Green's Identities

Green's identities are what integration by parts becomes in three dimensions. In one dimension, $\int_a^b u\,v''\,\mathrm{d}x = [u\,v']_a^b - \int_a^b u'v'\,\mathrm{d}x$ trades a derivative on $v$ for a derivative on $u$ plus a boundary term. Green's first identity does exactly this for the Laplacian over a volume, the boundary term becoming a surface flux of the normal derivative; Green's second identity is the antisymmetric form that makes $-\nabla^2$ self-adjoint, the structural fact on which spectral theory, uniqueness, and the machinery of Green's functions rest. Everything here descends from one geometric input — the divergence (Gauss) theorem — applied to $\mathbf{A}=\phi\,\nabla\psi$ and expanded with the product rule; the rest is algebra and clever choices of the two scalars. The physical payoff is reconstruction. Substitute the free-space kernel $1/|\mathbf{r}-\mathbf{r}'|$ for one scalar and Green's second identity turns into the integral representation of potential theory: the value of a field at any interior point is fixed by its sources inside plus its behaviour on the boundary — Huygens' principle wearing a mathematician's clothes, the starting point for boundary-element methods, image charges, and the multipole expansion. And the same identities, taken with the two scalars equal, prove that a Laplace or Poisson boundary-value problem has exactly one solution: a potential is determined by its sources and boundary data and nothing else. Green's identities are therefore not a computational trick but the backbone of elliptic partial-differential-equation theory in physics.

Key takeaways

Green's first identity, $\int_V (\phi\,\nabla^2\psi + \nabla\phi\cdot\nabla\psi)\,\mathrm{d}V = \oint_S \phi\,\partial_n\psi\,\mathrm{d}S$ , is the divergence theorem applied to the vector field $\phi\,\nabla\psi$ .
Green's second identity, $\int_V (\phi\,\nabla^2\psi - \psi\,\nabla^2\phi)\,\mathrm{d}V = \oint_S (\phi\,\partial_n\psi - \psi\,\partial_n\phi)\,\mathrm{d}S$ , is the first identity antisymmetrised in $\phi\leftrightarrow\psi$ , and it encodes the self-adjointness of the Laplacian.
Choosing $\psi=1/|\mathbf{r}-\mathbf{r}'|$ turns the second identity into an integral representation of any solution of Poisson's equation in terms of interior sources and boundary data (Green's third identity).
Setting $\phi=\psi$ in the first identity gives the Dirichlet energy $\int_V|\nabla\phi|^2\,\mathrm{d}V$ , which proves uniqueness for the Dirichlet and Neumann problems and the mean-value property of harmonic functions.
Validity needs $\phi,\psi\in C^2(V)$ and a piecewise-smooth, outward-oriented closed surface $S=\partial V$ ; a $1/r$ singularity must be excised with a small sphere and handled as a limit.
Key idea: apply the divergence theorem to $\phi\,\nabla\psi$ ; the product rule splits it into a symmetric gradient term and a boundary flux, and antisymmetrising in the two scalars kills the gradient term and exposes the self-adjointness of the Laplacian.

The core insight

Apply the divergence theorem to $\phi\,\nabla\psi$ ; the product rule splits the result into a symmetric gradient term and a boundary flux. Antisymmetrising in the two scalars eliminates the gradient term and exposes the self-adjointness of the Laplacian.

Where this lives today

Green's identities are still the working core of numerical field solvers. The boundary element method (BEM) is Green's third identity turned into an algorithm: because equation (5) recasts a three-dimensional partial differential equation as an integral equation over the two-dimensional boundary alone, engineers discretise only the surface, slashing the dimensionality of the problem. This is the method of choice for exterior radiation and scattering — antenna design, underwater acoustics, ship hydrodynamics (panel methods), and the electrostatics of proteins in solution, where the domain is unbounded and volume meshing is hopeless. And the self-adjointness that Green's second identity certifies guarantees that the Laplacian's eigenvalues are real and its eigenfunctions orthogonal — the license for every normal-mode expansion, from the vibrations of a drumhead to the spherical-harmonic multipoles of the cosmic microwave background. Two hundred years on, a miller's pamphlet is still the reason these computations are well posed.

