= Laplacian
{c}
{title2=$\Delta$}
{wiki}
The <Laplacian> takes the <divergence> of a <scalar field>'s <gradient>. For a <real-valued function> $f$ on an open region of <Euclidean space>, with continuous <partial derivatives> through order two, its expression in a <Cartesian coordinate system> is
$$
\Delta f=\nabla\cdot\nabla f=\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2}=\operatorname{tr}(D^2f).
$$
Thus the <Laplacian>, also called the <Laplace operator>, adds the coordinate <second derivatives>; equivalently it is the <matrix trace> of the <Hessian matrix>. Its output is another <scalar field>.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others>[Volume II, §2–7, equations 2.52–2.54: divergence of a gradient and the Cartesian Laplacian]{ref}
Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite <second derivatives> cancel. It is a <harmonic function>, although its <graph of a function> is not a <plane>. For the <Gaussian function> $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation gives
$$
\Delta h=4(r^2-1)e^{-r^2}.
$$
Its <Laplacian> is negative inside the <circle> of <radius> one, zero on that <circle>, and positive outside it. The <Gaussian function> itself is positive throughout: its value and its <Laplacian> describe different properties.
\Image[/laplacian-bowl-saddle-and-gaussian.png]
{title=The same <Laplacian> measures a bowl, a saddle and a <Gaussian function>}
{description=Original analytic surfaces. Height shows $f$; the common color scale shows $\Delta f$, with blue negative, neutral gray zero and red positive. Vertical scales differ. The black <circle> on the <Gaussian function> is $r=1$, where its <Laplacian> changes sign.}
The sum does not favor a particular orientation. Under a change of <Cartesian coordinate system> represented by an <orthogonal matrix> $Q$, the <Hessian matrix> becomes $Q^{\mathsf T}(D^2f)Q$. The <cyclic property of the trace> and $QQ^{\mathsf T}=I$ preserve its <matrix trace>. This makes the <Laplacian> useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.
For a <vector field>, the Euclidean <Laplacian> acts separately on components expressed in a fixed <Cartesian coordinate system>. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others>[Volume II, component formula following equation 2.54 and final paragraph of §2–8: fixed rectangular components]{ref} The scalar formula in <Laplacian in polar coordinates> is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.
Back to article page
