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In the book Mathematics of Classical and Quantum Physics, the authors give an (enlightening) interpretation of the Laplace Operator $\nabla^{2}$ of a field $f(\mathbf{x})$,

$\nabla^{2}f(\mathbf{x})$ as the difference between the average value of $f$ over a infinitesimal element around the point $\mathbf{x}$ and the value of $f$ at the point.

$$\nabla^{2}f(\mathbf{x})=\bar{f}_{dv}-f(\mathbf{x})$$

This interpretation clarifies, for example, the heat equation as saying that the temperature at a point increases if the average around it is larger and vice versa.

What about the biharmonic operator $\nabla^{4}\equiv\nabla^{2}\nabla^{2}$? The obvious answer would be to interpret it like the Laplace operator but recursively (the difference between [the average of the difference between the average of $f$ and the value of $f$ at $\mathbf{x}$] and [the difference between the average of $f$ and the value of $f$ at $\mathbf{x}$]) but this would be very convoluted... is there a simpler interpretation?

I know that the biharmonic is classified along the Laplace operator as a diffusion-type term, so maybe the interpretation is similar, but I could not find anythin online.

  • Possibly related to the averaging aspect: https://physics.stackexchange.com/q/310447/25301. [math.se] might also provide an answer, if you don't get one here. – Kyle Kanos Aug 12 '19 at 11:34

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