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8. Surfaces, Surface Integrals and Integration by Parts Definition 8.1. a Subset M ⊂ R N Is a N−1 Dimensional C K-Hypersurfa

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8. Surfaces, Surface Integrals and Integration by Parts Definition 8.1. a Subset M ⊂ R N Is a N−1 Dimensional C K-Hypersurfa 94 BRUCE K. DRIVER† 8. Surfaces, Surface Integrals and Integration by Parts 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 open set 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 function 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 normal 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 integral 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 ∂1Σ ∂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 calculus (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 <r − | | 2 2 Z + Z| | q r y − | | n 1 A similar computation shows, with S − being the lowerq hemisphere, that − r gdσ = g(y, r2 y 2) dy. n 1 S − y <r − − | | 2 2 Z − Z| | q r y − | | q Lemma 8.8. If Σ : D M is another Ck — parametrization of M, then → ˜ f Σ(y)ρΣ(y)dy = f Σ(y)ρΣ(y)dy. e e ◦ ◦ DZ DZ e e 98 BRUCE K. DRIVER† 1 k Proof. By Proposition 8.3, φ := Σ− Σ : D˜ D is a C —diffeomorphism. n◦ 1 → By the change of variables 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 surface integral 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 ∈ sequence 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).
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