Exploring the Meaning Behind $\nabla^2 f(x) = -4$

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In summary, the first problem involves an integral with a function f that depends on the absolute value of xi and the partial derivatives of a function phi with respect to xi and eta. The result of this integral is 0, which is apparently obvious after only doing the eta integral. In the second problem, we are asked to find the Laplacian of a function f(x)=-|x|^2, with conflicting answers of -4 and -6. The discrepancy arises from the number of dimensions of x, with the Laplacian being 2 for two dimensions and 6 for three dimensions.
  • #1
1, apparently

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

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

2, what is [itex]\nabla^2 f(x)[/itex] where [itex]f(x)=-|x|^2[/itex]
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
[itex]\partial_i \partial_i r_j r_j = \partial_i (2 \delta_{ij} r_j)=2 \partial_i r_i = 2 \delta_{ii} =6[/itex]

what is going on here?
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  • #2
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.)
  • #3
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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