Intuition behind $\nabla \cdot \frac{1}{\rho}\nabla$?

Question

Oftentimes instead of the Laplacian I notice the very similar operator

$$\nabla \cdot \frac{1}{\rho}\nabla$$

What is the intuition behind this operator? How does it differ intuitively from the Laplace operator $\Delta$?

Answer 1

Take for example the heat equation where the temperature $T$ is being studied. The simplest model of heat is where the flow of heat through a material is in a direction parallel to the gradient of temperature (but oppositely aligned because it flows from hot to cold) with a proportional index $C$ that is the heat conduction of the material. If the material properties remain constant in time, then $C=C(x)$ does not depend on time. For a small volume $V$ within the material, the time rate of change of heat in the volume is the negative of the heat flow out through the surface: \begin{align} \frac{dH}{dt} & = -\frac{d}{dt}\int_{\partial V}C(x)\nabla T(x,t)\cdot \hat{n} dS \\ & = -\frac{d}{dt}\int_{V}\nabla\cdot\left(C(x)\nabla T\right)dV. \end{align} So, the volume rate of heat flow into a volume is $\nabla\cdot(C(x)\nabla T)$. The Laplacian occurs when the heat conduction rate throughout the material is constant. Otherwise, $C(x)$ is the heat conduction function; the reciprocal of conduction would be a resistance, which is your index $\rho$. This is one way to interpret your index $\rho$, which may depend of $x$.