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Vector forms: the boundary integrals of fn and of n×F

Statement

Let a finite gluing of elementary solid regions be given, with union E and outer boundary presentation Σout=((D1,φ1),,(DM,φM)). Vector-valued integrals below are taken coordinatewise, so that for a continuous R3-valued W on E the symbol EW denotes the vector whose kth coordinate is EWk, and for a continuous R3-valued Z on the boundary the symbol EZ denotes the vector whose kth coordinate is j=1MDjZk(φj), where n inside such an integrand is read as the oriented area vector φj,u×φj,v of the patch, exactly as in the scalar flux.

Then, for f of class C1 on an open set containing E and F of class C1 on an open set containing E,

Ef=Efn,EcurlF=En×F.

Facts & Assumptions

Given: The finite gluing with union E and outer presentation Σout, the C1 scalar f and the C1 field F, both on open sets containing E, and the coordinatewise reading of the vector integrals fixed in the Statement.

[F1]

The divergence of a C1 field is divG=i<niGi and the curl of a C1 field on an open subset of R3 is curlG=(yGzzGy, zGxxGz, xGyyGx) (Divergence and curl of a C1 vector field).

[F2]

For u=(ux,uy,uz) and v=(vx,vy,vz) in R3, u×v=(uyvzuzvy,uzvxuxvz,uxvyuyvx) (The cross product in R3).

[F4]

For scalar-valued f the gradient is f=(0f,,m1f) (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F5]

In a finite gluing the outer patches form a compatible finite patch presentation of E, over which flux is the sum of the patch values (Finite gluings of elementary solid regions and their outward boundary presentation, Finitely patched regular surfaces, their area, scalar integrals, and flux).

[L1]

For C1 fields and a C1 scalar on an open subset of Rn, div(gG)=g,G+gdivG (Divergence and curl are linear and satisfy the scalar product rules).

[L2]

For C1 fields G,H on an open subset of R3, div(G×H)=curlG,HG,curlH (The divergence and curl of a cross product).

[L3]

For a finite gluing with union E and outer presentation Σout and a C1 field G on an open set containing E, EdivG=EG,n (The divergence theorem for finite gluings of elementary solid regions).

Proof

technique · direct
1.1

Let cR3 and let c be the constant field with value c on the open set where f is C1. Its partial derivatives all vanish, so it is C1 with divc=0 and curlc=0 by [F1]. The field fc is C1 and [L1] gives div(fc)=f,c+fdivc=f,c, while its flux integrand against a vector ν is fc,ν=fc,ν by [F3].

givenF1F3L1
1.2

For all a,b,dR3, expanding both sides by [F2] and [F3] gives a×b,d=(aybzazby)dx+(azbxaxbz)dy+(axbyaybx)dz, d×a,b=(dyazdzay)bx+(dzaxdxaz)by+(dxaydyax)bz, and the six monomials of the first list are the six of the second with the same signs, matched as aybzdx with dxaybz, azbydx with dxazby, azbxdy with dyazbx, axbzdy with dyaxbz, axbydz with dzaxby and aybxdz with dzaybx. Hence a×b,d=d×a,b.

F2F3algebra
2.1

With c as in step 1.1 on the open set where F is C1, the field F×c is C1 and [L2] gives div(F×c)=curlF,cF,curlc=curlF,c.

givenF1F2L2
2.2

Apply [L3] to the field fc of step 1.1: Ef,c=j=1MDjf(φj)c,φj,u×φj,v, using [F5] to read the right side patch by patch. Take c=ek: by [F3] and [F4] the left side becomes Ekf, the kth coordinate of Ef, and the right side becomes jDjf(φj)(φj,u×φj,v)k, the kth coordinate of Efn. As k ranges over the three directions this is the first identity.

step 1.1F3F4F5L3
3.1

Apply [L3] to the field F×c of step 2.1: EcurlF,c=j=1MDjF(φj)×c,φj,u×φj,v. Step 1.2 with a=F(φj), b=c and d=φj,u×φj,v rewrites each integrand as (φj,u×φj,v)×F(φj),c. Take c=ek: by [F3] the left side becomes E(curlF)k and the right side becomes jDj((φj,u×φj,v)×F(φj))k, so as k ranges over the three directions this is the second identity.

step 2.1step 1.2F3F5L3
4.1

Steps 2.2 and 3.1 are the two asserted identities.

step 2.2step 3.1

Remarks

  • Why a constant vector is the right device. Both clauses assert an equality of vectors, and the divergence theorem produces only scalars. Pairing with a fixed c turns each vector identity into a scalar one; running c over the standard basis recovers the vector identity coordinate by coordinate, and nothing else about c is used.

  • The triple-product identity of step 1.2 is the determinant identity in disguise. By The cross product is bilinear, alternating, and orthogonal to both factors each of a×b,d and d×a,b is the determinant of the matrix with the three vectors as columns, in the orders a,b,d and d,a,b; those two orders differ by a cyclic permutation of three columns. The coordinate expansion above is the same fact written out, and it is what the proof uses.

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