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
Unformatted text preview: 62 BRUCE K. DRIVER 6. Elliptic Ordinary Differential Operators Let o R n be a bounded connected open region. A function u C 2 ( ) is said to satisfy Laplaces equation if 4 u = 0 in . More generally if f C ( ) is given we say u solves the Poisson equation if 4 u = f in . In order to get a unique solution to either of these equations it is necessary to impose boundary" conditions on u. Example 6.1. For Dirichlet boundary conditions we impose u = g on and for Neumann boundary conditions we impose u = g on , where g : R is a given function. Lemma 6.2. Suppose f : C R , is C 2 and g : R is continuous. Then if there exists a solution to 4 u = f with u = g on such that u C 2 ( ) C 1 ( ) then the solution is unique. De f nition 6.3. Given an open set R n we say u C 1 ( ) if u C 1 ( ) C ( ) and u extends to a continuous function on . Proof. If e u is another solution then v = e u u solves 4 v = 0 , v = 0 on . By the divergence theorem, 0 = Z 4 v vdm = Z  v  2 dm + Z v v nd = Z  v  2 dm, where the boundary terms are zero since v = 0 on . This identity implies R  u  2 dx = 0 which then shows v and since is connected we learn v is constant on . Because v is zero on we conclude v , that is u = u. For the rest of this section we will now restrict to n = 1 . However we will allow for more general operators than in this case. 6.1. Symmetric Elliptic ODE. Let a C 1 ([0 , 1] , (0 , )) and (6.1) Lf = ( af ) = af 00 a f for f C 2 ([0 , 1]) . In the following theorem we will impose Dirichlet boundary conditions on L by restricting the domain of L to D ( L ) := { f C 2 ([0 , 1] , R ) : f (0) = f (1) = 0 } . Theorem 6.4. The linear operator L : D ( L ) C ([0 , 1] , R ) is invertible and L 1 : C ([0 , 1] , R ) D ( L ) C 2 ([0 , 1] , R ) is a bounded operator. Proof. (1) (Uniqueness) If f, g D ( L ) then by integration by parts (6.2) ( Lf, g ) := Z 1 ( Lf )( x ) g ( x ) dx = Z 1 a ( x ) f ( x ) g ( x ) dx. Therefore if Lf = 0 then 0 = ( Lf, f ) = Z 1 a ( x ) f ( x ) 2 dx PDE LECTURE NOTES, MATH 237AB 63 and hence f and since f (0) = 0 , f . This shows L is injective. (2) (Existence) Given g C ([0 , 1] , R ) we are looking for f D ( L ) such that Lf = g, i.e. ( af ) = g. Integrating this equation implies a ( x ) f ( x ) = C + Z x g ( y ) dy. Therefore f ( z ) = C a ( z ) Z 1 y z 1 a ( z ) g ( y ) dy which upon integration and using f (0) = 0 gives f ( x ) = Z x C a ( z ) dz Z 1 y z x 1 a ( z ) g ( y ) dz dy....
View
Full
Document
This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.
 Three '10
 Smith

Click to edit the document details