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.