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.

On $0<x<L_x$, $0<y<L_y$ with positive side lengths and zero Dirichlet boundary conditions, separation of variables for $-\Delta u=\lambda u$ giveswith the displayed eigenvalues. The one-dimensional endpoint conditions exclude constant and hyperbolic factors and force the positive integer sine frequencies. The product Fourier sine series basis is complete in the square-integrable functions on the rectangle. Consequently an eigenfunction with prescribed $\lambda$ is a linear combination of exactly those modes with ${\lambda }_{mn}=\lambda$; if none match, only the zero solution exists. Distinct pairs with the same sum give an eigenvalue with geometric multiplicity greater than one. On a square of side $\pi$, for instance, $\lambda =25$ has precisely the pairs $(3,4)$ and $(4,3)$, so its eigenspace has dimension two.

If ${\beta }_1,{\beta }_2,{\beta }_3$ are the affine barycentric coordinates of an equilateral triangle of side $\ell$, then $h={\prod }_j\sin (\pi {\beta }_j)=\frac{1}{4}{\sum }_j\sin (2\pi {\beta }_j)$ vanishes on the boundary and satisfies $\Delta h=-16{\pi }^{2}h/(3{\ell }^2)$. Thus the problem $\Delta q-4\lambda q=0$ with zero Dirichlet boundary data has distinct solutions zero and $h$ at the displayed negative parameter. Their normal traces differ, demonstrating failure of a uniquely determined Dirichlet-to-Neumann map at resonance.

For the positive Dirichlet Laplacian $A$ on a bounded smooth domain, with normalized eigenfunctions $e_j$ and positive eigenvalues ${\mu }_j$, define $A^{\alpha }u={\sum }_j{\mu }_j^{\alpha }(u,e_j)e_j$ on the vectors for which ${\sum }_j{\mu }_j^{2\alpha }|(u,e_j){|}^2<\infty$. The resulting Hilbert space embeds in the Sobolev space $H^{2\alpha }$, with the boundary conditions appropriate to the exponent. In particular $H_{\alpha }$ embeds in $C^{0,\beta }$ whenever $0<\beta <\min (1,2\alpha -n/2)$. High powers impose further boundary compatibility; ordinary smoothness alone does not imply membership in every fractional domain.

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.

For $0\le r<1$, this Poisson kernel for the unit disk has the uniformly convergent Fourier seriesThe identity follows by taking the real part of the geometric series $1+2{\sum }_{n\ge 1}(r{e}^{i\theta }{)}^n$. For integrable $2\pi$-periodic boundary data $f$, its Poisson extension iswhere $a_n,b_n$ are the usual Fourier coefficients. Each term is a harmonic function in the open disk. For continuous boundary data, the extension solves the Dirichlet problem with those boundary values.

On a disk of radius $a$, prescribe $V_1$ on the upper open semicircle and $V_2$ on the lower one. The bounded harmonic extension has the formula in the title for $r<a$. The boundary data have only a constant term and odd sine Fourier coefficients, giving $u=(V_1+V_2)/2+[2(V_1-V_2)/\pi ]{\sum }_{m\ge 0}(r/a{)}^{2m+1}\sin ((2m+1)\theta )/(2m+1)$. The complex series is half of $\log [(1+w)/(1-w)]$, where $w=(r/a)e^{i\theta }$. Its quotient has positive real part, so taking the imaginary part yields the displayed principal arctangent without an argument ambiguity. At the two boundary jumps, the radial limit is the mean; the harmonic function has no continuous extension matching both adjacent constant arcs there.

For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.

In the upper half-plane outside a circle of radius $R$ centered at $i{y}_0$, with $y_0>R$, set $b=\sqrt{y_0^2-R^2}$. The displayed harmonic function is zero on the real axis and equals $V_0$ on the circle. Indeed, the Apollonius circle with ratio $P>1$ has center $ib(P^2+1)/(P^2-1)$ and radius $2bP/(P^2-1)$; hence $\log P=\mathrm{arcosh}(y_0/R)$. The logarithmic singularity at $ib$ is inside the excluded circle, and the one at $-ib$ is below the real axis. The real part of a local holomorphic logarithm is harmonic even if its imaginary part is multivalued. The potential tends to zero at infinity. Finite boundary values alone do not impose a growth condition at infinity; uniqueness in the unbounded domain needs such an additional condition, for example boundedness with the prescribed limit.

