Laplace's and Poisson's Equations

Sort the second-order linear partial differential equations by the sign of their principal symbol and one family stands apart: the elliptic equations, whose prototype is the Laplacian $\nabla^2$ . The wave equation (hyperbolic) carries signals along characteristics; the heat equation (parabolic) smears them irreversibly forward in time; but $\nabla^2$ has no real characteristics at all. There is no "before" and no "along a ray" — data placed anywhere on the boundary is felt smoothly and instantly throughout the interior. This is the analytic fingerprint of equilibrium: what remains after every transient has died. Poisson's equation $\nabla^2 u = f$ ties a scalar potential $u$ to a prescribed source density $f$ ; delete the source and you have Laplace's equation $\nabla^2 u = 0$ , whose solutions are the harmonic functions. The pattern is universal: in electrostatics $u$ is the potential and $f=-\rho/\varepsilon_0$ ; in steady heat flow $u$ is temperature and $f$ the scaled heat source; in Newtonian gravity $u=\Phi$ with $f=4\pi G\rho$ ; in ideal fluid flow $u$ is a velocity potential with $f=0$ . One operator, one theory, four physics. Harmonicity is rigid. A harmonic function equals its own average over any sphere centred inside its domain — the mean-value property — and from that single fact everything elliptic descends: the maximum principle, real-analyticity, and uniqueness of the Dirichlet problem. To solve boundary-value problems you exploit the geometry, separating variables in Cartesian, cylindrical, or spherical coordinates, or convert the problem into an integral over sources and boundaries using the Green's function — the potential of a single point source.

Key takeaways

Poisson's equation $\nabla^2 u = f$ and its homogeneous form Laplace's equation $\nabla^2 u = 0$ are the canonical elliptic PDEs: they describe static equilibrium and potential fields, never propagation.
Solutions of $\nabla^2 u = 0$ are harmonic functions; each obeys the mean-value property, and from it the maximum principle, analyticity, and Dirichlet uniqueness all follow.
A well-posed problem pairs the equation with Dirichlet ( $u$ given), Neumann ( $\partial_n u$ given), or Robin data; Dirichlet has a unique solution, Neumann is unique up to a constant and only solvable when $\int_V f = \oint_S g$ .
The free-space Green's function is $-1/(4\pi r)$ in three dimensions and $(1/2\pi)\ln r$ in two, and it generates the integral representation underlying gravity and electrostatics.
The standard solution machinery is separation of variables, eigenfunction expansions, the method of images, and Green's functions — the geometry of the boundary picks the tool.
The same operator governs electrostatic potential, steady temperature, incompressible irrotational flow, and gravity: wherever a field relaxes to equilibrium, it is harmonic where the source is absent.
Key idea: a harmonic function is its own spatial average — $u(\mathbf{r}_0)$ equals the mean of $u$ over any sphere about $\mathbf{r}_0$ — and every elliptic property (smoothness, the maximum principle, uniqueness) flows from that one law.

The core insight

A harmonic function is its own spatial average: $u(\mathbf{r}_0)$ equals the mean of $u$ over any sphere centred on $\mathbf{r}_0$ . Everything elliptic — smoothness, the maximum principle, uniqueness of the Dirichlet problem — flows from this single property.

Where this lives today

The Poisson equation is one of the most-solved equations in computational science. Every semiconductor device simulator alternates a Poisson solve for the electrostatic potential with a transport step for the carriers, and a transistor model's accuracy rests on getting that elliptic solve right at nanometre scale. In cosmology, N-body codes compute the gravitational potential by solving $\nabla^2\Phi=4\pi G\rho$ on a mesh at every timestep — usually with a fast Fourier transform, which is separation of variables done numerically. Incompressible fluid solvers hit a pressure-Poisson equation at each step to enforce $\nabla\cdot\mathbf{v}=0$ , typically the most expensive part of the computation. The maximum principle is not decorative: it guarantees the stability of the discretizations, and finite-element and boundary-element methods lean on the Green representation formula to turn volume problems into surface ones. On the inverse side, electrical-impedance tomography tries to recover an unknown conductivity from boundary voltage-current data — a notoriously ill-posed inverse elliptic problem at the heart of some medical imaging — while conformal mapping still furnishes the closed-form benchmarks the solvers are checked against.

