Green's Functions for Ordinary Differential Equations

A linear differential operator $\mathcal{L}$ acting on functions over an interval, together with homogeneous boundary conditions, behaves exactly like a matrix on a finite-dimensional vector space. Solving $\mathcal{L}u=f$ is then the continuous analogue of inverting a matrix, $u=\mathcal{L}^{-1}f$ , and the Green's function $G(x,\xi)$ is precisely the integral kernel of that inverse operator: the entry of $\mathcal{L}^{-1}$ labelled by a row $x$ and a column $\xi$ . Construct $G$ once and the response to every source $f$ follows by a single integral, because superposition does the rest — this is the payoff that makes the kernel worth building. Physically $G(x,\xi)$ is the response measured at $x$ to a unit point source placed at $\xi$ : the deflection of a loaded string, the potential of a charged sheet between grounded plates, the motion of an oscillator struck by a sharp impulse. The delta-function source $\mathcal{L}_xG=\delta(x-\xi)$ makes this literal, and it is the same statement as $\mathcal{L}\mathcal{L}^{-1}=I$ with the Dirac delta standing in for the identity's Kronecker delta. An arbitrary load is a weighted continuum of point sources, so the total response is their integral against $G$ — the bridge between the static (boundary-value) and dynamic (causal, initial-value) faces of the theory. For a regular second-order operator the construction is explicit and algorithmic: pick a homogeneous solution obeying the left boundary condition and one obeying the right, glue them at the source with continuity and a Wronskian-fixed jump in slope, and the answer organizes itself around the same machinery that drives variation of parameters. The self-adjoint (Sturm–Liouville) case is more beautiful still — $G$ is symmetric, real, and carries an eigenfunction expansion whose poles are the system's resonances. This article builds all of it, with two exactly-solvable examples where every number checks by hand.

Key takeaways

A Green's function $G(x,\xi)$ is the kernel that inverts a linear differential operator: it solves $\mathcal{L}_xG(x,\xi)=\delta(x-\xi)$ under the same homogeneous boundary conditions as the problem.
Once $G$ is known, the inhomogeneous equation $\mathcal{L}u=f$ is solved in one stroke by superposition, $u(x)=\int G(x,\xi)\,f(\xi)\,\mathrm{d}\xi$ .
For a second-order operator, $G$ is built from two boundary-adapted homogeneous solutions $u_1,u_2$ , glued at $x=\xi$ by continuity plus a slope jump fixed to $1/p(\xi)$ by integrating the delta across the source.
The closed form is $G=u_1(x_<)\,u_2(x_>)/[p(\xi)W(\xi)]$ , tying Green's functions directly to the Wronskian and to variation of parameters.
For self-adjoint (Sturm–Liouville) operators $G$ is symmetric, $G(x,\xi)=G(\xi,x)$ — reciprocity — and expands spectrally as $G=\sum_n\phi_n(x)\phi_n(\xi)/(\lambda-\lambda_n)$ , whose poles at $\lambda=\lambda_n$ are resonances.
When zero is an eigenvalue the ordinary $G$ diverges ( $W\to0$ ); the modified Green's function plus a Fredholm solvability condition $\int f\,\phi_0=0$ save the day.
Key idea: treat $\mathcal{L}$ as an invertible operator on functions obeying homogeneous boundary conditions; its inverse is an integral operator whose kernel $G$ solves $\mathcal{L}_xG(x,\xi)=\delta(x-\xi)$ , built piecewise from boundary-adapted solutions joined by continuity and a Wronskian-fixed slope jump.

The core insight

Treat $\mathcal{L}$ as an invertible linear operator on a space of functions satisfying homogeneous boundary conditions; its inverse is an integral operator whose kernel $G(x,\xi)$ solves $\mathcal{L}_xG(x,\xi)=\delta(x-\xi)$ with the same boundary conditions, built piecewise from boundary-adapted homogeneous solutions joined by continuity and a Wronskian-fixed slope jump.

Where this lives today

The impulse-response kernel Green wrote down for grounded conductors is now one of the most reused objects in quantitative science. In structural engineering the loaded-string and Euler–Bernoulli-beam Green's functions (piecewise cubic, with continuity of $G,G',G''$ and a unit jump in $G'''$ ) underlie the influence-line and stiffness methods that size bridges and aircraft spars. In signals and control the retarded oscillator kernel is the impulse response $h(t)$ : convolving it with an input is exactly $u=\int Gf$ , and its Laplace transform is the transfer function every filter and feedback loop is designed around. In many-body physics the resolvent $G(x,\xi;\lambda)=\sum_n\phi_n\phi_n/(\lambda_n-\lambda)$ becomes the single-particle Green's function whose poles are excitation energies and whose residues are spectral weights measured by photoemission. And in quantum scattering the Lippmann–Schwinger equation is nothing but the superposition formula (2) with the free-particle Green's function as kernel — the same $\mathcal{L}^{-1}$ -as-integral idea, promoted to operators, that turns a differential equation into an integral equation solvable by iteration. The open frontier is numerical: boundary-element and Green's-function Monte Carlo methods trade a volume PDE for a surface or stochastic integral against $G$ .

