Showing $\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k-1}(r^{2k-1}\phi(r))=\left(\frac{1}{r}\frac{d}{dr}\right)^{k}(r^{2k}\phi'(r))$

Question

How to show that $\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k-1}(r^{2k-1}\phi(r))=\left(\frac{1}{r}\frac{d}{dr}\right)^{k}(r^{2k}\phi'(r))$ for $k\ge 1, r>0$ and $\phi$ sufficiently differentiable

$\left(\frac{1}{r}\frac{d}{dr}\right)^{k}$ means $\left(\frac{1}{r}\frac{d}{dr}\right)\circ\left(\frac{1}{r}\frac{d}{dr}\right)\circ\dots$ (k times) and not $\frac{1}{r^k}\frac{d^k}{dr^k}$

I tried induction but it didn't work for $k=1$ it is OK on both sides I've $2\phi'+r\phi''$, then I have to take LHS I think, and normally in induction proofs one begins with $n=k+1$ and uses the the fact that it was true for $n=k$, here I think I have to begin with $n=k+2$ or not ? Otherwise I get something like,

$\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k}(r^{2k+1}\phi(r))=\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k-1}\left(\frac{1}{r}\frac{d}{dr}\right)(r^{2k+1}\phi(r))$

$=\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}((2k+1)r^{2k-1}\phi(r)+r^{2k}\phi'(r))$

$=(2k+1)\left(\frac1r\frac{d}{dr}\right)^{k}((r^{2k}\phi'(r))+\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+2}\phi'(r))$

Answer 1

Consider the case $\underline{k=1}$

On the LHS

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{{0}}(r\phi))=\frac{d^2}{dr^2}(r\phi)=\frac{d}{dr}(r'\phi+r\phi')=2\phi'+r\phi''$

On the RHS

$\left(\frac1r\frac{d}{dr}\right)(r^{2}\phi')=\frac1r(2r\phi'+r^2\phi'')=2\phi'+r\phi''$

so they match

Assume now $\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)=\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+2}\phi')\tag1$

holds. We want to show that,

$\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi(r))=\left(\frac1r\frac{d}{dr}\right)^{k+2}(r^{2k+4}\phi'(r))$

$(1)$ implies;

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)=\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+2}\phi')$

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)=\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+2}(\phi'+r\phi''))-\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')$

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)=\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}r\phi')-\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')$

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)=\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+2}\phi')-\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')$

$\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+1}\phi)-\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+2}\phi')-\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')=0$

$\left(\left(\frac1r\frac{d}{dr}\right)^{k+1}((2k+4)r^{2k+2}\phi')-\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}((2k+3)r^{2k+1}\phi)\right)-\left(\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}(r^{2k+2}\phi')-\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')\right)=0$

$\left(\left(\frac1r\frac{d}{dr}\right)^{k+1}((2k+4)r^{2k+2}\phi')+\left(\frac1r\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi'')\right)-\left(\frac{d^2}{dr^2}\left(\frac1r\frac{d}{dr}\right)^{k}((2k+3)r^{2k+1}\phi+r^{2k+2}\phi')\right)=0$

$\left(\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi)\right)-\left(\left(\frac1r\frac{d}{dr}\right)^{k+2}(r^{2k+4}\phi')\right)=0$

$\frac{d^2}{dr^2}\left(\frac{1}{r}\frac{d}{dr}\right)^{k+1}(r^{2k+3}\phi)=\left(\frac1r\frac{d}{dr}\right)^{k+2}(r^{2k+4}\phi')$