rayquaza17
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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.
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Hodor
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(Original post by rayquaza17)
Basically: am I correct to rewrite the LHS as I did on the 4th line of text?
Yes, I think so.

and how could I then work out the second partial derivative below?
Product rule, and use the chain rule to differentiate f'(r).
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rayquaza17
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(Original post by Hodor)
Yes, I think so.



Product rule, and use the chain rule to differentiate f'(r).
Thank you very much.
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TeeEm
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(Original post by rayquaza17)
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_..._operators.pdf

page 13
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rayquaza17
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(Original post by TeeEm)
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_..._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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TeeEm
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(Original post by rayquaza17)
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
I am glad you found it useful.
Very long and tidious.
(I remember when I wrote the solution it also took me ages, this question a good test for concentration)
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DFranklin
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(Original post by rayquaza17)
..
(Original post by TeeEm)
..
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.
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TeeEm
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(Original post by DFranklin)
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.
I am grateful for your time.

As I mentioned to Rayquaza there are shortcuts for all these operators which like yourself I did a long time ago (longer in my case).
Unfortunately I have forgotten most of this material and my efforts to remember often hit a brick wall. (Brain stalls and that is it)

I will look at your proof when my brain "feels better" and I hope that I can follow it.
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rayquaza17
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(Original post by DFranklin)
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.
Thank you for taking the time to post this.

I am going to give this a proper look through this weekend.
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