30.5. GREEN’S THEOREM 1075

Define P1 : Rn→ Rn−1 byP1u≡ (u2, · · · ,un)

and Σ : Rn→ Rn given byΣu≡ u−g(P1u)e1

≡ u−g(u2, · · · ,un)e1

≡ (u1−g(u2, · · · ,un) ,u2, · · · ,un)

Thus Σ is invertible andΣ−1u = u+g(P1u)e1

≡ (u1 +g(u2, · · · ,un) ,u2, · · · ,un)

For x∈ ∂Ω∩Q, it follows the first component of Rx is g(P1 (Rx)). Now define R :W →Rn≤

asu≡ Rx≡ Rx−g(P1 (Rx))e1 ≡ ΣRx

and so it followsR−1 = R∗Σ−1.

These mappings R involve first a rotation followed by a variable sheer in the direction ofthe u1 axis.

Since ∂Ω is compact, there are finitely many of these open sets, Q1, · · · ,Qp whichcover ∂Ω. Let the orthogonal transformations and other quantities described above also beindexed by k for k = 1, · · · , p. Also let Q0 be an open set with Q0 ⊆ Ω and Ω is coveredby Q0,Q1, · · · ,Qp. Let u≡ R0x≡ x− ke1 where k is large enough that R0Q0 ⊆ Rn

<. Thusin this case, the orthogonal transformation R0 equals I and Σ0x ≡ x− ke1. I claim Ω is anoriented manifold with boundary and the charts are (Wi,Ri) .

Letting A be an open set contained in Ri (Wi∩Wj) such that ∂A has measure 0 and ∂Aseparates Rn into two components, consider

d(u,A,R j ◦R−1

i), u /∈ R j ◦R−1

i (∂A)

By convolving g with a mollifier, there exists a sequence of infinitely differentiable func-tions gε which converge uniformly to g on all of Rn−1 as ε → 0. Therefore, letting Σε bethe corresponding functions defined above with g replaced with gε , it follows the Σε willconverge uniformly to Σ and Σ−1

ε will converge uniformly to Σ−1. Thus from the abovedescriptions of R−1

j , it follows R−1jε converges uniformly to R−1

j for each j. Therefore,if ε is small enough, u /∈

(tR jε ◦R−1

iε +(1− t)R j ◦R−1i

)(∂A) and so from properties of

the degree, the mappings R j and R−1i can be replaced with smooth ones in computing the

degree. To save on notation, I will drop the ε . The mapping involved is

Σ jR jR∗i Σ−1i

and it is a one to one mapping. What is the determinant of its derivative? By the chain rule,

D(Σ jR jR∗i Σ

−1i)= DΣ j

(R jR∗i Σ

−1i)

DR j(R∗i Σ

−1i)

DR∗i(Σ−1i)

DΣ−1i

30.5. GREEN’S THEOREM 1075Define P; : R” > R"! byP\u = (u2,°+ . Un)and X: R” > R” given byLu =u-—g(P\u)eSu g(ur,-* Un) e1(uy) —g(u2,°°* Un) ,U2,°** Un)Thus © is invertible andYlu=u+g (P,u)e,=(u,+g(u2,-:: ,Un) U2, °° ,Un)For x € dQNQ, it follows the first component of Rx is g (P) (Rx)). Now define R :W — Rtasu = Rx = Rx — g (P| (Rx)) e; = URxand so it followsR'=R*r I.These mappings R involve first a rotation followed by a variable sheer in the direction ofthe uw, axis.Since dQ is compact, there are finitely many of these open sets, Q1,--- ,Q,) whichcover OQ. Let the orthogonal transformations and other quantities described above also beindexed by k for k = 1,--- ,p. Also let Qg be an open set with Qp C Q and Q is coveredby Qo, Q1,°-:,Qp. Let u= Rox = x— ke, where k is large enough that RoQo C R”. Thusin this case, the orthogonal transformation Rp equals J and Xox = x — ke,. I claim Q is anoriented manifold with boundary and the charts are (W;,R;).Letting A be an open set contained in R; (W; Wj) such that 0A has measure 0 and dAseparates IR” into two components, considerd(u,A,R;oR;'), u¢ RjoR;! (dA)By convolving g with a mollifier, there exists a sequence of infinitely differentiable func-tions ge which converge uniformly to g on all of R”~! as € + 0. Therefore, letting LZ, bethe corresponding functions defined above with g replaced with ge, it follows the Y¢ willconverge uniformly to £ and £;! will converge uniformly to 2~'. Thus from the abovedescriptions of R5', it follows Rie converges uniformly to Rj! for each j. Therefore,if € is small enough, u ¢ (*Rje oR;,' + (1—1)RjoR;') (0A) and so from properties ofthe degree, the mappings R; and R; ' can be replaced with smooth ones in computing thedegree. To save on notation, I will drop the ¢. The mapping involved is*y— |and it is a one to one mapping. What is the determinant of its derivative? By the chain rule,D (ZjRjR;E; |) = DE; (RjR7E; |) DR; (RFE; ') DR; (Z;') DEF"