Green's Functions via Eigenfunction Expansion and the Method of Images

Solving an inhomogeneous linear PDE on a bounded domain $\Omega$ — Poisson's equation with a source, the driven Helmholtz equation in a cavity, heat flow from a hot spot — collapses to a single integral once you know one object: the Green's function $G(\mathbf{r},\mathbf{r}')$ , the field at $\mathbf{r}$ from a unit point source at $\mathbf{r}'$ . It is defined by $\mathcal{L}_{\mathbf{r}} G=\delta(\mathbf{r}-\mathbf{r}')$ together with homogeneous boundary conditions of the same type — Dirichlet, Neumann, or Robin — as the physical problem. Once $G$ is in hand, $\mathcal{L}$ has been inverted and the solution is quadrature: convolve $G$ against the source, add a surface term for the boundary data. This article is about the two great constructions of $G$ , which live in opposite regimes yet must agree. The first is spectral: Sturm–Liouville theory gives self-adjoint $\mathcal{L}$ a complete orthonormal eigenbasis $\{\phi_n\}$ , so both $\delta$ and $G$ expand in them and the coefficients fall out as $1/(\lambda_n-\lambda)$ . This works wherever separation of variables gives the spectrum and exposes resonances as poles, but it converges slowly, building a delta-singularity out of smooth modes. The second is geometric: the method of images adds fictitious sources in the forbidden region whose symmetry forces $G$ to vanish (or have zero normal derivative) on the boundary — one image charge for a plane, a fractional Kelvin image for a sphere, a ring of $2m$ images for a wedge of opening $\pi/m$ . It is closed-form and exact where the series struggles, but exists only with enough reflection or inversion symmetry. Because both produce the unique $G$ of a well-posed self-adjoint problem, equating them turns geometry into an eigenfunction-series identity: images are a resummation of the spectrum.

Key takeaways

A boundary-value Green's function solves $\mathcal{L}G=\delta(\mathbf{r}-\mathbf{r}')$ with homogeneous boundary conditions, converting any inhomogeneous PDE plus boundary data into a volume integral over sources plus a surface integral.
The eigenfunction method gives the spectral form $G=\sum_n \phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')/(\lambda_n-\lambda)$ — exact, with a pole at every eigenvalue, but slowly convergent near the source.
The method of images places fictitious sources so the free-space kernel automatically satisfies $G=0$ (Dirichlet) or $\partial_n G=0$ (Neumann); it is closed-form but restricted to planes, spheres, and wedges of angle $\pi/m$ .
Both yield the same $G$ ; equating a Kelvin image with a spherical-harmonic series shows images are a geometric resummation of the eigenfunction sum.
Reciprocity $G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}',\mathbf{r})$ follows from Green's second identity whenever $\mathcal{L}$ is self-adjoint — the face of source–detector exchange.
The pure-Neumann Laplacian has no ordinary $G$ : the constant zero mode diverges the $n=0$ term, forcing a modified Green's function with a uniform background subtracted.
Key idea: a Green's function is the inverse of a differential operator restricted to functions obeying the homogeneous boundary conditions, and you can build that inverse spectrally — sum over eigenmodes weighted by $1/(\lambda_n-\lambda)$ — or geometrically — superpose free-space sources whose symmetry kills the boundary term. Both give the same unique object.

The core insight

A Green's function is the inverse of a differential operator restricted to functions satisfying the homogeneous boundary conditions; you can build that inverse spectrally (sum over eigenmodes, $1/(\lambda_n-\lambda)$ ) or geometrically (superpose free-space sources whose symmetry kills the boundary term). Both give the same unique object.

