Green's Functions for Ordinary Differential Equations

A remarkable fraction of physics reduces to one sentence: solve a linear differential equation Lu(x)=f(x)\mathcal{L}u(x)=f(x) for the field uu produced by a source ff, subject to boundary conditions. The source might be a charge density feeding an electrostatic potential, a driving force pushing an oscillator, or a heat input warming a rod. Solving this afresh for every new ff is wasteful, because linearity offers a shortcut: decompose the source into an infinite comb of point sources, solve once for the response to a single unit point source δ(x−ξ)\delta(x-\xi), then rebuild the answer for any ff by superposition. That universal response is the Green's function G(x,ξ)G(x,\xi) — the closest thing mathematical physics has to a master key, turning a differential equation into an integral and the abstract operator L\mathcal{L} into an explicit inverse. The defining property is LxG(x,ξ)=δ(x−ξ)\mathcal{L}_x G(x,\xi)=\delta(x-\xi), the subscript recording that L\mathcal{L} differentiates the field variable xx while ξ\xi merely labels where the source sits. Because L\mathcal{L} is a differential operator, GG solves the homogeneous equation everywhere except at x=ξx=\xi, obeys the boundary conditions, and carries a precise discontinuity there matched to the delta — for a second-order operator, GG stays continuous but G′G' jumps by a known amount. Written in Sturm–Liouville form the structure becomes symmetric, G(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x), an exact statement of reciprocity, and the operator's eigenfunctions furnish a spectral sum that exposes resonances at a glance. The one-dimensional case fixes every idea — jump conditions, symmetry, spectral expansion, solvability — in its cleanest form, and it is the direct ancestor of the propagators and resolvents that run through electromagnetism, quantum mechanics, and field theory.

Key takeaways
  • A Green's function G(x,ξ)G(x,\xi) is the response of a linear operator L\mathcal{L} to a unit point source at x=ξx=\xi: it satisfies LxG(x,ξ)=δ(x−ξ)\mathcal{L}_x G(x,\xi)=\delta(x-\xi) together with the problem's boundary conditions.
  • Once GG is known, the solution of Lu=f\mathcal{L}u=f is a single integral, u(x)=∫G(x,ξ) f(ξ) dξu(x)=\int G(x,\xi)\,f(\xi)\,\mathrm{d}\xi — superposition turns the differential equation into an integral one.
  • For a second-order operator GG is patched from two homogeneous solutions, one obeying each boundary; continuity plus a derivative jump of 1/p(ξ)1/p(\xi) fix it, and the Wronskian normalizes it.
  • For self-adjoint L\mathcal{L} the Green's function is symmetric, G(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x) — reciprocity — and admits the bilinear eigenfunction expansion G=∑nϕn(x)ϕn(ξ)/(λ−λn)G=\sum_n \phi_n(x)\phi_n(\xi)/(\lambda-\lambda_n).
  • The construction breaks down exactly when the spectral parameter hits an eigenvalue; the Fredholm alternative then decides solvability and forces a modified Green's function.
  • Key idea: Solve Lu=f\mathcal{L}u=f in two stages — first find the response G(x,ξ)G(x,\xi) to a unit point source via LxG=δ(x−ξ)\mathcal{L}_xG=\delta(x-\xi) with the boundary conditions, then superpose, u(x)=∫G(x,ξ)f(ξ) dξu(x)=\int G(x,\xi)f(\xi)\,\mathrm{d}\xi. For a second-order L\mathcal{L}, GG is stitched from two homogeneous solutions, one per boundary, joined at x=ξx=\xi by continuity plus a derivative jump set by the Wronskian.
The core insight

Solve Lu=f\mathcal{L}u=f in two stages: first find the response G(x,ξ)G(x,\xi) to a unit point source via LxG=δ(x−ξ)\mathcal{L}_xG=\delta(x-\xi) subject to the boundary conditions, then superpose, u(x)=∫G(x,ξ)f(ξ) dξu(x)=\int G(x,\xi)f(\xi)\,\mathrm{d}\xi. For a second-order L\mathcal{L}, GG is patched from two homogeneous solutions — one per boundary — joined at x=ξx=\xi by continuity plus a derivative jump set by the Wronskian.

Where this lives today

The ODE Green's function is the seed of the propagators and resolvents that organize modern physics. In quantum field theory the Feynman propagator is literally the Green's function of the Klein–Gordon or Dirac operator, and every Feynman diagram is a product of such propagators tied at vertices; lattice QCD extracts hadron masses from the decay of quark propagators on a discretized spacetime. In condensed-matter physics the many-body Green's function obeys Dyson's equation G=G0+G0ΣGG=G_0+G_0\Sigma G — the analogue of Lippmann–Schwinger with the self-energy Σ\Sigma in place of the scattering potential — and the GWGW approximation built on it now predicts band gaps to tenths of an electronvolt in materials databases. The non-equilibrium Green's function formalism is the workhorse of nanoscale device simulation, modelling transport through sub-10-nanometre transistors by solving coupled Green's-function and Poisson equations self-consistently, while dynamical mean-field theory attacks strongly correlated systems — high-temperature superconductors, Mott insulators — through a local Green's function that flags metal-to-insulator transitions band theory misses. In every case the one-dimensional lessons — jump conditions, poles as spectrum, reciprocity, the Fredholm alternative — survive intact.

