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Em Elementary Mathematical Prerequisites

1 · Prerequisites

2 · Summary

The elementary mathematical suppliers of the classical electromagnetism collection, all of them mathematics: complex notation for time-harmonic fields, moving parametrized patches, surface distributions and interface jumps, localization, multipole remainders, parameter integrals, point-support densities, quadratic stress identities, joint regularity of radial potentials, and the algebraic interface system. Complex plane-wave fields: real parts of complex exponentials fixes the encoding of real vector fields as the real parts of complex exponentials and the conventions under which complex linear equations may be projected, Period averages of products of real parts of complex exponentials computes the period averages 1T∫0TRe⁡(ae−iωt)Re⁡(de−iωt)dt=12Re⁡(ad‾) with the conjugation required for quadratic expressions, and The two-equation interface system at normal incidence records the purely algebraic two-equation interface system with its balance R+T=1. Moving parametrized surface patches and their flux defines a time-dependent parametrized family, its compatible spacetime velocity field and its multiplicity-counted flux, and Transport of flux through a moving patch proves the transport identity Φ′(t)=∫S(t)∂tB⋅n dS−∮∂S(t)(v×B)⋅dr on nondegenerate time intervals for divergence-free fields, using this normal notation at rank-two slices and the parameter-integral version at degenerate endpoint slices, with the additional term ∫S(t)(v⋅n)div⁡B dS in general; the companion example page checks its sign convention and the necessity of the motion term. Surface distributions and interfacial jumps defines the surface distribution δS(ψ)=∫Sψ dS and the jump [F]=F+−F− of a two-sided field, and Distributional divergence and curl across a stationary interface proves the distributional identities div⁡F={div⁡F}+n⋅[F]δS and curl⁡F={curl⁡F}+n×[F]δS on the defined compact hypersurface, which has no manifold boundary. Point-support spacetime densities: distributions, continuity equation and endpoint fluxes computes the distributional continuity equation for point-support spacetime densities, including the endpoint fluxes qδ(z(a),a)−qδ(z(b),b) of the nonproper branch, and Localization: vanishing integrals or fluxes force a continuous function to vanish supplies the localization converse used by variational arguments and potential theory. Monopole-dipole expansion with an explicit remainder for compactly supported L1 densities gives the monopole-dipole expansion with an explicit 16a2∥ρ∥1/r3 remainder for compactly supported L1 densities, Differentiation under the integral sign over a compact measurable set and over a compact C1 surface differentiates under the integral sign over a compact measurable set and a compact C1 surface, Joint regularity of the radial vector and scalar potentials on a star-shaped domain upgrades the radial vector and scalar potentials to joint Ck regularity on star-shaped domains, and The quadratic stress identity and the local angular-balance identity records the coordinate identities div⁡Q=Fdiv⁡F−F×curl⁡F and the local angular-balance identity used by the momentum and stress items.

All items on this page carry domain: mathematics, depend only on mathematical suppliers, and assert no physical conclusion. Countable Choice is declared and used exactly where the surface integrals, localization estimates, dipole remainder, parameter integrals and transport identity require it, as each item records; the period-average, point-current and algebraic items are explicitly choice-free or carry their declared hypotheses. The physical consumers are the classical electromagnetism and special relativity pages, which use these items as stated mathematical premises under the physics schema's dependency roles.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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Complex plane-wave fields: real parts of complex exponentials

Definition

Fix a vector k∈R3, a parameter ω>0 and amplitudes e,b∈C3. Here C3 is the set of functions {0,1,2}→C under componentwise operations, the complex analogue of the convention for Rn in The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn; we relabel the function values at 0,1,2 by 1,2,3 and write the coordinates of e as (e1,e2,e3) and those of b as (b1,b2,b3). For x∈R3, ⟨k,x⟩ is the Euclidean inner product of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, the real number ∑j=13kjxj; for t∈R put

θ(x,t):=⟨k,x⟩−ωt∈R.

Let exp⁡ be the complex exponential of The complex exponential by its power series (the power series ∑n≥0zn/n!, convergent for every z∈C), let Re⁡ be the real part of Real and imaginary parts, complex conjugation, and modulus, and let i be the imaginary unit of The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i.

For each t∈R define maps E(⋅,t),B(⋅,t):R3→R3 componentwise by

Ej(x,t):=Re⁡(ejexp⁡(i θ(x,t))),Bj(x,t):=Re⁡(bjexp⁡(i θ(x,t))),j∈{1,2,3}.

Equivalently, E(⋅,t) is the real part, taken componentwise, of the complex vector eexp⁡(iθ(x,t)), and likewise for B(⋅,t) with b in place of e.

Well-definedness and conventions

For every (x,t)∈R3×R the number θ(x,t) is real, so i θ(x,t) is a complex number and exp⁡(i θ(x,t)) is defined by the everywhere-convergent exponential series; multiplying by the complex number ej or bj and taking real parts is likewise defined. Both amplitudes may be zero, in which case the corresponding field vanishes identically. The joint map (x,t)↦(E(x,t),B(x,t)) has R3×R, an open subset of R4, as its domain, so the Ck conventions of Ck Euclidean maps and diffeomorphisms apply to any regularity statement made about it; this definition asserts no regularity of its own beyond well-definedness. The exponential is entire with exp⁡′=exp⁡ (The complex exponential is entire and its complex derivative is itself), which is what makes the encoded expressions differentiable in the plane-wave form; derivations that differentiate them use Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a) and the chain rule for holomorphic maps where they are applied.

Complex notation here is an encoding of real fields and of real linear equations, not a complex physical field. The symbols e and b are constant complex amplitudes, not fields; the complex quantity eexp⁡(iθ(x,t)) is an auxiliary object whose real part is the field. Real parts, imaginary parts and complex conjugation are the explicit operations of Real and imaginary parts, complex conjugation, and modulus. Since Re⁡ is additive and commutes with real scalar multiples, any identity obtained from the encoded expressions by sums, real scalar multiples, partial differentiation or integration may be projected by Re⁡ term by term; identities that multiply two encoded real fields may not be read off from a product of complex expressions without conjugation, and the correct second-order average rule is proved separately. No electromagnetic interpretation, equation of motion or physical identification is part of this definition.

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Moving parametrized surface patches and their flux

Definition

Let D⊆R2 be a nonempty finite elementary Green region in the sense of Type I, Type II, and elementary regions for Green's theorem, written as D=D1∪⋯∪DN with its supplied decomposition into elementary Green regions of pairwise disjoint interiors and with the shared-arc structure of that definition, and let ∂D be its positive boundary chain (Positive orientation of elementary-region boundaries). We require in addition that D be Jordan measurable; this is explained below and holds for every finite elementary Green region. Let I⊆R be an interval and let

r:D×I→R3

be a map that extends to a C2 map (Ck Euclidean maps and diffeomorphisms) on an open neighbourhood of D×I⊆R3. Write p=(u,v) for the parameter point, write ru(p,t),rv(p,t),rt(p,t)∈R3 for the partial derivatives of that C2 extension at (p,t), and require

ru(p,t)×rv(p,t)≠0for every (p,t)∈D∘×I∘,

the cross product being that of The cross product in R3. For every t∈I∘ this gives rank two on D∘. It does not imply the connectedness or interior injectivity required by Regular parametrized surface patches on compact Jordan parameter regions; neither is a standing hypothesis here. At endpoint or singleton times the rank condition is not imposed.

The swept set of the family is

S:=⋃t∈Ir(D×{t})⊆R3,Σ:={(r(p,t),t):p∈D, t∈I}⊆R4.

A velocity field for r is a pair (W,v) consisting of an open set W⊆R4 with Σ⊆W and a map v:W→R3 of class C1 such that

v(r(p,t),t)=rt(p,t)for every (p,t)∈D×I.

Such a pair is required to exist: the standing hypothesis of this definition is that a velocity field for r is given, and v denotes it. The equality above is what single-valuedness means here: whenever two parameter points at the same time map to the same spatial point, their time derivatives agree. Values at different times may differ.

Finally let B:V→R3 be a C1 field on an open set V⊆R4 containing Σ. The flux of B through the patch at time t∈I is

Φ(t):=∫D⟨B(r(p,t),t), ru(p,t)×rv(p,t)⟩ du dv,

the integral being the Riemann integral over the compact Jordan measurable set D of the displayed continuous integrand, and the pairing being the Euclidean inner product of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. This is a parametrized flux, counted with the supplied parametrization. When (D,r(⋅,t)) satisfies the regular-patch definition, it agrees with Unit normal fields, orientations, and flux through a regular surface patch. In general different parameter points may represent the same spatial point with different normals. At a rank-two slice the unit normal means N(p,t)=(ru×rv)/∥ru×rv∥2 on D∘, not necessarily a function of the spatial image alone.

For an injectively parametrized material-sheet interpretation, require in addition that r(⋅,t):D→R3 be injective for every t∈I. Injectivity alone does not imply ru×rv≠0: a regular-slice interpretation also requires this condition on D∘×I, including endpoint and singleton times. These additional requirements are not standing hypotheses and are not used by the flux formula above.

Well-definedness and conventions

Jordan measurability of D. Each piece is compact and Jordan measurable by Type I, Type II, and elementary regions for Green's theorem. The topological boundary of their finite union is contained in the finite union of their content-zero boundaries: a point outside all those boundaries is either interior to one piece or exterior to every piece. Combining finite cube covers proves content zero without Countable Choice. Thus D is compact and Jordan measurable by A bounded set in Rm is Jordan measurable iff its boundary is null, equivalently of content zero; the positive boundary chain ∂D is separate supplied orientation data.

Extension independence. Each elementary piece is the closure of its interior by its strict graph inequalities, so D=D∘‾. Two C2 extensions agree on D×I, and their spatial difference quotients agree at interior parameter points. Their continuous spatial partials therefore agree on all of D×I: a nonzero difference at a boundary point would remain nonzero nearby and contradict the dense-interior equality. For nondegenerate I, the same argument on the dense set D∘×I∘ makes rt extension-independent. For singleton I={t0} it does not: the same zero restriction admits extensions 0 and (t−t0)χ(p) with different time derivatives. In that case the chosen extension is part of the velocity data; the flux still depends only on the spatial partials.

Integrability. For fixed t∈I, the map p↦B(r(p,t),t) is continuous on D as a composition of continuous maps, the maps p↦ru(p,t) and p↦rv(p,t) are continuous on D, and the cross product and inner product are continuous bilinear operations; hence the integrand is continuous on the compact set D. A continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set), so Φ(t) is defined for every t∈I.

Degenerate time domains. If I is empty then D×I is empty, the family is vacuous and no flux value is asserted. If I={t0} is a singleton, the compatibility condition still reads v(r(p,t0),t0)=rt(p,t0), but it can always be met: choosing the extension r~(p,t):=r(p,t0), which is constant in the time variable, gives rt≡0 and admits the compatible C1 velocity field v≡0. A different extension may require a different velocity field or may fail compatibility; as explained above, rt(p,t0) is not determined by r alone in this case. The flux Φ(t0) involves only ru×rv, so it is the parametrized flux of the fixed map and does not depend on that choice. The condition ru×rv≠0 is imposed only on D∘×I∘, so rank failures on the boundary of the parameter region (seams) are permitted exactly as in Regular parametrized surface patches on compact Jordan parameter regions, and at an endpoint time of a nondegenerate I the definition asserts no regularity of the slice.

No physical assertion. This is a definition of a mathematical object: a time-dependent parametrized family, a velocity field compatible with it, and the flux integral. It asserts nothing about electromagnetic fields, about which vector fields occur in nature, or about the derivative of Φ; the transport identity for Φ′ is proved separately.

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Surface distributions and interfacial jumps

Definition

Throughout, assume the Axiom of Countable Choice ACω (The Axiom of Countable Choice (ACω)); the surface integral of Surface integration on compact C1 hypersurfaces and the domains of Bounded C1 domains and their outward normals are developed under that convention.

Let Ω⊆R3 be open. Let S⊂Ω be a compact embedded C2 hypersurface (Codimension and hypersurfaces, Embedded submanifolds and slice charts): every p∈S has an open neighbourhood U⊆R3 and a C2 diffeomorphism φ:U→φ(U) onto an open subset of R3 with φ(S∩U)=φ(U)∩(R2×{0}), in the sense of Ck Euclidean maps and diffeomorphisms. Such an S is a compact embedded C1 hypersurface, so the surface integral and the surface measure S(A)=∫S1A dS of Surface integration on compact C1 hypersurfaces and Chart and partition independence of surface measure apply; that measure is finite. The slice-chart definition has no manifold boundary; S has no edge. If S is empty, all surface functionals below are zero and the estimates are read separately without taking a supremum over the empty set.

A two-sided presentation of S consists of a continuous unit normal field n:S→R3 together with sides: n(x) is orthogonal to S at x in the sense that ⟨n(x),γ′(0)⟩=0 for every C1 path γ:(−1,1)→S with γ(0)=x and ∥n(x)∥2=1; and Ω−,Ω+⊆Ω are open sets with

Ω∖S=Ω−∪Ω+,Ω−∩Ω+=∅,

such that for every x∈S there is ε=ε(x)>0 with x+sn(x)∈Ω+ for all 0<s<ε and x+sn(x)∈Ω− for all −ε<s<0. The normal n points from the side Ω− to the side Ω+; the orientation convention for normals is that of Unit normal fields, orientations, and flux through a regular surface patch.

The surface distribution δS. For ψ∈D(Ω) (Test function space d of an open set) define

δS(ψ):=∫Sψ dS,

where the right-hand side is the surface integral of the restriction of ψ to S, taken in the sense of Surface integration on compact C1 hypersurfaces; complex-valued ψ are integrated componentwise in real and imaginary parts. For a continuous density a:S→R define

aδS(ψ):=∫Sa ψ dS.

