# Laplacian of the Euclidean Norm

Let $$r:\mathbb{R}^n \rightarrow \mathbb{R}$$ be defined by $$r(x)=\|x\|,$$ where $$\large\|x\|=(x_1^2+\dots+x_n^2)^{\frac{1}{2}}.$$ Compute $$\nabla^2r,$$ the Laplacian of $$r.$$

I perform the following computations, $$\large\partial_{x_i}r(x)=\frac{1}{2} \cdot \frac{2x_i}{(x_1^2+\dots+x_n^2)^{\frac{1}{2}}}=\frac{x_i}{r(x)},$$ so that we get $$\large\partial_{x_i}\partial_{x_i}r(x)=\frac{r(x)-\frac{x_i^2}{r(x)}}{r(x)^2},$$ and putting these together yields \begin{align*}\large\nabla^2r(x) &=\large\partial_{x_1}^2r(x)+\dots+\partial_{x_n}^2r(x) \\ &=\large\frac{r(x)-x_1^2\cdot r(x)^{-1}}{r(x)^2}+\dots+\frac{r(x)-x_n^2\cdot r(x)^{-1}}{r(x)^2}\\ &=\large\frac{n\cdot r(x)-r(x)^{-1}\cdot r(x)^2}{r(x)^2}\\ &=\frac{n-1}{r(x)}.\\ \end{align*}

Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere

## 1 Answer

If $$u(x)=v(|x|)$$ then $$\Delta u=\frac{n-1}{|x|}v'(|x|)+v''(|x|).$$ See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf