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Home , Integration by parts, Surface integral

94 BRUCE K. DRIVER†

8. Surfaces, and Definition 8.1. A subset M Rn is a n 1 dimensional Ck-Hypersurface if for ⊂ − n k all x0 M there exists >0 an 0 D R and a C -diffeomorphism ∈ ∈ ⊂ ψ : D B(x0,) such that ψ(D xn =0 )=B(x0,) M. See Figure 16 below. → ∩ { } ∩

Figure 16. An embedded submanifold of R2.

k n 1 C Example 8.2. Suppose V 0 R − and g : V R. Then M := Γ(g)= k ⊂ −→ (y, g(y)) : y V is a C hypersurface. To verify this assertion, given x0 = { ∈ } (y0,g(y0)) Γ(g) define ∈ ψ(y, z):=(y + y0,g(y + y0) z). − Ck Then ψ : V y0) R V R diffeomorphism { − × −→ × ψ((V y0) 0 )= (y + y0,g(y + y0)) : y V y0 = Γ(g). − ×{ } { ∈ − } n 1 Proposition 8.3 (Parametrized Surfaces). Let k 1,D 0 R − and Σ ≥ ⊂ ∈ Ck(D, Rn) satisfy (1) Σ : D M := Σ(D) is a homeomorphism and → n 1 n k (2) Σ0(y):R − R is injective for all y D. (We will call M a C — parametrized→ surface and Σ : D M a∈parametrization of M.) k n → d n Then M is a C -hypersurface in R . Moreover if f C(W 0 R , R ) is a ∈k n⊂ 1 continuous such that f(W ) M, then f C (W, R ) iff Σ− f Ck(U, D). ⊂ ∈ ◦ ∈

Proof. Let y0 D and x0 = Σ(y0) and n0 be a vector to M at x0, i.e. ∈ n0 Ran (Σ0(y0)) , and let ⊥ ψ(t, y):=Σ(y0 + y)+tn0 for t R and y D y0, ∈ ∈ − ∂ψ seeFigure17below.SinceDyψ(0, 0) = Σ0(y0) and (0, 0) = n0 / Ran (Σ0(y0)) , ∂t ∈ ψ0(0, 0) is invertible. so by the inverse function theorem there exists a neighborhood n n k V of (0, 0) R such that ψ V : V R is a C —diffeomorphism. ∈ | → PDE LECTURE NOTES, MATH 237A-B 95

Figure 17. Showing a parametrized surface is an embedded hyper-surface.

Choose an >0 such that B(x0,) M Σ(V t =0 ) and B(x0,) ψ(V ). 1 ∩ ⊂ ∩ { } ⊂ Then set U := ψ− (B(x0,)).Onefinds ψ U : U B(x0,) has the desired properties. | → d n Now suppose f C(W 0 R , R ) such that f(W ) M, a W and x0 = ∈ ⊂ ⊂ ∈ f(a) M. By shrinking W if necessary we may assume f(W ) B(x0,) where ∈ ⊂ B(x0,) is the ball used previously. (This is where we used the continuity of f.) Then

1 1 Σ− f = π ψ− f ◦ ◦ ◦ 1 where π is projection onto t =0 . Form this identity it clearly follows Σ− f is k k { } 1 k 1◦ C if f is C . The converse is easier since if Σ− f is C then f = Σ (Σ− f) is Ck as well. ◦ ◦ ◦

8.1. Surface Integrals.

n 1 n 1 Definition 8.4. Suppose Σ : D 0 R − M R is a C - parameterized n ⊂ → ⊂ hypersurface of R and f Cc(M,R). Then the surface of f over M, fdσ,is defined by ∈ M R ∂Σ(y) ∂Σ(y) fdσ= f Σ(y) det[ ,..., n(y)] dy ◦ ∂y1 | ∂yn 1 | Z Z ¯ − ¯ M D ¯ ¯ ¯ ¯ = f Σ(y) ¯det[Σ0(y)e1 ... Σ0(y)en 1 n(¯y)] dy ◦ | | | − | | DZ

n where n(y) R is a unit normal vector perpendicular of ran(Σ0(y)) for each y D. We will abbreviate∈ this formula by writing ∈

∂Σ(y) ∂Σ(y) (8.1) dσ = det[ ,..., n(y)] dy, ∂y1 | ∂yn 1 | ¯ − ¯ ¯ ¯ ¯ ¯ see Figure 18 below for the¯ motivation. ¯ 96 BRUCE K. DRIVER†

Figure 18. The approximate area spanned by Σ([y,y+dy]) should ∂Σ(y) ∂Σ(y) be equal to the area spaced by dy1 and dy2 which is equal ∂y1 ∂y2 ∂Σ(y) ∂Σ(y) to the volume of the parallelepiped spanned by dy1, dy2 ∂y1 ∂y2 and n(Σ(y)) and hence the formula in Eq. (8.1).

Remark 8.5. Let A = A(y):=[Σ0(y)e1,...,Σ0(y)en 1,n(y)]. Then − t ∂1Σ t ∂2Σ tr  .  A A = . [∂1Σ ... ∂n 1Σ n] . −  t  | | |  ∂n 1Σ   −   nt     ∂1Σ ∂1Σ ∂1Σ ∂2Σ ... ∂1Σ ∂n 1Σ 0 · · · − ∂2Σ ∂1Σ ∂2Σ ∂2Σ ... ∂2Σ ∂n 1Σ 0  ·. ·. . · . − .  = . . . . .    ∂n 1Σ ∂1Σ ∂n 1Σ ∂2Σ ... ∂n 1Σ ∂n 1Σ 0   − · − · − · −   00... 01     and therefore ∂Σ(y) ∂Σ(y) det[ ,..., n(y)] = det(A) dy = det (AtrA)dy ∂y1 | ∂yn 1 | | | ¯ − ¯ ¯ ¯ p ¯ ¯ n 1 tr ¯ ¯ = det (∂iΣ ∂jΣ) − = det (Σ0) Σ0 . · i,j=1 r h i r h i This implies dσ = ρΣ(y)dy or more precisely that

fdσ= f Σ(y)ρΣ(y)dy ◦ MZ DZ where

Σ n 1 tr ρ (y):= det (∂iΣ ∂jΣ) − = det (Σ ) Σ . · i,j=1 0 0 r h i r h i PDE LECTURE NOTES, MATH 237A-B 97