The physical picture

Take a string clamped at both ends and press it at a single interior point $\xi$ with a unit force. The string does not bow into a smooth curve — it takes two straight segments rising to a sharp kink directly under the finger, and that shape is $G(x,\xi)$ . Everywhere off the source the string is unforced, so it satisfies the homogeneous equation (straight lines, for a uniform string); the kink at $\xi$ is the delta-function source made visible as a discontinuity in slope. A general load $f$ is a sum of such pokes, so the loaded string's shape is the integral of $G$ against $f$ . The first figure is this picture with the numbers filled in.

Figure

The Green's function G(x, ξ) for −u″ = δ(x − ξ) on the interval 0 ≤ x ≤ 1 with u(0) = u(1) = 0, drawn for a source at ξ = 0.5: two straight segments meeting in a kink of height ξ(1 − ξ) = 0.25, with a unit downward jump in slope from +0.5 to −0.5 that encodes the point source

Two features of that tent are the whole theory in miniature. The height $\xi(1-\xi)=0.25$ at the centre is continuous, because a jump in $G$ itself would make $G''$ far too singular to match a mere delta. The slope, by contrast, falls abruptly from $+0.5$ to $-0.5$ , a jump of exactly $-1=-1/p$ : that discontinuity is where the delta lives, since a second derivative of a corner is a delta. Reciprocity — the symmetry $G(x,\xi)=G(\xi,x)$ — says the deflection at $x$ from a source at $\xi$ equals the deflection at $\xi$ from a source at $x$ ; it is the ODE shadow of Betti–Rayleigh reciprocity. The same kernel has a dynamical twin. Replace the clamped string by a mass on a spring and the point source by a sharp kick, and $G$ becomes the impulse response: zero until the kick arrives (causality), then free oscillation forever after. The second figure shows the retarded Green's function of an undamped oscillator — the initial-value face of exactly the same $\mathcal{L}^{-1}=$ kernel idea, now with the "right boundary" sitting at $t=+\infty$ .

Figure

The causal retarded impulse response G(t, t′) = (1/ω₀) sin ω₀(t − t′) for a unit impulse delivered at t′ = 0 to an oscillator with ω₀ = 2 rad/s: exactly zero before the kick, then a sine of amplitude 1/ω₀ = 0.5 s and period 2π/ω₀ = π ≈ 3.14 s

The core equations

The defining relation is the continuous version of "invert the matrix": $G$ is the response of $\mathcal{L}$ to the sharpest possible source — a unit spike at $\xi$ — and the boundary conditions fix it uniquely. Everything else in this article either builds $G$ from this equation or uses it.

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

where: $\mathcal{L}_x$ — linear differential operator acting in the variable $x$ ; $G(x,\xi)$ — Green's function, the response at $x$ to a unit source at $\xi$ ; $\delta(x-\xi)$ — Dirac delta, the unit point source that plays the role of the identity kernel. Read it as $\mathcal{L}\mathcal{L}^{-1}=I$ . Because $\mathcal{L}$ is linear, a general source is handled by superposition: weight each point-source response by the local strength $f(\xi)$ and integrate. This is the equation you actually use — one integral produces the solution for any forcing.

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

where: $u(x)$ — solution of $\mathcal{L}u=f$ obeying the homogeneous boundary conditions; $f(\xi)$ — inhomogeneous source or forcing term; $[a,b]$ — the interval carrying the boundary conditions. Applying $\mathcal{L}_x$ under the integral and using (1) returns $f(x)$ exactly. Almost any regular second-order operator can be written in self-adjoint (Sturm–Liouville) form, which is what makes the symmetric, spectral theory available. An integrating factor turns a generic $a_2u''+a_1u'+a_0u$ into a single exact derivative $(pu')'$ .