Both are complex-linear functionals on D(Ω), and both are distributions: δS,aδS∈D′(Ω) (Distribution).

Jumps across S. Let F:Ω∖S→R3 be a field that is of class C1 on Ω− and on Ω+ separately and has one-sided continuous traces: there are continuous maps F+,F−:S→R3 with

F+(x)=lim⁡y→xy∈Ω+F(y),F−(x)=lim⁡y→xy∈Ω−F(y)

for every x∈S, the limits taken in R3. The jump of F across S is the continuous map

[F]:=F+−F−:S→R3.

No continuity of F across S is assumed or implied; F need not even be defined on S, and the two traces may differ.

Well-definedness and finite order

The surface functionals are distributions. Fix a compact K⊆Ω and ψ∈DK={ψ∈D(Ω):supp⁡ψ⊆K} with the seminorm p0(ψ)=sup⁡K∣ψ∣ of Fixed support test function frechet space. Since ψ vanishes outside K and is continuous, sup⁡S∣ψ∣≤p0(ψ); and ∫S∣aψ∣ dS≤∥a∥∞,Ssup⁡S∣ψ∣ S(S) with ∥a∥∞,S<∞ by compactness of S and continuity of a. Hence

∣δS(ψ)∣≤S(S) p0(ψ),∣aδS(ψ)∣≤∥a∥∞,S S(S) p0(ψ),

a finite-order estimate of order 0 with constants independent of K; by the local finite-order characterization (Local finite order characterization of distributions) both functionals are distributions. This is the sense in which boundedness on the compact interface and finiteness of the surface measure give continuity of finite order. Linearity is immediate from linearity of the integral; the pairing is complex bilinear with no conjugation, as in Distribution.

The field has locally integrable components. The hypersurface S is a Lebesgue-null set of R3: by the slice-chart property it is covered by finitely many compact pieces each of which is, in suitable C2 coordinates, a graph over a compact subset of a coordinate plane, hence has content zero (The graph of a continuous function on a compact Euclidean set has content zero), and each piece has content zero in the ambient coordinates because it is the image under the C1 chart inverse of a compact content-zero set (A C1 map sends a compact set of content zero to a set of content zero); a finite union of compact content-zero sets is compact and null (Subsets and countable unions of null subsets of Rm are null), hence has content zero (For compact subsets of Rm, measure zero and content zero coincide), so S has content zero in the sense of Measure zero and content zero in Rm by countable and finite cube covers. Now F is continuous on Ω∖S=Ω−∪Ω+. For each x∈S, the two one-sided limits are finite; hence there is a neighbourhood Ux of x on which ∣F∣ is bounded on each side (use the defining limit with tolerance 1 for each of F+(x) and F−(x)). Away from S, continuity also gives a neighbourhood with a finite bound at every point. For a compact K⊆Ω, these neighbourhoods cover K; take a finite subcover and the maximum of its finitely many bounds. This gives a finite MK such that ∣F∣≤MK on K∖S. Extend F by zero on the null closed set S; the extension is measurable because the two side restrictions are continuous, and

∫K∥F∥2≤MKλ3(K)<∞,

the contributions from K∩S being null. Thus the components of F are locally integrable on Ω in the sense of Regular distribution from a locally integrable function, and define a regular distribution; the trace condition is exactly what makes the boundary values F± well-defined continuous functions on S, so the jump [F] is continuous and bounded on S.

Conventions and exclusions

The sign convention is fixed by the normal: n points from Ω− to Ω+, so [F]=F+−F− is measured in the direction of n. This convention is used by the interface-derivative lemma on this page, where div⁡F and curl⁡F acquire the surface terms n⋅[F] δS and n×[F] δS; reversing n reverses the signs of both sides consistently. The definition assumes a fixed stationary interface S: no time dependence of S is part of the data, and the stationary jump formulas are not asserted for moving interfaces. The density a is continuous, so aδS is defined by the displayed integral; no product of two arbitrary distributions, and no product of δS with a merely locally integrable field, is defined here. The definition is stated under ACω because the surface measure it invokes is; the positivity of the surface measure of nonempty open pieces of S, and the finite-order estimates above, are its only uses here.

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Localization: vanishing integrals or fluxes force a continuous function to vanish

Statement

Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)).

Integral form. Let n,m≥1, let U⊆Rn be open, and let g:U→Rm be continuous. Write gi(x):=g(x)(i) for 0≤i<m, using the canonical function coordinates of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. For a measurable V⊆U write, componentwise,

∫Vg dλn:=(∫g0χV dλn, …, ∫gm−1χV dλn),

each component 0≤i<m, when integrable, being the Lebesgue integral of gi on V, extended by zero outside V (Integrable real and complex functions, and their integrals, Integral over a measurable subset). A ball V⊆U is compactly contained in U when its closure is a compact subset of U; for a Euclidean ball this means V‾⊆U. If ∫Vg dλn=0 for every ball V compactly contained in U, then g=0 on U.

Disc form. Let W⊆R3 be open and let F:W→R3 be continuous. Write Fi(x):=F(x)(i) for 0≤i<3 and e0=(1,0,0), e1=(0,1,0), e2=(0,0,1), the canonical standard vectors of The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0. Extend the basis labels cyclically by ej:=ej mod 3 for j≥3. Thus ei+1×ei+2=ei for each i∈{0,1,2}. For such i, q∈W and r>0 put

φi,q(s,t):=q+s ei+1+t ei+2,Dr:={(s,t)∈R2:s2+t2<r2},

with Dr‾ the closed disc. If for every i, every q∈W and every r>0 with φi,q(Dr‾)⊆W one has

∫Dr‾(Fi∘φi,q) dλ2=0,

then F=0 on W.

Conventions. The displayed disc condition is the vanishing of the flux of F through the flat disc φi,q(Dr‾) in the orientation with unit normal ei: the affine parametrization is a regular patch whose tangent vectors ei+1,ei+2 have cross product ei of norm one (Regular parametrized surface patches on compact Jordan parameter regions, The cross product in R3, The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0), so the flux of Unit normal fields, orientations, and flux through a regular surface patch is the Riemann integral ∫Dr‾(Fi∘φi,q) d(s,t); that integral equals the displayed Lebesgue integral because the integrand is continuous on the compact Jordan region Dr‾ (A closed disc of radius r≥0 has Jordan content πr2, The Riemann integral of a bounded function over a bounded Jordan measurable set, A continuous real function on a compact Jordan measurable set is Riemann integrable over that set, Borel Darboux integrands in finite dimension). Both clauses are pointwise conclusions on their open domains. The first clause is the converse from universal integral relations to a pointwise equation; a single vanishing integral over one ball or one disc supplies no conclusion, and the hypothesis ranges over the full family described.

Facts & Assumptions

Given: The Axiom of Countable Choice; integers n,m≥1; an open U⊆Rn and a continuous g:U→Rm with ∫Vg dλn=0 for every ball V compactly contained in U; an open W⊆R3 and a continuous F:W→R3 satisfying the disc condition of the Statement, with canonical indices 0≤i<m for gi and 0≤i<3 for Fi,ei; and, inside the contradiction argument below, either a point p∈U with g(p)≠0 or a point x∈W with F(x)≠0.

[F1]

Metric balls and continuity at a point: BX(a,ρ)={y∈X:dX(y,a)<ρ}, and f is continuous at a when for every real ε>0 there is δ>0 with f[BX(a,δ)]⊆BY(f(a),ε) (Open ball, closed ball and sphere in a metric space, Continuity of a map between metric spaces, at a point and globally, in the ε-δ form).

[F2]

Open subsets of a metric space: U is open exactly when every x∈U has r>0 with B(x,r)⊆U (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[F3]

Under ACω, every Euclidean ball B(x,r) with r>0 is Lebesgue measurable with 0<λ(B(x,r))<∞ (Euclidean balls have positive finite Lebesgue measure).

[F4]

Integral of signed and nonnegative functions: for measurable real u with at most one of ∫u+, ∫u− equal to +∞ one sets ∫u dλ:=∫u+dλ−∫u−dλ, and u is integrable when both are finite, equivalently ∫∣u∣ dλ<∞; then ∫(−u) dλ=−∫u dλ; and for nonnegative measurable f the integral over E is ∫Ef dλ=∫fχE dλ (Integrable real and complex functions, and their integrals, Integral over a measurable subset).

[F5]

Monotonicity and nonnegative homogeneity for the nonnegative integral: if 0≤f≤h are measurable then ∫f dλ≤∫h dλ, and if c>0 then ∫cf dλ=c∫f dλ (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F6]

Simple integral: for a nonnegative simple measurable function s=∑j≤NcjχEj one has ∫s dλ=∑j≤Ncjλ(Ej) with 0⋅∞=0, and the nonnegative integral of a nonnegative simple measurable function equals its simple integral; hence ∫cχE dλ=cλ(E) for c≥0 and measurable E (The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions).

[F7]

Measurability: every continuous map Rn→Rm is Borel measurable under ACω; the Borel σ-algebra is generated by the open sets, so indicators of open sets are Borel (Continuous functions on Euclidean spaces are Borel measurable, The Borel sigma-algebra of a topological space).

[F9]

Canonical coordinates: Rm consists of functions on m={0,…,m−1}, zi=z(i) for i<m, and ⟨z,w⟩=∑i<mziwi with ∥z∥2=⟨z,z⟩ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn). The standard vectors ei, i<m, have coordinate 1 at i and 0 at the other coordinates (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0). Hence ∣zi∣≤∥z∥2, since zi2≤∑k<mzk2, and ⟨z,ei⟩=zi; applying the bound to differences shows that every component of a continuous vector map is continuous.

Proof

technique · contradiction; assume the Statement fails, and contradict in turn a nonzero value of $g$ (cases (A)) and, under the disc hypothesis, a nonzero value of $F$ (case (B))
1.1givenassume-contra

Suppose the Statement fails. Then either (A) g(p)≠0 for some p∈U, or (B) the disc condition holds and F(x)≠0 for some x∈W; if neither holds then g=0 on U and, whenever the disc condition holds, F=0 on W, which is exactly the Statement. We contradict (A) and (B) in turn.

2.1step 1.1F1F2F4F9givenalgebra

In case (A): choose a component index i<m with gi(p)≠0 and put u:=gi, continuous by [F9]. Replacing the i-th component of g by its negative preserves the hypothesis, since ∫V(−gi) dλ=−∫Vgi dλ by [F4], and does not change whether g vanishes; hence we may suppose u(p)=2c with c:=u(p)/2>0. By [F1] applied to the continuous u at p with ε=c there is δ0>0 with ∣u(y)−u(p)∣<c for all y∈B(p,δ0); by [F2] and p∈U there is δ2>0 with B(p,δ2)⊆U; put δ:=min⁡{δ0,δ2}>0. Then the ball V:=B(p,δ/2) satisfies V‾⊆B(p,δ)⊆U, so V is compactly contained in U, and u(y)>u(p)−c=c for every y∈V. The hypothesis gives ∫Vg dλn=0, whose i-th component is ∫uχV dλ=0.

2.2step 1.1F1F2F9givenalgebra

In case (B): choose i∈{0,1,2} with Fi(x)≠0 and put φ:=φi,x, W′:=φ−1[W]={(s,t):φ(s,t)∈W} and h:=Fi∘φ:W′→R, so that h(0,0)=Fi(x)≠0. Then W′ is open in R2, h is continuous, and for every r>0 with B‾((0,0),r)⊆W′ the disc hypothesis with centre x gives ∫B‾((0,0),r)h dλ2=0. Indeed, the standard unit vectors are orthonormal by [F9], so ∥φ(s,t)−φ(s′,t′)∥2=∥(s,t)−(s′,t′)∥2 for all (s,t),(s′,t′); hence φ is continuous, and the inverse image of a ball contained in W at φ(s0,t0) contains the parameter ball of the same radius at (s0,t0), so W′ is open by [F2], while for (s0,t0)∈W′ and ε>0, continuity of Fi by [F9] at φ(s0,t0)∈W gives ρ>0 with ∣Fi(y)−Fi(φ(s0,t0))∣<ε whenever d(y,φ(s0,t0))<ρ, and then d((s,t),(s0,t0))<ρ gives ∣h(s,t)−h(s0,t0)∣<ε by the isometry identity; finally, when B‾((0,0),r)⊆W′ the disc φ(B‾((0,0),r)) lies in W, so the Statement's condition applies with centre x and radius r.

3.1step 2.1F3F4F5F6F7F8algebra

In case (A): the function uχV is Borel and nonnegative, and 0≤cχV≤uχV pointwise, because u>c>0 on V by step 2.1 while uχV=0 off V, and u is continuous while V is open [F7]. More explicitly, for a continuous function on open U, inverse images of open real sets are relatively open, hence open in Rn; after multiplication by χV and zero extension, inverse images are Borel (adding Rn∖V when the target contains zero). Moreover V‾ is a nonempty compact metric space, so M:=max⁡V‾∣u∣ is finite by [F8], and ∣uχV∣≤MχB(p,δ) with ∫MχB(p,δ) dλ=Mλ(B(p,δ))<∞ by [F5], [F6] and [F3]; hence uχV is integrable and ∫uχV dλ is a finite real number by [F4]. Monotonicity [F5] and the simple-integral rule [F6] give ∫uχV dλ≥∫cχV dλ=c λ(V)>0 by [F3], contradicting ∫uχV dλ=0 from step 2.1. Therefore case (A) is impossible: the integral hypothesis forces g=0 on U.

