help with vector calculus

Question

the question is : how do I prove that: $\nabla^2 (r^n\vec r)=n(n+3)r^{n-2}\vec r$

Answer 1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Use $\ds{\quad\left\lbrace\begin{array}{rcl} \ds{\nabla^{2}\pars{ab}} & \ds{=} & \ds{b\nabla^{2}a + 2\nabla a\cdot\nabla b + a\nabla^{2}b} \\[1mm] \ds{\nabla\,\mathrm{f}\pars{r}} & \ds{=} & \ds{\,\mathrm{f}\,'\pars{r}\,{\vec{r} \over r}} \\[1mm] \ds{\nabla\cdot\pars{c\,\vec{d}}} & \ds{=} & \ds{\nabla c\cdot\vec{d} + c\nabla\cdot\vec{d}} \end{array}\right.}$

\begin{align} \nabla^{2}\pars{r^{n}\,x} & = \nabla^{2}\pars{r^{n}}x + 2\nabla\pars{r^{n}}\cdot \overbrace{\nabla x}^{\ds{\hat{x}}}\ +\ r^{n}\, \overbrace{\nabla^{2}x}^{\ds{0}} \\[3mm] \nabla r^{n} & = n\,r^{n - 1}\,\,{\vec{r} \over r} = n\,r^{n - 2}\,\,\,\vec{r} \\[3mm] \nabla^{2}r^{n} & = \nabla\cdot\nabla r^{n} = n\pars{n - 2}r^{n - 3}\,\,\, {\vec{r} \over r}\cdot\vec{r} + n\,r^{n - 2}\,\times 3 = n\pars{n + 1}r^{n - 2} \\[3mm] \color{#f00}{\nabla^{2}\pars{r^{n}\,x}} & = n\pars{n + 1}r^{n - 2}\,\,x + 2nr^{n - 2}\,\,x = \color{#f00}{n\pars{n + 3}r^{n - 2}\,\,x} \end{align}

---

Similarly, for the $y$ and $z$: $$ \color{#f00}{\nabla^{2}\pars{r^{n}\,\vec{r}}} = \color{#f00}{n\pars{n + 3}r^{n - 2}\,\,\vec{r}} $$