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The planar divergence theorem: the flux form of Green's theorem

Statement

Let D be a finite elementary Green region with its supplied decomposition, positively oriented, and let F=(Fx,Fy) be C1 on an open UR2 containing D. Then

D(Fy)dx+Fxdy=D(xFx+yFy)dA,

the right-hand integrand being the divergence of F as a field on an open subset of R2.

Moreover, if σ:[α,β]R2 is one of the arcs of the positive boundary chain and its derivative is nowhere zero on a piece with continuous derivative extension v, then on that piece

σ(Fy)dx+Fxdy=σF,νds,ν:=(v2,v1)v2,

where ν is the unit vector obtained from the tangent v by a quarter turn clockwise.

Facts & Assumptions

Given: The finite elementary Green region D with its supplied decomposition and positive orientation, and the C1 field F=(Fx,Fy) on the open UD.

[F1]

The divergence of a C1 field on an open subset of Rn is divG=i<niGi; for n=2 and coordinates named x,y this is xGx+yGy (Divergence and curl of a C1 vector field).

[F2]

For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, written DPdx+Qdy for the field (P,Q) (Positive orientation of elementary-region boundaries).

[F3]

For a piecewise-C1 path with admissible partition and continuous derivative extensions vi, γGdr=ititi+1G(γ(t)),vi(t)dt and γhds=ititi+1h(γ(t))vi(t)2dt (Scalar line integrals with respect to arc length and vector-field line integrals, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[F4]

A finite elementary Green region is a nonempty finite union of elementary Green regions with pairwise disjoint interiors and the stated shared-arc conditions, supplied as data (Type I, Type II, and elementary regions for Green's theorem).

[F5]

For x,yRm, x,y=i<mxiyi and x2=x,x (The Euclidean inner product x,y=k<nxkyk on Rn); a map is Ck when each component is (Ck Euclidean maps and diffeomorphisms).

[L1]

Let D=D1DN be a finite elementary Green region with its supplied decomposition, oriented positively, and let P,Q be C1 on an open neighbourhood of D. Then DPdx+Qdy=D(xQyP)dA (Green's theorem for finite unions of elementary regions).

Proof

technique · direct
1.1

Put P:=Fy and Q:=Fx on U. These are C1 on U by [F5], since the components of F are, so [L1] applies with the supplied decomposition of [F4] and gives DPdx+Qdy=D(xQyP)dA.

givenF4F5L1
1.2

Fix such an arc σ of the positive boundary chain, a piece of it carrying a continuous derivative extension v=(v1,v2) with v nowhere zero there, and set ν:=(v2,v1)/v2. By [F5] the vector ν has norm 1, since v22+v12=v22, and ν,v=(v2v1v1v2)/v2=0. Writing v=v2(cosτ,sinτ) for the direction of v is not needed: the map (s,t)(t,s) is the quarter turn clockwise, as it carries (1,0) to (0,1) and (0,1) to (1,0).

givenF3F5
2.1

By step 1.1 and [F1], xQyP=xFxy(Fy)=xFx+yFy=divF on U; substituting into step 1.1 and reading the left side by [F2] gives the first asserted identity.

step 1.1F1F2
3.1

By [F3] the integral of Pdx+Qdy over that piece of σ is (P(σ(t))v1(t)+Q(σ(t))v2(t))dt=(Fy(σ(t))v1(t)+Fx(σ(t))v2(t))dt, while by step 1.2 and [F5] F(σ(t)),ν(t)v(t)2=Fx(σ(t))v2(t)Fy(σ(t))v1(t). The two integrands are equal, so by [F3] the two integrals over that piece agree, which is the second asserted identity.

step 1.1step 1.2F3F5

Remarks

  • The first identity needs no regularity of the boundary arcs; the second does. The positive boundary chain of an elementary Green region is built from continuous piecewise-C1 graphs, whose derivative may vanish, and where it vanishes there is no unit tangent and hence no ν. That is why the normal reading is a separate clause under an extra hypothesis, and why the identity that Green's theorem actually delivers is stated in the dx,dy form.

  • Outwardness of ν is not claimed here. For a positively oriented boundary the quarter turn clockwise of the tangent does point out of the region, but establishing that at a boundary point requires the same kind of local analysis that At interior base points, the graph faces of an adapted presentation induce the outward unit normal carries out in space, and it is not carried out for plane regions on this page. What is proved is the equality of the two integrals for the stated ν.

  • Why this is called a divergence theorem. The right-hand integrand is the divergence of a field on an open subset of R2, and the left-hand side is the boundary integral of the normal component. The three-dimensional statement of The divergence theorem for finite gluings of elementary solid regions has the same shape; neither is derived from the other on this page.

Depends on

Used by

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Sources