Laplacian ($\Delta$)

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 isThus 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.^[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 givesIts 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.

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 $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ 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.^[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.

For a constant orthogonal matrix $R$, the Laplacian is unchanged by the coordinate transformation $\xi =Rx$. If $u$ is twice continuously differentiable, the chain rule givesbecause ${\sum }_i{R}_{ji}{R}_{ki}={\delta }_{jk}$ by orthogonality of the rows. Here ${\delta }_{jk}$ is the Kronecker delta. In two dimensions, the rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$ makes the mixed partial derivatives cancel explicitly. The same argument works for reflections and translations, so the Laplacian depends on the Euclidean geometry rather than the orientation or origin of the coordinates.

The Laplacian detects whether a value lies below or above nearby values, to second order. For a scalar field whose partial derivatives through order four are continuous functions, take the arithmetic mean of four equally spaced neighbors:Applying Taylor's theorem in each coordinate cancels the first- and third-order terms between opposite neighbors. The two pairs leaveThis is an asymptotic expansion, rather than an equality for arbitrary $h$. Rearranging gives the familiar central finite difference approximationThis finite difference replaces partial derivatives by five sampled values. The arithmetic mean lies above the center for a sufficiently small $h$ if the Laplacian is strictly positive there, and below if it is strictly negative. A zero Laplacian cancels this second-order discrepancy without requiring equality of individual values.

At $(0,0)$, the bowl $x^2+y^2$ has center value zero and four neighbor values $h^2$, so the discrepancy is exactly $h^2$. The saddle $x^2-y^2$ has neighbor values $h^2,h^2,-h^2,-h^2$, whose arithmetic mean is exactly zero. For the Gaussian function $e^{-x^2-y^2}$ the discrepancy is $e^{-h^2}-1=-h^2+O(h^4)$, agreeing with $\Delta f(0,0)=-4$.

A harmonic function need not equal this four-neighbor arithmetic mean for a finite $h$. The polynomial $f(x,y)=x^4-6x^2{y}^2+y^4$ has $\Delta f=0$ everywhere, but $A_{h}f(0,0)=h^4$. Higher-order terms remain even when the second-order terms cancel.

At a local minimum, each coordinate second derivative is at least zero, hence the Laplacian is at least zero. At a local maximum the inequality reverses. The sum alone does not characterize an extremum: cancellation at a saddle point of a scalar function can also produce zero. The local-average interpretation explains why the heat equation smooths differences, and helps motivate the heat equation maximum principle.

For diffusion with a constant diffusion coefficient $D>0$ and no sources, the model flux is $J=-D\nabla u$. A continuity equation gives $u_t=-\nabla \cdot J=D\Delta u$. This proportionality is a constitutive assumption about transport; the continuity equation expresses conservation. If $D$ varies with position, it must remain inside the divergence, giving $u_t=\nabla \cdot (D\nabla u)$ instead.^[ref]

An exact constructed heat equation example on the whole plane uses dimensionless coordinates with $D=1$ and the smooth initial Gaussian function $u(x,y,0)=e^{-r^2}$. Set $s=1+4t$. For $t\ge 0$,Direct differentiation verifies both the initial condition andThe center decreases, while the region $r^2>s$ increases at that instant. The sign-change circle expands with radius $\sqrt{s}$. A fixed outer location can initially increase and later decrease as this circle passes it.

At time $t$, the peak is $1/s$, and the radius where the profile is $1/e$ times its peak is $\sqrt{s}$. The width grows as the height falls. Integration in polar coordinates givesThus a decreasing peak is compatible with a conserved spatial integral. Dividing $u$ by $\pi$ makes a probability density function with coordinate variances $s/2=1/2+2t$, connecting the same spreading to Brownian motion with diffusion coefficient one. Equivalently the profile is evolved by the Gaussian heat kernel with time parameter $2t$, since that article uses generator $\frac{1}{2}\Delta$.

For a twice differentiable scalar field on the plane,

For $x=r\cos \theta$, $y=r\sin \theta$ and $r>0$, set $c=\cos \theta$, $s=\sin \theta$. The differential operators are ${\partial }_x=c{\partial }_r-(s/r){\partial }_{\theta }$ and ${\partial }_y=s{\partial }_r+(c/r){\partial }_{\theta }$. Applying them twice givesTheir sum is the Laplacian in polar coordinates. The first-derivative terms come from differentiating the coordinate coefficients, as required for second derivatives under a two-variable coordinate change. The origin is excluded from the coordinate formulas, even when the underlying Cartesian function is smooth there.

For twice differentiable scalar fields,

The biharmonic operator applies the Laplacian twice. Under clamped boundary conditions, two integrations by parts give the positive quadratic form

The clamped problem uses the weak solution identity $\int \Delta u\Delta v=\int fv$ for every $v$ in the clamped second-order Sobolev space. For $f\in L^2(U)$, the clamped Hessian identity and the Poincaré inequality make this a bounded coercive bilinear form. The Lax-Milgram theorem gives a unique weak solution without a mean-zero condition on $f$.

The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

For angular Fourier modes of integer order $n\ge 2$, the biharmonic equation in polar coordinates has radial basis $r^n,r^{-n},r^{n+2},r^{2-n}$. Apply the Laplacian twice to obtain the displayed indicial equation. For $n=0$, repeated roots give the radial basis $1,\log r,r^2,r^2\log r$; for $n=1$, the basis is $r^{-1},r,r\log r,r^3$. These are Cauchy-Euler equations. Regularity at the origin and boundary conditions select the admissible powers. For $n=2$, a constant radial factor still multiplies $\cos 2\theta$ or $\sin 2\theta$, so it is bounded but fails continuity at the origin.

If $H,G$ are holomorphic functions, both real and imaginary parts of $\bar{z}H(z)+G(z)$ solve the biharmonic equation. In Wirtinger derivatives, $\Delta =4{\partial }_z{\partial }_{\bar{z}}$, so $\Delta (\bar{z}H+G)=4H^{\prime }$ is harmonic and its next Laplacian is zero. This construction gives local representations of planar solutions of the biharmonic equation; it is useful for the biharmonic stream function for planar Stokes flow.

For a smooth three-dimensional biharmonic function regular throughout a ball, its surface average has exactly this form. Taylor expansion and rotationally invariant spherical moments turn each even term into a multiple of an iterated Laplacian; terms from the fourth derivative onward vanish when ${\Delta }^{2}f=0$. This is the mean-value identity that converts the reciprocal surface average into Faxén's first law.