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One-dimensional Dirichlet Green kernel on an interval

Statement

Assume the Axiom of Countable Choice. For a<b and x,y∈(a,b), the one-dimensional Dirichlet kernel for −d2/dx2 is G(x,y)=((min⁡{x,y}−a)(b−max⁡{x,y}))/(b−a). It is symmetric, nonnegative, vanishes at a,b, and its x-derivative has jump ∂xG(y+,y)−∂xG(y−,y)=−1, so −∂x2G(⋅,y)=δy in D′(a,b). Its difference from −∣x−y∣/2 is affine in x. The stated Countable Choice assumption is used for the named Lebesgue-measure and regular-distribution interfaces below; no full Axiom of Choice is used.

Facts & Assumptions

Given: Assume ACω, let a<b, fix y∈(a,b), and put L=b−a>0. The differential operator is −d2/dx2.

[A1]

Countable Choice is written ACω (The Axiom of Countable Choice (ACω)). It enters through the Borel-to-Lebesgue, finite-interval measure, Riemann-to-Lebesgue, and published locally-integrable-to-distribution interfaces used in steps 2.2, 3.2 and 6.1. The piecewise slope and integration-by-parts calculations themselves make no choice.

[F1]

For two real numbers the minimum and maximum select the lesser and greater values (Maximum and minimum of a set).

[F2]

The one-dimensional normalized kernel is Φ1(x)=−∣x∣/2; the assigned Laplace definition separately verifies −Φ1′′=δ0 (Fundamental solution for the positive operator minus Laplacian). A fundamental solution of L is a distribution E with LE=δ0, and its translated distribution has source δy (Fundamental solution of a constant-coefficient operator).

[F3]

A test function on (a,b) has compact support in that open interval, so its zero extension is smooth on R and it vanishes near both endpoints (Test function space d of an open set).

[F4]

The distributional second derivative satisfies ⟨∂x2T,φ⟩=⟨T,φ′′⟩ (Distributional derivative).

[F5]

The Dirac distribution satisfies δy(φ)=φ(y) (Dirac delta and its derivatives).

[F6]

If u,v are differentiable on a closed interval and their derivatives are integrable, integration by parts gives ∫uv′=[uv]−∫u′v (If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v). The statement is real-valued; apply it to the real and imaginary parts of a complex test separately.

[F7]

A continuous real function on a closed bounded interval is Riemann integrable (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F9]

Under ACω, every bounded Riemann-integrable function on a closed bounded interval is Lebesgue integrable there with the same integral (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[F10]

A continuous map has Borel preimages of Borel sets, and under ACω Borel functions on R are Lebesgue measurable (A continuous map has Borel preimages of Borel sets, Borel measurable and Lebesgue measurable functions on Rn).

[F11]

Under ACω, the interval [a,b] is measurable and λ1([a,b])=b−a (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included). The nonnegative integral is monotone and positively homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F12]

A measurable complex function is integrable when its absolute value has finite integral (Integrable real and complex functions, and their integrals); local integrability is defined by finite absolute integrals on balls or, on an open set, on each compact subset (A locally integrable function on Rn, Regular distribution from a locally integrable function).

[F13]

A locally integrable function defines the regular functional ⟨uf,φ⟩=∫fφ (Regular distribution from a locally integrable function).

[F14]

Under ACω, this regular functional is a distribution (Locally integrable functions embed in distributions).

Proof

technique · direct
1.1givenF1algebra

Let Gy(x)=G(x,y) on (a,b) and extend it by zero off (a,b), including the endpoint values. By [F1], its branches on [a,y] and [y,b] are (x−a)(b−y)/L and (y−a)(b−x)/L. They agree at x=y, both giving (y−a)(b−y)/L, and vanish at a,b, so the extension is continuous and compactly supported. Swapping x,y leaves the min/max formula unchanged. Both factors are nonnegative on either branch, and their sum is at most L, so their product is at most L2/4; hence Gy is symmetric, nonnegative, bounded by L/4, and Gy(y)>0.

2.1step 1.1algebra

On x<y and x>y, respectively, the affine branches from step 1.1 have one-sided derivatives s−:=∂xG(y−,y)=(b−y)/L and s+:=∂xG(y+,y)=−(y−a)/L. Therefore s+−s−=−((y−a)+(b−y))/L=−1.

2.2A1F10F11F12step 1.1algebra

The continuous compactly supported extension Gy is Borel by [F10] and hence Lebesgue measurable under [A1]. It is supported in [a,b] and bounded by L/4 by step 1.1. By [F11], ∫R∣Gy∣ dλ1≤(L/4)λ1([a,b])=L2/4<∞. Thus it is integrable and its restriction is locally integrable on (a,b) by [F12] and monotonicity of the nonnegative integral. This is the exact measure-side use of ACω.

3.1F2step 1.1step 2.1algebra

By [F2], the one-dimensional free-space fundamental profile translated to y is −∣x−y∣/2. For x<y, Gy(x)+∣x−y∣/2 is affine with slope s−−1/2; for x>y it is affine with slope s++1/2. Step 2.1 makes those slopes equal, and step 1.1 gives continuity at y, so the two pieces form one affine function on (a,b).

3.2A1F13F14step 2.2

The locally integrable function Gy defines the regular functional Ty(φ)=∫(a,b)Gy(x)φ(x) dx by [F13], and [F14] makes Ty a distribution under ACω. No full Axiom of Choice is used.

4.1F3F6F7step 3.2step 2.1

Fix a real-valued φ∈D(a,b). By [F3], φ vanishes near a. On [a,y], apply [F6] first with u=Gy,v=φ′ and then with u=s−,v=φ; the functions and derivatives involved are continuous and integrable by [F7]. This gives ∫ayGyφ′′=[Gyφ′]ay−s−[φ]ay=Gy(y)φ′(y)−s−φ(y).

5.1F3F6F7step 4.1step 2.1

Since φ also vanishes near b, the same two applications of [F6] on [y,b], with slope s+, give ∫ybGyφ′′=[Gyφ′]yb−s+[φ]yb=−Gy(y)φ′(y)+s+φ(y).

6.1A1F4F5F8F9F13step 2.1step 4.1step 5.1∎

By [F8], the two Riemann integrals from steps 4.1 and 5.1 sum to the Riemann integral on [a,b]; by [F9] and [A1], this equals the Lebesgue integral that gives Ty(φ′′), since φ vanishes near the endpoints. The terms at y cancel, and step 2.1 gives Ty(φ′′)=(s+−s−)φ(y)=−φ(y). Apply this real calculation to the real and imaginary parts of a complex test. By [F4], [F5] and [F13], ⟨−∂x2Ty,φ⟩=−⟨Ty,φ′′⟩=φ(y)=⟨δy,φ⟩. Thus −∂x2G(⋅,y)=δy in D′(a,b).

Source notes

Hunter, §2.6 equation (2.13), printed p. 33 (PDF p. 39), gives the free-space one-dimensional profile Γ(x)=−∣x∣/2; the discussion on printed p. 40 (PDF p. 46) writes the corresponding whole-line potential for −u′′=f. Teschl, §5.4 Problem 5.21, printed p. 129 (PDF p. 142), asks the reader to find the Green function for an interval but supplies neither its formula nor a solution. The piecewise Dirichlet formula, its endpoint and symmetry checks, and the distributional jump computation above are derived directly; the sources supply context and normalization, not this proof.

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