The physical picture

Picture a taut string pinned at both ends, pressed down at an interior point ξ\xi with a needle. It straightens into two segments — solving the source-free equation −u′′=0-u''=0 everywhere the needle is not — meeting at a sharp kink under the needle. That triangular tent is the Green's function, and every feature is dictated by three demands: it must vanish at the supports (boundary conditions), be a homogeneous solution on each side of the needle (straight segments), and kink just steeply enough that the second derivative supplies a unit delta at ξ\xi (the jump condition). The first figure makes the corner and its unit jump one picture.

Figure
The point-source response for −u″ = δ(x−ξ) on the interval 0 to 1 with ξ = 0.4: a tent that vanishes at both ends, peaks at G = ξ(1−ξ) = 0.24, and kinks so that its slope drops from +0.6 to −0.4, a unit jump G′(ξ⁺) − G′(ξ⁻) = −1 set by the source
The kink: a unit point source bends the string into a tent with a fixed corner x G ξ = 0.4 G(ξ,ξ) = ξ(1−ξ) = 0.24 slope +0.6 slope −0.4 0 1 G = 0 G = 0 jump: G′(ξ⁺) − G′(ξ⁻) = −0.4 − 0.6 = −1 = −1/p

Two homogeneous solutions do all the work. Call y1y_1 the one that obeys the left boundary and y2y_2 the right; the Green's function is y1y_1 left of the source and y2y_2 right, glued at ξ\xi. The source never changes the shape of either piece — only their relative weighting and the size of the corner. Superposition finishes the story: an arbitrary load is a dense comb of point presses, so the total displacement is the integral of the tents weighted by the load. The second figure shows this assembly and delivers a hard number — a uniform unit load turns the family of tents into a parabola with a definite peak.

Figure
Superposition in action: adding the point-source tents G(x,ξ) weighted by a uniform load f = 1 builds the smooth solution of −u″ = 1, namely u(x) = x(1−x)/2, which peaks at u(½) = 1/8 = 0.125 — the whole differential equation solved by one integral
Superposition: summing tents against a uniform load builds the solution u(x) x u point-source tents G(x,ξ) u(½) = 1/8 = 0.125 u(x) = x(1−x)/2 0 1 solving −u″ = 1 by one integral over the source

Reciprocity is the last idea the picture carries. For a self-adjoint operator the tent produced at xx by a source at ξ\xi equals the tent produced at ξ\xi by a source at xx: G(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x), the same symmetry that makes mutual inductances, acoustic transfer functions, and antenna gain patterns identical when source and receiver swap. Driven near a natural frequency, the spectral sum ∑nϕn(x)ϕn(ξ)/(λ−λn)\sum_n \phi_n(x)\phi_n(\xi)/(\lambda-\lambda_n) shows one term ballooning: GG diverges, announcing a resonance and the failure of an ordinary inverse.

The core equations

Everything begins with a single definition: the Green's function is what the operator returns when fed a unit point source. The subscript on Lx\mathcal{L}_x is not decoration — the differentiation acts on the field point xx, treating the source point ξ\xi as a fixed parameter.

Lx G(x,ξ)=δ(x−ξ)(1)\mathcal{L}_x\,G(x,\xi)=\delta(x-\xi) \qquad (1)

where: Lx\mathcal{L}_x — linear differential operator acting on xx; G(x,ξ)G(x,\xi) — Green's function, the response at xx to a unit source at ξ\xi; δ(x−ξ)\delta(x-\xi) — Dirac delta, the unit point source. Reading: GG solves the homogeneous problem away from ξ\xi and absorbs the delta through a controlled singularity there. The reason GG is worth computing is that it inverts the operator once and for all. Superposing the point responses against the actual source recovers the full solution, and the boundary conditions ride along for free because every G(⋅,ξ)G(\cdot,\xi) already obeys them.

u(x)=∫abG(x,ξ) f(ξ) dξ(2)u(x)=\int_a^b G(x,\xi)\,f(\xi)\,\mathrm{d}\xi \qquad (2)

where: u(x)u(x) — solution of Lu=f\mathcal{L}u=f obeying the boundary conditions; f(ξ)f(\xi) — prescribed source; [a,b][a,b] — domain. Reading: the differential equation has become a definite integral, with GG the kernel that represents L−1\mathcal{L}^{-1}. The structure is cleanest when L\mathcal{L} is self-adjoint — which any real second-order operator becomes once its coefficients are absorbed into Sturm–Liouville form. Here pp multiplies the leading derivative and sets the scale of the jump, while qq plays the role of a potential.