The next lemma shows that fdσis well defined, i.e. independent of how M M is parametrized. R

k n 1 C Example 8.6. Suppose V 0 R − and g : V R and M := Γ(g)= (y, g(y)) : y V as in Example 8.2. We⊂ now compute dσ−→in the parametrization Σ{: V M de∈fined} by Σ(y)=(y, g(y)). To simplify notation, let →

g(y):=(∂1g(y),...,∂n 1g(y)) . ∇ − As is standard from multivariable (and is easily verified), ( g(y), 1) n(y):= ∇ − 1+ g(y) 2 |∇ | q is a normal vector to M at Σ(y), i.e. n(y) ∂kΣ(y)=0for all k =1, 2 ...,n 1. Therefore, · −

dσ = det [∂1Σ ... ∂n 1Σ n] dy | | | − | | tr 1 In 1 ( g) = det − ∇ dy 2 g 1 1+ g(y) ¯ · ∇ − ¸¯ |∇ | ¯ ¯ ¯ ¯ q ¯ ¯ 1 In 1 0 = det − 2 dy 2 g 1 g 1+ g(y) ¯ · ∇ − − |∇ | ¸¯ |∇ | ¯ ¯ q 1 ¯ ¯ = ¯ 1+ g(y) 2 dy = 1+¯ g(y) 2dy. 2 |∇ | |∇ | 1+ g(y) q |∇ | ³ ´ Hence if g : M qR, we have → gdσ = g(Σ(y)) 1+ g(y) 2dy. |∇ | ZM ZV q Example 8.7. Keeping the same notation as in Example 8.6, but now taking n 1 2 2 n 1 V := B(0,r) R − and g(y):= r y . In this case M = S+− , the upper- ⊂ n 1 − | | hemisphere of S − , g(y)= y/gq(y), ∇ − r dσ = 1+ y 2 /g2(y)dy = dy | | g(y) q and so r gdσ = g(y, r2 y 2) dy. n 1 S − y

1 k Proof. By Proposition 8.3, φ := Σ− Σ : D˜ D is a C —diffeomorphism. n◦ 1 → By the theorem on R − with y = φ(˜y) (using Σ = Σ φ, see Figure 19) we find e ◦

˜ e f Σ(˜y)ρΣ(˜y)dy˜ = f Σ det Σ trΣ dy˜ ◦ ◦ 0 0 DZ DZ p e e e e tr e= f Σ φ det (Σe φ)0 (Σ φ)0dy˜ ◦ ◦ ◦ ◦ DZ q

= e f Σ φ det (Σ (φ)φ )tr Σ (φ)φ dy˜ ◦ ◦ 0 0 0 0 r DZ h i

= e f Σ φ det[φ0tr [Σ (φ)trΣ (φ)]φ dy˜ ◦ ◦ 0 0 0 DZ q

tr = e (f Σ φ) √det Σ0 Σ0 φ det φ0 dy˜ D ◦ ◦ · ◦ ·| | Z ³ ´ = e f Σ√det Σ0tr Σ dy. ◦ 0 ZD

Figure 19. Verifying surface integrals are independent of parametrization.

1 Definition 8.9. Let M be a C -embedded hypersurface and f Cc(M). Then we define the of f over M as ∈ n fdσ= φifdσ i=1 MZ XMZi 1 where φi C (M,[0, 1]) are chosen so that iϕi 1 with equality on supp(f) ∈ c ≤ and the supp(φif) Mi M where Mi is a subregion of M which may be viewed as a parametrized surface.⊂ ⊂ P 1 Remark 8.10. The integral fdσis well defined for if ψj Cc (M,[0, 1]) is another M ∈ satisfying the propertiesR of φi with supp(ψj) M 0 M then (using { } ⊂ j ⊂ PDE LECTURE NOTES, MATH 237A-B 99

Lemma 8.8 implicitly)

φifdσ= ψjφifdσ= ψjφifdσ i i j ij MZi MZi MiZM X X X X ∩ j0 with a similar computation showing

ψifdσ= ψjφifdσ= ψjφifdσ. j ji ij MZ Mi ZM Mi ZM X j0 X ∩ j0 X ∩ j0 Remark 8.11. By the Reisz theorem, there exists a unique Radon measure σ on M such that fdσ= fdσ. ZM ZM This σ is called surface measure on M.

Lemma 8.12 (Surface Measure). Let M be a C2 —embeddedhypersurfaceinRn and B M be a measurable set such that B¯ is compact and contained inside Σ(D) ⊂ where Σ : D M Rn is a parametrization. Then → ⊂ d σ(B) = lim m(B )= 0+m(B ) 0 d| ↓ where B := x + tn(x):x B,0 t { ∈ ≤ ≤ } and n(x) isaunitnormaltoM at x M, see Figure 20. ∈

Figure 20. Computing the surface area of B asthevolumeofan - fattened neighborhood of B.

1 k 1 n Proof. Let A := Σ− (B) and ν(y):=n(Σ(y)) so that ν C − (D, R ) if ∈ Σ Ck(D, Rn). Define ∈ ψ(y, t)=Σ(y)+tn(Σ(y)) = Σ(y)+tν(y) so that B = ψ(A [0,]). Hence by the change of variables formula × (8.2) m(B )= det ψ0(y, t) dy dt = dt dy det ψ0(y,t) | | 0 A | | A Z[0,] Z Z × 100 BRUCE K. DRIVER† so that by the fundamental theorem of calculus, d d 0+m(B )= 0+ dt dy det ψ0(y, t) = det ψ0(y,0) dy. d| d| | | | | Z0 ZA ZA But det ψ0(y,0) = det[Σ0(y) n(Σ(y))] = ρΣ(y) | | | | | which shows d 0+m(B )= ρΣ(y)dy = 1B(Σ(y))ρΣ(y)dy =: σ(B). d| ZA ZD

n 1 n Example 8.13. Let Σ = rS − bethesphereofradiusr>0 contained in R and for B Σ and α>0 let ⊂ Bα := tω : ω B and 0 t α = αB1. { ∈ ≤ ≤ } n 1 Assuming N(ω)=ω/r is the outward pointing normal to S − , we have B = B B1 =[(1+/r)B1] B1 (1+/r) \ \ and hence n m(B )=m ([(1 + /r)B1] B1)=m ([(1 + /r)B1]) m(B1)=[(1+/r) 1] m(B1). \ − − Therefore,

d n n n 1 1 n 1 1 σ(B)= 0 [(1 + /r) 1] m(B1)= m(B1)=nr − m r− B1 = r − σ(r− B), d| − r i.e. n ¡ ¢ σ(B)= m(B )=nrn 1m r 1B = rn 1σ(r 1B). r 1 − − 1 − − n Theorem 8.14. If f : R [0, ] is a ( ¡Rn , )—measurable¢ function then → ∞ B B n 1 (8.3) f(x)dm(x)= f(rω) r − drdσ(ω). Rn Z [0, )ZSn 1 ∞ × − In particular if f : R+ R+ is measurable then → ∞ (8.4) f( x )dx = f(r)dV (r) | | 0 RZn Z n 1 n 1 n where V (r)=m (B(0,r)) = r m (B(0, 1)) = n− σ S − r . n 1 Proof. Let B S − , 0

K n 1 n 1 let S − = Bi be a partition of S − with Bi small and choose wi Bi.Let i=1 ∈ 0

f(x)dx f(r ω )m((B ) (B ) ) = j i i rj+1 i rj n ∼ \ ZR X n n = f rjωi)(r r m((Bi) ) j+1 − j 1 rj+1 X ¡ n 1¢ = f(rjωi) r − dr nm((Bi)1) Zrj X rj+1 n 1 = f(rjωi)r − dr σ(Bi) rj X Z rj+1 n 1 ∼= f(rjω)dσ(ω) r − dr rj   ij Z nZ 1 X S −   ∞ n 1 ∼= f(rω)dσ(ω) r − dr. 0   n 1 Z S Z−   Eq. (8.4) is a simple special case of Eq. (8.3). It can also be proved directly as follows. Suppose first f C1([0, )) then ∈ c ∞ ∞ f( x )dx = dx drf 0(r)= dx 1 x rf 0(r)dr | |≤ | | − x − R RZn RZn Z| | RZn Z ∞ ∞ = V (r)f 0(r)dr = V 0(r)f(r)dr. − Z0 Z0 The result now extends to general f by a density argument. We are now going to work out some integrals using Eq. (8.3). The first we leave as an exercise.

