advanced28 min studyLesson 24 in Mathematics

Distributions and Green functions

Replace pointwise singularities with meaningful actions on test functions and construct responses to localized sources.

Before you begin
On this page

What you’ll understand

  • Define the Dirac delta by its action rather than an infinite point value.
  • Differentiate a nonsmooth object using duality and integration by parts.
  • Explain why boundary or causal conditions belong to the definition of a Green function.

An ideal point charge and an instantaneous impulse are useful models, but their mathematical objects are not ordinary functions with extremely large values. Distribution theory describes them by how they act when averaged against smooth probes. This makes derivatives of discontinuous functions precise and supports the propagators used in field theory. It also sets limits: distributions cannot generally be multiplied with one another, and solving a differential equation still requires a choice of boundary or causal behavior.

An object defined by what it does to a probe

A test function is smooth and compactly supported. A distribution is a continuous linear functional on the space of these probes, with continuity understood in the test-function topology. Every locally integrable function defines a distribution by integration against a probe. The delta at a instead returns the probe's value at a. Writing δ(x−a) is convenient notation for this functional; it does not create a finite-valued function that vanishes everywhere except at one point.

A family of narrow, normalized smooth bumps can approach delta distributionally: every fixed smooth probe sees its weighted average approach the probe's value at the center. The bumps need not converge to an ordinary integrable function with area one. This distinction resolves the apparent paradox of an infinitely narrow source. In quantum field theory even a field may be operator-valued distributional data, so smearing against probes is mathematically more basic than assigning an operator independently at every spacetime point.

⟨δa,φ⟩=φ(a),⟨Tf,φ⟩=∫Rf(x)φ(x) dx\langle\delta_a,\varphi\rangle=\varphi(a),\qquad \langle T_f,\varphi\rangle=\int_{\mathbb R}f(x)\varphi(x)\,dx
φ is smooth with compact support, and f is locally integrable. Delta is the evaluation functional, not an ordinary function with an infinite value.↔ Scroll wide equations horizontally.

Move derivatives onto the smooth side

Integration by parts suggests the definition of a distributional derivative: apply the original distribution to the derivative of the test function and insert a minus sign. Compact support removes boundary terms. This extends ordinary differentiation and permits derivatives of jumps. The Heaviside step has derivative delta because the integral of the probe's derivative over the positive half-line is minus its value at zero. The arbitrary assigned value of the step at the single point zero is irrelevant to its associated distribution.

A distribution can be multiplied by a smooth function because the smooth function times a test function is still a valid test function. A product of two arbitrary distributions has no such general definition. Expressions such as δ² require extra structure or a specified regularization and limiting procedure, and different procedures may disagree. Renormalization is not simply permission to treat singular objects as ordinary numbers; it supplies controlled prescriptions and consistency conditions for the products arising in particular theories.

⟨T′,φ⟩=−⟨T,φ′⟩,H′=δ0,⟨δ0′,φ⟩=−φ′(0)\langle T',\varphi\rangle=-\langle T,\varphi'\rangle,\qquad H'=\delta_0,\qquad \langle\delta_0',\varphi\rangle=-\varphi'(0)
All equalities are distributional on the real line. The minus sign follows from integration by parts; it is essential for consistent source equations.↔ Scroll wide equations horizontally.

A Green function is an inverse with conditions attached

For a linear differential operator L, a fundamental solution satisfies LG=δ in a translation-invariant full-space setting. Convolving G with a suitable source then constructs a solution, when the convolution exists. On a bounded domain, a Green kernel generally depends separately on the observation and source positions and must satisfy the selected boundary conditions. Adding any homogeneous solution leaves the source equation unchanged, which is why the equation LG=δ alone rarely selects a unique physical response.

For time-dependent problems, retarded and advanced Green functions answer different questions. The retarded solution has no response before the source acts. A Feynman propagator uses a time-ordering prescription appropriate to quantum correlation functions; it is not identical to a classical retarded response. In frequency space these distinctions often appear as pole prescriptions. They carry physical information that is lost if one merely writes an algebraic reciprocal of the differential operator's symbol.

d2dx2(∣x∣2)=δ0,u(x)=∫R∣x−y∣2f(y) dy\frac{d^2}{dx^2}\left(\frac{|x|}{2}\right)=\delta_0,\qquad u(x)=\int_{\mathbb R}\frac{|x-y|}{2}f(y)\,dy
The first equality holds distributionally. For smooth compactly supported f, the displayed convolution solves u''=f on the line; it imposes no decay or finite-interval boundary condition.↔ Scroll wide equations horizontally.

Why a kink represents a point source

Verify that G(x)=∣x∣/2G(x)=|x|/2 is a fundamental solution of the one-dimensional second derivative.

  1. Away from zero, G is linear: its derivative is −1/2 on the left and +1/2 on the right. Its ordinary second derivative is zero there, so any source must be concentrated at the kink.

  2. For a compactly supported smooth φ, evaluate the distributional second derivative as ⟨G′′,φ⟩=∫Gφ′′ dx\langle G'',\varphi\rangle=\int G\varphi''\,dx. Split the integral into negative and positive half-lines.

  3. Integrate by parts on each half-line. Outer boundary terms vanish. The jump in G' at zero is (+1/2)−(−1/2)=1, so the surviving term is φ(0).

  4. Since the result equals φ(0) for every probe, G''=δ. Adding ax+b produces another fundamental solution because its second derivative vanishes; extra conditions select among these choices.

The result

The source strength equals the jump in the first derivative. Pointwise differentiation away from the kink would miss the entire source.

Explore the model

Interactive lab

Interactive lab · fields

A source and its surrounding response

Change source strength q and rotation Ω in a softened two-dimensional field q r/(r²+ε)+Ω(−y,x), with fixed positive ε. Softening keeps the drawn source finite and rotation adds a separate circulating field. This is a source-response analogy, not the one-dimensional Green function derived here or a delta, retarded, or Feynman solver.

A source and its surrounding responseThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA singular source is understood through integrated response and boundary prescriptions, not a giant displayed pixel.

Check your understanding

What is the action of δ′ on a smooth compactly supported test function φ?

Try it yourself

Let v(x)=∣x−a∣v(x)=|x-a|. Find v'' as a distribution and explain why the source coefficient is two.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare the one-sided first derivatives at a.

Compare with the solution

The first derivative jumps from −1 to +1 at a, a jump of two. Testing its derivative against φ and integrating on the two sides gives ⟨v′′,φ⟩=2φ(a)\langle v'',\varphi\rangle=2\varphi(a), hence v′′=2δav''=2\delta_a. The value assigned to v' at a single point does not change the distribution.

Take it with you

  • A distribution is a continuous linear action on smooth probes.
  • Weak differentiation captures jumps while retaining integration-by-parts identities.
  • An inverse source equation needs boundary or causal conditions, and singular products need justification.

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