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Given $f: \mathbb{R}^n \backslash \{0\} \rightarrow \mathbb{R}$, a twice differentiable, rotationally symmetric function, that is to say $\exists \varphi: \mathbb{R}_{>0} \rightarrow \mathbb{R}$ with $f(x) = \varphi(\|x\|)$, where $\|.\|$ is the Euclidean norm. I want to express $\Delta f$ in terms of derivatives of $\varphi$.

I first looked at the partial derivatives of $f$:

$$\partial_i f = \frac{x_i}{\|x\|} \cdot \varphi ' (\|x\|)$$

$$\partial_i^2 f = \left ( \frac{1}{\|x\|} - \frac{x_i^2}{\|x\|^3} \right ) \cdot \varphi ' (\|x\|) + \frac{x_i^2}{\|x\|^2} \cdot \varphi '' (\|x\|)$$

That's what I get applying chain- and product rule. So:

$$\Delta f = \sum_{i=1}^n \partial_i^2 f$$

but this is still a very ugly sum. I tried to simplify it as far as possible and if I didn't make any mistake, I ended up with:

$$\Delta f = \frac{1}{\|x\|^2} \cdot \left ( \varphi ' \cdot \left ( \sum \left ( \frac{\|x\|}{x_i^2} \right ) - \frac{1}{\|x\|} \right ) + \varphi '' \right )$$

And I think I can rewrite the sum with something like:

$$n \cdot \|x\| \cdot (x_1^2 x_2^2 ... x_n^2)^{n-2}$$

But this doesn't look any better. I know that I have to arrive at something similar to:

$$\Delta f = \varphi '' (\|x\|) + \frac{n-1}{\|x\|} \cdot \varphi ' (\|x\|)$$

Does anyone have an idea how I could further simplify my term or where I went wrong, or whether there is an easier way to simplify the first sum such that I arrive at the wished result? If so, I'd be very happy to see how to. Thanks in advance.

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You can use \| to get the symbol for the norm, $\|$, two vertical lines that are closer together than two separate lines would be set. –  joriki Mar 19 '11 at 20:21
    
@joriki: I will remember to use it in the future. –  Huy Mar 19 '11 at 20:36

3 Answers 3

up vote 3 down vote accepted

After the second and third equation you are almost there. Just use $\sum_{i=1}^n 1 = n$ and $x_1^2 + \cdots + x_n^2 = \|x\|^2$.

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Which exactly is the second/the third equation? –  Huy Mar 19 '11 at 20:19

Let $r$ denote the norm of $x$. As you computed, $\nabla f=\phi' x/r$. By Gauss theorem we therefore have $\int_{B^n(r)}\Delta f \,dV=\int_{S^{n-1}(r)}\phi'\,dS$. If we differentiate wrt. $r$ we get $A r^{n-1} \Delta f=d(\phi' A r^{n-1})/dr$ where $A$ is the volume of the unit $n-1$-dim sphere. From here we finally get $\Delta f=\phi''+(n-1)\phi'/r$.

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We did not yet introduce or even prove Gauss' theorem yet. There must be a way to get to the result without using it, then... –  Huy Mar 19 '11 at 21:40

You have already shown

$$\partial_i f = \frac{x_i}{|x|} \cdot \varphi ' (|x|),$$ and $$\partial_i^2 f = \left( \frac{1}{|x|} - \frac{x_i^2}{|x|^3} \right ) \cdot \varphi ' (|x|) + \frac{x_i^2}{|x|^2} \cdot \varphi '' (|x|).$$

Taking a sum, we have

$$\Delta f = \left(\frac{n}{|x|} - \frac{|x|^2}{|x|^3} \right) \cdot \varphi ' (|x|) + \frac{|x|^2}{|x|^2} \cdot \varphi '' (|x|).$$

Simplifying and rewriting,

$$\Delta f = \varphi''(|x|) + \frac{n-1}{|x|}\varphi'(|x|),$$

as desired. (This is essentially a clarification of user3148's answer.)

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Thank you very much, it's actually pretty easy but somehow I just didn't see it. –  Huy Mar 20 '11 at 10:47

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