Exercise 8.1. Use the results of Example 8.7 and Theorem 8.14 to show,

1 n 1 n 2 1 n 2 σ(S − )=2σ(S − ) ρ − dρ. 0 1 ρ2 Z − The result in Exercise 8.1 may be used to computep the volume of in any dimension. This method will be left to the reader. We will do this in another way. The first step will be to directly compute the following Gaussian integrals. The result will also be needed for later purposes.

Lemma 8.15. Let a>0 and

a x 2 (8.5) In(a):= e− | | dm(x).

RZn

n/2 Then In(a)=(π/a) . 102 BRUCE K. DRIVER†

Proof. By Tonelli’s theorem and induction,

a y 2 at2 In(a)= e− | | e− mn 1(dy) dt n 1 − R − R Z × n (8.6) = In 1(a)I1(a)=I1 (a). − So it suffices to compute:

a x 2 a(x2+x2) I2(a)= e− | | dm(x)= e− 1 2 dx1dx2.

RZ2 R2 Z 0 \{ } We now make the change of variables,

x1 = r cos θ and x2 = r sin θ for 0

(3) For n N, ∈ 2πn+1 2(2π)n (8.9) σ(S2n+1)= and σ(S2n)= . n! (2n 1)!! − Proof. Let In be as in Lemma 8.15. Using Theorem 8.14 we may alternatively n/2 compute π = In(1) as

n/2 ∞ n 1 r2 n 1 ∞ n 1 r2 π = In(1) = dr r − e− dσ = σ(S − ) r − e− dr. 0 0 n 1 Z S Z− Z We simplify this last integral by making the change of variables u = r2 so that 1/2 1 1/2 r = u and dr = 2 u− du. The result is

∞ n 1 r2 ∞ n 1 u 1 1/2 r e dr = u −2 e u du − − − 2 − Z0 Z0 1 ∞ n 1 u 1 (8.10) = u 2 e du = Γ(n/2). 2 − − 2 Z0 Collecting these observations implies that 1 πn/2 = I (1) = σ(Sn 1)Γ(n/2) n 2 − which proves Eq. (8.7). The computation of Γ(1) is easy and is left to the reader. By Eq. (8.10),

∞ r2 ∞ r2 Γ(1/2) = 2 e− dr = e− dr 0 Z Z−∞ = I1(1) = √π. The relation, Γ(x +1)=xΓ(x) is the consequence of integration by parts:

∞ u x+1 du ∞ x d u Γ(x +1)= e− u = u e− du u −du Z0 Z0 µ ¶ ∞ x 1 u = x u − e− du = x Γ(x). Z0 Eq. (8.8) follows by induction from the relations just proved. Eq. (8.9) is a consequence of items 1. and 2. as follows: 2π(2n+2)/2 2πn+1 2πn+1 σ(S2n+1)= = = Γ((2n +2)/2) Γ(n +1) n! and (2n+1)/2 n+1/2 n+1/2 n 2n 2π 2π 2π 2(2π) σ(S )= = = (2n 1)!! = . Γ((2n +1)/2) Γ(n +1/2) − √π (2n 1)!! 2n · −

8.2. More spherical coordinates. In this section we will define spherical coor- dinates in all dimensions. Along the way we will develop an explicit method for computing surface integrals on spheres. As usual when n =2define spherical coordinates (r, θ) (0, ) [0, 2π) so that ∈ ∞ × x1 r cos θ = = ψ2(θ, r). x2 r sin θ µ ¶ µ ¶ 104 BRUCE K. DRIVER†

For n =3we let x3 = r cos φ1 and then

x1 = ψ2(θ, r sin φ1), x2 µ ¶ as can be seen from Figure 21, so that

Figure 21. Setting up polar coordinates in two and three dimensions.

x1 r sin φ1 cos θ ψ2(θ, r sin φ1) x2 = = r sin φ1 sin θ =: ψ3(θ, φ1,r,).   r cos φ1   x3 µ ¶ r cos φ1 We continue to work inductively this way to define 

x1 . . ψn(θ, φ1,...,φn 2,rsin φn 1, )   = − − = ψn+1(θ, φ1,...,φn 2,φn 1,r). x r cos φn 1 − −  n  µ − ¶  xn+1    So for example,

x1 = r sin φ2 sin φ1 cos θ

x2 = r sin φ2 sin φ1 sin θ

x3 = r sin φ2 cos φ1

x4 = r cos φ2 and more generally,

x1 = r sin φn 2 ...sin φ2 sin φ1 cos θ − x2 = r sin φn 2 ...sin φ2 sin φ1 sin θ − x3 = r sin φn 2 ...sin φ2 cos φ1 − . .

xn 2 = r sin φn 2 sin φn 3 cos φn 4 − − − − xn 1 = r sin φn 2 cos φn 3 − − − xn = r cos φn 2. − By the change of variables formula, (8.11) ∞ f(x)dm(x)= dr dφ1 ...dφn 2dθ∆n(θ, φ1,...,φn 2,r)f(ψn(θ, φ1,...,φn 2,r)) n − − − R 0 0 φi π,0 θ 2π Z Z Z ≤ ≤ ≤ ≤ PDE LECTURE NOTES, MATH 237A-B 105 where

∆n(θ, φ1,...,φn 2,r):= det ψn0 (θ, φ1,...,φn 2,r) . − | − | Proposition 8.17. The Jacobian, ∆n is given by n 1 n 2 2 (8.12) ∆n(θ, φ1,...,φn 2,r)=r − sin − φn 2 ...sin φ2 sin φ1. − − n 1 n If f is a function on rS − — the sphere of radius r centered at 0 inside of R , then

n 1 f(x)dσ(x)=r − f(rω)dσ(ω) n 1 n 1 ZrS − ZS − (8.13) n 1 n 2 2 = r − f(ψn(θ, φ1,...,φn 2,r)) sin − φn 2 ...sin φ2 sin φ1dφ1 ...dφn 2dθ − − − 0 φi π,0 θ 2π Z ≤ ≤ ≤ ≤ Proof. We are going to compute ∆n inductively. Letting ρ := r sin φn 1 and ∂ψn ∂ψn − writing for (θ, φ1,...,φn 2,ρ) we have ∂ξ ∂ξ − ∆n+1(θ,φ1,...,φn 2,φn 1,r) − − ∂ψn ∂ψn ∂ψn ∂ψn ∂ψn ... r cos φn 1 sin φn 1 ∂θ ∂φ1 ∂φn 2 ∂ρ ∂ρ = − − − 00... 0 r sin φn 1 cos φn 1 ¯· − − − ¸¯ ¯ 2 2 ¯ = ¯r cos φn 1 +sin φn 1 ∆n(,θ,φ1,...,φn 2,ρ) ¯ ¯ − − − ¯ = r∆n(θ, φ1,...,φn 2,rsin φn 1), ¡ − ¢ − i.e.