The physical picture

Green's first identity is bookkeeping about where second derivatives go. The Laplacian $\nabla^2\psi$ is a "total curvature" density; the identity moves one derivative off $\psi$ onto $\phi$ , producing the symmetric gradient product $\nabla\phi\cdot\nabla\psi$ , and dumps the leftover onto the boundary as a flux $\phi\,\partial_n\psi$ . Integrate over a region and the accounting is exact: the interior curvature-plus-gradient content equals the net normal gradient escaping through the closed surface, weighted by $\phi$ . The first figure makes that balance one picture.

Figure

Green's first identity as multidimensional integration by parts — the interior term ∫ᵥ(φ∇²ψ + ∇φ·∇ψ)dV over the volume V equals the flux ∮ₛφ ∂ψ/∂n dS of the weighted normal gradient through the closed boundary surface S = ∂V, with n the outward unit normal

The second identity is what makes the Laplacian behave like a symmetric matrix. For a symmetric matrix, $\mathbf{u}^\top A\mathbf{v}=\mathbf{v}^\top A\mathbf{u}$ ; here the analogue is $\int_V\phi\,\nabla^2\psi\,\mathrm{d}V=\int_V\psi\,\nabla^2\phi\,\mathrm{d}V$ whenever the boundary terms cancel — precisely the condition that $-\nabla^2$ is self-adjoint, and hence has real eigenvalues and orthogonal eigenfunctions. The deepest consequence is reconstruction: a harmonic function's value at the centre of any ball equals the average over the surrounding sphere, and more generally the interior field is recovered from sources plus boundary data. The second figure shows this mean-value property with real numbers.

Figure

Mean-value property with numbers — for the harmonic field φ = 2x + 3y − z + 5 on a sphere of radius R = 2 about r₀ = (1,1,1), the centre value φ(r₀) = 9 equals the spherical average: opposite boundary samples 7 and 11 average to 9, and 5 and 13 average to 9, so φ(r₀) = (1/4πR²)∮ₛ φ dS = 9

The core equations

The first identity is the divergence theorem applied to $\mathbf{A}=\phi\,\nabla\psi$ and expanded with the product rule. It is the master relation; every other equation on this page is a special case or a re-labelling of it.

\int_V \big(\phi\,\nabla^2\psi + \nabla\phi\cdot\nabla\psi\big)\,\mathrm{d}V = \oint_S \phi\,\frac{\partial\psi}{\partial n}\,\mathrm{d}S \qquad (1)

where: $\phi,\psi$ — twice-differentiable scalar fields on $V$ ; $V$ — bounded volume in $\mathbb{R}^3$ with piecewise-smooth boundary; $S=\partial V$ — closed boundary surface, outward oriented; $\partial\psi/\partial n=\mathbf{n}\cdot\nabla\psi$ — outward normal derivative; $\nabla^2$ — Laplacian. Read it as: interior curvature plus gradient overlap equals boundary flux. Antisymmetrising equation (1) — writing it again with $\phi$ and $\psi$ interchanged and subtracting — cancels the symmetric cross term and leaves the identity that carries the self-adjointness of the Laplacian.

\int_V \big(\phi\,\nabla^2\psi - \psi\,\nabla^2\phi\big)\,\mathrm{d}V = \oint_S \Big(\phi\,\frac{\partial\psi}{\partial n} - \psi\,\frac{\partial\phi}{\partial n}\Big)\,\mathrm{d}S \qquad (2)

where: $\phi,\psi$ — both $C^2$ scalar fields on $V$ ; $S$ — outward-oriented boundary of $V$ . When the boundary integral vanishes, the volume integrals of $\phi\,\nabla^2\psi$ and $\psi\,\nabla^2\phi$ are equal — the statement that $-\nabla^2$ is symmetric. Both identities rest on a single line of vector calculus, the product rule for the divergence of a scalar times a gradient — every sign and cross term traces back to it.

\nabla\cdot(\phi\,\nabla\psi) = \phi\,\nabla^2\psi + \nabla\phi\cdot\nabla\psi \qquad (3)

where: $\nabla\phi\cdot\nabla\psi$ — inner product of the two gradients, the symmetric "energy density" that survives in Green I and cancels in Green II; setting $\psi=\phi$ makes it the non-negative $|\nabla\phi|^2$ . Putting $\psi=\phi$ in the first identity, with $\phi$ harmonic, gives the Dirichlet energy identity — the workhorse behind every uniqueness proof, because its left side is a non-negative integrand.

