Intuitive interpretation of the Laplacian

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: $$\Delta f({\bf 0}) =\lim_{r\to 0+}\ {2n\over r^2}{1\over \omega(S_r)} \int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})\ .\qquad (*)$$ 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 $${\partial u\over\partial t}=a^2\ \Delta u\ ,$$ 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 $$f({\bf x})-f({\bf 0})= \sum_{i=1}^n f_{.i} x_i +{1\over2}\sum_{i,k} f_{.ik} x_i x_k + o(|{\bf x}|^2)\qquad({\bf x}\to{\bf 0}) .$$ 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 $$\int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})={1\over2}\sum_{i}f_{.ii}\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x}) +o\bigl(r^{2+(n-1)}\bigr) \qquad(r\to0+)\ ,$$ 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 $$\int\nolimits_{S_r}x_i^2\ {\rm d}\omega({\bf x})={1\over n}\int\nolimits_{S_r}\sum_k x_k^2\ {\rm d}\omega({\bf x})={r^2\over n}\omega(S_r)\qquad(1\leq i\leq n)\ .$$ Putting it all together we obtain $$\int\nolimits_{S_r}\bigl(f({\bf x})-f({\bf 0})\bigr)\ {\rm d}\omega({\bf x})={r^2\over 2n}\omega(S_r)\Delta f({\bf 0}) +o(r^{n+1})\qquad(r\to 0+)\ ,$$ 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?

It behaves like a local averaging operator. This can be seen easily when considering a finite-difference approximation to the Laplacian:

$$\nabla^2 f(x,y) \approx \frac{f(x+h,y) + f(x-h,y) + f(x,y+h) + f(x,y - h) - 4f(x,y)}{h^2}$$

You can rewrite this in vector notation as:

$$\nabla^2 f(x,y) \approx \frac{1}{h^2} \sum_{\mathbf{h}} f(\mathbf{x+h}) - f(\mathbf{x})$$

where the sum is over vectors in the $x$, $-x$, $y$ and $-y$ directions, and $h=||\mathbf{h}||$. Higher order approximations to the Laplacian will involve averaging the rates of change in more directions.

So you can think of the Laplacian as behaving like an 'average rate of change'. As pointed out in Glen Wheeler's answer, the average rate of change can be zero even when there is significant curvature at a point, for example as in the function $f(x,y)=x^2-y^2$.

In image processing, a discrete Laplacian (where $h$ is one pixel in the definition I gave above) can be used as a crude edge-detection filter. It is close to zero in regions where the image is varying smoothly, and has large values in regions where the image has sharp transitions from low to high intensity.

In physics, the Laplacian is interpreted as a diffusion operator, as in the equation

$$\frac{\partial u}{\partial t} = \nabla^2 u$$

This says that the rate of change of $u$ in time is given by the average rate of change of $u$ in space. If we interpret $u$ as a temperature (and hence $\partial u/\partial t$ is the rate of change of temperature) then we see that there is more heat exchange in regions where the temperature is highly variable, and less heat exchange when the temperature varies smoothly.

The laplacian is also the trace of the hessian matrix (the matrix of second-order partial derivatives). Since the trace of a matrix is invariant under a change of basis, then the laplacian does not change if you do a change of basis. For instance if you work in $\mathbb{R}^2$, then $$\Delta f=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}$$ in cartesian coordinates and $$\Delta f=\frac{1}{r}\frac{\partial f}{\partial r}+\frac{\partial^2 f}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}$$ in polar coordinates but at each point $(x,y)$ or $(r,\theta)$ this is the same value.

[The hessian matrix contains information about the way your function bends around some point. Since it is symmetric (by the Schwarz theorem) and real, you can diagonalize it and so the laplacian is also the sum of the eigenvalues of the hessian, and these eigenvalues are important to detect which kind of bending you have locally.]