# Intuitive interpretation of the Laplacian Operator

Just as the gradient is “the direction of steepest ascent”, and the divergence is “amount of stuff created at a point”, is there a nice interpretation of the Laplacian Operator (a.k.a. divergence of gradient)?

Assume the function $f$ is $C^2$ in a neighbourhood of ${\bf 0}\in{\mathbb R}^n$. Using the Taylor expansion of $f$ at ${\bf 0}$ one can prove the following:

This formula says that $\Delta f({\bf 0})$ is “essentially” (i.e., up to the scaling factor ${2n \over r^2}$) equal to the average difference $f({\bf x})-f({\bf 0})$ over small spheres around ${\bf 0}$.

Using this interpretation one gets, e.g., an intuitive understanding of the heat equation

namely: If averaged over small spheres around a point ${\bf p}$ it is hotter than at ${\bf p}$ itself, then in the next second the temperature at ${\bf p}$ will rise.

Given the interest in the above formula $(*)$, here are some hints for the proof: By Taylor’s theorem one has

Here the $f_{.i}$ and the $f_{.ik}$ are the partial derivatives of $f$ evaluated at ${\bf 0}$, whence constants. Now we integrate this over $S_r$ with respect to the surface measure ${\rm d}\omega$ and obtain

because all other terms are odd in at least one variable. The integrals $\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x})$ are all equal; therefore we have

Putting it all together we obtain

and solving for $\Delta f({\bf 0})$ we get the stated formula.

For a proof using Gauss’ theorem see here:

Nice way of thinking about the Laplace operator… but what’s the proof?