(8.14) ∆n+1(θ, φ1,...,φn 2,φn 1,r)=r∆n(θ, φ1,...,φn 2,rsin φn 1). − − − − To arrive at this result we have expanded the determinant along the bottom row. Staring with the well known and easy to compute fact that ∆2(θ, r)=r, Eq. (8.14) implies 2 ∆3(θ, φ1,r)=r∆2(θ, r sin φ1)=r sin φ1 3 2 ∆4(θ, φ1,φ2,r)=r∆3(θ, φ1,rsin φ2)=r sin φ2 sin φ1 . . n 1 n 2 2 ∆n(θ, φ1,...,φn 2,r)=r − sin − φn 2 ...sin φ2 sin φ1 − − which proves Eq. (8.12). Eq. (8.13) now follows from Eqs. (8.3), (8.11) and (8.12).

As a simple application, Eq. (8.13) implies

n 1 n 2 2 σ(S − )= sin − φn 2 ...sin φ2 sin φ1dφ1 ...dφn 2dθ − − 0 φi π,0 θ 2π Z ≤ ≤ ≤ ≤ n 2 − n 2 (8.15) =2π γk = σ(S − )γn 2 − k=1 Y π k where γk := sin φdφ. If k 1, we have by integration by parts that, 0 ≥ π π π Rk k 1 k 2 2 γk = sin φdφ = sin − φdcos φ =2δk,1 +(k 1) sin − φ cos φdφ 0 − 0 − 0 Z πZ Z k 2 2 =2δk,1 +(k 1) sin − φ 1 sin φ dφ =2δk,1 +(k 1) [γk 2 γk] − − − − − Z0 ¡ ¢ 106 BRUCE K. DRIVER† and hence γk satisfies γ0 = π, γ1 =2and the recursion relation k 1 γk = − γk 2 for k 2. k − ≥ Hence we may conclude 1 2 3 1 4 2 5 3 1 γ = π, γ =2,γ = π, γ = 2,γ = π, γ = 2,γ = π 0 1 2 2 3 3 4 4 2 5 5 3 6 6 4 2 and more generally by induction that (2k 1)!! (2k)!! γ = π − and γ =2 . 2k (2k)!! 2k+1 (2k +1)!! Indeed, 2k +2 2k +2 (2k)!! [2(k +1)]!! γ = γ = 2 =2 2(k+1)+1 2k +3 2k+1 2k +3 (2k +1)!! (2(k +1)+1)!! and 2k +1 2k +1 (2k 1)!! (2k +1)!! γ = γ = π − = π . 2(k+1) 2k +1 2k 2k +2 (2k)!! (2k +2)!! The recursion relation in Eq. (8.15) may be written as n n 1 (8.16) σ(S )=σ S − γn 1 − which combined with σ S1 =2π implies¡ ¢ σ S1¡ =2¢ π, 2 σ(S )=2π γ1 =2π 2, ¡ ¢ · · 2 2 3 1 2 π σ(S )=2π 2 γ2 =2π 2 π = , · · · · 2 2!! 2 2 2 2 3 2 4 2 π 2 π 2 2 π σ(S )= γ3 = 2 = 2!! · 2!! · 3 3!! 1 2 3 1 23π3 σ(S5)=2π 2 π 2 π = , · · 2 · 3 · 4 2 4!! 1 2 3 1 4 2 24π3 σ(S6)=2π 2 π 2 π 2= · · 2 · 3 · 4 2 · 5 3 5!! and more generally that 2(2π)n (2π)n+1 (8.17) σ(S2n)= and σ(S2n+1)= (2n 1)!! (2n)!! − which is verified inductively using Eq. (8.16). Indeed, 2(2π)n (2n 1)!! (2π)n+1 σ(S2n+1)=σ(S2n)γ = π − = 2n (2n 1)!! (2n)!! (2n)!! − and (2π)n+1 (2n)!! 2(2π)n+1 σ(S(n+1))=σ(S2n+2)=σ(S2n+1)γ = 2 = . 2n+1 (2n)!! (2n +1)!! (2n +1)!! Using (2n)!! = 2n (2(n 1)) ...(2 1) = 2nn! − · PDE LECTURE NOTES, MATH 237A-B 107

2n+1 2πn+1 we may write σ(S )= n! which shows that Eqs. (8.9) and (8.17) are in agreement. We may also write the formula in Eq. (8.17) as

n/2 2(2π) for n even n (n 1)!! σ(S )= −n+1  (2π) 2 (n 1)!! for n odd.  − 8.3. n — dimensional manifolds with boundaries. Definition 8.18. AsetΩ Rn is said to be a Ck — manifold with boundary ⊂ o o if for each x0 ∂Ω := Ω Ω (here Ω is the interior of Ω) there exists >0 ∈ n \ k an open set 0 D R and a C -diffeomorphism ψ : D B(x0,) such that ∈ ⊂ → ψ(D yn 0 )=B(x0,) Ω. SeeFigure22below.Wecall∂Ω the manifold boundary∩ { ≥of Ω}. ∩

Figure 22. Flattening out a neighborhood of a boundary point.

Remarks 8.19. (1) In Definition 8.18 we have defined ∂Ω = Ω Ωo which is not the topological boundary of Ω, defined by bd(Ω):=Ω¯ Ω\0. Clearly we always have ∂Ω bd(Ω) with equality iff Ω is closed. \ ⊂ (2) It is easily checked that if Ω Rn is a Ck — manifold with boundary, then ⊂ ∂Ω is a Ck — hypersurface in Rn. The reader is left to verify the following examples. n n Example 8.20. Let H = x R : xn > 0 . ¯ n { ∈ } (1) H is a C∞ — manifold with boundary and ¯ n ¯ n n 1 ∂H =bd H = R − 0 . ×{ } (2) Ω = B(ξ,r) is a C∞ — manifold¡ ¢ with boundary and ∂Ω =bd(B(ξ,r)), as the reader should verify. See Exercise 8.2 for a general result containing this statement. n 1 n (3) Let U be the open unit ball in R − ,thenΩ = H (U 0 ) is a C∞ — ∪ ×{n} 1 manifold with boundary and ∂Ω = U 0 while bd(Ω)=R − 0 . ×{ } ×{ } 108 BRUCE K. DRIVER†

