Laplacian

Intuition: higher values of $\Delta f(x,y)$ correspond to $f(x,y)$ being more like a valley as opposed to a hill. Basically tells you how much a point is like a local minimum. Analogous to second derivative in 1D calculus

$\Delta ={\nabla }^2={\partial }_i^2$

• ${\partial }_i^2$ is in Einstein Notation
• For 3 dimensions, $\Delta =\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}\right)$

$\nabla \cdot (\nabla \vec{f})=\Delta f\ne \nabla (\nabla \cdot \vec{v})$

• Divergence of gradient of $\vec{f}$ is the laplacian of $\vec{f}$

$\Delta f={\partial }_i^{2}f$

$\Delta \vec{F}={\partial }_i^2{F}_i$

Other Coordinates

How to remember the Laplacian in spherical and cylindrical coordinates

${\nabla }^2=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\left(\frac{{\partial }^2}{\partial {\varphi }^2}\right)$

• Laplacian in spherical coordinates

${\nabla }^2=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{{\partial }^2}{\partial z^2}$

• Laplacian in cylindrical coordinates