\mathcal{L}=\frac{\mathrm{d}}{\mathrm{d}x}\!\left(p(x)\frac{\mathrm{d}}{\mathrm{d}x}\right)+q(x),\qquad p(x)=\exp\!\left(\int\frac{a_1}{a_2}\,\mathrm{d}x\right) \qquad (3)

where: $p(x)$ — leading coefficient, strictly positive on $(a,b)$ ; $q(x)$ — zeroth-order coefficient; the integrating factor $p/a_2$ is what collapses $a_2u''+a_1u'$ into $(pu')'$ . In this form $\mathcal{L}$ is formally self-adjoint. Integrating the defining equation across an infinitesimal window around $\xi$ isolates what the delta does to $G$ : it leaves $G$ continuous but forces a definite jump in the flux $pG'$ . With continuity, this pins down the two free constants in the piecewise construction.

\left[\,p(x)\,\partial_x G(x,\xi)\,\right]_{x=\xi^-}^{x=\xi^+}=1 \qquad (4)

where: $\xi^\pm$ — limits approaching the source from above and below; the bounded $qG$ term integrates to zero across the shrinking window, leaving only the flux jump. The jump in slope itself is $1/p(\xi)$ . Solving (1) off the source and matching with (4) gives the closed form. Build $G$ from $u_1$ (satisfying the left boundary condition) below the source and $u_2$ (satisfying the right) above it, and the constants collapse onto the Wronskian.

G(x,\xi)=\frac{u_1(x_<)\,u_2(x_>)}{p(\xi)\,W(\xi)},\qquad x_<=\min(x,\xi),\ \ x_>=\max(x,\xi) \qquad (5)

where: $u_1$ — homogeneous solution obeying the left boundary condition at $a$ ; $u_2$ — homogeneous solution obeying the right boundary condition at $b$ ; $W=u_1u_2'-u_1'u_2$ — their Wronskian, and $p\,W$ is constant by Abel's theorem for Sturm–Liouville $\mathcal{L}$ . The product $x_<,x_>$ notation encodes the piecewise swap automatically. When $\mathcal{L}$ is self-adjoint its orthonormal eigenfunctions resolve the delta by completeness, and $G$ becomes a sum over modes — each carrying a pole at an eigenvalue, so the spectral form reads off exactly where the operator becomes singular: its resonances.

G(x,\xi;\lambda)=\sum_{n}\frac{\phi_n(x)\,\phi_n(\xi)}{\lambda_n-\lambda} \qquad (6)

where: $\phi_n$ — orthonormal eigenfunctions, $\mathcal{L}\phi_n=\lambda_n\phi_n$ ; $\lambda_n$ — eigenvalues; $\lambda$ — spectral parameter in $(\mathcal{L}-\lambda)G=\delta$ . As $\lambda\to\lambda_n$ the $n$ -th term diverges — resonance — unless the source is orthogonal to $\phi_n$ .

Derivation AConstructing the Green's function for a regular second-order BVP

The goal is to turn the abstract statement " $G$ inverts $\mathcal{L}$ " into a formula you can write down from two homogeneous solutions. We work on $[a,b]$ with $\mathcal{L}=(p\,\mathrm{d}/\mathrm{d}x)'+q$ and homogeneous boundary conditions at both ends, exploiting the fact that the delta source vanishes everywhere except at $\xi$ . Off that point $G$ solves the homogeneous equation, so the problem reduces to choosing the right homogeneous solution on each side and stitching the two together with the matching conditions the delta demands — continuity of $G$ and a fixed jump in its slope, with the Wronskian emerging as the determinant of the joining system.

Step 1. For $x\neq\xi$ the delta vanishes, so $\mathcal{L}_xG=0$ : on each side of the source $G$ is a homogeneous solution. Why: the point source can only act at $\xi$ ; away from it the equation is the free one, so $G$ must be a combination of the two independent homogeneous solutions.

Step 2. Impose the boundary conditions by choosing $G=A\,u_1(x)$ for $x<\xi$ and $G=B\,u_2(x)$ for $x>\xi$ , where $u_1$ satisfies the left condition and $u_2$ the right. Why: $G$ must inherit the homogeneous boundary conditions, so below the source it can only be the boundary-adapted solution $u_1$ , and above it only $u_2$ ; the constants $A,B$ may depend on $\xi$ .

Step 3. Demand continuity at the source, $A\,u_1(\xi)=B\,u_2(\xi)$ . Why: if $G$ jumped, $G'$ would contain a delta and $G''$ a derivative of a delta — a source far more singular than the plain $\delta(x-\xi)$ on the right-hand side, which $\mathcal{L}G$ could never match.

Step 4. Impose the slope jump from equation (4), $p(\xi)\big[B\,u_2'(\xi)-A\,u_1'(\xi)\big]=1$ . Why: integrating $\mathcal{L}G=\delta$ across the source shows the flux $pG'$ must jump by exactly one, which is the only trace the delta leaves once $G$ is continuous.

