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One of the definitions of the discrete (weighted) $p$-Laplacian is the following:

$$\Delta_{p,w}u(x):=\sum_y |u(y)-u(x)|^{p-2}(u(y)-u(x))w(x,y).$$

Consider the one dimensional case. Then the free Laplacian corresponds $w(x,y)=1$ if and only if $|y-x|=1$.

In the the continuous case one sees that the bi-Laplacian, that is the operator $$\Delta^2:=\Delta \circ \Delta,$$ is $\textbf{equal}$ to the the 4-Laplacian (defined in the standard way http://en.wikipedia.org/wiki/P-Laplacian ).

It seems that it is $\textbf{not true in the the discrete case}$, i.e., the discrete 4-Laplacian is not equal to the discrete bi-Laplacian. Am I right or do I make a mistake somewhere? If indeed there is no equality between the bi-Laplacian and the 4-Laplacian in the discrete case , how should one understand it (what is the reason, how does it relate to the continuous case etc.)?

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    $\begingroup$ I am not at all convinced the continuous non-linear 4-Laplacian is equal to the linear operator $\Delta^2$. $\endgroup$ Feb 6, 2015 at 21:47
  • $\begingroup$ The continuous 4-Laplacian $\Delta_4$ is not the same as the bi-Laplacian $\Delta^2$. $\Delta_4$ is nonlinear and second order, $\Delta^2$ is linear and fourth order. $\endgroup$ Feb 7, 2015 at 14:13
  • $\begingroup$ You are right. Thanks. By the way, do you know any literature about the discrete bi-Laplacian? $\endgroup$
    – scouser
    Feb 7, 2015 at 21:13
  • $\begingroup$ You're welcome. Unfortunately I'm not familiar with the bi-Laplacian or the $p$-Laplacian in the discrete setting. $\endgroup$ Feb 7, 2015 at 21:49
  • $\begingroup$ @scouser You can find some properties here: arxiv.org/abs/1712.07370 $\endgroup$ Nov 7, 2023 at 16:17

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