compute laplacian of x/|x|

Question

How do I compute

$$\triangle u$$ for $u:\mathbb{R}^n/\{0\} \longrightarrow \mathbb{S}^{n-1}\subset \mathbb{R}^n$ given by $u(x) = \frac{x}{|x|}$

I thought of using $\triangle = \nabla \cdot \nabla u$ but I just got confused.

Answer 1

I would start from the identity $\Delta=\sum_{i=1}^n \partial_{ii}$ so that $$\Delta u(x)=\left(\sum_{i=1}^n \partial_{ii}\right)\left(\sum_{j=1}^n \frac{x_j}{|x|}e_j \right)=\sum_i \sum_j\left(\partial_{ii} \frac{x_j}{|x|} \right)e_j.$$

Now if $j=i$ then $$\partial_{ii}\frac{x_j}{|x|}=\partial_i\left(\frac{|x|-x_i(x_i/|x|)}{|x|^2} \right)=\partial_i \left(\frac{|x|^2-x_i^2}{|x|^3} \right)=\partial_i \left(\frac{\sum_{k\neq i}x_k^2}{|x|^3} \right)=\frac{|x|^3\cdot 0-\left(\sum_{k\neq i}x_k^2\right)\left(3x_i|x|\right)}{|x|^6}=-\frac{\left(\sum_{k\neq i}x_k^2\right)3x_i}{|x|^5}.$$

If $j \neq i$ then $$\partial_{ii}\frac{x_j}{|x|}=\partial_i\left(\frac{|x|\cdot 0 - x_j(x_i/|x|)}{|x|^2}\right)=\partial_i\left(\frac{-x_ix_j}{|x|^3}\right)=\frac{|x|^3\cdot(-x_j)-(-x_ix_j)(3x_i|x|)}{|x|^6}=-\frac{|x|^2x_j-3x_i^2x_j}{|x|^5}=-\frac{x_j(|x|^2-3x_i^2)}{|x|^5}. $$

Now we can hold $j$ fixed and sum over $i$ (this is a finite sum so no convergence issues) to get $$\begin{align}\sum_{i=1}^n\partial_{ii}\frac{x_j}{|x|}&=-\frac{x_j(|x|^2-3x_1^2)}{|x|^5}-\cdots-\frac{x_j(|x|^2-3x_{j-1}^2)}{|x|^5}-\frac{(\sum_{k\neq j}x_k^2)3x_j}{|x|^5}-\ldots-\frac{x_j(|x|^2-3x_n^2)}{|x|^5}\\&=-\frac{x_j}{|x|^5}\left((|x|^2-3x_1^2)+\cdots+(|x|^2-3x_{j-1}^2)+3\left(\sum_{k\neq j}x_k^2\right)+\cdots+(|x|^2-3x_n^2)\right)\\ &=-\frac{x_j}{|x|^5}\left((n-1)|x|^2+3\left(\sum_{k\neq j}x_k^2\right)-3\left(\sum_{k\neq j}x_k^2\right)\right)\\&=-\frac{(n-1)x_j}{|x|^3}.\end{align}$$

Each of these terms is multiplied by $e_j$, so that's what the $j$-th component of $\Delta u(x)$ is, giving a final answer of

$$\Delta u(x)=-\frac{(n-1)x_j}{|x|^3}e_j=-\frac{(n-1)}{|x|^3}x$$