4.1step 2.1step 2.2step 3.1F3F4F5F6F8algebra

In case (B), replace h by −h if necessary, which preserves every zero disc integral by [F4], so h(0,0)=2c′>0. Continuity and openness as in step 2.1 give r>0 with B‾((0,0),r)⊆W′ and h>c′ throughout that closed disc C. It is compact by [F8] and contained in the open ball B((0,0),2r), of finite measure by [F3], so hχC is integrable by the boundedness and Borel argument of step 3.1. Since 0≤c′χB((0,0),r)≤hχC, [F5] and [F6] give ∫Ch dλ2≥c′λ2(B((0,0),r))>0 by [F3]. This contradicts the closed-disc integral hypothesis of step 2.2, without replacing it by an unstated open-ball hypothesis. Thus F=0 on W.

5.1step 3.1step 4.1discharge-contradiction∎

Cases (A) and (B) both end in contradiction, so no failure of the Statement occurs: the integral hypothesis forces g=0 on U, and the disc hypothesis forces F=0 on W.

Remarks

  • The proof uses, for each clause, the hypothesis only on balls centred at the point where the function is assumed nonzero and with radius half of a continuity threshold; the full family in the Statement is what makes the lemma a localization statement, but no choice from the family is made, and the argument is choice-free apart from the declared ACω used by the surface and measurability suppliers.
  • The disc clause is not a statement about a single loop or box: one special disc carries no information, and the components must be tested in all three coordinate directions at all admissible centres and radii.
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Monopole-dipole expansion with an explicit remainder for compactly supported L1 densities

Statement

Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)). Let a>0 and let ρ∈L1(λ3) be a real or complex valued Lebesgue integrable function on R3 that vanishes λ3-almost everywhere outside the closed ball Ba:={y∈R3:∥y∥2≤a} (The class L1(μ) of integrable functions, Open ball, closed ball and sphere in a metric space). Put

Q:=∫ρ dλ3,p:=∫y ρ(y) dλ3∈C3,∥ρ∥1:=∫∣ρ∣ dλ3,

the vector p being defined by pk:=∫ykρ(y) dλ3(y) for the canonical coordinates k<3, namely k∈{0,1,2}, and ⟨p,x⟩:=∑k<3pkxk being the complex-bilinear coordinate contraction, with no complex conjugation, which agrees with the real Euclidean pairing when p is real; if ρ is real-valued then Q is real and p∈R3, while R(x) below is real in that case and complex in general. Then for every x∈R3 with r:=∥x∥2≥2a>0,

∫ρ(y)∥x−y∥2 dλ3(y)=Qr+⟨p,x⟩r3+R(x),R(x):=∫Baρ(y)(1∥x−y∥2−1r−⟨x,y⟩r3)dλ3(y),

and the remainder satisfies the explicit bound

∣R(x)∣≤16 a2 ∥ρ∥1r3.

In the all-space kernel integral define Kx(y)=∥x−y∥2−1 for y≠x and Kx(x)=0, so its integrand means ρKx and is an everywhere-defined measurable function. This convention cannot affect the integral: x∉Ba and the density vanishes almost everywhere outside Ba. All integrals are Lebesgue integrals; the closed ball Ba is Borel and χBaρ is integrable. The constant 16 is permissible but not sharp, and Q, p and ∥ρ∥1 are finite. No derivative of the remainder in x is asserted.

Facts & Assumptions

Given: The Axiom of Countable Choice; a real a>0; a real or complex ρ∈L1(λ3) vanishing λ3-almost everywhere outside Ba={y:∥y∥2≤a}; and a fixed x∈R3 with r=∥x∥2≥2a.

[F1]

Euclidean inner product and norm: ⟨x,y⟩=∑k<3xkyk and ∥x∥2=⟨x,x⟩; Cauchy-Schwarz gives ∣⟨x,y⟩∣≤∥x∥2∥y∥2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation).

[F3]

Fundamental theorem with interior derivative: if G:[0,1]→R is continuous and differentiable on (0,1) and f is Riemann integrable on [0,1] with f=G′ on (0,1), then ∫01f=G(1)−G(0) (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative).

[F4]

Monotonicity and two-sided bounds for the Riemann integral: if f≤g on [0,1] then ∫01f≤∫01g, and m≤f≤M gives m≤∫01f≤M (If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[F5]

L1(λ3) is a complex vector space and the Lebesgue integral is complex-linear on it: ∫(αf+βg)=α∫f+β∫g; integrable functions are measurable (The class L1(μ) of integrable functions, The Lebesgue integral is linear on L1(μ), Integrable real and complex functions, and their integrals).

[F6]

Triangle inequality for the Lebesgue integral: ∣∫f dλ∣≤∫∣f∣ dλ for f∈L1(λ3) (The modulus of an integral is bounded by the integral of the modulus).

[F7]

Monotonicity and nonnegative homogeneity of the nonnegative integral, and the integral over a measurable set: for measurable 0≤f≤h one has ∫f≤∫h, for c>0 one has ∫cf=c∫f, and ∫Ef dλ=∫fχE dλ for nonnegative measurable f (Monotonicity and nonnegative homogeneity of the nonnegative integral, Integral over a measurable subset).

[F8]

Almost everywhere equality: the definition of a property holding λ3-almost everywhere, and the equivalence, for f,g∈L1(λ3), of f=g a.e. with ∫Af=∫Ag for every measurable A (Measure-null sets and almost-everywhere statements relative to a measure, Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[F9]

Measurability: continuous maps Rn→Rm are Borel measurable under ACω, and the Borel σ-algebra contains the open sets and hence the closed sets (Continuous functions on Euclidean spaces are Borel measurable, The Borel sigma-algebra of a topological space).

Proof

technique · direct; reduce the kernel to a one-variable function on the parameter segment, apply the fundamental theorem twice to obtain a Taylor identity with explicit second-derivative bound, then integrate against the density
1.1givenF1F2algebra

Fix y∈Ba and define w(s):=x−sy for s∈R, the open set Jy:={s∈R:w(s)≠0}, and h:Jy→R by h(s):=∥w(s)∥2−1. For 0≤s≤1 one has ∥w(s)∥2≥r−s∥y∥2≥r−a≥r/2>0 by [F1], so [0,1]⊆Jy and h is of class C2 near [0,1]. To justify the root regularity, for u>0 rationalization gives ∣u+η−u∣≤∣η∣/u and (u+η−u)/η=1/(u+η+u) for small nonzero η. Thus the root is continuous and has derivative 1/(2u); the quotient rule makes this derivative continuously differentiable, so the reciprocal root is C2. Composing with the positive polynomial ⟨w,w⟩ and using [F2] now gives h′(s)=⟨w(s),y⟩∥w(s)∥2−3 and h′′(s)=−∥y∥22∥w(s)∥2−3+3⟨w(s),y⟩2∥w(s)∥2−5.

2.1step 1.1F1algebra

By Cauchy-Schwarz [F1], ∣⟨w(s),y⟩∣≤∥w(s)∥2∥y∥2, so the terms of step 1.1 give ∣h′′(s)∣≤∥y∥22∥w(s)∥2−3+3∥y∥22∥w(s)∥2−3=4∥y∥22∥w(s)∥2−3. Since ∥w(s)∥2≥r−a and r−a≥r/2 for r≥2a, this is at most 4∥y∥22(r−a)−3≤32∥y∥22r−3 for every s∈[0,1].

3.1step 2.1F3F4algebra

Put G(s):=(1−s)h′(s)+h(s) on [0,1]. G is continuous on [0,1] and differentiable on (0,1) with G′(s)=(1−s)h′′(s), a continuous hence Riemann integrable function on [0,1]; the fundamental theorem [F3] applied to G and f:=(1−s)h′′(s) therefore gives Ry(x):=∫01(1−s)h′′(s) ds=G(1)−G(0)=h(1)−h(0)−h′(0). Consequently ∥x−y∥2−1=h(1)=∥x∥2−1+⟨x,y⟩r−3+Ry(x), with h(0)=r−1 and h′(0)=⟨x,y⟩r−3 by step 1.1; and monotonicity [F4] applied to −∣h′′∣≤(1−s)h′′≤∣h′′∣ together with step 2.1 gives ∣Ry(x)∣≤∫01(1−s)∣h′′(s)∣ ds≤32∥y∥22r−3∫01(1−s) ds=16∥y∥22r−3≤16a2r−3.

4.1givenstep 3.1F1F5F7F9algebra

The numbers Q, p, ∥ρ∥1 and the remainder are finite and well defined. Indeed, for each k<3 one has ∣ρyk∣≤a∣ρ∣ on Ba and ρyk=0 almost everywhere off Ba, so every scalar component ykρ lies in L1(λ3) and p∈C3 is finite by [F5] and [F7]; for the remainder, the function obtained by extending y↦(∥x−y∥2−1−r−1−⟨x,y⟩r−3) from Ba by zero is Borel on R3, because the bracket is continuous on the open set {∥y∥2<r}⊇Ba by [F9] (the denominator ∥x−y∥2≥r−a>0 is bounded away from zero on Ba) and χBa is the indicator of a closed hence Borel set, and its product with ρ is bounded in modulus by ∣ρ∣ times the finite constant C:=a−1+r−1+ar−2, since ∥x−y∥2≥r−a≥a and ∣⟨x,y⟩∣r−3≤ar−2 on Ba; hence the product with ρ lies in L1(λ3) by [F5] and [F7].

5.1step 3.1step 4.1F5F7F8algebra

The stated expansion holds. Replacing ρ by its representative χBaρ changes Q, p, ∥ρ∥1 and the left-hand integral by nothing, by [F8], so it suffices to prove the identity for χBaρ, for which the pointwise identity of step 3.1 holds at every y∈Ba. For that representative, writing the left-hand integrand as ρ(y)χBa(y)Kx(y) and using χBa=χBa2, linearity [F5] of the Lebesgue integral on the space of integrable functions gives ∫ρ(y)∥x−y∥2−1dλ3(y)=r−1∫ρ dλ3+r−3∑k<3xk∫ykρ(y) dλ3+R(x)=Q/r+⟨p,x⟩r−3+R(x), with R(x) as displayed in the Statement; the three summands are integrable by step 4.1 and the finite complex-linear combination is legitimate. The contraction is exactly ∑k<3xkpk in the Statement, so no conjugation or coordinate relabelling is used. For real ρ, each pk and Q is real and the real kernel makes R(x) real; the same identity and modulus estimate hold for complex ρ by complex linearity.

6.1step 3.1step 5.1F6F7algebra∎

The bound holds: by the triangle inequality [F6] and monotonicity [F7], ∣R(x)∣≤∫Ba∣ρ(y)∣∣∥x−y∥2−1−r−1−⟨x,y⟩r−3∣dλ3(y)≤16a2r−3∫∣ρ∣ dλ3=16a2∥ρ∥1r−3, the middle inequality being the pointwise bound 16a2r−3 for the bracket on Ba from step 3.1 and the final equality the definition of ∥ρ∥1. Hence the expansion and the explicit remainder bound hold for every r≥2a.

Remarks

  • The hypothesis is almost-everywhere support in the closed ball, not continuity of ρ and not compact support in the topological sense; the moments and the integral are unchanged by modification on null sets, and the remainder bound is valid for every admissible representative.
  • Only the explicit kernel and a one-variable second-order Taylor identity are used; no decay theorem for Newtonian potentials and no derivative estimate of the remainder in x is invoked, and none is proved here.
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Differentiation under the integral sign over a compact measurable set and over a compact C1 surface

Statement

Set form. Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)), used to establish the Lebesgue measure space of Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume. Let K⊆Rn be compact, Lebesgue measurable and of finite Lebesgue measure λn(K)<∞, let a<b, and let f be a real-valued function, defined and continuous together with its partial derivative ∂tf on an open neighbourhood of K×[a,b] in Rn+1. Then t↦∫Kf(x,t) dλn(x) is differentiable at every t∈(a,b) with

ddt∫Kf(x,t) dλn(x)=∫K∂tf(x,t) dλn(x),

and at t=a and t=b it has relative (one-sided) derivatives, taken within [a,b], both equal to the same displayed integral. Equivalently, for every t∈[a,b] and every sequence hk→0 with hk≠0 and t+hk∈[a,b],

lim⁡k→∞1hk(∫Kf(x,t+hk) dλn(x)−∫Kf(x,t) dλn(x))=∫K∂tf(x,t) dλn(x).

Surface form. Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)) for the surface measure of Surface integration on compact C1 hypersurfaces. Let S be a fixed compact embedded C1 hypersurface in Rm, m≥2, and let f be real-valued, defined and continuous together with ∂tf on an open neighbourhood of S×[a,b]. Then t↦∫Sf(σ,t) dS(σ) is differentiable in the same relative sense with

ddt∫Sf(σ,t) dS(σ)=∫S∂tf(σ,t) dS(σ).

The statements are pointwise in t and make no claim about moving parameter sets K(t) or S(t), about singularities crossing K or S, about differentiability of ∂tf, or about differentiation in the spatial variables; only the one-dimensional parameter t is differentiated. Integrals are Lebesgue integrals, and the surface integrals are those of the cited surface measure; the case K=∅ or λn(K)=0 is included and the two sides vanish.

Facts & Assumptions

Given: The Axiom of Countable Choice ACω for both forms; an integer n≥1; a compact measurable K⊆Rn with λn(K)<∞; reals a<b; an open N⊆Rn+1 with K×[a,b]⊆N and a real-valued f on N such that f and its partial derivative ∂tf are continuous on N; and, for the surface form, a compact embedded C1 hypersurface S in Rm with a corresponding f on a neighbourhood of S×[a,b].

