Let $B$ be the unit ball and $S$ the unit sphere in $\mathbb{R}^3$. Consider the map $u: B\rightarrow S$ defined as: \begin{equation} u^j(x)=\frac{x_j}{|x|}\quad\forall \ j =1, 2, 3. \end{equation}I would like to show that for each $j=1, 2, 3,$ we have \begin{equation} -\triangle u^j=|Du|^2u^j\quad \text{in }B\setminus\{0\}. \end{equation}

My working is as below but I can't exactly establish the relationship; I am missing a factor of two and I don't know where I am going wrong.

Firstly, we have \begin{equation} \frac{\partial u^j}{\partial x_i}=\frac{|x|\delta_{ij}-x_jx_i|x|^{-1}}{|x|^2}=\frac{\delta_{ij}}{|x|}-\frac{x_ix_j}{|x|^3} \end{equation}where $\delta_{ij}=1$ if $i=j$ and zero otherwise. Consequently, \begin{align} |Du|^2=\sum_{j=1}^3\sum_{i=1}^{3}\left(\frac{\partial u^j}{\partial x_i}\right)^2&=\sum_{j=1}^3\sum_{i=1}^{3}\frac{|x|^2\delta_{ij}-2\delta_{ij}x_ix_j+|x|^{-2}x_j^2x_i^2}{|x|^4}\\ &=\frac{3}{|x|^2}-\frac{2}{|x|^2}+\frac{1}{|x|^2}\\ &=\frac{2}{|x|^2}. \end{align}

So \begin{equation} |Du|^2u^j=\frac{2x_j}{|x|^3}. \end{equation}

For the left hand side, however, I end up with \begin{align} -\sum_{i=1}^3\frac{\partial}{\partial x_i}\left(\frac{\partial u^j}{\partial x_i}\right)&=\sum_{i=1}^3-\frac{\partial}{\partial x_i}\left(\frac{\delta_{ij}}{|x|}\right)+\frac{\partial}{\partial x_i}\left(\frac{x_ix_j}{|x|^3}\right)\\ &=\sum_{i=1}^3\frac{\delta_{ij}x_i}{|x|^3}+\frac{|x|^3x_j-3|x|x_i^2x_j}{|x|^6}\\ &=\frac{x_j}{|x|^3} \end{align}

Somewhere I am missing a factor of two and I can't spot it.


The first part is correct. In the second, you did not correctly account for the derivative of $x_ix_j$ with respect to $x_i$, in the case $i=j$. Let's separate the two cases: $$ \begin{align} -\frac{\partial}{\partial x_j}\left(\frac{\partial u^j}{\partial x_i}\right) &=-\frac{\partial}{\partial x_j}\left(\frac{1}{|x|}\right)+\frac{\partial}{\partial x_j}\left(\frac{x_j^2}{|x|^3}\right) \\ &=\frac{x_j}{|x|^3}+\frac{2x_j}{|x|^3}+x_j^2 \frac{\partial}{\partial x_j}\left(\frac{1}{|x|^3}\right) \end{align} $$ and for $i\ne j$, \begin{align} -\frac{\partial}{\partial x_i}\left(\frac{\partial u^j}{\partial x_i}\right) & = \frac{\partial}{\partial x_i}\left(\frac{x_ix_j}{|x|^3}\right) \\ & = \frac{x_j}{|x|^3} +x_ix_j \frac{\partial}{\partial x_i}\left(\frac{1}{|x|^3}\right) \end{align} Adding everything and using the identity $$ x_j\sum_{i=1}^3 x_i \frac{\partial}{\partial x_i}\left(\frac{1}{|x|^3}\right) = -\frac{3x_j}{|x|^3} $$ we end up with $1+2+2-3=2$ copies of $x_j/|x|^3$.


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