Laplacian of f(x) = -|x|² giving -4 not -2 or -6

[latentcorpse]

1, apparently

$\frac{1}{2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(-|\xi|) 4 \phi_{\xi \eta} d \xi d \eta=0$

also apparently this is obvious after only doing the eta integral. any ideas why?

2, what is $\nabla^2 f(x)$ where $f(x)=-|x|^2$
the answers say its -4
when i just do the second derivative with respect to x i get -2
and when i use index notation i get -6 since
$\partial_i \partial_i r_j r_j = \partial_i (2 \delta_{ij} r_j)=2 \partial_i r_i = 2 \delta_{ii} =6$

what is going on here?

 

[HallsofIvy]

Latex still isn't working so I had to look at your LaTex code.

For problem 1, what is "phi_{xi, eta}"?

For the second, "nabla^2" is usually defined for functions of several variables. If f(x) really is |x|²= x², then the second derivative is 2 as you say. If x is a two dimensional vector with x= <x, y>, then |x|= x²+ y² and its Laplacian is 4. If x is a three dimensional vector with x= <x, y, z> then |x|= x²+ y² and its Laplacian is 6. (More generally, if x is an n-dimensional vector, its Laplacian is 2n.)

 

[latentcorpse]

ok. thanks i follow the laplacian thing now.

phi_xi,eta is the derivative of phi wrt xi and then wrt eta