# Singularity in Laplacian operator

1. Aug 12, 2012

### MrHigh

I am trying to understand the following basic problem,

$\partial_{xx} f^\alpha (x) = \alpha (\alpha-1) \frac{1}{f^{2-\alpha}} \partial_x f + \alpha \frac{1}{f^{1-\alpha}} \partial_{xx} f$

So it is not hard to see that if $f$ tends to zero the laplacian becomes undefined (im not sure if i am using proper terminology).

that should happen for any $\alpha < 2$.

I don't understand why the singulary happens and how the trivial f(x) = 0 for all x can be applied.

thanks for the attention.

Last edited: Aug 12, 2012