Timeline for Laplace operator's interpretation
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 3, 2022 at 17:31 | comment | added | Galen | Suppose that $f(x_1, \cdots, x_n) = \prod_{j=1}^n x_j$, then $\nabla \cdot \nabla f = 0$. So it is not just linear functions, but also multilinear functions that will have this property. | |
| Aug 28, 2019 at 5:43 | comment | added | Royi | You mention 1D and 3D, what about 2D? | |
| Feb 8, 2012 at 19:38 | comment | added | joseph f. johnson | It can be thought of as a kind of curvature. Our Euclidean intuition of flat or straight is, going back to Euclid himself or earlier, « lying evenly between its extremes.» (Admittedly, to make sense of this in general was a Hilbert problem...) A harmonic function obeys the averaging property, so it lies evenly between its extremes.... | |
| Feb 8, 2012 at 18:25 | comment | added | Džuris | Yes, the curvature interpretation seems natural if you look at 1D Laplacian. May I also look at 2D or 3D case? If I take any of the wikimentioned harmonic functions and plot it I get this: wolframalpha.com/input/?i=log%28x%5E2+%2B+y%5E2%29 Harmonicity of a function means it's Laplacian is equal to zero almost everywhere. However, it doesn't appear to me that the curvature (or stress) of this field is equal to zero. It doesn't even seem to be constant, actually. | |
| Feb 8, 2012 at 17:08 | history | answered | joseph f. johnson | CC BY-SA 3.0 |