[A1]

Lebesgue measure-space premise: assume ACω (The Axiom of Countable Choice (ACω)). For n≥1, (Rn,L(Rn),λn) is a complete measure space; the definition alone introduces only a family of sets and a restricted set function (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn, Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume). All set-form integrals below use this measure space. The surface form retains the same choice premise for its declared finite surface measure.

[F1]

Mean value theorem: for a<c<d<b-type real intervals, a function continuous on a closed interval and differentiable in its interior attains its average rate of change at an interior point; in particular, for fixed x and s,u∈[a,b] with s≠u there is τ strictly between s and u with f(x,u)−f(x,s)=∂tf(x,τ)(u−s) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[F2]

On the Lebesgue measure space established by [A1], integrability of bounded measurable functions on finite-measure sets, and the triangle inequality for the integral: a bounded measurable real function g on measurable K with λn(K)<∞ is integrable, and ∣∫Kg dλn∣≤∫K∣g∣ dλn (Integral over a measurable subset, Integrable real and complex functions, and their integrals, The modulus of an integral is bounded by the integral of the modulus). A continuous function on the compact set K is Borel measurable on K by continuous preimages; its Borel preimages are traces of Borel sets of Rn, and these are Lebesgue measurable under [A1] (A continuous map has Borel preimages of Borel sets, The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra, Assuming countable choice, every Borel subset of Rn is Lebesgue measurable). Boundedness and the simple-integral/order rules in [F3] give integrability on the finite-measure K.

[F3]

On the measure space established by [A1], linearity, monotonicity and nonnegative homogeneity of the Lebesgue integral: ∫K(αg+βh) dλn=α∫Kg dλn+β∫Kh dλn for integrable g,h and real α,β, and for measurable 0≤g≤h one has ∫g dλ≤∫h dλ, with ∫c g dλ=c∫g dλ for c>0; moreover ∫Ec dλn=c λn(E) for c≥0 and measurable E, in particular ∫Kη dλn=η λn(K) (The Lebesgue integral is linear on L1(μ), Monotonicity and nonnegative homogeneity of the nonnegative integral, The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions).

[F5]

Surface measure: the surface integral on a compact embedded C1 hypersurface defines a finite Borel measure, independent of the finite charts and partitions used, continuous and monotone on nonnegative functions, and linear on signed functions integrable against it (Surface integration on compact C1 hypersurfaces, Chart and partition independence of surface measure, The Axiom of Countable Choice (ACω)).

Proof

technique · direct; represent the difference quotient by the mean value theorem as an average of $\partial_tf$ and estimate it by uniform continuity on the compact product
1.1givenA1F2algebra

The function F(t):=∫Kf(x,t) dλn(x) is defined and finite for every t∈[a,b]: for fixed t the map x↦f(x,t) is continuous on K, hence bounded and Borel measurable, and λn(K)<∞, so it is integrable by [F2].

2.1step 1.1F1F2F3algebra

Fix s,t∈[a,b] with s≠t and put g(x):=f(x,s)−f(x,t)s−t for x∈K; then g is continuous on K, hence bounded, Borel measurable and integrable by [F2]. For every x∈K the mean value theorem [F1] applied to the one-variable function u↦f(x,u) on the closed interval with endpoints t and s supplies a point τ strictly between t and s with f(x,s)−f(x,t)=∂tf(x,τ)(s−t), that is, g(x)=∂tf(x,τ); no point τ is selected as a function of x anywhere below, only the pointwise inequality ∣g(x)−∂tf(x,t)∣<η that such an existential τ yields, together with the separately defined continuous function g. By linearity of the integral [F3] applied to the integrable functions f(⋅,s) and f(⋅,t) of step 1.1, F(s)−F(t)=∫K(f(x,s)−f(x,t))dλn(x)=(s−t)∫Kg dλn(x), so the difference quotient is ∫Kg dλn.

3.1step 2.1F2F3F4algebra

Given ε>0, put η:=ε/(1+λn(K))>0; the function ∂tf is uniformly continuous on the compact set K×[a,b] by [F4], so there is δ>0 such that ∣∂tf(x,u)−∂tf(x,v)∣<η whenever (x,u),(x,v)∈K×[a,b] with ∣u−v∣<δ. Fix s,t∈[a,b] with 0<∣s−t∣<δ and let g be the continuous function of step 2.1. For every x∈K the mean value point τ supplied by [F1] lies strictly between t and s, so ∣τ−t∣<∣s−t∣<δ and therefore ∣g(x)−∂tf(x,t)∣=∣∂tf(x,τ)−∂tf(x,t)∣<η; the function h:=g−∂tf(⋅,t) is continuous on K, hence bounded and Borel measurable, so by [F2] and [F3] ∣F(s)−F(t)s−t−∫K∂tf(x,t) dλn(x)∣=∣∫Kh dλn∣≤∫K∣h∣ dλn≤∫Kη dλn=η λn(K)=ελn(K)1+λn(K)<ε; the case λn(K)=0 is covered by the same chain, both sides vanishing. No mean-value point is selected as a function of x in this estimate; ACω has already supplied the Lebesgue measure-space premise [A1].

4.1step 3.1algebra

Step 3.1 shows that for every t∈[a,b] and every sequence hk→0 with hk≠0 and t+hk∈[a,b] the difference quotients converge to ∫K∂tf(x,t) dλn(x): for interior t this is differentiability with the stated derivative, and at t=a (respectively t=b), where only hk>0 (respectively hk<0) is admissible, it is the relative derivative there; because N is open, ∂tf(x,t) is the genuine two-sided partial derivative, so the relative endpoint derivative exists and equals the displayed integral. This proves the set form.

5.1step 3.1step 4.1F1F4F5algebra∎

Surface form: let G(t):=∫Sf(σ,t) dS(σ), a finite real number for each t by [F5]. Fix s,t∈[a,b] with s≠t and put g(σ):=f(σ,s)−f(σ,t)s−t; g is continuous on the compact S, hence σ↦g(σ)−∂tf(σ,t) is continuous, bounded and integrable against dS, and by linearity of the surface integral [F5] on the integrable functions G(s)−G(t)=∫S(f(σ,s)−f(σ,t))dS(σ)=(s−t)∫Sg dS(σ). For every σ∈S the mean value theorem [F1] supplies τ strictly between t and s with g(σ)=∂tf(σ,τ), and again no τ is selected as a function; uniform continuity of ∂tf on the compact set S×[a,b] by [F4], the finiteness of the surface measure S(S) and the monotonicity and triangle inequality for the surface integral [F5] then give, for 0<∣s−t∣<δ with the δ of step 3.1 applied to η:=ε/(1+S(S)), the pointwise bound ∣g(σ)−∂tf(σ,t)∣<η and hence ∣[G(s)−G(t)]/(s−t)−∫S∂tf(σ,t) dS(σ)∣=∣∫S(g−∂tf(⋅,t))dS∣≤η S(S)<ε, exactly as in steps 2.1 and 3.1 with S in place of K and dS in place of dλn. Passing to the limit as in step 4.1 gives the surface form, including the relative endpoint derivatives.

Remarks

  • The proof is the difference-quotient argument of the published rectangle theorem in the more general setting of a compact measurable parameter set and of a fixed compact C1 surface; the finite measure (or finite surface measure) of the parameter set is what makes the uniform error bound convert into an integral error bound.
  • No moving set is treated. For a moving surface the parametrization itself depends on t and additional terms appear; that transport identity is proved separately using the moving-patch definition on this page.
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Point-support spacetime densities: distributions, continuity equation and endpoint fluxes

Statement

Proper-graph branch. Let I⊆R be an open interval, let z:I→R3 be a C1 trajectory, put γ(t):=(z(t),t) and let Ω⊆R4 be open with γ[I]⊆Ω. Assume the graph map γ:I→Ω is proper: γ−1[K] is a compact subset of I for every compact K⊆Ω. For q∈R and ψ∈D(Ω) (Test function space d of an open set) define

⟨ρ,ψ⟩:=q∫Iψ(z(t),t) dt,⟨J,ψ⟩:=q∫Iz′(t) ψ(z(t),t) dt,

the second integral being componentwise, so J=(J1,J2,J3) is a triple of distributions, with components ⟨Ji,ψ⟩=q∫Izi′(t)ψ(z(t),t) dt. Then ρ and J are distributions on Ω (Distribution) and, with the distributional derivative of Distributional derivative,

∂tρ+div⁡J=0in D′(Ω).

Complex integrals mean ∫w:=∫Re⁡w+i∫Im⁡w, with the same componentwise convention for improper limits. In formulas involving complex tests, ⟨u,w⟩:=∑j=13ujwj for u∈R3 and w∈C3, with no conjugation. This extends the real coordinate pairing; estimates on J mean estimates on each component.

In the proper branch, an integral over I means the proper integral over any compact subinterval containing the pulled-back support, with the integrand zero near its ends; additivity makes this independent of the subinterval. For I=∅ both functionals are defined to be zero. Properness is automatic if Ω=R3×I, and also if I=R.

Endpoint branch. Suppose instead that I=(a,b) is bounded with finite endpoints, that z extends continuously to [a,b], that the absolute improper integral ∫ab∥z′(t)∥2 dt is finite, and that both endpoint events (z(a),a) and (z(b),b) belong to Ω. Then the same formulas, with the integrals understood as improper integrals over (a,b), define distributions ρ and J on Ω, and

∂tρ+div⁡J=q δ(z(a),a)−q δ(z(b),b),

the Dirac distributions being those of Dirac delta and its derivatives.

Mere graph containment γ[I]⊆Ω does not make the pulled-back tests compactly supported, and it does not by itself give the source-free identity: in the endpoint branch the endpoint terms are exactly as displayed. Each individual endpoint contribution vanishes on a test if and only if q=0 or that test vanishes at the corresponding endpoint event. Their total pairing vanishes if and only if q=0 or ψ(z(a),a)=ψ(z(b),b). No Maxwell equation, no point self-force and no physical interpretation of ρ or J is asserted; only the distributional identities for the prescribed support are proved.

Facts & Assumptions

Given: An open interval I⊆R; a C1 trajectory z:I→R3; the graph map γ(t)=(z(t),t) mapping I into an open set Ω⊆R4, proper in the first branch; a real q; and, in the endpoint branch, I=(a,b) with continuous extension of z to [a,b], finite ∫ab∥z′∥2, and (z(a),a),(z(b),b)∈Ω.

[F1]

Test functions and distributions: D(Ω)=Cc∞(Ω;C) consists of smooth functions with support compact in Ω; a distribution is a continuous complex-linear functional on it; a complex-linear functional is a distribution if and only if on each DK={ψ∈D(Ω):supp⁡ψ⊆K} it obeys a finite-order estimate ∣u(ψ)∣≤Cpm(ψ), in particular a bound by p0(ψ)=sup⁡K∣ψ∣ (Test function space d of an open set, Distribution, Fixed support test function frechet space, Local finite order characterization of distributions).

[F2]

Distributional derivative and Dirac distributions: ⟨∂αu,φ⟩=(−1)∣α∣⟨u,∂αφ⟩; δa(φ)=φ(a) and its derivatives are distributions (Distributional derivative, Dirac delta and its derivatives).

[F5]

All real integral estimates use nonnegativity and the two-sided bound of If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a). The complex modulus of Real and imaginary parts, complex conjugation, and modulus satisfies ∣a+ib∣≤∣a∣+∣b∣ by Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive. Applying the real bounds to the real and imaginary parts gives ∣∫w∣≤2∫∣w∣ on proper intervals and, by limits, on each convergent improper piece. Fundamental theorem and integral algebra: if G is continuous on [u,v] and differentiable on (u,v) with Riemann integrable derivative, then ∫uvG′=G(v)−G(u); the integral is additive over subintervals and linear on integrable functions (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative, For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c, Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg).

[F6]

Improper integrals at finite endpoints: one-ended improper integrals as limits of proper truncations, with a mixed interval split finitely and each singular-end piece required to converge separately; the comparison test: if ∣f∣≤g eventually and ∫g converges, then ∫f converges absolutely, and absolute convergence implies convergence with ∣∫f∣≤∫∣f∣ on a one-ended interval (Improper integrals at a finite singular endpoint, Improper integrals with several singular ends, Comparison tests for improper integrals, Absolute convergence implies improper convergence).

Proof

technique · direct; write the pairing of the divergence as minus the integral of the derivative of the pulled-back test along the trajectory, and evaluate that integral by the fundamental theorem, first with compact support and then with the truncated improper integral in the endpoint branch
1.1givenF1F3F4algebra

If I=∅, the stated functionals and all their derivatives are zero, proving the proper branch. Otherwise assume I nonempty. The graph map γ(t)=(z(t),t) is C1 on I and injective, with values in Ω; for ψ∈D(Ω) the pullback φ:=ψ∘γ:I→C is C1, and supp⁡φ⊆γ−1[supp⁡ψ]: if φ(t)≠0 then γ(t)∈supp⁡ψ. In the proper branch γ−1[supp⁡ψ] is compact, so supp⁡φ is a closed subset of a compact set, hence compact in I; thus φ∈Cc1(I) and φ′ is continuous.

2.1step 1.1F1F3F5algebra

In the proper branch, ρ is a distribution. Let K⊆Ω be compact and ψ∈DK, so that supp⁡φ⊆CK:=γ−1[K], a compact subset of I; if CK=∅ then φ=0. Otherwise u:=min⁡CK and v:=max⁡CK lie in I with CK⊆[u,v]⊆I, and ∣φ∣≤p0(ψ) on [u,v] because ψ vanishes off K and p0(ψ)=sup⁡K∣ψ∣; hence ∣⟨ρ,ψ⟩∣≤2∣q∣∫uv∣φ∣ dt≤2∣q∣(v−u)p0(ψ) by [F5], a finite-order estimate of order zero, so ρ is a distribution by [F1].