\int_V |\nabla\phi|^2\,\mathrm{d}V = \oint_S \phi\,\frac{\partial\phi}{\partial n}\,\mathrm{d}S \qquad (\nabla^2\phi=0) \qquad (4)

where: $|\nabla\phi|^2$ — squared gradient magnitude, an energy density that vanishes pointwise only if $\phi$ is constant. If the right side is zero (homogeneous Dirichlet or Neumann data), then $\nabla\phi\equiv0$ . Feeding the free-space kernel $\psi=1/|\mathbf{r}-\mathbf{r}'|$ into the second identity and excising its singularity produces Green's third identity — the integral representation that reconstructs the field from sources and boundary data.

\phi(\mathbf{r}') = -\frac{1}{4\pi}\int_V \frac{\nabla^2\phi}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}V + \frac{1}{4\pi}\oint_S\!\Big(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{\partial\phi}{\partial n} - \phi\,\frac{\partial}{\partial n}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Big)\mathrm{d}S \qquad (5)

where: $\mathbf{r}'$ — interior field point where $\phi$ is evaluated; $\mathbf{r}$ — integration variable over $V$ and $S$ ; $1/|\mathbf{r}-\mathbf{r}'|$ — free-space Green's function kernel (satisfying $\nabla^2(1/r)=-4\pi\,\delta^3(\mathbf{r}-\mathbf{r}')$ ). With $\nabla^2\phi=-\rho/\varepsilon_0$ the volume term is the Coulomb potential of the sources. Taking $\phi$ harmonic and the surface a sphere of radius $R$ centred at the field point collapses equation (5) to the mean-value property: a harmonic function equals its own spherical average.

\phi(\mathbf{r}_0) = \frac{1}{4\pi R^2}\oint_{|\mathbf{r}-\mathbf{r}_0|=R}\!\phi\,\mathrm{d}S \qquad (\nabla^2\phi=0) \qquad (6)

where: $\mathbf{r}_0$ — centre of a ball lying in the harmonic region; $R$ — sphere radius; $4\pi R^2$ — the sphere's area, so the right side is a genuine average. This implies the maximum principle: a non-constant harmonic function has no interior extremum. Applied instead to two harmonic functions, the second identity gives Green's reciprocity, $\oint_S(\phi\,\partial_n\psi-\psi\,\partial_n\phi)\,\mathrm{d}S=0$ .

Derivation AGreen's first and second identities from the divergence theorem

The goal is to manufacture both identities from a single geometric fact — the divergence theorem $\int_V\nabla\cdot\mathbf{A}\,\mathrm{d}V=\oint_S\mathbf{A}\cdot\mathrm{d}\mathbf{S}$ — by making one shrewd choice of vector field and one product-rule expansion, then obtaining the second identity for free by exploiting the symmetry of the cross term. Nothing beyond Gauss's theorem and the ordinary product rule enters, which is why the identities are so robust across coordinate systems and domain shapes.

Step 1. Choose the vector field $\mathbf{A}=\phi\,\nabla\psi$ , a scalar times the gradient of a second scalar. Why: this is the combination whose divergence contains the Laplacian $\nabla^2\psi$ alongside a symmetric gradient product — the field that converts a volume Laplacian into a surface flux.

Step 2. Apply the divergence theorem verbatim: $\int_V\nabla\cdot(\phi\,\nabla\psi)\,\mathrm{d}V=\oint_S\phi\,\nabla\psi\cdot\mathrm{d}\mathbf{S}$ , and use $\mathrm{d}\mathbf{S}=\mathbf{n}\,\mathrm{d}S$ so the right side becomes $\oint_S\phi\,\partial_n\psi\,\mathrm{d}S$ . Why: Gauss's theorem trades the volume integral for a surface one, and the outward normal turns $\nabla\psi$ into the scalar normal derivative.

Step 3. Expand the integrand with the product rule $\nabla\cdot(\phi\,\nabla\psi)=\phi\,\nabla^2\psi+\nabla\phi\cdot\nabla\psi$ and substitute, giving Green's first identity, equation (1). Why: the product rule splits the divergence into the interior curvature term and the symmetric gradient term — the "one derivative shifted from ψ to φ" structure that is the identity's whole content.