Where this lives today

The two constructions are working machinery, not museum pieces. In electrostatics and MEMS, the image force on a charge near a conductor sets the pull-in behaviour of actuators (worked example 2). In condensed matter the spectral Green's function is the observable: the local density of states seen by a scanning tunnelling microscope is $\rho(\mathbf{r},E)=-\tfrac{1}{\pi}\,\mathrm{Im}\,G(\mathbf{r},\mathbf{r};E+i0^{+})$ , its poles at $\lambda_{mn}$ read straight off quantum-dot spectra. In acoustics the Helmholtz Green's function's poles are the room modes a hall designer must place and damp, while the image method builds the early-reflection impulse response every geometric-acoustics and VR audio engine simulates. The heat kernel with an absorbing wall — free kernel minus Dirichlet image — gives the first-passage and survival distributions behind stochastic-process and barrier-option calculations. Across numerical PDE, boundary-element and fast-multipole methods build on the free-space Green's function, replacing a volume mesh with a surface integral. An open problem is computing $G$ efficiently in disordered or near-resonant geometries where neither images close nor the series converges quickly — the domain of modern hybrid and machine-learned solvers.

The physical picture

Picture $G(\mathbf{r},\mathbf{r}')$ as the shape of a stretched drumhead pushed up by a unit needle at $\mathbf{r}'$ while its rim is pinned to zero. Two ways describe that dent, and they are the whole subject. The spectral way writes it as a superposition of normal modes: each $\phi_n$ contributes $\phi_n(\mathbf{r}')/\lambda_n$ , so soft modes dominate and the sum rebuilds the pointed dent term by term. The image way sums nothing: it continues the drumhead past its rim and adds a phantom needle pushing down at the mirror point, so by symmetry the displacement is zero along the rim — the boundary condition enforced by geometry, not by a cancelling series. The first figure shows both routes converging on the same pinned dent.

Figure

Two routes to the same G. Left: the eigenfunction expansion builds the pinned dent as a sum of normal modes φ₁ + φ₂ + φ₃ + ⋯, each weighted by 1/λ_n. Right: the method of images places a mirror source of opposite sign across the boundary so that G = 0 on the wall by symmetry — one image replaces the whole series

The two methods are strong in opposite regimes, which is why both survive. On the source the boundary looks infinitely far away, so $G$ is dominated by its free-space singularity — $-1/4\pi|\mathbf{r}-\mathbf{r}'|$ in 3D, $\ln|\mathbf{r}-\mathbf{r}'|/2\pi$ in 2D — which the image form delivers instantly while the series converges painfully; but a box needing reflections off six walls breeds an infinite image lattice, and the triple-sine series wins. Either way the homogeneous condition on $G$ is non-negotiable: $G$ inverts $\mathcal{L}$ on the functions that vanish (or have zero flux) on $\partial\Omega$ , and the nonzero data re-enter through the surface term of Green's second identity. The second figure makes "series resums to a closed form" quantitative on the simplest case: the pinned string.

Figure

The 1D box Green's function on 0 ≤ x ≤ 1 with a source at x′ = 0.3. The exact closed form is the triangle G = x_< (1 − x_>) with peak 0.15 at x = x′; the eigenfunction series 2 Σ sin(nπx) sin(nπx′)/(nπ)² converges to it — one term overshoots to 0.164, five trace the tent, twenty give 0.1499

Reciprocity is the last and deepest piece. Because $\mathcal{L}$ is self-adjoint and $G$ obeys homogeneous boundary conditions, $G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}',\mathbf{r})$ : the field at A from a source at B equals the field at B from a source at A — swap loudspeaker and microphone in a room and the transfer function is unchanged. It is self-adjointness expressed through Green's second identity.

The core equations

Everything begins with the definition: $G$ is the response to a unit point source, carrying the homogeneous version of whatever boundary condition the physical problem imposes. This one line is what makes $\mathcal{L}$ invertible on the constrained function space.

