Proving that $\nabla ^2 r^n = n(n+1)r^{n-2}$ using index notation

Question

With additional parameters that: \begin{align} \ r^2 &= x_j x_j \\ \end{align} and \begin{align} \ \frac{\partial x_i}{\partial x_j} = \delta_{ij}\\ \end{align}

So, what I managed to figure so far is that: \begin{align} \nabla ^2 r^n &= \nabla \cdot (\nabla r^n ) \\ &=\nabla \cdot (nr^{n-1} \frac{\partial r}{\partial x_i}) \\ \end{align}

where I found that: \begin{align} \ \frac{\partial r}{\partial x_i} = \frac{x_i}{r} \\ \end{align} using the above parameters, but beyond that, I am stuck on how to proceed with the rest of the equation for the proof.

Answer 1

We have $r=|\textbf{x}|$ and $$\frac{\partial r^n}{\partial x_i}=nr^{n-1}\frac{\partial r}{\partial x_i}=x_inr^{n-2}$$ $$\frac{\partial^2 r^n}{\partial x_{i}^2}=n\left(r^{n-2}+(n-2)x_{ i }^2r^{n-4}\right)$$

Thus $$\nabla^2 r^{n}=n\left(3r^{n-2}+(n-2)r^{n-2}\right)=n(n+1)r^{n-2}$$

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Or note that if $f=f(r)$ and $r=|\textbf{x}|$, then $\nabla f=\frac{\partial f}{\partial r}\nabla r=\frac{\partial f}{\partial r}\frac{\textbf{x}}{r}.$ So we have

$$ \nabla ^2 r^n = \nabla \cdot (\nabla r^n )=\nabla \cdot (nr^{n-1} \frac{\textbf{x}}{r})$$ $$=n\left(\textbf{x}\nabla\cdot \left(\frac{r^{n-1}}{r}\right)+r^{n-2}\nabla\cdot \textbf{x}\right)$$ $$=n\left(r(n-2)r^{n-3}+3r^{n-2}\right)=n(n+1)r^{n-2}$$

Answer 2

$ \def\l{r}\def\n{\nabla}\def\o{{\tt1}}\def\p{\partial} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\hess#1#2#3{\frac{\p^2 #1}{\p #2\,\p #3}} $Use multiplication by the transpose to calculate the inner product, i.e. $$\eqalign{ \l^2 &= x^Tx \\ d\l^2 &= 2\l\,d\l = 2x^Tdx \\ d\l &= \l^{-1}\LR{x^Tdx} \\ }$$ Square the objective function before calculating the gradient. $$\eqalign{ \phi &= \l^n \\ \phi^2 &= \LR{\l^2}^n \\ 2\phi\;d\phi &= n\LR{\l^2}^{n-1}\;d\!\LR{\l^2} \\ 2\l^n\,d\phi &= n\l^{2n-2}\,\LR{2x^Tdx} \\ d\phi &= \LR{n\l^{n-2}\;x}^Tdx \\ \grad{\phi}{x} &= n\l^{n-2}\;x \;\;\doteq\; g &\big({\rm the\;gradient}\big) \\ }$$ Now calculate the Hessian (i.e. the gradient of the gradient) $$\eqalign{ dg &= n(n-2)\LR{\l^{n-3}\,d\l}x + n\l^{n-2}\;dx \\ &= n(n-2)\l^{n-4}\LR{xx^T}dx + n\l^{n-2}\LR{I}\;dx \\ &= \BR{n(n-2)\l^{n-4}xx^T + n\l^{n-2}I}\,dx \\ \grad{g}{x} &= \BR{n(n-2)\l^{n-4}xx^T + n\l^{n-2}I} \;\doteq\; H &\big({\rm the\;Hessian}\big) \\ }$$ The Laplacian is the trace of the Hessian, therefore $$\eqalign{ \nabla^2\phi &= \trace{H} \\ &= {n(n-2)\l^{n-4}\trace{xx^T} + n\l^{n-2}}\trace{I} \\ &= {n(n-2)\l^{n-4}\LR{\l^2} + n\l^{n-2}}\LR{3} \\ &= n(n+1)\l^{n-2} \\ }$$ This calculation may have been longer than strictly necessary, but you now have an expression for the Hessian of the function as a bonus.