Step 5. Solve the two-by-two system. The determinant is $p(\xi)\big[u_1(\xi)u_2'(\xi)-u_1'(\xi)u_2(\xi)\big]=p(\xi)W(\xi)$ , giving $A=u_2(\xi)/[p(\xi)W(\xi)]$ and $B=u_1(\xi)/[p(\xi)W(\xi)]$ . Why: the Wronskian $W$ is nonzero precisely when $u_1,u_2$ are independent, i.e. when the homogeneous problem has no nontrivial solution and the inverse exists.

Step 6. Assemble the two pieces into $G(x,\xi)=u_1(x_<)\,u_2(x_>)/[p(\xi)W(\xi)]$ . Why: below the source $x_<=x$ , $x_>=\xi$ and $G=A\,u_1(x)$ ; above it the roles swap; and Abel's theorem makes $pW$ constant, so the formula is manifestly symmetric under $x\leftrightarrow\xi$ . *The assembled $G$ is continuous, solves $\mathcal{L}G=0$ off the source, jumps by $1/p$ in slope, and obeys both boundary conditions — identical to the variation-of-parameters particular solution.*

Go deeper · Derivation B · 1 minSymmetry of the self-adjoint Green's function

Derivation BSymmetry of the self-adjoint Green's function

The goal is to prove reciprocity, $G(x,\xi)=G(\xi,x)$ , directly from self-adjointness rather than by inspecting a formula, so the result holds for every self-adjoint boundary condition — Dirichlet, Neumann, Robin, periodic — at once. The engine is Lagrange's identity, which converts $v\mathcal{L}u-u\mathcal{L}v$ into a total derivative of a bilinear boundary term (the concomitant). Integrating over the interval turns that into boundary values the self-adjoint conditions annihilate, and feeding in the delta equations for two source points collapses the rest to the symmetry statement by sifting.

Step 1. Write Lagrange's identity for $\mathcal{L}=(p\,\mathrm{d}/\mathrm{d}x)'+q$ : $v\,\mathcal{L}u-u\,\mathcal{L}v=\frac{\mathrm{d}}{\mathrm{d}x}\big[p(vu'-uv')\big]$ . Why: this exact-derivative structure is the defining property of formal self-adjointness — the $q$ terms cancel and the $p$ terms bundle into a single boundary flux.

Step 2. Choose $u=G(x,x_1)$ and $v=G(x,x_2)$ and integrate over $[a,b]$ , giving $\int_a^b\!\big[G(x,x_2)\mathcal{L}_xG(x,x_1)-G(x,x_1)\mathcal{L}_xG(x,x_2)\big]\mathrm{d}x=\big[p(vu'-uv')\big]_a^b$ . Why: both Green's functions obey the same homogeneous boundary conditions in $x$ , which is exactly what will kill the right-hand side.

Step 3. The boundary term vanishes, $\big[p(vu'-uv')\big]_a^b=0$ . Why: self-adjoint boundary conditions are defined so that the bilinear concomitant annihilates any pair of functions satisfying them, at both endpoints.

Step 4. Substitute the defining equations $\mathcal{L}_xG(x,x_i)=\delta(x-x_i)$ to get $\int_a^b\!\big[G(x,x_2)\delta(x-x_1)-G(x,x_1)\delta(x-x_2)\big]\mathrm{d}x=0$ . Why: each Green's function is by construction the response to a delta at its second argument, so the operator can be traded for a delta.

Step 5. Sift the deltas: $G(x_1,x_2)-G(x_2,x_1)=0$ . Why: the sifting property evaluates each integral at the delta's location, and the vanishing difference is precisely reciprocity — the response at $x_1$ to a source at $x_2$ equals the response at $x_2$ to a source at $x_1$ . *This holds whenever the boundary conditions make the concomitant vanish — Dirichlet, Neumann, Robin, or periodic — and is consistent with the explicit $x_<,x_>$ symmetry of equation (5) once $pW$ is constant.*

Go deeper · Derivation C · 1 minThe eigenfunction expansion and the origin of resonance

Derivation CThe eigenfunction expansion and the origin of resonance

The goal is to build $G$ a third way — not piecewise, but as a sum over the natural modes of $\mathcal{L}$ — and to show that this spectral form makes resonance a visible pole rather than a hidden accident. We work with the resolvent problem $(\mathcal{L}-\lambda)G=\delta$ , expand both $G$ and the delta in the orthonormal eigenfunctions $\phi_n$ , and match coefficients mode by mode. Completeness lets the delta be written as a sum, orthonormality lets each coefficient be read off independently, and the denominator $\lambda_n-\lambda$ that emerges is the whole story of driven response — it blows up exactly when the drive parameter hits an eigenvalue.