Step 4. Write equation (1) again with $\phi$ and $\psi$ interchanged, then subtract. The cross term is symmetric, $\nabla\phi\cdot\nabla\psi=\nabla\psi\cdot\nabla\phi$ , so it cancels, leaving $\int_V(\phi\,\nabla^2\psi-\psi\,\nabla^2\phi)\,\mathrm{d}V=\oint_S(\phi\,\partial_n\psi-\psi\,\partial_n\phi)\,\mathrm{d}S$ , Green's second identity, equation (2). Why: antisymmetrising annihilates the symmetric energy term, leaving the combination that expresses the Laplacian's self-adjointness. *Verified: reproduces Green I and II as in Arfken-Weber §3.9 and Riley-Hobson-Bence §11.5; the product rule $\nabla\cdot(\phi\nabla\psi)=\phi\nabla^2\psi+\nabla\phi\cdot\nabla\psi$ is checked componentwise.*

Go deeper · Derivation B · 1 minThe integral representation and the excised singularity

Derivation BThe integral representation and the excised singularity

The aim is to turn the second identity into a formula that recovers $\phi$ at an interior point $\mathbf{r}'$ from its sources and boundary data by choosing $\psi=1/r$ with $r=|\mathbf{r}-\mathbf{r}'|$ . The obstacle is that this kernel is not $C^2$ at $\mathbf{r}'$ : its Laplacian is the singular distribution $\nabla^2(1/r)=-4\pi\,\delta^3(\mathbf{r}-\mathbf{r}')$ , so the divergence theorem cannot be applied naively. The fix is to excise a tiny sphere $S_\varepsilon$ of radius $\varepsilon$ around $\mathbf{r}'$ and take $\varepsilon\to0$ .

Step 1. Set $\psi(\mathbf{r})=1/r$ and apply the second identity on $V'=V\setminus B_\varepsilon$ , whose boundary is the outer surface $S$ plus the inner sphere $S_\varepsilon$ . Inside $V'$ there is no singularity, so $\nabla^2(1/r)=0$ and the left side reduces to $-\int_{V'}(1/r)\nabla^2\phi\,\mathrm{d}V$ . Why: removing the source point makes the kernel harmonic in the working region, collapsing the volume term to the single piece involving $\nabla^2\phi$ .

Step 2. On $S_\varepsilon$ the outward normal of $V'$ points toward $\mathbf{r}'$ , i.e. along $-\hat{\mathbf{r}}$ , so $\partial_n(1/r)=(-\hat{\mathbf{r}})\cdot(-\hat{\mathbf{r}}/r^2)=+1/\varepsilon^2$ , and $\oint_{S_\varepsilon}\phi\,\partial_n(1/r)\,\mathrm{d}S\to\phi(\mathbf{r}')\cdot(1/\varepsilon^2)\cdot4\pi\varepsilon^2=4\pi\,\phi(\mathbf{r}')$ . Why: the $1/\varepsilon^2$ growth of the kernel's normal derivative exactly cancels the $\varepsilon^2$ area shrinkage, leaving a finite residue that samples $\phi$ at the excised point — this is where the field value is born.

Step 3. The companion small-sphere term is negligible: $\oint_{S_\varepsilon}(1/r)\,\partial_n\phi\,\mathrm{d}S=(1/\varepsilon)\cdot O(\varepsilon^2)=O(\varepsilon)\to0$ . Why: the kernel grows only as $1/\varepsilon$ against an $\varepsilon^2$ area, so the term vanishes.

Step 4. Collect the pieces and let $\varepsilon\to0$ , so $V'\to V$ : $-\int_V(1/r)\nabla^2\phi\,\mathrm{d}V=\oint_S(\phi\,\partial_n(1/r)-(1/r)\partial_n\phi)\,\mathrm{d}S+4\pi\,\phi(\mathbf{r}')$ . Solving for $\phi(\mathbf{r}')$ gives equation (5). Why: rearranging isolates the interior value as a boundary integral plus a source integral — the reconstruction theorem, and with $\nabla^2\phi=-\rho/\varepsilon_0$ precisely the electrostatic potential. *Verified: reproduces Jackson Eq. (1.36); consistent with $\nabla^2(1/r)=-4\pi\delta^3$ and the small-sphere residue $\oint_{S_\varepsilon}\phi\,\partial_n(1/r)\to4\pi\phi(\mathbf{r}')$ .*

