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)?

**Answer**

Assume the function f is C2 in a neighbourhood of 0∈Rn. Using the Taylor expansion of f at 0 one can prove the following:

Δf(0)=lim

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?

**Attribution***Source : Link , Question Author : koletenbert , Answer Author : Community*