L=ddx ⁣(p(x)ddx)+q(x)(3)\mathcal{L}=\frac{\mathrm{d}}{\mathrm{d}x}\!\left(p(x)\frac{\mathrm{d}}{\mathrm{d}x}\right)+q(x) \qquad (3)

where: p(x)p(x) — positive coefficient of the leading derivative; q(x)q(x) — zeroth-order (potential-like) coefficient. Reading: in this form the operator is formally self-adjoint, which is what makes GG symmetric and its eigenfunctions orthogonal. For a second-order operator the whole Green's function collapses to one closed form: the two homogeneous solutions multiplied together — smaller argument into y1y_1, larger into y2y_2 — divided by a single constant.

G(x,ξ)={y1(x) y2(ξ)p(ξ) W(ξ),a≤x≤ξy1(ξ) y2(x)p(ξ) W(ξ),ξ≤x≤b(4)G(x,\xi)=\begin{cases}\dfrac{y_1(x)\,y_2(\xi)}{p(\xi)\,W(\xi)},& a\le x\le\xi\\[2mm]\dfrac{y_1(\xi)\,y_2(x)}{p(\xi)\,W(\xi)},& \xi\le x\le b\end{cases} \qquad (4)

where: y1y_1 — homogeneous solution obeying the left boundary at aa; y2y_2 — obeying the right boundary at bb; W(ξ)=y1y2′−y1′y2W(\xi)=y_1y_2'-y_1'y_2 — the Wronskian. Reading: the numerator is symmetric under x↔ξx\leftrightarrow\xi, so GG is manifestly reciprocal, and the denominator (below) is a genuine constant. The number that normalizes GG comes from integrating the defining equation across the source. Only the highest-derivative term survives the shrinking interval, so it carries the entire delta, forcing a jump in p G′p\,G'.

[ p(x) ∂xG(x,ξ) ]x=ξ−x=ξ+=1(5)\Big[\,p(x)\,\partial_x G(x,\xi)\,\Big]_{x=\xi^-}^{x=\xi^+}=1 \qquad (5)

where: ξ±\xi^{\pm} — approach to the source from above and below. Reading: with GG continuous, the first derivative jumps by exactly 1/p(ξ)1/p(\xi) — large pp a gentle kink, small pp a sharp one. The denominator p Wp\,W looks as though it should vary with ξ\xi, but Abel's identity for the self-adjoint form kills its derivative, so one evaluation fixes it everywhere.

W(x)=W(x0)exp⁡ ⁣(−∫x0xp′(s)p(s) ds)  ⇒  p(x) W(x)=const(6)W(x)=W(x_0)\exp\!\left(-\int_{x_0}^{x}\frac{p'(s)}{p(s)}\,\mathrm{d}s\right)\;\Rightarrow\; p(x)\,W(x)=\text{const} \qquad (6)

where: W(x)W(x) — Wronskian of the two homogeneous solutions; p(x)p(x) — Sturm–Liouville leading coefficient. Reading: (pW)′=0(pW)'=0, so the constant in equation (4) is evaluated at whichever point is convenient. When the eigenfunctions are known, GG has a second face — a bilinear sum over the spectrum, each mode weighted inversely by its distance from the driving parameter, so the sum diverges the moment λ\lambda meets an eigenvalue.

G(x,ξ)=∑nϕn(x) ϕn(ξ)λ−λn(7)G(x,\xi)=\sum_n \frac{\phi_n(x)\,\phi_n(\xi)}{\lambda-\lambda_n} \qquad (7)

where: ϕn\phi_n — orthonormal eigenfunctions, Lϕn=λnϕn\mathcal{L}\phi_n=\lambda_n\phi_n; λn\lambda_n — eigenvalues; λ\lambda — spectral parameter in (L−λ)G=−δ(\mathcal{L}-\lambda)G=-\delta. Reading: this is the resolvent (L−λ)−1(\mathcal{L}-\lambda)^{-1} in the eigenbasis, its poles the spectrum. Finally, self-adjointness delivers reciprocity as an exact theorem, not a lucky symmetry of one example.

G(x,ξ)=G(ξ,x)(8)G(x,\xi)=G(\xi,x) \qquad (8)

where: G(x,ξ)G(x,\xi) — Green's function of a self-adjoint operator with self-adjoint boundary conditions. Reading: swapping source and observation point leaves the response unchanged — the content of every reciprocity theorem.

Derivation ABuilding the second-order Green's function from continuity and the jump

The goal is to show that the closed form of equation (4) is forced, not guessed: for a second-order self-adjoint operator on [a,b][a,b] with homogeneous boundary conditions, GG must be y1y_1 times a constant on one side of the source and y2y_2 times a constant on the other, the two constants pinned uniquely by continuity plus the derivative jump extracted from the delta. The Wronskian appears as the natural normalization, and Abel's identity makes the denominator a single number, not a function of source location.