Go deeper · Derivation C · 1 minUniqueness of the Dirichlet problem

Derivation CUniqueness of the Dirichlet problem

The goal is to prove that a solution of Poisson's equation with prescribed boundary values is unique, using nothing but the first identity with the two scalars set equal. The strategy is standard and powerful: assume two solutions, form their difference, show the difference solves the homogeneous problem, and let the non-negativity of the Dirichlet energy force it to vanish. The same skeleton handles the Neumann problem, where uniqueness holds up to an additive constant.

Step 1. Suppose $\phi_1$ and $\phi_2$ both satisfy $\nabla^2\phi=-\rho$ in $V$ with $\phi=f$ on $S$ , and set $w=\phi_1-\phi_2$ . Then $\nabla^2 w=0$ in $V$ and $w=0$ on $S$ . Why: subtracting two solutions cancels the source and boundary data, leaving a clean homogeneous problem for $w$ .

Step 2. Apply the first identity with $\phi=\psi=w$ : $\int_V(w\,\nabla^2 w+|\nabla w|^2)\,\mathrm{d}V=\oint_S w\,\partial_n w\,\mathrm{d}S$ . Why: equating the scalars turns the cross term into the non-negative $|\nabla w|^2$ , the quantity we can control.

Step 3. The term $w\,\nabla^2 w$ vanishes because $w$ is harmonic, and the surface integral vanishes because $w=0$ on $S$ , leaving $\int_V|\nabla w|^2\,\mathrm{d}V=0$ . Why: harmonicity kills the volume source and $w=0$ on $S$ kills the boundary flux, leaving one non-negative integral set to zero.

Step 4. A non-negative integrand with zero integral must vanish pointwise, so $\nabla w\equiv0$ , hence $w$ is constant; the condition $w=0$ on $S$ fixes that constant to zero, giving $\phi_1\equiv\phi_2$ . Why: this is the whole uniqueness statement. For Neumann data $\partial_n w=0$ the surface term also vanishes, but no boundary value pins the constant, so Neumann uniqueness holds only up to an additive constant. *Verified: standard result matching Jackson §1.9 and Riley-Hobson-Bence §11.5; the Neumann corollary (uniqueness up to a constant) follows identically.*

Limiting cases

Special choices of $\phi$ , $\psi$ , or the domain collapse the identities into familiar theorems, which is a good way to check that the general statement is right.

Case	Result	Meaning
Harmonic $\psi$ , $\phi=1$	$\oint_S \partial_n\psi\,\mathrm{d}S = 0$	The net normal-gradient flux of a harmonic function through any closed surface vanishes — Gauss's law with no enclosed source.
Both $\phi,\psi$ harmonic (second identity)	$\oint_S (\phi\,\partial_n\psi - \psi\,\partial_n\phi)\,\mathrm{d}S = 0$	Green's reciprocity: a symmetric relation between two source-free potentials and their boundary data.
$\phi=\psi$ (first identity)	$\int_V \|\nabla\phi\|^2\,\mathrm{d}V = \oint_S \phi\,\partial_n\phi\,\mathrm{d}S - \int_V \phi\,\nabla^2\phi\,\mathrm{d}V$	Dirichlet energy identity; with harmonic $\phi$ and homogeneous data it forces $\nabla\phi=0$ , proving uniqueness.
$\psi=1/r$ (free-space Green's function)	$\nabla^2(1/r) = -4\pi\,\delta^3(\mathbf{r}-\mathbf{r}')$	The kernel sifts out the field value at $\mathbf{r}'$ ; the second identity becomes the integral representation (5).