Step 1. Expand $G(x,\xi)=\sum_n c_n(\xi)\phi_n(x)$ and use completeness to write the source as $\delta(x-\xi)=\sum_n\phi_n(x)\phi_n(\xi)$ . Why: the eigenfunctions of a self-adjoint operator form a complete orthonormal basis, so any function of $x$ — including the delta and $G$ itself — can be resolved onto them.

Step 2. Apply $(\mathcal{L}-\lambda)$ term by term: since $\mathcal{L}\phi_n=\lambda_n\phi_n$ , this gives $\sum_n c_n(\xi)(\lambda_n-\lambda)\phi_n(x)=\sum_n\phi_n(x)\phi_n(\xi)$ . Why: each eigenfunction is left untouched in shape by $\mathcal{L}$ and merely scaled by $\lambda_n$ , so the operator turns into simple multiplication in this basis.

Step 3. Match coefficients of each $\phi_n(x)$ to get $c_n(\xi)=\phi_n(\xi)/(\lambda_n-\lambda)$ , hence $G(x,\xi;\lambda)=\sum_n\phi_n(x)\phi_n(\xi)/(\lambda_n-\lambda)$ . Why: orthonormality decouples the modes, so the coefficient of each is fixed on its own, producing the spectral representation of equation (6).

Step 4. Read off resonance: as $\lambda\to\lambda_m$ the $m$ -th term diverges unless $\phi_m(\xi)=0$ or the source is orthogonal to $\phi_m$ . Why: driving a self-adjoint system at one of its natural eigenvalues produces an unbounded response — the pole of the resolvent — which is the mathematical content of resonance and the seed of the Fredholm alternative when $\lambda=0$ is itself an eigenvalue. *The spectral form reduces to equation (5) when the mode sum is resummed for a specific operator, and the poles at $\lambda=\lambda_n$ match the resolvent structure of Sturm–Liouville theory.*

Limiting cases

Green's functions specialize and degenerate in instructive ways — each row below is a regime where the general machinery simplifies or breaks:

Case	Result	Meaning
Self-adjoint (Sturm–Liouville) operator	$G(x,\xi)=G(\xi,x)$	With $\mathcal{L}=(pu')'+q$ and self-adjoint boundary conditions, $G$ is symmetric and real — reciprocity holds and the spectral form (6) is available.
Zero eigenvalue / singular operator	$W(u_1,u_2)=0$	The homogeneous BVP has a nontrivial solution $\phi_0$ , so $u_1,u_2$ become proportional and (5) diverges; use the modified Green's function and impose $\int f\,\phi_0=0$ (Fredholm alternative).
Causal / initial-value (retarded) limit	$G(t,t')=0$ for $t<t'$	For dynamical problems the "right boundary" sits at $t=+\infty$ ; the retarded Green's function is the causal impulse response used for driven oscillators and linear response.
Constant-coefficient operator	$G(x,\xi)=G(x-\xi)$	Translation invariance makes $G$ depend only on $x-\xi$ (until boundaries break it), enabling Fourier construction $\tilde G(k)=1/\tilde{\mathcal{L}}(k)$ .

Common trapA Green's function always exists, so you can always write u = ∫Gf.

Not so. If the homogeneous boundary-value problem $\mathcal{L}u=0$ has a nontrivial solution $\phi_0$ satisfying the boundary conditions, then $\mathcal{L}$ is singular — the continuous analogue of a matrix with a zero eigenvalue and no inverse. Concretely, the boundary-adapted solutions $u_1,u_2$ become proportional, their Wronskian $W\to0$ , and the closed form (5) blows up. The equation $\mathcal{L}u=f$ is then solvable only if the source is orthogonal to the zero mode, $\int_a^b f\,\phi_0\,\mathrm{d}\xi=0$ — the Fredholm alternative — and even then the solution is unique only up to multiples of $\phi_0$ . The repair is the modified Green's function, which inverts $\mathcal{L}$ on the subspace orthogonal to $\phi_0$ and satisfies $\mathcal{L}_xG_{\text{mod}}=\delta(x-\xi)-\phi_0(x)\phi_0(\xi)$ . A second, quieter trap: $G$ inherits the homogeneous versions of the original boundary conditions, even when the original problem had inhomogeneous data — the data are carried by separate boundary terms, not by $G$ .

Worked examples

1 · The clamped string under uniform load (fully worked)

Solve $-u''=f$ on $[0,1]$ with $u(0)=u(1)=0$ for the uniform load $f(x)=1$ , using the Green's function, and check the answer against direct integration.