2.2step 1.1F3algebra

The automatic properness claims hold. If Ω=R3×I and K⊆Ω is compact, then the time projection prt[K] is a compact subset of I by [F3] and γ−1[K] is closed in I (preimage of the closed set K under the continuous map γ) with γ−1[K]⊆prt[K]; being closed in I and contained in prt[K], it is closed in the subspace prt[K], hence compact as a closed subset of a compact metric space by [F3]. If I=R, then γ−1[K] is closed and contained in the bounded set prt[K], hence compact in R by [F3].

3.1step 1.1F1F3F5algebra

In the proper branch, J is a distribution: with the notation of step 2.1, ∣z′∣ is continuous on [u,v], hence bounded there by MK<∞, and ∣⟨Ji,ψ⟩∣≤2∣q∣∫uv∥z′(t)∥2∣φ(t)∣ dt≤2∣q∣MK(v−u)p0(ψ), again a finite-order estimate of order zero.

3.2step 1.1step 2.1F1F5F6algebra

Endpoint branch, well-definedness: let I=(a,b), with z extending continuously to [a,b], ∫ab∥z′∥2<∞ and both endpoint events in Ω. Then φ=ψ∘γ extends continuously to [a,b] with ∣φ∣≤p0(ψ) and ∣zi′φ∣≤p0(ψ)∥z′∥2 on (a,b); by the comparison test [F6] the improper integrals ∫abφ and ∫abz′φ converge absolutely (the first against the constant p0(ψ), the second against p0(ψ)∥z′∥2), so ⟨ρ,ψ⟩:=q∫abφ and ⟨J,ψ⟩:=q∫abz′φ are well-defined finite values; they are linear in ψ by linearity of the truncated integrals and passage to the limit [F5, F6], and for ψ∈DK they obey ∣⟨ρ,ψ⟩∣≤2∣q∣p0(ψ)(b−a) and ∣⟨Ji,ψ⟩∣≤2∣q∣p0(ψ)∫ab∥z′∥2, so they are distributions by [F1].

4.1step 2.1step 3.1F2F4algebra

For ψ∈D(Ω) in the proper branch, the pairing of the divergence is ⟨∂tρ+div⁡J,ψ⟩=−⟨ρ,∂tψ⟩−⟨J,∇ψ⟩=−q∫I(∂tψ(γ(t))+⟨z′(t),∇ψ(γ(t))⟩)dt, the components of ∇ψ being the spatial partial derivatives and div⁡J the sum of the coordinate distributional derivatives of the components of J by [F2].

5.1step 4.1F4algebra

By the chain rule [F4], ddtφ(t)=Dψ(γ(t))γ′(t)=∂tψ(γ(t))+⟨∇ψ(γ(t)),z′(t)⟩, since γ′=(z′,1); hence the integrand of step 4.1 is exactly φ′.

6.1step 4.1step 5.1F5algebra

Since φ∈Cc1(I) by step 1.1, there are u<v in I with supp⁡φ⊆[u,v] and φ(u)=φ(v)=0; by [F5] applied to the subinterval and by additivity of the integral over subintervals, ∫Iφ′=∫uvφ′=φ(v)−φ(u)=0. Therefore ⟨∂tρ+div⁡J,ψ⟩=0 for every ψ∈D(Ω), which is the identity ∂tρ+div⁡J=0 of distributions.

7.1step 5.1step 6.1step 3.2F2F5F6algebra∎

Endpoint branch, identity: applying [F2] and the chain rule as in steps 4.1 and 5.1 gives ⟨∂tρ+div⁡J,ψ⟩=−q∫abφ′(t) dt, where φ′ is continuous on (a,b) and ∣φ′∣≤C(1+∥z′∥2) for a constant C depending on the fixed test, so the improper integral converges absolutely by the comparison test [F6]; for 0<ε<(b−a)/2 the proper fundamental theorem [F5] gives ∫a+εb−εφ′=φ(b−ε)−φ(a+ε), and the mixed improper integral is the limit of these truncations as ε↓0 [F6]; continuity of φ at the endpoints gives φ(b−ε)−φ(a+ε)→φ(b)−φ(a)=ψ(z(b),b)−ψ(z(a),a). Hence ⟨∂tρ+div⁡J,ψ⟩=qψ(z(a),a)−qψ(z(b),b)=⟨qδ(z(a),a)−qδ(z(b),b),ψ⟩ for every ψ∈D(Ω) by [F2], which is the endpoint identity. If q=0, each endpoint contribution and their sum vanish for every test; if q≠0, division by q shows that an individual contribution vanishes exactly when its endpoint test value is zero, and the sum vanishes exactly when the two endpoint test values are equal.

Remarks

  • The endpoint terms are real. For q≠0, I=(0,1), z≡0 and Ω=R4 the trajectory is constant, ∫01∥z′∥2=0 and the identity reads ∂tρ+div⁡J=qδ(0,0)−qδ(0,1): the static point charge on a finite time window is not source-free as a spacetime distribution, and the two endpoint contributions have opposite signs. The claimed free identity for this configuration would be false.
  • Where properness is used. In the proper branch properness makes every pulled-back test compactly supported, which is what allows the fundamental theorem without endpoint terms and gives the order-zero estimates. Neither graph containment alone nor compactness of the ambient test alone replaces this hypothesis, and no physical existence of the sources is asserted.
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The quadratic stress identity and the local angular-balance identity

Statement

Let U⊆R3 be open and let F:U→R3 be of class C1 (Ck Euclidean maps and diffeomorphisms), with coordinates x1,x2,x3, partial derivatives ∂j=∂/∂xj, Euclidean inner product ⟨⋅,⋅⟩ and norm ∣⋅∣2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn), divergence and curl as in Divergence and curl of a C1 vector field, and cross product as in The cross product in R3. Put

Qij:=FiFj−12δij∣F∣22,δij:={1,i=j,0,i≠j,(div⁡Q)i:=∑j=13∂jQij.

Then Q is a C1 tensor field on U and

div⁡Q=F div⁡F−F×curl⁡F.

Moreover, writing ϵijk for the Levi-Civita symbol (ϵijk=+1 if (i,j,k) is an even permutation of (1,2,3), −1 if it is an odd permutation, and 0 if two indices coincide), every symmetric tensor Q (that is, Qkj=Qjk for all j,k) satisfies

ϵijkQkj=0(i∈{1,2,3}),

the sum over j,k being the nine-term sum with each pair counted separately. Finally, let I⊆R be open. A tensor field here is a matrix-valued map Π:U×I→R3×3; assume its entries and all spatial first partial derivatives are continuous and Πkj=Πjk. Let g,f:U×I→R3 be continuous, assume the componentwise time derivative ∂tg exists, and suppose ∂tg+div⁡Π=−f on U×I, where (div⁡Π)i=∑ℓ=13∂ℓΠiℓ and (x×Π)iℓ:=ϵijkxjΠkℓ with the sum over j,k. Then the vector field x×g and the tensor field x×Π satisfy

∂t(x×g)+div⁡(x×Π)=−x×f.

No symmetry of Q beyond the displayed form is assumed, and no boundary integral is asserted: the first two identities are pointwise on U and the angular identity is pointwise on U×I.

Facts & Assumptions

Given: The open set U⊆R3; the C1 field F:U→R3 with Qij=FiFj−12δij∣F∣22; a symmetric tensor Q in the second claim and a spatially C1 symmetric matrix field Π on U×I, continuous g,f there, and the existing time derivative ∂tg satisfying the stated balance.

[F1]

Cross product: for u,v∈R3, u×v=(u2v3−u3v2, u3v1−u1v3, u1v2−u2v1) (The cross product in R3).

[F2]

Divergence and curl: div⁡F=∑i=13∂iFi and curl⁡F=(∂2F3−∂3F2, ∂3F1−∂1F3, ∂1F2−∂2F1) (Divergence and curl of a C1 vector field).

[F3]

Inner product and norm: ⟨u,v⟩=∑i=13uivi and ∣F∣22=⟨F,F⟩ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

Proof

technique · direct; expand the divergence of the quadratic tensor in coordinates, match the resulting expression with $-F\times\operatorname{curl}F$, cancel symmetric terms in the Levi-Civita contraction, and differentiate the cross-product products in the angular-balance identity
1.1givenF3F4algebra

The entries Qij=FiFj−12δij⟨F,F⟩ are sums of products of C1 functions, hence C1 on U by [F3] and [F4], so both sides of the first identity are defined; moreover ∂j⟨F,F⟩=∂j∑l=13FlFl=2∑l=13Fl ∂jFl by the product rule [F4].

2.1step 1.1F2F4algebra

For fixed i, ∑j=13∂jQij=∑j=13∂j(FiFj)−12∑j=13δij∂j⟨F,F⟩=∑j=13(Fj∂jFi+Fi∂jFj)−∑l=13Fl∂iFl=Fidiv⁡F+∑j=13(Fj∂jFi−Fj∂iFj) by step 1.1, the product rule [F4] and the divergence formula [F2].

2.2step 1.1algebra

For a symmetric tensor and fixed i, the nine-term sum ϵijkQkj vanishes: terms with j=k have ϵijk=0, and for j≠k the pair (j,k) and the pair (k,j) contribute ϵijkQkj+ϵikjQjk=ϵijkQkj−ϵijkQkj=0, because ϵ changes sign under exchange of its last two indices and Q is symmetric. This is the second claim.

3.1step 2.1F1F2algebra

The first coordinate of F×curl⁡F is, by [F1] and [F2], F2(∂1F2−∂2F1)−F3(∂3F1−∂1F3)=F2∂1F2+F3∂1F3−F2∂2F1−F3∂3F1; adding and subtracting F1∂1F1 shows that −(F×curl⁡F)1=∑j=13(Fj∂jF1−Fj∂1Fj), and the same expansion in the second and third coordinates gives −(F×curl⁡F)i=∑j=13(Fj∂jFi−Fj∂iFj) for every i. Substituting this into step 2.1 gives div⁡Q=Fdiv⁡F−F×curl⁡F, which is the first identity.

3.2step 2.2F4algebra

At each time the field (x×Π)iℓ=ϵijkxjΠkℓ is spatially C1, and differentiating the product in xℓ and summing over ℓ gives (div⁡(x×Π))i=∑ℓ=13∂ℓ(ϵijkxjΠkℓ)=ϵijk∑ℓ=13(δjℓΠkℓ+xj∂ℓΠkℓ)=ϵijkΠkj+ϵijkxj∑ℓ=13∂ℓΠkℓ=ϵijkxj(div⁡Π)k, where the term ϵijkΠkj vanishes by step 2.2 applied to the symmetric field Π; here δjℓ is the Kronecker symbol and the product rule [F4] was used in each summand.

4.1step 3.2F4algebra∎

Substituting ∂tgk=−fk−(div⁡Π)k and differentiating the product with the time-independent factor xj gives ∂t(x×g)i=ϵijkxj∂tgk=−ϵijkxjfk−ϵijkxj(div⁡Π)k=−(x×f)i−(div⁡(x×Π))i by step 3.2; hence ∂t(x×g)+div⁡(x×Π)=−x×f, which is the third claim.

Remarks

  • The identity div⁡Q=Fdiv⁡F−F×curl⁡F requires only first derivatives and no interchange of mixed partials, so the C1 hypothesis is the natural one; the symmetry claim for ϵijkQkj is a purely algebraic statement about symmetric tensors.
  • The angular-balance identity is local; passing to an integral balance over a region requires a divergence theorem and boundary terms, which are not asserted here. For any time-dependent data meeting the displayed balance, the angular identity follows; this item adopts no evolution equation for F.
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Distributional divergence and curl across a stationary interface

Statement

Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)). Let Ω⊆R3 be open and let (S,Ω−,Ω+,n) be a two-sided presentation of a compact embedded C2 interface in Ω in the sense of Surface distributions and interfacial jumps: Ω∖S=Ω−∪Ω+ are the two open sides and n:S→R3 is the continuous unit normal pointing from Ω− to Ω+. Let F:Ω∖S→R3 be of class C1 on Ω− and on Ω+ separately, with one-sided C1 traces on S: F and each first partial derivative ∂iF extend continuously from each side to S, and the traces F+,F−∈C0(S;R3) give the jump [F]=F+−F−.

Let {∂iFj}, {div⁡F} and {curl⁡F} denote the locally integrable fields equal on Ω± to the corresponding classical derivative of F; these are regular distributions (Regular distribution from a locally integrable function). The slice-chart definition makes S a hypersurface without manifold boundary, so no edge is part of these data. Then, in the distributional sense,

∂iFj={∂iFj}+ni [Fj] δS(i,j∈{1,2,3}),

and consequently

div⁡F={div⁡F}piecewise+n⋅[F] δS,curl⁡F={curl⁡F}piecewise+n×[F] δS,

as distributions on Ω: their pairings agree for every ψ∈D(Ω). Here δS is the surface distribution of Surface distributions and interfacial jumps, n⋅[F]=∑ini[Fi], and n×[F] is the cross product of The cross product in R3. A moving interface with ∂tS≠0 requires additional terms and is not covered; the stationary jump formulas may not be reused for a moving interface. Interfaces with a manifold boundary require separate definitions and are not included.

Facts & Assumptions

Given: The open set Ω⊆R3; a two-sided compact embedded C2 interface S with sides Ω± and unit normal n from Ω− to Ω+; a field F of class C1 on each side with one-sided C1 traces; and a test function ψ∈D(Ω) with compact support in Ω.