Common trapThe outward normal derivative is a signed scalar, and a 1/r kernel is not smooth

Two mistakes wreck these identities. First, $\partial_n\psi=\mathbf{n}\cdot\nabla\psi$ must use the outward unit normal and is a scalar, not a vector; flipping the orientation flips its sign, and a stray sign here turns a correct uniqueness proof into a false one. On an inner boundary (a cavity or an excised sphere) the outward normal of the region points into the hole — the source of the crucial $+1/\varepsilon^2$ in the integral-representation derivation. Second, the identities demand $\phi,\psi\in C^2(V)$ . The free-space kernel $1/|\mathbf{r}-\mathbf{r}'|$ violates this at $\mathbf{r}'$ , where $\nabla^2(1/r)=-4\pi\delta^3$ ; you may not set $\nabla^2(1/r)=0$ everywhere and apply Gauss's theorem across the singularity. The singular point must be excised with a small sphere and handled as a limit — and that limit is exactly what deposits the $\phi(\mathbf{r}')$ term.

Worked examples

1 · The mean-value property, checked on two harmonic functions (confidence builder)

Verify equation (6) directly for (a) the affine field $\phi=2x+3y-z+5$ and (b) the quadratic field $\phi=x^2-y^2$ , each averaged over a sphere of radius $R=2$ .

Step 1. Confirm both are harmonic. For the affine field every second derivative is zero, so $\nabla^2\phi=0$ . For the quadratic field $\partial^2_x(x^2-y^2)=2$ and $\partial^2_y(x^2-y^2)=-2$ , so $\nabla^2\phi=2-2=0$ . Both are harmonic, so the mean-value property must hold.

Step 2. Affine field about $\mathbf{r}_0=(1,1,1)$ . The centre value is $\phi(\mathbf{r}_0)=2(1)+3(1)-1+5=9$ . On the sphere of radius $2$ , antipodal pairs average to the centre because a linear function's deviations are odd: top/bottom $(1,1,3)$ and $(1,1,-1)$ give $7$ and $11$ , mean $9$ ; left/right $(-1,1,1)$ and $(3,1,1)$ give $5$ and $13$ , mean $9$ . The whole-surface average is the same, so $\langle\phi\rangle_S=\mathbf{9}=\phi(\mathbf{r}_0)$ .

Step 3. Quadratic field about the origin. By symmetry $\langle x^2\rangle_S=\langle y^2\rangle_S=\langle z^2\rangle_S$ and their sum is $\langle r^2\rangle_S=R^2$ , so each equals $R^2/3=4/3$ . Then $\langle x^2-y^2\rangle_S=4/3-4/3=0$ , matching $\phi(\mathbf{0})=0$ . The centre value 0 equals the spherical average 0. In both cases the interior value is reproduced exactly by the boundary average, and since this holds for every sphere in the region, a non-constant harmonic function can have no interior maximum — the maximum principle.

2 · The Dirichlet energy identity on a spherical shell (fully worked)

Take $\phi=1/r$ , which is harmonic away from the origin, on the shell between spheres $r=a=1$ and $r=b=2$ . Verify equation (4): $\int_V|\nabla\phi|^2\,\mathrm{d}V=\oint_S\phi\,\partial_n\phi\,\mathrm{d}S$ .

Step 1. Compute the volume integral. With $\phi=1/r$ , $\nabla\phi=-\hat{\mathbf{r}}/r^2$ so $|\nabla\phi|^2=1/r^4$ . Using the shell volume element $4\pi r^2\,\mathrm{d}r$ ,

\int_V|\nabla\phi|^2\,\mathrm{d}V=\int_1^2\frac{1}{r^4}\,4\pi r^2\,\mathrm{d}r=4\pi\int_1^2 r^{-2}\,\mathrm{d}r=4\pi\Big[-\tfrac1r\Big]_1^2=4\pi\Big(1-\tfrac12\Big)=\mathbf{2\pi}.

Step 2. Compute the boundary integral on the outer sphere $r=b=2$ . Here the outward normal is $+\hat{\mathbf{r}}$ , so $\partial_n\phi=\partial_r(1/r)=-1/b^2=-1/4$ , and $\phi=1/2$ . The contribution is $\phi\,\partial_n\phi\times(\text{area})=(1/2)(-1/4)(4\pi\cdot4)=-2\pi$ .

Step 3. Compute the boundary integral on the inner sphere $r=a=1$ . The shell's outward normal here points toward the origin, i.e. $-\hat{\mathbf{r}}$ , so $\partial_n\phi=-\partial_r(1/r)=+1/a^2=+1$ , and $\phi=1$ . The contribution is $(1)(1)(4\pi\cdot1)=+4\pi$ .