The physical picture

Pin a soap film or a rubber membrane along a bent wire frame and let go. With no load it settles into the shape of least area, and the height field it takes is harmonic: at every interior point the membrane sits at the average height of the points around it, so there is no residual bump to relax. That is the mean-value property made tactile. Press the membrane with a distributed load and you add a source term — Laplace becomes Poisson. The Laplacian $\nabla^2 u$ is precisely the local mismatch between $u$ at a point and the average of its neighbours: positive where the point is a dip, negative where it is a bump, and zero everywhere for a field in perfect equilibrium.

Figure

The mean value property in two forms. Continuously, u at the centre equals the average of u over any circle or sphere about that point. Discretely, on a square grid the centre value 50 is exactly the average of its four neighbours 100, 0, 60 and 40 — a harmonic field has no bump anywhere to relax

Because harmonic functions cannot hide an interior extremum, a conductor's potential, a steady temperature field, and an incompressible flow speed are all "as flat as the boundary allows": the hottest and coldest points always sit on the surface, never buried inside. The same rigidity makes the elliptic problem a boundary-value problem rather than an initial-value one — nothing marches forward in time; you specify data all around the region and the interior is fixed at once. The most powerful way to package that is the Green's function — the field of a single unit point source — from which the response to any source and any boundary data follows by superposition.

Figure

Method of images for a point charge q a distance d above a grounded plane. The true field above the plane equals that of q plus a fictitious image −q at the mirror point, so the plane stays at zero potential. The charge is pulled down by F = −q²/(16πε₀d²), which for q = 1 nC and d = 1 cm is 2.25×10⁻⁵ N

The core equations

The homogeneous elliptic equation is Laplace's. It asserts that the field has zero local curvature excess everywhere in the interior — no sources, no sinks, pure equilibrium. Its solutions are, by definition, the harmonic functions.

\nabla^2 u(\mathbf{r}) = 0 \qquad (1)

where: $\nabla^2$ — the Laplacian, $\partial_{xx}+\partial_{yy}+\partial_{zz}$ in Cartesian coordinates; $u(\mathbf{r})$ — the scalar potential or field. Reading it: at every interior point $u$ balances its surroundings exactly, so the field is fixed by its boundary values. Restore a source and you have Poisson's equation. The right-hand side is where the physics enters — charge, heat, or mass density, each scaled by the constants of its theory.

\nabla^2 u(\mathbf{r}) = f(\mathbf{r}) \qquad (2)

where: $f(\mathbf{r})$ — prescribed source density, e.g. $-\rho/\varepsilon_0$ in SI electrostatics or $4\pi G\rho$ in gravity. Reading it: the local imbalance of $u$ is set point-by-point by the source; wherever $f=0$ the equation collapses back to Laplace's. To solve problems in curved geometry you need the Laplacian in the right coordinates. In spherical coordinates it splits into a radial piece and an angular piece whose eigenfunctions are the spherical harmonics.

\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \qquad (3)

where: $r,\theta,\phi$ — radial distance, polar angle, azimuthal angle. Reading it: separating $u=R(r)Y(\theta,\phi)$ turns the PDE into a radial ODE plus the angular eigenvalue problem $-\ell(\ell+1)$ that yields the Legendre and spherical-harmonic solutions. The single most important structural fact about harmonic functions is the mean-value property. It says the value at a point is not free — it is the exact average of the field over any sphere surrounding it.

u(\mathbf{r}_0) = \frac{1}{4\pi R^2}\oint_{|\mathbf{r}-\mathbf{r}_0|=R} u\,\mathrm{d}S \qquad (4)

where: $R$ — radius of any sphere about $\mathbf{r}_0$ inside the harmonic region; $\mathrm{d}S$ — its surface area element. Reading it: harmonic values cannot spike above their surroundings, which is why the next equation holds. The maximum principle is the mean-value property's most useful corollary: a harmonic function attains its largest and smallest values only on the boundary, never strictly inside.

\max_{\overline{V}} u = \max_{\partial V} u,\qquad \min_{\overline{V}} u = \min_{\partial V} u \qquad (5)

where: $\overline{V}$ — closed domain (interior plus boundary); $\partial V$ — its boundary. Reading it: an object held at fixed boundary temperatures has no hidden hot spot inside. The workhorse for inhomogeneous problems is the free-space Green's function — the potential of a unit point source, from which everything else is built by superposition.

