Technopedia

The technologist’s encyclopedia

Distributions

Distributions (generalized functions) extend the notion of a function so that differentiation always exists. They let us write down objects like the Dirac delta and solve differential equations — via weak derivatives and fundamental solutions — in situations where classical derivatives do not exist. This page covers test functions, the definition of a distribution, the Dirac delta, differentiation (weak derivatives), and fundamental solutions.

Authoritative sources

Test functions

Definition. A test function is a smooth function with compact support, i.e. an element of D(Rn)=Cc∞(Rn)\mathcal{D}(\mathbb{R}^n) = C_c^\infty(\mathbb{R}^n).

Details

Why smooth and compactly supported

A test function is the “probe” we use to sample a distribution. We want two things: (1) it must be differentiable to all orders, so that we can define derivatives of distributions by pushing derivatives onto the test function; and (2) it must vanish outside some bounded region, so that integrals against it are always finite and no boundary terms appear. Smooth + compact support gives exactly both.

Example. The function φ(x)=e−1/(1−x2)\varphi(x) = e^{-1/(1-x^2)} for ∣x∣<1|x| < 1 and φ(x)=0\varphi(x) = 0 for ∣x∣≥1|x| \geq 1 is a standard test function on R\mathbb{R}: it is smooth everywhere (all derivatives vanish at ±1\pm 1) and supported in [−1,1][-1, 1].

Convergence of test functions

A sequence φk→0\varphi_k \to 0 in D\mathcal{D} if all the φk\varphi_k are supported in one fixed compact set and every derivative ∂αφk→0\partial^\alpha \varphi_k \to 0 uniformly. This topology is what “continuity” of a distribution means.

Example problem

Problem. Show that the function above, φ(x)=e−1/(1−x2)\varphi(x) = e^{-1/(1-x^2)} on (−1,1)(-1, 1), is smooth at x=1x = 1.

Solution

For x<1x < 1, φ\varphi is a composition of smooth functions, hence smooth. At the boundary point 11, every derivative of e−1/(1−x2)e^{-1/(1-x^2)} tends to 00 as x→1−x \to 1^- (the exponential decay beats any polynomial growth from the chain rule), matching the value 00 on the other side. So all one-sided derivatives agree and φ\varphi is smooth at the boundary. ∎

Distributions as continuous functionals

Definition. A distribution on Rn\mathbb{R}^n is a continuous linear functional T:D(Rn)→CT : \mathcal{D}(\mathbb{R}^n) \to \mathbb{C}. The space of all distributions is D′(Rn)\mathcal{D}'(\mathbb{R}^n). Every locally integrable function ff defines one by Tf(φ)=∫Rnf(x) φ(x) dxT_f(\varphi) = \int_{\mathbb{R}^n} f(x)\,\varphi(x)\,dx.

Details

Functions are distributions

The first thing to notice is that every “ordinary” (locally integrable) function is a distribution: it acts on a test function by integration. Distributions are therefore a generalization of functions, not a replacement — the classical functions sit inside D′\mathcal{D}' as a special case.

Example. The constant function f=1f = 1 gives the distribution Tf(φ)=∫φT_f(\varphi) = \int \varphi, which just integrates the test function.

Continuity is automatic for functions, not for all distributions

For a locally integrable ff, the map φ↦∫fφ\varphi \mapsto \int f\varphi is automatically continuous in the test-function topology. But a general distribution need not be given by any function — it can be a “more singular” object. The order of a distribution measures how many derivatives of φ\varphi it depends on: a function has order 00, the Dirac delta has order 00, and δ′\delta' has order 11.

Example problem

Problem. Show that the locally integrable function f(x)=1/∣x∣f(x) = 1/|x| on R\mathbb{R} defines a distribution.

