Laplace equation

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.