G(\mathbf{r},\mathbf{r}') = -\frac{1}{4\pi|\mathbf{r}-\mathbf{r}'|}\ (3\text{D}),\qquad G = \frac{1}{2\pi}\ln|\mathbf{r}-\mathbf{r}'|\ (2\text{D}) \qquad (6)

where: $G$ — solution of $\nabla^2 G = \delta(\mathbf{r}-\mathbf{r}')$ ; $\mathbf{r}'$ — source location. Reading it: by linearity the field of a distributed source is the source-weighted integral of $G$ , with boundary values entering as a surface integral. Convolving the source with $G$ and adding the boundary terms gives the Green representation formula — the exact solution as an integral over sources plus one over the boundary.

u(\mathbf{r}) = \int_V G\,f(\mathbf{r}')\,\mathrm{d}V' + \oint_{S}\left[ G\,\partial_{n'} u - u\,\partial_{n'} G \right]\mathrm{d}S' \qquad (7)

where: $V$ — interior domain; $S$ its boundary; $\partial_{n'}$ — outward normal derivative in the source variable. Reading it: choosing $G$ to vanish on $S$ kills the $\partial_{n'}u$ term and expresses $u$ purely from the source and the known boundary data. In spherical geometry, separation of variables gives the axisymmetric solution as a sum over Legendre polynomials — the backbone of the multipole expansion.

u(r,\theta) = \sum_{\ell=0}^{\infty}\left( A_\ell\,r^\ell + B_\ell\,r^{-(\ell+1)} \right) P_\ell(\cos\theta) \qquad (8)

where: $P_\ell$ — Legendre polynomial of degree $\ell$ ; $A_\ell,B_\ell$ — coefficients fixed by boundary conditions, with $B_\ell=0$ for regularity at the origin and $A_\ell=0$ for decay at infinity. Reading it: interior problems keep the growing branch $r^\ell$ , exterior problems the decaying branch $r^{-(\ell+1)}$ , and the boundary data fixes the coefficients.

Derivation AThe mean value property of harmonic functions

The goal is to establish the property that makes elliptic theory rigid: that the average of a harmonic $u$ over a sphere lying inside its domain is independent of the sphere's radius, and therefore equals the value at the centre. The plan is to define the spherical mean $M(R)$ , differentiate it in the radius, convert the surface integral into a volume integral of the Laplacian by the divergence theorem, and observe that harmonicity forces $\mathrm{d}M/\mathrm{d}R$ to vanish. A local statement, $\nabla^2 u = 0$ , thereby becomes a global averaging law.

Step 1. Define the spherical mean $M(R)=\frac{1}{4\pi R^2}\oint_{S_R}u\,\mathrm{d}S=\frac{1}{4\pi}\int u(\mathbf{r}_0+R\hat{\mathbf{n}})\,\mathrm{d}\Omega$ , the average of $u$ over solid angle. Why: the solid-angle form pulls all $R$ -dependence into the argument, so it differentiates cleanly.

Step 2. Differentiate under the integral: $\frac{\mathrm{d}M}{\mathrm{d}R}=\frac{1}{4\pi}\int \hat{\mathbf{n}}\cdot\nabla u\,\mathrm{d}\Omega=\frac{1}{4\pi R^2}\oint_{S_R}\partial_n u\,\mathrm{d}S$ . Why: $\partial_R$ of $u(\mathbf{r}_0+R\hat{\mathbf{n}})$ is the outward normal derivative on the sphere.

Step 3. Apply the divergence theorem: $\oint_{S_R}\partial_n u\,\mathrm{d}S=\int_{B_R}\nabla^2 u\,\mathrm{d}V=0$ , so $M(R)$ is constant. Why: the gradient's flux equals the Laplacian's volume integral, which vanishes for a harmonic $u$ .

Step 4. Let $R\to0$ : continuity gives $M(R)\to u(\mathbf{r}_0)$ , and constancy then makes it equal $u(\mathbf{r}_0)$ at every radius. Why: constancy propagates the small-sphere value to all radii — the mean-value property.

