Laplace operator's interpretation (Laplacian)

Question

What is your interpretation of Laplace operator? When evaluating Laplacian of some scalar field at a given point one can get a value. What does this value tell us about the field or it's behaviour in the given spot?

I can grasp the meaning of gradient and divergence. But viewing Laplace operator as divergence of gradient gives me interpretation "sources of gradient" which to be honest doesn't make sense to me.

It seems a bit easier to interpret Laplacian in certain physical situations or to interpret Laplace's equation, that might be a good place to start. Or misleading. I seek an interpretation that would be as universal as gradients interpretation seems to me - applicable, correct and understandable on any scalar field.

PS The text of this question is taken from Physics StackExchange. I found it useful for people who search in the Math StackExchange forum.

Answer 1

I came across this question during my research today! Let me share with you my understanding of the problem with an example.

First off, the Laplacian operator is the application of the divergence operation on the gradient of a scalar quantity. $$ \Delta q = \nabla^2q = \nabla . \nabla q$$

Lets assume that we apply Laplacian operator to a physical and tangible scalar quantity namely the water pressure (analogous to the electric potential).

You can think of the gradient of the water pressure like a time steady current direction of water caused by direct contact with other molecules like a filed of direction arrows (a lame analog for the electric field (there doesn't exist a good analogy)).

So the divergence of the gradient of a water pressure is the same thing as the divergence of the field of water current direction arrows. If this field has zero divergence (i.e. Laplace's equation) then the current is not being converged (compressed) or diverged (expanded) (i.e. water maintaining constant density)

In this context, the Laplace's equation perfectly matches the incompressible fluids (water is a good example).

PS I did not worry about the sign of equations.

Answer 2

It's enlightening to note that the adjoint of $\nabla$ is $-\text{div}$, so that $-\text{div} \nabla$ has the familiar pattern $A^T A$, which recurs throughout linear algebra. Hence you would expect (or hope) $-\text{div} \nabla$ to have the properties enjoyed by a symmetric positive definite matrix -- namely, the eigenvalues are nonnegative and there is an orthonormal basis of eigenvectors.

$-\text{div}$ is sometimes called the "convergence", and this viewpoint suggests that the Laplacian should have been defined as the convergence of the gradient.