\mathcal{L}_{\mathbf{r}}\,G(\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}'),\qquad G\ \text{obeys the homogeneous BC on}\ \partial\Omega \qquad (1)

where: $\mathcal{L}$ — a self-adjoint linear differential operator, e.g. $-\nabla^2$ or $\nabla^2+k^2$ ; $G(\mathbf{r},\mathbf{r}')$ — the field at $\mathbf{r}$ from a unit point source at $\mathbf{r}'$ ; $\delta(\mathbf{r}-\mathbf{r}')$ — the unit point source; $\partial\Omega$ — the boundary of $\Omega$ . Read it as " $G$ is the inverse $\mathcal{L}^{-1}$ written as a kernel." Once $G$ exists, the full problem $\mathcal{L}u=f$ with prescribed data on $\partial\Omega$ is solved by a volume convolution plus a surface term — the representation formula, where the inhomogeneous boundary data live.

u(\mathbf{r})=\int_\Omega G(\mathbf{r},\mathbf{r}')\,f(\mathbf{r}')\,\mathrm{d}^3r' + \oint_{\partial\Omega}\big[u\,\partial_{n'}G - G\,\partial_{n'}u\big]\,\mathrm{d}S' \qquad (2)

where: $u$ — the solution of $\mathcal{L}u=f$ ; $f$ — the source; $\partial_{n'}$ — outward normal derivative in the primed variable. For Dirichlet data, build $G$ with $G=0$ on $\partial\Omega$ and only $u\,\partial_{n'}G$ survives — known data times a known kernel. The spectral construction inverts $\mathcal{L}$ mode by mode: because the eigenfunctions are complete, $G$ is a sum over them, each weighted by the reciprocal of its distance from the spectral parameter.

G(\mathbf{r},\mathbf{r}')=\sum_n \frac{\phi_n(\mathbf{r})\,\phi_n^*(\mathbf{r}')}{\lambda_n-\lambda} \qquad (3)

where: $\phi_n$ — orthonormal eigenfunctions, $\mathcal{L}\phi_n=\lambda_n\phi_n$ , obeying the homogeneous BC; $\lambda_n$ — eigenvalues of $\mathcal{L}$ on $\Omega$ ; $\lambda$ — spectral parameter in $(\mathcal{L}-\lambda)G=\delta$ ( $\lambda=0$ for the bare operator). It is the partial-fraction expansion of $\mathcal{L}^{-1}$ : a simple pole at every eigenvalue. The image method needs the free-space kernel as raw material — the Coulomb potential in 3D, logarithmic in 2D.

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

where: $|\mathbf{r}-\mathbf{r}'|$ — Euclidean distance; $\ln$ — natural logarithm. Both solve $\nabla^2 G=\delta$ in free space, signs fixed by Gauss's theorem. For a grounded plane one mirror image of opposite sign forces $G=0$ on the wall; for a grounded sphere Kelvin's fractional image at the inverse point does the same by similar triangles.

G_D^{\text{plane}}=-\frac{1}{4\pi}\!\left[\frac{1}{|\mathbf{r}-\mathbf{r}'|}-\frac{1}{|\mathbf{r}-\mathbf{r}'^{*}|}\right],\qquad G_D^{\text{sphere}}=-\frac{1}{4\pi}\!\left[\frac{1}{|\mathbf{r}-\mathbf{r}'|}-\frac{a/r'}{|\mathbf{r}-\mathbf{r}'^{*}|}\right] \qquad (5)

where: plane $\mathbf{r}'^{*}=(x',y',-z')$ is the mirror point; sphere $\mathbf{r}'^{*}=(a^2/r'^2)\mathbf{r}'$ is the inverse point at radius $a^2/r'$ , charge $a/r'$ ; $a$ — radius; $r'=|\mathbf{r}'|$ . Each image sits outside the domain, so $\nabla^2 G=\delta$ is untouched inside. Finally, self-adjointness makes $G$ symmetric under exchange of its arguments — reciprocity.

G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}',\mathbf{r}) \qquad (6)

where: the identity holds for any self-adjoint $\mathcal{L}$ with self-adjoint boundary conditions — the kernel of a symmetric operator is a symmetric kernel.

Derivation AInverting the operator mode by mode

The goal is to derive the spectral form (3) from scratch and see why a pole appears at every eigenvalue. We assume only the Sturm–Liouville guarantee — that the self-adjoint $\mathcal{L}$ has a complete orthonormal eigenbasis $\{\phi_n\}$ obeying the homogeneous boundary conditions — and expand both $\delta$ and $G$ in it, matching coefficients to invert the operator one mode at a time.