(4) Now let Ω = Hn (U¯ 0 ), then Ω is not a C1 — manifold with boundary. The bad points are∪ bd(×{U)} 0 . ×{ } n 1 k (5) Suppose V is an open subset of R − and g : V R is a C — function and set → n Ω := (y,z) V R R : z g(y) , { ∈ × ⊂ ≥ } then Ω is a Ck — manifold with boundary and ∂Ω = Γ(g) — the graph of g. Again the reader should check this statement. (6) Let Ω =[(0, 1) (0, 1)] [( 1, 0) ( 1, 0)] [( 1, 1) 0 ] × ∪ − × − ∪ − ×{ } in which case Ωo =[(0, 1) (0, 1)] [( 1, 0) ( 1, 0)] × ∪ − × − and hence ∂Ω =( 1, 1) 0 is a Ck —hypersurfaceinR2. Nevertheless Ω is not a Ck — manifold− ×{ with} boundary as can be seen by looking at the point (0, 0) ∂Ω. n ∈1 n (7) If Ω = S − R , then ∂Ω = Ω is a C∞ - hypersurface. However, as in the previous⊂ example Ω is not an n — dimensional Ck — manifold with boundary despite the fact that Ω is now closed. (Warning: there is a clash of notation here with that of the more general theory of manifolds where n 1 n 1 ∂S − = when viewing S − as a manifold in its own right.) ∅ n k Lemma 8.21. Suppose Ω o R such that bd(Ω) is a C — hypersurface, then Ω¯ is Ck — manifold with boundary.⊂ (It is not necessarily true that ∂Ω¯ =bd(Ω). For example, let Ω := B(0, 1) x Rn :1< x < 2 . In this case Ω¯ = B(0, 2) so ∪ { ∈ | | } ∂Ω¯ = x Rn : x =2 while bd(Ω)= x Rn : x =2or x =1 .) { ∈ | | } { ∈ | | | | } n n Proof. Claim: Suppose U =( 1, 1) o R and V o U such that bd(V ) n − n⊂ ⊂ ∩ U = ∂H U. Then V is either, U+ := U H = U xn > 0 or U := U xn < 0 ∩n ∩ ∩{ } − ∩{ } or U ∂H = U+ U . \ ∪ − To prove the claim, first observe that V U ∂Hn and V is not empty, so either ⊂ \ V U+ or V U is not empty. Suppose for example there exists ξ V U+. Let ∩ ∩ − ∈ ∩ σ :[0, 1) U Hn be a continuous path such that σ(0) = ξ and → ∩ T =sup t<1:σ([0,t]) V . { ⊂ } If T =1, then η := σ(T ) is a point in U+ which is also in bd(V )=V¯ V. But this 6 n \ contradicts bd(V ) U = ∂H U and hence T =1. Because U+ is path connected, ∩ ∩ we have shown U+ V. Similarly if V U = , then U V as well and this completestheproofoftheclaim.⊂ ∩ − 6 ∅ − ⊂ We are now ready to show Ω¯ is a Ck — manifold with boundary. To this end, suppose ξ ∂Ω¯ =bd(Ω¯)=Ω¯ Ω¯ o Ω¯ Ω =bd(Ω). ∈ \ ⊂ \ Since bd(Ω) is a Ck — hypersurface, we may find an open neighborhood O of ξ such that there exists a Ck —diffeomorphism ψ : U O such that ψ (O bd(Ω)) = → ∩ U Hn. Recall that ∩ c O O bd(Ω)=O Ω¯ Ω = Ω O (O Ω)=bdO (Ω O) ∩ ∩ ∩ ∩ \ \ ∩ O where A and bdO(A) denotes the closure and boundary of a set A O in the relative topology on A. Since ψ is a Ck —diffeomorphism, it follows⊂ that V := PDE LECTURE NOTES, MATH 237A-B 109

ψ (O Ω) is an open set such that ∩ n bd(V ) U =bdU (V )=ψ (bdO (Ω O)) = ψ (O bd(Ω)) = U H . ∩ ∩ ∩ ∩ Therefore by the claim, we learn either V = U+ of U or U+ U . However the latter case can not occur because in this case ξ would− be in the∪ interior− of Ω¯ and hence not in bd(Ω¯). This completes the proof, since by changing the sign on the th n coordinate of ψ if necessary, we may arrange it so that ψ Ω¯ O = U+. ∩ Exercise 8.2. Suppose F : Rn R is a Ck — function and assume¡ ¢ → n F<0 := x R : F (x) < 0 = { } { ∈ } 6 ∅ n and F 0(ξ):R R is surjective (or equivalently F (ξ) =0)for all → ∇ 6 n ξ F =0 := x R : F (x)=0 . ∈ { } { ∈ } Then Ω := F 0 is a Ck — manifold with boundary and ∂Ω = F =0 . { ≤ } n n 1 { } Hint: For ξ F =0 , let A : R R − be a linear transformation such that A ∈:Nul({ F (ξ})) n 1 is→ invertible and A 0 and then Nul(F 0(ξ)) 0 R − Nul(F 0(ξ))⊥ define | → | ≡ n 1 n φ(x):=(A (x ξ) , F (x)) R − R = R . − − ∈ × Now use the inverse function theorem to construct ψ. Definition 8.22 (Outward pointing unit normal vector). Let Ω be a C1 — manifold with boundary, the outward pointing unit normal to ∂Ω istheuniquefunction n : ∂Ω Rn satisfying the following requirements. → (1) (Unit length.) n(x) =1for all x ∂Ω. | | ∈ (2) (Orthogonality to ∂Ω.) If x0 ∂Ω and ψ : D B(x0,) is as in the ∈ n → Definition 8.18, then n(xo) ψ0(0) (∂H ) , i.e. n(x0) is perpendicular of ∂Ω. ⊥ 1 (3) (Outward Pointing.) If φ := ψ− , then φ0(0)n(xo) en < 0 or equivalently · put ψ0(0)en n(x0) < 0, seeFigure23below. · 8.4. Divergence Theorem.

Theorem 8.23 ( Theorem). Let Ω Rn be a manifold with C2 — ⊂ boundary and n : ∂Ω Rn be the unit outward pointing normal to Ω. If Z n 1 o n → ∈ Cc(Ω, R ) C (Ω , R ) and ∩ (8.18) Z dm < |∇ · | ∞ ZΩ then

(8.19) Z(x) n(x)dσ(x)= Z(x) dx. · ∇ · ∂ZΩ ΩZ The proof of Theorem 8.23 will be given after stating a few corollaries and then a number preliminary results. Example 8.24. Let x sin 1 on [0, 1], f(x)= x 0 if x =0 ½ ¡ ¢ 110 BRUCE K. DRIVER†

1 1 1 then f C([0, 1]) C∞((0, 1)) and f 0(x)=sin sin for x>0. Since ∈ ∩ x − x x 1 ¡ ¢ ¡ ¢ 1 1 ∞ 1 ∞ sin(u) sin dx = u sin(u) du = | |du = , x x | |u2 u ∞ Z ¯ µ ¶¯ Z Z 0 ¯ ¯ 1 1 ¯ ¯ 1 ¯ ¯ f 0(x) dx = and the integrability assumption, Z dx < , in Theorem 0 | | ∞ Ω |∇ · | ∞ R8.23 is necessary. R Corollary 8.25. Let Ω Rn be a closed manifold with C2 — boundary and n : ⊂ ∂Ω Rn be the outward pointing unit normal to Ω. If Z C(Ω, Rn) C1(Ωo, Rn) and→ ∈ ∩

(8.20) Z + Z dm + Z n dσ < {| | |∇ · |} | · | ∞ ZΩ Z∂Ω then Eq. (8.19) is valid, i.e.