Step 1. Away from x=ξx=\xi the defining equation reads LxG=0\mathcal{L}_xG=0, so on each side GG is a homogeneous solution. The left piece must be proportional to y1y_1 (which obeys the condition at aa) and the right piece to y2y_2 (which obeys the condition at bb): write G=A(ξ) y1(x)G=A(\xi)\,y_1(x) for x<ξx<\xi and G=B(ξ) y2(x)G=B(\xi)\,y_2(x) for x>ξx>\xi. Why: choosing the boundary-adapted solution on each side discharges the boundary conditions before the source enters, leaving only two unknowns.

Step 2. Impose continuity at the source, A y1(ξ)=B y2(ξ)A\,y_1(\xi)=B\,y_2(\xi). This is mandatory: if GG itself jumped, G′′G'' would contain a δ′\delta', too strong for the single δ\delta on the right. Why: the operator's order dictates which derivative absorbs the delta; for order two it is the first derivative, so GG stays continuous.

Step 3. Extract the jump by integrating LG=δ\mathcal{L}G=\delta over [ξ−ϵ,ξ+ϵ][\xi-\epsilon,\xi+\epsilon]. The q Gq\,G term vanishes as ϵ→0\epsilon\to0 (GG bounded), leaving [p G′]ξ−ξ+=1[p\,G']_{\xi^-}^{\xi^+}=1, i.e. B y2′(ξ)−A y1′(ξ)=1/p(ξ)B\,y_2'(\xi)-A\,y_1'(\xi)=1/p(\xi). Why: only the leading-derivative term can produce a delta, so it alone carries the unit source, giving a jump of exactly 1/p(ξ)1/p(\xi).

Step 4. Solve the two-by-two system from Steps 2 and 3. Its determinant is the Wronskian W(ξ)=y1y2′−y1′y2W(\xi)=y_1y_2'-y_1'y_2, so A=y2(ξ)/[p(ξ)W(ξ)]A=y_2(\xi)/[p(\xi)W(\xi)] and B=y1(ξ)/[p(ξ)W(ξ)]B=y_1(\xi)/[p(\xi)W(\xi)], which substituted back give equation (4). Why: the Wronskian is nonzero precisely when y1y_1 and y2y_2 are independent, exactly the condition for a unique Green's function.

Step 5. Confirm the denominator is constant. Differentiating pWpW and using (pyi′)′=−qyi(py_i')'=-qy_i gives (pW)′=y1(py2′)′−y2(py1′)′=−qy1y2+qy2y1=0(pW)'=y_1(py_2')'-y_2(py_1')'=-qy_1y_2+qy_2y_1=0, Abel's identity (6). Hence p(ξ)W(ξ)p(\xi)W(\xi) is one number for the whole interval, and the symmetry G(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x) is visible directly in the numerator. Why: a constant denominator makes GG a clean product of two solutions and guarantees reciprocity with no further argument.

Go deeper · Derivation B · 1 minReciprocity from Green's identity
Derivation BReciprocity from Green's identity

The aim is to promote the symmetry G(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x) from an observation about one example to a theorem holding for every self-adjoint operator with self-adjoint boundary conditions. The tool is Lagrange's identity, which converts the difference of two operator actions into a boundary term that vanishes under exactly the conditions making L\mathcal{L} self-adjoint; feeding two source points into it leaves precisely the statement of reciprocity.

Step 1. Take two source locations ξ1\xi_1 and ξ2\xi_2 and form the bilinear combination G(x,ξ1)LG(x,ξ2)−G(x,ξ2)LG(x,ξ1)G(x,\xi_1)\mathcal{L}G(x,\xi_2)-G(x,\xi_2)\mathcal{L}G(x,\xi_1), then integrate over [a,b][a,b]. Why: this antisymmetric pairing is engineered so that Lagrange's identity can turn the bulk integral into a pure boundary contribution.

Step 2. Lagrange's identity makes the integrand the exact derivative ddx[p (G1G2′−G2G1′)]\frac{\mathrm{d}}{\mathrm{d}x}\big[p\,(G_1 G_2'-G_2 G_1')\big], so the integral equals the boundary term [p(G1G2′−G2G1′)]ab[p(G_1G_2'-G_2G_1')]_a^b, which vanishes for self-adjoint boundary conditions (Dirichlet, Neumann, Robin, periodic). Why: self-adjointness is defined precisely as the vanishing of this bilinear boundary form, so the whole integral is zero.

Step 3. Evaluate the same integral with LG(x,ξi)=δ(x−ξi)\mathcal{L}G(x,\xi_i)=\delta(x-\xi_i): the two deltas sift out G(ξ2,ξ1)−G(ξ1,ξ2)G(\xi_2,\xi_1)-G(\xi_1,\xi_2). Setting this to the zero from Step 2 gives G(ξ2,ξ1)=G(ξ1,ξ2)G(\xi_2,\xi_1)=G(\xi_1,\xi_2), reciprocity for arbitrary source points. Why: the delta sources turn the integral into the plain difference of GG with arguments swapped, so zero means symmetric.

