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In Physics SO the intuition of the Laplace operator (divergence of the gradient) is explained by resorting to the finite difference version: the Laplace equation is satisfied as long as the value at a vertex is the average of the surrounding values, akin to the explanation in Wikipedia of harmonic functions.

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Or the more enjoyable blog explanation, motivating them through minimal surfaces: soap bubbles,

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or the 3Blue1Brown video: Much like a minimum with a positive second derivative, a positive Laplacian in a 2D surface would indicate local minimum (concave up) in the way that the neighboring points on average are higher in value.

Professor Strang gives a rather artistic impromptu intuition right here proposing a grid with unweighted edges, and noticing that for internal vertices, the degree would be $4,$ corresponding to second differences with the surrounding unit-value edges, but this is a bit shaky in what it is really happening in the example. Clearly the average of adjacent entries in a Laplacian matrix doesn't really work because of the embedded degree matrix in the diagonal:

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Others approach the graph Laplacian by noticing the connection to Newton's law of cooling.

Is there a better "visual" to see the intuition of the graph Laplacian?

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  • $\begingroup$ Both this question and the answers are truly awesome. Thank you very much! $\endgroup$ Sep 2, 2021 at 21:20
  • $\begingroup$ @BorisBurkov I appreciate your kind words. $\endgroup$ Sep 3, 2021 at 0:55

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The same intuition, that the Laplacian describes how a function differs from its average locally, holds for graph Laplacians. To write it out clearly, let $W$ be an $n\times n$ adjacency matrix for a graph. This means $W_{ji}=W_{ij}=1$ if there is an edge between $i$ and $j$, and $W_{ji}=W_{ij}=0$ otherwise. The degree matrix $D$ is the $n\times n$ diagonal matrix with $(i,i)$ entry $D_{ii} = \sum_{j=1}^n W_{ij}$. Then the graph Laplacian matrix is $L=D-W$.

If $x\in \mathbb{R}^n$ is a vector with $Lx=0$ (i.e., a harmonic function), then we have

$$0 = [Lx]_i = D_{ii}x_i - \sum_{j=1}^n W_{ij} x_j$$

for each $i=1,\dots,n$. Rearranging we get

$$x_i = \frac{\sum_{j=1}^n W_{ij} x_j}{D_{ii}}= \frac{\sum_{j=1}^n W_{ij} x_j}{\sum_{j=1}^n W_{ij}}.$$

This says that $x_i$ is the weighted average of its neighboring values $x_j$, weighted by the adjacency matrix. Note that the weights don't need to be binary 0/1, and any nonnegative and symmetric weights can be used.

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Perhaps trying to encourage better answers, I want to at least share / promote an excellent intuition of a graph Laplacian and second derivatives presented here.

The explanation might have skipped over some physics details, but it is great in its purpose. The idea is a system of cylinders of mass $1$ that can bob up and down without moving sideways, and springs connecting adjacent cylinders:

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The force calculated by Newton's law $F=m\times a$ should equal the force in Hooke's law $F=-k x_i:$

$$\begin{align} \frac{d^2}{dt^2} x_i &= \underset{\text{sum of spring forces}}{\underbrace{-k(x_i-x_{i+1 }) - k(x_i-x_{i-1})}}\\[2ex] &=-k(-x_{i-1}+2x_i -x_{i-1}) \end{align}$$

Placing all the individual displacements into a vector $\vec x(t) =\begin{bmatrix}x_0(t)& x_1(t) & x_2(t) \cdots x_{n-1}(t)\end{bmatrix}^\top$ the system of differential equations can be written as

$$\frac{d^2}{dt^2} \vec x =-k \underset{\text{graph Laplacian of a line graph}}{\underbrace{\begin{bmatrix}1&-1&&&&&\\-1&2&-1&&&&\\&-1&2&-1&&&\\&&&\ddots\\&&&-1&2&-1 \\&&&&-1&1\end{bmatrix}}}\vec x$$

At some steady state the graph Laplacian is a discrete equivalent of the relative position of the cylinders otherwise calculated as the sum of sine and cosine waves (harmonics in a Fourier series).

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