Step 1. Write the completeness relation, $\delta(\mathbf{r}-\mathbf{r}')=\sum_n \phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')$ : any point source is a superposition of all modes evaluated at the source point. Why: the delta is the hardest thing to represent, and completeness is exactly the statement that the eigenbasis can do it.

Step 2. Expand $G=\sum_n c_n(\mathbf{r}')\,\phi_n(\mathbf{r})$ , substitute into $(\mathcal{L}-\lambda)G=\delta$ , and use $\mathcal{L}\phi_n=\lambda_n\phi_n$ : the operator becomes multiplication, $(\mathcal{L}-\lambda)G=\sum_n(\lambda_n-\lambda)c_n\,\phi_n$ . Why: expanding in eigenfunctions diagonalizes $\mathcal{L}$ — the differential operator becomes the number $\lambda_n$ on each mode.

Step 3. Match term by term. Orthonormality forces $(\lambda_n-\lambda)c_n=\phi_n^*(\mathbf{r}')$ , so $G=\sum_n \phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')/(\lambda_n-\lambda)$ . Why: an infinite diagonal linear system, each mode inverted independently — the small denominators are where resonance lives.

Step 4. Read off the structure. As a function of $\lambda$ , $G$ has a simple pole with residue $-\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')$ at each $\lambda_n$ , and for real eigenfunctions the exchange symmetry is plain reciprocity. Why: the poles are the spectrum, so where $G$ blows up is where the eigenvalues sit — the content of resonance and the local density of states.

Go deeper · Derivation B · 1 minThe pinned-string Green's function two ways

Derivation BThe pinned-string Green's function two ways

The goal is to build $G$ for the simplest boundary-value problem — $-G''=\delta(x-x')$ on $0\le x\le L$ with $G(0)=G(L)=0$ — by both routes and prove they are the same function, making "images resum the series" concrete and checkable. The operator is $\mathcal{L}=-\mathrm{d}^2/\mathrm{d}x^2$ , eigenfunctions $\phi_n=\sqrt{2/L}\,\sin(n\pi x/L)$ , eigenvalues $\lambda_n=(n\pi/L)^2$ ; we compute the spectral sum, then the closed form, then check numerically that they meet.

Step 1. Spectral form. Insert the eigendata into (3) at $\lambda=0$ : $G=\frac{2}{L}\sum_{n=1}^{\infty}\dfrac{\sin(n\pi x/L)\sin(n\pi x'/L)}{(n\pi/L)^2}$ — exact, but a Fourier sine series building a corner at $x=x'$ from smooth sines, converging like $1/n^2$ . Why: the $1/\lambda_n\sim 1/n^2$ weighting makes the series converge but slow — the corner is the delta's fingerprint.

Step 2. Direct construction. Away from the source $-G''=0$ , so $G$ is piecewise linear: $G=Ax$ for $x<x'$ (giving $G(0)=0$ ), $G=B(L-x)$ for $x>x'$ (giving $G(L)=0$ ), with continuity $Ax'=B(L-x')$ . Why: the homogeneous equation between sources is trivial, so all the physics sits in the matching at $x'$ .

Step 3. Jump condition. Integrate $-G''=\delta$ across $x'$ : the slope jumps by $-1$ , i.e. $-B-A=-1$ . With continuity, $A=(L-x')/L$ , $B=x'/L$ , giving $G=x_<(L-x_>)/L$ with $x_<=\min(x,x')$ , $x_>=\max(x,x')$ — a symmetric, manifestly reciprocal tent. Why: the unit slope-jump is the delta integrated once, what distinguishes a Green's function from a homogeneous solution.

