Show that if $\chi(r)=rR(r)$ then $\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=r\frac{d^2 \chi(r)}{dr^2}$

Question

My Attempt:

I first applied the product rule to $$\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}\tag{1}\label1$$ to obtain $$2r\frac{dR(r)}{dr}+r^2\frac{d^2R(r)}{dr^2}\tag{2}\label2,$$ and then wrote out the chain rule for $$\frac{dR(r)}{dr}$$ as $$\frac{dR(r)}{dr}=\frac{dR(r)}{d \chi(r)}\frac{d\chi(r)}{dr}\tag{3}\label3.$$

Then from $\chi(r)=rR(r)$ and using the product rule again $$\frac{d\chi(r)}{dr}=R(r)+r\frac{d^2R(r)}{dr^2}\tag{4}\label4,$$ and since $R(r)=\dfrac{\chi(r)}{r},$ then $$\frac{dR(r)}{d \chi(r)}=\frac{1}{r}\tag{5}\label5.$$

Substituting $\eqref4$ and $\eqref5$ into $\eqref3$, I find that $$\frac{dR(r)}{dr}=\frac{1}{r}\left(R(r)+r\frac{d^2R(r)}{dr^2}\right)=\frac{1}{r}R(r)+\frac{d^2R(r)}{dr^2}\tag{6}\label6$$

Substituting $\eqref6$ into $\eqref2$ yields $$\begin{align}\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}&=2r\frac{dR(r)}{dr}+r^2\frac{d^2R(r)}{dr^2}\\&=2r\left(\frac{1}{r}R(r)+\frac{d^2R(r)}{dr^2}\right)+r^2\frac{d^2R(r)}{dr^2}\\&=2R(r)+(2r+r^2)\frac{d^2R(r)}{dr^2}.\end{align}$$

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As you can see the expression I desire for $$\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}$$ is getting every-more complicated and I haven't even computed the second derivative of $R(r)$ yet. I fear that I am either making a mistake or missing a much simpler approach. In my notes it said that

It is straightforward to show that $$\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=r\frac{d^2 \chi(r)}{dr^2}.$$

So this implies that I am missing something very simple.

If someone would care to point out my errors and/or give me any hints/tips on how to reach the desired result I would be most thankful. Regards.

Answer 1

I write $f$ instead of $R$ and $g$ instead of $\chi$. Then it is to show:

$\frac{d}{dr}(r^2f'(r))=rg''(r)$

From $g(r)=rf(r)$ we get $g'(r)=f(r)+rf'(r)$, hence $r^2f'(r)=rg'(r)-rf(r)=rg'(r)-g(r)$,

thus

$\frac{d}{dr}(r^2f'(r))=g'(r)+rg''(r)-g'(r)=rg''(r)$, as desired.

Answer 2

Well, we have that:

$$\mathcal{Z}=\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}\left(\text{r}\right)\right)\right)=\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\cdot\text{R}'\left(\text{r}\right)\right)=\text{r}^2\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}'\left(\text{r}\right)\right)+\text{R}'\left(\text{r}\right)\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\right)\tag1$$

Using the product rule.

So, we get:

$$\mathcal{Z}=\text{r}^2\cdot\text{R}''\left(\text{r}\right)+\text{R}'\left(\text{r}\right)\cdot2\text{r}=\text{r}^2\text{R}''\left(\text{r}\right)+2\text{r}\text{R}'\left(\text{r}\right)\tag2$$

And we also have:

$$\text{r}\cdot\chi\space''\left(\text{r}\right)=\text{r}\cdot\frac{\text{d}^2}{\text{d}\text{r}^2}\left(\text{r}\cdot\text{R}\left(\text{r}\right)\right)=\text{r}\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}\left(\text{r}\right)\right)+\text{R}\left(\text{r}\right)\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\right)\right)\tag3$$