Step 1. Build the kernel. Here $p=1$ , $q=0$ , so the homogeneous solutions are linear; the boundary-adapted choices are $u_1(x)=x$ (vanishes at $0$ ) and $u_2(x)=1-x$ (vanishes at $1$ ). Their Wronskian is $W=u_1u_2'-u_1'u_2=x(-1)-(1)(1-x)=-1$ , constant as Abel's theorem requires. Because $\mathcal{L}=-\mathrm{d}^2/\mathrm{d}x^2$ carries an overall minus, the formula gives $G=-u_1(x_<)u_2(x_>)/(pW)=-x_<(1-x_>)/(-1)$ , the positive triangular kernel $G(x,\xi)=x_<(1-x_>)$ .

Step 2. Sanity-check the kernel. At the source $G(\xi,\xi)=\xi(1-\xi)$ — continuous, and equal to $0.25$ at $\xi=0.5$ , matching the first figure. The slope just left of $\xi$ is $\partial_xG=1-\xi$ and just right is $-\xi$ , so the jump is $-\xi-(1-\xi)=\mathbf{-1}=-1/p$ , exactly the unit impulse for $-G''=\delta$ . Symmetry $G(x,\xi)=G(\xi,x)$ is manifest.

Step 3. Integrate against the load. With $f=1$ ,

u(x)=\int_0^1 G(x,\xi)\,\mathrm{d}\xi=(1-x)\!\int_0^x\xi\,\mathrm{d}\xi+x\!\int_x^1(1-\xi)\,\mathrm{d}\xi.

The first integral is $x^2/2$ . The second is $\big[\xi-\xi^2/2\big]_x^1=\tfrac12-(x-x^2/2)$ .

Step 4. Collect the pieces:

u(x)=(1-x)\frac{x^2}{2}+x\!\left(\frac12-x+\frac{x^2}{2}\right)=\frac{x^2}{2}-\frac{x^2}{2}+\frac{x}{2}=\frac{x(1-x)}{2}.

So $u(x)=\tfrac12 x(1-x)$ , a parabola pinned at both ends with maximum sag $u(1/2)=\mathbf{1/8=0.125}$ at the centre.

Step 5. Verify directly. Differentiate: $u'=\tfrac12-x$ , $u''=-1$ , so $-u''=\mathbf{1}=f$ , and $u(0)=u(1)=0$ . The Green's-function route and the two-line direct integration agree exactly, and the same $u=\int Gf$ now delivers the sag for any load $f$ without redoing the calculus.

2 · The driven oscillator via its retarded Green's function (fully worked)

A mass $m=1\ \text{kg}$ on a spring has natural frequency $\omega_0=2\ \text{rad/s}$ (so $k=m\omega_0^2=4\ \text{N/m}$ ). Starting from rest at equilibrium, it is driven by a constant force $f_0=8\ \text{N}$ switched on at $t=0$ . Find the motion using $G(t,t')=\omega_0^{-1}\sin\omega_0(t-t')\,\theta(t-t')$ .

Step 1. Confirm the kernel. For $\ddot x+\omega_0^2x=f/m$ the retarded Green's function must satisfy $(\partial_t^2+\omega_0^2)G=\delta(t-t')$ with $G=0$ for $t<t'$ . Take $G=\omega_0^{-1}\sin\omega_0(t-t')$ for $t>t'$ : it vanishes at $t=t'$ (continuity) and has $\partial_tG\to\cos\omega_0(t-t')\to\mathbf{1}$ as $t\to t'^+$ , the unit slope jump for a unit-coefficient second derivative. Amplitude $1/\omega_0=0.5\ \text{s}$ , period $2\pi/\omega_0=\pi\approx3.14\ \text{s}$ — the second figure.

Step 2. Superpose. The causal solution is $x(t)=\int_{-\infty}^t G(t,t')\,\frac{f(t')}{m}\,\mathrm{d}t'$ . With $f(t')/m=f_0/m$ constant for $t'>0$ ,

x(t)=\frac{f_0}{m}\int_0^t\frac{1}{\omega_0}\sin\omega_0(t-t')\,\mathrm{d}t'.

Step 3. Do the integral. Substitute $s=t-t'$ ( $\mathrm{d}t'=-\mathrm{d}s$ , limits $t\to0$ ): $\int_0^t\sin\omega_0 s\,\mathrm{d}s=[1-\cos\omega_0 t]/\omega_0$ . Hence

x(t)=\frac{f_0}{m\omega_0^2}\big(1-\cos\omega_0 t\big)=\frac{f_0}{k}\big(1-\cos\omega_0 t\big).

