What Does $\nabla^2 f(x) = -4$ Reveal About Its Meaning?

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The discussion centers on the mathematical expression $\nabla^2 f(x) = -4$, specifically analyzing the Laplacian of the function $f(x) = -|x|^2$. Participants clarify that for a two-dimensional vector, the Laplacian results in 4, while for a three-dimensional vector, it yields 6. The confusion arises from the interpretation of the function's dimensionality and the application of the Laplacian operator. Additionally, the term "phi_xi,eta" refers to the second derivative of the function phi with respect to the variables xi and eta.

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

For problem 1, what is "phixi, 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.)
 
ok. thanks i follow the laplacian thing now.

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

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