Step 4. Confront the two. Take $L=1$ , $x'=0.3$ : the closed form gives peak $G(0.3,0.3)=0.3\times0.7=\mathbf{0.15}$ , and the sine series climbs to it from below — one term overshoots to $0.164$ , five trace the tent, twenty give $0.1499$ . Why: the resummation made numerical — the same $G$ , exact and approached term by term, as the second figure shows. *The closed form gives peak $0.15$ exactly; the partial sums converge as $0.164\,(N=1)\to 0.14994\,(N=20)\to 0.150000\,(N=20000)$ .*

Go deeper · Derivation C · 1 minThe Kelvin image as a resummed spherical-harmonic series

Derivation CThe Kelvin image as a resummed spherical-harmonic series

The goal is to prove the deepest claim of the article: that the single closed-form Kelvin image in (5) is identical, mode by mode, to the infinite spherical-harmonic eigenfunction series for the interior Dirichlet Green's function of a grounded sphere — the cleanest instance of "geometry resums the spectrum," explaining why one fractional image stands in for an entire multipole expansion.

Step 1. Expand the source with the Legendre generating function, $\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}=\sum_{l=0}^{\infty}\dfrac{r_<^{\,l}}{r_>^{\,l+1}}P_l(\cos\gamma)$ , with $\gamma$ the angle between $\mathbf{r},\mathbf{r}'$ ; the addition theorem rewrites $P_l(\cos\gamma)$ as $\sum_m Y_{lm}(\hat r)Y_{lm}^*(\hat r')$ . Why: this is the eigenfunction expansion of the free-space kernel in the sphere's natural basis — the first half of $G$ .

Step 2. Expand the image the same way. Its strength $a/r'$ at radius $a^2/r'$ gives, for $r,r'<a$ , $\dfrac{a/r'}{|\mathbf{r}-\mathbf{r}'^{*}|}=\sum_{l}\dfrac{(rr')^{l}}{a^{2l+1}}P_l(\cos\gamma)$ . Why: the inverse-point geometry is exactly the substitution that turns the image into the regular-at-origin partner of each multipole.

Step 3. Subtract, following (5): $G_D=-\dfrac{1}{4\pi}\sum_{l}P_l(\cos\gamma)\Big[\dfrac{r_<^{\,l}}{r_>^{\,l+1}}-\dfrac{(rr')^{l}}{a^{2l+1}}\Big]$ . At $r=a$ the bracket is $\dfrac{r'^{\,l}}{a^{l+1}}-\dfrac{a^{l}r'^{l}}{a^{2l+1}}=0$ for every $l$ . Why: the boundary condition holds term by term, so image and mode sum agree multipole by multipole.

Step 4. Conclude. The single Kelvin image is the exact closed-form sum of the spherical-harmonic series — one geometric object equal to an infinite spectral sum. Why: the general principle in miniature — with enough symmetry, images perform the resummation the series can only approach term by term.

Limiting cases

How the two constructions behave in the characteristic regimes:

Case	Result	Meaning
Near the source	$G\sim G_{\text{free}}(\mathbf{r}-\mathbf{r}')$	the boundary is far compared to $\vert\mathbf{r}-\mathbf{r}'\vert$ , so $G$ reduces to the free-space singularity; images give it instantly, the series crawls
Resonance $\lambda\to\lambda_n$	$G\approx\dfrac{\phi_n(\mathbf{r})\phi_n^*(\mathbf{r}')}{\lambda_n-\lambda}$	for $\mathcal{L}=\nabla^2+\lambda$ , a simple pole at each eigenvalue: driving a lossless cavity at a normal mode makes the response diverge
Neumann zero mode	$\lambda_0=0,\ \phi_0=\text{const}$	the $n=0$ term blows up, no ordinary $G$ ; use the modified $\tilde G=\sum_{n\ne0}\phi_n\phi_n^*/(-\lambda_n)$ solving $\nabla^2\tilde G=\delta-1/V$
Wedge of angle $\pi/m$ (integer $m$ )	$G=\sum_{k=0}^{m-1}\big[G_{\text{free}}(\mathbf{r},R_k\mathbf{r}')-G_{\text{free}}(\mathbf{r},\bar R_k\mathbf{r}')\big]$	images close into $2m$ charges only for integer $\pi/\alpha$ ; otherwise they proliferate and the Fourier–Bessel series is required
Unbounded domain, $L\to\infty$	$G\to G_{\text{free}}$ , sum $\to$ integral	the discrete spectrum condenses to a continuum; the sum becomes a Fourier integral over free-space modes