A single-valued harmonic function in a planar annulus has radial Fourier modes $1,\log r$ and $r^n,r^{-n}$ multiplying $\cos (n\theta ),\sin (n\theta )$ for integers $n\ge 1$. Substitute $R(r)\Theta (\theta )$ in Laplace's equation: periodicity gives ${\Theta }^{\prime \prime }+n^2\Theta =0$, while $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Regularity at the origin excludes $\log r,r^{-n}$; boundedness at infinity excludes $\log r,r^n$. Coefficients must give locally convergent series, with the boundary regularity required by the particular Dirichlet problem. Mere finiteness at each finite exterior point does not imply boundedness at infinity.

For a harmonic function on the planar annulus $a<r<b$, the radial amplitude of a sine or cosine Fourier mode of positive integer index $n$ is a linear combination of $r^n$ and $r^{-n}$. If its values at $a,b$ are $A_n,B_n$, respectively, the displayed hyperbolic-sine formula gives that combination. It follows either by solving the two boundary equations or by putting $s=\log (r/a)$, which changes the radial equation to $R_{ss}-n^{2}R=0$. The zero angular mode is instead affine in $\log r$:Superposing the modes gives the harmonic function with suitable prescribed Dirichlet boundary conditions. For a fixed interior radius, high angular modes from either boundary are exponentially suppressed. Piecewise smooth boundary data are attained at continuity points, while jumps have the usual mean-value Fourier limit.

For $0<x<L$, $0<y<H$, choose $k=m\pi /H>0$ and seek a harmonic function with zero values on the horizontal edges, boundary value $\sin (ky)$ at $x=L$, and $\varphi +\alpha {\partial }_n\varphi =0$ at $x=0$, with constant $\alpha \ge 0$. Separation of variables gives $\varphi =[\alpha k\cosh (kx)+\sinh (kx)]\sin (ky)/[\alpha k\cosh (kL)+\sinh (kL)]$. The outward derivative at the left edge is $-{\partial }_x$, giving $F(0)-\alpha F^{\prime }(0)=0$. The denominator is positive and Poisson uniqueness with a nonnegative Robin normal coefficient proves uniqueness in the regular solution class.

For a regular interior and a decaying exterior harmonic potential with continuous values on the unit sphere, the same coefficient multiplies the two radial branches of each Legendre polynomial mode. Its exterior-minus-interior radial derivative is $-(2n+1)a_n{P}_n$. Thus an axisymmetric prescribed jump $j_n{P}_n(\cos \theta )$ produces coefficient $a_n=-j_n/(2n+1)$. This diagonal matching rule reduces a spherical interface problem to its angular-mode coefficients.

For the Laplace equation on $0<y<1$, prescribe bottom data $g$ and zero top data. A horizontal Fourier transform reduces the equation to ${\hat{u}}_{yy}-s^2\hat{u}=0$, giving the displayed multiplier and its limit $1-y$ at $s=0$. Integrable transformed data give a bounded inverse-transform solution. The bounded solution is unique by the maximum principle for harmonic functions, using barriers $\epsilon \cosh (\delta x)\cos (\delta (y-1/2))$ for $0<\delta <\pi$ on expanding rectangles. Without a growth condition, one can add $e^{\pi x}\sin (\pi y)$, so the two boundary traces alone do not determine an arbitrary harmonic function on an infinite strip.

For a single-valued harmonic function tending to zero both at the origin and at infinity, match an angular Fourier series across the unit circle. A nonzero mode with continuous interface value $K_n\cos (n\theta )$ has interior field $K_n{r}^n\cos (n\theta )$ and exterior field $K_n{r}^{-n}\cos (n\theta )$. Its exterior-minus-interior radial derivative is $-2n{K}_n\cos (n\theta )$. Thus a prescribed jump coefficient $j_n$ determines $K_n=-j_n/(2n)$. Sine modes obey the same relation. The constant mode must have zero jump under these decay conditions.

The imaginary part of the entire function $e^{z^2}$ is a harmonic function. On the upper half-plane it vanishes at $y=0$ and tends to zero as $y\to \infty$ for every fixed $x$, but is nonzero and grows rapidly in the horizontal direction. Adding it to a Poisson integral preserves those boundary and vertical-decay conditions. A boundedness or suitable transform-growth condition is therefore needed to make the half-plane Dirichlet problem unique.

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

This is the free-space Green function for the negative Laplace operator.