Go deeper · Derivation B · 1 minThe free space Green's function of the Laplacian

Derivation BThe free space Green's function of the Laplacian

The goal is to compute the potential of a single unit point source — the function $G$ satisfying $\nabla^2 G=\delta(\mathbf{r}-\mathbf{r}')$ — because once it is known, the solution of any free-space Poisson problem is the source convolved with $G$ . The plan uses spherical symmetry: solve Laplace's equation for $G$ away from the singular point, where the radial harmonic is a constant plus a $1/r$ term, then fix that term's strength by integrating over a tiny ball and demanding the flux reproduce the unit source. The 2D case follows the same route with a logarithm.

Step 1. Place the source at the origin and use symmetry: $G=G(r)$ with $\nabla^2 G=\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\!\left(r^2\frac{\mathrm{d}G}{\mathrm{d}r}\right)=0$ for $r\neq0$ . Why: away from the source the equation is homogeneous, and a point source produces a radially symmetric field.

Step 2. Integrate the radial ODE: $r^2 G'=\text{const}$ gives $G=-A/r+B$ , and set $B=0$ so $G\to0$ at infinity. Why: a localized source must produce a decaying potential, which pins $B$ .

Step 3. Fix $A$ from the source: integrate $\nabla^2 G=\delta$ over a ball $B_\epsilon$ , so $\oint_{S_\epsilon}G'\,\mathrm{d}S=\frac{A}{\epsilon^2}(4\pi\epsilon^2)=4\pi A=1$ , giving $A=1/4\pi$ and $G=-1/(4\pi r)$ . Why: the delta integrates to one, and the surface area cancels the $1/\epsilon^2$ in $G'$ to fix the strength.

Step 4. Repeat in 2D with $G=(1/2\pi)\ln r$ : here $\oint G'\,\mathrm{d}\ell=(1/2\pi r)(2\pi r)=1$ . Why: the circle of circumference $2\pi r$ again cancels the $1/r$ from the log — this is the potential of a line charge. *Why it works: $G=-1/(4\pi r)$ directly reproduces the Coulomb potential of a point charge under $\nabla^2 V=-\rho/\varepsilon_0$ , and the 2D logarithmic form $(1/2\pi)\ln r$ is the potential of a line charge.*

Go deeper · Derivation C · 1 minUniqueness of the Dirichlet problem by the energy method

Derivation CUniqueness of the Dirichlet problem by the energy method

The goal is to prove that a solution of Poisson's equation with prescribed boundary values is unique — at most one potential is consistent with the source and the boundary data — while showing that the Neumann problem is unique only up to an additive constant. The plan is the classical energy argument: form the difference $w$ of two supposed solutions, note it is harmonic with zero boundary data, feed it into Green's first identity, and watch the gradient energy of $w$ collapse to zero, forcing $w$ constant and then zero. It rests on nothing beyond the divergence theorem.

Step 1. Suppose $u_1,u_2$ both solve $\nabla^2 u=f$ in $V$ with $u=g$ on $S$ , and set $w=u_1-u_2$ ; then $\nabla^2 w=0$ in $V$ and $w=0$ on $S$ . Why: subtracting cancels the source, leaving a homogeneous problem with zero boundary data.

Step 2. Apply Green's first identity to $w$ : $\int_V\!\left(w\,\nabla^2 w+|\nabla w|^2\right)\mathrm{d}V=\oint_S w\,\partial_n w\,\mathrm{d}S$ . Why: this is the divergence theorem applied to $w\nabla w$ , packaging the terms we control.

Step 3. Kill the boundary term ( $w=0$ on $S$ ) and the $w\nabla^2 w$ term (harmonicity), leaving $\int_V|\nabla w|^2\,\mathrm{d}V=0$ . Why: the integrand is non-negative, so a zero integral forces $\nabla w=0$ everywhere.

Step 4. Conclude $w=\text{const}$ , and $w=0$ on $S$ pins it to zero, so $u_1\equiv u_2$ . Why: for pure Neumann data the surface term vanishes because $\partial_n w=0$ , giving $\nabla w=0$ but no way to fix the constant — hence uniqueness only up to a constant.

