Poisson's equation

On a periodic domain, integrating $\Delta \Phi =f$ gives $\int f=0$ because boundary fluxes cancel. On a periodic grid the row sum of the discrete Laplacian is zero, so ${\sum }_j{f}_j=0$ is necessary. The constant gravitational potential remains an arbitrary gauge.

Let $Y_{\ell }$ be a spherical harmonic of degree $\ell$ on the unit sphere. The Laplacian in spherical coordinates separates its angular part from a radial power, giving the displayed identity for $r>0$. Thus a Poisson equation $\Delta \Phi =r^q{Y}_{\ell }$ has the particular solution $\Phi =r^{q+2}{Y}_{\ell }/[(q+2)(q+3)-\ell (\ell +1)]$ when the denominator is nonzero. Add $A{r}^{\ell }{Y}_{\ell }+B{r}^{-\ell -1}{Y}_{\ell }$ to impose the radial boundary conditions. If the denominator vanishes at $p=q+2$, differentiating the displayed identity with respect to $p$ gives instead $\Delta (r^p\log r{Y}_{\ell })=(2p+1)r^{p-2}{Y}_{\ell }$. Regularity at the origin and decay at infinity must be checked separately; the radial identity alone is local away from zero.

For integer angular frequency $m\ge 0$, the Laplacian in polar coordinates gives the displayed identity on $r>0$ by direct differentiation. Consequently a source $\beta r^q\cos (m\theta )$ has particular solution $\beta r^{q+2}\cos (m\theta )/[(q+2{)}^2-m^2]$ when the denominator is nonzero. The sine case is identical. If $s=q+2=\pm m\ne 0$, differentiating the identity with respect to $s$ gives the resonant particular solution $\beta r^s\ln r\cos (m\theta )/(2s)$. For $s=m=0$, the radial source $\beta r^{-2}$ instead has particular solution $\beta (\ln r{)}^2/2$. These are local statements on the punctured plane; boundedness, smoothness at the origin and boundary data must be imposed separately by adding suitable harmonic functions.

For $\beta >0$, solve ${\nabla }^2\varphi =-\rho$ in three dimensions with $\rho (r)=A{e}^{-\beta r}/r$, finite potential at the origin and decay at infinity. The Green function is $1/(4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|)$ in this sign convention. The angular integral supplied by the spherical shell theorem givesThe value at the origin is the limit $A/\beta$. A second derivation sets $u=r\varphi$: the radial Poisson equation becomes $u^{\prime \prime }=-A{e}^{-\beta r}$. Decay excludes the constant-potential term and boundedness at the origin cancels the remaining $1/r$ singularity. Although the source is locally integrable, it is singular at the origin; the resulting potential solves the equation there in the sense of distributions, without an extra Dirac delta.

For the Poisson equation $\Delta u={\sum }_{n\ge 1}{b}_n\sin (n\theta )$ on the unit disk with zero Dirichlet boundary condition, separation of variables in polar coordinates gives $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. For $n\ne 2$, boundedness at $r=0$ and $R_n(1)=0$ select $R_n=(r^n-r^2)/(n^2-4)$. The resonant mode is $R_2=r^2\log r/4$. For discontinuous angular data, the Poisson equation is interpreted in its weak formulation; its Dirichlet energy gives uniqueness. Orthogonality of the angular Fourier modes gives $\int ufdA=-{\sum }_n\pi b_n^2/[4(n+2{)}^2]$, with the resonant case obtained directly or by the limit $n\to 2$.

On a smooth bounded connected domain, the homogeneous Neumann problem seeks $u\in H^1(U)$ such that $\int \nabla u\cdot \nabla v=\int fv$ for every $v\in H^1(U)$. For $f\in L^2(U)$ it is solvable exactly when $\int f=0$, and the weak solution is unique up to constants. The Poincare-Wirtinger inequality and the Riesz representation theorem give existence on the mean-zero Sobolev space. The test function one gives necessity of the compatibility condition.

In the upper half-space $z>0$, the displayed decaying harmonic function has outward normal derivative ${\partial }_n\Phi =-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$. This is the Poisson kernel for the upper half-space, whose integral on a horizontal plane is one and whose limit as $z\downarrow 0$ is the two-dimensional Dirac delta. Thus ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional boundary sense. The compensating flux through a large hemisphere is minus one. This illustrates why a nonzero total boundary source is possible on an unbounded domain even when the interior source is zero.