Z(x) n(x)dσ(x)= Z(x) dx. · ∇ · ∂ZΩ ΩZ n Proof. Let ψ Cc∞(R , [0, 1]) such that ψ =1in a neighborhood of 0 and set ∈ ψk(x):=ψ(x/k) and Zk := ψkZ. We have supp(Zk) supp(ψk) Ω —whichisa 1 ⊂ ∩ compact set since Ω is closed. Since ψk(x)= ( ψ)(x/k) is bounded, ∇ k ∇

Zk dm = ψk Z + ψk Z dm C Z dm + Z dm < . |∇ · | |∇ · ∇ · | ≤ | | |∇ · | ∞ ZΩ ZΩ ZΩ ZΩ Hence Theorem 8.23 implies

(8.21) Zkdm = Zk ndσ. ∇ · · ZΩ Z∂Ω By the D.C.T., 1 Zkdm = ( ψ)(x/k) Z(x)+ψ(x/k) Z(x) dx Zdm ∇ · k ∇ · ∇ · → ∇ · ZΩ ZΩ · ¸ ZΩ and

Zk ndσ= ψkZ ndσ Z ndσ, · · → · Z∂Ω Z∂Ω Z∂Ω which completes the proof by passing the in Eq. (8.21).

Corollary 8.26 (Integration by parts I). Let Ω Rn be a closed manifold with ⊂ C2 — boundary, n : ∂Ω Rn be the outward pointing normal to Ω,Z C(Ω, Rn) → ∈ ∩ C1(Ωo, Rn) and f C(Ω, R) C1(Ωo, R) such that ∈ ∩ f [ Z + Z ]+ f Z dm + f Z n dσ < {| | | | |∇ · | |∇ || |} | || · | ∞ ZΩ Z∂Ω then

f(x) Z(x) dx = f(x) Z(x) dx + Z(x) n(x)dσ(x). ∇ · − ∇ · · ΩZ ΩZ ∂ZΩ Proof. Apply Corollary 8.25 with Z replaced by fZ. PDE LECTURE NOTES, MATH 237A-B 111

Corollary 8.27 (Integration by parts II). Let Ω Rn be a closed manifold with ⊂ C2 — boundary , n : ∂Ω Rn be the outward pointing normal to Ω and f,g → ∈∈ C(Ω, R) C1(Ωo, R) such that ∩

f g + ∂if g + f ∂ig dm + fgni dσ < Ω {| || | | || | | || |} | | ∞ Z ∂ZΩ then

f(x)∂ig(x) dm = ∂if(x) g(x) dm + f(x)g(x)ni(x)dσ(x). − · ΩZ ΩZ ∂ZΩ

Proof. Apply Corollary 8.26 with Z chosen so that Zj =0if j = i and Zi = g, (i.e. Z =(0,...,g,0 ...,0)). 6 Proposition 8.28. Let Ω be as in Corollary 8.25 and suppose u, v C2(Ωo) C1(Ω) such that u, v, u, v, ∆u, ∆v L2(Ω) and u, v, ∂u, ∂v L2(∂Ω∈,dσ) then∩ ∇ ∇ ∈ ∂n ∂n ∈ ∂u (8.22) u vdm = u vdm + v dσ 4 · − ∇ · ∇ ∂n ΩZ ΩZ ∂ZΩ and ∂u ∂v (8.23) ( uv vu)dm = v u dσ. 4 − 4 ∂n − ∂n ΩZ ∂ZΩ µ ¶ Proof. Eq. (8.22) follows by applying Corollary 8.26 with f = v and Z = u. Similarly applying Corollary 8.26 with f = u and Z = v implies ∇ ∇ ∂v v udm = u vdm + u dσ 4 · − ∇ · ∇ ∂n ΩZ ΩZ ∂ZΩ and subtracting this equation from Eq. (8.22) implies Eq. (8.23).

8.5. The proof of Theorem 8.23.

n 1 n 1 Lemma 8.29. Suppose Ω o R and Z C (Ω, R ) and f Cc (Ω, R), then ⊂ ∈ ∈ f Zdx= f Zdx. ∇ · − ∇ · ΩZ ΩZ c n n Proof. Let W := fZ on Ω and W =0on Ω , then W Cc(R , R ). By Fubini’s theorem and the fundamental theorem of calculus, ∈ n ∂Wi (fZ) dx = ( W )dx = i dx1 ...dxn =0. ∇ · ∇ · n ∂x i=1 R ΩZ RZn X Z This completes the proof because (fZ)= f Z + f Z. ∇ · ∇ · ∇ · Corollary 8.30. If Ω Rn,Z C1(Ω, Rn) and g C(Ω, R) then g = Z iff ⊂ ∈ ∈ ∇ · (8.24) gf dx = Z fdxfor all f C1(Ω). − · ∇ ∈ c ΩZ ΩZ 112 BRUCE K. DRIVER†

Proof. By Lemma 8.29, Eq. (8.24) holds iff

gf dx = Zfdxfor all f C1(Ω) ∇ · ∈ c ΩZ ΩZ which happens iff g = Z. ∇ · Proposition 8.31 (Behavior of under coordinate transformations). Let ψ : ∇ W Ω is a C2 —diffeomorphism where W and Ω and open subsets of Rn. Given → f C1(Ω, R) and Z C1(Ω, Rn) let f ψ = f ψ C1(W, R) and Zψ C1(W, Rn) ∈ ψ ∈ 1 ◦ ∈ ∈ be defined by Z (y)=ψ0(y)− Z(ψ(y)). Then ψ tr (1) f = (f ψ)=(ψ0) ( f) ψ and ∇ ∇ ◦ψ ∇ ◦ 2 (2) [det ψ0 Z ]=( Z) ψ det ψ0. (Notice that we use ψ is C at this ∇point.)· ∇ · ◦ ·