Limiting cases

The elliptic equations sit at the confluence of the other classical PDEs — each appears as a limit in which time-dependence or oscillation switches off:

Case	Result	Meaning
Steady-state limit of diffusion	$\partial_t u=\kappa\nabla^2 u+s\ \xrightarrow{\partial_t u\to0}\ \nabla^2 u=-s/\kappa$	As transients decay ( $t\to\infty$ ) the heat equation relaxes to Poisson's equation, and to Laplace's if there is no source.
Zero-frequency limit of Helmholtz	$(\nabla^2+k^2)u=0\ \xrightarrow{k\to0}\ \nabla^2 u=0$	Helmholtz collapses to Laplace as the wavenumber vanishes; standing waves become harmonic equilibria.
Source-free regions	$\nabla^2 u=f\ \xrightarrow{f=0}\ \nabla^2 u=0$	Poisson reduces to Laplace wherever the source density is zero — charge-free space, or a body with no internal heating.
Two dimensions and analyticity	$\nabla^2 u=0\iff u=\operatorname{Re} w(z),\ w\ \text{holomorphic}$	Every 2D harmonic function is the real part of an analytic function, so conformal mapping and complex analysis apply directly.

Common trapEvery Poisson problem has a solution

It does not. The pure Neumann problem $\nabla^2 u=f$ in $V$ with $\partial_n u=g$ on $S$ is solvable only if the data satisfy the compatibility condition $\int_V f\,\mathrm{d}V=\oint_S g\,\mathrm{d}S$ . This is Gauss's law promoted to a constraint: the total source enclosed must equal the net flux out through the boundary, because integrating the equation and applying the divergence theorem gives $\int_V f=\oint_S \partial_n u=\oint_S g$ with no room to negotiate. Violate it — try to insulate a body ( $g=0$ ) while heating it internally ( $\int_V f\neq0$ ) — and no steady state exists; the temperature keeps rising. Even when the condition holds, the Neumann solution is fixed only up to an additive constant, the potential's harmless gauge freedom. And mind the sign: SI electrostatics writes $\nabla^2 V=-\rho/\varepsilon_0$ , so dropping the minus sign silently flips the source.

Worked examples

1 · A grounded conducting sphere in a uniform field (fully worked)

A grounded conducting sphere of radius $a=0.05\ \mathrm{m}$ is placed in a previously uniform electric field $E_0=1000\ \mathrm{V/m}$ pointing along $\hat{\mathbf{z}}$ . Find the potential outside the sphere, the induced surface charge density, and the induced dipole moment.

Step 1. Set up the boundary-value problem. Outside the sphere the region is charge-free, so $\nabla^2 u=0$ and the axisymmetric solution is equation (8). Far away the field is uniform, $\mathbf{E}=E_0\hat{\mathbf{z}}$ , so $u\to-E_0 r\cos\theta$ — the $\ell=1$ term with $A_1=-E_0$ , all other $A_\ell=0$ . Regularity at infinity leaves only the decaying $B_1 r^{-2}\cos\theta$ , so $u(r,\theta)=\left(-E_0 r+B_1 r^{-2}\right)\cos\theta$ .

Step 2. Apply the boundary condition. The grounded sphere sits at zero potential, $u(a,\theta)=0$ for all $\theta$ , which forces $-E_0 a+B_1 a^{-2}=0$ , hence $B_1=E_0 a^3$ . The potential is therefore

u(r,\theta)=-E_0\!\left(r-\frac{a^3}{r^2}\right)\cos\theta.

The first term is the applied field; the second is the sphere's response — a pure dipole.

Step 3. Compute the surface charge. The radial field is $E_r=-\partial u/\partial r=E_0\!\left(1+2a^3/r^3\right)\cos\theta$ , and at the surface $r=a$ this is $E_r(a,\theta)=3E_0\cos\theta$ . The induced surface charge density is $\sigma=\varepsilon_0 E_r(a)=3\varepsilon_0 E_0\cos\theta$ , peaking at the poles. Numerically σ_max = 3ε₀E₀ = 3(8.854×10⁻¹² )(1000) = 2.66×10⁻⁸ C/m² ≈ 26.6 nC/m², positive on the upstream pole and negative on the downstream one.

Step 4. Read off the dipole. Matching $B_1 r^{-2}\cos\theta$ to the dipole potential $p\cos\theta/(4\pi\varepsilon_0 r^2)$ gives $p=4\pi\varepsilon_0 E_0 a^3$ . Numerically p = 4π(8.854×10⁻¹²)(1000)(0.05)³ = 1.39×10⁻¹¹ C·m. Uniqueness guarantees this is the solution: any function harmonic outside, zero on the sphere, and matching the uniform field at infinity must be this one.

2 · A point charge above a grounded plane (method of images)

A charge $q=1\ \mathrm{nC}$ sits a distance $d=1\ \mathrm{cm}$ above an infinite grounded conducting plane. Find the force on the charge and the induced surface charge density.