Go deeper · Derivation C · 1 minThe spectral expansion and the Fredholm alternative
Derivation CThe spectral expansion and the Fredholm alternative

This derivation builds the eigenfunction representation of equation (7) and reads off, from a single divergent term, exactly when the construction fails and what repairs it. The payoff is conceptual: the poles of GG are the operator's spectrum, resonance is the collision of the driving parameter with an eigenvalue, and the Fredholm alternative — a solvability condition met again in perturbation theory — falls out of demanding the source not overlap the offending mode.

Step 1. Expand GG and the delta in the orthonormal eigenbasis {ϕn}\{\phi_n\}, using completeness δ(x−ξ)=∑nϕn(x)ϕn(ξ)\delta(x-\xi)=\sum_n\phi_n(x)\phi_n(\xi). Writing G=∑ncn(ξ)ϕn(x)G=\sum_n c_n(\xi)\phi_n(x) and demanding (L−λ)G=−δ(\mathcal{L}-\lambda)G=-\delta gives (λn−λ)cn=−ϕn(ξ)(\lambda_n-\lambda)c_n=-\phi_n(\xi), hence G=∑nϕn(x)ϕn(ξ)/(λ−λn)G=\sum_n\phi_n(x)\phi_n(\xi)/(\lambda-\lambda_n). Why: in the eigenbasis the operator is diagonal, so inverting it is term-by-term division, exposing every eigenvalue as a potential pole.

Step 2. Let λ→λm\lambda\to\lambda_m. The term n=mn=m blows up like ϕm(x)ϕm(ξ)/(λ−λm)\phi_m(x)\phi_m(\xi)/(\lambda-\lambda_m) and dominates — GG diverges. Physically L−λm\mathcal{L}-\lambda_m now has a nontrivial kernel ϕm\phi_m, so it is not invertible and no ordinary GG exists. Why: a zero eigenvalue of L−λ\mathcal{L}-\lambda is a direction the operator annihilates, and you cannot divide by zero along it.

Step 3. When does a solution nonetheless exist? Pairing (L−λm)u=f(\mathcal{L}-\lambda_m)u=f with ϕm\phi_m and using self-adjointness gives ⟨ϕm,f⟩=⟨(L−λm)ϕm,u⟩=0\langle\phi_m,f\rangle=\langle(\mathcal{L}-\lambda_m)\phi_m,u\rangle=0: a solution exists only if the source is orthogonal to the kernel eigenfunction — the Fredholm alternative. Why: the operator reaches only source components in its range, which is exactly the orthogonal complement of the kernel.

Step 4. When it holds, invert on the orthogonal complement by dropping the singular term: Gmod=∑n≠mϕn(x)ϕn(ξ)/(λm−λn)G_{\text{mod}}=\sum_{n\ne m}\phi_n(x)\phi_n(\xi)/(\lambda_m-\lambda_n) satisfies LxGmod=δ(x−ξ)−ϕm(x)ϕm(ξ)\mathcal{L}_xG_{\text{mod}}=\delta(x-\xi)-\phi_m(x)\phi_m(\xi), unique up to a multiple of ϕm\phi_m. Why: subtracting the projector onto ϕm\phi_m removes the un-invertible direction, leaving a well-defined inverse on everything the operator can reach.

Limiting cases

The single object GG specializes to a menagerie of named kernels depending on the operator and its boundary conditions:

CaseResultMeaning
On resonance (λ→λn\lambda\to\lambda_n)G∼ϕn(x)ϕn(ξ)λ−λnG\sim\dfrac{\phi_n(x)\phi_n(\xi)}{\lambda-\lambda_n}one term dominates and diverges; no ordinary GG exists — use the modified Green's function and impose ⟨f,ϕn⟩=0\langle f,\phi_n\rangle=0 (Fredholm alternative)
Causal / retardedG(t,τ)=0G(t,\tau)=0 for t<τt<\tauthe initial-value Green's function: the response vanishes before the source acts, built from the solution obeying the initial conditions
Free space (whole line)L=−d2dx2+k2, G=12ke−k∣x−ξ∣\mathcal{L}=-\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}+k^2,\ G=\dfrac{1}{2k}e^{-k|x-\xi|}with decay at ±∞\pm\infty, GG is the screened (Yukawa-like) kernel, symmetric and a function of ∣x−ξ∣|x-\xi| only
Self-adjoint operatorG(x,ξ)=G(ξ,x)G(x,\xi)=G(\xi,x)reciprocity holds; for a non-self-adjoint L\mathcal{L} one instead gets GG built from the adjoint L†\mathcal{L}^{\dagger}, and symmetry is replaced by GL(x,ξ)=GL†(ξ,x)G_{\mathcal{L}}(x,\xi)=G_{\mathcal{L}^\dagger}(\xi,x)

Worked examples

1 · The 1D Poisson Green's function (fully worked)

Construct G(x,ξ)G(x,\xi) for −u′′=f-u''=f on [0,1][0,1] with u(0)=u(1)=0u(0)=u(1)=0, evaluate it at a sample point, and verify continuity, the jump, and symmetry.