Proof. 1. Let v Rn, then by definition of the and using the , ∈ tr (f ψ) v = ∂v(f ψ)= f(ψ) ψ0v =(ψ0) f(ψ) v. ∇ ◦ · ◦ ∇ · ∇ · 2. Let f C1(Ω). By the change of variables formula, ∈ c

f Zdm = f ψ ( Z) ψ det ψ0 dm ∇ · ◦ ∇ · ◦ | | ZΩ ZW ψ (8.25) = f ( Z) ψ det ψ0 dm. ∇ · ◦ | | ZW On the other hand

f Zdm = f Zdm = f(ψ) Z(ψ) det ψ0 dm Ω ∇ · − Ω ∇ · − W ∇ · | | Z Z 1 Z tr − ψ = (ψ0) f Z(ψ) det ψ0 dm − W ∇ · | | Z h i ψ 1 = f (ψ0)− Z(ψ) det ψ0 dm − ∇ · | | ZW ψ ψ = f Z det ψ0 dm − ∇ · | | ZW ψ¡ ¢ ψ (8.26) = f det ψ0 Z dm. ∇ · | | ZW ¡ ¢ Since Eqs. (8.25) and (8.26) hold for all f C1(Ω) we may conclude ∈ c ψ det ψ0 Z =( Z) ψ det ψ0 ∇ · | | ∇ · ◦ | | and by linearity this proves¡ item 2. ¢

Lemma 8.32. Eq. (8.19 of the 8.23 holds when Ω = H¯ n = n ¯ n n 1 n n x R : xn 0 and Z Cc(H , R ) C (H , R ) satisfies { ∈ ≥ } ∈ ∩ Z dx < |∇ · | ∞ HZn PDE LECTURE NOTES, MATH 237A-B 113

n 1 Proof. In this case ∂Ω = R − 0 and n(x)= en for x ∂Ω is the outward pointing normal to Ω. By Fubini’s×{ } theorem and the− fundamental∈ theorem of calculus, n 1 − ∂Zi dx =0 ∂xi i=1 Z Xxn>δ and ∂Zn dx = Zn(y, δ)dy. ∂xn − n 1 xnZ>δ R Z− Therefore

D.C.T. Zdx = lim Zdx= lim Zn(y, δ)dy ∇ · δ 0 ∇ · − δ 0 n ↓ ↓ n 1 HZ xnZ>δ R Z−

= Zn(y,0)dy = Z(x) n(x) dσ(x). − · n 1 n R Z− ∂HZ

Remark 8.33. The same argument used in the proof of Lemma 8.32 shows Theorem 8.23 holds when ¯ n n Ω = R+ := x R : xi 0 for all i . { ∈ ≥ } ¯ n Notice that R+ has a corners and edges, etc. and so ∂Ω is not smooth in this case. 8.5.1. The Proof of the Divergence Theorem 8.23. Proof. First suppose that supp(Z) is a compact subset of B(x0,) Ω for some x0 ∂Ω and >0 is sufficiently n ∩ 2 ∈ small so that there exists V o R and C —diffeomorphism ψ : V B(x0,) ⊂ o −→ (see Figure 23) such that ψ(V yn > 0 )=B(x0,) Ω and ∩ { } ∩ ψ(V yn =0 )=B(x0,) ∂Ω. ∩ { } ∩ Because n is the outward pointing normal, n(ψ(y)) ψ0(y)en < 0 on yn =0. Since ·

Figure 23. Reducing the divergence theorem for general Ω to Ω = Hn.

V is connected and det ψ0(y) is never zero on V, ς := sgn (det ψ0(y)) 1 is ∈ {± } 114 BRUCE K. DRIVER† constant independent of y V. For y ∂H¯ n, ∈ ∈

(Z n)(ψ(y)) det[ψ0(y)e1 ... ψ0(y)en 1 n(ψ(y))] · | | | − | | = ς(Z n)(ψ(y)) det[ψ0(y)e1 ... ψ0(y)en 1 n(ψ(y))] − · | | − | = ς det[ψ0(y)e1 ... ψ0(y)en 1 Z(ψ(y))] − | | − | ψ = ς det[ψ0(y)e1 ... ψ0(y)en 1 ψ0(y)Z (y)] − | | − | ψ = ς det ψ0(y) det[e1 ... en 1 Z (y)] − · | | − | ψ = det ψ0(y) Z (y) en, − | | · wherein the second equality we used the linearity properties of the determinant and the identity

n 1 − Z(ψ(y)) = Z n(ψ(y)) + αiψ0(y)ei for some αi. · i=1 X Starting with the definition of the surface integral we find

Z ndσ = (Z n)(ψ(y)) det[ψ0(y)e1 ... ψ0(y)en 1 n(ψ(y))] dy · · | | | − | | ∂ZΩ ∂HZ¯ n ψ = det ψ0(y)Z (y) ( en) dy · − ∂HZ¯ n ψ = det ψ0Z dm (by Lemma 8.32) ∇ · Zn H £ ¤

= [( Z) ψ]detψ0dm (by Proposition 8.31) ∇ · ◦ HZn = ( Z) dm (by the Change of variables theorem). ∇ · ΩZ

n 1 o n 2) We now prove the general case where Z Cc(Ω, R ) C (Ω , R ) and Ω ∈ ∩ n |∇ · Z dm < . Using Theorem 7.23, we may choose φi Cc∞(R ) such that | ∞ ∈ R N (1) φi 1 with equality in a neighborhood of K =Supp(Z). i=1 ≤ (2) ForP all i either supp(φi) Ω or supp(φi) B(x0,) where x0 ∂Ω and >0 are as in the previous⊂ paragraph. ⊂ ∈ Then by special cases proved in the previous paragraph and in Lemma 8.29,

Zdx= ( φi Z) dx = (φiZ)dx = (φiZ) ndσ ∇ · ∇ · ∇ · · i i i ΩZ ΩZ X X ΩZ X∂ZΩ

= φiZ ndσ= Z ndσ. · · i ∂ZΩ X ∂ZΩ PDE LECTURE NOTES, MATH 237A-B 115

2 8.6. Application to Holomorphic functions. Let Ω C ∼= R be a compact manifold with C2 — boundary. ⊂ 2 Definition 8.34. Let Ω C = R be a compact manifold with C2 — boundary ⊂ ∼ and f C(∂Ω, C). The contour integral, f(z)dz, of f along ∂Ω is defined by ∈ ∂Ω f(z)dz :=R i fndσ Z∂Ω Z∂Ω where n := (Re n, Im n) is the outward pointing normal, see Figure 24.

Figure 24. The induced direction for countour integrals along boundaries of regions.

