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pde8 - 94 BRUCE K DRIVER † 8 Surfaces Surface Integrals...

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Unformatted text preview: 94 BRUCE K. DRIVER † 8. Surfaces, Surface Integrals and Integration by Parts De f nition 8.1. A subset M ⊂ R n is a n − 1 dimensional C k- Hypersurface if for all x ∈ M there exists ¡ > an open set ∈ D ⊂ R n and a C k-di f eomorphism ψ : D → B ( x , ¡ ) such that ψ ( D ∩ { x n = 0 } ) = B ( x , ¡ ) ∩ M. See Figure 16 below. Figure 16. An embedded submanifold of R 2 . Example 8.2. Suppose V ⊂ R n − 1 and g : V C k −→ R . Then M := Γ ( g ) = { ( y, g ( y )) : y ∈ V } is a C k hypersurface. To verify this assertion, given x = ( y , g ( y )) ∈ Γ ( g ) de f ne ψ ( y, z ) := ( y + y , g ( y + y ) − z ) . Then ψ : { V − y ) × R C k −→ V × R di f eomorphism ψ (( V − y ) × { } ) = { ( y + y , g ( y + y )) : y ∈ V − y } = Γ ( g ) . Proposition 8.3 (Parametrized Surfaces) . Let k ≥ 1 , D ⊂ R n − 1 and Σ ∈ C k ( D, R n ) satisfy (1) Σ : D → M := Σ ( D ) is a homeomorphism and (2) Σ ( y ) : R n − 1 → R n is injective for all y ∈ D. (We will call M a C k — parametrized surface and Σ : D → M a parametrization of M. ) Then M is a C k-hypersurface in R n . Moreover if f ∈ C ( W ⊂ R d , R n ) is a continuous function such that f ( W ) ⊂ M, then f ∈ C k ( W, R n ) i f Σ − 1 ◦ f ∈ C k ( U, D ) . Proof. Let y ∈ D and x = Σ ( y ) and n be a normal vector to M at x , i.e. n ⊥ Ran ( Σ ( y )) , and let ψ ( t, y ) := Σ ( y + y ) + t n for t ∈ R and y ∈ D − y , see Figure 17 below. Since D y ψ (0 , 0) = Σ ( y ) and ∂ψ ∂t (0 , 0) = n / ∈ Ran ( Σ ( y )) , ψ (0 , 0) is invertible. so by the inverse function theorem there exists a neighborhood V of (0 , 0) ∈ R n such that ψ | V : V → R n is a C k — di f eomorphism. PDE LECTURE NOTES, MATH 237A-B 95 Figure 17. Showing a parametrized surface is an embedded hyper-surface. Choose an ¡ > such that B ( x , ¡ ) ∩ M ⊂ Σ ( V ∩ { t = 0 } ) and B ( x , ¡ ) ⊂ ψ ( V ) . Then set U := ψ − 1 ( B ( x , ¡ )) . One f nds ψ | U : U → B ( x , ¡ ) has the desired properties. Now suppose f ∈ C ( W ⊂ R d , R n ) such that f ( W ) ⊂ M, a ∈ W and x = f ( a ) ∈ M. By shrinking W if necessary we may assume f ( W ) ⊂ B ( x , ¡ ) where B ( x , ¡ ) is the ball used previously. (This is where we used the continuity of f. ) Then Σ − 1 ◦ f = π ◦ ψ − 1 ◦ f where π is projection onto { t = 0 } . Form this identity it clearly follows Σ − 1 ◦ f is C k if f is C k . The converse is easier since if Σ − 1 ◦ f is C k then f = Σ ◦ ( Σ − 1 ◦ f ) is C k as well. 8.1. Surface Integrals. De f nition 8.4. Suppose Σ : D ⊂ R n − 1 → M ⊂ R n is a C 1- parameterized hypersurface of R n and f ∈ C c ( M, R ) . Then the surface integral of f over M, R M f dσ, is de f ned by Z M f dσ = Z D f ◦ Σ ( y ) ¯ ¯ ¯ ¯ det[ ∂ Σ ( y ) ∂y 1 | , . . . , ∂ Σ ( y ) ∂y n − 1 | n ( y )] ¯ ¯ ¯ ¯ dy = Z D f ◦ Σ ( y ) | det[ Σ ( y ) e 1 | . . . | Σ ( y ) e n − 1 | n...
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pde8 - 94 BRUCE K DRIVER † 8 Surfaces Surface Integrals...

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