For three noncompact spatial coordinates of radius $r$ and a circle of circumference $L$, the positive Laplace Green function on the source slice is $G(r,0)={\sum }_{n\in \mathbb{Z}}[4{\pi }^2(r^2+n^2{L}^2){]}^{-1}=(4\pi Lr{)}^{-1}\coth (\pi r/L)$. At $r\ll L$ this is $1/(4{\pi }^2{r}^2)$; at $r\gg L$ it is $1/(4\pi Lr)$ with exponentially small corrections. The long-distance coefficient also follows by replacing the dense image sum with $L^{-1}{\int }_{-\infty }^{\infty }dy/(r^2+y^2)=\pi /(Lr)$. Electric and gravitational potentials inherit this crossover after multiplying by their signed source couplings.

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.

A real harmonic function on the unit disk has the displayed expansion, convergent locally uniformly with its derivatives in the interior. If its continuous boundary data have Fourier coefficients $A_0,A_n,B_n$ in the convention $f(\theta )=A_0/2+{\sum }_{n\ge 1}(A_n\cos (n\theta )+B_n\sin (n\theta ))$, its harmonic extension has $a_0=A_0/2$, $a_n=A_n$, $b_n=B_n$.

To obtain the radial factors, separation of variables in the Laplace equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-\lambda R=0$ and ${\Theta }^{\prime \prime }+\lambda \Theta =0$. Angular periodicity restricts nonconstant modes to $\lambda =n^2$ for positive integers $n$. The radial factors are $r^n,r^{-n}$, with $1,\log r$ for the constant angular mode. Boundedness at the origin removes $r^{-n}$ and $\log r$. Each surviving factor $r^n\cos (n\theta )$ or $r^n\sin (n\theta )$ is a harmonic polynomial, the real or imaginary part of $(x+iy{)}^n$. The expansion can also be derived from the Poisson integral on the unit disk for the disk.

For $0<\theta <\gamma$ and $0<r<R$, a bounded harmonic function that vanishes on the straight edges and has outer boundary data $\phi$ has coefficientsThis follows by separation of variables and Fourier orthogonality. Negative radial powers are excluded by boundedness at the vertex. Reflection-symmetric data $\phi (\theta )=\phi (\gamma -\theta )$ have only odd Fourier modes. Knowing the resulting solution on the bisector determines all these odd coefficients by its radial power series. Without reflection symmetry, the even angular modes vanish on the bisector and cannot be recovered from that restriction alone.

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

For $r>0$, the substitution $y=ru$ removes the first derivative in the three-dimensional radial Laplacian, since $u^{\prime \prime }+2u^{\prime }/r=y^{\prime \prime }/r$. Bounded $u$ at zero implies $y(0)=0$. Bounded $u$ at infinity means $y=O(r)$; it does not require bounded $y$. For example, constant forcing $g=1$ has the bounded particular solution $u=-1$, whose reduced function is $y=-r$. A Dirichlet half-line Green function for d2 minus 1 can still represent this solution when its integral converges, but the resulting reduced function has the growth demanded by the forcing.

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.

For boundary data $f(u)$ on the sphere $r=a$, with $u=\cos \theta$, define its Legendre polynomial coefficients by $f_{\ell }=(2\ell +1){\int }_{-1}^{1}f(u)P_{\ell }(u)du/2$. Separation of variables in the Laplace equation gives radial factors $r^{\ell }$ and $r^{-\ell -1}$. Regularity at the origin selects the first for the interior Dirichlet problem; decay at infinity selects the second for the exterior problem. Matching each boundary coefficient gives the displayed series. For continuous boundary data the resulting harmonic functions attain those data, with convergence interpreted through their harmonic extension rather than requiring uniform convergence of the unweighted boundary series. The maximum principle for harmonic functions proves uniqueness in the bounded interior and decaying exterior classes.

Suppose the exterior of a sphere also has zero Dirichlet boundary conditions on the part of its equatorial plane outside the sphere. The plane separates the upper and lower domains. Extend the sphere data of each hemisphere oddly across the equator, and apply the axisymmetric harmonic extension from a sphere separately to these two odd extensions. Only odd-degree Legendre polynomials occur, and every such polynomial vanishes at $u=\cos \theta =0$. Thus the resulting harmonic function vanishes on the plane. For upper data $f(u)$ on $0\le u\le 1$, the exterior coefficients are $(2\ell +1){\int }_0^{1}f(u)P_{\ell }(u)du$ for odd $\ell$. If $f(0)=0$, the plane and sphere data agree at their shared edge. The same construction with the lower data gives the lower solution; their normal derivatives need not match through a conducting sheet.

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

A scalar $\lambda$ for which a nonzero Laplacian eigenfunction satisfies this equation, with specified boundary conditions or on a closed manifold, is an eigenvalue of the positive Laplacian. A Dirichlet Laplacian eigenvalue imposes zero boundary values; other boundary conditions generally change the spectrum.

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.