Step 1. Here L=−d2/dx2\mathcal{L}=-\mathrm{d}^2/\mathrm{d}x^2, so p=1p=1, q=0q=0. The homogeneous equation −y′′=0-y''=0 has linear solutions; pick the boundary-adapted pair y1(x)=xy_1(x)=x (vanishes at 00) and y2(x)=1−xy_2(x)=1-x (vanishes at 11). Their Wronskian is W=x(−1)−(1)(1−x)=−1W=x(-1)-(1)(1-x)=-1, so pW=−1pW=-1, constant as Abel promised.

Step 2. Construct directly from continuity and the jump [−G′]ξ−ξ+=1[-G']_{\xi^-}^{\xi^+}=1 (minding the overall minus in L\mathcal{L}): this gives G(x,ξ)=x(1−ξ)G(x,\xi)=x(1-\xi) for x≤ξx\le\xi and ξ(1−x)\xi(1-x) for x≥ξx\ge\xi, i.e. the compact G(x,ξ)=min⁡(x,ξ) [1−max⁡(x,ξ)]G(x,\xi)=\min(x,\xi)\,[1-\max(x,\xi)].

Step 3. Evaluate at x=0.3, ξ=0.6x=0.3,\ \xi=0.6: since x<ξx<\xi, G=x(1−ξ)=0.3×0.4=0.12G=x(1-\xi)=0.3\times0.4=\mathbf{0.12}. Swapping arguments, G(0.6,0.3)G(0.6,0.3) has x>ξx>\xi, so G=ξ(1−x)=0.3×0.4=0.12G=\xi(1-x)=0.3\times0.4=\mathbf{0.12} — identical, confirming reciprocity numerically.

Step 4. Check the jump. Left slope ∂xG=1−ξ\partial_xG=1-\xi; right slope −ξ-\xi; jump ∂xG(ξ+)−∂xG(ξ−)=−ξ−(1−ξ)=−1\partial_xG(\xi^+)-\partial_xG(\xi^-)=-\xi-(1-\xi)=\mathbf{-1}, and since −G′′=δ-G''=\delta this is [−G′]=+1[-G']=+1 as required. At ξ=0.4\xi=0.4 the tent peaks at ξ(1−ξ)=0.24\xi(1-\xi)=\mathbf{0.24} with slopes +0.6+0.6 and −0.4-0.4 — precisely the first figure. G=min⁡(x,ξ)[1−max⁡(x,ξ)]G=\min(x,\xi)[1-\max(x,\xi)], symmetric, with a unit derivative jump.

2 · Solving a real source with the kernel

Use the Green's function of Example 1 to solve −u′′=1-u''=1 on [0,1][0,1] (a uniformly loaded string) and confirm the result independently.

Step 1. Superpose via equation (2), splitting at the source so each piece uses the correct branch: u(x)=∫01G(x,ξ) dξ=∫0xξ(1−x) dξ+∫x1x(1−ξ) dξu(x)=\int_0^1 G(x,\xi)\,\mathrm{d}\xi=\int_0^x \xi(1-x)\,\mathrm{d}\xi+\int_x^1 x(1-\xi)\,\mathrm{d}\xi.

Step 2. The first integral is (1−x)x22(1-x)\tfrac{x^2}{2}, the second x(1−x)22x\tfrac{(1-x)^2}{2}. Adding and factoring x(1−x)2[x+(1−x)]\tfrac{x(1-x)}{2}[x+(1-x)] collapses to u(x)=12x(1−x)u(x)=\tfrac{1}{2}x(1-x), a downward parabola pinned at both ends.

Step 3. Verify directly: u=12(x−x2)u=\tfrac12(x-x^2) gives u′′=−1u''=-1, so −u′′=1-u''=1 as required, and u(0)=u(1)=0u(0)=u(1)=0 holds. The maximum sag is at the centre, u(12)=12⋅12⋅12=0.125=1/8u(\tfrac12)=\tfrac12\cdot\tfrac12\cdot\tfrac12=\mathbf{0.125=1/8} — the green curve and peak of the second figure. The kernel reproduced the exact parabolic deflection, peak 1/81/8, in one integration.

3 · The retarded oscillator (impulse and step response)

Find the causal Green's function of L=d2/dt2+ω2\mathcal{L}=\mathrm{d}^2/\mathrm{d}t^2+\omega^2, then use it to compute the response of an initially resting oscillator to a suddenly applied constant force, with ω=2 s−1\omega=2\ \mathrm{s}^{-1}.