Common trapThe Green's function does not carry the boundary data

The most common error is to bake the physical boundary values into $G$ . They do not go there. $G$ always satisfies the homogeneous condition — $G=0$ for a Dirichlet problem even when the real wall is at a fixed nonzero voltage — and the actual data re-enter only through the surface term $\oint u\,\partial_{n'}G\,\mathrm{d}S'$ in (2). Two related traps: (i) the pure-Neumann Laplacian has no naive $G$ — the constant zero mode makes the source unsolvable unless it is charge-neutral, so use the modified Green's function that subtracts a uniform background $1/V$ ; and (ii) images only terminate for special symmetric geometries — a box needs an infinite lattice, an irrational wedge never closes — so reaching for images outside their symmetry class wastes effort the eigenfunction series spends better.

Worked examples

1 · The pinned-string Green's function, series versus closed form (fully worked)

On the unit interval $0\le x\le 1$ solve $-G''=\delta(x-x')$ with $G(0)=G(1)=0$ for a source at $x'=0.3$ . Find the closed form, evaluate the peak, and check the eigenfunction series converges to it.

Step 1. Closed form. From Derivation 2 the tent is $G(x,x')=x_<(1-x_>)$ . With $x'=0.3$ the branches are $G=0.7\,x$ for $x<0.3$ and $G=0.3(1-x)$ for $x>0.3$ ; they meet at the source and the slope jumps by $-0.3-0.7=-1$ , confirming unit-source normalization.

Step 2. Peak value. The maximum sits at the source, $G(0.3,0.3)=0.3\times0.7=\mathbf{0.15}$ — the diagonal of the Green's function, handed over in closed form with no summation.

Step 3. Series check. The spectral form is $G=2\sum_{n=1}^{\infty}\sin(n\pi x)\sin(0.3n\pi)/(n\pi)^2$ . Its first term $0.16394\sin(\pi x)$ is a smooth arch peaking at $x=0.5$ (not $x=0.3$ ) with height $0.164$ — it overshoots and misplaces the peak, because one mode cannot make a corner. Adding terms fixes both: $N=5\to0.1489$ , $N=20\to0.14994$ , $N=100\to0.149999$ .

Step 4. Lesson. The closed form gives 0.15 exactly and instantly; the series approaches it as $1/N$ , needing hundreds of terms near the singular diagonal. The whole trade in one problem — direct construction wins on the source, the spectrum for smooth global quantities, and the two are the same $G$ .

2 · Image force on a charge above a grounded plane (fully worked)

A point charge $q=1\ \mathrm{nC}$ sits a height $d=1\ \mathrm{cm}$ above a grounded conducting plane. Use the image Green's function to find the attractive force and the total induced surface charge.

Step 1. Image. The Dirichlet $G$ for the half-space is the plane result of (5): a real charge at height $d$ plus a mirror image $-q$ at depth $d$ . In the physical region $z>0$ the field is exactly that of this pair, and the plane carries the induced charge the image mimics.

Step 2. Force. The real charge feels only the image (no net self-force), a distance $2d$ away: $F=\dfrac{q^2}{16\pi\varepsilon_0 d^2}$ , toward the plane. With $q=10^{-9}\ \mathrm{C}$ , $d=10^{-2}\ \mathrm{m}$ : $F=\dfrac{(10^{-9})^2}{16\pi(8.854\times10^{-12})(10^{-2})^2}=\mathbf{2.25\times10^{-5}\ N}$ , about $22.5\ \mu\mathrm{N}$ of attraction.

Step 3. Induced charge. Gauss's theorem guarantees the total induced surface charge equals the image charge: $q_{\text{ind}}=\oint\sigma\,\mathrm{d}S=-q=\mathbf{-1\ nC}$ — every field line that would have escaped to infinity terminates on the plane instead.

