{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 25

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online