[F1]

Surface distributions and jumps: δS(φ)=∫Sφ dS, the density functional and the jump [F]=F+−F− are those of Surface distributions and interfacial jumps; a field C1 on each side with one-sided continuous traces has locally integrable components, and the traces extend continuously to S.

[F2]

Distributional derivative and regular distributions: ⟨∂iu,φ⟩=−⟨u,∂iφ⟩ for u∈D′(Ω) (Distributional derivative), and a locally integrable function defines a distribution by integration (Regular distribution from a locally integrable function, Distribution).

[F3]

Divergence theorem and domains: under ACω, a bounded open set with a specified finite piecewise C1 presentation admits the divergence theorem ∫Ωdiv⁡G=∑j∫SjG⋅νj dS for G∈C1(Ω‾) (Specified finite piecewise C1 boundary presentations, Divergence for finite piecewise C1 presentations, Surface integration on compact C1 hypersurfaces).

[F4]

Localization: compactly supported smooth cutoffs and smooth partitions of unity subordinate to a given open cover of an open set exist (Test function cutoffs and euclidean localization).

[F5]

Derivative conventions: divergence and curl are those of Divergence and curl of a C1 vector field; the cross product satisfies (u×v)k=ϵkijuivj in the right-handed convention of The cross product in R3, where ϵkij is +1 for an even permutation of (1,2,3), −1 for an odd permutation and zero for repeated indices (all repeated indices are summed), and the Euclidean inner product is that of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn; products and one-sided derivatives follow the usual product and chain rules (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Ck Euclidean maps and diffeomorphisms).

Proof

technique · direct; reduce to a local single-patch computation by a partition of unity, integrate by parts on each side with the divergence theorem, and read off the jump terms
1.1givenF1F2algebra

By [F1] the components of F and of its one-sided first derivatives are locally integrable on Ω, the piecewise fields {∂iFj} are locally integrable, and [Fj] is continuous on S; hence all the distributions displayed in the Statement are defined, and ni[Fj]δS is the surface distribution of the continuous density ni[Fj] of [F1].

1.2givenF1F2F3F4F5algebra

Reduction to finite chart boxes. At a point of S use the C2 slice diffeomorphism ϕ from [F1], with ϕ(S)={s=0} locally. Choose a closed rectangular product box inside its open image, centred at the image point, and write B for its open inverse image. Shrink the box so its closure is compactly contained in the chart and in Ω. The two sets B±=B∩Ω± are inverse images of the two open half-boxes: the local side designation is determined by the normal convention and continuity, and the half-boxes are connected. Each has a specified finite piecewise C1 presentation, with the interface rectangle at s=0 and the finitely many outer rectangular faces transported by the C2 chart inverse. Faces are regular parametrized compact patches. Their seams are transported line segments and vertices, hence null in ambient volume: the chart inverse has bounded first derivatives on the compact box, so a subdivision of each segment into O(δ−1) pieces has images covered by balls of radius Cδ, whose total three-dimensional content is O(δ2)→0. For the required surface-null condition, parametrize each face by its two free box coordinates. Its seam preimage is a finite union of line segments, covered by O(δ−1) parameter squares of area O(δ2); thus it has two-dimensional Lebesgue measure zero. The chart surface density is bounded on the compact face, so its surface integral on those seams is zero. This verifies the finite-face and seam obligations in [F3], without asserting any property of a spherical chart cut. Compactness of the test support and [F4] now give finitely many such B1,…,BN covering supp⁡ψ∩S, with compactly supported smooth cutoffs χk inside Bk summing to one near that intersection. Write ψ=∑kχkψ+(1−∑kχk)ψ; the last test vanishes near S, where the identity is the ordinary classical integration-by-parts identity. Thus it suffices to prove the displayed component identity for a test supported in one Bk.

2.1step 1.2F2F3algebra

Local computation: let φ∈D(Ω) have support in a chart box B=Bk as in step 1.2. By [F2], ⟨∂iFj,φ⟩=−∫ΩFj∂iφ=−∫Ω+Fj∂iφ−∫Ω−Fj∂iφ. On each side apply the divergence theorem [F3] to G±:=Fjφ ei. Its values and first spatial derivatives extend continuously to the closed half-box by the one-sided trace hypothesis and the product rule, so it belongs to C1(B±‾) in the stated up-to-boundary convention, which requires no ambient extension. It vanishes near every outer box face because supp⁡φ⊆B. For complex φ, apply the real divergence theorem separately to its real and imaginary parts; the outward normal of Ω+ on the interface face is −n and that of Ω− is +n, so ∫Ω+∂i(Fjφ)=−∫S∩B(Fj)+φ ni dS and ∫Ω−∂i(Fjφ)=∫S∩B(Fj)−φ ni dS. Substituting ∫Ω±Fj∂iφ=∫Ω±∂i(Fjφ)−∫Ω±(∂iFj)φ gives ⟨∂iFj,φ⟩=∫Ω{∂iFj}φ+∫S((Fj)+−(Fj)−)φ ni dS=∫Ω{∂iFj}φ+⟨ni[Fj]δS,φ⟩, which is the componentwise identity.

3.1step 1.2step 2.1F5algebra

Summing step 2.1 over i (and using the definitions of Divergence and curl of a C1 vector field and of the piecewise fields) gives ⟨div⁡F,φ⟩=∫Ω{div⁡F}φ+⟨n⋅[F]δS,φ⟩ for every test supported in some Bk, hence, by the finite partition of step 1.2, for every ψ∈D(Ω); this is the divergence identity on Ω.

4.1step 3.1F1F5algebra∎

Antisymmetrizing step 2.1 gives the curl identity: by [F5], (curl⁡F)k=ϵkij∂iFj=ϵkij{∂iFj}+ϵkijni[Fj]δS={curl⁡F}k+(n×[F])kδS for each k, since ϵkij{∂iFj} is the classical k-th component of the piecewise curl and ϵkijni[Fj]=(n×[F])k in the stated convention; hence, for every ψ∈D(Ω), curl⁡F={curl⁡F}piecewise+n×[F]δS.

Remarks

  • The sign convention is the one fixed by the normal: because n points from Ω− to Ω+, the outward normal of Ω+ on the interface is −n and that of Ω− is +n, which produces the difference F+−F−. Reversing n reverses both n and [F], leaving n⋅[F] and n×[F] unchanged, as it must.
  • The computation uses local charts on the edge-free interface specified here; interfaces with boundary are outside this definition. A moving interface contributes an additional term proportional to the normal velocity, not covered here.
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Joint regularity of the radial vector and scalar potentials on a star-shaped domain

Statement

Regularity convention for a possibly non-open parameter interval. Let U⊆R3 be open, let I⊆R be an interval and let k∈N. A map on U×I is jointly Ck when it is the restriction of a Ck map on some open neighbourhood N⊆R4 of U×I, in the sense of Ck Euclidean maps and diffeomorphisms; it is jointly C∞ when it is the restriction of one smooth map on such an open neighbourhood. For open I these are exactly the usual open-domain classes; at included endpoints of a nondegenerate interval, time derivatives are the corresponding relative one-sided derivatives. For empty or singleton I the convention still asserts extension existence and the spatial identities below, but it does not determine time jets uniquely from the restricted values alone. The convention does not require a fixed uniform enlargement of an unbounded I.

Vector witness. Let U⊆R3 be open and star-shaped with centre a∈U (Star-shaped open subsets of Euclidean space), let I⊆R be an interval, let k≥1, and let B:U×I→R3 be jointly Ck with div⁡B(⋅,t)=0 on U for every t∈I (Divergence and curl of a C1 vector field). Then

A(x,t):=∫01s B(a+s(x−a),t)×(x−a) ds

is jointly Ck on U×I and

curl⁡A(⋅,t)=B(⋅,t)(t∈I),

the curl being taken in the spatial variables and the cross product being that of The cross product in R3.

Scalar witness. If in addition C:U×I→R3 is jointly Ck with curl⁡C(⋅,t)=0 on U for every t∈I, then

ϕ(x,t):=−∫01⟨C(a+s(x−a),t), x−a⟩ ds

is jointly Ck on U×I and

∇ϕ(⋅,t)=−C(⋅,t)(t∈I),

the gradient and inner product being those of Directional derivatives and partial derivatives of a map U⊆Rm→Rn and The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. For jointly C∞ data both potentials are jointly C∞. No global potential on a non-star-shaped open set is claimed, and the restrictions are continuous at included endpoints by joint regularity.

Facts & Assumptions

Given: An open star-shaped U⊆R3 with centre a; an interval I⊆R; an integer k≥1; a jointly Ck field B:U×I→R3 with div⁡B(⋅,t)=0 for every t; and, for the scalar witness, a jointly Ck field C:U×I→R3 with curl⁡C(⋅,t)=0 for every t.

[F1]

Star-shapedness: a+s(x−a)∈U for every x∈U and 0≤s≤1 (Star-shaped open subsets of Euclidean space).

[F2]

Radial vector potential: if U⊆R3 is open and star-shaped with centre a and B:U→R3 is C1 divergence-free, then A(x)=∫01t B(a+t(x−a))×(x−a) dt is a vector potential for B in the sense of Vector potentials of a continuous field on an open subset of R3: it is C1 and curl⁡A=B (A divergence-free C1 field on a star-shaped open subset of R3 has a vector potential).

[F3]

Poincaré lemma and closedness: on a star-shaped open set every closed C1 field F (that is, ∂jFi=∂iFj) is exact, with C2 potential ϕ(x)=∫01⟨F(a+t(x−a)),x−a⟩ dt; exactness means F=∇ϕ (Poincare's lemma on a star-shaped domain: every closed C1 field is exact, Exact and closed C1 vector fields).

[F5]

Differentiation under the integral over a compact parameter interval: if g,h:[c,d]×[α,β]→R are continuous and g(⋅,s) is differentiable with derivative h(⋅,s) for each s, then x↦∫αβg(x,s) ds is differentiable with derivative ∫αβh(x,s) ds, relative and one-sided at the endpoints (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral).

[F6]

Uniform bounds: continuous real functions on compact metric spaces are uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous), closed bounded subsets of Rm are compact (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and ∣∫uvf−∫uvg∣≤η∣v−u∣ when ∣f−g∣≤η on the interval (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[F7]

Fundamental theorem: if G is continuous on [u,v] and differentiable on (u,v) with Riemann integrable derivative, then ∫uvG′=G(v)−G(u) (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative).

Proof

technique · direct; extend the data to a neighbourhood of $U\times I$, prove joint regularity of the radial integrals on an open radial-admissibility set by induction on the derivative order using differentiation under the integral sign, and verify the spatial field identities slice by slice
1.1givenF1F4F6algebra

Extension and radial-admissibility. By the joint Ck convention there are an open neighbourhood N⊆R4 of U×I and a Ck map B~:N→R3 with B~∣U×I=B; for the scalar witness take N to be the intersection of the two neighbourhoods and let C~ be the corresponding Ck extension, which exists on that intersection. Put P:={(x,t)∈N: (a+s(x−a),t)∈N for every s∈[0,1]}; then U×I⊆P by [F1], and P is open: if (x,t)∈P, the radial image R:={(a+s(x−a),t):0≤s≤1} is compact and contained in the open set N, use the indexed cover of all balls B(q,ε) with q∈R, ε>0 and B(q,2ε)⊆N. Openness supplies at least one at each point, without selecting a radius function. Take a finite subcover and put ρ equal to the minimum of its finitely many positive radii. Every point within distance ρ of a point of R then belongs to N. The map Φ(x′,t′,s)=(a+s(x′−a),t′) satisfies ∥Φ(x′,t′,s)−Φ(x,t,s)∥2≤∥x′−x∥2+∣t′−t∣ uniformly in s, so for ∥x′−x∥2+∣t′−t∣<ρ every point of the radial image of (x′,t′) lies within distance ρ of R, hence in N, and (x′,t′)∈P.

2.1step 1.1F4F5F6algebra

Joint regularity of the radial integrals on P: for every compact box Q×J⊆P and every multi-index α in the variables (x,t) with ∣α∣≤k, the partial derivative ∂α of the integral A~(x,t):=∫01s B~(a+s(x−a),t)×(x−a) ds exists on Q×J and equals ∫01∂α[s B~(a+s(x−a),t)×(x−a)]ds, and this is jointly continuous on Q×J; the same holds for Ψ~(x,t):=∫01⟨C~(a+s(x−a),t),x−a⟩ ds. Indeed, on the compact set (Q×J)×[0,1] every radial argument (a+s(x−a),t) lies in a compact subset of N, and each integrand is a sum of products of the Ck functions B~, C~ composed with the smooth map (x,t,s)↦(a+s(x−a),t), the polynomial factor s and the polynomial vector x−a, hence is Ck jointly in (x,t,s) by [F4]. Induction on ∣α∣: for ∣α∣=0 joint continuity of the s-integral follows from uniform continuity of the integrand on the compact product and [F6]; if the claim holds for α, then differentiating the integral in one further spatial or time variable under the integral sign is legitimate by [F5] applied with parameter s∈[0,1] and the other variables frozen at values in Q×J, and the resulting integral is again jointly continuous by the same uniform-continuity argument; at an included endpoint of a nondegenerate J this gives the relative derivative, as stated there.

3.1step 1.1step 2.1algebra

Restriction to U×I: since U×I⊆P and the extensions agree with B and C there, the maps A=A~∣U×I and −Ψ~∣U×I are the restrictions of the Ck maps A~ and −Ψ~ on the open set P, hence are jointly Ck on U×I; the restricted integrals are exactly the formulas displayed in the Statement because B~∣U×I=B and C~∣U×I=C.

