TITLE: Intuitive interpretation of the Laplacian Operator
QUESTION [141 upvotes]: 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)?

REPLY [2 votes]: Although this thread is quite old, I think a quick picture would add up to the excellent answers here. An explanation is also below. I've tried to be as non-technical as I could get, since this would also add up intuition to the other more technical answers. Anyway, let's see how come that the Laplacian is a local averaging operator.

So, let's imagine the points $A$, $B$, $C$, $D$ and $O$ are some points on a metal sheet and we have an associated temperature at each (call it $a$, $b$, $c$, $d$, $o$).
Cool.
Now we want to learn something about how this temperature is distributed on the sheet and specifically on these five points.
Specifically, how fast the temperature changes from $A$ to $O$? Well, this should be the total change in temperature from $A$ to $O$ (call it $o-a$) divided by the distance from $A$ to $O$ (call it $h$, and the distance from $A$ to $C$ - $2h$). Thus, as we are going from $A$ to $O$, the temperature changes with a rate of $(o-a)/h$ per unit distance. Similarly, we can calculate $(b-o)/h$, $(c-o)/h$ and $(o-d)/h$. For brevity, let these be $a', b', c'$ and $d'$.
What these tell us is whether the temperature is increasing in a given direction (and how fast). For example, if $a'>0$ then the temperature at $O$ is higher than the temperature at $A$, and it is increasing as we go from $A$ to $O$ at a rate of $(o-a)/h$. Obviously, as $h$ tends to zero, this becomes the first derivative.
...but we can do more.
What if we wanted to know how the change of temperature changes as we move on the metal sheet? In other words, the "second derivatives"? (Formally, we are still "discrete", so these are not exactly derivatives, but let's call them so for convenience.)
Now let's find this for the temperature between $A$ and $C$ by taking the average change of $a'$ and $c'$: $$\frac{(a'+b')}{2h} = \frac{1}{2}\frac{\frac{c-o}{h}-\frac{o-a}{h}}{h} = \frac{1}{2}\frac{a-2o+c}{h^2}$$
We can calculate the second derivative between $B$ and $D$ analogously.
This doesn't seem too big of a step toward the Laplacian, but let's continue by adding what happens in the $y$ direction. Which is, let's add the second derivative in the y-direction to the second derivative in the x-direction, giving: $$\frac{1}{2}\frac{a-2o+c}{h^2}+\frac{1}{2}\frac{b-2o+d}{h^2}=\frac{1}{2h^2}(a+b+c+d-4o)$$
The last thing is the sum of the two second partial derivatives, which is - the Laplacian. Let's see what happens when it is equal to zero... $$\frac{1}{2h^2}(a+b+c+d-4o)=0$$discarding the $1/2h^2$ term: $$a+b+c+d-4o=0$$
...which gives:$$o=\frac{a+b+c+d}{4}$$
Wait... but this means that saying that the Laplacian being equal to zero is the same as saying that the temperature at $O$ is equal to the average of the temperatures at $A,B,C$ and $D$! Exactly as we wanted.