Step 4. Reciprocity. The potential at height $z$ from a charge at height $d$ equals that at height $d$ from a charge at height $z$ — the manifest $\mathbf{r}\leftrightarrow\mathbf{r}'$ symmetry of (5).

3 · Grounded-sphere Kelvin image (numbers)

A unit charge sits at $r'=0.2\ \mathrm{m}$ from the centre of a grounded sphere of radius $a=0.1\ \mathrm{m}$ . Locate the image, give its strength, verify the boundary condition.

Step 1. Location and strength. The inverse point is at radius $a^2/r'=0.01/0.2=\mathbf{0.05\ m}$ along the same ray (inside the sphere), carrying fractional charge $-a/r'=\mathbf{-0.5}$ of the real charge — a half-strength negative image reproduces the grounded sphere outside.

Step 2. Boundary check. Similar triangles give $|\mathbf{r}-\mathbf{r}'^{*}|=(a/r')|\mathbf{r}-\mathbf{r}'|$ on the surface; at a surface point 0.7 rad off axis, both sides equal $\mathbf{0.0697\ m}$ , so the two Coulomb terms cancel and $G_D=0$ on the sphere.

Physics of the portrait

The Green's function is named for a self-taught miller's son whose one privately printed pamphlet, unread for twenty years, founded a large part of mathematical physics.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841, English mathematician and physicist · potential theory and the origin of the Green's function

George Green left school at nine to work in his father's Nottingham windmill and taught himself mathematics from library books, in near-total isolation. In 1828, aged thirty-five and with no university training, he printed at his own expense An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism — a work that introduced the term "potential," proved what we now call Green's theorem, and constructed for the first time the auxiliary function that solves a boundary-value problem by superposing point-source responses: the Green's function. Fewer than a hundred copies sold, and the Essay sank almost without trace; Green entered Cambridge only at forty and died in 1841, still obscure. The rescue came from William Thomson (later Lord Kelvin), who found a copy in 1845 and reprinted it in Crelle's Journal (1850–1854), spreading Green's methods to the Continent. Thomson also invented, around 1845, the method of electrical images and the grounded-sphere image — the geometric counterpart to Green's analytic construction, and the other half of this article. Green's functions now pervade electrostatics, quantum field theory (as propagators), and many-body physics.

History

1828

— George Green's Essay introduces the potential, Green's theorem, and the Green's function

1845

— William Thomson (Kelvin) devises the method of electrical images and the grounded-sphere image

1850

— Thomson republishes Green's forgotten Essay in Crelle's Journal, spreading it to the Continent

1877

— Rayleigh states the acoustic reciprocity theorem in The Theory of Sound

1926

— Courant and Hilbert systematize eigenfunction expansions of Green's functions

Going further

Texts. Stone & Goldbart, Mathematics for Physics, ch. 11 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., ch. 10 · Hassani, Mathematical Physics, ch. 21–22 · Byron & Fuller, Mathematics of Classical and Quantum Physics, ch. 7 · Riley, Hobson & Bence, Mathematical Methods, 3rd ed., ch. 21 · Jackson, Classical Electrodynamics, 3rd ed., ch. 2–3 · Morse & Feshbach, Methods of Theoretical Physics, ch. 7. Key results to keep in view: the spectral form (3) has poles at each $\lambda_n$ and reproduces completeness; the 1D tent is $G=x_<(L-x_>)/L$ with peak $0.15$ at $x'=0.3,\ L=1$ , and the sine-series converges as $0.164\to0.14994\to0.150000$ ; the free-space kernels $-1/4\pi r$ and $\ln r/2\pi$ follow from Gauss's theorem; plane and Kelvin-sphere images satisfy $G=0$ on the boundary, with the sphere image $-a/r'$ placed at $a^2/r'^2\,\mathbf{r}'$ and boundary distances $0.0697\ \mathrm{m}$ at $a=0.1,\ r'=0.2\ \mathrm{m}$ ; the image force is $F=q^2/16\pi\varepsilon_0 d^2=2.25\times10^{-5}\ \mathrm{N}$ for $q=1\ \mathrm{nC},\ d=1\ \mathrm{cm}$ ; and reciprocity follows from Green's second identity.