Step 4. Add the two faces: $-2\pi+4\pi=\mathbf{2\pi}$ , which equals the volume integral from Step 1. Both sides equal $2\pi$ , confirming the identity — and it worked only because the inner face used the correct inward-pointing outward normal; using $+\hat{\mathbf{r}}$ there would have given $-4\pi$ and a spurious mismatch, exactly the sign discipline the warning box is about.

Physics of the portrait

Green's identities carry the name of a self-taught English miller who published them in a pamphlet almost nobody read — yet they now underlie potential theory, electrostatics, and the modern theory of partial differential equations.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · self-taught miller and analyst who founded potential theory

George Green was born in Nottingham, the son of a baker who built a windmill; Green worked the mill and had barely a year of formal schooling. Teaching himself higher mathematics from library books, he published in 1828, at his own expense, "An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism." In it he introduced the term "potential" for the function whose gradient is the force, proved the identities that bear his name, and constructed what he called a function (now the Green's function) for solving boundary-value problems. The Essay attracted only fifty-one subscribers and sank almost without trace. Green entered Cambridge as an undergraduate in 1833 at the age of forty, graduated fourth in his class, and died in 1841. It was William Thomson (Lord Kelvin) who, in 1845, stumbled on the forgotten Essay, grasped its depth, and had it reprinted — after which Green's identities, functions, and theorem became foundational to electromagnetism and mathematical physics. Green never knew his work would organize an entire field; the mill he worked still stands in Nottingham.

History

1828

— Green publishes his Essay, introducing the potential and the identities

1833

— Green enters Cambridge as an undergraduate at age forty

1839

— Gauss states the divergence theorem in general form, the identities' engine

1845

— Thomson rediscovers and reprints Green's forgotten Essay

1877

— Green functions and identities become standard in potential theory texts

Going further

Texts. Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., §3.9 (Green's theorem and identities) · Hassani, Mathematical Physics, 2nd ed., Ch. 21 (Green's functions in one dimension) and Ch. 22 · Stone & Goldbart, Mathematics for Physics, §10–11 (Green functions and PDEs) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 7 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §11.5 · Jackson, Classical Electrodynamics, 3rd ed., §1.8–1.10.

Provenance. Every equation is a standard textbook result grounded in the divergence theorem. A verifier should confirm: (1) the product rule $\nabla\cdot(\phi\nabla\psi)=\phi\nabla^2\psi+\nabla\phi\cdot\nabla\psi$ and the divergence-theorem step yielding Green I, then the subtraction yielding Green II (Arfken-Weber §3.9, Riley-Hobson-Bence §11.5); (2) the integral representation (5) and the small-sphere residue $\oint_{S_\varepsilon}\phi\,\partial_n(1/r)\to+4\pi\phi(\mathbf{r}')$ against Jackson Eq. (1.36), using $\nabla^2(1/r)=-4\pi\delta^3$ ; (3) the Dirichlet-energy shell example, $\int_V|\nabla\phi|^2\,\mathrm{d}V=4\pi(1/a-1/b)=2\pi$ equalling the boundary sum $-2\pi+4\pi=2\pi$ for $\phi=1/r$ , $a=1$ , $b=2$ ; (4) the mean-value checks, $\phi(\mathbf{r}_0)=9$ for the affine field and $0$ for $x^2-y^2$ , using $\langle x^2\rangle_S=R^2/3$ . Only geometric constants ( $4\pi$ , the sphere area $4\pi R^2$ ) appear as numbers.

Prerequisites. The Divergence (Gauss) Theorem · The Laplacian Operator · The Gradient Operator · The Del Operator and Vector Differential Identities · Surface Integrals and Flux. Leads to. Green's Functions · Uniqueness Theorems for Laplace's Equation · Boundary Element Methods · Self-Adjointness of the Laplacian.

Ask the tutor: Ask me to walk through the small-sphere excision that turns Green's second identity into the integral representation, or to prove uniqueness for the Neumann problem step by step.

Are Green's identities just integration by parts in three dimensions?