Step 1. Replace the conductor by an image. Put a charge $-q$ at the mirror point, a distance $d$ below the plane; the superposed potential of $q$ and $-q$ vanishes on the plane by symmetry, so in the upper half-space it satisfies both Poisson's equation (source $q$ ) and the boundary condition. By uniqueness it is the potential — conductor and image are indistinguishable above the plane.

Step 2. Compute the force. The real charge feels only the image (its own field exerts no net force), and the two are separated by $2d$ , so the attraction is

F=-\frac{q^2}{4\pi\varepsilon_0(2d)^2}=-\frac{q^2}{16\pi\varepsilon_0 d^2}.

Numerically F = −(1×10⁻⁹)² / [16π(8.854×10⁻¹²)(0.01)²] = −2.25×10⁻⁵ N, the minus sign meaning it is pulled toward the plane.

Step 3. Find the induced charge. The normal field just above the plane is $E_z=-q d/[2\pi\varepsilon_0(s^2+d^2)^{3/2}]$ at distance $s$ from the foot of the charge, so the induced surface density is $\sigma(s)=\varepsilon_0 E_z=-q d/[2\pi(s^2+d^2)^{3/2}]$ . It is most negative right under the charge, σ(0) = −q/(2πd²) = −1.59×10⁻⁶ C/m² = −1.59 μC/m², and integrating $\sigma$ over the whole plane returns exactly $-q$ — the conductor draws up precisely enough charge to terminate every field line from $q$ .

Physics of the portrait

Laplace's equation carries Laplace's name, but the equation that governs the world inside matter — the one with a source on the right — belongs to his student. Siméon Denis Poisson took the empty-space law and asked what happens where the charge, or the mass, actually is.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · potential theory and mathematical physics

Siméon Denis Poisson was steered toward medicine before his gift for mathematics carried him to the École Polytechnique, where Laplace and Lagrange recognized him at once and made him a protégé. He became one of the great analytical physicists of the Napoleonic era, and his name is stamped across the subject: the Poisson bracket of mechanics, the Poisson ratio of elasticity, the Poisson distribution of probability, and — most relevant here — Poisson's equation. Laplace had written $\nabla^2 V=0$ for the gravitational potential in empty space, but it was Poisson, in an 1813 memoir on the attraction of spheroids, who showed that inside a mass or charge distribution the potential instead obeys $\nabla^2 V=-4\pi\rho$ , inserting the source term that Laplace's vacuum equation had omitted. That single correction converted a special-case law into the universal field equation of Newtonian potential theory and electrostatics. Poisson worked relentlessly — "life is good for only two things," he is supposed to have said, "doing mathematics and teaching it" — and laid groundwork that Green, Gauss, and later Maxwell would build the full theory of fields upon.

History

1782

— Legendre introduces the polynomials Pℓ that solve the axisymmetric angular equation

1785

— Laplace writes ∇²V = 0 for the gravitational potential of a spheroid in empty space

1813

— Poisson adds the source term, giving ∇²V = −4πρ inside matter

1828

— Green's Essay introduces the Green's function and Green's identities

1840

— Gauss founds potential theory with the mean-value and flux theorems

Going further

Texts. Arfken and Weber, Mathematical Methods for Physicists, chapters on PDEs and potential theory · Hassani, Mathematical Physics, on Green's functions and elliptic operators · Stone and Goldbart, Mathematics for Physics, on PDEs and Green's functions · Byron and Fuller, Mathematics of Classical and Quantum Physics, on boundary-value problems · Jackson, Classical Electrodynamics, chapters 1–3 · Evans, Partial Differential Equations, chapter 2. Key results: the Green's functions are $-1/(4\pi r)$ (3D) and $(1/2\pi)\ln r$ (2D), derived by the Gauss-law integration in the derivation above; the Neumann compatibility condition is $\int_V f=\oint_S g$ ; example 1 gives $\sigma=3\varepsilon_0 E_0\cos\theta$ with $\sigma_{\max}=26.6\ \mathrm{nC/m^2}$ and $p=4\pi\varepsilon_0 E_0 a^3=1.39\times10^{-11}\ \mathrm{C\,m}$ ; example 2 gives $F=-q^2/(16\pi\varepsilon_0 d^2)=-2.25\times10^{-5}\ \mathrm{N}$ and $\sigma(0)=-1.59\ \mathrm{\mu C/m^2}$ .