Prerequisites. Sturm–Liouville Theory and Eigenfunction Expansions · Laplace's and Poisson's Equations · Boundary Conditions and Well-Posedness · Separation of Variables in Cartesian Coordinates · The Helmholtz Equation. Leads to. The Heat (Diffusion) Equation · The Wave Equation · Separation of Variables in Spherical Coordinates.

Ask the tutor: Want to watch the sine series build the triangular Green's function term by term, or work out the image force for your own conductor geometry? Ask and we'll do it live.

What actually is a Green's function, in one sentence?

It is the inverse of a differential operator written as a kernel: if $\mathcal{L}u=f$ , then $u=\int G f$ , the role $\mathcal{L}^{-1}$ plays for a matrix equation $Lx=b$ . Concretely $G(\mathbf{r},\mathbf{r}')$ is the field at $\mathbf{r}$ from a unit point source at $\mathbf{r}'$ , and since any source is a superposition of point sources, superposing the responses solves the whole problem. Everything else — spectral series, images, reciprocity — is technique for constructing this one object under the boundary conditions.

Why is the eigenfunction series so slow near the source when it is exact everywhere?

Because at $\mathbf{r}=\mathbf{r}'$ the Green's function must reproduce a delta-like singularity (a corner in 1D, a $1/r$ blow-up in 3D) out of smooth eigenfunctions. Fourier's theorem says you can, but only with infinitely many terms; truncating leaves overshoot and slow $1/N$ convergence exactly where the field is most singular. Away from the source $G$ is smooth and the series converges fast. This is why images help: they capture the singular free-space part in closed form, so you never sum it.

Why must the Green's function satisfy the homogeneous boundary condition?

Because $G$ inverts $\mathcal{L}$ on the space of functions obeying the homogeneous condition, which is where the operator is self-adjoint and invertible. Loading the real nonzero data into $G$ would break its linear response to the source alone, and superposition would fail. The representation formula (2) separates the two: the volume integral handles the source $f$ , the surface integral the boundary data, each weighted by the homogeneous $G$ or its normal derivative.

When should I reach for images and when for the eigenfunction expansion?

Use images when the geometry has strong reflection or inversion symmetry — a plane, a sphere, a wedge of angle $\pi/m$ , a half-space heat problem — because a finite set of images then gives an exact closed form, especially accurate near the source. Use the eigenfunction expansion when you can separate variables but the geometry lacks image symmetry (a box, a cylinder, an annulus), or when you want the modal/resonance information, since the poles are the eigenvalues. Near-resonant problems almost demand the spectral form; near the singular diagonal, images win.

What goes wrong for the pure-Neumann problem, and how is it fixed?

The Neumann Laplacian has a constant eigenfunction with $\lambda_0=0$ , so the spectral term $\phi_0\phi_0^*/\lambda_0$ diverges — no ordinary $G$ . Physically this is the solvability condition: a flux source can only be balanced if its total flux is zero, just as a closed insulated region cannot absorb net charge or heat. The fix is the modified $\tilde G=\sum_{n\ne0}\phi_n\phi_n^*/(-\lambda_n)$ , solving $\nabla^2\tilde G=\delta-1/V$ : the source is neutralized by a uniform background $1/V$ , and $\tilde G$ inverts $\mathcal{L}$ on the space orthogonal to the constant mode.

Is reciprocity ever broken?

Yes — reciprocity relies on $\mathcal{L}$ being self-adjoint. Introduce a term that breaks self-adjointness — a magnetic field (the vector potential enters linearly), a moving medium, a gyrotropic or magneto-optical material, or dissipation with a preferred time direction — and $G$ loses its symmetry. This is no curiosity: non-reciprocal Green's functions are exactly what optical isolators and circulators exploit to pass a signal one way but not the other. Engineering non-reciprocity means deliberately violating the self-adjointness hypothesis.