Laplace operator

The Laplace operator is the divergence of the gradient, $\Delta =\nabla \cdot \nabla$.

The nonnegative Neumann Laplacian is the self-adjoint realization of $-\mathrm{div}\mathrm{grad}$ with vanishing outward normal derivative. Its weak quadratic form is the Dirichlet energy on $H^1$. On a connected compact domain the constants give a one-dimensional zero eigenvalue space.

The nonnegative Dirichlet Laplacian is the self-adjoint realization of $-\mathrm{div}\mathrm{grad}$ with zero boundary values. On a bounded domain, its weak form is the Dirichlet energy on $H_0^1$. For polygonal domains the weak operator domain is preferable to a requirement of smoothness at corners.

In ${\mathbb{R}}^{d+1}$, $g=d{r}^2+r^2{g}_{S^d}$ and the volume density is $r^d$. With the nonnegative Laplacian convention, $\tilde{\Delta }=-{\partial }_r^2-(d/r){\partial }_r+r^{-2}{\Delta }_{S^d}$. The identity separates radial homogeneity from the angular Laplace-Beltrami operator and derives the spectrum of the Laplacian on a sphere.

The infinitesimal rotation $R_{ij}=x_i{\partial }_j-x_j{\partial }_i$ commutes with the Laplacian: the extra terms $2{\partial }_i{\partial }_j$ cancel. It is tangent to every sphere, so it preserves zero Dirichlet traces on spherical shells. If $\Delta u=f$ and $f$ is radial, uniqueness applied to $R_{ij}u$ forces all these rotation derivatives to vanish, hence $u$ is radial.

Laplace's equation isIts solutions are harmonic functions.

An axisymmetric harmonic potential outside a sphere of radius $r_0$, with far field $v_{0}r\cos \theta$ and zero radial derivative on the sphere, isThe regular spherical expansion in Legendre polynomials fixes all coefficients. A spatial constant remains arbitrary if only the far-field gradient is specified, and is fixed by prescribing the potential difference at infinity.

In three dimensions,Consequently the decaying solution of $-{\nabla }^{2}u=f$ in free space is

Up to normalization and sign convention, a fundamental solution of the Laplace equation is $\Phi (x)=|x{|}^{2-n}$ for $n\ge 3$ and $\Phi (x)=\log |x|$ for $n=2$. It is harmonic away from the origin and satisfies $\Delta \Phi =c_n{\delta }_0$ distributionally.

In plane polar coordinates, Laplace's equation isSeparation of variables gives radial powers $r^n,r^{-n}$ for positive angular modes and $1,\log r$ for the zero mode.

In cylindrical coordinates $(r,\theta ,z)$, Laplace's equation is

In the unit half-cylinder $z\ge 0$, specifying only the curved-side boundary values does not determine a bounded harmonic function. The displayed nonzero function, where $j_{0,1}$ is a positive zero of the Bessel function of the first kind $J_0$, is regular at the axis, bounded, harmonic, and zero on $r=1$. It may be added to any particular solution without changing the side data, and even decays at infinity. Boundary information at the base $z=0$ is still required for uniqueness.

For radial $u(r)$ in $n$ dimensions, $\Delta u=u\prime \prime +(n-1)u\prime /r$.

The spherical-coordinate Laplacian splits into a radial operator and the sphere’s angular Laplacian divided by $r^2$.

Axisymmetric harmonic functions separate into radial powers times Legendre polynomials.

A Laplacian eigenfunction satisfies $-\Delta u=\lambda u$ together with specified boundary conditions.

A Dirichlet Laplacian eigenfunction vanishes on the boundary and satisfies $-\Delta u=\lambda u$.

For $n\ge 3$ and a bounded smooth domain, a Dirichlet Laplacian eigenfunction with eigenvalue $\lambda$ satisfies the displayed estimate. A signed power test for a Laplacian eigenfunction and the zero-boundary Sobolev inequality give a Moser iteration with exponents $p_j=2[n/(n-2){]}^j$. The geometric series ${\sum }_{j}1/p_j=n/4$ gives the exact power of $\lambda$, while ${\sum }_j(\log p_j)/p_j<\infty$ controls the remaining constant. The constant depends only on dimension, because the zero-boundary Sobolev inequality follows by extension by zero to the whole space.

A Dirichlet Laplacian eigenvalue is a number $\lambda$ for which $-\Delta u=\lambda u$ has a nonzero solution with zero boundary trace. At such a value, the shifted operator $\Delta +\lambda$ is not invertible.