pde6 - 62 BRUCE K. DRIVER 6. Elliptic Ordinary Differential...

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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 237A-B 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....
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This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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pde6 - 62 BRUCE K. DRIVER 6. Elliptic Ordinary Differential...

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