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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 (a.k.a. divergence of gradient)?

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I'd suggest including the word (laplacian operator or laplace operator, in fact both). Currently the title is hard to search because of the different names people give this mathematical concept. –  Pinocchio May 3 at 17:25

5 Answers 5

up vote 52 down vote accepted

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?

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This point of view is one of the best for not just understanding the heat equation, but also the connection of the Laplacian to Brownian motion, or mathematical models of flows in porous media. –  Willie Wong Jul 8 '11 at 8:56
It's also related to the mean value property of harmonic functions. –  anon Jul 8 '11 at 10:18
Very nice, this is exactly the kind of answer I was hoping for! Thanks to all other posters, too –  koletenbert Jul 8 '11 at 20:15
Nice answer brother. –  Alexander Grothendieck Nov 17 '13 at 0:07

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.

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One important thing to mention is that because the first term in the Taylor expansion of a function is odd, it disappears when you integrate over a symmetric region. Hence when you measure the "average difference" the lowest order term is in fact the Laplacian, and not some first order derivative. (Unless, of course, your function is not continuously differentiable at that point.) –  Willie Wong Jul 8 '11 at 8:59
Inmediate corollary: in the first order aproximation, over a discrete grid, the laplacian is zero if and only iff the function at each point equals the average of its four neighbours. –  leonbloy Jul 8 '11 at 18:28
+1, imho this answer is more intuitive than the currently accepted one. –  Calmarius Nov 15 '14 at 17:32

Sure. One way is to interpret the laplacian of a function at a point, $\Delta f(p)$, as a measurement of the "average bending" of the image at $f(p)$.

Note that in higher dimensions, this is not really very intuitive, due to the "average" part of the "average bending": a function may have all second derivatives negative except one, which is very large and positive, and nevertheless have positive Laplacian at that point. In such situations one would be better off considering the full Hessian of the function, or the eigenvalues of it, in turn.

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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.]

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In addition to other answers, you also should pay attention to the fact that there is a connection between the Laplacian and the Gaussian kernel of the form $$ \frac{1}{\sqrt{2\pi t}}e^{-\|x\|^2/2t} $$ which in fact one of solutions of the equation $u_t = \Delta u$ mentioned by Chris. This kernel has a lot of applications to the averaging since it has a well-known "Bell shape" and is a density function for a normal (Gaussian) distribution.

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