In order to carry out the integral in Definition 8.34 more effectively, suppose that z = γ(t) with a t b is a parametrization of a part of the boundary of Ω and γ is chosen so that T≤:=≤γ˙ (t)/ γ˙ (t) = in(γ(t)). That is to say T is gotten from n by a 90o rotation in the counterclockwise| | direction. Combining this with dσ = γ˙ (t) dt we see that | | indσ= T γ˙ (t) dt = γ˙ (t)dt =: dz | | so that b f(z)dz = f(γ(t))γ˙ (t)dt. Zγ Za 1 ¯ ¯ 1 Proposition 8.35. Let f C (Ω, C) and ∂ := (∂x + i∂y) , then ∈ 2 (8.27) f(z)dz =2i ∂fdm.¯ Z∂Ω ZΩ Now suppose w Ω, then ∈ 1 f(z) 1 ∂f¯ (z) (8.28) f(w)= dz dm(z). 2πi z w − π z w Z∂Ω − ZΩ − Proof. By the divergence theorem, 1 1 ∂fdm¯ = (∂ + i∂ ) fdm = f (n + in ) dσ 2 x y 2 1 2 ZΩ ZΩ Z∂Ω 1 i = fndσ = f(z)dz. 2 −2 Z∂Ω Z∂Ω 116 BRUCE K. DRIVER†

Given >0 small, let Ω := Ω B(w, ). Eq. (8.27) with Ω = Ω and f being f(z) \ replaced by z w implies − f(z) ∂f¯ (8.29) dz =2i dm z w z w Z∂Ω − ZΩ − ¯ 1 wherein we have used the and the fact that ∂(z w)− =0to conclude − f(z) ∂f¯ (z) ∂¯ = . z w z w · − ¸ − Noting that ∂Ω = ∂Ω ∂B(w, ) and ∂B(w, ) may be parametrized by z = iθ ∪ w + e− with 0 θ 2π, we have ≤ ≤ 2π iθ f(z) f(z) f(w + e− ) iθ dz = dz + ( i) e− dθ z w z w e iθ − Z∂Ω − Z∂Ω − Z0 − 2π f(z) iθ = dz i f(w + e− )dθ z w − Z∂Ω − Z0 and hence 2π ¯ f(z) iθ ∂f(z) (8.30) dz i f(w + e− )dθ =2i dm(z) z w − z w Z∂Ω − Z0 ZΩ − Since 2π iθ lim f(w + e− )dθ =2πf(w) 0 ↓ Z0 and ∂f¯ ∂f¯ (z) lim dm = dm(z). 0 z w z w ↓ ZΩ − ZΩ − we may pass to the limit in Eq. (8.30) to find f(z) ∂f¯ (z) dz 2πif(w)=2i dm(z) z w − z w Z∂Ω − ZΩ − which is equivalent to Eq. (8.28).

Exercise 8.3. Let Ω be as above and assume f C1(Ω¯, C) satisfies g := ∂f¯ ∈ ∈ C∞(Ω, C). Show f C∞(Ω, C). Hint, let w0 Ω and >0 be small and choose ∈ ∈ φ Cc∞(B(z0,)) such that φ =1in a neighborhood of w0 and let ψ =1 φ. Then by∈ Eq. (8.28), − 1 f(z) 1 g(z) 1 g(z) f(w)= dz φ(z)dm(z) ψ(z)dm(z). 2πi z w − π z w − π z w Z∂Ω − ZΩ − ZΩ − Now show each of the three terms above are smooth in w for w near w0. To handle themiddletermnoticethat g(z) g(z + w) φ(z)dm(z)= φ(z + w)dm(z) z w z ZΩ − ZC for w near w0.

Definition 8.36. A function f C1(Ω, C) is said to be holomorphic if ∂f¯ =0. ∈ PDE LECTURE NOTES, MATH 237A-B 117

By Proposition 8.35, if f C1(Ω¯, C) and ∂f¯ =0on Ω, then Cauchy’s integral formula holds for w Ω, namely∈ ∈ 1 f(z) f(w)= dz 2πi z w Z∂Ω − and f C∞(Ω, C). For more details on Holomorphic functions, see the complex variable∈ appendix.

8.6.1. Appendix: More Proofs of Proposition 8.31.

1 Exercise 8.4. det0(A)B =det(A)tr(A− B). 8.4.

d d 1 1 det(A + tB)=det(A) det(A + tA− B)=det(A)tr(A− B). dt 0 dt 0 ¯ ¯ ¯ ¯ ¯ ¯ Proof. 2nd Proof of Proposition 8.31 by direct computation. Letting A = ψ0, 1 1 (det AZψ)= Zψ det A +detA Zψ det A∇ · det A{ · ∇ ∇ · } 1 ψ (8.31) =tr[A− ∂ ψ A]+ Z Z ∇ · and

ψ 1 1 Z = (A− Z ψ)=∂i(A− Z ψ) ei ∇ · ∇ · ◦ ◦ · 1 1 1 = ei ( A− ∂iAA− Z ψ)+ei A− (Z0 ψ)Aei · − ◦ · ◦ 1 1 1 = ei (A− ψ00 ei,A− Z ψ )+tr(A− (Z0 ψ)A) − · h ◦ i ◦ 1 1 = ei (A− ψ00 ei,A− Z ψ )+tr(Z0 ψ) − · h ◦ i ◦ 1 ψ = tr(A− ψ00 Z , )+( Z) ψ − h −i ∇ · ◦ 1 (8.32) = tr A− ∂ ψ A +( Z) ψ. − Z ∇ · ◦ Combining Eqs. (8.31) and£ (8.32) gives¤ the desired result:

ψ (det ψ0 Z )=detψ0( Z) ψ. ∇ · ∇ · ◦

Lemma 8.37 (Flow interpretation of the divergence). Let Z C1(Ω, Rn). Then ∈ d tZ Z = det(e )0 ∇ · dt 0 ¯ and ¯ ¯ d (fZ)dm = fdm. ∇ · dt 0 Z tZZ Ω ¯ e (Ω) ¯ Proof. By Exercise 8.4 and the change of variables formula,

d tZ d tZ det(e )0 =tr (e )0 =tr(Z0)= Z dt 0 dt 0 ∇ · ¯ µ ¯ ¶ ¯ ¯ ¯ ¯ 118 BRUCE K. DRIVER† and

d d tZ tZ f(x)dx = f(e (y)) det(e )0(y)dy dt 0 dt 0 ¯ etZZ(Ω) ¯ ΩZ ¯ ¯ ¯ ¯ = f(y) Z(y)+f(y) Z(y) dy {∇ · ∇ · } ΩZ = (fZ) dm. ∇ · ΩZ

Proof. 3rd Proof of Proposition 8.31. Using Lemma 8.37 with f =detψ0 and Z = Zψ and the change of variables formula,

ψ d d tZψ (det ψ0 Z )dm = det ψ0 dm = m(ψ e (Ω)) ∇ · dt 0 dt 0 ◦ ΩZ ¯ etZZ(Ω) ¯ ¯ ¯ d ¯ 1 tZ ¯ d tZ = m(ψ ψ− e ψ(Ω)) = m(e (ψ(Ω))) dt 0 ◦ ◦ ◦ dt 0 d ¯ ¯ = ¯ 1 dm = Zdm¯ dt¯0 ∇ · ¯ ¯ etZ (Zψ(Ω)) ψ(ZΩ) ¯ ¯ = ( Z) ψ det ψ0 dm. ∇ · ◦ ΩZ ψ ψ Since this is true for all regions Ω, it follows that (det ψ0 Z )=detψ0( Z ). ∇· ∇·