For a radial Dirichlet solution on a fixed three-dimensional shell $a<r<b$, $(r^2{U}^{\prime }{)}^{\prime }=r^{2}F$. Thus $r^2{U}^{\prime }=C_0+{\int }_a^r{s}^{2}F(s)ds$, and $U(a)=U(b)=0$ bounds $|C_0|$ by the supremum of that integral. Cauchy-Schwarz inequality in the radial volume measure bounds the integral by $C\Vert f{\Vert }_2$. Since $r\ge a>0$, the displayed gradient bound follows. The estimate exploits radiality and does not extend to arbitrary forcing in three dimensions.

Inside any three-dimensional domain containing a ball, let $u_{\epsilon }(x)={\epsilon }^{1/2}\varphi ((x-x_0)/\epsilon )$ for a fixed smooth compactly supported nonconstant function. The right-hand side $f_{\epsilon }=\Delta u_{\epsilon }$ has constant $L^2$ norm, whereas $\Vert \nabla u_{\epsilon }{\Vert }_{\infty }={\epsilon }^{-1/2}\Vert \nabla \varphi {\Vert }_{\infty }$ diverges. The functions have zero boundary trace, giving a scaling counterexample to a uniform $L^2$-forcing gradient estimate.

A Poisson kernel represents the harmonic extension of Dirichlet boundary data by an integral over the boundary. When the Dirichlet Green function is normalized by $-\Delta G_{\Omega }=\delta$, its negative outward normal derivative gives the displayed kernel. The Poisson kernel for the upper half-space and the Poisson kernel for the upper half-plane give explicit examples with different boundary dimensions. Boundedness or other stated growth conditions are needed when selecting a solution in an unbounded domain.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

Take boundary data one on a disk of radius $a>0$ and zero outside. The Poisson kernel for the upper half-space gives on the axisThe axis value tends to one at the boundary and behaves as $a^2/(2z^2)$ far away. The boundary data at the circular rim are understood almost everywhere: changing a value on the rim, a set of planar measure zero, leaves the integral unchanged. This does not assert a continuous boundary extension across the jump.

For the complex upper half-plane, the Poisson kernel isIt is the density of harmonic measure on the real boundary as viewed from $x+iy$.

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

For bounded continuous boundary data $f$, the half-plane Poisson kernel is the law of $x+yC$. On one common probability space, $x+yC\to x_0$ almost surely as $(x,y)\to (x_0,0)$ with $y>0$. Boundedness of $f$ and dominated convergence theorem imply $\varphi (x+iy)\to f(x_0)$. This proves continuity at every finite boundary point even for tangential approach. No moment of the Cauchy distribution is needed.

The displayed Poisson integral is the bounded harmonic function on the upper half-plane with continuous boundary values $f(x,0)=e^{-|x|}$. Taking a Fourier transform in $x$ reduces Laplace's equation to ${\hat{f}}_{yy}-s^2\hat{f}=0$, and the bounded solution has multiplier $e^{-y|s|}$. The convolution theorem gives the displayed expression. At frequency zero the transformed function is constant in $y$, not zero: its value is the total boundary integral, two.

The function decays uniformly as $x^2+y^2\to \infty$. For large $y$, use $P_y\le 1/(\pi y)$ and the boundary L1 norm two. When $y$ is bounded and $|x|$ is large, split the convolution into $|t|\le |x|/2$ and its complement: the kernel is small on the first part, and the exponential boundary value is uniformly small on the second. This also distinguishes spatial decay of the harmonic function from decay of its zero-frequency Fourier coefficient.

For $f$ holomorphic on a neighbourhood of the radius-$R$ closed disk and $|z|<R$, the displayed representation holds even though $f$ may be complex-valued. For $z\ne 0$, put $w=R^2/\bar{z}$. On $|\zeta |=R$, the Poisson kernel equals $\zeta /(\zeta -z)-\zeta /(\zeta -w)$. Replacing $d\theta$ by $d\zeta /(i\zeta )$ reduces the formula to the Cauchy integral formula minus a zero contour integral whose pole lies outside. The case $z=0$ is the mean value property for holomorphic functions.