Step 4. Put in numbers. $f_0/k=8/4=\mathbf{2\ \text{m}}$ , so $x(t)=2\,(1-\cos 2t)$ : the mass oscillates about its static displacement $2\ \text{m}$ with amplitude $2\ \text{m}$ , swinging between $\mathbf{0}$ and $\mathbf{4\ \text{m}}$ , and reaching the far point $4\ \text{m}$ first at $2t=\pi$ , i.e. $t=\pi/2\approx1.57\ \text{s}$ .

Step 5. Verify. $\dot x=(f_0/m\omega_0)\sin\omega_0 t$ , $\ddot x=(f_0/m)\cos\omega_0 t$ , so $\ddot x+\omega_0^2x=(f_0/m)\cos\omega_0 t+(f_0/m)(1-\cos\omega_0 t)=\mathbf{f_0/m}$ , and $x(0)=\dot x(0)=0$ . The impulse-response kernel has turned "switch on a steady force" into a one-line integral — the static offset $f_0/k$ plus a cosine transient, the textbook signature of a suddenly loaded oscillator.

Physics of the portrait

The Green's function is named for a self-taught English miller whose 1828 pamphlet introduced both the object and the word "potential" — and then sank so completely that the mathematics had to be rediscovered before it could organize an entire field.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · self-taught miller who invented the Green's function in an unread 1828 essay

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 analysis from borrowed 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 word "potential," proved the reciprocal identities that bear his name, and — the piece relevant here — constructed what he called simply "a function" for solving boundary-value problems by superposing the responses to point sources. That object is the Green's function: the kernel of $\mathcal{L}^{-1}$ , built to satisfy the boundary conditions and to reduce any inhomogeneous problem to an integral. The Essay drew only fifty-one subscribers and vanished. Green entered Cambridge as an undergraduate in 1833 at the age of forty, graduated fourth in his year, and died in 1841. It was William Thomson (Lord Kelvin) who in 1845 came upon the forgotten Essay, saw its depth, and had it reprinted — after which "Green's function" passed into the working vocabulary of electromagnetism, quantum theory, and mathematical physics. The kernel that now carries perturbation theory, scattering amplitudes, and lattice sums began as a miller's private idea for inverting the Laplacian between grounded conductors.

History

1828

— Green publishes his Essay, introducing the potential and the Green's function

1832

— Liouville and Sturm develop the eigenfunction theory that will give G its spectral form

1833

— Green enters Cambridge as an undergraduate at age forty

1845

— Thomson rediscovers and reprints Green's forgotten Essay

1894

— Heaviside and contemporaries spread operational/impulse-response methods for ODEs

1948

— Feynman and Schwinger place Green's functions at the heart of quantum field theory

Going further

Texts. Stone & Goldbart, Mathematics for Physics (CUP, 2009), Ch. 5 — the cleanest operator-theoretic treatment · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 10 · Hassani, Mathematical Physics, 2nd ed., Ch. 20–22 (Green's functions in one and more dimensions) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 7 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §15.2.5 · Morse & Feshbach, Methods of Theoretical Physics, Vol. I, Ch. 7 · Courant & Hilbert, Methods of Mathematical Physics, Vol. I, Ch. V. The key results to keep in mind: the construction (5) follows from continuity plus the jump condition (4), with system determinant $pW$ . Worked Example 1 gives $G=x_<(1-x_>)$ , slope jump $-1$ , and $u=\tfrac12x(1-x)$ with peak $0.125$ under $f=1$ . Worked Example 2 gives $G=\omega_0^{-1}\sin\omega_0(t-t')\theta(t-t')$ satisfying $(\partial_t^2+\omega_0^2)G=\delta$ , with a step force yielding $x=(f_0/k)(1-\cos\omega_0 t)$ , i.e. $2(1-\cos 2t)$ for the stated numbers. All quoted numbers — slope jump $=1/p$ , Wronskian $=-1$ , peak sag $=1/8$ , static offset $f_0/k=2\ \text{m}$ — follow directly from the constructions above.

Prerequisites. Linear Second-Order ODEs · The Wronskian and Linear Independence · Variation of Parameters · Two-Point Boundary-Value Problems · Existence and Uniqueness Theorems for ODEs. Leads to. Sturm–Liouville Theory and Eigenfunction Expansions · Green's Functions for Partial Differential Equations · Fredholm Alternative and Integral Equations.

Ask the tutor: Ask me to build the Green's function for any operator and boundary conditions you like — a beam, a ring, a driven RLC circuit — and we will construct it, check the jump and symmetry, and solve it for a source of your choosing.

Why must G be continuous at the source but have a kink in its slope?