4.1step 3.1F2F3F4F5F7algebra

Field identities at each fixed time. Fix t∈I. The slice B(⋅,t) is C1 on U (as k≥1) and divergence-free, and A(⋅,t) coincides with the radial witness of [F2] for that slice; hence curl⁡A(⋅,t)=B(⋅,t) by [F2]. For the scalar witness put ΨC(x,t):=∫01⟨C(a+s(x−a),t),x−a⟩ ds, so that ϕ=−ΨC; the slice C(⋅,t) is C1 and closed on U. Differentiating ΨC in xj under the integral sign, legitimate by step 2.1 with ∣α∣=1, gives ∂jΨC(x,t)=∫01[Cj(a+s(x−a),t)+s∑i(x−a)i ∂jCi(a+s(x−a),t)]ds; closedness of the slice replaces ∂jCi by ∂iCj, so the integrand is dds[s Cj(a+s(x−a),t)] by the chain and product rules [F4], and the fundamental theorem [F7] gives ∂jΨC(x,t)=Cj(x,t)−0⋅Cj(a,t)=Cj(x,t); hence ∇ϕ(⋅,t)=−C(⋅,t).

5.1step 4.1algebra

Jointly C∞ data: if B and C are jointly C∞, choose one smooth extension of each on a common neighbourhood, repeat steps 1.1 through 4.1 for every k, and conclude that the same radial integrals are jointly Ck on U×I for every k, hence jointly C∞, with the same field identities at every t.

6.1step 3.1step 4.1step 5.1algebra∎

Degenerate and unbounded time domains: for I=∅ the statements over U×I are vacuous; for I={t0} the constant-in-time extensions B~(x,t)=B(x,t0) and C~(x,t)=C(x,t0) are Ck on U×R and the construction of steps 1.1 through 4.1 produces the two potentials with the stated spatial identities at t0, while the values of the extension away from t0 are not determined by the original data, so no unique time jets are asserted; for unbounded I the set P is constructed from the given neighbourhood N of U×I without any fixed uniform enlargement, and the argument applies verbatim. Hence both witnesses are jointly Ck with the stated identities, which is the Statement.

Remarks

  • The corrected sign convention is the one proved here: the positive radial integral ∫01⟨C,x−a⟩ ds has gradient +C, so the scalar witness is its negative; the vector identity curl⁡A=B keeps the positive sign because the cross product and curl conventions are the ones of the cited definitions. For C=e1 the formula gives ϕ=−(x1−a1) and ∇ϕ=−e1, as required.
  • The construction is local on a star-shaped U even when the surrounding open set has holes: on a ball the radial segment stays inside the ball. No global potential on an arbitrary open set and no uniqueness of the potentials are asserted.
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Transport of flux through a moving patch

Statement

Assume the Axiom of Countable Choice (ACω, The Axiom of Countable Choice (ACω)). In the setting of Moving parametrized surface patches and their flux: D⊆R2 is a finite elementary Green region with positive boundary chain ∂D, I⊆R is an interval, r:D×I→R3 extends to a C2 map on a neighbourhood of D×I with ru×rv≠0 on D∘×I∘, (W,v) is a C1 velocity field with v(r(p,t),t)=rt(p,t), and B is a C1 field on an open set containing the swept graph Σ={(r(p,t),t):p∈D,t∈I}⊆R4, with flux

Φ(t)=∫D⟨B(r(p,t),t), ru(p,t)×rv(p,t)⟩ dp.

For nondegenerate I, Φ is differentiable at interior times and relatively differentiable at included endpoints. Empty I gives no flux values, and singleton I gives one flux value but no derivative determined by that value. If the spatial divergence div⁡B=0, then for every time of such a nondegenerate I whose slice has rank two, that is ru(p,t)×rv(p,t)≠0 for all p∈D∘,

Φ′(t)=∫S(t)∂tB⋅n dS−∮∂S(t)(v×B)⋅dr.

By the standing hypothesis the slice has rank two at every interior time. At an included endpoint the displayed normal notation is used only if its slice also has rank two. Here S(t) means the parametrized family (D,r(⋅,t)), and n means the parameter normal N(p,t)=(ru×rv)/∥ru×rv∥2 on D∘. Define ∫S(t)H⋅n dS:=∫D⟨H(r(p,t),t),ru×rv⟩ dp, including multiplicity; similarly ∫S(t)(v⋅n)div⁡B dS:=∫D⟨v(r(p,t),t),ru×rv⟩div⁡B(r(p,t),t) dp. These agree with the ordinary flux definitions when the slice meets Regular parametrized surface patches on compact Jordan parameter regions (through Moving parametrized surface patches and their flux); no image-wide normal or injectivity is assumed. If ∂D=(σ1,…,σm), define ∮∂S(t)G⋅dr:=∑ℓ=1m∫r(σℓ,t)G(⋅,t)⋅dr using Positive orientation of elementary-region boundaries and Scalar line integrals with respect to arc length and vector-field line integrals. Without the hypothesis div⁡B=0 the additional term

∫S(t)(v⋅n) div⁡B dS

appears on the right-hand side. The identity is local in t and is a statement about the parametrized mathematical object; no electromagnetic interpretation is asserted.

Facts & Assumptions

Given: The data (D,I,r,v,B,Φ) of the Statement, a nondegenerate I and times t∈I that are interior or whose slice has rank two (the two cases needed below), and the notation w:=ru×rv, N:=w/∥w∥2 on D∘ at a rank-two slice, and v~(p,t):=rt(p,t)=v(r(p,t),t).

[F1]

Moving patch: the flux is the integral displayed in the Statement over the compact Jordan measurable region D; the velocity field is C1 on an open spacetime neighbourhood of the swept graph and satisfies v(r(p,t),t)=rt(p,t); r extends C2 to a neighbourhood of D×I (Moving parametrized surface patches and their flux).

[F2]

Differentiation under the integral sign over the compact measurable set D: if a real-valued function and its partial t-derivative are continuous on a neighbourhood of D×[a,b], then t↦∫Df(p,t) dp is differentiable with the stated relative endpoint derivatives and derivative ∫D∂tf(p,t) dp (Differentiation under the integral sign over a compact measurable set and over a compact C1 surface). To apply its Lebesgue formulation to this Riemann flux, use Borel Darboux integrands in finite dimension on the bounded Borel zero extension from the compact Jordan set D to a bounding box; it is Riemann integrable by the flux definition and the same is true of the continuous derivative integrand.

[F4]

Cross product and mixed product: u×v is bilinear and alternating and ⟨u×v,w⟩=det⁡[u v w], so the scalar triple product is multilinear and alternating in its three arguments (The cross product in R3, The cross product is bilinear, alternating, and orthogonal to both factors); the Euclidean inner product is that of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

[F5]

Rigidity of alternating column-multilinear matrix functions: if F:M3(R)→R is column-multilinear in the sense of Column-multilinear, alternating, normalized and antisymmetric functions on square matrices over a commutative ring and alternating (it vanishes whenever two columns are equal), then F(M)=F(I3)det⁡(M) for every M; in particular a column-multilinear alternating F with F(I3)=0 vanishes identically (Every alternating multilinear F satisfies F(A)=F(I)∑σ∈Snsgn⁡(σ)∏iaσ(i),i).

[F6]

Cross-product differential identity: for C1 fields F,G, curl⁡(F×G)=(div⁡G)F−(div⁡F)G+DF G−DG F with divergence and curl of Divergence and curl of a C1 vector field (The divergence and curl of a cross product).

[F7]

Parametric Stokes prerequisites, with no rank or injectivity hypothesis: The curl flux integrand of a C2 patch is a two-dimensional curl of the pulled-back field gives ∂uQ∗−∂vP∗=⟨curl⁡G∘φ,φu×φv⟩ for P∗=⟨G∘φ,φu⟩, Q∗=⟨G∘φ,φv⟩ on an open parameter neighbourhood. Green's theorem for finite unions of elementary regions integrates this over any supplied finite elementary Green region, and A vector line integral along an image arc is the parameter line integral of the pulled-back field identifies each resulting boundary arc integral with its image line integral.

[F8]

Surface integrals: for a regular patch, ∫Sq dS=∫D(q∘φ)Jφ with area density Jφ=∥φu×φv∥2, and the flux in the induced orientation is ∫D⟨F∘φ,φu×φv⟩ (Surface area and scalar surface integrals on a regular patch, The surface area density is the norm of the cross product of the parameter tangents, Unit normal fields, orientations, and flux through a regular surface patch).

Proof

technique · direct; differentiate the parametrized flux under the integral sign, express the derivative of the area vector by the determinant identity, convert the resulting curl term into a boundary circulation by Stokes, and identify each term with the corresponding surface integral
1.1givenF1F3algebra

Preliminaries. The integrand h(p,t):=⟨B(r(p,t),t),w(p,t)⟩ is defined for (p,t) in a neighbourhood of D×I and is continuous there together with ∂th: r and its first derivatives are continuous, B is C1 on an open spacetime neighbourhood of the swept graph, and sums, products and compositions of continuous maps are continuous. Moreover v~(p,t)=rt(p,t)=v(r(p,t),t) is C1 in (p,t), so Dxv denotes the spatial Jacobian of the velocity field, evaluated at r(p,t), and div⁡v its spatial divergence; at an interior time or a rank-two endpoint slice the vector w=ru×rv does not vanish on D∘, so the unit normal N=w/∥w∥2 is defined there.

1.2givenF4F5algebra

Algebraic identity for the area vector. For every 3×3 real matrix A and all B0,u1,u2∈R3, ⟨B0,(Au1)×u2⟩+⟨B0,u1×(Au2)⟩=tr⁡(A)⟨B0,u1×u2⟩−⟨AB0,u1×u2⟩. To prove it, let M=[m1 m2 m3] range over 3×3 matrices and put E(M):=⟨m1,(Am2)×m3⟩+⟨m1,m2×(Am3)⟩−tr⁡(A)⟨m1,m2×m3⟩+⟨Am1,m2×m3⟩. Each of the four terms is linear in each column separately: the triple product is trilinear by [F4] and multiplication by the fixed matrix A is linear, so E is column-multilinear. Writing the four terms as the determinants T1=det⁡[m1,Am2,m3], T2=det⁡[m1,m2,Am3], T3=−tr⁡(A)det⁡[m1,m2,m3] and T4=det⁡[Am1,m2,m3] by [F4], each interchange of two of the columns m1,m2,m3 carries every Ti to ± one of the four terms in such a way that the total reverses sign: for example exchanging the first two columns sends T1 to −T4, T4 to −T1, and T2,T3 to their own negatives, so E↦−E; the other two interchanges are verified by the same determinant expansion. Hence E is alternating as well. Therefore [F5] gives E(M)=E(I3)det⁡(M) for every M, and E(I3)=A22+A33−tr⁡(A)+A11=0, because ⟨e1,Ae2×e3⟩=A22, ⟨e1,e2×Ae3⟩=A33 and ⟨Ae1,e2×e3⟩=A11 computed from the coordinate formula of [F4]. Hence E≡0, which is the displayed identity at (B0,u1,u2)=(m1,m2,m3).

2.1step 1.1F2algebra

Differentiation of the flux. Since h and ∂th are continuous on a neighbourhood of D×I, use [F2] on any local compact time segment with K=D; its Riemann/Lebesgue bridge identifies both parameter integrals and gives that Φ is differentiable on I, with relative one-sided derivatives at included endpoints, and Φ′(t)=∫D∂th(p,t) dp.

2.2step 1.2F3F4algebra

The derivative of the integrand. Since rt=v~ and v~(p,t)=v(r(p,t),t), the chain rule [F3] gives ∂tru=∂urt=(Dxv)ru and ∂trv=(Dxv)rv, so, using equality of mixed partials for the C2 map r, ∂tw=((Dxv)ru)×rv+ru×((Dxv)rv); and ∂t[B(r(p,t),t)]=∂tB(r(p,t),t)+(DxB)rt by the chain rule. Hence, multiplying out and evaluating the spatial derivatives at r(p,t), ∂th=⟨∂tB+(DxB)v~,w⟩+⟨B,∂tw⟩=⟨∂tB+(DxB)v+(div⁡v)B−(Dxv)B, w⟩, the last equality by step 1.2 with A:=Dxv, B0:=B, (u1,u2):=(ru,rv).

3.1step 2.2F6algebra

Conversion by the cross-product identity. By [F6] with F:=v and G:=B, curl⁡(v×B)=(div⁡B)v−(div⁡v)B+(Dv)B−(DB)v, where (Dv)B=(Dxv)B and (DB)v=(DxB)v; substituting into step 2.2 gives ∂th=⟨∂tB+v div⁡B−curl⁡(v×B), w⟩.

4.1step 2.1step 3.1F7F8algebra∎

Parametric Stokes and conclusion. Fix a time t of the nondegenerate I. The spatial slices of v,B are C1 on the open slice of W∩V containing r(D,t); hence G=(v×B)(⋅,t) is C1 there. Restrict the C2 extension of φ=r(⋅,t) to an open neighbourhood O⊇D mapping into that slice. The first supplier in [F7] makes P∗,Q∗ C1 on O with the displayed curl integrand. Green then gives its integral over D as the positive boundary integral, and the pullback supplier identifies each arc integral with ∫φ∘σℓG⋅dr. Thus ∫D⟨curl⁡(v×B)(r,t),w⟩ dp=∮∂S(t)(v×B)⋅dr without requiring a published regular patch. Substituting into steps 2.1 and 3.1 gives the always-defined parameter identity Φ′(t)=∫D⟨∂tB(r,t)+v(r,t)div⁡B(r,t),w⟩ dp−∮∂S(t)(v×B)⋅dr. At a rank-two slice its first two integrals are exactly the surface shorthand defined in the Statement (and, for an ordinary regular patch, [F8]); if div⁡B=0 the middle term vanishes. This proves the stated identity with the minus sign, including relative endpoint derivatives whenever the normal shorthand applies.