Solution

ff is locally integrable because near 00, ∫−11∣x∣−1 ∣φ(x)∣ dx≤∥φ∥∞∫−11∣x∣−1 dx<∞\int_{-1}^{1} |x|^{-1}\,|\varphi(x)|\,dx \leq \|\varphi\|_\infty \int_{-1}^{1} |x|^{-1}\,dx < \infty (the integral of 1/∣x∣1/|x| over a neighborhood of 00 is finite in one dimension). Since φ\varphi has compact support, the integral ∫fφ\int f\varphi is finite, and the map φ↦∫fφ\varphi \mapsto \int f\varphi is continuous in the test-function topology. Hence ff defines a distribution. ∎

The Dirac delta

Definition. The Dirac delta δ\delta is the distribution defined by δ(φ)=φ(0)\delta(\varphi) = \varphi(0) for every test function φ\varphi.

Details

A point mass, not a function

The delta is “infinite at 00 and zero elsewhere” in the informal picture, but it is not a function: no function satisfies ∫fφ=φ(0)\int f\varphi = \varphi(0) for all test functions φ\varphi. It is a distribution of order 00 that “picks out” the value of φ\varphi at the origin. Physically it models a point charge or an instantaneous impulse.

Properties.

0 narrow bumps, area 1, converge to delta
Example problem

Problem. Show that no (locally integrable) function ff satisfies ∫f(x)φ(x) dx=φ(0)\int f(x)\varphi(x)\,dx = \varphi(0) for every test function φ\varphi.

Solution

Suppose such ff existed. Take a nonnegative test function φ≥0\varphi \geq 0 with φ(0)=0\varphi(0) = 0 (e.g. one supported away from 00). Then ∫fφ=φ(0)=0\int f\varphi = \varphi(0) = 0 for all such φ\varphi. Now take ψ≥0\psi \geq 0 with ψ(0)>0\psi(0) > 0. Writing ψ=ψ(0)⋅1+(ψ−ψ(0))\psi = \psi(0) \cdot 1 + (\psi - \psi(0)) and using that ψ−ψ(0)\psi - \psi(0) vanishes at 00, one gets ∫fψ=ψ(0)\int f\psi = \psi(0). But choosing ψ\psi with arbitrarily small support around 00 forces ff to concentrate all its mass at a single point, contradicting local integrability (a locally integrable function has ∫∣x∣<ϵ∣f∣→0\int_{|x|<\epsilon} |f| \to 0 as ϵ→0\epsilon \to 0, whereas the delta requires the integral to stay =φ(0)≠0= \varphi(0) \neq 0). Hence no such function exists. ∎

Differentiation of distributions

Definition. The derivative of a distribution TT is the distribution T′T' defined by T′(φ)=−T(φ′)T'(\varphi) = -T(\varphi'). The weak derivative of a locally integrable function ff is a locally integrable function gg satisfying ∫g φ=−∫f φ′\int g\,\varphi = -\int f\,\varphi' for all test functions φ\varphi.

Details

Where the minus sign comes from

The definition T′(φ)=−T(φ′)T'(\varphi) = -T(\varphi') is just integration by parts in disguise. If T=TfT = T_f comes from a smooth function ff, then

Tf′(φ)=−∫f φ′=∫f′ φ=Tf′(φ),T_f'(\varphi) = -\int f\,\varphi' = \int f'\,\varphi = T_{f'}(\varphi),

where the boundary term vanishes because φ\varphi has compact support. So the distributional derivative of a smooth function is its ordinary derivative — the definition is engineered to agree with the classical one.

Every distribution is infinitely differentiable

This is the payoff. Because the definition only ever differentiates the test function (which is smooth), it works for any distribution, no matter how singular. Distributions can be differentiated any number of times, and the result is again a distribution. This is impossible for ordinary functions.

Weak derivatives catch jumps

A function with a jump has no classical derivative at the jump, but it has a weak derivative that includes a delta. The Heaviside step function H(x)=1H(x) = 1 for x>0x > 0 and 00 for x<0x < 0 has weak derivative δ\delta: indeed ∫Hφ′=−φ(0)=−δ(φ)\int H\varphi' = -\varphi(0) = -\delta(\varphi), so H′=δH' = \delta.

H(x) jump at 0 -> H' = delta
Example problem

Problem. Compute the distributional second derivative of the Heaviside function HH.