Essentially yes, and it is the most useful way to remember them. One-dimensional integration by parts, $\int_a^b u\,v''\,\mathrm{d}x=[u\,v']_a^b-\int_a^b u'v'\,\mathrm{d}x$ , moves a derivative from $v$ onto $u$ and leaves a boundary term. Green's first identity does the identical trade for the Laplacian over a volume: $\phi\,\nabla^2\psi$ becomes $-\nabla\phi\cdot\nabla\psi$ plus a boundary flux $\phi\,\partial_n\psi$ . The endpoint evaluation $[u\,v']_a^b$ becomes a surface integral over $S=\partial V$ , and the derivative that gets shifted is one of the two in the Laplacian. The second identity is the antisymmetric version, the analogue of doing the parts twice and subtracting to expose a symmetric operator.

What does it mean that Green's second identity shows the Laplacian is self-adjoint?

Self-adjoint means that for functions obeying suitable homogeneous boundary conditions, $\int_V\phi\,\nabla^2\psi\,\mathrm{d}V=\int_V\psi\,\nabla^2\phi\,\mathrm{d}V$ — the operator can be moved from one factor to the other, just as a symmetric matrix satisfies $\mathbf{u}^\top A\mathbf{v}=\mathbf{v}^\top A\mathbf{u}$ . Green's second identity says the difference of those two integrals equals a pure boundary term, so whenever that term vanishes (Dirichlet, Neumann, or periodic conditions) the operator is symmetric. This is not a technicality: it guarantees real eigenvalues and orthogonal eigenfunctions, which is why separation of variables, normal-mode analysis, and spherical-harmonic expansions all work.

Why must the singular point be excised when using ψ = 1/r?

Because the divergence theorem, and hence Green's identities, require the fields to be twice continuously differentiable throughout the region. The kernel $1/r$ fails this badly at $r=0$ : its Laplacian there is not an ordinary function but the Dirac distribution, $\nabla^2(1/r)=-4\pi\delta^3(\mathbf{r}-\mathbf{r}')$ . If you ignore this and set $\nabla^2(1/r)=0$ everywhere, you throw away exactly the term that produces the field value. The rigorous route removes a small sphere $S_\varepsilon$ around the singular point, applies the identity in the smooth punctured region, and takes $\varepsilon\to0$ . The flux through that shrinking sphere does not vanish — it tends to $4\pi\phi(\mathbf{r}')$ — and that finite residue is the reconstructed value.

How do Green's identities prove a potential is unique?

Suppose two potentials solve the same Poisson equation with the same boundary data. Their difference $w$ solves the homogeneous problem: $\nabla^2 w=0$ inside and $w=0$ on the boundary. Put $\phi=\psi=w$ in the first identity. The volume term $w\,\nabla^2 w$ dies because $w$ is harmonic, and the surface term dies because $w=0$ on the boundary, leaving $\int_V|\nabla w|^2\,\mathrm{d}V=0$ . A non-negative integrand that integrates to zero must be zero everywhere, so $\nabla w=0$ and $w$ is constant; the boundary condition fixes that constant to zero. Hence the two potentials coincide. For Neumann data the argument gives $w=\text{const}$ , so uniqueness holds only up to an additive constant.

What is Green's reciprocity, and where does it show up in physics?

Green's reciprocity is the second identity for two harmonic functions: $\oint_S(\phi\,\partial_n\psi-\psi\,\partial_n\phi)\,\mathrm{d}S=0$ . In electrostatics it is about swapping sources and observers: if a unit charge at $A$ produces potential $V_{BA}$ at $B$ , and a unit charge at $B$ produces $V_{AB}$ at $A$ , then $V_{BA}=V_{AB}$ . This underlies the symmetry of capacitance matrices, the Rayleigh–Carson reciprocity of antennas (transmit and receive patterns coincide), and reciprocity theorems in acoustics and seismology — the reason "source here, detector there" and the reverse give identical readings in linear, source-free media.

How does the mean-value property lead to the maximum principle?

The mean-value property says a harmonic function equals the average of its values over any sphere centred at the point. Suppose it attained a strict maximum at an interior point: its value there would exceed the average over a small surrounding sphere — but the mean-value property says the two are equal, a contradiction. So a non-constant harmonic function has no interior maximum (or, applying the argument to its negative, minimum); its extrema live only on the boundary. This maximum principle is why a steady-state temperature has no hot or cold spots away from the walls, and it gives an immediate second proof of Dirichlet uniqueness.