Vector laplacian and del squared confusion.

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SUMMARY

The discussion centers on the mathematical relationship between the Laplacian operator and the divergence of the gradient in the context of the equation z = a - (∇²E(t))⁻¹ΔE(t). The confusion arises from the treatment of these operations as separate, leading to questions about the implications of division resulting in 1. The equation involves a twice differentiable and convex function E, indicating that the operations may not yield a trivial result in all cases.

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darkpsi
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Hello all,

I am reading a research paper and have found the equation below:
[URL]http://latex.codecogs.com/gif.latex?\mathbf{z}%20=%20\mathbf{a}%20-%20%28\nabla^2E%28\mathbf{t}%29%29^{-1}\Delta%20E%28\mathbf{t}%29%29[/URL]

in which E is some function with the variable t being the vector input, and a and z being unimportant other vectors.

Now, it's my understanding that the Laplacian of a vector is the same as the divergence of the gradient, but in this equation they are being treated as separate operations, namely because division would result in 1 which would be meaningless. I'm not sure if the first part is taking an inverse, if the two operations are not the same, and/or something else entirely. I can provide more information if it turns out the equation or vectors make the difference.

Thanks for any help!
 
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To clarify...

To be more specific, is the division going to give a value of 1 or is there ever a case where this doesn't happen? If it helps, the equation E is both twice differentiable and convex.
 

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