Solution

We have H′=δH' = \delta. Differentiating again, H′′=δ′H'' = \delta', where δ′(φ)=−φ′(0)\delta'(\varphi) = -\varphi'(0). So the second distributional derivative of HH is δ′\delta'. (Classically, HH has no second derivative at all; distributionally it is perfectly well defined.) ∎

Fundamental solutions

Definition. A fundamental solution of a linear differential operator LL is a distribution EE satisfying LE=δL E = \delta.

Details

Why fundamental solutions solve equations

If EE is a fundamental solution of LL, then for any (nice) source ff, the convolution u=E∗fu = E * f solves Lu=fL u = f. Indeed, L(E∗f)=(LE)∗f=δ∗f=fL(E * f) = (L E) * f = \delta * f = f. So finding one fundamental solution reduces the whole class of inhomogeneous equations to a single convolution.

Example. For the Laplacian Δ\Delta on Rn\mathbb{R}^n (n≥3n \geq 3), a fundamental solution is

E(x)=1n(2−n) ωn 1∣x∣n−2,E(x) = \frac{1}{n(2-n)\,\omega_n}\, \frac{1}{|x|^{n-2}},

where ωn\omega_n is the volume of the unit ball. In two dimensions the fundamental solution is E(x)=12πln⁡∣x∣E(x) = \frac{1}{2\pi}\ln|x|. These are the Green’s functions behind the Newtonian potential and the solution of Poisson’s equation.

Fundamental solutions tie distributions back to Operators and Applications: the operator LL is (usually) an unbounded operator on an L2L^2 space, and the existence of EE is what lets one prove existence of weak solutions to PDEs.

Example problem

Problem. Verify that E(x)=12πln⁡∣x∣E(x) = \frac{1}{2\pi}\ln|x| on R2\mathbb{R}^2 is a fundamental solution of the Laplacian, i.e. ΔE=δ\Delta E = \delta.

Solution

For x≠0x \neq 0, EE is smooth and Δln⁡∣x∣=0\Delta \ln|x| = 0, so ΔE=0\Delta E = 0 away from the origin. To find the contribution at 00, integrate ΔE\Delta E against a test function φ\varphi with φ(0)=1\varphi(0) = 1 near 00: by Green’s identity over the disk BRB_R minus a small disk BϵB_\epsilon around 00,

∫BR∖BϵΔE φ=∫∂Bϵ∂E∂n dS+o(1).\int_{B_R \setminus B_\epsilon} \Delta E\,\varphi = \int_{\partial B_\epsilon} \frac{\partial E}{\partial n}\,dS + o(1).

Since ∂E/∂n=12π1ϵ\partial E/\partial n = \frac{1}{2\pi}\frac{1}{\epsilon} on ∂Bϵ\partial B_\epsilon and ∣∂Bϵ∣=2πϵ|\partial B_\epsilon| = 2\pi\epsilon, the boundary integral is 12π⋅1ϵ⋅2πϵ=1\frac{1}{2\pi}\cdot\frac{1}{\epsilon}\cdot 2\pi\epsilon = 1. Letting ϵ→0\epsilon \to 0 gives ⟨ΔE,φ⟩=φ(0)=δ(φ)\langle \Delta E, \varphi \rangle = \varphi(0) = \delta(\varphi). Hence ΔE=δ\Delta E = \delta. ∎

Quiz

Example problem

Problem. Let ff be a piecewise C1C^1 function on R\mathbb{R} with a single jump of size J=f(0+)−f(0−)J = f(0^+) - f(0^-) at x=0x = 0. Show that the distributional derivative f′f' equals the classical derivative (as a function) plus J δJ\,\delta.

Solution

Write f=freg+J Hf = f_{\text{reg}} + J\,H, where fregf_{\text{reg}} is ff with the jump removed (so fregf_{\text{reg}} is continuous and piecewise C1C^1) and HH is the Heaviside step. The distributional derivative is linear, so f′=freg′+J H′f' = f_{\text{reg}}' + J\,H'. The function fregf_{\text{reg}} has no jump, so its distributional derivative equals its classical (piecewise) derivative as a function. And H′=δH' = \delta. Therefore f′=fclassical′+J δf' = f'_{\text{classical}} + J\,\delta. ∎