Step 1. Solve G′′+ω2G=δ(t−τ)G''+\omega^2G=\delta(t-\tau) with G=0G=0 for t<τt<\tau (causality). For t>τt>\tau, G=Ccos⁡ω(t−τ)+Dsin⁡ω(t−τ)G=C\cos\omega(t-\tau)+D\sin\omega(t-\tau); continuity at t=τt=\tau forces C=0C=0 and the unit jump in G′G' (here p=1p=1) forces Dω=1D\omega=1. Hence G(t,τ)=sin⁡[ω(t−τ)]ω θ(t−τ)G(t,\tau)=\dfrac{\sin[\omega(t-\tau)]}{\omega}\,\theta(t-\tau), an oscillation that switches on at the impulse with onset slope G′(τ+)=1G'(\tau^+)=1.

Step 2. Drive with a unit step force f(t)=F0 θ(t)f(t)=F_0\,\theta(t), F0=1F_0=1, from rest. Convolve: u(t)=∫0tsin⁡ω(t−τ)ωF0 dτ=F0ω∫0tsin⁡ωs ds=F0ω2(1−cos⁡ωt)u(t)=\int_0^t \dfrac{\sin\omega(t-\tau)}{\omega}F_0\,\mathrm{d}\tau=\dfrac{F_0}{\omega}\int_0^t\sin\omega s\,\mathrm{d}s=\dfrac{F_0}{\omega^2}\big(1-\cos\omega t\big).

Step 3. Read the physics. A slowly applied F0F_0 gives static deflection F0/ω2=1/4=0.25F_0/\omega^2=1/4=\mathbf{0.25}; the dynamic response oscillates about it between 00 and 2F0/ω22F_0/\omega^2, so a suddenly applied load overshoots to twice the static value — dynamic amplification factor exactly 2\mathbf{2}. With ω=2\omega=2 the peak u=0.5u=\mathbf{0.5} first occurs at ωt=π\omega t=\pi, i.e. t=π/2≈1.57 st=\pi/2\approx1.57\ \mathrm{s}. u(t)=14(1−cos⁡2t)u(t)=\tfrac14(1-\cos2t), static 0.250.25, peak 0.50.5, amplification factor 2.

Physics of the portrait

The Green's function carries the name of a man who had almost no formal education and whose masterwork sold to a few dozen subscribers before vanishing for two decades — that it now underlies quantum field theory is one of the great improbabilities in the history of mathematics.

Origins · 1 minGeorge Green and how this was discovered
GG
George Green
1793–1841 · self-taught Nottingham miller, founder of British potential theory

George Green was born in 1793 in Sneinton, near Nottingham, and worked his father's windmill for most of his life with, by surviving records, only about a year of formal schooling as a child. How he taught himself the continental mathematics of Laplace, Lagrange, and Poisson — none of it then taught at English universities — remains a genuine mystery; the Nottingham Subscription Library and a lone patron, Sir Edward Bromhead, are the only documented threads. In 1828 Green privately printed An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, which introduced the word "potential" in its modern sense, stated the integral identity now called Green's theorem, and constructed the reciprocal function GG that solves a boundary-value problem by representing the field of a point source. It sold to 51 subscribers. Encouraged by Bromhead, he entered Gonville and Caius College, Cambridge, as an undergraduate in 1833 at the age of forty, graduating fourth wrangler in 1837 (the Green strain tensor of elasticity is also his). He died in 1841, still largely unknown. Only in 1845 did the young William Thomson (later Lord Kelvin) find a copy, grasp its importance, and republish it — after which Green's function became, and remains, the central object of every field theory from electrostatics to the Standard Model.

History
1828
— Green privately prints his Essay, introducing the potential and the reciprocal (Green's) function
1836
— Sturm and Liouville develop the eigenvalue theory that later grounds the spectral expansion of G
1845
— William Thomson rediscovers Green's forgotten Essay and champions it
1903
— Fredholm's theory of integral equations delivers the solvability alternative
1910
— Hilbert and Schmidt put the bilinear spectral expansion on rigorous footing

Going further

Texts. Stone and Goldbart, Mathematics for Physics, Ch. 5 (Green's Functions) · Arfken, Weber and Harris, Mathematical Methods for Physicists, 7th ed., Ch. 10 · Hassani, Mathematical Physics, 2nd ed., Ch. 21 · Byron and Fuller, Mathematics of Classical and Quantum Physics, Ch. 7 · Riley, Hobson and Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 17 · Courant and Hilbert, Methods of Mathematical Physics, Vol. I · Morse and Feshbach, Methods of Theoretical Physics, Part I, Ch. 7. All equations are standard results. Key checks: the jump [pG′]=1[pG']=1 follows from integrating LG=δ\mathcal{L}G=\delta across ξ\xi; the 1D Poisson kernel G=min⁡(x,ξ)[1−max⁡(x,ξ)]G=\min(x,\xi)[1-\max(x,\xi)] satisfies −G′′=δ-G''=\delta, is symmetric, and peaks at ξ(1−ξ)=0.24\xi(1-\xi)=0.24 when ξ=0.4\xi=0.4; integrating that kernel against f=1f=1 recovers u=12x(1−x)u=\tfrac12 x(1-x) with maximum 1/81/8; the retarded oscillator kernel sin⁡[ω(t−τ)]/ω⋅θ(t−τ)\sin[\omega(t-\tau)]/\omega\cdot\theta(t-\tau) yields the step response F0ω2(1−cos⁡ωt)\tfrac{F_0}{\omega^2}(1-\cos\omega t) with amplification factor 2; and Abel's identity (pW)′=0(pW)'=0 underpins the modified-Green's-function construction. No numerical constants beyond exact analytic forms are asserted.

