\neq \boldsymbol{0}_n$, we obtain $$\Psi_n(\mathbf{x})\,(\nabla^2f)=-(\boldsymbol{\nabla}\cdot \mathbf{v})(\mathbf{x})\,.$$ From (*), we get $$L(f)=-\lim_{R\to0^+}\,\int_{\mathbb{R}^n\setminus B_R(\boldsymbol{0}_n)}\,(\boldsymbol{\nabla}\cdot\mathbf{v})(\mathbf{x})\,\text{d}\lambda_n(\mathbf{x})\,.$$ Using the Divergence Theorem, we obtain $$L(f)=\lim_{R\to 0^+}\,\int_{\partial B_R(\boldsymbol{0}_n)}\,\mathbf{v}(\mathbf{x})\cdot\mathbf{x}\,\|\mathbf{x}\|_2^{n-2}\,\text{d}\sigma_{n-1}(\Phi_{n-1})\,.$$ That is, $$\begin{align}L(f)&=\lim_{R\to 0^+}\,\int_{\partial B_R(\boldsymbol{0}_n)}\,\Big(f(\mathbf{x})\,(\boldsymbol{\nabla}\Psi_n)(\mathbf{x})-\Psi_n(\mathbf{x})\,(\boldsymbol{\nabla}f)(\mathbf{x})\Big)\cdot\mathbf{x}\,\|\mathbf{x}\|_2^{n-2}\,\text{d}\sigma_{n-1}(\Phi_{n-1})\\ &=\lim_{R\to0^+}\,\left(\int_{\partial B_R(\boldsymbol{0}_n)}\,(-n+2)\,f(\mathbf{x})\,\text{d}\sigma_{n-1}(\Phi_{n-1})-\int_{\partial B_R(\boldsymbol{0}_n)}\,\mathbf{x}\cdot(\boldsymbol{\nabla}f)(\mathbf{x})\,\text{d}\sigma_{n-1}(\Phi_{n-1})\right) \\ &=-(n-2)\,f(\boldsymbol{0}_n)\,\int_{\partial B_1(\boldsymbol{0}_n)}\,\text{d}\sigma_{n-1}(\Phi_{n-1})-0=-(n-2)\,\Sigma_{n-1}\,f(\boldsymbol{0}_n)\,,\end{align}$$ where $\Sigma_k$ denotes the hypersurface measure of the unit $k$-sphere. (For example, $\Sigma_1=2\pi$, $\Sigma_2=4\pi$, $\Sigma_3=2\pi^2$, and for every $k=1,2,3,\ldots$, $\Sigma_k=\dfrac{2\pi^{\frac{k+1}{2}}}{\Gamma\left(\frac{k+1}{2}\right)}$, where $\Gamma$ is the gamma function.)

In conclusion, $$\left(\nabla^2\Psi_n\right)(\mathbf{x})=-\frac{2(n-2)\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\,\delta_n(\mathbf{x})\,,\tag{#}$$ where $\delta_n$ is the $n$-dimensional Dirac delta distribution. In particular, $$\left(\nabla^2\Psi_3\right)(\mathbf{x})=-4\pi\,\delta_3(\mathbf{x})\text{ and }\left(\nabla^2\Psi_4\right)(\mathbf{x})=-4\pi^2\,\delta_4(\mathbf{x})\,.$$ In fact, it also makes sense to consider $\Psi_1$, where $\Psi_1(x)=|x|$ for all $x\in\mathbb{R}$. The same formula (#) works and we can readily check that $$\left(\nabla^2\Psi_1\right)(x)=2\,\delta_1(x)\text{ for all }x\in\mathbb{R}\,.$$ For $\Psi_2$, (#) also works trivially, noting that $\Psi_2\equiv 1$ almost everywhere, whence $\nabla^2\Psi_2\equiv 0$ almost everywhere.

If you want to obtain a similar result in the $2$-dimensional case, then you can take $$\Xi(\mathbf{x}):=\ln\big(\|\mathbf{x}\|_2\big)\text{ for all }\mathbf{x}\in\mathbb{R}^2\setminus\{\boldsymbol{0}_2\}\,.$$ Then, $$L(f):=\int_{\mathbb{R}^2}\,\Xi(\mathbf{x})\,(\nabla^2f)(\mathbf{x})\,\text{d}\lambda_2(\mathbf{x})=\int_{\mathbb{R}^2}\,f(\mathbf{x})\,(\nabla^2\Xi)(\mathbf{x})\,\text{d}\lambda_2(\mathbf{x})$$ for all well behaved functions $f:\mathbb{R}^2\to\mathbb{R}$. Note that $$(\boldsymbol{\nabla}\Xi)(\mathbf{x})=\frac{\mathbf{x}}{\|\mathbf{x}\|_2^2}\text{ and }(\nabla^2\Xi)(\mathbf{x})=0\text{ for all }\mathbf{x}\neq \boldsymbol{0}_2\,.$$ We perform the same trick as before to get $$\begin{align}L(f)&=\lim_{R\to0^+}\,\int_{\partial B_R(\boldsymbol{0}_2)}\,\Big(f(\mathbf{x})\,(\boldsymbol{\nabla}\Xi)(\mathbf{x})-\Xi(\mathbf{x})\,(\boldsymbol{\nabla}f)(\mathbf{x})\Big)\cdot\mathbf{x}\,\text{d}\sigma_1(\Phi_1) \\&=f(\boldsymbol{0}_2)\,\int_{\partial B_1(\boldsymbol{0}_2)}\,\text{d}\sigma_1(\Phi_1)-0=\Sigma_1\,f(\boldsymbol{0}_2)=2\pi\,f(\boldsymbol{0}_2)\,. \end{align}$$ That is, $$(\nabla^2\Xi)(\mathbf{x})=2\pi\,\delta_2(\mathbf{x})\,.$$

Answer 2

Since the Fourier transform of a radial function is also a radial function and the transform of $r^\lambda$ is a homogeneous function of degree $-\lambda - n$ for $(-\lambda - n)/2 \notin \mathbb N^0$, $$\mathcal F[r^\lambda] = (r^\lambda, e^{i \boldsymbol x \cdot \boldsymbol \xi}) = C_{\lambda, n} \rho^{-\lambda - n}, \\ \mathcal F[r^{2 - n}] = C_n \rho^{-2}, \\ \mathcal F[\nabla^2 r^{2 - n}] = -\rho^2 \mathcal F[r^{2 - n}] = -C_n, \\ \nabla^2 r^{2 - n} = -C_n \delta(\boldsymbol x).$$ The constant $C_n = 4 \pi^{n/2}/\Gamma(n/2 - 1)$ can be found by taking a simple test function and using the identity $(r^\lambda, \mathcal F[\phi]) = (\mathcal F[r^\lambda], \phi)$.