Kenneth Kuttler

492 CHAPTER 18. DIFFERENTIAL FORMS

and suppose det(Dr (u))≥ 0. Then∫r

dω =∫r([a,b])

∑j

∂α j

∂x j(x)dH p =

∫r(∂ [a,b])

α ·NdH p−1 (18.10)

whereN is an outer unit normal in the sense that the angle between the vector xul andN

is no more than 90 degrees if∂(x1,··· ,xp)∂(u1,··· ,up)

≥ 0.

Also, if you know the divergence theorem, then you can directly give the usual Calculusversion of Green’s and Stoke’s theorems from Calculus. This is developed in the exercises.

18.9 The Reynolds Transport FormulaThe Reynolds transport formula is a generalization of the formula for taking the derivativeunder an integral. It depends on the divergence theorem. I will use the chain rule ofTheorem 17.3.5 as needed.

ddt

∫ b(t)

a(t)f (x, t)dx =

∫ b(t)

a(t)

∂ f∂ t

(x, t)dx+ f (b(t) , t)b′ (t)− f (a(t) , t)a′ (t)

First is an interesting lemma about the determinant. A p× p matrix can be thought ofas a vector in Cp2

. Just imagine stringing it out into one long list of numbers. In fact, away to give the norm of a matrix is just ∑i ∑ j

∣∣Ai j∣∣2 ≡ ∥A∥2. This is called the Frobenius

norm for a matrix. It makes no difference since all norms are equivalent, but this one isconvenient in what follows. Also recall that det maps p× p matrices to C. It makes senseto ask for the derivative of det on the set of invertible matrices, an open subset of Cp2

withthe norm measured as just described because A→ det(A) is continuous, so the set wheredet(A) ̸= 0 would be an open set. Recall from linear algebra that the sum of the entrieson the main diagonal satisfies trace(AB) = trace(BA) whenever both products make sense.Indeed, trace(AB)≡ ∑i ∑ j Ai jB ji = trace(BA)

This next lemma is a very interesting observation about the determinant of a matrixadded to the identity.

Lemma 18.9.1 det(I +U) = 1+ trace(U)+ o(U) where o(U) is defined in terms ofthe Frobenius norm for p× p matrices.

Proof:

det(I +U)≡ ∑i1,i2,...,ip

sgn(i1, i2, ..., ip)(δ i11 +Ui11) · · ·(δ ip p +Uip p

)which equals det(I) added to trace(U) added to a sum of higher order terms of products ofthe Ui j. The trace(U) comes from using only one Ui j in the above product. The resultingterm will be 0 unless i = j and so the end result of these will be trace(U). Of course ifmore of the Ui j are included in the product, this yields the o(U) term. ■

Of course, by equivalence of norms, one could use any other norm for the p× p matri-ces.

492 CHAPTER 18. DIFFERENTIAL FORMSand suppose det (Dr (u)) > 0. Then[ao= [ are (w)a.” = | a:-Ndx?"! (18.10)[a,b])Xj /r(d[a,b))where N is an outer unit normal in the sense that the angle between the vector x,, and Nis no more than 90 degrees if pe > 0.Also, if you know the divergence theorem, then you can directly give the usual Calculusversion of Green’s and Stoke’s theorems from Calculus. This is developed in the exercises.18.9 The Reynolds Transport FormulaThe Reynolds transport formula is a generalization of the formula for taking the derivativeunder an integral. It depends on the divergence theorem. I will use the chain rule ofTheorem 17.3.5 as needed.d ott) bt) Ofae fonar= | 9, tax t F(t) ,t)6 (t)—f (a(t) ,t)a' (t)First is an interesting lemma about the determinant. A p x p matrix can be thought ofas a vector in CP”. Just imagine stringing it out into one long list of numbers. In fact, away to give the norm of a matrix is just 1; |Aii|” = ||A||?. This is called the Frobeniusnorm for a matrix. It makes no difference since all norms are equivalent, but this one isconvenient in what follows. Also recall that det maps p x p matrices to C. It makes senseto ask for the derivative of det on the set of invertible matrices, an open subset of cP’ withthe norm measured as just described because A — det (A) is continuous, so the set wheredet (A) 4 0 would be an open set. Recall from linear algebra that the sum of the entrieson the main diagonal satisfies trace (AB) = trace (BA) whenever both products make sense.Indeed, trace (AB) = ¥; 1; AijB ji = trace (BA)This next lemma is a very interesting observation about the determinant of a matrixadded to the identity.Lemma 18.9.1 det(/+U) = 1 +trace(U) + 0(U) where o(U) is defined in terms ofthe Frobenius norm for p x p matrices.Proof:det(I+U)= y sgn (i1,i2,..-,Up) (Si,1 +Ui,1) ee (5:,»+Ui,p)i ,2,-5tpwhich equals det (7) added to trace (U) added to a sum of higher order terms of products ofthe Uj;. The trace (U) comes from using only one Uj; in the above product. The resultingterm will be 0 unless i = j and so the end result of these will be trace(U). Of course ifmore of the U;; are included in the product, this yields the o(U) term.Of course, by equivalence of norms, one could use any other norm for the p x p matri-ces.