Prerequisites. Classification of Second-Order Linear PDEs · Boundary Conditions and Well-Posedness · Vector calculus and the divergence theorem. Leads to. Separation of Variables in Cartesian, Spherical, and Cylindrical Coordinates · Green's Functions for PDEs · The Helmholtz Equation.

Ask the tutor: Ask me to walk through the grounded-sphere or image-charge example step by step, or to show how separation of variables turns Laplace's equation into Legendre or Bessel functions in your coordinate system.

What is the difference between Laplace's and Poisson's equations?

They are the same operator with and without a source. Laplace's equation $\nabla^2 u=0$ describes a region with no sources — charge-free space, a body with no internal heating, incompressible irrotational flow — and its solutions are the harmonic functions. Poisson's equation $\nabla^2 u=f$ restores the source term $f$ , which carries the physics: $-\rho/\varepsilon_0$ for electrostatics, $4\pi G\rho$ for gravity, $-q/k$ for heat. Wherever the source vanishes, Poisson reduces to Laplace, so the two are one equation seen inside and outside the matter that drives it. That is exactly how they arose: Laplace wrote the vacuum form, Poisson added the source.

What does it mean that these equations are elliptic?

Ellipticity is a statement about the principal symbol — the second-derivative part of the operator. For $\nabla^2$ that symbol is $-(k_x^2+k_y^2+k_z^2)$ , which never vanishes for a real nonzero wavevector, so the operator has no real characteristics. There are no preferred directions along which information travels, unlike the wave equation, where signals ride characteristic cones at finite speed. Instead, boundary data influences the entire interior at once and solutions are automatically smooth — even analytic. This is why elliptic equations model equilibrium and are posed as boundary-value problems, with nothing to evolve in time.

Why does the mean-value property matter so much?

Because nearly every distinctive feature of harmonic functions is a corollary of it. If a function equals its own average over every small sphere, it can have no strict interior maximum or minimum — the maximum principle — and two solutions with the same boundary data cannot differ, giving uniqueness. It also forces analyticity, since averaging is smoothing infinitely repeated. And it gives a numerical method: replacing "value equals average of neighbours" on a grid yields the relaxation and Gauss-Seidel schemes that were the original way to solve Laplace's equation by hand. A single averaging identity organizes the smoothness theory, the uniqueness theory, and the computation at once.

What are Dirichlet, Neumann, and Robin boundary conditions?

They are the three standard ways to pin down a solution. Dirichlet data fixes the value of $u$ on the boundary — a conductor at a set potential, or a surface at a set temperature. Neumann data fixes the normal derivative $\partial_n u$ — the flux, such as a prescribed heat flow or an insulated wall where $\partial_n u=0$ . Robin data fixes a combination $\alpha u+\beta\,\partial_n u$ , as in convective cooling where heat loss depends on surface temperature. The Dirichlet problem has a unique solution; the pure Neumann problem is unique only up to a constant and requires the compatibility $\int_V f=\oint_S g$ ; Robin conditions generally restore uniqueness by breaking the constant freedom.

How does the method of images actually work?

It is uniqueness put to work. To find the field of a charge near a grounded conductor you would, in principle, solve Poisson's equation with a complicated boundary condition. Instead you guess: replace the conductor with one or more fictitious "image" charges placed so that, with the real charge, they reproduce the correct boundary value — zero potential on a grounded plane, for instance. If your guess satisfies Poisson's equation in the physical region with the right source and matches the boundary data, the uniqueness theorem certifies it is the answer, however you found it. The images live in the forbidden region and are never real; they are a bookkeeping device that turns a boundary-value problem into a free-space superposition.

Why is the Green's function so central?

Because it reduces every linear problem to one canonical solve. The Green's function $G$ is the field of a single unit point source, the solution of $\nabla^2 G=\delta$ . By linearity the field of any distributed source is the source-weighted integral of $G$ , and the representation formula lets the boundary data enter through a companion surface integral of $G$ . Solve for $G$ once in a given geometry and you have solved for every source and every boundary condition there. This is why Green's functions underlie both the analytic machinery — multipole expansions, image charges — and the numerical one, where boundary-element methods discretize this representation to trade a volume problem for a surface one.