TITLE: Understanding the Laplace operator conceptually
QUESTION [65 upvotes]: The Laplace operator: those of you who now understand it, how would you explain what it "does" conceptually? How do you wish you had been taught it?
Any good essays (combining both history and conceptual understanding) on the Laplace operator, and its subsequent variations (e.g. Laplace-Bertrami) that you would highly recommend?

REPLY [20 votes]: Another view along the lines of the answers above:
Suppose you have some region in the plane $\Omega$, and you are given the value of some scalar function $f$ along the boundary $\partial \Omega$. You now want to fill in $f$ on the interior of $\Omega$ "as smoothly as possible." (A common physical interpretation is that $f$ is the heat of the region: you are fixing the temperature of the boundary of $\Omega$ and want to know what the temperature on the interior will be at steady state.)
What does "as smooth as possible" mean? Well, one measure of the smoothness of $f$ is to look at its gradient $\nabla f$ and measure
$$E(f) = \int_{\Omega} \|\nabla f\|^2\,dA.$$
Notice that this integral, called the Dirichlet energy of $f$, achieves its lowest possible value of $0$ when $f$ is constant. The less smooth (to first order) that $f$ is, the higher the Dirichlet energy will be. Making $f$ as smooth as possible means finding the $f$ that satisfies the boundary conditions and minimizes $E$.
How do we minimize $E$? We "take the derivative and set it to zero":
$$\nabla_f E = 0.$$
It may look a little weird to differentiate a scalar (the Dirichlet energy) with respect to a function, but the idea is the same as when you work with the ordinary gradient. Recall that for an ordinary scalar function $g(x,y,z):\mathbb{R}^3\to\mathbb{R}$, the gradient $\nabla g$ at a point is the unique vector that, when you dot it with any direction $v$, tells you the directional derivative of $g$ in that direction:
$$\nabla g(x,y,z)\cdot v = \frac{d}{dt}g[(x,y,z)+t v]\biggr\vert_{t\to 0}.$$
The gradient of $E$ works the same way: it gives you the unique function over $\Omega$ that, when you take the inner product of $\nabla E(f)$ with any variation $\delta f$ of $f$, gives you the directional derivative of $E$ in that "direction":
$$\int_{\Omega} \nabla E(f) \delta f\,dA = \frac{d}{dt}E(f+t \delta f)\biggr\vert_{t\to 0}.$$
You can do the multivariable calculus and after some integration by parts, you will see that $\nabla E(f) = -\Delta f.$ Several takeaways from this:

The function $f$ that interpolates the boundary conditions as smoothly as possible (in the sense of minimizing the Dirichlet energy) is the solution to the Laplace equation $-\Delta f = 0$.
Given some function $f$ that interpolates the boundary conditions but does not minimize the Dirichlet energy, the gradient of $E$, $-\Delta f$, is the "direction of steepest ascent" of $E$ -- the direction to change $f$ if you want to most quickly increase $E$. The negative of this, $\Delta f$, is the direction that most quickly decreases $E$: if you are trying to smooth $f$, this is then the direction that you want to flow $f$ in. This insight leads to the heat equation $$\frac{df}{dt} = \Delta f$$
which, given initial temperatures on $\Omega$, flows in the direction that best decreases the Dirichlet energy until the heat has diffused as smoothly as possible over the surface.
Nowhere in the above discussion was it essential that $\Omega$ was a piece of a plane: as long as you can define functions on $\Omega$ and take gradients of $f$ to get the Dirichlet energy, the above works equally well, and is one way of motivating the Laplace-Beltrami operator on arbitrary manifolds in $\mathbb{R}^3$. The physical picture here is that you have some conductive plate in empty space, and heat up the boundary of the plate, and look at how the heat equalizes over the plate.