For boundary value $c$ on $(-a,a)$ and zero outside, integrating the Poisson kernel for the upper half-plane gives the displayed bounded harmonic function for $y>0$. It approaches the assigned values at boundary continuity points and the midpoint value $c/2$ at either jump. Changing the value at an isolated endpoint does not alter the integral. This is the unique bounded harmonic extension; pointwise decay in the vertical direction alone, without a spatial growth restriction, does not give uniqueness.

The conformal map $w=z^2$ sends $x,y>0$ to the complex upper half-plane. For bounded continuous boundary values $g_2$ on the horizontal axis and $g_1$ on the vertical axis with equal corner values, set $f(s)=g_2(\sqrt{s})$ for $s>0$ and $f(s)=g_1(\sqrt{-s})$ for $s<0$. The bounded harmonic function with those Dirichlet boundary conditions isEquivalently, if $z^2=a+ic$ with $c>0$, this is $\frac{2c}{\pi }{\int }_0^{\infty }r\left[g_2(r)/((r^2-a{)}^2+c^2)+g_1(r)/((r^2+a{)}^2+c^2)\right]dr$. The Poisson kernel is a positive approximate identity, so the prescribed edge values are attained. Boundedness is essential for uniqueness on this unbounded domain.

Two bounded harmonic functions continuous to the quadrant edges with the same Dirichlet boundary conditions are equal. Their difference is bounded and zero on both axes. Odd reflection across each axis gives a bounded harmonic extension away from the corner; the removable singularity for a bounded harmonic function fills in that corner. The harmonic Liouville theorem makes that extension constant, and oddness makes the constant zero. This does not exclude unbounded examples such as the zero-boundary harmonic growth in a quadrant.

The harmonic function $v(x,y)=xy$ vanishes on both axes but is nonzero in the quadrant, and $v_z=-iz/2$. More generally $\mathrm{Im}{z}^{2m}$, for a positive integer $m$, vanishes on both axes and is harmonic. Adding these unbounded functions leaves all edge Dirichlet boundary conditions unchanged. Thus decay of the boundary data does not imply decay of every solution, and an edge-data-only formula cannot determine an arbitrary solution's Wirtinger derivative. A boundedness or suitable growth condition must select the solution.

For the bounded Dirichlet Poisson integral in a quadrant, the Wirtinger derivative isThis follows by differentiating the Schwarz integral formula composed with $z^2$, remembering that for real boundary data $u_z$ is half the derivative of a holomorphic function whose real part is $u$. For complex boundary data, apply the construction to the real and imaginary parts separately and use linearity. If the data and their derivatives decay sufficiently, integration by parts givesThe endpoint terms cancel exactly when $g_1(0)=g_2(0)$. Interior denominators do not vanish because $\mathrm{Im}{z}^2>0$.

The Schwarz integral formula reconstructs a holomorphic function from boundary values of its real part. For a real boundary function $f$ on the complex upper half-plane satisfying ${\int }_{\mathbb{R}}|f(s)|/(1+s^2)ds<\infty$, the regularized form isIts real part is the Poisson integral of $f$. The subtraction makes the integral converge and changes only the imaginary normalization. Integration by parts, when the boundary terms vanish, gives $W^{\prime }(w)=(\pi i{)}^{-1}\int f^{\prime }(s)/(s-w)ds$; jumps of $f$ are included as Dirac delta function terms in its distributional derivative.

For real boundary data $u$, the displayed holomorphic integral has real part equal to the disk Poisson integral of $u$ and imaginary part its normalized harmonic conjugate. If $u\ge 0$ is not identically zero, the real part is strictly positive inside the disk. A holomorphic logarithm then has imaginary part between $-\pi /2$ and $\pi /2$, while its real part records the logarithm of the magnitude. This separates uniform imaginary-part control from large real parts on small boundary sets.

For a harmonic function in the quadrant with boundary derivatives $u_y(0,y)=g_1(y)$ and $u_x(x,0)=g_2(x)$, integrating the data to equal corner values and applying the Schwarz integral formula after the conformal map $w=z^2$ givesThese are tangential boundary derivatives, so uniqueness requires an additional normalization or growth restriction. A homogeneous contribution has the form $ih(z^2)/z$ with $h$ holomorphic in the complex upper half-plane and real on its boundary away from zero. For example, $u=xy$ has zero prescribed derivatives on both edges.