It is a matching of singularities. The right-hand side of $\mathcal{L}G=\delta$ is a single delta function — the mildest genuine singularity. If $G$ itself jumped at $\xi$ , then $G'$ would contain a delta and $G''$ a derivative of a delta, which is far too singular to equal a plain $\delta$ . If instead $G$ is continuous but its slope jumps, then $G''$ contains exactly one delta and nothing worse — a perfect match. So continuity is forced from above (no worse singularity allowed) and the slope jump is forced from below (some singularity is needed to produce the delta at all). Integrating the equation across the source pins the jump magnitude to $1/p(\xi)$ .

How is the Green's-function solution related to variation of parameters?

They are the same particular solution written two ways. Variation of parameters produces $u_p=-u_1\int^x\! u_2 f/(pW)\,\mathrm{d}\xi+u_2\int_x\! u_1 f/(pW)\,\mathrm{d}\xi$ , with the two integrals split at the observation point $x$ . Collecting those split limits into a single integral with a kernel that changes form at $\xi=x$ gives exactly $u(x)=\int G(x,\xi)f(\xi)\,\mathrm{d}\xi$ with $G=u_1(x_<)u_2(x_>)/(pW)$ . The Green's-function viewpoint simply names that kernel and computes it once, so that a new source needs one integral rather than a fresh application of the variation-of-parameters bookkeeping.

What exactly is "self-adjoint" and why does it matter so much here?

An operator is formally self-adjoint if $\int(v\,\mathcal{L}u-u\,\mathcal{L}v)\,\mathrm{d}x$ reduces to boundary terms alone; the Sturm–Liouville form $(pu')'+q$ achieves this automatically, and boundary conditions are "self-adjoint" when they make those boundary terms vanish. This matters for three reasons: it makes $G$ symmetric (reciprocity); it guarantees real eigenvalues and a complete orthonormal set of eigenfunctions, which the spectral expansion (6) needs; and it makes the theory the continuous mirror of Hermitian matrices, so finite-dimensional linear-algebra intuition transfers intact. Almost every operator of physics — the Laplacian, the Schrödinger operator, the string and beam operators — is self-adjoint under physical boundary conditions, which is why the beautiful case is also the usual one.

Does a Green's function always exist?

No, and the exception is important. If the homogeneous boundary-value problem $\mathcal{L}u=0$ has a nontrivial solution $\phi_0$ that satisfies the boundary conditions, then $\mathcal{L}$ has a zero eigenvalue and is not invertible — just like a singular matrix. The boundary-adapted solutions become proportional, their Wronskian vanishes, and the closed form (5) diverges. The inhomogeneous problem $\mathcal{L}u=f$ then has a solution only when $f$ is orthogonal to the zero mode, $\int f\,\phi_0=0$ (the Fredholm alternative), and the fix is the modified Green's function, which inverts $\mathcal{L}$ on the subspace orthogonal to $\phi_0$ . This is not an exotic edge case: it is exactly the situation of a free (Neumann) rod or a periodic ring, where the constant function is a zero mode.

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

Both solve $\mathcal{L}G=\delta$ ; they differ only in which "boundary" data are imposed. The boundary-value kernel lives on a finite interval $[a,b]$ and is fixed by conditions at both ends — it is symmetric and typically a tent-like shape, as in the clamped string. The causal kernel lives on the time axis and is fixed by conditions at one end only: $G=0$ before the source, because effects cannot precede their cause, with the "other boundary" pushed to $t=+\infty$ . This retarded Green's function is generally not symmetric in $t\leftrightarrow t'$ — causality breaks the reciprocity that self-adjoint spatial boundary conditions would enforce — and it is the object convolved with a driving force to give the response of an oscillator, a circuit, or any linear time-invariant system.

How does the eigenfunction expansion connect Green's functions to resonance?

Write the resolvent Green's function as $G(x,\xi;\lambda)=\sum_n\phi_n(x)\phi_n(\xi)/(\lambda_n-\lambda)$ . Each term is a simple pole in the drive parameter $\lambda$ , located at an eigenvalue $\lambda_n$ . As you tune $\lambda$ toward $\lambda_n$ that term's denominator shrinks to zero and the response diverges — this is resonance, the unbounded reaction of a system driven at one of its natural frequencies. The residue $\phi_n(x)\phi_n(\xi)$ tells you the shape of the resonant response: it is largest where the eigenmode $\phi_n$ is largest. The only escape from the divergence is orthogonality — if the source has no overlap with $\phi_n$ (so $\int f\phi_n=0$ ) that mode is never excited, which is precisely the solvability condition that reappears as the Fredholm alternative when $\lambda_n=0$ .