del operator question

Basically: am I correct to rewrite the LHS as I did on the 4th line of text?

and how could I then work out the second partial derivative below?

Yes, I think so.



Product rule, and use the chain rule to differentiate f'(r).

I need to show:[triangle is del operator]
$\bigtriangledown \cdot \bigtriangledown f(r)=f''+\frac{2}{r}f'$
where r=|r|, r=(x,y,z), and f(r) is a twice differentiable scalar function with f'=df/dr, f''=d^2f/dr^2

So I was thinking I could write this as $(\bigtriangledown \cdot \bigtriangledown )f(r)=\bigtriangleup f(r)$ [other triangle is the laplace operator], but then how could I work out:
$\frac{\partial^2 }{\partial x^2}f(r)$?

I know that:
$\frac{\partial }{\partial x}f(r)=\frac{\mathrm{df} }{\mathrm{dr} }\frac{\partial r}{\partial x}=\frac{\mathrm{df} }{\mathrm{dr} }\frac{x}{r}$.

Basically: am I correct to rewrite the LHS as I did on the 4th line of text, and how could I then work out the second partial derivative below it?

Thanks in advance.

painful...

these types of questions usually have tricks which cut the work out, but I can never see them...

the question can be found on the link

http://madasmaths.com/archive/maths_booklets/advanced_topics/vector_operators.pdf

page 13

Much appreciated.

I used this to check my answer, and I'm so glad that I got it right after spending close to 2 hours trying to do this!

prsom

<3 <3 <3

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There are analogues to the chain and product rules that you can use with the del operator. The problem is that it takes a fair bit of experience to know which analogues are valid and which aren't. (I used to be able to do this easily, but it's 20 years since I had exams in this and I struggle a bit now...)

Anyhow, here's my proof:

$\nabla f(r) = f'(r) \nabla r$ (chain rule analogue for del)

$= f'(r) \frac{\bf{r}}{r} = \dfrac{f'(r)}{r} {\bf {r}}$ (*) (from knowing $\nabla r = \frac{\bf{r}}{r}$ which you really should know.)

$\nabla.\nabla f(r) = \left(\nabla \dfrac{f'(r)}{r}\right). {\bf {r}}+ \dfrac{f'(r)}{r} (\nabla.{\bf {r}})$ (product rule analogue for del)

Now f'(r) / r is a function of r, so we can reapply the result from (*) to say:

$\nabla \dfrac{f'(r)}{r} = \left(\dfrac{d}{dr}\dfrac{f'(r)}{r} \right) \frac{\bf{r}}{r} = \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) \frac{\bf{r}}{r}$

and so $\left(\nabla \dfrac{f'(r)}{r}\right). {\bf {r}} = \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) \dfrac{\bf{r.r}}{r}$

$= r \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) = f''(r) - f'(r) / r$ (**)

Meanwhile, $\nabla.{\bf r}$ = 3 (again, something you should know!), so

$\dfrac{f'(r)}{r} (\nabla.{\bf {r}}) = 3 f'(r) / r$ (***)

Adding (**) and (***) gives the desired result.

There are analogues to the chain and product rules that you can use with the del operator. The problem is that it takes a fair bit of experience to know which analogues are valid and which aren't. (I used to be able to do this easily, but it's 20 years since I had exams in this and I struggle a bit now...)

Anyhow, here's my proof:

$\nabla f(r) = f'(r) \nabla r$ (chain rule analogue for del)

$= f'(r) \frac{\bf{r}}{r} = \dfrac{f'(r)}{r} {\bf {r}}$ (*) (from knowing $\nabla r = \frac{\bf{r}}{r}$ which you really should know.)

$\nabla.\nabla f(r) = \left(\nabla \dfrac{f'(r)}{r}\right). {\bf {r}}+ \dfrac{f'(r)}{r} (\nabla.{\bf {r}})$ (product rule analogue for del)

Now f'(r) / r is a function of r, so we can reapply the result from (*) to say:

$\nabla \dfrac{f'(r)}{r} = \left(\dfrac{d}{dr}\dfrac{f'(r)}{r} \right) \frac{\bf{r}}{r} = \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) \frac{\bf{r}}{r}$

and so $\left(\nabla \dfrac{f'(r)}{r}\right). {\bf {r}} = \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) \dfrac{\bf{r.r}}{r}$

$= r \left(\dfrac{f''(r)}{r} - \dfrac{f'(r)}{r^2} \right) = f''(r) - f'(r) / r$ (**)

Meanwhile, $\nabla.{\bf r}$ = 3 (again, something you should know!), so

$\dfrac{f'(r)}{r} (\nabla.{\bf {r}}) = 3 f'(r) / r$ (***)

Adding (**) and (***) gives the desired result.