Ethan's Notes

    Laplacian

    #Math
    Intuition: higher values of $\displaystyle \Delta f(x,y)$ correspond to $\displaystyle f(x,y)$ 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

    $\displaystyle \Delta=\nabla ^{2}=\partial_{i}^{2}$

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

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

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

    $\displaystyle \Delta f=\partial ^{2}_{i}f$

    $\displaystyle \Delta \vec{F}=\partial_{i}^{2}F_{i}$

    Other Coordinates

    How to remember the Laplacian in spherical and cylindrical coordinates

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

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