Remarks

  • Sign bookkeeping. The boundary term appears with a minus sign because Stokes converts the flux of the curl into the circulation in the induced orientation; the additional term (v⋅n)div⁡B vanishes whenever the field is divergence-free. The example on the companion page checks both signs by direct computation.
  • Hypotheses used. The C2 regularity of r is used for the equality of mixed partials, the single-valued C1 velocity makes v~ well defined, and the nonvanishing of ru×rv on the interior identifies the orientation; the velocity is given on one open neighbourhood of the whole swept graph, and boundary chains may have the corners permitted by the finite elementary decomposition.
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Period averages of products of real parts of complex exponentials

Statement

Let ω>0 and T:=2π/ω>0.

Scalar form. Let a,d∈C. Then

1T∫0TRe⁡(ae−iωt)Re⁡(de−iωt) dt=12Re⁡(ad‾).

Vector forms. Let a,d∈C3, write the same symbols for the componentwise complex-bilinear extensions of the Euclidean inner product and of the cross product of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn and The cross product in R3 to C3, and let Re⁡ act componentwise. In coordinate formulas, ϵijk is +1 for even permutations of (1,2,3), −1 for odd permutations and zero for repeated indices; all displayed repeated-index sums run from 1 to 3. Then

1T∫0T⟨Re⁡(ae−iωt),Re⁡(de−iωt)⟩ dt=12Re⁡(⟨a,d‾⟩),1T∫0TRe⁡(ae−iωt)×Re⁡(de−iωt) dt=12Re⁡(a×d‾),

the integrals of the real vector-valued integrands being componentwise. The real-valued functions Re⁡(ae−iωt) are the time-dependent factors of the plane-wave fields of Complex plane-wave fields: real parts of complex exponentials at a fixed position, with the phase absorbed into a and d; no electromagnetic interpretation is asserted. Real quadratic expressions may not be replaced by complex quadratic expressions without conjugation: the average is not 12Re⁡(ad) in general. For a=d=1+i the proved average is 12∣1+i∣2=1, whereas 12Re⁡((1+i)2)=0.

Facts & Assumptions

Given: A real ω>0, T=2π/ω; complex numbers a,d∈C and, in the vector forms, a,d∈C3 with the componentwise conventions of the Statement.

[F1]

Complex exponential and its calculus: exp⁡z=∑n≥0zn/n! has exp⁡0=1, is entire with exp⁡′=exp⁡, and satisfies the addition law exp⁡(z+w)=exp⁡zexp⁡w; for real x,y one has exp⁡(x+iy)=ex(cos⁡y+isin⁡y) and ∣exp⁡(x+iy)∣=ex, so ∣eiθ∣=1 for real θ (The complex exponential by its power series, The complex exponential is entire and its complex derivative is itself, exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential, exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0).

[F2]

Coordinates, conjugation and modulus: z=a+bi has Re⁡z=a, Im⁡z=b, z‾=a−bi (Real and imaginary parts, complex conjugation, and modulus); conjugation is a real-field automorphism with z+w‾=z‾+w‾, zw‾=z‾ w‾, z‾‾=z, and zz‾=∣z∣2 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive); in particular i‾=−i, r‾=r for real r, and the field axioms of C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2) hold. Consequently, for real θ, eiθe−iθ=e0=1 by [F1] and ∣eiθ∣=1 by [F1], so eiθ‾=eiθ‾⋅eiθe−iθ=∣eiθ∣2e−iθ=e−iθ; replacing θ by −θ gives e−iθ‾=eiθ.

[F3]

Kernel of the exponential: ker⁡(exp⁡)=2πiZ, so e2πin=1 for every n∈Z; in particular e−4πi=1 (ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ).

[F4]

Real and imaginary parts of differentiable paths: identifying C with R2, the maps Re⁡,Im⁡:C→R are real-linear, so for a differentiable path z:J→C on an interval J the functions t↦Re⁡z(t) and t↦Im⁡z(t) are differentiable with derivatives Re⁡z′(t) and Im⁡z′(t) by the chain rule (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0); in particular ddte−2iωt=−2iω e−2iωt, since t↦−2iωt is real-linear with derivative −2iω and exp⁡′=exp⁡ by [F1].

[F5]

Fundamental theorem of calculus: if G:[α,β]→R is continuous on [α,β] and differentiable on (α,β), and f is Riemann integrable on [α,β] with f=G′ on (α,β), then ∫αβf=G(β)−G(α) (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative).

[F6]

Additivity of the real part: Re⁡(u+v)=Re⁡u+Re⁡v for u,v∈C, hence the same for every finite sum, and Re⁡(λu)=λRe⁡u for real λ; these are the coordinate rules of the real-linear map Re⁡ of [F2].

Proof

technique · direct; expand each real part with $\operatorname{Re}z=(z+\overline z)/2$, exhibit an explicit antiderivative whose real part has the integrand as derivative, and evaluate it by the fundamental theorem using the period $2\pi i$ of the exponential
1.1givenF2F6algebra

Scalar algebraic identity. Let a,d∈C and t∈R. By [F2], Re⁡(ae−iωt)=12(ae−iωt+a‾ e−iωt‾)=12(ae−iωt+a‾ eiωt), and likewise Re⁡(de−iωt)=12(de−iωt+d‾ eiωt). Multiplying the two real expressions and using the conjugation rules of [F2] throughout gives 4Re⁡(ae−iωt)Re⁡(de−iωt)=ad e−2iωt+ad‾+a‾ d+a‾d‾ e2iωt. Since ad e−2iωt‾=a‾d‾ e2iωt and ad‾‾=a‾ d by [F2], the first and fourth terms are conjugate, as are the second and third, so the last display equals 2Re⁡(ad e−2iωt)+2Re⁡(ad‾)=2Re⁡(ad e−2iωt+ad‾) by [F6]; hence Re⁡(ae−iωt)Re⁡(de−iωt)=12Re⁡(ad e−2iωt+ad‾).

2.1step 1.1F1F4algebra

Antiderivative. Define Φ(t):=ad e−2iωt−2iω+ad‾ t for t∈R; the division is by the nonzero complex number −2iω, and Φ is differentiable with Φ′(t)=ad e−2iωt+ad‾ by [F1], [F4] and the algebra of derivatives: the derivative of e−2iωt is −2iω e−2iωt because exp⁡′=exp⁡ ([F1]) and the parameter map is real-linear. Hence the real function G:=Re⁡Φ is continuous on [0,T] and differentiable on (0,T) with derivative G′(t)=Re⁡Φ′(t)=Re⁡(ad e−2iωt+ad‾)=2Re⁡(ae−iωt)Re⁡(de−iωt) by [F4] and step 1.1; in particular G′ is continuous, hence Riemann integrable on [0,T].

3.1step 2.1F3F5F6algebra

Scalar identity. Since 2ωT=4π, the kernel computation of [F3] gives e−2iωT=e−4πi=1; therefore Φ(T)=ad−2iω+ad‾ T and Φ(0)=ad−2iω, so Φ(T)−Φ(0)=ad‾ T. The fundamental theorem [F5] applied to G and G′ gives ∫0TG′=G(T)−G(0)=Re⁡(Φ(T)−Φ(0))=TRe⁡(ad‾) by [F6]. Since G′=2Re⁡(ae−iωt)Re⁡(de−iωt) by step 2.1, dividing the identity ∫0TG′=TRe⁡(ad‾) by 2T>0 gives the scalar form of the Statement.

4.1step 2.1step 3.1F5F6algebra∎

Vector forms. Let a,d∈C3 and F(t):=Re⁡(ae−iωt), G(t):=Re⁡(de−iωt). For every j,k∈{1,2,3}, the already proved scalar identity of step 3.1 gives T−1∫0TFj(t)Gk(t) dt=12Re⁡(ajdk‾). The coordinate inner product and cross product use finite sums: ⟨F,G⟩=∑j=13FjGj and (F×G)i=∑j,k=13ϵijkFjGk. Each product has the explicit real antiderivative constructed in step 2.1 for its pair (aj,dk); summing those finitely many antiderivatives and applying [F5] proves that their integrals sum with the same coefficients. Hence T−1∫0T⟨F,G⟩ dt=12Re⁡∑j=13ajdj‾=12Re⁡⟨a,d‾⟩, and T−1∫0T(F×G)i dt=12Re⁡∑j,k=13ϵijkajdk‾=12Re⁡(a×d‾)i by real linearity of [F6]. This proves both componentwise vector formulas, including all zero amplitudes.

Remarks

  • The conjugation in 12Re⁡(ad‾) is essential and cannot be absorbed into the amplitudes: at t=0 with a=d=1+i one has Re⁡(a)Re⁡(d)=1 while 12Re⁡(ad)=12Re⁡(2i)=0, so the real quadratic form and the complex square differ even pointwise.
  • The proof uses no choice principle: only the power-series exponential, its kernel, the fundamental theorem and real-linear algebra of Re⁡ enter, and the integrand is a continuous real function on a compact interval.
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The two-equation interface system at normal incidence

Statement

Let εj,μj>0 be real for j=1,2, put Zj:=μj/εj for the nonnegative square root of Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, and let Ei,Er,Et∈C with Ei≠0 satisfy the two linear equations

Ei+Er=Et,Ei−ErZ1=EtZ2.

Then

ErEi=Z2−Z1Z2+Z1,EtEi=2Z2Z1+Z2,

and the real numbers R:=∣Er/Ei∣2 and T:=(Z1/Z2)∣Et/Ei∣2 satisfy

R+T=1.

The two displayed equations are assumptions, not derived from any wave or field model; the ratio definitions of R and T are algebraic, and no electromagnetic validity is asserted. Oblique incidence, complex impedance and anisotropic media require further systems and branch conventions and are outside this statement.

Facts & Assumptions

Given: Real numbers εj,μj>0 for j=1,2, the square roots Zj=μj/εj, and complex numbers Ei,Er,Et with Ei≠0 satisfying Ei+Er=Et and (Ei−Er)/Z1=Et/Z2.

[L1]

Square roots and positivity: for real x≥0 there is a unique s≥0 with s2=x, written x, and x>0 implies x>0; also x2=x for x≥0 (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}). Order and field arithmetic in R are those of Complete ordered field (least-upper-bound property): sums and products of positive reals are positive, a positive real is nonzero, and R⊆C under the embedding of The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i, on which the field operations of C restrict to those of R (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2)).

[L2]

For a real r, the coordinate definition ∣r+0i∣=r2 and [L1] give ∣r∣2=r2 (Real and imaginary parts, complex conjugation, and modulus). This is the only modulus property needed below.

Proof

technique · direct; solve the two linear equations by substitution, then substitute the resulting ratios into the algebraic expressions defining $R$ and $T$
1.1givenL1algebra

Positivity and invertibility. Since μj>0 and εj>0, the quotients μj/εj are positive by [L1], so Zj=μj/εj>0 for j=1,2; in particular Z1,Z2≠0 and Z1+Z2>0, so Z1+Z2, Z1Z2 and Ei are invertible complex numbers.

2.1step 1.1L1algebra

Solving the system. Substituting the first equation Et=Ei+Er into the second gives Ei−ErZ1=Ei+ErZ2; multiplying by the nonzero number Z1Z2 and collecting terms yields Z2Ei−Z2Er=Z1Ei+Z1Er, that is, Er(Z1+Z2)=Ei(Z2−Z1). Dividing by the nonzero numbers Ei and Z1+Z2 gives ErEi=Z2−Z1Z1+Z2, and then EtEi=1+ErEi=Z1+Z2+Z2−Z1Z1+Z2=2Z2Z1+Z2 by the field arithmetic of [L1]. Both ratios are real numbers, since all Zj are real.

3.1step 2.1L1L2algebra∎

The balance R+T=1. Because the ratios of step 2.1 are real, [L2] gives ∣Er/Ei∣2=(Z2−Z1Z1+Z2)2 and ∣Et/Ei∣2=(2Z2Z1+Z2)2, so R=(Z2−Z1)2(Z1+Z2)2 and T=Z1Z2⋅4Z22(Z1+Z2)2=4Z1Z2(Z1+Z2)2, and therefore R+T=(Z2−Z1)2+4Z1Z2(Z1+Z2)2=Z22−2Z1Z2+Z12+4Z1Z2(Z1+Z2)2=(Z1+Z2)2(Z1+Z2)2=1 by the expansion (Z1+Z2)2=Z12+2Z1Z2+Z22 and the invertibility of Z1+Z2 from step 1.1.

Remarks

  • In the period-averaged reading of the amplitudes as time-harmonic complex phasors, Period averages of products of real parts of complex exponentials with a=d=A gives 1T0∫0T0Re⁡(Ae−iωt)2 dt=12∣A∣2 for ω>0 and T0=2π/ω, so ∣Er/Ei∣2 is the ratio of the mean squares of the real reflected and incident fields and T differs from the corresponding transmitted mean-square ratio by the real factor Z1/Z2; the identity R+T=1 itself is the algebraic consequence proved above and does not rest on any field model.
  • Neither the system nor the conclusion involves AC or any choice principle; it is finite algebra in R and C. The equations are satisfied by Et=Ei, Er=0 exactly when Z1=Z2, in which case R=0, T=1.

5 · Examples, experiments and counterexamples

None yet.

Sources