{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

elliptic-pde-1

# elliptic-pde-1 - 1 1 2nd order differential operators...

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

1 1. 2 nd order differential operators Notations 1.1. Let be a precompact open subset of R d ,A ij = A ji ,A i ,A 0 BC ( ) for i, j =1 ,...,d, p ( x, ξ ):= d X i,j =1 A ij ξ i ξ j + d X i =1 A i ξ i + A 0 and L = p ( x, ∂ )= d X i,j =1 A ij i j + d X i =1 A i i + A 0 . We also let L = d X i,j =1 i j M A ij d X i =1 i M A i + A 0 . Remark 1.2 . The operators L and L have the following properties. (1) The operator L is the formal adjoint of L, i.e. ( Lu, v )=( u, L v ) for all u, v D ( )= C c ( ) . (2) We may view L as an operator on D 0 ( ) via the formula u D 0 ( ) Lu D 0 ( ) where h Lu, φ i := h u, L φ i for all φ C c ( ) . (3) The restriction of L to H k +2 ( ) gives a bounded linear transformation L : H k +2 ( ) H k ( ) for k N 0 . Indeed, L may be written as L = d X i,j =1 M A ij i j + d X i =1 M A i i + M A 0 . Now i : H k ( ) H k +1 ( ) is bounded and M ψ : H k ( ) H k ( ) is bounded where ψ BC ( ) . Therefore, for k N 0 ,L : H k +2 ( ) H k ( ) is bounded. De f nition 1.3. For u D 0 ( ) , let k u k H 1 ( ) := sup 0 6 = φ D ( ) |h u, φ i| k φ k H 1 0 ( ) and H 1 ( ):= © u D 0 ( ): k u k H 1 ( ) < ª . Example 1.4. Let = R d and S be the unit sphere in R d . Then de f ne σ D 0 ( ) by h σ, φ i := Z S φdσ. Let us shows that σ H 1 ( ) . For this let T : H 1 ( ) L 2 ( S, dσ ) denote the trace operator, i.e. the unique bounded linear operator such that = φ | S for all φ C c ¡ R d ¢ . Since T is bounded, |h σ, φ i| σ ( S ) 1 / 2 k k L 2 ( S ) σ ( S ) 1 / 2 k T k L ( H 1 ( ) ,L 2 ( S )) k φ k H 1 ( ) .

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

View Full Document
2 This shows σ H 1 ( ) and k σ k H 1 ( ) σ ( S ) 1 / 2 k T k L ( H 1 ( ) ,L 2 ( S )) . Lemma 1.5. Suppose is an open subset of R d such that ¯ is a manifold with C 0 — boundary and = ¯ o , then the map u £ H 1 0 ( ) ¤ u | D ( ) H 1 ( ) is a unitary map of Hilbert spaces. Proof. By de f nition C c ( ) is dense in H 1 0 ( ) , and hence it follows that the map u £ H 1 0 ( ) ¤ u | D ( ) H 1 ( ) is isometric. If u H 1 ( ) , it has a unique extension to H 1 0 ( )= C c ( ) H 1 ( ) and this provides the inverse map. If we identify L 2 ( )= H 0 ( ) with elements of D 0 ( ) via u ( u, · ) L 2 ( ) , then D 0 ( ) H 1 ( ) H 0 ( )= L 2 ( ) H 1 ( ) H 2 ( ) ...
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

elliptic-pde-1 - 1 1 2nd order differential operators...

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

View Full Document
Ask a homework question - tutors are online