Builds on: Sturm–Liouville Theory · Dirac Delta Distribution · Wronskian and Linear Independence · Second-Order Linear ODEs. Leads to: WKB Approximation · Connection Formulae and Turning Points · Bohr–Sommerfeld Quantization · Asymptotic Series and Poincaré Expansions.

Ask the tutor: Name any operator — a damped oscillator, a Bessel equation, the finite-well Schrödinger problem — and I will build its Green's function with you from the jump condition to the full superposed solution.

Why does the Green's function turn a differential equation into an integral?

Because a differential operator is linear, and a linear map is fully known once you know what it does to a basis. The deltas {δ(x−ξ)}\{\delta(x-\xi)\} form a continuous basis, so knowing the response G(x,ξ)G(x,\xi) to each one, the response to a source f=∫f(ξ)δ(x−ξ) dξf=\int f(\xi)\delta(x-\xi)\,\mathrm{d}\xi is the same superposition: u(x)=∫G(x,ξ)f(ξ) dξu(x)=\int G(x,\xi)f(\xi)\,\mathrm{d}\xi. In operator language GG is the integral kernel of L−1\mathcal{L}^{-1}, so solving once for a point source buys the solution for every source at the cost of one integration.

Why is G continuous but G′ discontinuous for a second-order operator?

The right-hand side of LG=δ\mathcal{L}G=\delta is a delta — singular, but mildly so, since a delta comes from differentiating a step exactly once. In a second-order operator the most-differentiated term is G′′G'', so a jump in G′G' produces the δ\delta in G′′G'' and balances the equation. If GG itself jumped, G′′G'' would contain δ′\delta', a dipole-like singularity far too strong for the single δ\delta on the right. So the value stays continuous and the discontinuity is pushed into the first derivative, its size fixed at 1/p(ξ)1/p(\xi) by integrating across the source.

What actually is the Wronskian doing in the formula?

The Wronskian W=y1y2′−y1′y2W=y_1y_2'-y_1'y_2 measures how independent the two homogeneous solutions are; it is the determinant of the two-by-two system fixing the coefficients on each side of the source, so it lands in the denominator and normalizes GG. If W=0W=0 the two solutions are proportional — they satisfy the same boundary condition — and no Green's function exists, the algebraic face of a resonance. Abel's identity then makes p(ξ)W(ξ)p(\xi)W(\xi) a single constant, evaluated once and reused.

What is the difference between the boundary-value and causal Green's functions?

Both satisfy LG=δ\mathcal{L}G=\delta but carry different side conditions. The boundary-value Green's function lives on a finite interval [a,b][a,b] and obeys conditions at both ends — symmetric, built from y1y_1 (left) and y2y_2 (right), describing a static field like a loaded string. The causal or retarded Green's function lives on a time axis and obeys initial conditions, G=0G=0 before the source acts. It is not symmetric — cause precedes effect — and is built from the single solution starting from rest. The retarded oscillator kernel of Worked Example 3 is the canonical case, the impulse response engineers convolve for any driven transient.

Why does the Green's function blow up at a resonance?

The spectral form G=∑nϕn(x)ϕn(ξ)/(λ−λn)G=\sum_n\phi_n(x)\phi_n(\xi)/(\lambda-\lambda_n) makes it plain: each eigenmode contributes a term inversely proportional to how far λ\lambda sits from its eigenvalue λn\lambda_n. Drive the system exactly at an eigenvalue and one denominator vanishes, so that term — and the whole sum — diverges. Physically L−λ\mathcal{L}-\lambda then annihilates ϕm\phi_m and has no inverse; you are pushing a frictionless swing at its natural frequency and the amplitude runs away. The cure is the Fredholm alternative: a finite solution exists only if ⟨f,ϕm⟩=0\langle f,\phi_m\rangle=0, whereupon the modified Green's function with that mode projected out does the job.

How does this ODE object become the propagator of quantum field theory?

The same construction, applied to the wave, Schrödinger, or Klein–Gordon operator instead of an ODE, produces a Green's function of space and time — precisely the propagator, the amplitude for a disturbance created at one spacetime point to be detected at another, the reciprocity idea promoted to four dimensions. Boundary conditions in time select the flavour: retarded (causal), advanced, or — Feynman's choice, positive frequencies forward and negative backward in time — the Feynman propagator. Perturbation theory then strings these free propagators together with interaction vertices, which is what Feynman diagrams draw. Every ODE lesson (jump conditions, poles as spectrum, the resolvent, solvability) reappears, just dressed in more indices.

Before this

Where this leads